calc 501-1000 by James Bardo

6.9 Hyperbolic Functions and Hanging Cables

477

represent a portion of the curve x 2 − y 2 = 1, as may be seen by writing x 2 − y 2 = cosh2 t − sinh2 t = 1

and observing that x = cosh t > 0. This curve, which is shown in Figure 6.9.3b, is the right half of a larger curve called the unit hyperbola; this is the reason why the functions in this section are called hyperbolic functions. It can be shown that if t ≥ 0, then the parameter t can be interpreted as twice the shaded area in Figure 6.9.3b. (We omit the details.) DERIVATIVE AND INTEGRAL FORMULAS

Derivative formulas for sinh x and cosh x can be obtained by expressing these functions in terms of ex and e−x : d ex − e−x ex + e−x d [sinh x] = = cosh x = dx dx 2 2 d d ex + e−x ex − e−x [cosh x] = = sinh x = dx dx 2 2

Derivatives of the remaining hyperbolic functions can be obtained by expressing them in terms of sinh and cosh and applying appropriate identities. For example, d d [tanh x] = dx dx

sinh x cosh x

cosh x =

d d [sinh x] − sinh x [cosh x] dx dx cosh2 x

cosh2 x − sinh2 x 1 = = sech2 x cosh2 x cosh2 x The following theorem provides a complete list of the generalized derivative formulas and corresponding integration formulas for the hyperbolic functions. =

6.9.3

theorem d du [sinh u] = cosh u dx dx du d [cosh u] = sinh u dx dx d du [tanh u] = sech2 u dx dx du d [coth u] = −csch2 u dx dx d du [sech u] = −sech u tanh u dx dx du d [csch u] = −csch u coth u dx dx

cosh u du = sinh u + C sinh u du = cosh u + C sech2 u du = tanh u + C csch2 u du = − coth u + C sech u tanh u du = −sech u + C csch u coth u du = −csch u + C

Example 2 d d 3 [cosh(x 3 )] = sinh(x 3 ) · [x ] = 3x 2 sinh(x 3 ) dx dx d 1 d sech2 x [ln(tanh x)] = · [tanh x] = dx tanh x dx tanh x

Chapter 6 / Applications of the Definite Integral in Geometry, Science, and Engineering

478

Example 3

sinh5 x cosh x dx = tanh x dx =

1 6

sinh6 x + C

u = sinh x du = cosh x dx

sinh x dx cosh x

= ln |cosh x| + C = ln(cosh x) + C

u = cosh x du = sinh x dx

We were justiďŹ ed in dropping the absolute value signs since cosh x > 0 for all x. Example 4 A 100 ft wire is attached at its ends to the tops of two 50 ft poles that are positioned 90 ft apart. How high above the ground is the middle of the wire?

Solution. From above, the wire forms a catenary curve with equation y = a cosh

+c a where the origin is on the ground midway between the poles. Using Formula (4) of Section 6.4 for the length of the catenary, we have 2 45 dy 1+ dx 100 = dx â&#x2C6;&#x2019;45 2 45 dy By symmetry dx 1+ =2 about the y-axis dx 0 45 x =2 1 + sinh2 dx a 0 45 x By (1) and the fact cosh dx =2 that cosh x > 0 a 0 x 45 45 = 2a sinh = 2a sinh a 0 a Using a calculating utilityâ&#x20AC;&#x2122;s numeric solver to solve 45 100 = 2a sinh a

y 50 40 30 20 10 â&#x2C6;&#x2019;45

x 45

y = 56.01 cosh ĺ&#x2020;¸ x ĺ&#x2020;š â&#x2C6;&#x2019; 25.08 56.01 Figure 6.9.4

for a gives a â&#x2030;&#x2C6; 56.01. Then

45 50 = y(45) = 56.01 cosh 56.01

+ c â&#x2030;&#x2C6; 75.08 + c

so c â&#x2030;&#x2C6; â&#x2C6;&#x2019;25.08. Thus, the middle of the wire is y(0) â&#x2030;&#x2C6; 56.01 â&#x2C6;&#x2019; 25.08 = 30.93 ft above the ground (Figure 6.9.4). INVERSES OF HYPERBOLIC FUNCTIONS

Referring to Figure 6.9.1, it is evident that the graphs of sinh x, tanh x, coth x, and csch x pass the horizontal line test, but the graphs of cosh x and sech x do not. In the latter case, restricting x to be nonnegative makes the functions invertible (Figure 6.9.5). The graphs of the six inverse hyperbolic functions in Figure 6.9.6 were obtained by reďŹ&#x201A;ecting the graphs of the hyperbolic functions (with the appropriate restrictions) about the line y = x.

6.9 Hyperbolic Functions and Hanging Cables y

479

Table 6.9.1 summarizes the basic properties of the inverse hyperbolic functions. You should conﬁrm that the domains and ranges listed in this table agree with the graphs in Figure 6.9.6. y = cosh x

y = sech x

With the restriction that x ≥ 0, the curves y = cosh x and y = sech x pass the horizontal line test.

x −1

Figure 6.9.5

y = sinh−1 x

y = cosh−1 x

x −1

Figure 6.9.6

y = tanh−1 x

y = coth−1 x

y = sech−1 x

y = csch−1 x

Table 6.9.1 properties of inverse hyperbolic functions

function

domain

range

sinh−1 x

(− ∞, + ∞)

cosh−1 x

[1, + ∞)

[0, + ∞)

tanh−1 x

(−1, 1)

(− ∞, + ∞)

coth−1 x

(− ∞, −1) 傼 (1, + ∞)

(− ∞, 0) 傼 (0, + ∞)

sech−1 x

(0, 1]

[0, + ∞)

csch−1 x

(− ∞, 0) 傼 (0, + ∞)

basic relationships sinh−1 (sinh x) = x sinh(sinh−1 x) = x

if – ∞ < x < + ∞ if – ∞ < x < + ∞

cosh−1 (cosh x) = x if x ≥ 0 cosh(cosh−1 x) = x if x ≥ 1 tanh−1 (tanh x) = x if – ∞ < x < + ∞ tanh(tanh−1 x) = x if –1 < x < 1 coth−1 (coth x) = x coth(coth−1 x) = x

if x < 0 or x > 0 if x < –1 or x > 1

sech−1 (sech x) = x if x ≥ 0 sech(sech−1 x) = x if 0 < x ≤ 1 csch−1 (csch x) = x if x < 0 or x > 0 csch(csch−1 x) = x if x < 0 or x > 0

480

Chapter 6 / Applications of the Definite Integral in Geometry, Science, and Engineering

LOGARITHMIC FORMS OF INVERSE HYPERBOLIC FUNCTIONS

Because the hyperbolic functions are expressible in terms of ex , it should not be surprising that the inverse hyperbolic functions are expressible in terms of natural logarithms; the next theorem shows that this is so.

6.9.4 theorem The following relationships hold for all x in the domains of the stated inverse hyperbolic functions: sinh−1 x = ln(x + x 2 + 1 ) 1 1+x tanh−1 x = ln 2 1−x 2 1 + 1 − x sech−1 x = ln x

√ cosh−1 x = ln(x + x 2 − 1 ) 1 x+1 coth−1 x = ln 2 x−1 2 1 1 + x + csch−1 x = ln x |x|

We will show how to derive the ﬁrst formula in this theorem and leave the rest as exercises. The basic idea is to write the equation x = sinh y in terms of exponential functions and solve this equation for y as a function of x. This will produce the equation y = sinh−1 x with sinh−1 x expressed in terms of natural logarithms. Expressing x = sinh y in terms of exponentials yields ey − e−y x = sinh y = 2 which can be rewritten as ey − 2x − e−y = 0 Multiplying this equation through by ey we obtain e2y − 2xey − 1 = 0 and applying the quadratic formula yields √ 2x ± 4x 2 + 4 y = x ± x2 + 1 e = 2

Since ey > 0, the solution involving the minus sign is extraneous and must be discarded. Thus, ey = x + x 2 + 1

Taking natural logarithms yields y = ln(x + x 2 + 1 )

sinh−1 x = ln(x +

x2 + 1 )

Example 5 √

√ 12 + 1 ) = ln(1 + 2 ) ≈ 0.8814 1 + 21 1 1 −1 1 = ln 3 ≈ 0.5493 tanh = ln 2 2 2 1 − 21

sinh−1 1 = ln(1 +

6.9 Hyperbolic Functions and Hanging Cables

481

DERIVATIVES AND INTEGRALS INVOLVING INVERSE HYPERBOLIC FUNCTIONS Show that the derivative of the function sinh−1 x can also be obtained by letting y = sinh−1 x and then differentiating x = sinh y implicitly.

Formulas for the derivatives of the inverse hyperbolic functions can be obtained from Theorem 6.9.4. For example, d d 1 x −1 2 [sinh x] = [ln(x + x + 1 )] = 1+ dx dx x + x2 + 1 x2 + 1 x2 + 1 + x 1 = = (x + x 2 + 1 )( x 2 + 1 ) x2 + 1

This computation leads to two integral formulas, a formula that involves sinh−1 x and an equivalent formula that involves logarithms: dx = sinh−1 x + C = ln(x + x 2 + 1 ) + C x2 + 1 The following two theorems list the generalized derivative formulas and corresponding integration formulas for the inverse hyperbolic functions. Some of the proofs appear as exercises.

6.9.5

theorem

1 d du (sinh−1 u) = 2 dx dx 1+u d 1 du −1 (cosh u) = , dx u2 − 1 dx d 1 du (tanh−1 u) = , dx 1 − u2 dx

6.9.6

d 1 du (coth−1 u) = , dx 1 − u2 dx

u>1 |u| < 1

|u| > 1

d 1 du (sech−1 u) = − , dx u 1 − u2 dx

d 1 du (csch−1 u) = − , dx |u| 1 + u2 dx

0<u<1 u = 0

theorem If a > 0, then du

a 2 + u2 du

u2 − a 2

= sinh−1 = cosh−1

a u

+ C or ln(u +

a u

u2 + a 2 ) + C u2 − a 2 ) + C, u > a

⎧ 1 ⎪ ⎪ + C, |u| < a tanh−1 ⎨ a + u 1 du a a + C, |u| = a or = ln u ⎪ a 2 − u2 2a a − u 1 ⎪ ⎩ coth−1 + C, |u| > a a a du 1 a + a 2 − u2 1 −1 u + C, 0 < |u| < a = − sech + C or − ln a a a |u| u a 2 − u2 du 1 1 a + a 2 + u2 −1 u = − csch + C or − ln + C, u = 0 a a a |u| u a 2 + u2

482

Chapter 6 / Applications of the Definite Integral in Geometry, Science, and Engineering

Example 6

Evaluate

4x 2 − 9

,x >

3 . 2

Solution. Let u = 2x. Thus, du = 2 dx and

2 dx du 1 1 = = 2 2 2 2 2 u − 32 4x − 9 4x − 9 1 1 u 2x = cosh−1 + C = cosh−1 +C 2 3 2 3 dx

Alternatively, we can use the logarithmic equivalent of cosh−1 (2x /3), 2x = ln(2x + 4x 2 − 9 ) − ln 3 cosh−1 3 (verify), and express the answer as dx 1 = ln(2x + 4x 2 − 9 ) + C 2 4x 2 − 9

✔QUICK CHECK EXERCISES 6.9 1. cosh x = tanh x =

(See page 485 for answers.)

sinh x =

2. Complete the table. cosh x sinh x tanh x coth x sech x csch x domain range

3. The parametric equations x = cosh t,

y = sinh t

(−⬁ < t < +⬁)

represent the right half of the curve called a inating the parameter, the equation of this curve is

EXERCISE SET 6.9

d [cosh x] = dx d [tanh x] = dx 5. cosh x dx = tanh x dx = 4.

. Elim.

d [cosh−1 x] = dx d [tanh−1 x] = dx

d [sinh x] = dx

sinh x dx =

d [sinh−1 x] = dx

Graphing Utility

1–2 Approximate the expression to four decimal places. ■

1. (a) sinh 3 (d) sinh−1 (−2)

(b) cosh(−2) (e) cosh−1 3

2. (a) csch(−1) (d) sech−1 21

(b) sech(ln 2) (e) coth−1 3

3. Find the exact numerical value of each expression. (a) sinh(ln 3) (b) cosh(− ln 2) (c) tanh(2 ln 5) (d) sinh(−3 ln 2) 4. In each part, rewrite the expression as a ratio of polynomials. (a) cosh(ln x) (b) sinh(ln x) (c) tanh(2 ln x) (d) cosh(− ln x) 5. In each part, a value for one of the hyperbolic functions is given at an unspeciﬁed positive number x0 . Use appropri-

ate identities to ﬁnd the exact values of the remaining ﬁve hyperbolic functions at x0 . (a) sinh x0 = 2 (b) cosh x0 = 45 (c) tanh x0 = 45 6. Obtain the derivative formulas for csch x, sech x, and coth x from the derivative formulas for sinh x, cosh x, and tanh x. 7. Find the derivatives of cosh−1 x and tanh−1 x by differentiating the formulas in Theorem 6.9.4. 8. Find the derivatives of sinh−1 x, cosh−1 x, and tanh−1 x by differentiating the equations x = sinh y, x = cosh y, and x = tanh y implicitly. 9–28 Find dy/dx. ■

9. y = sinh(4x − 8)

10. y = cosh(x 4 )

6.9 Hyperbolic Functions and Hanging Cables

11. y = coth(ln x) 13. y = csch(1/x) 15. y = 4x + cosh2 (5x) √ 17. y = x 3 tanh2 ( x )

19. y = sinh−1 31 x

21. y = ln(cosh−1 x) 1 23. y = tanh−1 x 25. y = cosh−1 (cosh x) √ 27. y = ex sech−1 x

12. y = ln(tanh 2x) 14. y = sech(e2x )

16. y = sinh3 (2x)

18. y = sinh(cos 3x) 20. y = sinh−1 (1/x)

22. y = cosh−1 (sinh−1 x)

24. y = (coth

−1

26. y = sinh−1 (tanh x)

28. y = (1 + x csch−1 x)10

is 2 square units. Express your answer to at least ﬁve decimal places. 53. Find the arc length of the catenary y = cosh x between x = 0 and x = ln 2. 54. Find the arc length of the catenary y = a cosh(x /a) between x = 0 and x = x1 (x1 > 0).

55. In parts (a)–(f ) ﬁnd the limits, and conﬁrm that they are consistent with the graphs in Figures 6.9.1 and 6.9.6. (a) lim sinh x (b) lim sinh x (c) (e)

x → +⬁

x → −⬁

lim tanh x

(d) lim tanh x

lim sinh−1 x

(f ) lim− tanh−1 x

x → +⬁

x → −⬁

x → +⬁

x →1

29–44 Evaluate the integrals. ■

29. 31. 33. 35. 37. 39. 41. 43.

sinh6 x cosh x dx

√ tanh x sech2 x dx tanh x dx ln 3 tanh x sech3 x dx ln 2 dx 1 + 9x 2 dx (x < 0) 1 − e2x dx x 1 + 4x 2 1/2 dx 1 − x2 0

30. 32. 34.

F O C U S O N C O N C E P TS

cosh(2x − 3) dx csch2 (3x) dx coth2 x csch2 x dx ln 3

ex − e−x dx ex + e−x 0 √ dx 38. (x > 2 ) x2 − 2 sin θ dθ 40. 1 + cos2 θ dx 42. (x > 5/3) 9x 2 − 25 √3 dt 44. 0 t2 + 1 36.

483

45–48 True–False Determine whether the statement is true or false. Explain your answer. ■

45. The equation cosh x = sinh x has no solutions.

46. Exactly two of the hyperbolic functions are bounded. 47. There is exactly one hyperbolic function f(x) such that for all real numbers a, the equation f(x) = a has a unique solution x. 48. The identities in Theorem 6.9.2 may be obtained from the corresponding trigonometric identities by replacing each trigonometric function with its hyperbolic analogue. 49. Find the area enclosed by y = sinh 2x, y = 0, and x = ln 3. 50. Find the volume of the solid that is generated when the region enclosed by y = sech x, y = 0, x = 0, and x = ln 2 is revolved about the x-axis.

51. Find the volume of the solid that is generated when the region enclosed by y = cosh 2x, y = sinh 2x, x = 0, and x = 5 is revolved about the x-axis. 52. Approximate the positive value of the constant a such that the area enclosed by y = cosh ax, y = 0, x = 0, and x = 1

56. Explain how to obtain the asymptotes for y = tanh x from the curvilinear asymptotes for y = cosh x and y = sinh x.

57. Prove that sinh x is an odd function of x and that cosh x is an even function of x, and check that this is consistent with the graphs in Figure 6.9.1.

58–59 Prove the identities. ■

58. (a) (b) (c) (d) (e) (f ) (g) (h)

cosh x + sinh x = ex cosh x − sinh x = e−x sinh(x + y) = sinh x cosh y + cosh x sinh y sinh 2x = 2 sinh x cosh x cosh(x + y) = cosh x cosh y + sinh x sinh y cosh 2x = cosh2 x + sinh2 x cosh 2x = 2 sinh2 x + 1 cosh 2x = 2 cosh2 x − 1

59. (a) 1 − tanh2 x = sech2 x tanh x + tanh y (b) tanh(x + y) = 1 + tanh x tanh y 2 tanh x (c) tanh 2x = 1 + tanh2 x 60. Prove: √ (a) cosh−1 x = ln(x + x 2 − 1 ), x ≥ 1 1+x 1 −1 (b) tanh x = ln , −1 < x < 1. 2 1−x 61. Use Exercise 60 to obtain the derivative formulas for cosh−1 x and tanh−1 x. 62. Prove: sech−1 x = cosh−1 (1/x), coth

−1

csch

−1

x = tanh (1/x), −1

x = sinh (1/x),

0<x≤1

|x| > 1

x = 0

63. Use Exercise 62 to express the integral du 1 − u2 entirely in terms of tanh−1 .

484

Chapter 6 / Applications of the Definite Integral in Geometry, Science, and Engineering

(b) Find the length of the centerline to four decimal places. (c) For what values of x is the height of the arch 100 ft? Round your answers to four decimal places. (d) Approximate, to the nearest degree, the acute angle that the tangent line to the centerline makes with the ground at the ends of the arch.

64. Show that d 1 (a) [sech−1 |x|] = − dx x 1 − x2 d 1 −1 (b) [csch |x|] = − . dx x 1 + x2 65. In each part, ﬁnd the limit.

cosh x ex 66. Use the ﬁrst and second derivatives to show that the graph of y = tanh−1 x is always increasing and has an inﬂection point at the origin. √ 67. The integration formulas for 1/ u2 − a 2 in Theorem 6.9.6 are valid for u > a. Show that the following formula is valid for u < −a: (a)

lim (cosh−1 x − ln x) (b) lim

x → +⬁

u du = − cosh−1 − + C a u2 − a 2

ln u + u2 − a 2 + C

68. Show that (sinh x + cosh x)n = sinh nx + cosh nx. 69. Show that

−a

etx dx =

2 sinh at t

70. A cable is suspended between two poles as shown in Figure 6.9.2. Assume that the equation of the curve formed by the cable is y = a cosh(x /a), where a is a positive constant. Suppose that the x-coordinates of the points of support are x = −b and x = b, where b > 0. (a) Show that the length L of the cable is given by b L = 2a sinh a (b) Show that the sag S (the vertical distance between the highest and lowest points on the cable) is given by b S = a cosh − a a 71–72 These exercises refer to the hanging cable described in

Exercise 70. ■ 71. Assuming that the poles are 400 ft apart and the sag in the cable is 30 ft, approximate the length of the cable by approximating a. Express your ﬁnal answer to the nearest tenth of a foot. [Hint: First let u = 200/a.]

72. Assuming that the cable is 120 ft long and the poles are 100 ft apart, approximate the sag in the cable by approximating a. Express your ﬁnal answer to the nearest tenth of a foot. [Hint: First let u = 50/a.] 73. The design of the Gateway Arch in St. Louis, Missouri, by architect Eero Saarinan was implemented using equations provided by Dr. Hannskarl Badel. The equation used for the centerline of the arch was y = 693.8597 − 68.7672 cosh(0.0100333x) ft

for x between −299.2239 and 299.2239. (a) Use a graphing utility to graph the centerline of the arch.

74. Suppose that a hollow tube rotates with a constant angular velocity of ω rad/s about a horizontal axis at one end of the tube, as shown in the accompanying ﬁgure. Assume that an object is free to slide without friction in the tube while the tube is rotating. Let r be the distance from the object to the pivot point at time t ≥ 0, and assume that the object is at rest and r = 0 when t = 0. It can be shown that if the tube is horizontal at time t = 0 and rotating as shown in the ﬁgure, then r=

g [sinh(ωt) − sin(ωt)] 2ω2

during the period that the object is in the tube. Assume that t is in seconds and r is in meters, and use g = 9.8 m/s2 and ω = 2 rad/s. (a) Graph r versus t for 0 ≤ t ≤ 1. (b) Assuming that the tube has a length of 1 m, approximately how long does it take for the object to reach the end of the tube? (c) Use the result of part (b) to approximate dr/dt at the instant that the object reaches the end of the tube.

r v

Figure Ex-74

75. The accompanying ﬁgure (on the next page) shows a person pulling a boat by holding a rope of length a attached to the bow and walking along the edge of a dock. If we assume that the rope is always tangent to the curve traced by the bow of the boat, then this curve, which is called a tractrix, has the property that the segment of the tangent line between the curve and the y-axis has a constant length a. It can be proved that the equation of this tractrix is y = a sech−1

x 2 − a − x2 a

(a) Show that to move the bow of the boat to a point (x, y), the person must walk a distance D = a sech−1

x a

from the origin. (b) If the rope has a length of 15 m, how far must the person walk from the origin to bring the boat 10 m from the dock? Round your answer to two decimal places. (c) Find the distance traveled by the bow along the tractrix as it moves from its initial position to the point where it is 5 m from the dock.

Chapter 6 Review Exercises

(i) P (x, y) is on the right branch of the unit hyperbola x 2 − y 2 = 1; (ii) t and y have the same sign (or are both 0); (iii) the area of the region bounded by the x-axis, the right branch of the unit hyperbola, and the segment from the origin to P is |t|/2.” Discuss what properties would first need to be verified in order for this to be a legitimate definition.

(x, y) x Dock

(a, 0)

Initial position

485

77. Writing Investigate what properties of cosh t and sinh t can be proved directly from the geometric deﬁnition in Exercise 76. Write a short description of the results of your investigation.

Figure Ex-75

76. Writing Suppose that, by analogy with the trigonometric functions, we deﬁne cosh t and sinh t geometrically using Figure 6.9.3b: “For any real number t, deﬁne x = cosh t and y = sinh t to be the unique values of x and y such that

✔QUICK CHECK ANSWERS 6.9 1. 2.

ex + e−x ex − e−x ex − e−x ; ; x 2 2 e + e−x cosh x

sinh x

tanh x

coth x

sech x

csch x

domain

(− ∞, + ∞)

(− ∞, 0) ∪ (0, + ∞)

(− ∞, + ∞)

(− ∞, 0) ∪ (0, + ∞)

range

[1, + ∞)

(− ∞, + ∞)

(−1, 1)

(− ∞, −1) ∪ (1, + ∞)

(0, 1]

(− ∞, 0) ∪ (0, + ∞)

3. unit hyperbola; x 2 − y 2 = 1 4. sinh x; cosh x; sech2 x 1 1 1 6. √ ; √ ; x2 − 1 1 + x2 1 − x2

5. sinh x + C; cosh x + C; ln(cosh x) + C

CHAPTER 6 REVIEW EXERCISES 1. Describe the method of slicing for finding volumes, and use that method to derive an integral formula for finding volumes by the method of disks. 2. State an integral formula for finding a volume by the method of cylindrical shells, and use Riemann sums to derive the formula. 3. State an integral formula for finding the arc length of a smooth curve y = f(x) over an interval [a, b], and use Riemann sums to derive the formula. 4. State an integral formula for the work W done by a variable force F (x) applied in the direction of motion to an object moving from x = a to x = b, and use Riemann sums to derive the formula. 5. State an integral formula for the fluid force F exerted on a vertical flat surface immersed in a fluid of weight density ρ, and use Riemann sums to derive the formula. 6. Let R be the region in the first quadrant enclosed by y = x 2 , y = 2 + x, and x = 0. In each part, set up, but do not eval-

uate, an integral or a sum of integrals that will solve the problem. (a) Find the area of R by integrating with respect to x. (b) Find the area of R by integrating with respect to y. (c) Find the volume of the solid generated by revolving R about the x-axis by integrating with respect to x. (d) Find the volume of the solid generated by revolving R about the x-axis by integrating with respect to y. (e) Find the volume of the solid generated by revolving R about the y-axis by integrating with respect to x. (f ) Find the volume of the solid generated by revolving R about the y-axis by integrating with respect to y. (g) Find the volume of the solid generated by revolving R about the line y = −3 by integrating with respect to x. (h) Find the volume of the solid generated by revolving R about the line x = 5 by integrating with respect to x.

7. (a) Set up a sum of deﬁnite integrals that represents the total shaded area between the curves y = f(x) and y = g(x) in the accompanying ﬁgure on the next page. (cont.)

486

Chapter 6 / Applications of the Definite Integral in Geometry, Science, and Engineering

(b) Find the total area enclosed between y = x 3 and y = x over the interval [â&#x2C6;&#x2019;1, 2]. y

y = f (x)

17. Consider the solid generated by revolving the region enclosed by y = sec x, x = 0, x = Ď&#x20AC;/3, and y = 0 about the x-axis. Find the average value of the area of a cross section of this solid taken perpendicular to the x-axis.

y = g(x)

Figure Ex-7

8. The accompanying ďŹ gure shows velocity versus time curves for two cars that move along a straight track, accelerating from rest at a common starting line. (a) How far apart are the cars after 60 seconds? (b) How far apart are the cars after T seconds, where 0 â&#x2030;¤ T â&#x2030;¤ 60? v (ft /s)

180 v1(t) = 3t v2 (t) = t 2 /20 t (s) 60

(b) Find the area of the surface generated by revolving C about the y-axis by integrating with respect to y. (c) Find the area of the surface generated by revolving C about the line y = â&#x2C6;&#x2019;2 by integrating with respect to y.

Figure Ex-8

9. Let R be the region enclosed by the curves y = x 2 + 4, y = x 3 , and the y-axis. Find and evaluate a deďŹ nite integral that represents the volume of the solid generated by revolving R about the x-axis. 10. A football has the shape of the solid generated by revolving the region bounded between the x-axis and the parabola y = 4R(x 2 â&#x2C6;&#x2019; 41 L2 )/L2 about the x-axis. Find its volume. 11. Find the volume of the solid whose â&#x2C6;&#x161; base is the â&#x2C6;&#x161;region bounded between the curves y = x and y = 1/ x for 1 â&#x2030;¤ x â&#x2030;¤ 4 and whose cross sections perpendicular to the x-axis are squares. 12. Consider the region enclosed by y = sinâ&#x2C6;&#x2019;1 x, y = 0, and x = 1. Set up, but do not evaluate, an integral that represents the volume of the solid generated by revolving the region about the x-axis using (a) disks (b) cylindrical shells. 13. Find the arc length in the second quadrant of the curve x 2/3 + y 2/3 = 4 from x = â&#x2C6;&#x2019;8 to x = â&#x2C6;&#x2019;1.

14. Let C be the curve y = ex between x = 0 and x = ln 10. In each part, set up, but do not evaluate, an integral that solves the problem. (a) Find the arc length of C by integrating with respect to x. (b) Find the arc length of C by integrating with respect to y. 15. Find â&#x2C6;&#x161; the area of the surface generated by revolving the curve y = 25 â&#x2C6;&#x2019; x, 9 â&#x2030;¤ x â&#x2030;¤ 16, about the x-axis. 16. Let C be the curve 27x â&#x2C6;&#x2019; y 3 = 0 between y = 0 and y = 2. In each part, set up, but do not evaluate, an integral or a sum of integrals that solves the problem. (a) Find the area of the surface generated by revolving C about the x-axis by integrating with respect to x.

18. Consider the solid â&#x2C6;&#x161; generated by revolving the region enclosed by y = a 2 â&#x2C6;&#x2019; x 2 and y = 0 about the x-axis. Without performing an integration, ďŹ nd the average value of the area of a cross section of this solid taken perpendicular to the x-axis. 19. (a) A spring exerts a force of 0.5 N when stretched 0.25 m beyond its natural length. Assuming that Hookeâ&#x20AC;&#x2122;s law applies, how much work was performed in stretching the spring to this length? (b) How far beyond its natural length can the spring be stretched with 25 J of work? 20. A boat is anchored so that the anchor is 150 ft below the surface of the water. In the water, the anchor weighs 2000 lb and the chain weighs 30 lb/ft. How much work is required to raise the anchor to the surface? 21â&#x20AC;&#x201C;22 Find the centroid of the region. â&#x2013;

21. The region bounded by y 2 = 4x and y 2 = 8(x â&#x2C6;&#x2019; 2). 22. The upper half of the ellipse (x /a)2 + (y /b)2 = 1.

23. In each part, set up, but do not evaluate, an integral that solves the problem. (a) Find the ďŹ&#x201A;uid force exerted on a side of a box that has a 3 m square base and is ďŹ lled to a depth of 1 m with a liquid of weight density Ď N/m3 . (b) Find the ďŹ&#x201A;uid force exerted by a liquid of weight density Ď lb/ft3 on a face of the vertical plate shown in part (a) of the accompanying ďŹ gure. (c) Find the ďŹ&#x201A;uid force exerted on the parabolic dam in part (b) of the accompanying ďŹ gure by water that extends to the top of the dam. 1 ft 2 ft

25 m

10 m 4 ft

(a)

(b)

Figure Ex-23

24. Show that for any constant a, the function y = sinh(ax) satisďŹ es the equation y â&#x20AC;˛â&#x20AC;˛ = a 2 y.

25. In each part, prove the identity. (a) cosh 3x = 4 cosh3 x â&#x2C6;&#x2019; 3 cosh x (b) cosh 21 x =

1 (cosh x + 1) 2 (c) sinh 21 x = Âą 21 (cosh x â&#x2C6;&#x2019; 1)

Chapter 6 Making Connections

487

CHAPTER 6 MAKING CONNECTIONS 1. Suppose that f is a nonnegative function deﬁned on [0, 1] such that the area between the graph of f and the interval [0, 1] is A1 and such that the area of the region R between the graph of g(x) = f(x 2 ) and the interval [0, 1] is A2 . In each part, express your answer in terms of A1 and A2 . (a) What is the volume of the solid of revolution generated by revolving R about the y-axis? (b) Find a value of a such that if the xy-plane were horizontal, the region R would balance on the line x = a.

2. A water tank has the shape of a conical frustum with radius of the base 5 ft, radius of the top 10 ft and (vertical) height 15 ft. Suppose the tank is filled with water and consider the problem of finding the work required to pump all the water out through a hole in the top of the tank. (a) Solve this problem using the method of Example 5 in Section 6.6. (b) Solve this problem using Definition 6.6.3. [Hint: Think of the base as the head of a piston that expands to a watertight fit against the sides of the tank as the piston is pushed upward. What important result about water pressure do you need to use?]

3. A disk of radius a is an inhomogeneous lamina whose density is a function f(r) of the distance r to the center of the

lamina. Modify the argument used to derive the method of cylindrical shells to find a formula for the mass of the lamina. 4. Compare Formula (10) in Section 6.7 with Formula (8) in Section 6.8. Then give a plausible argument that the force on a flat surface immersed vertically in a fluid of constant weight density is equal to the product of the area of the surface and the pressure at the centroid of the surface. Conclude that the force on the surface is the same as if the surface were immersed horizontally at the depth of the centroid. 5. Archimedes’ Principle states that a solid immersed in a fluid experiences a buoyant force equal to the weight of the fluid displaced by the solid. (a) Use the results of Section 6.8 to verify Archimedes’ Principle in the case of (i) a box-shaped solid with a pair of faces parallel to the surface of the fluid, (ii) a solid cylinder with vertical axis, and (iii) a cylindrical shell with vertical axis. (b) Give a plausible argument for Archimedes’ Principle in the case of a solid of revolution immersed in fluid such that the axis of revolution of the solid is vertical. [Hint: Approximate the solid by a union of cylindrical shells and use the result from part (a).]

7 PRINCIPLES OF INTEGRAL EVALUATION © AP/Wide World Photos

The ﬂoating roof on the Stade de France sports complex is an ellipse. Finding the arc length of an ellipse involves numerical integration techniques introduced in this chapter.

7.1

In earlier chapters we obtained many basic integration formulas as an immediate consequence of the corresponding differentiation formulas. For example, knowing that the derivative of sin x is cos x enabled us to deduce that the integral of cos x is sin x. Subsequently, we expanded our integration repertoire by introducing the method of u-substitution. That method enabled us to integrate many functions by transforming the integrand of an unfamiliar integral into a familiar form. However, u-substitution alone is not adequate to handle the wide variety of integrals that arise in applications, so additional integration techniques are still needed. In this chapter we will discuss some of those techniques, and we will provide a more systematic procedure for attacking unfamiliar integrals. We will talk more about numerical approximations of definite integrals, and we will explore the idea of integrating over infinite intervals.

AN OVERVIEW OF INTEGRATION METHODS In this section we will give a brief overview of methods for evaluating integrals, and we will review the integration formulas that were discussed in earlier sections.

METHODS FOR APPROACHING INTEGRATION PROBLEMS There are three basic approaches for evaluating unfamiliar integrals:

• Technology—CAS programs such as Mathematica, Maple, and the open source pro•

gram Sage are capable of evaluating extremely complicated integrals, and such programs are increasingly available for both computers and handheld calculators. Tables—Prior to the development of CAS programs, scientists relied heavily on tables to evaluate difﬁcult integrals arising in applications. Such tables were compiled over many years, incorporating the skills and experience of many people. One such table appears in the endpapers of this text, but more comprehensive tables appear in various reference books such as the CRC Standard Mathematical Tables and Formulae, CRC Press, Inc., 2002.

• Transformation Methods—Transformation methods are methods for converting unfamiliar integrals into familiar integrals. These include u-substitution, algebraic manipulation of the integrand, and other methods that we will discuss in this chapter. 488

7.1 An Overview of Integration Methods

489

None of the three methods is perfect; for example, CAS programs often encounter integrals that they cannot evaluate and they sometimes produce answers that are unnecessarily complicated, tables are not exhaustive and may not include a particular integral of interest, and transformation methods rely on human ingenuity that may prove to be inadequate in difﬁcult problems. In this chapter we will focus on transformation methods and tables, so it will not be necessary to have a CAS such as Mathematica, Maple, or Sage. However, if you have a CAS, then you can use it to conﬁrm the results in the examples, and there are exercises that are designed to be solved with a CAS. If you have a CAS, keep in mind that many of the algorithms that it uses are based on the methods we will discuss here, so an understanding of these methods will help you to use your technology in a more informed way. A REVIEW OF FAMILIAR INTEGRATION FORMULAS The following is a list of basic integrals that we have encountered thus far:

CONSTANTS, POWERS, EXPONENTIALS 1. du = u + C 2. a du = a du = au + C 3. 5.

ur du =

ur+1 + C, r = −1 r +1

eu du = eu + C

du = ln |u| + C u bu 6. bu du = + C, b > 0, b = 1 ln b

TRIGONOMETRIC FUNCTIONS 7. sin u du = − cos u + C 8. cos u du = sin u + C 9. sec2 u du = tan u + C 10. csc2 u du = − cot u + C 11. sec u tan u du = sec u + C 12. csc u cot u du = − csc u + C 13. tan u du = − ln | cos u| + C 14. cot u du = ln |sin u| + C HYPERBOLIC FUNCTIONS 15. sinh u du = cosh u + C

16.

cosh u du = sinh u + C csch2 u du = − coth u + C csch u coth u du = −csch u + C

17.

sech u du = tanh u + C

18.

19.

sech u tanh u du = −sech u + C

20.

ALGEBRAIC FUNCTIONS (a > 0) u du (|u| < a) = sin−1 + C 21. 2 2 a a −u du 1 u 22. = tan−1 + C 2 2 a +u a a u du 1 (0 < a < |u|) = sec−1 + C 23. 2 2 a a u u −a

490

Chapter 7 / Principles of Integral Evaluation

24. 25.

REMARK

a 2 + u2 du

= ln(u +

u2 + a 2 ) + C

= ln u + u2 − a 2 + C

(0 < a < |u|) u2 − a 2 a + u du 1 +C 26. = ln a 2 − u2 2a a − u du 1 a + a 2 − u2 27. (0 < |u| < a) = − ln +C a u u a 2 − u2 du 1 a + a 2 + u2 28. = − ln +C a u u a 2 + u2

Formula 25 is a generalization of a result in Theorem 6.9.6. Readers who did not cover Section 6.9 can ignore Formulas 24–28 for now, since we will develop other methods for obtaining them in this chapter.

✔QUICK CHECK EXERCISES 7.1

(See page 491 for answers.)

1. Use algebraic manipulation and (if necessary) u-substitution to integrate the function. x+1 (a) dx = x x+2 (b) dx = x+1 2x + 1 (c) dx = 2 x +1 (d)

xe3 ln x dx =

2. Use trigonometric identities and (if necessary) u-substitution to integrate the function. 1 (a) dx = csc x 1 (b) dx = cos2 x

(c)

(cot2 x + 1) dx =

1 dx = sec x + tan x 3. Integrate the function. √ x − 1 dx = (a) (b) e2x+1 dx = (c) (sin3 x cos x + sin x cos3 x) dx = 1 (d) dx = (ex + e−x )2 (d)

EXERCISE SET 7.1 1–30 Evaluate the integrals by making appropriate u-substi-

tutions and applying the formulas reviewed in this section. ■ √ 1. (4 − 2x)3 dx 2. 3 4 + 2x dx 3. x sec2 (x 2 ) dx 4. 4x tan(x 2 ) dx 1 sin 3x dx dx 6. 5. 2 + cos 3x 9 + 4x 2 sec(ln x) tan(ln x) 7. ex sinh(ex ) dx 8. dx x x 9. etan x sec2 x dx 10. dx 1 − x4

11.

13.

15.

cos5 5x sin 5x dx ex dx 4 + e2x √

e √

12. 14.

16.

x−1

x−1 √ cosh x 17. dx √ x dx 19. √ √x x3

18.

cos x dx sin x sin2 x + 1 −1

etan x dx 1 + x2

(x + 1) cot(x 2 + 2x) dx dx x(ln x)2

7.2 Integration by Parts

20. 21.

e−x dx 4 − e−2x ex dx 25. 1 − e2x x dx 27. csc(x 2 ) 2 29. x4−x dx 23.

32. (a) Derive the identity

sec(sin θ) tan(sin θ) cos θ dθ csch2 (2/x) dx x2

dx x2 − 4 cos(ln x) 24. dx x sinh(x −1/2 ) dx 26. x 3/2 ex dx 28. 4 − e2x 30. 2πx dx 22.

491

sech2 x = sech 2x 1 + tanh2 x (b) Use the result in part (a) to evaluate sech x dx. (c) Derive the identity sech x =

(d) Use the result in part (c) to evaluate sech x dx. (e) Explain why your answers to parts (b) and (d) are consistent. 33. (a) Derive the identity sec2 x 1 = tan x sin x cos x

F O C U S O N C O N C E P TS

31. (a) Evaluate the integral sin x cos x dx using the substitution u = sin x. (b) Evaluate the integral sin x cos x dx using the identity sin 2x = 2 sin x cos x. (c) Explain why your answers to parts (a) and (b) are consistent.

2ex +1

e2x

(b) Use the identity sin 2x = 2 sin x cos x along with the result in part (a) to evaluate csc x dx. (c) Use the identity cos x = sin[(π/2) − x] along with your answer to part (a) to evaluate sec x dx.

✔QUICK CHECK ANSWERS 7.1 x5 + C 2. (a) − cos x + C (b) tan x + C 5 2 1 2x+1 3/2 3. (a) 3 (x − 1) + C (b) 2 e + C (c) 21 sin2 x + C (d) 41 tanh x + C

1. (a) x + ln |x| + C (b) x + ln |x + 1| + C (c) ln(x 2 + 1) + tan−1 x + C (d) (c) − cot x + C (d) ln(1 + sin x) + C

7.2

INTEGRATION BY PARTS In this section we will discuss an integration technique that is essentially an antiderivative formulation of the formula for differentiating a product of two functions. THE PRODUCT RULE AND INTEGRATION BY PARTS

Our primary goal in this section is to develop a general method for attacking integrals of the form f(x)g(x) dx

As a ﬁrst step, let G(x) be any antiderivative of g(x). In this case G′ (x) = g(x), so the product rule for differentiating f(x)G(x) can be expressed as d [f(x)G(x)] = f(x)G′ (x) + f ′ (x)G(x) = f(x)g(x) + f ′ (x)G(x) (1) dx This implies that f(x)G(x) is an antiderivative of the function on the right side of (1), so we can express (1) in integral form as [f(x)g(x) + f ′ (x)G(x)] dx = f(x)G(x)

492

Chapter 7 / Principles of Integral Evaluation

or, equivalently, as

f(x)g(x) dx = f(x)G(x) −

f ′ (x)G(x) dx

(2)

This formula allows us to integrate f(x)g(x) by integrating f ′ (x)G(x) instead, and in many cases the net effect is to replace a difﬁcult integration with an easier one. The application of this formula is called integration by parts. In practice, we usually rewrite (2) by letting u = f(x), du = f ′(x) dx v = G(x), dv = G′ (x) dx = g(x) dx

This yields the following alternative form for (2): u dv = uv − v du Note that in Example 1 we omitted the constant of integration in calculating v from dv . Had we included a constant of integration, it would have eventually dropped out. This is always the case in integration by parts [Exercise 68(b)], so it is common to omit the constant at this stage of the computation. However, there are certain cases in which making a clever choice of a constant of integration to include with v can sim plify the computation of v du (Exercises 69–71).

Example 1

Use integration by parts to evaluate

(3)

x cos x dx.

Solution. We will apply Formula (3). The ﬁrst step is to make a choice for u and dv to put the given integral in the form u dv. We will let and

u=x

dv = cos x dx

(Other possibilities will be considered later.) The second step is to compute du from u and v from dv. This yields du = dx and v = dv = cos x dx = sin x

The third step is to apply Formula (3). This yields cos x dx − sin x dx = x sin x x u

= x sin x − (− cos x) + C = x sin x + cos x + C

GUIDELINES FOR INTEGRATION BY PARTS The main goal in integration by parts is to choose u and dv to obtain a new integral that is easier to evaluate than the original. In general, there are no hard and fast rules for doing this; it is mainly a matter of experience that comes from lots of practice. A strategy that often works is to choose u and dv so that u becomes “simpler” when differentiated, while leaving a dv that can be readily integrated to obtain v. Thus, for the integral x cos x dx in Example 1, both goals were achieved by letting u = x and dv = cos x dx. In contrast, u = cos x would not have been a good ﬁrst choice in that example, since du/dx = − sin x is no simpler than u. Indeed, had we chosen

u = cos x du = − sin x dx

dv = x dx x2 v = x dx = 2

then we would have obtained 2 x2 x x2 1 x cos x dx = x 2 sin x dx cos x − (− sin x) dx = cos x + 2 2 2 2 For this choice of u and dv, the new integral is actually more complicated than the original.

7.2 Integration by Parts

The LIATE method is discussed in the article “A Technique for Integration by Parts,” American Mathematical Monthly, Vol. 90, 1983, pp. 210–211, by Herbert Kasube.

493

There is another useful strategy for choosing u and dv that can be applied when the integrand is a product of two functions from different categories in the list Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential In this case you will often be successful if you take u to be the function whose category occurs earlier in the list and take dv to be the rest of the integrand. The acronym LIATE will help you to remember the order. The method does not work all the time, but it works often enough to be useful. Note, for example, that the integrand in Example 1 consists of the product of the algebraic function x and the trigonometric function cos x. Thus, the LIATE method suggests that we should let u = x and dv = cos x dx, which proved to be a successful choice. Example 2

Evaluate

xex dx.

Solution. In this case the integrand is the product of the algebraic function x with the exponential function ex . According to LIATE we should let and dv = ex dx

u=x so that du = dx

and

ex dx = ex

Thus, from (3) xex dx = u dv = uv − v du = xex − ex dx = xex − ex + C Example 3

Evaluate

ln x dx.

Solution. One choice is to let u = 1 and dv = ln x dx. But with this choice ﬁnding v is equivalent to evaluating reasonable choice is to let

ln x dx and we have gained nothing. Therefore, the only

u = ln x 1 du = dx x

dv = dx v = dx = x

With this choice it follows from (3) that ln x dx = u dv = uv − v du = x ln x − dx = x ln x − x + C REPEATED INTEGRATION BY PARTS

It is sometimes necessary to use integration by parts more than once in the same problem. Example 4

Evaluate

x 2 e−x dx.

Solution. Let 2

u=x ,

dv = e

−x

dx,

du = 2x dx,

e−x dx = −e−x

494

Chapter 7 / Principles of Integral Evaluation

so that from (3)

u dv = uv − v du = x 2 (−e−x ) − −e−x (2x) dx = − x 2 e−x + 2 xe−x dx

x 2 e−x dx =

(4)

The last integral is similar to the original except that we have replaced x 2 by x. Another integration by parts applied to xe−x dx will complete the problem. We let −x u = x, dv = e dx, du = dx, v = e−x dx = −e−x

so that xe−x dx = x(−e−x ) − −e−x dx = −xe−x + e−x dx = −xe−x − e−x + C Finally, substituting this into the last line of (4) yields x 2 e−x dx = −x 2 e−x + 2 xe−x dx = −x 2 e−x + 2(−xe−x − e−x ) + C = − (x 2 + 2x + 2)e−x + C

The LIATE method suggests that integrals of the form ax e sin bx dx and eax cos bx dx

can be evaluated by letting u = sin bx or u = cos bx and dv = eax dx. However, this will require a technique that deserves special attention.

Example 5

Evaluate

ex cos x dx.

Solution. Let u = cos x,

dv = ex dx,

du = − sin x dx,

ex dx = ex

Thus,

ex cos x dx =

u dv = uv −

ex sin x dx

ex cos x dx

v du = ex cos x +

(5)

Since the integral ex sin x dx is similar in form to the original integral ex cos x dx, it seems that nothing has been accomplished. However, let us integrate this new integral by parts. We let x u = sin x, dv = e dx, du = cos x dx, v = ex dx = ex Thus,

e sin x dx =

u dv = uv −

v du = e sin x −

Together with Equation (5) this yields x x x e cos x dx = e cos x + e sin x − ex cos x dx

(6)

7.2 Integration by Parts

495

which is an equation we can solve for the unknown integral. We obtain 2 ex cos x dx = ex cos x + ex sin x and hence

ex cos x dx = 21 ex cos x + 21 ex sin x + C

A TABULAR METHOD FOR REPEATED INTEGRATION BY PARTS

Integrals of the form More information on tabular integration by parts can be found in the articles “Tabular Integration by Parts,” College Mathematics Journal, Vol. 21, 1990, pp. 307–311, by David Horowitz and “More on Tabular Integration by Parts,” College Mathematics Journal, Vol. 22, 1991, pp. 407–410, by Leonard Gillman.

p(x)f(x) dx

where p(x) is a polynomial, can sometimes be evaluated using repeated integration by parts in which u is taken to be p(x) or one of its derivatives at each stage. Since du is computed by differentiating u, the repeated differentiation of p(x) will eventually produce 0, at which point you may be left with a simpliﬁed integration problem. A convenient method for organizing the computations into two columns is called tabular integration by parts.

Tabular Integration by Parts Step 1. Differentiate p(x) repeatedly until you obtain 0, and list the results in the ﬁrst column. Step 2. Integrate f(x) repeatedly and list the results in the second column. Step 3. Draw an arrow from each entry in the ﬁrst column to the entry that is one row down in the second column. Step 4. Label the arrows with alternating + and − signs, starting with a +. Step 5. For each arrow, form the product of the expressions at its tip and tail and then multiply that product by +1 or −1 in accordance with the sign on the arrow. Add the results to obtain the value of the integral.

This process is illustrated in Figure 7.2.1 for the integral (x 2 − x) cos x dx. repeated differentiation x2 − x

cos x

2x − 1

−

sin x

−cos x

Figure 7.2.1

repeated integration

−sin x

(x 2 − x) cos x dx = (x 2 − x) sin x + (2x − 1) cos x − 2 sin x + C = (x 2 − x − 2) sin x + (2x − 1) cos x + C

√ Example 6 In Example 11 of Section 5.3 we evaluated x 2 x − 1 dx using u-substitution. Evaluate this integral using tabular integration by parts.

496

Chapter 7 / Principles of Integral Evaluation

Solution. repeated differentiation

repeated integration

−

(x − 1)1/2 2_ (x 3 4 (x 15 8 (x 105

0 The result obtained in Example 6 looks quite different from that obtained in Example 11 of Section 5.3. Show that the two answers are equivalent.

Thus, it follows that √ x 2 x − 1 dx = 23 x 2 (x − 1)3/2 −

8 x(x 15

− 1)3/2 − 1)5/2 − 1)7/2

− 1)5/2 +

16 (x 105

− 1)7/2 + C

INTEGRATION BY PARTS FOR DEFINITE INTEGRALS

For deﬁnite integrals the formula corresponding to (3) is REMARK

u dv = uv

−

(7)

v du a

It is important to keep in mind that the variables u and v in this formula are functions of x and that the limits of integration in (7) are limits on the variable x . Sometimes it is helpful to emphasize this by writing (7) as

u dv = uv

x=a

−

(8)

v du x=a

The next example illustrates how integration by parts can be used to integrate the inverse trigonometric functions. Example 7

Evaluate

tan−1 x dx. 0

Solution. Let u = tan−1 x, Thus,

tan

−1

But

x dx =

u dv = uv

= x tan−1 x 1 x dx = 2 1+x 2

tan−1 x dx = x tan−1 x

dv = dx,

− 1

1 0

− 1

du =

1 dx, 1 + x2

v du 0

v=x

The limits of integration refer to x; that is, x = 0 and x = 1.

x dx 1 + x2

1 1 2x 1 2 dx = ln(1 + x ) = ln 2 2 1+x 2 2 0

−

1 √ 1 π ln 2 = − 0 − ln 2 = − ln 2 2 4 2 4

7.2 Integration by Parts

497

REDUCTION FORMULAS

Integration by parts can be used to derive reduction formulas for integrals. These are formulas that express an integral involving a power of a function in terms of an integral that involves a lower power of that function. For example, if n is a positive integer and n â&#x2030;Ľ 2, then integration by parts can be used to obtain the reduction formulas nâ&#x2C6;&#x2019;1 1 nâ&#x2C6;&#x2019;1 n x cos x + sinnâ&#x2C6;&#x2019;2 x dx sin x dx = â&#x2C6;&#x2019; sin (9) n n nâ&#x2C6;&#x2019;1 1 cosnâ&#x2C6;&#x2019;2 x dx cosn x dx = cosnâ&#x2C6;&#x2019;1 x sin x + (10) n n To illustrate how such formulas can be obtained, let us derive (10). We begin by writing cosn x as cosnâ&#x2C6;&#x2019;1 x Âˇ cos x and letting u = cosnâ&#x2C6;&#x2019;1 x

dv = cos x dx

du = (n â&#x2C6;&#x2019; 1) cosnâ&#x2C6;&#x2019;2 x(â&#x2C6;&#x2019; sin x) dx nâ&#x2C6;&#x2019;2

= â&#x2C6;&#x2019;(n â&#x2C6;&#x2019; 1) cos so that

cos x dx =

nâ&#x2C6;&#x2019;1

cos

nâ&#x2C6;&#x2019;1

= cos

nâ&#x2C6;&#x2019;1

= cos

nâ&#x2C6;&#x2019;1

= cos

v = sin x

x sin x dx

x cos x dx =

x sin x + (n â&#x2C6;&#x2019; 1) x sin x + (n â&#x2C6;&#x2019; 1) x sin x + (n â&#x2C6;&#x2019; 1)

u dv = uv â&#x2C6;&#x2019;

v du

sin2 x cosnâ&#x2C6;&#x2019;2 x dx (1 â&#x2C6;&#x2019; cos2 x) cosnâ&#x2C6;&#x2019;2 x dx nâ&#x2C6;&#x2019;2

cos

x dx â&#x2C6;&#x2019; (n â&#x2C6;&#x2019; 1)

cosn x dx

Moving the last term on the right to the left side yields n nâ&#x2C6;&#x2019;1 n cos x dx = cos x sin x + (n â&#x2C6;&#x2019; 1) cosnâ&#x2C6;&#x2019;2 x dx

from which (10) follows. The derivation of reduction formula (9) is similar (Exercise 63). Reduction formulas (9) and (10) reduce the exponent of sine (or cosine) by 2. Thus, if the formulas are applied repeatedly, the exponent can eventually be reduced to 0 if n is even or 1 if n is odd, at which point the integration can be completed. We will discuss this method in more detail in the next section, but for now, here is an example that illustrates how reduction formulas work. Example 8 Evaluate cos4 x dx.

Solution. From (10) with n = 4

cos4 x dx =

1 4

cos3 x sin x +

3 4

1 4

cos3 x sin x +

3 4

1 4

cos3 x sin x + 38 cos x sin x + 83 x + C

cos2 x dx 1 2

cos x sin x +

Now apply (10) with n = 2. 1 2

Chapter 7 / Principles of Integral Evaluation

498

✔QUICK CHECK EXERCISES 7.2

(See page 500 for answers.)

1. (a) If G′ (x) = g(x), then f(x)g(x) dx = f(x)G(x) −

(b) If u = f(x) and v = G(x), then the formula in part (a) can be written in the form u dv = .

2. Find an appropriate choice of u and dv for integration by parts of each integral. Do not evaluate the integral. (a)

(b)

x ln x dx; u =

, dv =

(x − 2) sin x dx; u =

sin−1 x dx; u =

, dv =

dx; u = , dv = x−1 3. Use integration by parts to evaluate the integral. √

xe2x dx

(a)

(c)

(b)

ln(x − 1) dx

π/6

x sin 3x dx 0

4. Use a reduction formula to evaluate

, dv =

sin3 x dx.

EXERCISE SET 7.2 1–38 Evaluate the integral. ■

1. 3. 5. 7. 9. 11. 13. 15. 17. 19. 21. 23. 25. 27. 29. 31.

xe−2x dx

x 2 ex dx

x 2 cos x dx

x ln x dx

10.

(ln x)2 dx

12.

ln(3x − 2) dx

14.

−1

x dx

16.

tan−1 (3x) dx

18.

ex sin x dx

20.

sin(ln x) dx

22.

x sec x dx 2

x 3 ex dx

x sin 3x dx

sin

24. 26.

xe2x dx

28.

0 e

x 2 ln x dx

30.

1 1

−1

ln(x + 2) dx

32.

33.

x 2 e−2x dx x cos 2x dx

θ dθ

34.

x sec−1 x dx

x sin 2x dx

36.

(x + x cos x) dx

37.

√ √ x tan−1 x dx

38.

ln(x 2 + 1) dx

√ x ln x dx

39. The main goal in integration by parts is to choose u and dv to obtain a new integral that is easier to evaluate than the original. 40. Applying the LIATE strategy to evaluate x 3 ln x dx, we should choose u = x 3 and dv = ln x dx. 41. To evaluate ln ex dx using integration by parts, choose dv = ex dx.

ln x √ dx x ln(x 2 + 4) dx cos−1 (2x) dx

e3x cos 2x dx

42. Tabular integration by parts is useful for integrals of the form p(x)f(x) dx, where p(x) is a polynomial and f(x) can be repeatedly integrated. 43–44 Evaluate the integral by making a u-substitution and

cos(ln x) dx

xex dx (x + 1)2

xe−5x dx 0 e

sin

−1

then integrating by parts. ■ √ 43. e x dx

44.

cos

√ x dx

45. Prove that tabular integration by parts gives the correct answer for p(x)f(x) dx where p(x) is any quadratic polynomial and f(x) is any function that can be repeatedly integrated.

ln x dx x2

√ e √3/2 0

√

39–42 True–False Determine whether the statement is true or false. Explain your answer. ■

x tan x dx

sec−1

x 2 sin x dx

35.

x tan−1 x dx

xe3x dx

x dx

46. The computations of any integral evaluated by repeated integration by parts can be organized using tabular integration by parts. Use this organization to evaluate ex cos x dx in

7.2 Integration by Parts

two ways: ﬁrst by repeated differentiation of cos x (compare Example 5), and then by repeated differentiation of ex . 47–52 Evaluate the integral using tabular integration by parts. ■

47. 49. 51.

(3x − x + 2)e

−x

48.

4x 4 sin 2x dx

50.

e sin bx dx

52.

(x + x + 1) sin x dx

e−3θ sin 5θ dθ

√ x 3 2x + 1 dx

53. Consider the integral sin x cos x dx. (a) Evaluate the integral two ways: ﬁrst using integration by parts, and then using the substitution u = sin x. (b) Show that the results of part (a) are equivalent. (c) Which of the two methods do you prefer? Discuss the reasons for your preference. 54. Evaluate the integral

√

x3 x2 + 1

using (a) integration by parts √ (b) the substitution u = x 2 + 1.

64. In each part, use integration by parts or other methods to derive the reduction formula. n−2 secn−2 x tan x (a) secn x dx = + secn−2 x dx n − 1 n − 1 tann−1 x − tann−2 x dx (b) tann x dx = n−1 (c)

x n ex dx = x n ex − n

x n−1 ex dx

65–66 Use the reduction formulas in Exercise 64 to evaluate the integrals. ■ 65. (a) tan4 x dx (b) sec4 x dx (c) x 3 ex dx 1 √ 66. (a) x 2 e3x dx xe− x dx (b)

[Hint: First make a substitution.]

67. Let f be a function whose second derivative is continuous on [−1, 1]. Show that 1 xf ′′(x) dx = f ′(1) + f ′(−1) − f(1) + f(−1) −1

F O C U S O N C O N C E P TS

68. (a) In the integral

55. (a) Find the area of the region enclosed by y = ln x, the line x = e, and the x-axis. (b) Find the volume of the solid generated when the region in part (a) is revolved about the x-axis. 56. Find the area of the region between y = x sin x and y = x for 0 ≤ x ≤ π/2.

57. Find the volume of the solid generated when the region between y = sin x and y = 0 for 0 ≤ x ≤ π is revolved about the y-axis. 58. Find the volume of the solid generated when the region enclosed between y = cos x and y = 0 for 0 ≤ x ≤ π/2 is revolved about the y-axis. 59. A particle moving along the x-axis has velocity function v(t) = t 3 sin t. How far does the particle travel from time t = 0 to t = π?

60. The study of sawtooth waves in electrical engineering leads to integrals of the form π/ω t sin(kωt) dt −π/ω

where k is an integer and ω is a nonzero constant. Evaluate the integral.

61. Use reduction formula (9) to evaluate π/2 (a) sin4 x dx sin5 x dx. (b) 0

62. Use reduction formula (10) to evaluate π/2 cos6 x dx. (b) (a) cos5 x dx 0

63. Derive reduction formula (9).

499

x cos x dx, let

u = x,

dv = cos x dx,

du = dx, v = sin x + C1 Show that the constant C1 cancels out, thus giving the same solution obtained by omitting C1 . (b) Show that in general uv − v du = u(v + C1 ) − (v + C1 ) du

thereby justifying the omission of the constant of integration when calculating v in integration by parts. 69. Evaluate ln(x + 1) dx using integration by parts. Simplify the computation of v du by introducing a constant of integration C1 = 1 when going from dv to v. 70. Evaluate ln(3x − 2) dx using integration by parts. Simplify the computation of v du by introducing a constant of integration C1 = − 23 when going from dv to v. Compare your solution with your answer to Exercise 13. 71. Evaluate x tan−1 x dx using integration by parts. Simplify the computation of v du by introducing a constant of integration C1 = 21 when going from dv to v.

72. What equation results if integration by parts is applied to the integral 1 dx x ln x with the choices 1 1 u= and dv = dx? ln x x In what sense is this equation true? In what sense is it false?

500

Chapter 7 / Principles of Integral Evaluation

73. Writing Explain how the product rule for derivatives and the technique of integration by parts are related.

techniques? Describe situations, with examples, where each of these techniques would be preferred over the other.

74. Writing For what sort of problems are the integration techniques of substitution and integration by parts â&#x20AC;&#x153;competingâ&#x20AC;?

â&#x153;&#x201D;QUICK CHECK ANSWERS 7.2 1 dx f â&#x20AC;˛ (x)G(x) dx (b) uv â&#x2C6;&#x2019; v du 2. (a) ln x; x dx (b) x â&#x2C6;&#x2019; 2; sin x dx (c) sinâ&#x2C6;&#x2019;1 x; dx (d) x; â&#x2C6;&#x161; xâ&#x2C6;&#x2019;1 x 1 2x 3. (a) â&#x2C6;&#x2019; e + C (b) (x â&#x2C6;&#x2019; 1) ln(x â&#x2C6;&#x2019; 1) â&#x2C6;&#x2019; x + C (c) 91 4. â&#x2C6;&#x2019; 13 sin2 x cos x â&#x2C6;&#x2019; 23 cos x + C 2 4 1. (a)

7.3

INTEGRATING TRIGONOMETRIC FUNCTIONS In the last section we derived reduction formulas for integrating positive integer powers of sine, cosine, tangent, and secant. In this section we will show how to work with those reduction formulas, and we will discuss methods for integrating other kinds of integrals that involve trigonometric functions. INTEGRATING POWERS OF SINE AND COSINE We begin by recalling two reduction formulas from the preceding section. 1 nâ&#x2C6;&#x2019;1 sinn x dx = â&#x2C6;&#x2019; sinnâ&#x2C6;&#x2019;1 x cos x + sinnâ&#x2C6;&#x2019;2 x dx n n

cosn x dx =

nâ&#x2C6;&#x2019;1 1 cosnâ&#x2C6;&#x2019;1 x sin x + n n

cosnâ&#x2C6;&#x2019;2 x dx

In the case where n = 2, these formulas yield sin2 x dx = â&#x2C6;&#x2019; 21 sin x cos x + 21 dx = 21 x â&#x2C6;&#x2019; 21 sin x cos x + C

cos2 x dx =

1 2

cos x sin x +

1 2

dx = 21 x + 21 sin x cos x + C

(1)

(2)

(3) (4)

Alternative forms of these integration formulas can be derived from the trigonometric identities sin2 x = 21 (1 â&#x2C6;&#x2019; cos 2x) and cos2 x = 21 (1 + cos 2x) (5â&#x20AC;&#x201C;6) which follow from the double-angle formulas cos 2x = 1 â&#x2C6;&#x2019; 2 sin2 x These identities yield sin2 x dx =

cos2 x dx =

and

cos 2x = 2 cos2 x â&#x2C6;&#x2019; 1

1 2

(1 â&#x2C6;&#x2019; cos 2x) dx = 21 x â&#x2C6;&#x2019; 41 sin 2x + C

(7)

1 2

(1 + cos 2x) dx = 21 x + 41 sin 2x + C

(8)

7.3 Integrating Trigonometric Functions

501

Observe that the antiderivatives in Formulas (3) and (4) involve both sines and cosines, whereas those in (7) and (8) involve sines alone. However, the apparent discrepancy is easy to resolve by using the identity sin 2x = 2 sin x cos x to rewrite (7) and (8) in forms (3) and (4), or conversely. In the case where n = 3, the reduction formulas for integrating sin3 x and cos3 x yield 3 1 2 2 sin x dx = â&#x2C6;&#x2019; 3 sin x cos x + 3 sin x dx = â&#x2C6;&#x2019; 13 sin2 x cos x â&#x2C6;&#x2019; 23 cos x + C (9)

T E C H N O LO GY M A ST E R Y The Maple CAS produces forms (11) and (12) when asked to integrate sin3 x and cos3 x , but Mathematica produces

sin3 x dx = â&#x2C6;&#x2019; 43 cos x

cos3 x dx =

+ 3 4

1 12

cos 3x + C

sin x 1 12

sin 3x + C

Use trigonometric identities to reconcile the results of the two programs.

1 3

cos2 x sin x +

2 3

cos x dx =

cos2 x sin x + 23 sin x + C

1 3

(10)

If desired, Formula (9) can be expressed in terms of cosines alone by using the identity sin2 x = 1 â&#x2C6;&#x2019; cos2 x, and Formula (10) can be expressed in terms of sines alone by using the identity cos2 x = 1 â&#x2C6;&#x2019; sin2 x. We leave it for you to do this and conďŹ rm that (11) sin3 x dx = 31 cos3 x â&#x2C6;&#x2019; cos x + C

cos3 x dx = sin x â&#x2C6;&#x2019; 13 sin3 x + C

(12)

We leave it as an exercise to obtain the following formulas by ďŹ rst applying the reduction formulas, and then using appropriate trigonometric identities. 1 (13) sin 4x + C sin4 x dx = 83 x â&#x2C6;&#x2019; 41 sin 2x + 32

cos4 x dx = 83 x + 41 sin 2x +

1 32

sin 4x + C

(14)

Example 1 Find the volume V of the solid that is obtained when the region under the curve y = sin2 x over the interval [0, Ď&#x20AC;] is revolved about the x-axis (Figure 7.3.1).

y = sin2 x

cos3 x dx =

Solution. Using the method of disks, Formula (5) of Section 6.2, and Formula (13) above yields V =

Ď&#x20AC;

Ď&#x20AC; sin4 x dx = Ď&#x20AC;

x â&#x2C6;&#x2019; 41 sin 2x +

1 32

sin 4x

Ď&#x20AC; 0

= 83 Ď&#x20AC;2

INTEGRATING PRODUCTS OF SINES AND COSINES Figure 7.3.1

If m and n are positive integers, then the integral sinm x cosn x dx

can be evaluated by one of the three procedures stated in Table 7.3.1, depending on whether m and n are odd or even.

Example 2

Evaluate (a) sin4 x cos5 x dx

(b)

sin4 x cos4 x dx

502

Chapter 7 / Principles of Integral Evaluation Table 7.3.1 integrating products of sines and cosines

sin m x cosn x dx

procedure

relevant identities

n odd

• Split off a factor of cos x. • Apply the relevant identity. • Make the substitution u = sin x.

cos 2 x = 1 − sin 2 x

m odd

• Split off a factor of sin x. • Apply the relevant identity. • Make the substitution u = cos x.

sin 2 x = 1 − cos 2 x

m even

sin 2 x = 12 (1 − cos 2x)

• Use the relevant identities to reduce the powers on sin x and cos x.

n even

cos 2 x = 12 (1 + cos 2x)

Solution (a). Since n = 5 is odd, we will follow the ﬁrst procedure in Table 7.3.1:

sin x cos x dx = = = =

sin4 x cos4 x cos x dx sin4 x(1 − sin2 x)2 cos x dx u4 (1 − u2 )2 du (u4 − 2u6 + u8 ) du

= 15 u5 − 27 u7 + 91 u9 + C =

1 5

sin5 x − 27 sin7 x + 91 sin9 x + C

Solution (b). Since m = n = 4, both exponents are even, so we will follow the third procedure in Table 7.3.1: 4 4 sin x cos x dx = (sin2 x)2 (cos2 x)2 dx =

1 16

sin4 2x dx

1 32

sin4 u du

u = 2x du = 2 dx or dx =

1 32

u − 41 sin 2u +

1 32

3 x 128

1 2

[1 − cos 2x]

2 1 2

[1 + cos 2x]

(1 − cos2 2x)2 dx

−

1 128

sin 4x +

Note that this can be obtained more directly from the original integral using the identity sin x cos x = 21 sin 2x.

sin 4u + C

1 1024

1 2

sin 8x + C

du Formula (13)

7.3 Integrating Trigonometric Functions

Integrals of the form sin mx cos nx dx,

sin mx sin nx dx,

can be found by using the trigonometric identities

cos mx cos nx dx

503

(15)

sin Îą cos Î˛ = 21 [sin(Îą â&#x2C6;&#x2019; Î˛) + sin(Îą + Î˛)]

(16)

sin Îą sin Î˛ = 21 [cos(Îą â&#x2C6;&#x2019; Î˛) â&#x2C6;&#x2019; cos(Îą + Î˛)]

(17)

cos Îą cos Î˛ = 21 [cos(Îą â&#x2C6;&#x2019; Î˛) + cos(Îą + Î˛)]

(18)

to express the integrand as a sum or difference of sines and cosines. Example 3

Evaluate

sin 7x cos 3x dx.

Solution. Using (16) yields

sin 7x cos 3x dx =

1 2

(sin 4x + sin 10x) dx = â&#x2C6;&#x2019; 81 cos 4x â&#x2C6;&#x2019;

1 20

cos 10x + C

INTEGRATING POWERS OF TANGENT AND SECANT The procedures for integrating powers of tangent and secant closely parallel those for sine and cosine. The idea is to use the following reduction formulas (which were derived in Exercise 64 of Section 7.2) to reduce the exponent in the integrand until the resulting integral can be evaluated: tannâ&#x2C6;&#x2019;1 x n â&#x2C6;&#x2019; tannâ&#x2C6;&#x2019;2 x dx (19) tan x dx = nâ&#x2C6;&#x2019;1 nâ&#x2C6;&#x2019;2 secnâ&#x2C6;&#x2019;2 x tan x (20) + secnâ&#x2C6;&#x2019;2 x dx secn x dx = nâ&#x2C6;&#x2019;1 nâ&#x2C6;&#x2019;1

In the case where n is odd, the exponent can be reduced to 1, leaving us with the problem of integrating tan x or sec x. These integrals are given by tan x dx = ln |sec x| + C (21)

sec x dx = ln |sec x + tan x| + C

Formula (21) can be obtained by writing sin x tan x dx = dx cos x = â&#x2C6;&#x2019; ln |cos x| + C = ln |sec x| + C

(22)

u = cos x du = â&#x2C6;&#x2019; sin x dx ln | cos x| = â&#x2C6;&#x2019; ln

1 | cos x|

To obtain Formula (22) we write sec2 x + sec x tan x sec x + tan x dx dx = sec x dx = sec x sec x + tan x sec x + tan x = ln |sec x + tan x| + C

u = sec x + tan x du = (sec2 x + sec x tan x) dx

504

Chapter 7 / Principles of Integral Evaluation

The following basic integrals occur frequently and are worth noting:

tan2 x dx = tan x − x + C

(23)

(24)

sec2 x dx = tan x + C

Formula (24) is already known to us, since the derivative of tan x is sec2 x. Formula (23) can be obtained by applying reduction formula (19) with n = 2 (verify) or, alternatively, by using the identity 1 + tan2 x = sec2 x to write

The formulas

tan2 x dx =

(sec2 x − 1) dx = tan x − x + C

tan3 x dx =

sec3 x dx =

1 2

tan2 x − ln |sec x| + C

(25)

sec x tan x + 21 ln |sec x + tan x| + C

(26)

can be deduced from (21), (22), and reduction formulas (19) and (20) as follows: tan3 x dx = 21 tan2 x − tan x dx = 21 tan2 x − ln |sec x| + C

sec x dx =

1 2

sec x tan x +

1 2

sec x dx =

1 2

sec x tan x + 21 ln |sec x + tan x| + C

INTEGRATING PRODUCTS OF TANGENTS AND SECANTS

If m and n are positive integers, then the integral tanm x secn x dx can be evaluated by one of the three procedures stated in Table 7.3.2, depending on whether m and n are odd or even.

Table 7.3.2 integrating products of tangents and secants

tan m x secn x dx

procedure

relevant identities

n even

• Split off a factor of sec 2 x. • Apply the relevant identity. • Make the substitution u = tan x.

sec 2 x = tan 2 x + 1

m odd

• Split off a factor of sec x tan x. • Apply the relevant identity. • Make the substitution u = sec x.

tan 2 x = sec 2 x − 1

• Use the relevant identities to reduce the integrand to powers of sec x alone. • Then use the reduction formula for powers of sec x.

tan 2 x = sec 2 x − 1

m even n odd

7.3 Integrating Trigonometric Functions

Example 4 Evaluate (a) tan2 x sec4 x dx

(b)

tan3 x sec3 x dx

(c)

505

tan2 x sec x dx

Solution (a). Since n = 4 is even, we will follow the ﬁrst procedure in Table 7.3.2:

tan2 x sec4 x dx = = =

tan2 x sec2 x sec2 x dx tan2 x(tan2 x + 1) sec2 x dx u2 (u2 + 1) du

= 15 u5 + 31 u3 + C =

1 5

tan5 x + 13 tan3 x + C

Solution (b). Since m = 3 is odd, we will follow the second procedure in Table 7.3.2:

tan3 x sec3 x dx = = =

tan2 x sec2 x(sec x tan x) dx (sec2 x − 1) sec2 x(sec x tan x) dx (u2 − 1)u2 du

= 51 u5 − 31 u3 + C =

1 5

sec5 x − 13 sec3 x + C

Solution (c). Since m = 2 is even and n = 1 is odd, we will follow the third procedure in Table 7.3.2: tan2 x sec x dx = (sec2 x − 1) sec x dx 3 = sec x dx − sec x dx

See (26) and (22)

1 2

sec x tan x + 21 ln |sec x + tan x| − ln |sec x + tan x| + C

1 2

sec x tan x − 21 ln |sec x + tan x| + C

AN ALTERNATIVE METHOD FOR INTEGRATING POWERS OF SINE, COSINE, TANGENT, AND SECANT With the aid of the identity

1 + cot 2 x = csc2 x the techniques in Table 7.3.2 can be adapted to evaluate integrals of the form

cotm x cscn x dx

It is also possible to derive reduction formulas for powers of cot and csc that are analogous to Formulas (19) and (20).

The methods in Tables 7.3.1 and 7.3.2 can sometimes be applied if m = 0 or n = 0 to integrate positive integer powers of sine, cosine, tangent, and secant without reduction formulas. For example, instead of using the reduction formula to integrate sin3 x, we can apply the second procedure in Table 7.3.1: sin3 x dx = (sin2 x) sin x dx u = cos x = (1 − cos2 x) sin x dx du = − sin x dx = − (1 − u2 ) du = 13 u3 − u + C =

which agrees with (11).

1 3

cos3 x − cos x + C

Chapter 7 / Principles of Integral Evaluation

506

MERCATOR’S MAP OF THE WORLD The integral of sec x plays an important role in the design of navigational maps for charting nautical and aeronautical courses. Sailors and pilots usually chart their courses along paths with constant compass headings; for example, the course might be 30 ◦ northeast or 135 ◦ southeast. Except for courses that are parallel to the equator or run due north or south, a course with constant compass heading spirals around the Earth toward one of the poles (as in the top part of Figure 7.3.2). In 1569 the Flemish mathematician and geographer Gerhard Kramer (1512–1594) (better known by the Latin name Mercator) devised a world map, called the Mercator projection, in which spirals of constant compass headings appear as straight lines. This was extremely important because it enabled sailors to determine compass headings between two points by connecting them with a straight line on a map (as in the bottom part of Figure 7.3.2). If the Earth is assumed to be a sphere of radius 4000 mi, then the lines of latitude at 1 ◦ increments are equally spaced about 70 mi apart (why?). However, in the Mercator projection, the lines of latitude become wider apart toward the poles, so that two widely spaced latitude lines near the poles may be actually the same distance apart on the Earth as two closely spaced latitude lines near the equator. It can be proved that on a Mercator map in which the equatorial line has length L, the vertical distance Dβ on the map between the equator (latitude 0 ◦ ) and the line of latitude β ◦ is L βπ/180 Dβ = sec x dx (27) 2π 0 Figure 7.3.2 A ﬂight path with constant compass heading from New York City to Moscow follows a spiral toward the North Pole but is a straight line segment on a Mercator projection.

✔QUICK CHECK EXERCISES 7.3

(See page 508 for answers.)

1. Complete each trigonometric identity with an expression involving cos 2x. (a) sin2 x = (b) cos2 x = 2 2 (c) cos x − sin x =

3. Use the indicated substitution to rewrite the integral in terms of u. Do not evaluate the integral. sin2 x cos x dx; u = sin x

(a)

(b)

2. Evaluate the integral. (a) sec2 x dx = (b) tan2 x dx = (c) sec x dx = (d) tan x dx =

sin3 x cos2 x dx; u = cos x

tan3 x sec2 x dx; u = tan x

tan3 x sec x dx; u = sec x

EXERCISE SET 7.3 1–52 Evaluate the integral. ■

1. 3.

cos3 x sin x dx

sin2 5θ dθ

sin5 3x cos 3x dx

cos2 3x dx

sin aθ dθ

sin ax cos ax dx

sin2 t cos3 t dt

10.

cos3 at dt sin3 x cos3 x dx sin3 x cos2 x dx

7.3 Integrating Trigonometric Functions

11. 13. 15. 17.

sin2 x cos2 x dx

12.

sin 2x cos 3x dx

14.

sin x cos(x /2) dx

16.

π/2

cos3 x dx

18.

19. 21. 23. 25. 27. 29. 31. 33. 35. 37. 39. 41. 43. 45.

π/3

sin4 3x cos3 3x dx

20.

sin 4x cos 2x dx

22.

sec2 (2x − 1) dx e−x tan(e−x ) dx

sec 4x dx 2

tan x sec x dx

tan 4x sec4 4x dx

sec5 x tan3 x dx tan4 x sec x dx tan t sec3 t dt

sec4 x dx

tan3 4x dx

√ tan x sec4 x dx

π/8

tan2 2x dx

24.

47. 49. 51.

sin 3θ cos 2θ dθ cos1/3 x sin x dx π/2

sin2 0

x x cos2 dx 2 2

cos2 5θ dθ

2π

sin2 kx dx

π/2

tan 5x dx

cot 3x dx √ sec( x) 28. dx √ x 30. tan5 x sec4 x dx 32. tan4 θ sec4 θ dθ 34. tan5 θ sec θ dθ 36. tan2 x sec3 x dx 38. tan x sec5 x dx 40. sec5 x dx 42. tan4 x dx 44. tan x sec3/2 x dx 26.

46.

sin2 x cos4 x dx

−π

π/6

x tan5 dx 2

48.

cot3 x csc3 x dx

50.

cot3 x dx

52.

π/6

sec3 2θ tan 2θ dθ 0

1/4

sec πx tan πx dx

507

55. The trigonometric identity sin α cos β = 21 [sin(α − β) + sin(α + β)] is useful for evaluating integrals of the form often sinm x cosn x dx. 56. The integral tan4 x sec5 x dx is equivalent to one whose integrand is a polynomial in sec x.

57. Let m, n be distinct nonnegative integers. Use Formulas (16) – (18) to prove: 2π (a) sin mx cos nx dx = 0 0 2π cos mx cos nx dx = 0 (b) 0 2π sin mx sin nx dx = 0. (c) 0

58. Evaluate the integrals in Exercise 57 when m and n denote the same nonnegative integer.

59. Find the arc length of the curve y = ln(cos x) over the interval [0, π/4].

60. Find the volume of the solid generated when the region enclosed by y = tan x, y = 1, and x = 0 is revolved about the x-axis. 61. Find the volume of the solid that results when the region enclosed by y = cos x, y = sin x, x = 0, and x = π/4 is revolved about the x-axis. 62. The region bounded below by the x-axis and above by the portion of y = sin x from x = 0 to x = π is revolved about the x-axis. Find the volume of the resulting solid.

63. Use Formula (27) to show that if the length of the equatorial line on a Mercator projection is L, then the vertical distance D between the latitude lines at α ◦ and β ◦ on the same side of the equator (where α < β) is L sec β ◦ + tan β ◦ D= ln 2π sec α ◦ + tan α ◦ 64. Suppose that the equator has a length of 100 cm on a Mercator projection. In each part, use the result in Exercise 63 to answer the question. (a) What is the vertical distance on the map between the equator and the line at 25 ◦ north latitude? (b) What is the vertical distance on the map between New Orleans, Louisiana, at 30 ◦ north latitude and Winnipeg, Canada, at 50 ◦ north latitude?

cot2 3t sec 3t dt F O C U S O N C O N C E P TS

csc4 x dx

53–56 True–False Determine whether the statement is true or

false. Explain your answer. ■ 53. To evaluate sin5 x cos8 x dx, use the trigonometric identity sin2 x = 1 − cos2 x and the substitution u = cos x. 54. To evaluate sin8 x cos5 x dx, use the trigonometric iden2 tity sin x = 1 − cos2 x and the substitution u = cos x.

65. (a) Show that csc x dx = − ln |csc x + cot x| + C

(b) Show that the result in part (a) can also be written as csc x dx = ln |csc x − cot x| + C and

csc x dx = ln tan 21 x + C

508

Chapter 7 / Principles of Integral Evaluation

66. Rewrite sin x + cos x in the form A sin(x + Ď&#x2020;) and use your result together with Exercise 65 to evaluate dx sin x + cos x 67. Use the method of Exercise 66 to evaluate dx (a, b not both zero) a sin x + b cos x 68. (a) Use Formula (9) in Section 7.2 to show that Ď&#x20AC;/2 n â&#x2C6;&#x2019; 1 Ď&#x20AC;/2 nâ&#x2C6;&#x2019;2 sinn x dx = sin x dx (n â&#x2030;Ľ 2) n 0 0 (b) Use this result to derive the Wallis sine formulas: Ď&#x20AC;/2 Ď&#x20AC; 1 Âˇ 3 Âˇ 5 Âˇ Âˇ Âˇ (n â&#x2C6;&#x2019; 1) n even sinn x dx = Âˇ and â&#x2030;Ľ 2 2 2 Âˇ 4 Âˇ 6ÂˇÂˇÂˇn 0 Ď&#x20AC;/2 2 Âˇ 4 Âˇ 6 Âˇ Âˇ Âˇ (n â&#x2C6;&#x2019; 1) n odd sinn x dx = and â&#x2030;Ľ 3 3 Âˇ 5 Âˇ 7ÂˇÂˇÂˇn 0 69. Use the Wallis formulas in Exercise 68 to evaluate Ď&#x20AC;/2 Ď&#x20AC;/2 3 (a) sin4 x dx sin x dx (b) 0

(c)

Ď&#x20AC;/2

sin5 x dx

(d)

Ď&#x20AC;/2

sin6 x dx. 0

70. Use Formula (10) in Section 7.2 and the method of Exercise 68 to derive the Wallis cosine formulas: Ď&#x20AC;/2 Ď&#x20AC; 1 Âˇ 3 Âˇ 5 Âˇ Âˇ Âˇ (n â&#x2C6;&#x2019; 1) n even cosn x dx = Âˇ and â&#x2030;Ľ 2 2 2 Âˇ 4 Âˇ 6ÂˇÂˇÂˇn 0 Ď&#x20AC;/2 2 Âˇ 4 Âˇ 6 Âˇ Âˇ Âˇ (n â&#x2C6;&#x2019; 1) n odd cosn x dx = and â&#x2030;Ľ 3 3 Âˇ 5 Âˇ 7ÂˇÂˇÂˇn 0 71. Writing Describe the various approaches for evaluating integrals of the form sinm x cosn x dx

Into what cases do these types of integrals fall? What procedures and identities are used in each case?

72. Writing Describe the various approaches for evaluating integrals of the form tanm x secn x dx

Into what cases do these types of integrals fall? What procedures and identities are used in each case?

â&#x153;&#x201D;QUICK CHECK ANSWERS 7.3 1 â&#x2C6;&#x2019; cos 2x 1 + cos 2x (b) (c) cos 2x 2. (a) tan x + C (b) tan x â&#x2C6;&#x2019; x + C (c) ln | sec x + tan x| + C (d) ln | sec x| + C 2 2 3. (a) u2 du (b) (u2 â&#x2C6;&#x2019; 1)u2 du (c) u3 du (d) (u2 â&#x2C6;&#x2019; 1) du

1. (a)

7.4

TRIGONOMETRIC SUBSTITUTIONS In this section we will discuss a method for evaluating integrals containing radicals by making substitutions involving trigonometric functions. We will also show how integrals containing quadratic polynomials can sometimes be evaluated by completing the square. THE METHOD OF TRIGONOMETRIC SUBSTITUTION To start, we will be concerned with integrals that contain expressions of the form a2 â&#x2C6;&#x2019; x 2, x 2 + a2, x 2 â&#x2C6;&#x2019; a2

in which a is a positive constant. The basic idea for evaluating such integrals is to make a substitution for x that will eliminate the radical. For example, to eliminate the radical in the expression a 2 â&#x2C6;&#x2019; x 2 , we can make the substitution x = a sin Î¸,

â&#x2C6;&#x2019;Ď&#x20AC;/2 â&#x2030;¤ Î¸ â&#x2030;¤ Ď&#x20AC;/2

which yields a 2 â&#x2C6;&#x2019; x 2 = a 2 â&#x2C6;&#x2019; a 2 sin2 Î¸ = a 2 (1 â&#x2C6;&#x2019; sin2 Î¸ ) â&#x2C6;&#x161; = a cos2 Î¸ = a|cos Î¸ | = a cos Î¸

cos Î¸ â&#x2030;Ľ 0 since â&#x2C6;&#x2019;Ď&#x20AC;/2 â&#x2030;¤ Î¸ â&#x2030;¤ Ď&#x20AC;/2

(1)

7.4 Trigonometric Substitutions

509

The restriction on θ in (1) serves two purposes—it enables us to replace |cos θ| by cos θ to simplify the calculations, and it also ensures that the substitutions can be rewritten as θ = sin−1 (x /a), if needed. Example 1

Evaluate

dx . 4 − x2

Solution. To eliminate the radical we make the substitution x = 2 sin θ, dx = 2 cos θ dθ This yields 2 cos θ dθ dx = 2 x2 4 − x (2 sin θ )2 4 − 4 sin2 θ 1 dθ 2 cos θ dθ = = (2 sin θ )2 (2 cos θ) 4 sin2 θ 1 1 csc2 θ dθ = − cot θ + C (2) = 4 4 At this point we have completed the integration; however, because the original integral was expressed in terms of x, it is desirable to express cot θ in terms of x as well. This can be done using trigonometric identities, but the expression can also be obtained by writing the substitution x = 2 sin θ as sin θ = x /2 and representing it geometrically as in Figure 7.4.1. From that ﬁgure we obtain 4 − x2 cot θ = x Substituting this in (2) yields dx 1 4 − x2 =− +C 4 x x2 4 − x2

2 u

√4 − x 2 x = 2 sin u Figure 7.4.1

Example 2

Evaluate

√

dx . 4 − x2

Solution. There are two possible approaches: we can make the substitution in the indefinite integral (as in Example 1) and then evaluate the deﬁnite integral using the x-limits of integration, or we can make the substitution in the deﬁnite integral and convert the x-limits to the corresponding θ-limits. Method 1. Using the result from Example 1 with the x-limits of integration yields √2 √ √2 √ 4 − x2 3−1 1 1 dx =− 1− 3 = =− 4 x 4 4 1 x2 4 − x2 1 Method 2. −1 The substitution x = 2 sin θ can be expressed √ as x /2 = sin θ or θ = sin (x /2), so the θ-limits that correspond to x = 1 and x = 2 are

x = 1: θ = sin−1 (1/2) = π/6 √ √ x = 2: θ = sin−1 ( 2/2) = π/4

510

Chapter 7 / Principles of Integral Evaluation

Thus, from (2) in Example 1 we obtain â&#x2C6;&#x161;2 dx 1 Ď&#x20AC;/4 2 csc Î¸ dÎ¸ Convert x-limits to Î¸ -limits. = 4 Ď&#x20AC;/6 1 x2 4 â&#x2C6;&#x2019; x2 â&#x2C6;&#x161; â&#x2C6;&#x161; Ď&#x20AC;/4 1 3â&#x2C6;&#x2019;1 1 = â&#x2C6;&#x2019; cot Î¸ Ď&#x20AC;/6 = â&#x2C6;&#x2019; 1 â&#x2C6;&#x2019; 3 = 4 4 4 Example 3

Find the area of the ellipse x2 y2 + =1 a2 b2

b x

x2 a

y2 b2

Solution. Because the ellipse is symmetric about both axes, its area A is four times the area in the ďŹ rst quadrant (Figure 7.4.2). If we solve the equation of the ellipse for y in terms of x, we obtain b 2 a â&#x2C6;&#x2019; x2 y=Âą a where the positive square root gives the equation of the upper half. Thus, the area A is a given by b 4b a 2 2 2 a â&#x2C6;&#x2019; x dx = a â&#x2C6;&#x2019; x 2 dx A=4 a 0 0 a

To evaluate this integral, we will make the substitution x = a sin Î¸ (so dx = a cos Î¸ dÎ¸) and convert the x-limits of integration to Î¸-limits. Since the substitution can be expressed as Î¸ = sinâ&#x2C6;&#x2019;1 (x /a), the Î¸-limits of integration are

Figure 7.4.2

Î¸ = sinâ&#x2C6;&#x2019;1 (0) = 0

x = 0:

x = a : Î¸ = sinâ&#x2C6;&#x2019;1 (1) = Ď&#x20AC;/2

Thus, we obtain A= y

y = â&#x2C6;&#x161;a 2 â&#x2C6;&#x2019; x 2

4b a 4b a

= 4ab x

â&#x2C6;&#x2019;a

Figure 7.4.3

REMARK

T E C H N O LO GY M A ST E R Y If you have a calculating utility with a numerical integration capability, use it and Formula (3) to approximate Ď&#x20AC; to three decimal places.

a 0

4b Ď&#x20AC;/2 2 a 2 â&#x2C6;&#x2019; x 2 dx = a â&#x2C6;&#x2019; a 2 sin2 Î¸ Âˇ a cos Î¸ dÎ¸ a 0

Ď&#x20AC;/2 0

a cos Î¸ Âˇ a cos Î¸ dÎ¸ Ď&#x20AC;/2

Ď&#x20AC;/2 1 = 2ab Î¸ + sin 2Î¸ 2 0

Ď&#x20AC;/2

1 (1 + cos 2Î¸ ) dÎ¸ 2 Ď&#x20AC; = 2ab â&#x2C6;&#x2019; 0 = Ď&#x20AC;ab 2

cos2 Î¸ dÎ¸ = 4ab

In the special case where a = b, the ellipse becomes a circle of radius a , and the area formula becomes A = Ď&#x20AC;a 2 , as expected. It is worth noting that

â&#x2C6;&#x2019;a

a 2 â&#x2C6;&#x2019; x 2 dx = 21 Ď&#x20AC;a 2

(3)

since this integral represents the area of the upper semicircle (Figure 7.4.3).

Thus far, we have focused on using the substitution x = a sin Î¸ to evaluate integrals involving radicals of the form a 2 â&#x2C6;&#x2019; x 2 . Table 7.4.1 summarizes this method and describes some other substitutions of this type.

7.4 Trigonometric Substitutions

511

Table 7.4.1 trigonometric substitutions

expression in the integrand

simplification

â&#x2C6;&#x161;a 2 â&#x2C6;&#x2019; x 2

x = a sin u

â&#x2C6;&#x2019;c/2 â&#x2030;¤ u â&#x2030;¤ c/2

a 2 â&#x2C6;&#x2019; x 2 = a 2 â&#x2C6;&#x2019; a 2 sin 2 u = a 2 cos 2 u

â&#x2C6;&#x161;a 2 + x 2

x = a tan u

â&#x2C6;&#x2019;c/2 < u < c/2

a 2 + x 2 = a 2 + a 2 tan 2 u = a 2 sec 2 u

â&#x2C6;&#x161;x 2 â&#x2C6;&#x2019; a 2

x = a sec u

Example 4 ure 7.4.4). 1

restriction on u

substitution

0 â&#x2030;¤ u < c/2 (if x â&#x2030;Ľ a) c/2 < u â&#x2030;¤ c (if x â&#x2030;¤ â&#x2C6;&#x2019;a)

x 2 â&#x2C6;&#x2019; a 2 = a 2 sec 2 u â&#x2C6;&#x2019; a 2 = a 2 tan 2 u

Find the arc length of the curve y = x 2 /2 from x = 0 to x = 1 (Fig-

y = x 2/ 2

Solution. From Formula (4) of Section 6.4 the arc length L of the curve is x

Figure 7.4.4

1 0

dy dx

dx =

1 + x 2 dx

â&#x2C6;&#x161; The integrand involves a radical of the form a 2 + x 2 with a = 1, so from Table 7.4.1 we make the substitution x = tan Î¸, â&#x2C6;&#x2019;Ď&#x20AC;/2 < Î¸ < Ď&#x20AC;/2 dx = sec2 Î¸ or dx = sec2 Î¸ dÎ¸ dÎ¸ Since this substitution can be expressed as Î¸ = tanâ&#x2C6;&#x2019;1 x, the Î¸-limits of integration that correspond to the x-limits, x = 0 and x = 1, are x = 0: Î¸ = tanâ&#x2C6;&#x2019;1 0 = 0 x = 1: Î¸ = tanâ&#x2C6;&#x2019;1 1 = Ď&#x20AC;/4 Thus, L=

dx = = = =

Ď&#x20AC;/4

1 + tan2 Î¸ sec2 Î¸ dÎ¸ â&#x2C6;&#x161;

sec2 Î¸ sec2 Î¸ dÎ¸

1 + tan2 Î¸ = sec2 Î¸

Ď&#x20AC;/4

|sec Î¸|sec2 Î¸ dÎ¸

Ď&#x20AC;/4

sec3 Î¸ dÎ¸

sec Î¸ > 0 since â&#x2C6;&#x2019;Ď&#x20AC;/2 < Î¸ < Ď&#x20AC;/2

Ď&#x20AC;/4 sec Î¸ tan Î¸ + 21 ln |sec Î¸ + tan Î¸ | 0 â&#x2C6;&#x161; â&#x2C6;&#x161; 1 =2 2 + ln( 2 + 1) â&#x2030;&#x2C6; 1.148 =

Example 5

1 2

2 x â&#x2C6;&#x2019; 25 Evaluate dx, assuming that x â&#x2030;Ľ 5. x

Formula (26) of Section 7.3

512

Chapter 7 / Principles of Integral Evaluation

Solution. The integrand involves a radical of the form

â&#x2C6;&#x161; x 2 â&#x2C6;&#x2019; a 2 with a = 5, so from

Table 7.4.1 we make the substitution x = 5 sec Î¸, 0 â&#x2030;¤ Î¸ < Ď&#x20AC;/2 dx = 5 sec Î¸ tan Î¸ or dx = 5 sec Î¸ tan Î¸ dÎ¸ dÎ¸ Thus, 2 x â&#x2C6;&#x2019; 25 25 sec2 Î¸ â&#x2C6;&#x2019; 25 dx = (5 sec Î¸ tan Î¸) dÎ¸ x 5 sec Î¸ 5|tan Î¸| (5 sec Î¸ tan Î¸) dÎ¸ = 5 sec Î¸ = 5 tan2 Î¸ dÎ¸ tan Î¸ â&#x2030;Ľ 0 since 0 â&#x2030;¤ Î¸ < Ď&#x20AC;/2 = 5 (sec2 Î¸ â&#x2C6;&#x2019; 1) dÎ¸ = 5 tan Î¸ â&#x2C6;&#x2019; 5Î¸ + C

x u 5 x = 5 sec u Figure 7.4.5

â&#x2C6;&#x161;x 2 â&#x2C6;&#x2019; 25

To express the solution in terms of x, we will represent the substitution x = 5 sec Î¸ geometrically by the triangle in Figure 7.4.5, from which we obtain x 2 â&#x2C6;&#x2019; 25 tan Î¸ = 5 From this and the fact that the substitution can be expressed as Î¸ = secâ&#x2C6;&#x2019;1 (x /5), we obtain 2 x

x â&#x2C6;&#x2019; 25 +C dx = x 2 â&#x2C6;&#x2019; 25 â&#x2C6;&#x2019; 5 secâ&#x2C6;&#x2019;1 x 5 INTEGRALS INVOLVING ax 2 + bx + c

Integrals that involve a quadratic expression ax 2 + bx + c, where a = 0 and b = 0, can often be evaluated by ďŹ rst completing the square, then making an appropriate substitution. The following example illustrates this idea. Example 6

Evaluate

x dx. x 2 â&#x2C6;&#x2019; 4x + 8

Solution. Completing the square yields x 2 â&#x2C6;&#x2019; 4x + 8 = (x 2 â&#x2C6;&#x2019; 4x + 4) + 8 â&#x2C6;&#x2019; 4 = (x â&#x2C6;&#x2019; 2)2 + 4

Thus, the substitution

u = x â&#x2C6;&#x2019; 2,

yields

x u+2 dx = du 2 (x â&#x2C6;&#x2019; 2) + 4 u2 + 4 u du du + 2 = u2 + 4 u2 + 4 du 2u 1 du + 2 = 2 u2 + 4 u2 + 4 1 1 u 2 = ln(u + 4) + 2 tanâ&#x2C6;&#x2019;1 + C 2 2 2 1 2 â&#x2C6;&#x2019;1 x â&#x2C6;&#x2019; 2 +C = ln[(x â&#x2C6;&#x2019; 2) + 4] + tan 2 2

x dx = 2 x â&#x2C6;&#x2019; 4x + 8

du = dx

7.4 Trigonometric Substitutions

✔QUICK CHECK EXERCISES 7.4

(See page 514 for answers.)

1. For each expression, give a trigonometric substitution that will √ eliminate the radical. √ (a) √a 2 − x 2 (b) a 2 + x 2 (c) x 2 − a 2 2. If x = 2 sec θ and 0 < θ < π/2, then (a) sin θ = (b) cos θ = (c) tan θ = .

3. In each part, state the trigonometric substitution that you would try ﬁrst to evaluate the integral. Do not evaluate the integral. (a) 9 + x 2 dx (b) 9 − x 2 dx (c) 1 − 9x 2 dx

EXERCISE SET 7.4

x2 dx 16 − x 2 dx 5. (4 + x 2 )2 2 x −9 7. dx x 3x 3 dx 9. 1 − x2 dx 11. 2 x 9x 2 − 4 dx 13. (1 − x 2 )3/2 dx 15. x2 − 9 dx 17. 2 (4x − 9)3/2 19. ex 1 − e2x dx 3.

21. 23.

0 2 √ 2

5x 3 1 − x 2 dx dx 2 x x2 − 1

x 2 − 9 dx (e) 9 + 3x 2 dx (f ) 1 + (9x)2 dx

(d)

4. In each u. part, determine the substitution 1 1 (a) du; dx = x 2 − 2x + 10 u2 + 3 2 u= (b) x 2 − 6x + 8 dx = u2 − 1 du;

u= (c) 12 − 4x − x 2 dx = 42 − u2 du; u=

CAS

1–26 Evaluate the integral. ■

1. 4 − x 2 dx

513

2. 1 − 4x 2 dx

dx 2 x 9 − x2 x2 dx 6. 5 + x2 dx 8. x 2 x 2 − 16 10. x 3 5 − x 2 dx 4.

12. 14. 16. 18. 20. 22. 24.

1 + t2 dt t dx 2 x x 2 + 25 dx 1 + 2x 2 + x 4 3x 3 dx x 2 − 25 cos θ dθ 2 − sin2 θ 1/2 dx (1 − x 2 )2 0 2 2 2x − 4 dx √ x 2

25.

3 1

dx 4 x x2 + 3

26.

3 0

x3 dx (3 + x 2 )5/2

27–30 True–False Determine whether the statement is true or false. Explain your answer. ■ √ 27. An integrand involving a radical of the form a 2 − x 2 suggests the substitution x = a sin θ .

28. The trigonometric substitution x = a sin θ is made with the restriction 0 ≤ θ ≤ π. √ 29. An integrand involving a radical of the form x 2 − a 2 suggests the substitution x = a cos θ. 30. The area enclosed by the ellipse x 2 + 4y 2 = 1 is π/2. F O C U S O N C O N C E P TS

31. The integral

x dx x2 + 4 can be evaluated either by a trigonometric substitution or by the substitution u = x 2 + 4. Do it both ways and show that the results are equivalent. 32. The integral x2 dx 2 x +4 can be evaluated either by a trigonometric substitution or by algebraically rewriting the numerator of the integrand as (x 2 + 4) − 4. Do it both ways and show that the results are equivalent. 33. Find the arc length of the curve y = ln x from x = 1 to x = 2. 34. Find the arc length of the curve y = x 2 from x = 0 to x = 1.

Chapter 7 / Principles of Integral Evaluation

514

(x cos x + sin x) 1 + x 2 sin2 x dx

50.

36. Find the volume of the solid generated when the region enclosed by x = y(1 − y 2 )1/4 , y = 0, y = 1, and x = 0 is revolved about the y-axis.

51. (a) Use the hyperbolic substitution x = 3 sinh u, the identity cosh2 u − sinh2 u = 1, and Theorem 6.9.4 to eval uate dx √ x2 + 9

37–48 Evaluate the integral. ■

37.

dx x 2 − 4x + 5

dx 3 + 2x − x 2 dx 41. 2 x − 6x + 10 3 − 2x − x 2 dx 43. 39.

35. Find the area of the surface generated when the curve in Exercise 34 is revolved about the x-axis.

dx 45. 2x 2 + 4x + 7 2 dx 47. 1 4x − x 2

38. 40. 42.

(b) Evaluate the integral in part (a) using a trigonometric substitution and show that the result agrees with that obtained in part (a).

dx 2x − x 2 dx 16x 2 + 16x + 5

x dx x 2 + 2x + 2

ex dx 1 + ex + e2x 2x + 3 46. dx 4x 2 + 4x + 5 4 48. x(4 − x) dx 44.

49–50 There is a good chance that your CAS will not be able to evaluate these integrals as stated. If this is so, make a substitution that converts the integral into one that your CAS can evaluate. ■ 49. cos x sin x 1 − sin4 x dx

52. Use the hyperbolic substitutionn x = cosh u, the identity sinh2 u = 21 (cosh 2u − 1), and the results referenced in Exercise 51 to evaluate x 2 − 1 dx, x ≥ 1

53. Writing The trigonometric substitution x = a sin θ, −π/2 ≤ θ ≤ π/2,√is suggested for an integral whose integrand involves a 2 − x 2 . Discuss the implications of restricting θ to π/2 ≤ θ ≤ 3π/2, and explain why the restriction −π/2 ≤ θ ≤ π/2 should be preferred. 54. Writing The trigonometric substitution x = a cos θ could also √ be used for an integral whose integrand involves a 2 − x 2 . Determine an appropriate restriction for θ with the substitution x = a cos θ , and discuss how to apply this substitution in appropriate integrals. Illustrate your discussion by evaluating the integral in Example 1 using a substitution of this type.

✔QUICK CHECK ANSWERS 7.4 √ √ 2 x2 − 4 x2 − 4 3. (a) x = 3 tan θ (b) x = 3 sin θ (b) (c) 1. (a) x = a sin θ (b) x = a tan θ (c) x = a sec θ 2. (a) x x 2 √ (c) x = 31 sin θ (d) x = 3 sec θ (e) x = 3 tan θ (f ) x = 19 tan θ 4. (a) x − 1 (b) x − 3 (c) x + 2

7.5

INTEGRATING RATIONAL FUNCTIONS BY PARTIAL FRACTIONS Recall that a rational function is a ratio of two polynomials. In this section we will give a general method for integrating rational functions that is based on the idea of decomposing a rational function into a sum of simple rational functions that can be integrated by the methods studied in earlier sections.

PARTIAL FRACTIONS

In algebra, one learns to combine two or more fractions into a single fraction by ﬁnding a common denominator. For example, 2 3 2(x + 1) + 3(x − 4) 5x − 10 + = = 2 x−4 x+1 (x − 4)(x + 1) x − 3x − 4

(1)

7.5 Integrating Rational Functions by Partial Fractions

515

However, for purposes of integration, the left side of (1) is preferable to the right side since each of the terms is easy to integrate: 3 2 5x − 10 dx = dx + dx = 2 ln |x − 4| + 3 ln |x + 1| + C x 2 − 3x − 4 x−4 x+1

Thus, it is desirable to have some method that will enable us to obtain the left side of (1), starting with the right side. To illustrate how this can be done, we begin by noting that on the left side the numerators are constants and the denominators are the factors of the denominator on the right side. Thus, to ﬁnd the left side of (1), starting from the right side, we could factor the denominator of the right side and look for constants A and B such that A B 5x − 10 = + (x − 4)(x + 1) x−4 x+1

(2)

One way to ﬁnd the constants A and B is to multiply (2) through by (x − 4)(x + 1) to clear fractions. This yields 5x − 10 = A(x + 1) + B(x − 4) (3) This relationship holds for all x, so it holds in particular if x = 4 or x = −1. Substituting x = 4 in (3) makes the second term on the right drop out and yields the equation 10 = 5A or A = 2; and substituting x = −1 in (3) makes the ﬁrst term on the right drop out and yields the equation −15 = −5B or B = 3. Substituting these values in (2) we obtain 2 3 5x − 10 = + (x − 4)(x + 1) x−4 x+1

(4)

which agrees with (1). A second method for finding the constants A and B is to multiply out the right side of (3) and collect like powers of x to obtain 5x − 10 = (A + B)x + (A − 4B) Since the polynomials on the two sides are identical, their corresponding coefficients must be the same. Equating the corresponding coefficients on the two sides yields the following system of equations in the unknowns A and B: A+ B = 5 A − 4B = −10 Solving this system yields A = 2 and B = 3 as before (verify). The terms on the right side of (4) are called partial fractions of the expression on the left side because they each constitute part of that expression. To find those partial fractions we first had to make a guess about their form, and then we had to find the unknown constants. Our next objective is to extend this idea to general rational functions. For this purpose, suppose that P (x)/Q(x) is a proper rational function, by which we mean that the degree of the numerator is less than the degree of the denominator. There is a theorem in advanced algebra which states that every proper rational function can be expressed as a sum P (x) = F1 (x) + F2 (x) + · · · + Fn (x) Q(x) where F1 (x), F2 (x), . . . , Fn (x) are rational functions of the form A (ax + b)k

Ax + B + bx + c)k

(ax 2

in which the denominators are factors of Q(x). The sum is called the partial fraction decomposition of P (x)/Q(x), and the terms are called partial fractions. As in our opening example, there are two parts to ﬁnding a partial fraction decomposition: determining the exact form of the decomposition and ﬁnding the unknown constants.

516

Chapter 7 / Principles of Integral Evaluation

FINDING THE FORM OF A PARTIAL FRACTION DECOMPOSITION The ﬁrst step in ﬁnding the form of the partial fraction decomposition of a proper rational function P (x)/Q(x) is to factor Q(x) completely into linear and irreducible quadratic factors, and then collect all repeated factors so that Q(x) is expressed as a product of distinct factors of the form

(ax + b)m

and

(ax 2 + bx + c)m

From these factors we can determine the form of the partial fraction decomposition using two rules that we will now discuss. LINEAR FACTORS

If all of the factors of Q(x) are linear, then the partial fraction decomposition of P (x)/Q(x) can be determined by using the following rule:

linear factor rule For each factor of the form (ax + b)m , the partial fraction decomposition contains the following sum of m partial fractions: A1 A2 Am + + ··· + 2 ax + b (ax + b) (ax + b)m where A1 , A2 , . . . , Am are constants to be determined. In the case where m = 1, only the ﬁrst term in the sum appears.

Example 1

Evaluate

dx . x2 + x − 2

Solution. The integrand is a proper rational function that can be written as x2

1 1 = +x−2 (x − 1)(x + 2)

The factors x − 1 and x + 2 are both linear and appear to the ﬁrst power, so each contributes one term to the partial fraction decomposition by the linear factor rule. Thus, the decomposition has the form 1 A B = + (x − 1)(x + 2) x−1 x+2

(5)

where A and B are constants to be determined. Multiplying this expression through by (x − 1)(x + 2) yields 1 = A(x + 2) + B(x − 1) (6) As discussed earlier, there are two methods for finding A and B: we can substitute values of x that are chosen to make terms on the right drop out, or we can multiply out on the right and equate corresponding coefficients on the two sides to obtain a system of equations that can be solved for A and B. We will use the first approach. Setting x = 1 makes the second term in (6) drop out and yields 1 = 3A or A = 31 ; and setting x = −2 makes the first term in (6) drop out and yields 1 = −3B or B = − 13 . Substituting these values in (5) yields the partial fraction decomposition 1 − 31 1 = 3 + (x − 1)(x + 2) x−1 x+2

7.5 Integrating Rational Functions by Partial Fractions

517

The integration can now be completed as follows: 1 1 dx dx dx = − (x − 1)(x + 2) 3 x−1 3 x+2 1 1 1 x − 1 = ln |x − 1| − ln |x + 2| + C = ln +C 3 3 3 x + 2

If the factors of Q(x) are linear and none are repeated, as in the last example, then the recommended method for finding the constants in the partial fraction decomposition is to substitute appropriate values of x to make terms drop out. However, if some of the linear factors are repeated, then it will not be possible to find all of the constants in this way. In this case the recommended procedure is to find as many constants as possible by substitution and then find the rest by equating coefficients. This is illustrated in the next example. Example 2

Evaluate

2x + 4 dx. x 3 − 2x 2

Solution. The integrand can be rewritten as 2x + 4 2x + 4 = 2 x 3 − 2x 2 x (x − 2)

Although x 2 is a quadratic factor, it is not irreducible since x 2 = xx. Thus, by the linear factor rule, x 2 introduces two terms (since m = 2) of the form B A + 2 x x and the factor x − 2 introduces one term (since m = 1) of the form C x−2 so the partial fraction decomposition is A B C 2x + 4 = + 2+ (7) x 2 (x − 2) x x x−2 Multiplying by x 2 (x − 2) yields

2x + 4 = Ax(x − 2) + B(x − 2) + Cx 2

(8)

which, after multiplying out and collecting like powers of x, becomes 2x + 4 = (A + C)x 2 + (−2A + B)x − 2B

(9)

Setting x = 0 in (8) makes the first and third terms drop out and yields B = −2, and setting x = 2 in (8) makes the first and second terms drop out and yields C = 2 (verify). However, there is no substitution in (8) that produces A directly, so we look to Equation (9) to find this value. This can be done by equating the coefficients of x 2 on the two sides to obtain A+C =0

or A = −C = −2

Substituting the values A = −2, B = −2, and C = 2 in (7) yields the partial fraction decomposition −2 −2 2 2x + 4 = + 2 + x 2 (x − 2) x x x−2 Thus, dx dx dx 2x + 4 dx = −2 − 2 + 2 2 2 x (x − 2) x x x−2 x − 2 2 2 + +C = −2 ln |x| + + 2 ln |x − 2| + C = 2 ln x x x

518

Chapter 7 / Principles of Integral Evaluation

QUADRATIC FACTORS

If some of the factors of Q(x) are irreducible quadratics, then the contribution of those factors to the partial fraction decomposition of P (x)/Q(x) can be determined from the following rule: quadratic factor rule For each factor of the form (ax 2 + bx + c)m , the partial fraction decomposition contains the following sum of m partial fractions: A 2 x + B2 Am x + B m A1 x + B 1 + + ··· + ax 2 + bx + c (ax 2 + bx + c)2 (ax 2 + bx + c)m where A1 , A2 , . . . , Am , B1 , B2 , . . . , Bm are constants to be determined. In the case where m = 1, only the ﬁrst term in the sum appears.

Example 3

Evaluate

x2 + x − 2 dx. 3x 3 − x 2 + 3x − 1

Solution. The denominator in the integrand can be factored by grouping: 3x 3 − x 2 + 3x − 1 = x 2 (3x − 1) + (3x − 1) = (3x − 1)(x 2 + 1) By the linear factor rule, the factor 3x − 1 introduces one term, namely, A 3x − 1

and by the quadratic factor rule, the factor x 2 + 1 introduces one term, namely,

Bx + C x2 + 1 Thus, the partial fraction decomposition is

A Bx + C x2 + x − 2 + 2 = 2 (3x − 1)(x + 1) 3x − 1 x +1

(10)

x 2 + x − 2 = A(x 2 + 1) + (Bx + C)(3x − 1)

(11)

Multiplying by (3x − 1)(x 2 + 1) yields 1 3

We could find A by substituting x = to make the last term drop out, and then find the rest of the constants by equating corresponding coefficients. However, in this case it is just as easy to find all of the constants by equating coefficients and solving the resulting system. For this purpose we multiply out the right side of (11) and collect like terms: x 2 + x − 2 = (A + 3B)x 2 + (−B + 3C)x + (A − C) Equating corresponding coefficients gives A + 3B

− B + 3C = 1 A − C = −2

To solve this system, subtract the third equation from the ﬁrst to eliminate A. Then use the resulting equation together with the second equation to solve for B and C. Finally, determine A from the ﬁrst or third equation. This yields (verify) 7 A=− , 5

4 , 5

3 5

7.5 Integrating Rational Functions by Partial Fractions

519

Thus, (10) becomes 4 − 57 x + 53 x2 + x − 2 5 = + (3x − 1)(x 2 + 1) 3x − 1 x2 + 1

and T E C H N O LO GY M A ST E R Y Computer algebra systems have builtin capabilities for finding partial fraction decompositions. If you have a CAS, use it to find the decompositions in Examples 1, 2, and 3.

x2 + x − 2 7 4 3 dx x dx dx = − + dx + 2 2 2 (3x − 1)(x + 1) 5 3x − 1 5 x +1 5 x +1 7 2 3 = − ln |3x − 1| + ln(x 2 + 1) + tan−1 x + C 15 5 5

Example 4

Evaluate

3x 4 + 4x 3 + 16x 2 + 20x + 9 dx. (x + 2)(x 2 + 3)2

Solution. Observe that the integrand is a proper rational function since the numerator has degree 4 and the denominator has degree 5. Thus, the method of partial fractions is applicable. By the linear factor rule, the factor x + 2 introduces the single term A x+2 and by the quadratic factor rule, the factor (x 2 + 3)2 introduces two terms (since m = 2): Dx + E Bx + C + 2 x2 + 3 (x + 3)2 Thus, the partial fraction decomposition of the integrand is 3x 4 + 4x 3 + 16x 2 + 20x + 9 A Bx + C Dx + E + 2 + 2 = 2 2 (x + 2)(x + 3) x+2 x +3 (x + 3)2

(12)

Multiplying by (x + 2)(x 2 + 3)2 yields 3x 4 + 4x 3 + 16x 2 + 20x + 9 = A(x 2 + 3)2 + (Bx + C)(x 2 + 3)(x + 2) + (Dx + E)(x + 2)

(13)

which, after multiplying out and collecting like powers of x, becomes 3x 4 + 4x 3 + 16x 2 + 20x + 9 = (A + B)x 4 + (2B + C)x 3 + (6A + 3B + 2C + D)x 2 + (6B + 3C + 2D + E)x + (9A + 6C + 2E)

(14)

Equating corresponding coefficients in (14) yields the following system of five linear equations in five unknowns: A+B = 3 2B + C = 4 6A + 3B + 2C + D = 16 (15) 6B + 3C + 2D + E = 20 9A + 6C + 2E = 9 Efficient methods for solving systems of linear equations such as this are studied in a branch of mathematics called linear algebra; those methods are outside the scope of this text. However, as a practical matter most linear systems of any size are solved by computer, and most computer algebra systems have commands that in many cases can solve linear systems exactly. In this particular case we can simplify the work by first substituting x = −2

520

Chapter 7 / Principles of Integral Evaluation

in (13), which yields A = 1. Substituting this known value of A in (15) yields the simpler system B= 2 2B + C = 4 3B + 2C + D = 10 6B + 3C + 2D + E = 20

(16)

6C + 2E = 0

This system can be solved by starting at the top and working down, ﬁrst substituting B = 2 in the second equation to get C = 0, then substituting the known values of B and C in the third equation to get D = 4, and so forth. This yields A = 1,

B = 2,

C = 0,

D = 4,

E=0

Thus, (12) becomes 3x 4 + 4x 3 + 16x 2 + 20x + 9 1 2x 4x = + 2 + 2 2 2 (x + 2)(x + 3) x + 2 x + 3 (x + 3)2

and so

3x 4 + 4x 3 + 16x 2 + 20x + 9 dx (x + 2)(x 2 + 3)2 dx x 2x = + dx + 4 dx 2 2 x+2 x +3 (x + 3)2 2 +C = ln |x + 2| + ln(x 2 + 3) − 2 x +3

INTEGRATING IMPROPER RATIONAL FUNCTIONS Although the method of partial fractions only applies to proper rational functions, an improper rational function can be integrated by performing a long division and expressing the function as the quotient plus the remainder over the divisor. The remainder over the divisor will be a proper rational function, which can then be decomposed into partial fractions. This idea is illustrated in the following example.

Example 5

Evaluate

3x 4 + 3x 3 − 5x 2 + x − 1 dx. x2 + x − 2

Solution. The integrand is an improper rational function since the numerator has degree 4 and the denominator has degree 2. Thus, we ﬁrst perform the long division 2

x +x−2

3x 2 +1 3x + 3x − 5x 2 + x − 1 3x 4 + 3x 3 − 6x 2 x2 + x − 1 x2 + x − 2 1 4

It follows that the integrand can be expressed as 3x 4 + 3x 3 − 5x 2 + x − 1 1 = (3x 2 + 1) + 2 2 x +x−2 x +x−2

and hence

3x 4 + 3x 3 − 5x 2 + x − 1 dx = x2 + x − 2

(3x + 1) dx +

dx x2 + x − 2

7.5 Integrating Rational Functions by Partial Fractions

521

The second integral on the right now involves a proper rational function and can thus be evaluated by a partial fraction decomposition. Using the result of Example 1 we obtain 3x 4 + 3x 3 − 5x 2 + x − 1 1 x − 1 3 +C dx = x + x + ln x2 + x − 2 3 x + 2

CONCLUDING REMARKS There are some cases in which the method of partial fractions is inappropriate. For example, it would be inefﬁcient to use partial fractions to perform the integration 3x 2 + 2 dx = ln |x 3 + 2x − 8| + C x 3 + 2x − 8

since the substitution u = x 3 + 2x − 8 is more direct. Similarly, the integration 2x dx 2x − 1 dx = dx − = ln(x 2 + 1) − tan−1 x + C x2 + 1 x2 + 1 x2 + 1

requires only a little algebra since the integrand is already in partial fraction form.

✔QUICK CHECK EXERCISES 7.5

(See page 523 for answers.)

(b) For each factor of Q(x) of the form (ax 2 + bx + c)m , where ax 2 + bx + c is an irreducible quadratic, the partial fraction decomposition of f contains the following sum of m partial fractions:

1. A partial fraction is a rational function of the form or of the form . 2. (a) What is a proper rational function? (b) What condition must the degree of the numerator and the degree of the denominator of a rational function satisfy for the method of partial fractions to be applicable directly? (c) If the condition in part (b) is not satisﬁed, what must you do if you want to use partial fractions? 3. Suppose that the function f(x) = P (x)/Q(x) is a proper rational function. (a) For each factor of Q(x) of the form (ax + b)m , the partial fraction decomposition of f contains the following sum of m partial fractions:

EXERCISE SET 7.5

4. Complete the partial fraction decomposition. −3 A 2 (a) = − (x + 1)(2x − 1) x + 1 2x − 1 2x 2 − 3x B 1 (b) − = 2 (x + 1)(3x + 2) 3x + 2 x 2 + 1 5. Evaluate the integral. 2x 2 − 3x 3 dx (b) dx (a) 2 (x + 1)(1 − 2x) (x + 1)(3x + 2)

CAS

1–8 Write out the form of the partial fraction decomposition.

11.

(Do not ﬁnd the numerical values of the coefﬁcients.) ■ 1. 3. 5. 7.

3x − 1 (x − 3)(x + 4) 2x − 3 x3 − x2 1 − x2 x 3 (x 2 + 2) 4x 3 − x (x 2 + 5)2

2. 4. 6. 8.

5 x(x 2 − 4) x2 (x + 2)3 3x (x − 1)(x 2 + 6) 1 − 3x 4 (x − 2)(x 2 + 1)2

13. 15. 17. 19. 21.

9–34 Evaluate the integral. ■

dx x 2 − 3x − 4

10.

dx x 2 − 6x − 7

23.

11x + 17 dx 2x 2 + 7x − 4 2x 2 − 9x − 9 dx x 3 − 9x x2 − 8 dx x+3 3x 2 − 10 dx 2 x − 4x + 4 2x − 3 dx x 2 − 3x − 10 x5 + x2 + 2 dx x3 − x 2x 2 + 3 dx x(x − 1)2

12. 14. 16. 18. 20. 22. 24.

5x − 5 dx − 8x − 3 dx x(x 2 − 1) x2 + 1 dx x−1 x2 dx 2 x − 3x + 2 3x + 1 dx 3x 2 + 2x − 1 x 5 − 4x 3 + 1 dx x 3 − 4x 3x 2 − x + 1 dx x3 − x2 3x 2

Chapter 7 / Principles of Integral Evaluation

522

25. 27. 29. 31. 33. 34.

2x 2 − 2x − 1 2x 2 − 10x + 4 dx 26. dx (x + 1)(x − 3)2 x3 − x2 x2 2x 2 + 3x + 3 dx dx 28. (x + 1)3 (x + 1)3 dx 2x 2 − 1 dx 30. (4x − 1)(x 2 + 1) x 3 + 2x 3 x + x2 + x + 2 x 3 + 3x 2 + x + 9 dx 32. dx (x 2 + 1)(x 2 + 3) (x 2 + 1)(x 2 + 2) x 3 − 2x 2 + 2x − 2 dx x2 + 1 x 4 + 6x 3 + 10x 2 + x dx x 2 + 6x + 10

35–38 True–False Determine whether the statement is true or

false. Explain your answer. ■ 35. The technique of partial fractions is used for integrals whose integrands are ratios of polynomials. 36. The integrand in

3x 4 + 5 dx (x 2 + 1)2 is a proper rational function. 37. The partial fraction decomposition of 2x + 3 2 3 is + 2 x2 x x 38. If f(x) = P (x)/(x + 5)3 is a proper rational function, then the partial fraction decomposition of f(x) has terms with constant numerators and denominators (x + 5), (x + 5)2 , and (x + 5)3 . 39–42 Evaluate the integral by making a substitution that converts the integrand to a rational function. ■ cos θ et 39. dt dθ 40. e2t − 4 sin2 θ + 4 sin θ − 5 5 + 2 ln x e3x dx dx 42. 41. e2x + 4 x(1 + ln x)2 43. Find the volume of the solid generated when the region enclosed by y = x 2 /(9 − x 2 ), y = 0, x = 0, and x = 2 is revolved about the x-axis. 44. Find the area of the region under the curve y = 1/(1 + ex ), over the interval [− ln 5, ln 5]. [Hint: Make a substitution that converts the integrand to a rational function.] C

45–46 Use a CAS to evaluate the integral in two ways: (i)

integrate directly; (ii) use the CAS to ﬁnd the partial fraction decomposition and integrate the decomposition. Integrate by hand to check the results. ■ x2 + 1 dx 45. 2 (x + 2x + 3)2 5 x + x 4 + 4x 3 + 4x 2 + 4x + 4 46. dx (x 2 + 2)3

47–48 Integrate by hand and check your answers using a CAS. ■

dx 47. x 4 − 3x 3 − 7x 2 + 27x − 18 dx 48. 3 16x − 4x 2 + 4x − 1 F O C U S O N C O N C E P TS

49. Show that

x π dx = +1 8 0 50. Use partial fractions to derive the integration formula a + x 1 1 +C dx = ln a2 − x 2 2a a − x 51. Suppose that ax 2 + bx + c is a quadratic polynomial and that the integration 1 dx ax 2 + bx + c produces a function with no inverse tangent terms. What does this tell you about the roots of the polynomial? x4

52. Suppose that ax 2 + bx + c is a quadratic polynomial and that the integration 1 dx 2 ax + bx + c produces a function with neither logarithmic nor inverse tangent terms. What does this tell you about the roots of the polynomial? 53. Does there exist a quadratic polynomial ax 2 + bx + c such that the integration x dx ax 2 + bx + c produces a function with no logarithmic terms? If so, give an example; if not, explain why no such polynomial can exist. 54. Writing Suppose that P (x) is a cubic polynomial. State the general form of the partial fraction decomposition for P (x) f(x) = (x + 5)4 and state the implications of this decomposition for evalu ating the integral f(x) dx. 55. Writing Consider the functions 1 x f(x) = 2 and g(x) = 2 x −4 x −4 Each of the integrals f(x) dx and g(x) dx can be evaluated using partial fractions and using at least one other integration technique. Demonstrate two different techniques for evaluating each of these integrals, and then discuss the considerations that would determine which technique you would use.

7.6 Using Computer Algebra Systems and Tables of Integrals

523

✔QUICK CHECK ANSWERS 7.5 Ax + B A ; 2. (a) A proper rational function is a rational function in which the degree of the numerator is (ax + b)k (ax 2 + bx + c)k less than the degree of the denominator. (b) The degree of the numerator must be less than the degree of the denominator. (c) Divide the denominator into the numerator, which results in the sum of a polynomial and a proper rational function. A1 Am A1 x + B1 Am x + Bm A2 A 2 x + B2 3. (a) + ··· + (b) + ··· + + + 2 m 2 2 2 ax + b (ax + b) (ax + b) ax + bx + c (ax + bx + c) (ax 2 + bx + c)m 4. (a) A = 1 (b) B = 2 x+1 2x 2 − 3x 2 3 + C (b) dx = ln dx = ln |3x + 2| − tan−1 x + C 5. (a) 2 (x + 1)(1 − 2x) 1 − 2x (x + 1)(3x + 2) 3 1.

7.6

USING COMPUTER ALGEBRA SYSTEMS AND TABLES OF INTEGRALS In this section we will discuss how to integrate using tables, and we will see some special substitutions to try when an integral doesn’t match any of the forms in an integral table. In particular, we will discuss a method for integrating rational functions of sin x and cos x. We will also address some of the issues that relate to using computer algebra systems for integration. Readers who are not using computer algebra systems can skip that material.

INTEGRAL TABLES Tables of integrals are useful for eliminating tedious hand computation. The endpapers of this text contain a relatively brief table of integrals that we will refer to as the Endpaper Integral Table; more comprehensive tables are published in standard reference books such as the CRC Standard Mathematical Tables and Formulae, CRC Press, Inc., 2002. All integral tables have their own scheme for classifying integrals according to the form of the integrand. For example, the Endpaper Integral Table classifies the integrals into 15 categories; Basic Functions, Reciprocals of Basic Functions, Powers of Trigonometric Functions, Products of Trigonometric Functions, and so forth. The first step in working with tables is to read through the classifications so that you understand the classification scheme and know where to look in the table for integrals of different types.

PERFECT MATCHES

If you are lucky, the integral you are attempting to evaluate will match up perfectly with one of the forms in the table. However, when looking for matches you may have to make an adjustment for the variable of integration. For example, the integral x 2 sin x dx is a perfect match with Formula (46) in the Endpaper Integral Table, except for the letter used for the variable of integration. Thus, to apply Formula (46) to the given integral we need to change the variable of integration in the formula from u to x. With that minor modiﬁcation we obtain x 2 sin x dx = 2x sin x + (2 − x 2 ) cos x + C Here are some more examples of perfect matches.

524

Chapter 7 / Principles of Integral Evaluation

Example 1

Use the Endpaper Integral Table to evaluate √ (a) sin 7x cos 2x dx (b) x 2 7 + 3x dx 2 − x2 (c) dx (d ) (x 3 + 7x + 1) sin πx dx x

Solution (a). The integrand can be classified as a product of trigonometric functions. Thus, from Formula (40) with m = 7 and n = 2 we obtain cos 5x cos 9x − +C sin 7x cos 2x dx = − 18 10 √ Solution (b). The integrand can be classified as a power of x multiplying a + bx. Thus, from Formula (103) with a = 7 and b = 3 we obtain √ 2 (135x 2 − 252x + 392)(7 + 3x)3/2 + C x 2 7 + 3x dx = 2835 √ Solution (c). The integrand√can be classified as a power of x dividing a 2 − x 2 . Thus, from Formula (79) with a = 2 we obtain √ 2 + √2 − x 2 √ 2 − x2 dx = 2 − x 2 − 2 ln +C x x Solution (d). The integrand can be classified as a polynomial multiplying a trigonometric function. Thus, we apply Formula (58) with p(x) = x 3 + 7x + 1 and a = π. The successive nonzero derivatives of p(x) are p ′ (x) = 3x 2 + 7,

p ′′ (x) = 6x,

p′′′ (x) = 6

and so (x 3 + 7x + 1) sin πx dx =−

x 3 + 7x + 1 3x 2 + 7 6x 6 cos πx + sin πx + 3 cos πx − 4 sin πx + C π π2 π π

MATCHES REQUIRING SUBSTITUTIONS

Sometimes an integral that does not match any table entry can be made to match by making an appropriate substitution.

Example 2

Use the Endpaper Integral Table to evaluate (b) x x 2 − 4x + 5 dx (a) eπx sin−1 (eπx ) dx

Solution (a). The integrand does not even come close to matching any of the forms in the table. However, a little thought suggests the substitution u = eπx ,

du = πeπx dx

7.6 Using Computer Algebra Systems and Tables of Integrals

525

from which we obtain

eπx sin−1 (eπx ) dx =

1 π

sin−1 u du

The integrand is now a basic function, and Formula (7) yields 1 u sin−1 u + 1 − u2 + C eπx sin−1 (eπx ) dx = π 1 πx −1 πx = e sin (e ) + 1 − e2πx + C π

Solution (b). Again, the integrand does not closely match any of the forms in the table. However, a√little thought suggests that it may be possible to bring the integrand closer to the form x x 2 + a 2 by completing the square to eliminate the term involving x inside the radical. Doing this yields (1) x x 2 − 4x + 5 dx = x (x 2 − 4x + 4) + 1 dx = x (x − 2)2 + 1 dx

√ At this point we are closer to the form x x 2 + a 2 , but we are not quite there because of the (x − 2)2 rather than x 2 inside the radical. However, we can resolve that problem with the substitution u = x − 2, du = dx With this substitution we have x = u + 2, so (1) can be expressed in terms of u as u2 + 1 du x x 2 − 4x + 5 dx = (u + 2) u2 + 1 du = u u2 + 1 du + 2

The ﬁrst integral on the right is now a perfect match with Formula (84) with a = 1, and the second is a perfect match with Formula (72) with a = 1. Thus, applying these formulas we obtain √ √ x x 2 − 4x + 5 dx = 31 (u2 + 1)3/2 + 2 21 u u2 + 1 + 21 ln(u + u2 + 1) + C If we now replace u by x − 2 (in which case u2 + 1 = x 2 − 4x + 5), we obtain x x 2 − 4x + 5 dx = 13 (x 2 − 4x + 5)3/2 + (x − 2) x 2 − 4x + 5

+ ln x − 2 + x 2 − 4x + 5 + C

Although correct, this form of the answer has an unnecessary mixture of radicals and fractional exponents. If desired, we can “clean up” the answer by writing (x 2 − 4x + 5)3/2 = (x 2 − 4x + 5) x 2 − 4x + 5 from which it follows that (verify) x x 2 − 4x + 5 dx = 13 (x 2 − x − 1) x 2 − 4x + 5

+ ln x − 2 + x 2 − 4x + 5 + C MATCHES REQUIRING REDUCTION FORMULAS

In cases where the entry in an integral table is a reduction formula, that formula will have to be applied ﬁrst to reduce the given integral to a form in which it can be evaluated.

526

Chapter 7 / Principles of Integral Evaluation

Example 3

Use the Endpaper Integral Table to evaluate

â&#x2C6;&#x161;

x3 1+x

dx.

Solution. The integrand can be classiďŹ ed as a power of x multiplying the reciprocal of â&#x2C6;&#x161; a + bx. Thus, from Formula (107) with a = 1, b = 1, and n = 3, followed by Formula (106), we obtain â&#x2C6;&#x161; 2x 3 1 + x x2 6 x3 dx = dx â&#x2C6;&#x2019; â&#x2C6;&#x161; â&#x2C6;&#x161; 7 7 1+x 1+x â&#x2C6;&#x161;

â&#x2C6;&#x161; 6 2 2x 3 1 + x 2 â&#x2C6;&#x2019; (3x â&#x2C6;&#x2019; 4x + 8) 1 + x + C = 7 7 15 3 2x 12x 2 16x 32 â&#x2C6;&#x161; = 1+x+C â&#x2C6;&#x2019; + â&#x2C6;&#x2019; 7 35 35 35 SPECIAL SUBSTITUTIONS The Endpaper Integral Table has numerous entries involving an exponent of 3/2 or involving square roots (exponent 1/2), but it has no entries with other fractional exponents. However, integrals involving fractional powers of x can often be simpliďŹ ed by making the substitution u = x 1/n in which n is the least common multiple of the denominators of the exponents. The resulting integral will then involve integer powers of u.

Example 4

Evaluate (a)

â&#x2C6;&#x161;

x â&#x2C6;&#x161; dx 1+ 3 x

(b)

â&#x2C6;&#x161; 1 + ex dx

Solution (a). The integrand contains x 1/2 and x 1/3 , so we make the substitution u = x 1/6 , from which we obtain

Thus,

x = u6 ,

dx = 6u5 du

â&#x2C6;&#x161; u8 x (u6 )1/2 5 (6u ) du = 6 du dx = â&#x2C6;&#x161; / 3 1 + u2 1+ x 1 + (u6 )1 3

By long division u8 1 = u6 â&#x2C6;&#x2019; u4 + u2 â&#x2C6;&#x2019; 1 + 2 1+u 1 + u2 from which it follows that â&#x2C6;&#x161; 1 x 6 4 2 â&#x2C6;&#x2019; u + u â&#x2C6;&#x2019; 1 + du u dx = 6 â&#x2C6;&#x161; 1 + u2 1+ 3 x = 67 u7 â&#x2C6;&#x2019; 56 u5 + 2u3 â&#x2C6;&#x2019; 6u + 6 tanâ&#x2C6;&#x2019;1 u + C = 76 x 7/6 â&#x2C6;&#x2019; 65 x 5/6 + 2x 1/2 â&#x2C6;&#x2019; 6x 1/6 + 6 tanâ&#x2C6;&#x2019;1 (x 1/6 ) + C

Solution (b). The integral does not match any of the forms inâ&#x2C6;&#x161;the Endpaper Integral Table.

However, the table does include several integrals containing substitution u = ex , from which we obtain x = ln u,

dx =

1 du u

a + bu. This suggests the

7.6 Using Computer Algebra Systems and Tables of Integrals

Try finding the antiderivative in Example 4(b) using the substitution

√ 1 + ex

527

Thus, from Formula (110) with a = 1 and b = 1, followed by Formula (108), we obtain √ √ 1+u x du 1 + e dx = u √ du =2 1+u+ √ u 1+u √ 1 + u − 1 √ = 2 1 + u + ln √ +C 1 + u + 1 √ x −1 √ 1 + e = 2 1 + ex + ln √ +C Absolute value not needed 1 + ex + 1 Functions that consist of ﬁnitely many sums, differences, quotients, and products of sin x and cos x are called rational functions of sin x and cos x. Some examples are sin x + 3 cos2 x sin x 3 sin5 x , , cos x + 4 sin x 1 + cos x − cos2 x 1 + 4 sin x The Endpaper Integral Table gives a few formulas for integrating rational functions of sin x and cos x under the heading Reciprocals of Basic Functions. For example, it follows from Formula (18) that 1 dx = tan x − sec x + C (2) 1 + sin x However, since the integrand is a rational function of sin x, it may be desirable in a particular application to express the value of the integral in terms of sin x and cos x and rewrite (2) as 1 sin x − 1 dx = +C 1 + sin x cos x Many rational functions of sin x and cos x can be evaluated by an ingenious method that was discovered by the mathematician Karl Weierstrass (see p. 102 for biography). The idea is to make the substitution u = tan(x /2),

−π/2 < x /2 < π/2

from which it follows that 2 du 1 + u2 To implement this substitution we need to express sin x and cos x in terms of u. For this purpose we will use the identities x = 2 tan−1 u,

dx =

sin x = 2 sin(x /2) cos(x /2) cos x = cos2 (x /2) − sin2 (x /2)

(3) (4)

and the following relationships suggested by Figure 7.6.1: √1 + u2 u x/2 1 Figure 7.6.1

sin(x /2) =

u 2

and

cos(x /2) =

1+u 1 + u2 Substituting these expressions in (3) and (4) yields u 1 2u sin x = 2 = 2 2 1 + u2 1+u 1+u 2 2 u 1 − u2 1 − = cos x = 1 + u2 1 + u2 1 + u2

528

Chapter 7 / Principles of Integral Evaluation

In summary, we have shown that the substitution u = tan(x /2) can be implemented in a rational function of sin x and cos x by letting sin x =

Example 5

2u , 1 + u2

Evaluate

cos x =

1 â&#x2C6;&#x2019; u2 , 1 + u2

dx =

2 du 1 + u2

(5)

dx . 1 â&#x2C6;&#x2019; sin x + cos x

Solution. The integrand is a rational function of sin x and cos x that does not match any of the formulas in the Endpaper Integral Table, so we make the substitution u = tan(x /2). Thus, from (5) we obtain

The substitution u = tan(x /2) will convert any rational function of sin x and cos x to an ordinary rational function of u. However, the method can lead to cumbersome partial fraction decompositions, so it may be worthwhile to consider other methods as well when hand computations are being used.

dx = 1 â&#x2C6;&#x2019; sin x + cos x

2 du 1 + u2 1 â&#x2C6;&#x2019; u2 2u + 1â&#x2C6;&#x2019; 1 + u2 1 + u2 2 du (1 + u2 ) â&#x2C6;&#x2019; 2u + (1 â&#x2C6;&#x2019; u2 ) du = â&#x2C6;&#x2019; ln |1 â&#x2C6;&#x2019; u| + C = â&#x2C6;&#x2019; ln |1 â&#x2C6;&#x2019; tan(x /2)| + C 1â&#x2C6;&#x2019;u

INTEGRATING WITH COMPUTER ALGEBRA SYSTEMS

Integration tables are rapidly giving way to computerized integration using computer algebra systems. However, as with many powerful tools, a knowledgeable operator is an important component of the system. Sometimes computer algebra systems do not produce the most general form of the indeďŹ nite integral. For example, the integral formula dx = ln |x â&#x2C6;&#x2019; 1| + C xâ&#x2C6;&#x2019;1

which can be obtained by inspection or by using the substitution u = x â&#x2C6;&#x2019; 1, is valid for x > 1 or for x < 1. However, not all computer algebra systems produce this form of the answer. Some typical answers produced by various implementations of Mathematica, Maple, and the CAS on a handheld calculator are ln(â&#x2C6;&#x2019;1 + x),

ln(x â&#x2C6;&#x2019; 1),

ln(|x â&#x2C6;&#x2019; 1|)

Observe that none of the systems include the constant of integrationâ&#x20AC;&#x201D;the answer produced is a particular antiderivative and not the most general antiderivative (indeďŹ nite integral). Observe also that only one of these answers includes the absolute value signs; the antiderivatives produced by the other systems are valid only for x > 1. All systems, however, are able to calculate the deďŹ nite integral 1 /2 dx = â&#x2C6;&#x2019; ln 2 x â&#x2C6;&#x2019;1 0 correctly. Now let us examine how these systems handle the integral x x 2 â&#x2C6;&#x2019; 4x + 5 dx = 13 (x 2 â&#x2C6;&#x2019; x â&#x2C6;&#x2019; 1) x 2 â&#x2C6;&#x2019; 4x + 5 + ln(x â&#x2C6;&#x2019; 2 + x 2 â&#x2C6;&#x2019; 4x + 5 )

(6)

7.6 Using Computer Algebra Systems and Tables of Integrals

529

which we obtained in Example 2(b) (with the constant of integration included). Some CAS implementations produce this result in slightly different algebraic forms, but a version of Maple produces the result x x 2 − 4x + 5 dx = 13 (x 2 − 4x + 5)3/2 + 21 (2x − 4) x 2 − 4x + 5 + sinh−1 (x − 2)

This can be rewritten as (6) by expressing the fractional exponent in radical form and expressing sinh−1 (x − 2) in logarithmic form using Theorem 6.9.4 (verify). A version of Mathematica produces the result x x 2 − 4x + 5 dx = 13 (x 2 − x − 1) x 2 − 4x + 5 − sinh−1 (2 − x) which can be rewritten in form (6) by using Theorem 6.9.4 together with the identity sinh−1 (−x) = − sinh−1 x (verify). Computer algebra systems can sometimes produce inconvenient or unnatural answers to integration problems. For example, various computer algebra systems produced the following results when asked to integrate (x + 1)7 :

Expanding the expression

(x + 1)8 8 produces a constant term of whereas the second expression in has no constant term. What is the planation?

1 8,

(7) ex-

1 8 7 35 7 (x + 1)8 , x + x 7 + x 6 + 7x 5 + x 4 + 7x 3 + x 2 + x (7) 8 8 2 4 2 The ﬁrst form is in keeping with the hand computation (x + 1)8 +C (x + 1)7 dx = 8 that uses the substitution u = x + 1, whereas the second form is based on expanding (x + 1)7 and integrating term by term. In Example 2(a) of Section 7.3 we showed that sin4 x cos5 x dx = 15 sin5 x − 27 sin7 x + 91 sin9 x + C However, a version of Mathematica integrates this as 3 128

sin x −

1 192

sin 3x −

1 320

sin 5x +

1 1792

sin 7x +

1 2304

sin 9x

whereas other computer algebra systems essentially integrate it as − 91 sin3 x cos6 x −

1 21

sin x cos6 x +

1 105

cos4 x sin x +

4 315

cos2 x sin x +

8 315

sin x

Although these three results look quite different, they can be obtained from one another using appropriate trigonometric identities.

T E C H N O LO GY M A ST E R Y Sometimes integrals that cannot be evaluated by a CAS in their given form can be evaluated by first rewriting them in a different form or by making a substitution. If you have a CAS, make a u-substitution in (8) that will enable you to evaluate the integral with your CAS. Then evaluate the integral.

COMPUTER ALGEBRA SYSTEMS HAVE LIMITATIONS A computer algebra system combines a set of integration rules (such as substitution) with a library of functions that it can use to construct antiderivatives. Such libraries contain elementary functions, such as polynomials, rational functions, trigonometric functions, as well as various nonelementary functions that arise in engineering, physics, and other applied ﬁelds. Just as our Endpaper Integral Table has only 121 indeﬁnite integrals, these libraries are not exhaustive of all possible integrands. If the system cannot manipulate the integrand to a form matching one in its library, the program will give some indication that it cannot evaluate the integral. For example, when asked to evaluate the integral (1 + ln x) 1 + (x ln x)2 dx (8)

all of the systems mentioned above respond by displaying some form of the unevaluated integral as an answer, indicating that they could not perform the integration. Sometimes computer algebra systems respond by expressing an integral in terms of 2 another integral. For example, if you try to integrate ex using Mathematica, Maple, or

Chapter 7 / Principles of Integral Evaluation

530

Sage, you will obtain an expression involving erf (which stands for error function). The function erf(x) is deﬁned as x 2 2 erf(x) = √ e−t dt π 0 so all three programs essentially rewrite the given integral in terms of a closely related integral. From one point of view this is what we did in integrating 1/x, since the natural logarithm function is (formally) deﬁned as x 1 ln x = dt 1 t (see Section 5.10). Example 6 A particle moves along an x-axis in such a way that its velocity v(t) at time t is v(t) = 30 cos7 t sin4 t (t ≥ 0) Graph the position versus time curve for the particle, given that the particle is at x = 1 when t = 0.

Solution. Since dx /dt = v(t) and x = 1 when t = 0, the position function x(t) is given by

x(t) = 1 +

v(s) ds

Some computer algebra systems will allow this expression to be entered directly into a command for plotting functions, but it is often more efﬁcient to perform the integration ﬁrst. The authors’ integration utility yields x = 30 cos7 t sin4 t dt

x 2

sin11 t + 10 sin9 t − = − 30 11

1.5 1 0.5 t 5

Figure 7.6.2

90 7

sin7 t + 6 sin5 t + C

where we have added the required constant of integration. Using the initial condition x(0) = 1, we substitute the values x = 1 and t = 0 into this equation to ﬁnd that C = 1, so x(t) = − 30 sin11 t + 10 sin9 t − 11

90 7

sin7 t + 6 sin5 t + 1 (t ≥ 0)

The graph of x versus t is shown in Figure 7.6.2.

✔QUICK CHECK EXERCISES 7.6

(See page 533 for answers.)

1. Find an integral formula in the Endpaper Integral Table that can be used to evaluate the integral. Do not evaluate the integral. 2x (a) dx 3x +4 1 (b) dx √ x 5x − 4 √ (c) x 3x + 2 dx (d) x 2 ln x dx

2. In each part, make the indicated u-substitution, and then ﬁnd an integral formula in the Endpaper Integral Table that

can be used to evaluate the integral. Do not evaluate the integral. x (a) dx; u = x 2 x2 1 + √ e √ (b) e x dx; u = x ex (c) dx; u = ex x 1 + sin(e ) 1 (d) dx; u = 2x (1 − 4x 2 )3/2 3. In each part, use the Endpaper Integral Table to evaluate the integral. (If necessary, ﬁrst make an appropriate substitution or complete the square.) (cont.)

7.6 Using Computer Algebra Systems and Tables of Integrals

1 dx = 4 − x2 (b) cos 2x cos x dx = (a)

EXERCISE SET 7.6 C

e3x

dx = 1 − e2x x dx = x 2 − 4x + 8 √

CAS

1–24 (a) Use the Endpaper Integral Table to evaluate the given

integral. (b) If you have a CAS, use it to evaluate the integral, and then conﬁrm that the result is equivalent to the one that you found in part (a). ■ 4x x 1. dx dx 2. 3x − 1 (4 − 5x)2 1 1 3. dx 4. dx 2 x(2x + 5) x (1 − 5x) √ x dx 6. 5. x 2x + 3 dx √ 2−x 1 1 7. dx 8. dx √ √ x 4 − 3x x 3x − 4 1 1 9. dx 10. dx 16 − x 2 x2 − 9 x2 − 5 11. x 2 − 3 dx 12. dx x2 x2 1 13. dx 14. dx 2 x +4 x2 x2 − 2 4 − x2 15. 9 − x 2 dx 16. dx x2 4 − x2 1 17. dx dx 18. x x 6x − x 2 19. sin 3x sin 4x dx 20. sin 2x cos 5x dx ln x 22. 21. x 3 ln x dx √ dx x3 23. e−2x sin 3x dx 24. ex cos 2x dx C

531

25–36 (a) Make the indicated u-substitution, and then use the Endpaper Integral Table to evaluate the integral. (b) If you have a CAS, use it to evaluate the integral, and then conﬁrm that the result is equivalent to the one that you found in part (a). ■ e4x 25. dx, u = e2x (4 − 3e2x )2 sin 2x dx, u = cos 2x 26. (cos 2x)(3 − cos 2x) √ 1 27. dx, u = 3 x √ x(9x + 4) cos 4x 28. dx, u = sin 4x 9 + sin2 4x

dx, u = 2x 4x 2 − 9 √ 30. x 2x 4 + 3 dx, u = 2x 2 4x 5 dx, u = 2x 2 31. 2 − 4x 4 1 dx, u = 2x 32. 2 x 3 − 4x 2 sin2 (ln x) 33. dx, u = ln x x 34. e−2x cos2 (e−2x ) dx, u = e−2x 35. xe−2x dx, u = −2x 36. ln(3x + 1) dx, u = 3x + 1 29.

37–48 (a) Make an appropriate u-substitution, and then use the Endpaper Integral Table to evaluate the integral. (b) If you have a CAS, use it to evaluate the integral (no substitution), and then conﬁrm that the result is equivalent to that in part (a). ■ cos 3x 37. dx (sin 3x)(sin 3x + 1)2 ln x dx 38. √ x 4 ln x − 1 ex x dx dx 40. 39. 16x 4 − 1 3 − 4e2x 4 − 9x 2 42. dx 41. ex 3 − 4e2x dx x2 1 43. dx 5x − 9x 2 dx 44. x x − 5x 2 √ 45. 4x sin 2x dx 46. cos x dx √ − x 47. e x ln(2 + x 2 ) dx dx 48.

49–52 (a) Complete the square, make an appropriate usubstitution, and then use the Endpaper Integral Table to evaluate the integral. (b) If you have a CAS, use it to evaluate the integral (no substitution or square completion), and then conﬁrm that the result is equivalent to that in part (a). ■ 1 49. 3 − 2x − x 2 dx dx 50. x 2 + 6x − 7

Chapter 7 / Principles of Integral Evaluation

532

51.

x dx 5 + 4x â&#x2C6;&#x2019; x 2

52.

x dx + 6x + 13

78. y =

79. y = eâ&#x2C6;&#x2019;x , y = 0, x = 0, x = 3

1/n

u=x or u = (x + a) , and then evaluate the integral. (b) If you have a CAS, use it to evaluate the integral, and then conďŹ rm that the result is equivalent to the one that you found in part (a). â&#x2013; â&#x2C6;&#x161; x dx 53. x x â&#x2C6;&#x2019; 2 dx 54. â&#x2C6;&#x161; x+1 1 55. x 5 x 3 + 1 dx 56. dx x x3 â&#x2C6;&#x2019; 1 dx dx 58. 57. â&#x2C6;&#x161; â&#x2C6;&#x161; â&#x2C6;&#x161; xâ&#x2C6;&#x2019; 3x x+ 3x â&#x2C6;&#x161; dx x 60. 59. dx x+1 x(1 â&#x2C6;&#x2019; x 1/4 ) â&#x2C6;&#x161; 1+ x dx 62. 61. â&#x2C6;&#x161; dx 1â&#x2C6;&#x2019; x x 1/2 â&#x2C6;&#x2019; x 1/3 x3 x 63. dx 64. dx â&#x2C6;&#x161; 2 (x + 3)1/5 1+x C

80. y = ln x, y = 0, x = 5

53â&#x20AC;&#x201C;64 (a) Make an appropriate u-substitution of the form 1/n

81â&#x20AC;&#x201C;82 Use any method to ďŹ nd the arc length of the curve. â&#x2013;

81. y = 2x 2 , 0 â&#x2030;¤ x â&#x2030;¤ 2

dx 1 + sin x + cos x dÎ¸ 67. 1 â&#x2C6;&#x2019; cos Î¸ dx 69. sin x + tan x

F O C U S O N C O N C E P TS

87. (a) Use the substitution u = tan(x /2) to show that 1 + tan(x /2) +C sec x dx = ln 1 â&#x2C6;&#x2019; tan(x /2)

and conďŹ rm that this is consistent with Formula (22) of Section 7.3. (b) Use the result in part (a) to show that Ď&#x20AC; x sec x dx = ln tan + +C 4 2 88. Use the substitution u = tan(x /2) to show that

1 1 â&#x2C6;&#x2019; cos x csc x dx = ln +C 2 1 + cos x

dx 2 + sin x dx 68. 4 sin x â&#x2C6;&#x2019; 3 cos x sin x 70. dx sin x + tan x 66.

71â&#x20AC;&#x201C;72 Use any method to solve for x. â&#x2013;

71. 72.

2 x 1

and conďŹ rm that this is consistent with the result in Exercise 65(a) of Section 7.3.

1 dt = 0.5, 2 < x < 4 t (4 â&#x2C6;&#x2019; t) 1 dt = 1, x > 21 â&#x2C6;&#x161; t 2t â&#x2C6;&#x2019; 1

73â&#x20AC;&#x201C;76 Use any method to ďŹ nd the area of the region enclosed by the curves. â&#x2013; 73. y = 25 â&#x2C6;&#x2019; x 2 , y = 0, x = 0, x = 4 74. y = 9x 2 â&#x2C6;&#x2019; 4, y = 0, x = 2 1 75. y = , y = 0, x = 0, x = 1 25 â&#x2C6;&#x2019; 16x 2 â&#x2C6;&#x161; 76. y = x ln x, y = 0, x = 4 77â&#x20AC;&#x201C;80 Use any method to ďŹ nd the volume of the solid generated when the region enclosed by the curves is revolved about the y-axis. â&#x2013; 77. y = cos x, y = 0, x = 0, x = Ď&#x20AC;/2

85â&#x20AC;&#x201C;86 Information is given about the motion of a particle moving along a coordinate line. (a) Use a CAS to ďŹ nd the position function of the particle for t â&#x2030;Ľ 0. (b) Graph the position versus time curve. â&#x2013; 85. v(t) = 20 cos6 t sin3 t, s(0) = 2

86. a(t) = eâ&#x2C6;&#x2019;t sin 2t sin 4t, v(0) = 0, s(0) = 10

â&#x2013;

82. y = 3 ln x, 1 â&#x2030;¤ x â&#x2030;¤ 3

83â&#x20AC;&#x201C;84 Use any method to ďŹ nd the area of the surface generated by revolving the curve about the x-axis. â&#x2013; 83. y = sin x, 0 â&#x2030;¤ x â&#x2030;¤ Ď&#x20AC; 84. y = 1/x, 1 â&#x2030;¤ x â&#x2030;¤ 4

65â&#x20AC;&#x201C;70 (a) Make u-substitution (5) to convert the integrand to a rational function of u, and then evaluate the integral. (b) If you have a CAS, use it to evaluate the integral (no substitution), and then conďŹ rm that the result is equivalent to that in part (a).

65.

â&#x2C6;&#x161; x â&#x2C6;&#x2019; 4, y = 0, x = 8

89. Find a substitution that can be used to integrate rational functions of sinh x and cosh x and use your substitution to evaluate dx 2 cosh x + sinh x without expressing the integrand in terms of ex and eâ&#x2C6;&#x2019;x .

90â&#x20AC;&#x201C;93 Some integrals that can be evaluated by hand cannot be evaluated by all computer algebra systems. Evaluate the integral by hand, and determine if it can be evaluated on your CAS. â&#x2013; x3 90. dx â&#x2C6;&#x161; 1 â&#x2C6;&#x2019; x8 91. (cos32 x sin30 x â&#x2C6;&#x2019; cos30 x sin32 x) dx â&#x2C6;&#x161; â&#x2C6;&#x161; x â&#x2C6;&#x2019; x 2 â&#x2C6;&#x2019; 4 dx [Hint: 21 ( x + 2 â&#x2C6;&#x2019; x â&#x2C6;&#x2019; 2)2 = ?] 92.

7.7 Numerical Integration; Simpsonâ&#x20AC;&#x2122;s Rule

93.

1 dx x 10 + x [Hint: Rewrite the denominator as x 10 (1 + x â&#x2C6;&#x2019;9 ).]

(a) Use a CAS to factor the denominator, and then write down the form of the partial fraction decomposition. You need not ďŹ nd the values of the constants. (b) Check your answer in part (a) by using the CAS to ďŹ nd the partial fraction decomposition of f . (c) Integrate f by hand, and then check your answer by integrating with the CAS.

94. Let f(x) =

533

â&#x2C6;&#x2019;2x 5 + 26x 4 + 15x 3 + 6x 2 + 20x + 43 x 6 â&#x2C6;&#x2019; x 5 â&#x2C6;&#x2019; 18x 4 â&#x2C6;&#x2019; 2x 3 â&#x2C6;&#x2019; 39x 2 â&#x2C6;&#x2019; x â&#x2C6;&#x2019; 20

â&#x153;&#x201D;QUICK CHECK ANSWERS 7.6 1. (a) Formula (60) (b) Formula (108) (c) Formula (b) Formula (51) (d) Formula (50) 2. (a) Formula (25)x (102) 1 x + 2 1 1 1 e (c) Formula (18) (d) Formula (97) 3. (a) ln + C (b) sin 3x + sin x + C (c) â&#x2C6;&#x2019; 1 â&#x2C6;&#x2019; e2x + sinâ&#x2C6;&#x2019;1 ex + C 4 x â&#x2C6;&#x2019; 2 6 2 2 2 1 xâ&#x2C6;&#x2019;2 (d) ln x 2 â&#x2C6;&#x2019; 4x + 8 + tanâ&#x2C6;&#x2019;1 +C 2 2

7.7

NUMERICAL INTEGRATION; SIMPSONâ&#x20AC;&#x2122;S RULE If it is necessary to evaluate a deďŹ nite integral of a function for which an antiderivative cannot be found, then one must settle for some kind of numerical approximation of the integral. In Section 5.4 we considered three such approximations in the context of areasâ&#x20AC;&#x201D;left endpoint approximation, right endpoint approximation, and midpoint approximation. In this section we will extend those methods to general deďŹ nite integrals, and we will develop some new methods that often provide more accuracy with less computation. We will also discuss the errors that arise in integral approximations.

A REVIEW OF RIEMANN SUM APPROXIMATIONS

Recall from Section 5.5 that the deďŹ nite integral of a continuous function f over an interval [a, b] may be computed as b n f(x) dx = lim f(xkâ&#x2C6;&#x2014; ) xk a

max xk â&#x2020;&#x2019; 0

k=1

where the sum that appears on the right side is called a Riemann sum. In this formula, xk is the width of the kth subinterval of a partition a = x0 < x1 < x2 < Âˇ Âˇ Âˇ < xn = b of [a, b] into n subintervals, and xkâ&#x2C6;&#x2014; denotes an arbitrary point in the kth subinterval. If we take all subintervals of the same width, so that xk = (b â&#x2C6;&#x2019; a)/n, then as n increases the Riemann sum will eventually be a good approximation to the deďŹ nite integral. We denote this by b writing bâ&#x2C6;&#x2019;a (1) f(x) dx â&#x2030;&#x2C6; [f(x1â&#x2C6;&#x2014; ) + f(x2â&#x2C6;&#x2014; ) + Âˇ Âˇ Âˇ + f(xnâ&#x2C6;&#x2014; )] n a If we denote the values of f at the endpoints of the subintervals by y0 = f(a),

y1 = f(x1 ),

y2 = f(x2 ), . . . , ynâ&#x2C6;&#x2019;1 = f(xnâ&#x2C6;&#x2019;1 ),

yn = f(xn )

and the values of f at the midpoints of the subintervals by ym1 , ym2 , . . . , ymn then it follows from (1) that the left endpoint, right endpoint, and midpoint approximations discussed in Section 5.4 can be expressed as shown in Table 7.7.1. Although we originally

Chapter 7 / Principles of Integral Evaluation

534

Table 7.7.1

left endpoint approximation b a

right endpoint approximation

f (x) dx â&#x2030;&#x2C6; b â&#x2C6;&#x2019; a [y0 + y1 + . . . + ynâ&#x2C6;&#x2019;1] n

ĺ&#x2020;¸

ĺ&#x2020;š

b a

midpoint approximation b

f (x) dx â&#x2030;&#x2C6; b â&#x2C6;&#x2019; a [ y1 + y2 + . . . + yn ] n

ĺ&#x2020;¸

ĺ&#x2020;š

f (x) dx â&#x2030;&#x2C6; b â&#x2C6;&#x2019; a [ ym1 + ym2 + . . . + ym n ] n

ĺ&#x2020;¸

ynâ&#x20AC;&#x201C;1

ĺ&#x2020;š

ynâ&#x20AC;&#x201C;1

ym1 ym2

yn x

ymn ...

obtained these results for nonnegative functions in the context of approximating areas, they are applicable to any function that is continuous on [a, b]. TRAPEZOIDAL APPROXIMATION

It will be convenient in this section to denote the left endpoint, right endpoint, and midpoint approximations with n subintervals by Ln , Rn , and Mn , respectively. Of the three approximations, the midpoint approximation is most widely used in applications. If we take the average of Ln and Rn , then we obtain another important approximation denoted by

Tn = 21 (Ln + Rn ) called the trapezoidal approximation:

y1 y2

ynâ&#x2C6;&#x2019;1 yn x

a Trapezoidal approximation

Figure 7.7.1

b a

Trapezoidal Approximation bâ&#x2C6;&#x2019;a [y0 + 2y1 + Âˇ Âˇ Âˇ + 2ynâ&#x2C6;&#x2019;1 + yn ] f(x) dx â&#x2030;&#x2C6; Tn = 2n

(2)

The name â&#x20AC;&#x153;trapezoidal approximationâ&#x20AC;? results from the fact that in the case where f is nonnegative on the interval of integration, the approximation Tn is the sum of the trapezoidal areas shown in Figure 7.7.1 (see Exercise 51). Example 1

In Table 7.7.2 we have approximated 2 1 ln 2 = dx x 1

using the midpoint approximation and the trapezoidal approximation. used n = 10 subdivisions of the interval [1, 2], so that bâ&#x2C6;&#x2019;a bâ&#x2C6;&#x2019;a 2â&#x2C6;&#x2019;1 2â&#x2C6;&#x2019;1 = = 0.1 and = = 0.05 20 10 n 2n Midpoint

â&#x2C6;&#x2014;

In each case we

Trapezoidal

Throughout this section we will show numerical values to nine places to the right of the decimal point. If your calculating utility does not show this many places, then you will need to make the appropriate adjustments. What is important here is that you understand the principles being discussed.

7.7 Numerical Integration; Simpson’s Rule

535

Table 7.7.2 2

midpoint and trapezoidal approximations for 1

midpoint approximation i

midpoint mi

1 2 3 4 5 6 7 8 9 10

1.05 1.15 1.25 1.35 1.45 1.55 1.65 1.75 1.85 1.95

2 1

trapezoidal approximation

ymi = f (mi ) = 1/mi

endpoint xi

yi = f (xi ) = 1/xi

0.952380952 0.869565217 0.800000000 0.740740741 0.689655172 0.645161290 0.606060606 0.571428571 0.540540541 0.512820513 6.928353603

0 1 2 3 4 5 6 7 8 9 10

1.0 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2.0

1.000000000 0.909090909 0.833333333 0.769230769 0.714285714 0.666666667 0.625000000 0.588235294 0.555555556 0.526315789 0.500000000

1 dx ≈ (0.1)(6.928353603) ≈ 0.692835360 x

By rewriting (3) and (4) in the form

1 dx x

f(x) dx = approximation + error

we see that positive values of EM and ET correspond to underestimates and negative values to overestimates.

multiplier wi y i

1.000000000 1.818181818 1.666666667 1.538461538 1.428571429 1.333333333 1.250000000 1.176470588 1.111111111 1.052631579 0.500000000 13.875428063

1 2 2 2 2 2 2 2 2 2 1

1 dx ≈ (0.05)(13.875428063) ≈ 0.693771403 x

COMPARISON OF THE MIDPOINT AND TRAPEZOIDAL APPROXIMATIONS We deﬁne the errors in the midpoint and trapezoidal approximations to be b b f(x) dx − Tn f(x) dx − Mn and ET = EM =

(3–4)

respectively, and we deﬁne |EM | and |ET | to be the absolute errors in these approximations. The absolute errors are nonnegative and do not distinguish between underestimates and overestimates. Example 2 The value of ln 2 to nine decimal places is 2 1 ln 2 = dx ≈ 0.693147181 1 x

(5)

so we see from Tables 7.7.2 and 7.7.3 that the absolute errors in approximating ln 2 by M10 and T10 are |E | = | ln 2 − M | ≈ 0.000311821 M

|ET | = | ln 2 − T10 | ≈ 0.000624222

Thus, the midpoint approximation is more accurate than the trapezoidal approximation in this case. Table 7.7.3

ln 2 (nine decimal places)

approximation

error

0.693147181 0.693147181

M10 ≈ 0.692835360 T10 ≈ 0.693771403

EM = ln 2 − M10 ≈ 0.000311821 ET = ln 2 − T10 ≈ −0.000624222

536

Chapter 7 / Principles of Integral Evaluation

mk The shaded triangles have equal areas.

Figure 7.7.2

It is not accidental in Example 2 that the midpoint approximation of ln 2 was more accurate than the trapezoidal approximation. To see why this is so, we first need to look at the midpoint approximation from another point of view. To simplify our explanation, we will assume that f is nonnegative on [a, b], though the conclusions we reach will be true without this assumption. If f is a differentiable function, then the midpoint approximation is sometimes called the tangent line approximation because for each subinterval of [a, b] the area of the rectangle used in the midpoint approximation is equal to the area of the trapezoid whose upper boundary is the tangent line to y = f(x) at the midpoint of the subinterval (Figure 7.7.2). The equality of these areas follows from the fact that the shaded areas in the figure are congruent. We will now show how this point of view about midpoint approximations can be used to establish useful criteria for determining which of Mn or Tn produces the better approximation of a given integral. In Figure 7.7.3a we have isolated a subinterval of [a, b] on which the graph of a function f is concave down, and we have shaded the areas that represent the errors in the midpoint and trapezoidal approximations over the subinterval. In Figure 7.7.3b we show a succession of four illustrations which make it evident that the error from the midpoint approximation is less than that from the trapezoidal approximation. If the graph of f were concave up, analogous figures would lead to the same conclusion. (This argument, due to Frank Buck, appeared in The College Mathematics Journal, Vol. 16, No. 1, 1985.) Midpoint error Trapezoidal error

Justify the conclusions in each step of Figure 7.7.3b.

Blue area

(a)

Blue area

< Yellow area

(b)

Figure 7.7.3

Figure 7.7.3a also suggests that on a subinterval where the graph is concave down, the midpoint approximation is larger than the value of the integral and the trapezoidal approximation is smaller. On an interval where the graph is concave up it is the other way around. In summary, we have the following result, which we state without formal proof: 7.7.1 theorem Let f be continuous on [a, b], and let |EM | and |ET | be the absolute b errors that result from the midpoint and trapezoidal approximations of a f(x) dx using n subintervals. (a) If the graph of f is either concave up or concave down on (a, b), then |EM | < |ET |, that is, the absolute error from the midpoint approximation is less than that from the trapezoidal approximation. (b) If the graph of f is concave down on (a, b), then b f(x) dx < Mn Tn < a

7.7 Numerical Integration; Simpson’s Rule

537

Example 3 Since the graph of f(x) = 1/x is continuous on the interval [1, 2] and concave up on the interval (1, 2), it follows from part (a) of Theorem 7.7.1 that Mn will always provide a better approximation than Tn for 2 1 dx = ln 2 x 1

WARNING Do not conclude that the midpoint approximation is always better than the trapezoidal approximation; the trapezoidal approximation may be better if the function changes concavity on the interval of integration.

Moreover, if follows from part (c) of Theorem 7.7.1 that Mn < ln 2 < Tn for every positive integer n. Note that this is consistent with our computations in Example 2. Example 4 The midpoint and trapezoidal approximations can be used to approximate 1 sin 1 by using the integral sin 1 = cos x dx 0

Since f(x) = cos x is continuous on [0, 1] and concave down on (0, 1), it follows from parts (a) and (b) of Theorem 7.7.1 that the absolute error in Mn will be less than that in Tn , and that Tn < sin 1 < Mn for every positive integer n. This is consistent with the results in Table 7.7.4 for n = 5 (intermediate computations are omitted). Table 7.7.4

sin 1 (nine decimal places)

approximation

error

0.841470985 0.841470985

M5 ≈ 0.842875074 T5 ≈ 0.838664210

EM = sin 1 − M5 ≈ − 0.001404089 ET = sin 1 − T5 ≈ 0.002806775

3 Example 5 Table 7.7.5 shows approximations for sin 3 = 0 cos x dx using the midpoint and trapezoidal approximations with n = 10 subdivisions of the interval [0, 3]. Note that |EM | < |ET | and T10 < sin 3 < M10 , although these results are not guaranteed by Theorem 7.7.1 since f(x) = cos x changes concavity on the interval [0, 3]. Table 7.7.5

sin 3 (nine decimal places)

approximation

error

0.141120008 0.141120008

M10 ≈ 0.141650601 T10 ≈ 0.140060017

EM = sin 3 − M10 ≈ −0.000530592 ET = sin 3 − T10 ≈ 0.001059991

SIMPSON’S RULE When the left and right endpoint approximations are averaged to produce the trapezoidal approximation, a better approximation often results. We now see how a weighted average of the midpoint and trapezoidal approximations can yield an even better approximation. The numerical evidence in Tables 7.7.3, 7.7.4, and 7.7.5 reveals that ET ≈ −2EM , so that 2EM + ET ≈ 0 in these instances. This suggests that b b b f (x) dx f (x) dx + f (x) dx = 2 3 a

= 2(Mk + EM ) + (Tk + ET )

= (2Mk + Tk ) + (2EM + ET )

≈ 2Mk + Tk

538

Chapter 7 / Principles of Integral Evaluation

This gives

WARNING Note that in (7) the subscript n in Sn is always even since it is twice the value of the subscripts for the corresponding midpoint and trapezoidal approximations. For example,

S10 = 13 (2M5 + T5 ) and

S20 =

1 3 (2M10

+ T10 )

b a

f (x) dx â&#x2030;&#x2C6; 31 (2Mk + Tk )

(6)

The midpoint approximation Mk in (6) requires the evaluation of f at k points in the interval [a, b], and the trapezoidal approximation Tk in (6) requires the evaluation of f at k + 1 points in [a, b]. Thus, 31 (2Mk + Tk ) uses 2k + 1 values of f , taken at equally spaced points in the interval [a, b]. These points are obtained by partitioning [a, b] into 2k equal subintervals indicated by the left endpoints, right endpoints, and midpoints used in Tk and Mk , respectively. Setting n = 2k, we use Sn to denote the weighted average of Mk and Tk in (6). That is, Sn = S2k = 13 (2Mk + Tk )

Sn = 13 (2Mn/2 + Tn/2 )

(7)

Table 7.7.6 displays the approximations Sn corresponding to the data in Tables 7.7.3 to 7.7.5. Table 7.7.6

function value (nine decimal places)

error

approximation

ln 2 â&#x2030;&#x2C6; 0.693147181

ĺ&#x2026;°12 (1/ x) dx â&#x2030;&#x2C6; S20 = 13 (2M10 + T10) â&#x2030;&#x2C6; 0.693147375

â&#x2C6;&#x2019;0.000000194

sin 1 â&#x2030;&#x2C6; 0.841470985

ĺ&#x2026;°01 cos x dx â&#x2030;&#x2C6; S10 = 13 (2M5 + T5) â&#x2030;&#x2C6; 0.841471453 ĺ&#x2026;°03 cos x dx â&#x2030;&#x2C6; S20 = 13 (2M10 + T10) â&#x2030;&#x2C6; 0.141120406

â&#x2C6;&#x2019;0.000000468

sin 3 â&#x2030;&#x2C6; 0.141120008

â&#x2C6;&#x2019;0.000000398

Using the midpoint approximation formula in Table 7.7.1 and Formula (2) for the trapezoidal approximation, we can derive a similar formula for Sn . We start by partitioning the interval [a, b] into an even number of equal subintervals. If n is the number of subintervals, then each subinterval has length (b â&#x2C6;&#x2019; a)/n. Label the endpoints of these subintervals successively by a = x0 , x1 , x2 , . . . , xn = b. Then x0 , x2 , x4 , . . . , xn deďŹ ne a partition of [a, b] into n/2 equal intervals, each of length 2(b â&#x2C6;&#x2019; a)/n, and the midpoints of these subintervals are x1 , x3 , x5 , . . . , xnâ&#x2C6;&#x2019;1 , respectively, as illustrated in Figure 7.7.4. Using yi = f (xi ), we have 2(b â&#x2C6;&#x2019; a) [y1 + y3 + Âˇ Âˇ Âˇ + ynâ&#x2C6;&#x2019;1 ] 2Mn/2 = 2 n bâ&#x2C6;&#x2019;a [4y1 + 4y3 + Âˇ Âˇ Âˇ + 4ynâ&#x2C6;&#x2019;1 ] = n Noting that (b â&#x2C6;&#x2019; a)/[2(n/2)] = (b â&#x2C6;&#x2019; a)/n, we can express Tn/2 as bâ&#x2C6;&#x2019;a Tn/2 = [y0 + 2y2 + 2y4 + Âˇ Âˇ Âˇ + 2ynâ&#x2C6;&#x2019;2 + yn ] n

Thus, Sn = 13 (2Mn/2 + Tn/2 ) can be expressed as 1 bâ&#x2C6;&#x2019;a Sn = [y0 + 4y1 + 2y2 + 4y3 + 2y4 + Âˇ Âˇ Âˇ + 2ynâ&#x2C6;&#x2019;2 + 4ynâ&#x2C6;&#x2019;1 + yn ] 3 n The approximation

(8)

b a

f (x) dx â&#x2030;&#x2C6; Sn

(9)

with Sn as given in (8) is known as Simpsonâ&#x20AC;&#x2122;s rule. We denote the error in this approximation b by f (x) dx â&#x2C6;&#x2019; Sn (10) ES = a

As before, the absolute error in the approximation (9) is given by |ES |.

7.7 Numerical Integration; Simpsonâ&#x20AC;&#x2122;s Rule n

539

a = x0

midpoint

...

midpoint

b = xn midpoint

... n

Figure 7.7.4

Example 6 In Table 7.7.7 we have used Simpsonâ&#x20AC;&#x2122;s rule with n = 10 subintervals to obtain the approximation 2 1 dx â&#x2030;&#x2C6; S10 = 0.693150231 ln 2 = 1 x

For this approximation,

1 3

bâ&#x2C6;&#x2019;a n

1 = 3

2â&#x2C6;&#x2019;1 10

1 30

Table 7.7.7 an approximation to ln 2 using simpson's rule

endpoint xi

yi = f (xi ) = 1/xi

0 1 2 3 4 5 6 7 8 9 10

1.0 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2.0

1.000000000 0.909090909 0.833333333 0.769230769 0.714285714 0.666666667 0.625000000 0.588235294 0.555555556 0.526315789 0.500000000 2 1

Thomas Simpson (1710â&#x20AC;&#x201C;1761) English mathematician. Simpson was the son of a weaver. He was trained to follow in his fatherâ&#x20AC;&#x2122;s footsteps and had little formal education in his early life. His interest in science and mathematics was aroused in 1724, when he witnessed an eclipse of the Sun and received two books from a peddler, one on astrology and the other on arithmetic. Simpson quickly absorbed their contents and soon became a successful local fortune teller. His improved ďŹ nancial situation enabled him to give up weaving and marry his landlady. Then in 1733 some mysterious â&#x20AC;&#x153;unfortunate incidentâ&#x20AC;? forced him to move. He settled in Derby, where he taught in an evening school and worked at weaving during the day. In 1736 he moved to London and published his

1 dx â&#x2030;&#x2C6; x

multiplier wi

1 4 2 4 2 4 2 4 2 4 1

wi yi

1.000000000 3.636363636 1.666666667 3.076923077 1.428571429 2.666666667 1.250000000 2.352941176 1.111111111 2.105263158 0.500000000 20.794506921

ĺ&#x2020;¸ 30 ĺ&#x2020;š(20.794506921) â&#x2030;&#x2C6; 0.693150231 1

ďŹ rst mathematical work in a periodical called the Ladiesâ&#x20AC;&#x2122; Diary (of which he later became the editor). In 1737 he published a successful calculus textbook that enabled him to give up weaving completely and concentrate on textbook writing and teaching. His fortunes improved further in 1740 when one Robert Heath accused him of plagiarism. The publicity was marvelous, and Simpson proceeded to dash off a succession of best-selling textbooks: Algebra (ten editions plus translations), Geometry (twelve editions plus translations), Trigonometry (ďŹ ve editions plus translations), and numerous others. It is interesting to note that Simpson did not discover the rule that bears his nameâ&#x20AC;&#x201D;it was a well-known result by Simpsonâ&#x20AC;&#x2122;s time. [Image: http://www-history.mcs.st-and.ac.uk/Posters/820.html]

540

Chapter 7 / Principles of Integral Evaluation

Although S10 is a weighted average of M5 and T5 , it makes sense to compare S10 to M10 and T10 , since the sums for these three approximations involve the same number of terms. Using the values for M10 and T10 from Example 2 and the value for S10 in Table 7.7.7, we have |EM | = | ln 2 − M10 | ≈ |0.693147181 − 0.692835360| = 0.000311821 |ET | = | ln 2 − T10 | ≈ |0.693147181 − 0.693771403| = 0.000624222 |ES | = | ln 2 − S10 | ≈ |0.693147181 − 0.693150231| = 0.000003050 Comparing these absolute errors, it is clear that S10 is a much more accurate approximation of ln 2 than either M10 or T10 .

y = f (x) y = Ax 2 + Bx + C

Y2 x

m Δx Figure 7.7.5

Δx

GEOMETRIC INTERPRETATION OF SIMPSON’S RULE The midpoint (or tangent line) approximation and the trapezoidal approximation for a deﬁnite integral are based on approximating a segment of the curve y = f(x) by line segments. Intuition suggests that we might improve on these approximations using parabolic arcs rather than line segments, thereby accounting for concavity of the curve y = f(x) more closely. At the heart of this idea is a formula, sometimes called the one-third rule. The one-third rule expresses a deﬁnite integral of a quadratic function g(x) = Ax 2 + Bx + C in terms of the values Y0 , Y1 , and Y2 of g at the left endpoint, midpoint, and right endpoint, respectively, of the interval of integration [m − x, m + x] (see Figure 7.7.5): m+ x x (11) [Y0 + 4Y1 + Y2 ] (Ax 2 + Bx + C) dx = 3 m− x

Veriﬁcation of the one-third rule is left for the reader (Exercise 53). By applying the one-third rule to subintervals [x2k−2 , x2k ], k = 1, . . . , n/2, one arrives at Formula (8) for Simpson’s rule (Exercise 54). Thus, Simpson’s rule corresponds to the integral of a piecewise-quadratic approximation to f(x). ERROR BOUNDS With all the methods studied in this section, there are two sources of error: the intrinsic or truncation error due to the approximation formula, and the roundoff error introduced in the calculations. In general, increasing n reduces the truncation error but increases the roundoff error, since more computations are required for larger n. In practical applications, it is important to know how large n must be taken to ensure that a speciﬁed degree of accuracy is obtained. The analysis of roundoff error is complicated and will not be considered here. However, the following theorems, which are proved in books on numerical analysis, provide upper bounds on the truncation errors in the midpoint, trapezoidal, and Simpson’s rule approximations.

7.7.2 theorem (Midpoint and Trapezoidal Error Bounds) If f ′′ is continuous on [a, b] and if K2 is the maximum value of | f ′′(x)| on [a, b], then b (b − a)3 K2 f(x) dx − Mn ≤ (12) (a) |EM | = 24n2 a b (b − a)3 K2 (b) |ET | = f(x) dx − Tn ≤ (13) 12n2 a

7.7 Numerical Integration; Simpson’s Rule

541

7.7.3 theorem (Simpson Error Bound) If f (4) is continuous on [a, b] and if K4 is the maximum value of | f (4) (x)| on [a, b], then b (b − a)5 K4 (14) f(x) dx − Sn ≤ |ES | = 180n4 a Example 7

Find an upper bound on the absolute error that results from approximating 2 1 ln 2 = dx 1 x

using (a) the midpoint approximation M10 , (b) the trapezoidal approximation T10 , and (c) Simpson’s rule S10 , each with n = 10 subintervals.

Solution. We will apply Formulas (12), (13), and (14) with f(x) =

1 , x

a = 1,

b = 2,

and

n = 10

We have f ′(x) = − Thus,

Note that the upper bounds calculated in Example 7 are consistent with the values |EM |, |ET |, and |ES | calculated in Example 6 but are considerably greater than those values. It is quite common that the upper bounds on the absolute errors given in Theorems 7.7.2 and 7.7.3 substantially exceed the actual absolute errors. However, that does not diminish the utility of these bounds.

1 , x2

f ′′(x) =

2 , x3

2 2 | f (x)| = 3 = 3 , x x ′′

6 24 , f (4) (x) = 5 x4 x 24 24 | f (4) (x)| = 5 = 5 x x

f ′′′(x) = −

where we have dropped the absolute values because f ′′(x) and f (4) (x) have positive values for 1 ≤ x ≤ 2. Since |f ′′(x)| and |f 4(x)| are continuous and decreasing on [1, 2], both functions have their maximum values at x = 1; for |f ′′(x)| this maximum value is 2 and for |f 4(x)| the maximum value is 24. Thus we can take K2 = 2 in (12) and (13), and K4 = 24 in (14). This yields |EM | ≤ |ET | ≤ |ES | ≤ Example 8

(b − a)3 K2 13 · 2 = ≈ 0.000833333 24n2 24 · 102

(b − a)3 K2 13 · 2 = ≈ 0.001666667 12n2 12 · 102

15 · 24 (b − a)5 K4 = ≈ 0.000013333 4 180n 180 · 104

How many subintervals should be used in approximating 2 1 ln 2 = dx x 1

by Simpson’s rule for ﬁve decimal-place accuracy?

Solution. To obtain ﬁve decimal-place accuracy, we must choose the number of subintervals so that

|ES | ≤ 0.000005 = 5 × 10−6

From (14), this can be achieved by taking n in Simpson’s rule to satisfy (b − a)5 K4 ≤ 5 × 10−6 180n4

542

Chapter 7 / Principles of Integral Evaluation

Taking a = 1, b = 2, and K4 = 24 (found in Example 7) in this inequality yields 15 · 24 ≤ 5 × 10−6 180 · n4 which, on taking reciprocals, can be rewritten as n4 ≥ Thus,

8 × 104 2 × 106 = 75 3

20 ≈ 12.779 n≥ √ 4 6

Since n must be an even integer, the smallest value of n that satisﬁes this requirement is n = 14. Thus, the approximation S14 using 14 subintervals will produce ﬁve decimal-place accuracy.

REMARK

In cases where it is difficult to find the values of K2 and K4 in Formulas (12), (13), and (14), these constants may be replaced by any larger constants. For example, suppose that a constant K can be easily found with the certainty that | f ′′(x)| < K on the interval. Then K2 ≤ K and

(b − a)3 K (b − a)3 K2 ≤ (15) 2 12n 12n2 so the right side of (15) is also an upper bound on the value of |ET |. Using K , however, will likely increase the computed value of n needed for a given error tolerance. Many applications involve the |ET | ≤

resolution of competing practical issues, illustrated here through the trade-off between the convenience of finding a crude bound for | f ′′(x)| versus the efficiency of using the smallest possible n for a desired accuracy.

Example 9

How many subintervals should be used in approximating 1 cos(x 2 ) dx 0

by the midpoint approximation for three decimal-place accuracy?

Solution. To obtain three decimal-place accuracy, we must choose n so that |EM | ≤ 0.0005 = 5 × 10−4

(16)

From (12) with f(x) = cos(x 2 ), a = 0, and b = 1, an upper bound on |EM | is given by

K2 24n2 where |K2 | is the maximum value of | f ′′(x)| on the interval [0, 1]. However, |EM | ≤

f ′(x) = −2x sin(x 2 )

f ′′(x) = −4x 2 cos(x 2 ) − 2 sin(x 2 ) = −[4x 2 cos(x 2 ) + 2 sin(x 2 )]

so that

3 2 1 x 1

y = | f ′′(x)| = |4x 2 cos(x 2) + 2 sin(x 2)| Figure 7.7.6

(17)

| f ′′(x)| = | 4x 2 cos(x 2 ) + 2 sin(x 2 )|

(18)

It would be tedious to look for the maximum value of this function on the interval [0, 1]. For x in [0, 1], it is easy to see that each of the expressions x 2 , cos(x 2 ), and sin(x 2 ) is bounded in absolute value by 1, so |4x 2 cos(x 2 ) + 2 sin(x 2 )| ≤ 4 + 2 = 6 on [0, 1]. We can improve on this by using a graphing utility to sketch | f ′′(x)|, as shown in Figure 7.7.6. It is evident from the graph that | f ′′(x)| < 4

for 0 ≤ x ≤ 1

7.7 Numerical Integration; Simpson’s Rule

543

Thus, it follows from (17) that 4 1 K2 < = 2 24n2 24n2 6n and hence we can satisfy (16) by choosing n so that |EM | ≤

1 < 5 × 10−4 6n2 which, on taking reciprocals, can be written as 104 30

102 or n > √ ≈ 18.257 30 The smallest integer value of n satisfying this inequality is n = 19. Thus, the midpoint approximation M19 using 19 subintervals will produce three decimal-place accuracy. n2 >

A COMPARISON OF THE THREE METHODS

Of the three methods studied in this section, Simpson’s rule generally produces more accurate results than the midpoint or trapezoidal approximations for the same amount of work. To make this plausible, let us express (12), (13), and (14) in terms of the subinterval width x =

b−a n

We obtain |EM | ≤

1 K2 (b − a)( x)2 24

(19)

|ET | ≤

1 K2 (b − a)( x)2 12

(20)

1 K4 (b − a)( x)4 (21) 180 (verify). For Simpson’s rule, the upper bound on the absolute error is proportional to ( x)4 , whereas the upper bound on the absolute error for the midpoint and trapezoidal approximations is proportional to ( x)2 . Thus, reducing the interval width by a factor of 10, for example, reduces the error bound by a factor of 100 for the midpoint and trapezoidal approximations but reduces the error bound by a factor of 10,000 for Simpson’s rule. This suggests that, as n increases, the accuracy of Simpson’s rule improves much more rapidly than that of the other approximations. As a ﬁnal note, observe that if f(x) is a polynomial of degree 3 or less, then we have f (4) (x) = 0 for all x, so K4 = 0 in (14) and consequently |ES | = 0. Thus, Simpson’s rule gives exact results for polynomials of degree 3 or less. Similarly, the midpoint and trapezoidal approximations give exact results for polynomials of degree 1 or less. (You should also be able to see that this is so geometrically.) |ES | ≤

✔QUICK CHECK EXERCISES 7.7

(See page 547 for answers.)

1. Let Tn be the trapezoidal approximation for the deﬁnite integral of f(x) over an interval [a, b] using n subintervals. (a) Expressed in terms of Ln and Rn (the left and right endpoint approximations), Tn = . (b) Expressed in terms of the function values y0 , y1 , . . . , yn . at the endpoints of the subintervals, Tn =

2. Let I denote the deﬁnite integral of f over an interval [a, b] with Tn and Mn the respective trapezoidal and midpoint approximations of I for a given n. Assume that the graph of f is concave up on the interval [a, b] and order the quantities Tn , Mn , and I from smallest to largest: < < .

544

Chapter 7 / Principles of Integral Evaluation

b 3. Let S6 be the Simpson’s rule approximation for a f(x) dx using n = 6 subintervals. (a) Expressed in terms of M3 and T3 (the midpoint and trapezoidal approximations), S6 = . (b) Using the function values y0 , y1 , y2 , . . . , y6 at the endpoints of the subintervals, S6 = .

4. Assume that f (4) is continuous on [0, 1] and that f (k) (x) satisﬁes |f (k) (x)| ≤ 1 on [0, 1], k = 1, 2, 3, 4. Find an upper

EXERCISE SET 7.7

bound on the absolute error that results from approximating the integral of f over [0, 1] using (a) the midpoint approximation M10 ; (b) the trapezoidal approximation T10 ; and (c) Simpson’s rule S10 . 3 1 5. Approximate dx using the indicated method. 2 x 1 (a) M1 = (b) T1 = (c) S2 =

CAS

1–6 Approximate the integral using (a) the midpoint approx-

imation M10 , (b) the trapezoidal approximation T10 , and (c) Simpson’s rule approximation S20 using Formula (7). In each case, ﬁnd the exact value of the integral and approximate the absolute error. Express your answers to at least four decimal places. ■ 3 9 π/2 √ 1 cos x dx 1. x + 1 dx 2. 3. √ dx x 0 0 4 2 3 3 1 4. sin x dx 5. e−2x dx 6. dx 0 1 0 3x + 1 7–12 Use inequalities (12), (13), and (14) to ﬁnd upper bounds

on the errors in parts (a), (b), and (c) of the indicated exercise. ■ 7. Exercise 1

8. Exercise 2

9. Exercise 3

10. Exercise 4

11. Exercise 5

12. Exercise 6

13–18 Use inequalities (12), (13), and (14) to ﬁnd a number n

of subintervals for (a) the midpoint approximation Mn , (b) the trapezoidal approximation Tn , and (c) Simpson’s rule approximation Sn to ensure that the absolute error will be less than the given value. ■ 13. Exercise 1; 5 × 10−4

14. Exercise 2; 5 × 10−4

15. Exercise 3; 10−3

16. Exercise 4; 10−3

17. Exercise 5; 10−4

18. Exercise 6; 10−4

19–22 True–False Determine whether the statement is true or

false. Explain your answer. ■ 19. The midpoint approximation, Mn , is the average of the left and right endpoint approximations, Ln and Rn , respectively. 20. If f(x) is concave down on the interval (a, b), then the b trapezoidal approximation Tn underestimates a f(x) dx. b 21. The Simpson’s rule approximation S50 for a f(x) dx is a weighted average of the approximations M50 and T50 , where M50 is given twice the weight of T50 in the average. b 22. Simpson’s rule approximation S50 for a f(x) dx correb sponds to a q(x) dx, where the graph of q is composed of 25 parabolic segments joined at points on the graph of f .

23–24 Find a function g(x) of the form

g(x) = Ax 2 + Bx + C

whose graph contains the points (m − x, f(m − x)), (m, f(m)), and (m + x, f(m + x)), for the given function f(x) and the given values of m and x. Then verify Formula m+ x (11): x g(x) dx = [Y0 + 4Y1 + Y2 ] 3 m− x where Y0 = f(m − x), Y1 = f(m), and Y2 = f(m + x). ■ 1 ; m = 3, x = 1 x 24. f(x) = sin2 (πx); m = 61 , x =

23. f(x) =

1 6

25–30 Approximate the integral using Simpson’s rule S10 and compare your answer to that produced by a calculating utility with a numerical integration capability. Express your answers to at least four decimal places. ■ 3 1 x 2 dx e−x dx 26. 25. √ 2x 3 + 1 0 −1 2 π x 27. x 2 + x 3 dx 28. dx −1 0 2 + sin x 1 2 29. (ln x)3/2 dx cos(x 2 ) dx 30. 0

31–32 The exact value of the given integral is π (verify). Approximate the integral using (a) the midpoint approximation M10 , (b) the trapezoidal approximation T10 , and (c) Simpson’s rule approximation S20 using Formula (7). Approximate the absolute error and express your answers to at least four decimal places. ■ 2 3 8 4 31. 9 − x 2 dx dx 32. 2 0 x +4 0 9 33. In Example 8 we showed that taking n = 14 subdivisions ensures that the approximation of 2 1 ln 2 = dx x 1 by Simpson’s rule is accurate to ﬁve decimal places. Conﬁrm this by comparing the approximation of ln 2 produced by Simpson’s rule with n = 14 to the value produced directly by your calculating utility.

34. In each part, determine whether a trapezoidal approximation would be an underestimate or an overestimate for the deﬁnite integral. 1 2 (a) cos(x 2 ) dx cos(x 2 ) dx (b) 3/2

35–36 Find a value of n to ensure that the absolute error in approximating the integral by the midpoint approximation will be less than 10−4 . Estimate the absolute error, and express your answers to at least four decimal places. ■ 1 2 x sin x dx 36. ecos x dx 35. 0

37–38 Show that the inequalities (12) and (13) are of no value in ﬁnding an upper bound on the absolute error that results from approximating the integral using either the midpoint approximation or the trapezoidal approximation. ■ 1 1 √ √ x x dx sin x dx 37. 38. 0

39–40 Use Simpson’s rule approximation S10 to approximate the length of the curve over the stated interval. Express your answers to at least four decimal places. ■

40. y = x −2 from x = 1 to x = 2 F O C U S O N C O N C E P TS

41. A graph of the speed v versus time t curve for a test run of a BMW 335i is shown in the accompanying ﬁgure. Estimate the speeds at times t = 0, 5, 10, 15, 20, 25, 30 s from the graph, convert to ft/s using 1 mi/h = 22 15 ft/s, and use these speeds and Simpson’s rule to approximate the number of feet traveled during the ﬁrst 30 s. Round your answer to the nearest foot. [Hint: Distance 30 traveled = 0 v(t) dt.] Source: Data from Car and Driver Magazine, September 2010.

160 140 120 100 80

0.5

40 20 0

Time t (s)

Figure Ex-41

42. A graph of the acceleration a versus time t for an object moving on a straight line is shown in the accompanying ﬁgure. Estimate the accelerations at t = 0, 1, 2, . . . , 8 seconds (s) from the graph and use Simpson’s rule to approximate the change in velocity from t = 0 to t = 8 s. Round your answer to the nearest tenth cm/s. [Hint: 8 Change in velocity = 0 a(t) dt.]

Time t (s)

Figure Ex-42

43–46 Numerical integration methods can be used in problems where only measured or experimentally determined values of the integrand are available. Use Simpson’s rule to estimate the value of the relevant integral in these exercises. ■

43. The accompanying table gives the speeds, in miles per second, at various times for a test rocket that was ﬁred upward from the surface of the Earth. Use these values to approximate the number of miles traveled during the ﬁrst 180 s. Round your answer to the nearest tenth of a mile. [Hint: 180 Distance traveled = 0 v(t) dt.] speed v (mi/s)

0 30 60 90 120 150 180

0.00 0.03 0.08 0.16 0.27 0.42 0.65

Table Ex-43

44. The accompanying table gives the speeds of a bullet at various distances from the muzzle of a riﬂe. Use these values to approximate the number of seconds for the bullet to travel 1800 ft. Express your answer to the nearest hundredth of a second. [Hint: If v is the speed of the bullet and x is the / / / distance 1800traveled, then v = dx dt so that dt dx = 1 v and t = 0 (1/v) dx.] distance x (ft)

545

1.0

time t (s)

39. y = sin x from x = 0 to x = π

Speed v (mi/ h)

Acceleration a (cm/s 2)

7.7 Numerical Integration; Simpson’s Rule

0 300 600 900 1200 1500 1800

speed v (ft/s) 3100 2908 2725 2549 2379 2216 2059

Table Ex-44

45. Measurements of a pottery shard recovered from an archaeological dig reveal that the shard came from a pot with a ﬂat bottom and circular cross sections (see the accompanying

546

Chapter 7 / Principles of Integral Evaluation

ﬁgure below). The ﬁgure shows interior radius measurements of the shard made every 4 cm from the bottom of the pot to the top. Use those values to approximate the interior volume of the pot to the nearest tenth of a liter (1 L = 1000 cm3 ). [Hint: Use 6.2.3 (volume by cross sections) to set up an appropriate integral for the volume.]

(b) How large must n be in the trapezoidal approximation 1 of 0 f(x) dx to ensure that the absolute error is less than 10−3 ? (c) Estimate the integral using the trapezoidal approximation with the value of n obtained in part (b). C

y (cm)

16.8 cm

16 12 8 4

15.4 cm 13.8 cm 11.5 cm C

8.5 cm Figure Ex-45

46. Engineers want to construct a straight and level road 600 ft long and 75 ft wide by making a vertical cut through an intervening hill (see the accompanying ﬁgure). Heights of the hill above the centerline of the proposed road, as obtained at various points from a contour map of the region, are shown in the accompanying ﬁgure. To estimate the construction costs, the engineers need to know the volume of earth that must be removed. Approximate this volume, rounded to the nearest cubic foot. [Hint: First set up an integral for the cross-sectional area of the cut along the centerline of the road, then assume that the height of the hill does not vary between the centerline and edges of the road.] horizontal distance x (ft)

height h (ft) x

0 100 200 300 400 500 600

0 7 16 24 25 16 0

600 ft

Centerline

75 ft

Figure Ex-46 C

47. Let f(x) = cos(x 2 ). (a) Use a CAS to approximate the maximum value of | f ′′(x)| on the interval [0, 1]. (b) How large must n be in the midpoint approximation of 1 0 f(x) dx to ensure that the absolute error is less than 5 × 10−4 ? Compare your result with that obtained in Example 9. (c) Estimate the integral using the midpoint approximation with the value of n obtained in part (b). √ 48. Let f(x) = 1 + x 3 . (a) Use a CAS to approximate the maximum value of | f ′′(x)| on the interval [0, 1].

49. Let f(x) = cos(x − x 2 ). (a) Use a CAS to approximate the maximum value of | f (4) (x)| on the interval [0, 1]. (b) How large must the value of n be in the approximation 1 Sn of 0 f(x) dx by Simpson’s rule to ensure that the absolute error is less than 10−4 ? (c) Estimate the integral using Simpson’s rule approximation Sn with the value of n obtained in part (b). √ 50. Let f(x) = 2 + x 3 . (a) Use a CAS to approximate the maximum value of | f (4) (x)| on the interval [0, 1]. (b) How large must the value of n be in the approximation 1 Sn of 0 f(x) dx by Simpson’s rule to ensure that the absolute error is less than 10−6 ? (c) Estimate the integral using Simpson’s rule approximation Sn with the value of n obtained in part (b). F O C U S O N C O N C E P TS

51. (a) Verify that the average of the left and right endpoint approximations as given in Table 7.7.1 gives Formula (2) for the trapezoidal approximation. (b) Suppose that f is a continuous nonnegative function on the interval [a, b] and partition [a, b] with equally spaced points, a = x0 < x1 < · · · < xn = b. Find the area of the trapezoid under the line segment joining points (xk , f(xk )) and (xk+1 , f(xk+1 )) and above the interval [xk , xk+1 ]. Show that the right side of Formula (2) is the sum of these trapezoidal areas (Figure 7.7.1). 52. Let f be a function that is positive, continuous, decreasing, and concave down on the interval [a, b]. Assuming that [a, b] is subdivided into n equal subintervals, b arrange the following approximations of a f(x) dx in order of increasing value: left endpoint, right endpoint, midpoint, and trapezoidal. 53. Suppose that x > 0 and g(x) = Ax 2 + Bx + C. Let m be a number and set Y0 = g(m − x), Y1 = g(m), and Y2 = g(m + x). Verify Formula (11): m+ x x [Y0 + 4Y1 + Y2 ] g(x) dx = 3 m− x 54. Suppose that f is a continuous nonnegative function on the interval [a, b], n is even, and [a, b] is partitioned using n + 1 equally spaced points, a = x0 < x1 < · · · < xn = b. Set y0 = f(x0 ), y1 = f(x1 ), . . . , yn = f(xn ). Let g1 , g2 , . . . , gn/2 be the quadratic functions of the form gi (x) = Ax 2 + Bx + C so that (cont.)

7.8 Improper Integrals

â&#x20AC;˘ the graph of g1 passes through the points (x0 , y0 ), (x1 , y1 ), and (x2 , y2 );

â&#x20AC;˘ the graph of g2 passes through the points (x2 , y2 ), (x3 , y3 ), and (x4 , y4 );

â&#x20AC;˘ ... â&#x20AC;˘ the graph of gn/2 passes through the points (xnâ&#x2C6;&#x2019;2 , ynâ&#x2C6;&#x2019;2 ), (xnâ&#x2C6;&#x2019;1 , ynâ&#x2C6;&#x2019;1 ), and (xn , yn ). Verify that Formula (8) computes the area under a piecewise quadratic function by showing that n/2 x2j gj (x) dx j =1

1 = 3

547

55. Writing Discuss two different circumstances under which numerical integration is necessary. 56. Writing For the numerical integration methods of this section, better accuracy of an approximation was obtained by increasing the number of subdivisions of the interval. Another strategy is to use the same number of subintervals, but to select subintervals of differing lengths. Discuss a scheme 4â&#x2C6;&#x161; for doing this to approximate 0 x dx using a trapezoidal approximation with 4 subintervals. Comment on the advantages and disadvantages of your scheme.

x2j â&#x2C6;&#x2019;2

bâ&#x2C6;&#x2019;a n

[y0 + 4y1 + 2y2 + 4y3 + 2y4 + Âˇ Âˇ Âˇ + 2ynâ&#x2C6;&#x2019;2 + 4ynâ&#x2C6;&#x2019;1 + yn ]

â&#x153;&#x201D;QUICK CHECK ANSWERS 7.7 bâ&#x2C6;&#x2019;a [y0 + 2y1 + Âˇ Âˇ Âˇ + 2ynâ&#x2C6;&#x2019;1 + yn ] 2. Mn < I < Tn 3. (a) 23 M3 + 31 T3 1. (a) 21 (Ln + Rn ) (b) 2n 1 1 1 bâ&#x2C6;&#x2019;a (b) (c) (y0 + 4y1 + 2y2 + 4y3 + 2y4 + 4y5 + y6 ) 4. (a) (b) 18 2400 1200 1,800,000

5. (a) M1 =

7.8

1 2

(b) T1 =

10 9

19 27

IMPROPER INTEGRALS Up to now we have focused on deďŹ nite integrals with continuous integrands and ďŹ nite intervals of integration. In this section we will extend the concept of a deďŹ nite integral to include inďŹ nite intervals of integration and integrands that become inďŹ nite within the interval of integration.

IMPROPER INTEGRALS

It is assumed in the deďŹ nition of the deďŹ nite integral b f(x) dx a

that [a, b] is a ďŹ nite interval and that the limit that deďŹ nes the integral exists; that is, the function f is integrable. We observed in Theorems 5.5.2 and 5.5.8 that continuous functions are integrable, as are bounded functions with ďŹ nitely many points of discontinuity. We also observed in Theorem 5.5.8 that functions that are not bounded on the interval of integration are not integrable. Thus, for example, a function with a vertical asymptote within the interval of integration would not be integrable. Our main objective in this section is to extend the concept of a deďŹ nite integral to allow for inďŹ nite intervals of integration and integrands with vertical asymptotes within the interval of integration. We will call the vertical asymptotes inďŹ nite discontinuities, and we will call

Chapter 7 / Principles of Integral Evaluation

548

integrals with inďŹ nite intervals of integration or inďŹ nite discontinuities within the interval of integration improper integrals. Here are some examples:

â&#x20AC;˘ Improper integrals with inďŹ nite intervals of integration:

+âŹ

dx , x2

0 x

e dx,

+âŹ

Ď&#x20AC;

â&#x2C6;&#x2019;âŹ

dx 1 + x2

â&#x20AC;˘ Improper integrals with inďŹ nite discontinuities in the interval of integration:

3 â&#x2C6;&#x2019;3

dx , x2

dx , xâ&#x2C6;&#x2019;1

tan x dx 0

â&#x20AC;˘ Improper integrals with inďŹ nite discontinuities and inďŹ nite intervals of integration:

+âŹ 0

dx â&#x2C6;&#x161; , x

+âŹ

â&#x2C6;&#x2019;âŹ

dx , x2 â&#x2C6;&#x2019; 9

+âŹ

sec x dx

INTEGRALS OVER INFINITE INTERVALS To motivate a reasonable deďŹ nition for improper integrals of the form +âŹ f(x) dx a

let us begin with the case where f is continuous and nonnegative on [a, +âŹ ), so we can think of the integral as the area under the curve y = f(x) over the interval [a, +âŹ ) (Figure 7.8.1). At ďŹ rst, you might be inclined to argue that this area is inďŹ nite because the region has inďŹ nite extent. However, such an argument would be based on vague intuition rather than precise mathematical logic, since the concept of area has only been deďŹ ned over intervals of ďŹ nite extent. Thus, before we can make any reasonable statements about the area of the region in Figure 7.8.1, we need to begin by deďŹ ning what we mean by the area of this region. For that purpose, it will help to focus on a speciďŹ c example. Suppose we are interested in the area A of the region that lies below the curve y = 1/x 2 and above the interval [1, +âŹ ) on the x-axis. Instead of trying to ďŹ nd the entire area at once, let us begin by calculating the portion of the area that lies above a ďŹ nite interval [1, b], where b > 1 is arbitrary. That area is

b 1 b 1 dx = â&#x2C6;&#x2019; =1â&#x2C6;&#x2019; 2 x 1 b 1 x

y +â&#x2C6;&#x17E;

f (x) dx a

a Figure 7.8.1

1 x2 b

Area = 1

dx 1 = 1â&#x2C6;&#x2019; b x2 x

(Figure 7.8.2). If we now allow b to increase so that b â&#x2020;&#x2019; +âŹ , then the portion of the area over the interval [1, b] will begin to ďŹ ll out the area over the entire interval [1, +âŹ ) (Figure 7.8.3), and hence we can reasonably deďŹ ne the area A under y = 1/x 2 over the interval [1, +âŹ ) to be +âŹ b 1 dx dx A= = lim = lim =1 1â&#x2C6;&#x2019; (1) b â&#x2020;&#x2019; +âŹ 1 x 2 b â&#x2020;&#x2019; +âŹ x2 b 1

Thus, the area has a ďŹ nite value of 1 and is not inďŹ nite as we ďŹ rst conjectured.

Figure 7.8.2 y

1 x2 Area =

1 2

1 x2 Area =

Figure 7.8.3

2 3

1 x2 Area =

x 1

3 4

Area = 1 x

1 x2 x

With the preceding discussion as our guide, we make the following deďŹ nition (which is applicable to functions with both positive and negative values).

7.8 Improper Integrals

7.8.1 to be

If f is nonnegative over the interval [a, +âŹ ), then the improper integral in Definition 7.8.1 can be interpreted to be the area under the graph of f over the interval [a, +âŹ ). If the integral converges, then the area is finite and equal to the value of the integral, and if the integral diverges, then the area is regarded to be infinite.

549

definition The improper integral of f over the interval [a, +â´Ľ) is deďŹ ned +âŹ b f(x) dx = lim f(x) dx b â&#x2020;&#x2019; +âŹ a

In the case where the limit exists, the improper integral is said to converge, and the limit is deďŹ ned to be the value of the integral. In the case where the limit does not exist, the improper integral is said to diverge, and it is not assigned a value.

Example 1

Evaluate (a)

+âŹ 1

dx x3

(b)

+âŹ 1

dx x

Solution (a). Following the deďŹ nition, we replace the inďŹ nite upper limit by a ďŹ nite upper limit b, and then take the limit of the resulting integral. This yields

b +âŹ 1 dx 1 1 b 1 dx = lim = lim â&#x2C6;&#x2019; 2 = lim â&#x2C6;&#x2019; 2 = 3 3 b â&#x2020;&#x2019; +âŹ b â&#x2020;&#x2019; +âŹ b â&#x2020;&#x2019; +âŹ x 2x 1 2 2b 2 1 x 1 Since the limit is ďŹ nite, the integral converges and its value is 1/2.

Solution (b).

+âŹ 1

dx = lim b â&#x2020;&#x2019; +âŹ x

b 1

b dx = lim ln x 1 = lim ln b = +âŹ b â&#x2020;&#x2019; +âŹ b â&#x2020;&#x2019; +âŹ x

In this case the integral diverges and hence has no value. y

1 y= x

1 x2

1 x3

2 1

x 1

Figure 7.8.4

Because the functions 1/x 3 , 1/x 2 , and 1/x are nonnegative over the interval [1, +âŹ ), it follows from (1) and the last example that over this interval the area under y = 1/x 3 is 1 , the area under y = 1/x 2 is 1, and the area under y = 1/x is inďŹ nite. However, on the 2 surface the graphs of the three functions seem very much alike (Figure 7.8.4), and there is nothing to suggest why one of the areas should be inďŹ nite and the other two ďŹ nite. One explanation is that 1/x 3 and 1/x 2 approach zero more rapidly than 1/x as x â&#x2020;&#x2019; +âŹ , so that the area over the interval [1, b] accumulates less rapidly under the curves y = 1/x 3 and y = 1/x 2 than under y = 1/x as b â&#x2020;&#x2019; +âŹ , and the difference is just enough that the ďŹ rst two areas are ďŹ nite and the third is inďŹ nite. +âŹ dx converge? Example 2 For what values of p does the integral xp 1

Solution. We know from the preceding example that the integral diverges if p = 1, so let us assume that p = 1. In this case we have

1â&#x2C6;&#x2019;p +âŹ b 1 dx x 1â&#x2C6;&#x2019;p b â&#x2C6;&#x2019;p = lim â&#x2C6;&#x2019; x dx = lim = lim b â&#x2020;&#x2019; +âŹ 1 b â&#x2020;&#x2019; +âŹ 1 â&#x2C6;&#x2019; p 1 b â&#x2020;&#x2019; +âŹ 1 â&#x2C6;&#x2019; p xp 1â&#x2C6;&#x2019;p 1

If p > 1, then the exponent 1 â&#x2C6;&#x2019; p is negative and b1â&#x2C6;&#x2019;p â&#x2020;&#x2019; 0 as b â&#x2020;&#x2019; +âŹ ; and if p < 1, then the exponent 1 â&#x2C6;&#x2019; p is positive and b1â&#x2C6;&#x2019;p â&#x2020;&#x2019; +âŹ as b â&#x2020;&#x2019; +âŹ . Thus, the integral converges if p > 1 and diverges otherwise. In the convergent case the value of the integral is

+âŹ dx 1 1 = 0â&#x2C6;&#x2019; (p > 1) = p x 1â&#x2C6;&#x2019;p pâ&#x2C6;&#x2019;1 1

550

Chapter 7 / Principles of Integral Evaluation

The following theorem summarizes this result. 7.8.2 theorem

+⬁

Example 3

Evaluate

⎧ ⎨

1 if p > 1 dx p − 1 = p ⎩ x diverges if p ≤ 1 (1 − x)e−x dx.

Solution. We begin by evaluating the indeﬁnite integral using integration by parts. Setting u = 1 − x and dv = e−x dx yields (1 − x)e−x dx = −e−x (1 − x) − e−x dx = −e−x + xe−x + e−x + C = xe−x + C

Thus,

y 1

y = (1 − x)e −x x 1

The net signed area between the graph and the interval [0, + ∞) is zero.

Figure 7.8.5

If f is nonnegative over the interval (−⬁, +⬁), then the improper integral +⬁ f(x) dx −⬁

can be interpreted to be the area under the graph of f over the interval (−⬁, +⬁). The area is finite and equal to the value of the integral if the integral converges and is infinite if it diverges.

+⬁

(1 − x)e−x dx = lim

b → +⬁ 0

(1 − x)e−x dx = lim

b → +⬁

b b xe−x = lim b b → +⬁ e 0

The limit is an indeterminate form of type ⬁/⬁, so we will apply L’Hôpital’s rule by differentiating the numerator and denominator with respect to b. This yields +⬁ 1 (1 − x)e−x dx = lim b = 0 b → +⬁ e 0

We can interpret this to mean that the net signed area between the graph of y = (1 − x)e−x and the interval [0, +⬁) is 0 (Figure 7.8.5). 7.8.3 definition The improper integral of f over the interval (−ⴥ, b] is deﬁned to be b b (2) f(x) dx f(x) dx = lim a → −⬁ a

−⬁

The integral is said to converge if the limit exists and diverge if it does not. The improper integral of f over the interval (−ⴥ, +ⴥ) is deﬁned as +⬁ +⬁ c f(x) dx f(x) dx + f(x) dx = −⬁

−⬁

(3)

where c is any real number. The improper integral is said to converge if both terms converge and diverge if either term diverges.

Example 4

Evaluate

+⬁ −⬁

dx . 1 + x2

Solution. We will evaluate the integral by choosing c = 0 in (3). With this value for c Although we usually choose c = 0 in (3), the choice does not matter because it can be proved that neither the convergence nor the value of the integral is affected by the choice of c.

we obtain +⬁ 0

−⬁

b b dx π dx −1 tan x = lim (tan−1 b) = = lim = lim 2 2 b → +⬁ b → +⬁ b → +⬁ 0 1+x 2 0 1+x 0 0 π dx dx = lim = lim tan−1 x = lim (− tan−1 a) = a → −⬁ a → −⬁ a 1 + x 2 a → −⬁ a 1 + x2 2

7.8 Improper Integrals

551

Thus, the integral converges and its value is y 1

Area = c

1 1 + x2

+⬁

−⬁

dx = 1 + x2

−⬁

dx + 1 + x2

+⬁ 0

dx π π = + =π 2 1+x 2 2

Since the integrand is nonnegative on the interval (−⬁, +⬁), the integral represents the area of the region shown in Figure 7.8.6.

Figure 7.8.6

INTEGRALS WHOSE INTEGRANDS HAVE INFINITE DISCONTINUITIES

f (x) dx a

Next we will consider improper integrals whose integrands have infinite discontinuities. We will start with the case where the interval of integration is a finite interval [a, b] and the infinite discontinuity occurs at the right-hand endpoint. To motivate an appropriate definition for such an integral let us consider the case where b f is nonnegative on [a, b], so we can interpret the improper integral a f(x) dx as the area of the region in Figure 7.8.7a. The problem of finding the area of this region is complicated by the fact that it extends indefinitely in the positive y-direction. However, instead of trying to find the entire area at once, we can proceed indirectly by calculating the portion of the area over the interval [a, k], where a ≤ k < b, and then letting k approach b to fill out the area of the entire region (Figure 7.8.7b). Motivated by this idea, we make the following definition.

(a)

7.8.4 definition If f is continuous on the interval [a, b], except for an inﬁnite discontinuity at b, then the improper integral of f over the interval [a, b] is deﬁned as

f (x) dx a

k b

b a

f(x) dx = lim− k →b

(4)

f(x) dx a

In the case where the indicated limit exists, the improper integral is said to converge, and the limit is deﬁned to be the value of the integral. In the case where the limit does not exist, the improper integral is said to diverge, and it is not assigned a value.

(b) Figure 7.8.7

Example 5

1 y= √1 − x

Evaluate

1 0

dx . √ 1−x

Solution. The integral is improper because the integrand approaches +⬁ as x approaches the upper limit 1 from the left (Figure 7.8.8). From (4),

1 0

√

dx 1−x

= lim−

√

= lim−

k →1 1−x 0 √ = lim− −2 1 − k + 2 = 2 k →1

k √ −2 1−x

k →1

1 Figure 7.8.8

Improper integrals with an inﬁnite discontinuity at the left-hand endpoint or inside the interval of integration are deﬁned as follows.

Chapter 7 / Principles of Integral Evaluation

552

7.8.5 definition If f is continuous on the interval [a, b], except for an inďŹ nite discontinuity at a, then the improper integral of f over the interval [a, b] is deďŹ ned as

f(x) dx = lim+ k â&#x2020;&#x2019;a

f(x) dx

(5)

The integral is said to converge if the indicated limit exists and diverge if it does not. If f is continuous on the interval [a, b], except for an inďŹ nite discontinuity at a point c in (a, b), then the improper integral of f over the interval [a, b] is deďŹ ned as

y b

f (x) dx

f(x) dx =

f(x) dx +

f(x) dx

(6)

where the two integrals on the right side are themselves improper. The improper integral on the left side is said to converge if both terms on the right side converge and diverge if either term on the right side diverges (Figure 7.8.9).

f (x) dx is improper.

Example 6

Evaluate

Figure 7.8.9

(a)

dx 1â&#x2C6;&#x2019;x

(b)

4 1

dx (x â&#x2C6;&#x2019; 2)2/3

Solution (a). The integral is improper because the integrand approaches â&#x2C6;&#x2019;âŹ as x ap-

x 1

proaches the lower limit 1 from the right (Figure 7.8.10). From DeďŹ nition 7.8.5 we obtain 2 2 2 dx dx = lim+ = lim+ â&#x2C6;&#x2019;ln |1 â&#x2C6;&#x2019; x| k â&#x2020;&#x2019;1 k â&#x2020;&#x2019;1 k k 1â&#x2C6;&#x2019;x 1 1â&#x2C6;&#x2019;x = lim+ â&#x2C6;&#x2019;ln | â&#x2C6;&#x2019;1| + ln |1 â&#x2C6;&#x2019; k| = lim+ ln |1 â&#x2C6;&#x2019; k| = â&#x2C6;&#x2019;âŹ k â&#x2020;&#x2019;1

k â&#x2020;&#x2019;1

1 1â&#x2C6;&#x2019;x

so the integral diverges.

Solution (b). The integral is improper because the integrand approaches +âŹ at x = 2, Figure 7.8.10

which is inside the interval of integration. From DeďŹ nition 7.8.5 we obtain 2 4 4 dx dx dx = + /3 /3 2 2 2 /3 1 (x â&#x2C6;&#x2019; 2) 1 (x â&#x2C6;&#x2019; 2) 2 (x â&#x2C6;&#x2019; 2)

and we must investigate the convergence of both improper integrals on the right. Since 2 k dx dx 1/3 1 /3 =3 = lim = lim 3(k â&#x2C6;&#x2019; 2) â&#x2C6;&#x2019; 3(1 â&#x2C6;&#x2019; 2) 2/3 k â&#x2020;&#x2019; 2â&#x2C6;&#x2019; 1 (x â&#x2C6;&#x2019; 2)2/3 k â&#x2020;&#x2019; 2â&#x2C6;&#x2019; 1 (x â&#x2C6;&#x2019; 2) 4 4 â&#x2C6;&#x161; dx dx 3 1/3 1 /3 2 = lim = lim = 3 3(4 â&#x2C6;&#x2019; 2) â&#x2C6;&#x2019; 3(k â&#x2C6;&#x2019; 2) / / 2 3 2 3 + + k â&#x2020;&#x2019; 2 k â&#x2020;&#x2019; 2 (x â&#x2C6;&#x2019; 2) (x â&#x2C6;&#x2019; 2) 2 k

we have from (7) that

4 1

â&#x2C6;&#x161; dx 3 = 3 + 3 2 (x â&#x2C6;&#x2019; 2)2/3

(7)

7.8 Improper Integrals

WARNING

553

It is sometimes tempting to apply the Fundamental Theorem of Calculus directly to an improper integral without taking the appropriate limits. To illustrate what can go wrong with this procedure, suppose we fail to recognize that the integral

2 0

dx (x â&#x2C6;&#x2019; 1)2

(8)

is improper and mistakenly evaluate this integral as

â&#x2C6;&#x2019;

1 xâ&#x2C6;&#x2019;1

2 0

= â&#x2C6;&#x2019;1 â&#x2C6;&#x2019; (1) = â&#x2C6;&#x2019;2

This result is clearly incorrect because the integrand is never negative and hence the integral cannot be negative! To evaluate (8) correctly we should first write

2 0

dx = (x â&#x2C6;&#x2019; 1)2

dx + (x â&#x2C6;&#x2019; 1)2

2 1

dx (x â&#x2C6;&#x2019; 1)2

and then treat each term as an improper integral. For the first term,

so (8) diverges.

1 0

dx = lim (x â&#x2C6;&#x2019; 1)2 k â&#x2020;&#x2019; 1â&#x2C6;&#x2019;

k 0

1 dx â&#x2C6;&#x2019; 1 = +âŹ = lim â&#x2C6;&#x2019; (x â&#x2C6;&#x2019; 1)2 kâ&#x2C6;&#x2019;1 k â&#x2020;&#x2019; 1â&#x2C6;&#x2019;

ARC LENGTH AND SURFACE AREA USING IMPROPER INTEGRALS

In DeďŹ nitions 6.4.2 and 6.5.2 for arc length and surface area we required the function f to be smooth (continuous ďŹ rst derivative) to ensure the integrability in the resulting formula. However, smoothness is overly restrictive since some of the most basic formulas in geometry involve functions that are not smooth but lead to convergent improper integrals. Accordingly, let us agree to extend the deďŹ nitions of arc length and surface area to allow functions that are not smooth, but for which the resulting integral in the formula converges. Example 7

Derive the formula for the circumference of a circle of radius r.

Solution. For convenience, let us assume that the circle is centered at the origin, in y

y = â&#x2C6;&#x161;r 2 â&#x2C6;&#x2019; x 2

â&#x2C6;&#x2019;r

Figure 7.8.11

which case its equation is x 2 + y 2 = r 2 . We will ďŹ nd the arc length of the portion of the circle that lies in the ďŹ rst quadrant and then multiply by 4 to obtain the total circumference (Figure 7.8.11). â&#x2C6;&#x161; Since the equation of the upper semicircle is y = r 2 â&#x2C6;&#x2019; x 2 , it follows from Formula (4) of Section 6.4 that the circumference C is 2 r r x 1 + (dy/dx)2 dx = 4 1+ â&#x2C6;&#x2019; dx C=4 2 2 r â&#x2C6;&#x2019;x 0 0 r dx = 4r 2 0 r â&#x2C6;&#x2019; x2 This integral is improper because of the inďŹ nite discontinuity at x = r, and hence we evaluate it by writing k dx C = 4r limâ&#x2C6;&#x2019; k â&#x2020;&#x2019;r 0 r 2 â&#x2C6;&#x2019; x2 x k Formula (77) in the = 4r limâ&#x2C6;&#x2019; sinâ&#x2C6;&#x2019;1 Endpaper Integral Table k â&#x2020;&#x2019;r r 0

k = 4r limâ&#x2C6;&#x2019; sinâ&#x2C6;&#x2019;1 â&#x2C6;&#x2019; sinâ&#x2C6;&#x2019;1 0 k â&#x2020;&#x2019;r r

Ď&#x20AC; â&#x2C6;&#x2019; 0 = 2Ď&#x20AC;r = 4r[sinâ&#x2C6;&#x2019;1 1 â&#x2C6;&#x2019; sinâ&#x2C6;&#x2019;1 0] = 4r 2

Chapter 7 / Principles of Integral Evaluation

554

✔QUICK CHECK EXERCISES 7.8

(See page 557 for answers.)

1. In each part, determine whether the integral is improper, and if so, explain why. Do not evaluate the integrals. 3π/4 π (a) cot x dx cot x dx (b) (c)

+⬁

1 dx x2 + 1

(d)

+⬁ 1

1 dx x2 − 1

2. Express each improper integral in Quick Check Exercise 1 in terms of one or more appropriate limits. Do not evaluate the limits.

EXERCISE SET 7.8

Graphing Utility

3–32 Evaluate the integrals that converge. ■

5. 7. 9. 11. 13. 15. 17.

+⬁ +⬁

2 dx 2−1 x 3 +⬁ 1 dx x ln3 x e 0 dx 3 −⬁ (2x − 1) 0 e3x dx −⬁ +⬁

x dx

+⬁

−⬁ 4

(x 2

x dx + 3)2

dx (x − 4)2

8. 10. 12. 14. 16. 18.

π/2

tan x dx

20.

21.

1 − x2

+⬁

−1 +⬁

x dx 1 + x2 xe

−x 2

−⬁

19.

e−2x dx

converges to

22.

+⬁

2 3 −⬁ 0

+⬁

x −p dx

provided

4. Evaluate the integrals that converge. +⬁ +⬁ (a) ex dx e−x dx (b) 0

(c)

1 dx x3

(d)

1 dx √ 3 2 x

CAS

1. In each part, determine whether the integral is improper, and if so, explain why. 5 5 1 dx dx (a) ln x dx (b) (c) 1 x +3 0 1 x −3 +⬁ π/4 +⬁ dx (d) (f ) tan x dx e−x dx (e) √ 3 x −1 0 −⬁ 1 2. In each part, determine all values of p for which the integral is improper. 1 2 1 dx dx (a) (b) e−px dx (c) p 0 x 1 x −p 0

π/4

3. The improper integral

1 dx √ x ln x dx x2 + 9

ex dx x −⬁ 3 − 2e +⬁ x dx 2 −⬁ x +2 +⬁ e−t dt −2t −⬁ 1 + e 8 dx √ 3 x 0 4 dx √ 4−x 0 1 x dx −3 9 − x2

23. 25. 27.

π/2

dx √ 1 − 2 cos x

π/3 3 0

26.

x −1/3 dx

28.

29. 31.

1 dx x2

1 0

24.

dx x−2

−1 +⬁ 0

sin x

√

dx x(x + 1)

30.

32.

π/4 0 2 −2 1

dx x2 dx (x − 1)2/3

+⬁ 1 +⬁ 0

sec2 x dx 1 − tan x

dx x x2 − 1 √

dx x(x + 1)

33–36 True–False Determine whether the statement is true or

false. Explain your answer. ■ +⬁ x −4/3 dx converges to 3. 33. 1

34. If f is continuous on [a, +⬁) and limx → +⬁ f(x) = 1, then +⬁ f(x) dx converges. a 2 1 dx is an improper integral. 35. x(x − 3) 1 1 1 dx = 0 36. 3 −1 x 37–40 Make the u-substitution and evaluate the resulting deﬁ-

nite integral. ■ +⬁ −√x √ e 37. √ dx; u = x [Note: u → +⬁ as x → +⬁.] x 0 +⬁ √ dx ; u = x [Note: u → +⬁ as x → +⬁.] 38. √ x(x + 4) 12 +⬁ e−x dx; u = 1 − e−x 39. 0 1 − e−x [Note: u → 1 as x → +⬁.]

7.8 Improper Integrals

40. C

+⬁

e−x

1 − e−2x

53–56 Use the results in Exercise 52. ■

dx; u = e−x

53. (a) Conﬁrm graphically and algebraically that

ate that limit with a CAS. Conﬁrm the answer by evaluating the integral directly with the CAS. ■ 41.

e−x cos x dx

+⬁

42.

xe−3x dx

e−x ≤ e−x

41–42 Express the improper integral as a limit, and then evalu+⬁

555

43. In each part, try to evaluate the integral exactly with a CAS. If your result is not a simple numerical answer, then use the CAS to ﬁnd a numerical approximation +⬁ of the integral. +⬁ 1 1 (a) dx dx (b) 8+x +1 x −⬁ 0 1 + x3 +⬁ +⬁ sin x ln x dx (d) dx (c) x e x2 1 1

44. In each part, conﬁrm the result with a CAS. +⬁ +⬁ √ sin x π 2 (a) (b) e−x dx = π √ dx = 2 x 0 −⬁ 1 ln x π2 dx = − (c) 12 0 1+x

45. Find the length of the curve y = (4 − x 2/3 )3/2 over the interval [0, 8]. √ 46. Find the length of the curve y = 4 − x 2 over the interval [0, 2].

47–48 Use L’Hôpital’s rule to help evaluate the improper integral. ■ +⬁ 1 ln x dx ln x dx 48. 47. x2 1 0

49. Find the area of the region between the x-axis and the curve y = e−3x for x ≥ 0. 50. Find the area of the region between the x-axis and the curve y = 8/(x 2 − 4) for x ≥ 4. 51. Suppose that the region between the x-axis and the curve y = e−x for x ≥ 0 is revolved about the x-axis. (a) Find the volume of the solid that is generated. (b) Find the surface area of the solid.

(b) Evaluate the integral

+⬁

(x ≥ 1)

e−x dx

54. (a) Confirm graphically and algebraically that ex 1 ≤ (x ≥ 0) 2x + 1 2x + 1 (b) Evaluate the integral +⬁ dx 2x + 1 0 (c) What does the result obtained in part (b) tell you about the integral +⬁ ex dx? 2x + 1 0 55. Let R be the region to the right of x = 1 that is bounded by the x-axis and the curve y = 1/x. When this region is revolved about the x-axis it generates a solid whose surface is known as Gabriel’s Horn (for reasons that should be clear from the accompanying figure). Show that the solid has a finite volume but its surface has an infinite area. [Note: It has been suggested that if one could saturate the interior of the solid with paint and allow it to seep through to the surface, then one could paint an infinite surface with a finite amount of paint! What do you think?] y

1 y= x

Figure Ex-55 F O C U S O N C O N C E P TS

52. Suppose that f and g are continuous functions and that 0 ≤ f(x) ≤ g(x) if x ≥ a. Give a reasonable informal argument using areas to explain why the following results are true. +⬁ +⬁ (a) If a f(x) dx diverges, then a g(x) dx diverges. +⬁ +⬁ (b) If a g(x) dx converges, then a f(x) dx con +⬁ +⬁ verges and a f(x) dx ≤ a g(x) dx. [Note: The results in this exercise are sometimes called comparison tests for improper integrals.]

56. In each part, use Exercise 52 to determine whether the integral converges or diverges. If it converges, then use part (b) of that exercise to ﬁnd an upper bound on the value of the integral. +⬁ 3 +⬁ x +1 x (a) dx (b) dx 5+1 x x 2 2 +⬁ xex dx (c) 2x + 1 0

556

Chapter 7 / Principles of Integral Evaluation

F O C U S O N C O N C E P TS

63. In Exercise 25 of Section 6.6, we determined the work required to lift a 6000 lb satellite to an orbital position that is 1000 mi above the Earthâ&#x20AC;&#x2122;s surface. The ideas discussed in that exercise will be needed here. (a) Find a deďŹ nite integral that represents the work required to lift a 6000 lb satellite to a position b miles above the Earthâ&#x20AC;&#x2122;s surface. (b) Find a deďŹ nite integral that represents the work required to lift a 6000 lb satellite an â&#x20AC;&#x153;inďŹ nite distanceâ&#x20AC;? above the Earthâ&#x20AC;&#x2122;s surface. Evaluate the integral. [Note: The result obtained here is sometimes called the work required to â&#x20AC;&#x153;escapeâ&#x20AC;? the Earthâ&#x20AC;&#x2122;s gravity.]

57. Sketch the region whose area is +âŹ dx 1 + x2 0 and use your sketch to show that 1 +âŹ dx 1â&#x2C6;&#x2019;y = dy 1 + x2 y 0 0

58. (a) Give a reasonable informal argument, based on areas, that explains why the integrals +âŹ +âŹ cos x dx sin x dx and 0

diverge.

(b) Show that

+âŹ

64â&#x20AC;&#x201C;65 A transform is a formula that converts or â&#x20AC;&#x153;transformsâ&#x20AC;?

â&#x2C6;&#x161; cos x dx diverges. â&#x2C6;&#x161; x

one function into another. Transforms are used in applications to convert a difďŹ cult problem into an easier problem whose solution can then be used to solve the original difďŹ cult problem. The Laplace transform of a function f(t), which plays an important role in the study of differential equations, is denoted by á?¸{f(t)} and is deďŹ ned by +âŹ eâ&#x2C6;&#x2019;st f(t) dt á?¸{f(t)} =

59. In electromagnetic theory, the magnetic potential at a point on the axis of a circular coil is given by 2Ď&#x20AC;NI r +âŹ dx u= 2 k (r + x 2 )3/2 a where N, I, r, k, and a are constants. Find u. C

60. The average speed, v, ÂŻ of the molecules of an ideal gas is given by / 4 M 3 2 +âŹ 3 â&#x2C6;&#x2019;Mv2 /(2RT ) vÂŻ = â&#x2C6;&#x161; dv v e Ď&#x20AC; 2RT 0 and the root-mean-square speed, vrms , by / M 3 2 +âŹ 4 â&#x2C6;&#x2019;Mv2 /(2RT ) 4 2 = â&#x2C6;&#x161; dv v e vrms Ď&#x20AC; 2RT 0 where v is the molecular speed, T is the gas temperature, M is the molecular weight of the gas, and R is the gas constant. (a) Use a CAS to show that +âŹ 1 2 2 x 3 eâ&#x2C6;&#x2019;a x dx = 4 , a > 0 2a 0 and use this result to show that vÂŻ = 8RT /(Ď&#x20AC;M). (b) Use a CAS to show that â&#x2C6;&#x161; +âŹ 3 Ď&#x20AC; 4 â&#x2C6;&#x2019;a 2 x 2 x e dx = , a>0 8a 5 0 â&#x2C6;&#x161; and use this result to show that vrms = 3RT /M. 61â&#x20AC;&#x201C;62 Medication can be administered to a patient using a vari-

ety of methods. For a given method, let c(t) denote the concentration of medication in the patientâ&#x20AC;&#x2122;s bloodstream (measured in mg/L) t hours after the dose is given. The area under the curve c = c(t) over the time interval [0, +âŹ ) indicates the â&#x20AC;&#x153;availabilityâ&#x20AC;? of the medication for the patientâ&#x20AC;&#x2122;s body. Determine which method provides the greater availability. â&#x2013; 61. Method 1: c1 (t) = 5(eâ&#x2C6;&#x2019;0.2t â&#x2C6;&#x2019; eâ&#x2C6;&#x2019;t ); Method 2: c2 (t) = 4(eâ&#x2C6;&#x2019;0.2t â&#x2C6;&#x2019; eâ&#x2C6;&#x2019;3t )

62. Method 1: c1 (t) = 6(eâ&#x2C6;&#x2019;0.4t â&#x2C6;&#x2019; eâ&#x2C6;&#x2019;1.3t ); Method 2: c2 (t) = 5(eâ&#x2C6;&#x2019;0.4t â&#x2C6;&#x2019; eâ&#x2C6;&#x2019;3t )

In this formula s is treated as a constant in the integration process; thus, the Laplace transform has the effect of transforming f(t) into a function of s. Use this formula in these exercises. â&#x2013;

64. Show that 1 1 (a) á?¸{1} = , s > 0 (b) á?¸{e2t } = , s>2 s sâ&#x2C6;&#x2019;2 1 (c) á?¸{sin t} = 2 , s>0 s +1 s (d) á?¸{cos t} = 2 , s > 0. s +1 65. In each part, ďŹ nd the Laplace transform. (a) f(t) = t, (b) f(t) = t 2 , s > 0 ! s>0 0, t < 3 , s>0 (c) f(t) = 1, t â&#x2030;Ľ 3 66. Later in the text, we will show that +âŹ â&#x2C6;&#x161; 2 eâ&#x2C6;&#x2019;x dx = 21 Ď&#x20AC; 0

ConďŹ rm that this is reasonable by using a CAS or a calculator with a numerical integration capability. 67. Use the result in Exercise 66 to show that +âŹ Ď&#x20AC; 2 (a) , a>0 eâ&#x2C6;&#x2019;ax dx = a â&#x2C6;&#x2019;âŹ +âŹ 1 2 2 eâ&#x2C6;&#x2019;x /2Ď&#x192; dx = 1, Ď&#x192; > 0. (b) â&#x2C6;&#x161; 2Ď&#x20AC;Ď&#x192; â&#x2C6;&#x2019;âŹ 68â&#x20AC;&#x201C;69 A convergent improper integral over an inďŹ nite interval can be approximated by ďŹ rst replacing the inďŹ nite limit(s) of integration by ďŹ nite limit(s), then using a numerical integration technique, such as Simpsonâ&#x20AC;&#x2122;s rule, to approximate the integral with ďŹ nite limit(s). This technique is illustrated in these exercises. â&#x2013;

Chapter 7 Review Exercises

(b) Use the result that you obtained in Exercise 52 and the fact that 1/(x 6 + 1) < 1/x 6 for x ≥ 4 to show that the truncation error for the approximation in part (a) satisﬁes 0 < E < 2 × 10−4 .

68. Suppose that the integral in Exercise 66 is approximated by ﬁrst writing it as +⬁ +⬁ K 2 −x 2 −x 2 e−x dx e dx + e dx = 0

then dropping the second term, and then applying Simpson’s rule to the integral K 2 e−x dx

+⬁

The resulting approximation has two sources of error: the error from Simpson’s rule and the error +⬁ 2 E= e−x dx K

that results from discarding the second term. We call E the truncation error. (a) Approximate the integral in Exercise 66 by applying Simpson’s rule with n = 10 subdivisions to the integral 3 2 e−x dx 0

Round√your answer to four decimal places and compare it to 21 π rounded to four decimal places. (b) Use the result that you obtained in Exercise 52 and the 2 2 fact that e−x ≤ 31 xe−x for x ≥ 3 to show that the truncation error for the approximation in part (a) satisﬁes 0 < E < 2.1 × 10−5 .

69. (a) It can be shown that +⬁

π 1 dx = +1 3 0 Approximate this integral by applying Simpson’s rule with n = 20 subdivisions to the integral 4 1 dx 6 0 x +1 Round your answer to three decimal places and compare it to π/3 rounded to three decimal places. x6

557

epx dx converge? 70. For what values of p does 1 0 dx /x p converges if p < 1 and diverges if 71. Show that 0 p ≥ 1. 72. It is sometimes possible to convert an improper integral into a “proper” integral having the same value by making an appropriate substitution. Evaluate the following integral by making the indicated substitution, and investigate what happens if you evaluate the integral directly using a CAS. 1 √ 1+x dx; u = 1 − x 1−x 0 73–74 Transform the given improper integral into a proper integral by making the stated u-substitution; then approximate the proper integral by Simpson’s rule with n = 10 subdivisions. Round your answer to three decimal places. ■ 1 √ cos x 73. √ dx; u = x x 0 1 √ sin x 74. dx; u = 1 − x √ 1−x 0 75. Writing What is “improper” about an integral over an infi b nite interval? Explain why Definition 5.5.1 for a f(x) dx +⬁ fails for a f(x) dx. Discuss a strategy for assigning a +⬁ value to a f(x) dx. 76. Writing What is “improper” about a definite integral over an interval on which the integrand has an infinite disconb tinuity? Explain why Definition 5.5.1 for a f(x) dx fails if the graph of f has a vertical asymptote b at x = a. Discuss a strategy for assigning a value to a f(x) dx in this circumstance.

✔QUICK CHECK ANSWERS 7.8 1. (a) proper (b) improper, since cot x has an infinite discontinuity at x = π (c) improper, since there is an infinite interval of integration (d) improper, since there is an infinite interval of integration and the integrand has an infinite discontinuity at x = 1 b 2 b b 1 1 1 1 2. (b) lim− dx (d) lim dx + lim dx 3. ; p>1 cot x dx (c) lim 2 2 2 + b → +⬁ b → +⬁ a → 1 b→π p−1 2 x −1 a x −1 0 x +1 π/4 4. (a) 1 (b) diverges (c) diverges (d) 3

CHAPTER 7 REVIEW EXERCISES 1–6 Evaluate the given integral with the aid of an appropriate

u-substitution. ■ √ 1. 4 + 9x dx √ cos x sin x dx 3. 5. x tan2 (x 2 ) sec2 (x 2 ) dx

1 dx sec πx dx 4. x ln x 9 √ x 6. dx x + 9 0

7. (a) Evaluate the integral

√

1 2x − x 2

√ three ways: using√the substitution u = x, using the substitution u = 2 − x, and completing the square. (b) Show that the answers in part (a) are equivalent.

Chapter 7 / Principles of Integral Evaluation

558

8. Evaluate the integral

dx x2 + 1 (a) using integration by parts √ (b) using the substitution u = x 2 + 1. 0

√

9–12 Use integration by parts to evaluate the integral. ■

−x

10.

x sin 2x dx 1/2

ln(2x + 3) dx 12. tan−1 (2x) dx 0 13. Evaluate 8x 4 cos 2x dx using tabular integration by parts. 11.

14. A particle moving along the x-axis has velocity function v(t) = t 2 e−t . How far does the particle travel from time t = 0 to t = 5?

15–20 Evaluate the integral. ■

15. 17. 19.

sin 5θ dθ

16.

sin x cos 2x dx

18.

sin4 2x dx

21–26 Evaluate the integral onometric substitution. ■ x2 21. dx √ 9 − x2 dx 23. √ x2 − 1 x2 25. dx √ 9 + x2

20.

sin3 2x cos2 2x dx

π/6

sin 2x cos 4x dx

x cos5 (x 2 ) dx

by making an appropriate trig

dx √ x 2 16 − x 2 x2 24. dx √ 2 x − 25 √ 1 + 4x 2 26. dx x 22.

27–32 Evaluate the integral using the method of partial fractions. ■ dx dx 27. 28. x 2 + 3x − 4 x 2 + 8x + 7 2 x +2 x 2 + x − 16 29. dx dx 30. x+2 (x − 1)(x − 3)2 x2 dx 31. dx 32. (x + 2)3 x3 + x 1 33. Consider the integral dx. x3 − x (a) Evaluate the integral using the substitution x = sec θ . For what values of x is your result valid? (b) Evaluate the integral using the substitution x = sin θ . For what values of x is your result valid? (c) Evaluate the integral using the method of partial fractions. For what values of x is your result valid?

34. Find the area of the region that is enclosed by the curves y = (x − 3)/(x 3 + x 2 ), y = 0, x = 1, and x = 2.

35–40 Use the Endpaper Integral Table to evaluate the integral. ■ 35. sin 7x cos 9x dx 36. (x 3 − x 2 )e−x dx dx 38. 37. x x − x 2 dx √ x 4x + 3 3x − 1 39. tan2 2x dx 40. dx 2 + x2 41–42 Approximate the integral using (a) the midpoint approximation M10 , (b) the trapezoidal approximation T10 , and (c) Simpson’s rule approximation S20 . In each case, ﬁnd the exact value of the integral and approximate the absolute error. Express your answers to at least four decimal places. ■ 1 3 1 1 dx 42. 41. dx √ 1 + x2 x + 1 1 −1 43–44 Use inequalities (12), (13), and (14) of Section 7.7 to ﬁnd upper bounds on the errors in parts (a), (b), or (c) of the indicated exercise. ■

43. Exercise 41

44. Exercise 42

45–46 Use inequalities (12), (13), and (14) of Section 7.7 to ﬁnd a number n of subintervals for (a) the midpoint approximation Mn , (b) the trapezoidal approximation Tn , and (c) Simpson’s rule approximation Sn to ensure the absolute error will be less than 10−4 . ■

45. Exercise 41

46. Exercise 42

47–50 Evaluate the integral if it converges. ■

47.

+⬁

e−x dx

48.

−⬁ 1

dx x2 + 4

dx 1 50. dx 9−x 0 2x − 1 51. Find the area that is enclosed between the x-axis and the curve y = (ln x − 1)/x 2 for x ≥ e. 49.

√

52. Find the volume of the solid that is generated when the region between the x-axis and the curve y = e−x for x ≥ 0 is revolved about the y-axis. 53. Find a positive value of a that satisﬁes the equation

+⬁ 0

1 dx = 1 x 2 + a2

54. Consider the following methods for evaluating integrals: u-substitution, integration by parts, partial fractions, reduction formulas, and trigonometric substitutions. In each part, state the approach that you would try ﬁrst to evaluate the integral. If none of them seems appropriate, then say so. You need not evaluate the integral. (a)

x sin x dx

(b)

cos x sin x dx

(cont.)

Chapter 7 Making Connections

(c)

tan7 x dx

(d)

tan7 x sec2 x dx

(f )

63.

3x 2 dx (x + 1)3 4 â&#x2C6;&#x2019; x 2 dx (h)

3x 2 dx 3 x +1 (g) tanâ&#x2C6;&#x2019;1 x dx (i) x 4 â&#x2C6;&#x2019; x 2 dx (e)

65. 67. 69.

55â&#x20AC;&#x201C;74 Evaluate the integral. â&#x2013;

55. 57.

dx (3 + x 2 )3/2

56.

tan7 Î¸ dÎ¸

58.

Ď&#x20AC;/4

59. 61.

sin2 2x cos3 2x dx e2x cos 3x dx

71.

x cos 3x dx

cos Î¸ dÎ¸ sin2 Î¸ â&#x2C6;&#x2019; 6 sin Î¸ + 12 4 1 dx 60. 2 0 (x â&#x2C6;&#x2019; 3) â&#x2C6;&#x161; 1/ 2 2 3/2 62. dx â&#x2C6;&#x161; (1 â&#x2C6;&#x2019; 2x )

73. 74.

dx 64. (x â&#x2C6;&#x2019; 1)(x + 2)(x â&#x2C6;&#x2019; 3) â&#x2C6;&#x161; 8 xâ&#x2C6;&#x2019;4 dx 66. x 4 1 dx 68. â&#x2C6;&#x161; ex + 1 1/2 sinâ&#x2C6;&#x2019;1 x dx 70. 0 x+3 dx 72. â&#x2C6;&#x161; x 2 + 2x + 2 +âŹ x dx 2 (x + 1)2 a +âŹ dx , a, b > 0 2 + b2 x 2 a 0

559

1/3

dx (4 â&#x2C6;&#x2019; 9x 2 )2 0 ln 2 â&#x2C6;&#x161; ex â&#x2C6;&#x2019; 1 dx 0 dx x(x 2 + x + 1) tan5 4x sec4 4x dx sec2 Î¸ dÎ¸ 3 tan Î¸ â&#x2C6;&#x2019; tan2 Î¸

â&#x2C6;&#x2019;1/ 2

CHAPTER 7 MAKING CONNECTIONS

CAS

1. Recall from Theorem 3.3.1 and the discussion preceding it that if f â&#x20AC;˛(x) > 0, then the function f is increasing and has an inverse function. Parts (a), (b), and (c) of this problem show that if this condition is satisďŹ ed and if f â&#x20AC;˛ is continuous, then a deďŹ nite integral of f â&#x2C6;&#x2019;1 can be expressed in terms of a deďŹ nite integral of f . (a) Use integration by parts to show that

f(x) dx = bf(b) â&#x2C6;&#x2019; af(a) â&#x2C6;&#x2019;

3. The Gamma function, Ĺ´(x), is deďŹ ned as +âŹ Ĺ´(x) = t xâ&#x2C6;&#x2019;1 eâ&#x2C6;&#x2019;t dt 0

It can be shown that this improper integral converges if and only if x > 0. (a) Find Ĺ´(1). (b) Prove: Ĺ´(x + 1) = xĹ´(x) for all x > 0. [Hint: Use integration by parts.] (c) Use the results in parts (a) and (b) to ďŹ nd Ĺ´(2), Ĺ´(3), and Ĺ´(4); and then make a conjecture about Ĺ´(n) for positive integer values of n. â&#x2C6;&#x161; (d) Show that Ĺ´ 21 = Ď&#x20AC;. [Hint: See Exercise 66 of Section 7.8.] (e) Use the results in parts â&#x2C6;&#x161; obtained â&#x2C6;&#x161; (b) and (d) to show that Ĺ´ 23 = 21 Ď&#x20AC; and Ĺ´ 25 = 43 Ď&#x20AC;.

xf â&#x20AC;˛(x) dx a

(b) Use the result in part (a) to show that if y = f(x), then

f(x) dx = bf(b) â&#x2C6;&#x2019; af(a) â&#x2C6;&#x2019;

f(b)

f â&#x2C6;&#x2019;1 (y) dy f(a)

4. Refer to the Gamma function deďŹ ned in Exercise 3 to show that 1

Î˛ Îą

f â&#x2C6;&#x2019;1 (x) dx = Î˛f â&#x2C6;&#x2019;1 (Î˛) â&#x2C6;&#x2019; Îąf â&#x2C6;&#x2019;1 (Îą) â&#x2C6;&#x2019;

f â&#x2C6;&#x2019;1 (Î˛)

f(x) dx

(b)

ln x dx = (2e2 â&#x2C6;&#x2019; e) â&#x2C6;&#x2019;

ex dx 1

f â&#x2C6;&#x2019;1 (Îą)

2. In each part, use the result in Exercise 1 to obtain the equation, and then conďŹ rm that the equation is correct by performing the integrations. 1/2 Ď&#x20AC;/6 â&#x2C6;&#x2019;1 1 â&#x2C6;&#x2019;1 1 (a) sin x dx â&#x2C6;&#x2019; sin x dx = 2 sin 2 0

(a)

(ln x)n dx = (â&#x2C6;&#x2019;1)n Ĺ´(n + 1), n > 0

[Hint: Let t = â&#x2C6;&#x2019; ln x.] +âŹ n+1 â&#x2C6;&#x2019;x n (b) e dx = Ĺ´ , n > 0. n 0 [Hint: Let t = x n . Use the result in Exercise 3(b).]

5. A simple pendulum consists of a mass that swings in a vertical plane at the end of a massless rod of length L, as shown in the accompanying ďŹ gure. Suppose that a simple pendulum is displaced through an angle Î¸0 and released from rest. It can be

560

Chapter 7 / Principles of Integral Evaluation

shown that in the absence of friction, the time T required for the pendulum to make one complete back-and-forth swing, called the period, is given by 1 8L θ0 T = (1) dθ √ g 0 cos θ − cos θ0 where θ = θ(t) is the angle the pendulum makes with the vertical at time t. The improper integral in (1) is difﬁcult to evaluate numerically. By a substitution outlined below it can be shown that the period can be expressed as 1 L π/2 T =4 dφ (2) g 0 1 − k 2 sin2 φ where k = sin(θ0 /2). The integral in (2) is called a complete elliptic integral of the ﬁrst kind and is more easily evaluated by numerical methods. (a) Obtain (2) from (1) by substituting cos θ = 1 − 2 sin2 (θ /2)

and then making the change of variable sin φ =

sin(θ /2) sin(θ /2) = sin(θ0 /2) k

(b) Use (2) and the numerical integration capability of your CAS to estimate the period of a simple pendulum for which L = 1.5 ft, θ0 = 20 ◦ , and g = 32 ft/s2 .

Figure Ex-5

cos θ0 = 1 − 2 sin2 (θ0 /2) k = sin(θ0 /2)

EXPANDING THE CALCULUS HORIZON To learn how numerical integration can be applied to the cost analysis of an engineering project, see the module entitled Railroad Design at: www.wiley.com/college/anton

8 Photo by Milton Bell, catalog # Lu1-mb1, Texas Archeological Research Labratory, The University of Texas at Austin

In the 1920s, excavation of an archeological site in Folsom, New Mexico, uncovered a collection of prehistoric stone spearheads now known as “Folsom points.” In 1950, carbon dating of charred bison bones found nearby conﬁrmed that human hunters lived in the area between 9000 B.C. and 8000 B.C. We will study carbon dating in this chapter.

8.1

MATHEMATICAL MODELING WITH DIFFERENTIAL EQUATIONS

Many fundamental laws of science and engineering can be expressed in terms of differential equations. We introduced the concept of a differential equation in Section 5.2, but in this chapter we will go into more detail. We will discuss some important mathematical models that involve differential equations, and we will discuss some methods for solving and approximating solutions of some of the basic types of differential equations. However, we will only be able to touch the surface of this topic, leaving many important topics in differential equations to courses that are devoted completely to the subject.

MODELING WITH DIFFERENTIAL EQUATIONS In this section we will introduce some basic terminology and concepts concerning differential equations. We will also discuss the general idea of modeling with differential equations, and we will encounter important models that can be applied to demography, medicine, ecology, and physics. In later sections of this chapter we will investigate methods that may be used to solve these differential equations.

TERMINOLOGY

Table 8.1.1

differential equation

order

dy = 3y dx

d2y dy −6 + 8y = 0 dx 2 dx

dy d 3y −t + (t 2 – 1)y = e t dt dt 3

y′− y = e2x

y′′+ y′ = cos t

Recall from Section 5.2 that a differential equation is an equation involving one or more derivatives of an unknown function. In this section we will denote the unknown function by y = y(x) unless the differential equation arises from an applied problem involving time, in which case we will denote it by y = y(t). The order of a differential equation is the order of the highest derivative that it contains. Some examples are given in Table 8.1.1. The last two equations in that table are expressed in “prime” notation, which does not specify the independent variable explicitly. However, you will usually be able to tell from the equation itself or from the context in which it arises whether to interpret y ′ as dy/dx or dy/dt. SOLUTIONS OF DIFFERENTIAL EQUATIONS A function y = y(x) is a solution of a differential equation on an open interval if the equation is satisﬁed identically on the interval when y and its derivatives are substituted 561

562

Chapter 8 / Mathematical Modeling with Differential Equations

into the equation. For example, y = e2x is a solution of the differential equation

dy − y = e2x (1) dx on the interval (−⬁, +⬁), since substituting y and its derivative into the left side of this equation yields dy d 2x −y = [e ] − e2x = 2e2x − e2x = e2x dx dx for all real values of x. However, this is not the only solution on (−⬁, +⬁); for example, the function y = e2x + Cex (2) is also a solution for every real value of the constant C, since

The first-order equation (1) has a single arbitrary constant in its general solution (2). Usually, the general solution of an nth-order differential equation will contain n arbitrary constants. This is plausible, since n integrations are needed to recover a function from its nth derivative.

dy d 2x −y = [e + Cex ] − (e2x + Cex ) = (2e2x + Cex ) − (e2x + Cex ) = e2x dx dx After developing some techniques for solving equations such as (1), we will be able to show that all solutions of (1) on (−⬁, +⬁) can be obtained by substituting values for the constant C in (2). On a given interval, a solution of a differential equation from which all solutions on that interval can be derived by substituting values for arbitrary constants is called a general solution of the equation on the interval. Thus (2) is a general solution of (1) on the interval (−⬁, +⬁). The graph of a solution of a differential equation is called an integral curve for the equation, so the general solution of a differential equation produces a family of integral curves corresponding to the different possible choices for the arbitrary constants. For example, Figure 8.1.1 shows some integral curves for (1), which were obtained by assigning values to the arbitrary constant in (2).

C=3 C=4

C=2 C=1 C=0

C = −1

8 6

y = e 2x + Ce x

C = −2

4 2

x −1

−2

Integral curves for

Figure 8.1.1

dy − y = e 2x dx

INITIAL-VALUE PROBLEMS When an applied problem leads to a differential equation, there are usually conditions in the problem that determine specific values for the arbitrary constants. As a rule of thumb, it requires n conditions to determine values for all n arbitrary constants in the general solution of an nth-order differential equation (one condition for each constant). For a firstorder equation, the single arbitrary constant can be determined by specifying the value of the unknown function y(x) at an arbitrary x-value x0 , say y(x0 ) = y0 . This is called an initial condition, and the problem of solving a first-order equation subject to an initial condition is called a first-order initial-value problem. Geometrically, the initial condition y(x0 ) = y0 has the effect of isolating the integral curve that passes through the point (x0 , y0 ) from the complete family of integral curves.

8.1 Modeling with Differential Equations

563

Example 1 The solution of the initial-value problem dy − y = e2x , dx

y(0) = 3

can be obtained by substituting the initial condition x = 0, y = 3 in the general solution (2) to ﬁnd C. We obtain 3 = e0 + Ce0 = 1 + C Thus, C = 2, and the solution of the initial-value problem, which is obtained by substituting this value of C in (2), is y = e2x + 2ex Geometrically, this solution is realized as the integral curve in Figure 8.1.1 that passes through the point (0, 3). Since many important principles in the physical and social sciences involve rates of change, it should not be surprising that such principles can often be modeled by differential equations. Here are some examples of the modeling process. UNINHIBITED POPULATION GROWTH

One of the simplest models of population growth is based on the observation that when populations (people, plants, bacteria, and fruit ﬂies, for example) are not constrained by environmental limitations, they tend to grow at a rate that is proportional to the size of the population—the larger the population, the more rapidly it grows. To translate this principle into a mathematical model, suppose that y = y(t) denotes the population at time t. At each point in time, the rate of increase of the population with respect to time is dy/dt, so the assumption that the rate of growth is proportional to the population is described by the differential equation dy = ky dt

When the number of bacteria is small, an uninhibited population growth model can be used to model the growth of bacteria in a petri dish.

(3)

where k is a positive constant of proportionality that can usually be determined experimentally. Thus, if the population is known at some point in time, say y = y0 at time t = 0, then a formula for the population y(t) can be obtained by solving the initial-value problem dy = ky, dt

y(0) = y0

INHIBITED POPULATION GROWTH; LOGISTIC MODELS The uninhibited population growth model was predicated on the assumption that the population y = y(t) was not constrained by the environment. While this assumption is reasonable as long as the size of the population is relatively small, environmental effects become increasingly important as the population grows. In general, populations grow within ecological systems that can only support a certain number of individuals; the number L of such individuals is called the carrying capacity of the system. When y > L, the population exceeds the capacity of the ecological system and tends to decrease toward L; when y < L, the population is below the capacity of the ecological system and tends to increase toward L; when y = L, the population is in balance with the capacity of the ecological system and tends to remain stable. To translate this into a mathematical model, we must look for a differential equation in which y > 0, L > 0, and

dy <0 dt

y > 1, L

dy >0 dt

y < 1, L

dy =0 dt

y =1 L

564

Chapter 8 / Mathematical Modeling with Differential Equations

Moreover, when the population is far below the carrying capacity (i.e., y /L ≈ 0), then the environmental constraints should have little effect, and the growth rate should behave like the uninhibited population model. Thus, we want dy y ≈ ky if ≈0 dt L A simple differential equation that meets all of these requirements is dy y y =k 1− dt L where k is a positive constant of proportionality. Thus if k and L can be determined experimentally, and if the population is known at some point, say y(0) = y0 , then a formula for the population y(t) can be determined by solving the initial-value problem dy y y, y(0) = y0 (4) =k 1− dt L This theory of population growth is due to the Belgian mathematician P. F. Verhulst ∗ (1804–1849), who introduced it in 1838 and described it as “logistic growth.” Thus, the differential equation in (4) is called the logistic differential equation, and the growth model described by (4) is called the logistic model. PHARMACOLOGY When a drug (say, penicillin or aspirin) is administered to an individual, it enters the bloodstream and then is absorbed by the body over time. Medical research has shown that the amount of a drug that is present in the bloodstream tends to decrease at a rate that is proportional to the amount of the drug present—the more of the drug that is present in the bloodstream, the more rapidly it is absorbed by the body. To translate this principle into a mathematical model, suppose that y = y(t) is the amount of the drug present in the bloodstream at time t. At each point in time, the rate of change in y with respect to t is dy/dt, so the assumption that the rate of decrease is proportional to the amount y in the bloodstream translates into the differential equation

dy (5) = −ky dt where k is a positive constant of proportionality that depends on the drug and can be determined experimentally. The negative sign is required because y decreases with time. Thus, if the initial dosage of the drug is known, say y = y0 at time t = 0, then a formula for y(t) can be obtained by solving the initial-value problem dy = −ky, dt

y(0) = y0

SPREAD OF DISEASE

Suppose that a disease begins to spread in a population of L individuals. Logic suggests that at each point in time the rate at which the disease spreads will depend on how many individuals are already affected and how many are not—as more individuals are affected, the opportunity to spread the disease tends to increase, but at the same time there are fewer individuals who are not affected, so the opportunity to spread the disease tends to decrease. Thus, there are two conﬂicting inﬂuences on the rate at which the disease spreads.

∗

Verhulst’s model fell into obscurity for nearly a hundred years because he did not have sufficient census data to test its validity. However, interest in the model was revived during the 1930s when biologists used it successfully to describe the growth of fruit fly and flour beetle populations. Verhulst himself used the model to predict that an upper limit of Belgium’s population would be approximately 9,400,000. In 2006 the population was about 10,379,000.

8.1 Modeling with Differential Equations

565

To translate this into a mathematical model, suppose that y = y(t) is the number of individuals who have the disease at time t, so of necessity the number of individuals who do not have the disease at time t is L − y. As the value of y increases, the value of L − y decreases, so the conﬂicting inﬂuences of the two factors on the rate of spread dy/dt are taken into account by the differential equation dy = ky(L − y) dt Show that the model for the spread of disease can be viewed as a logistic model with constant of proportionality kL by rewriting (6) appropriately.

where k is a positive constant of proportionality that depends on the nature of the disease and the behavior patterns of the individuals and can be determined experimentally. Thus, if the number of affected individuals is known at some point in time, say y = y0 at time t = 0, then a formula for y(t) can be obtained by solving the initial-value problem dy = ky(L − y), dt

y(0) = y0

(6)

NEWTON’S LAW OF COOLING

If a hot object is placed into a cool environment, the object will cool at a rate proportional to the difference in temperature between the object and the environment. Similarly, if a cold object is placed into a warm environment, the object will warm at a rate that is again proportional to the difference in temperature between the object and the environment. Together, these observations comprise a result known as Newton’s Law of Cooling. (Newton’s Law of Cooling appeared previously in the exercises of Section 2.2 and was mentioned brieﬂy in Section 5.8.) To translate this into a mathematical model, suppose that T = T (t) is the temperature of the object at time t and that Te is the temperature of the environment, which is assumed to be constant. Since the rate of change dT /dt is proportional to T − Te , we have dT = k(T − Te ) dt where k is a constant of proportionality. Moreover, since dT /dt is positive when T < Te , and is negative when T > Te , the sign of k must be negative. Thus if the temperature of the object is known at some time, say T = T0 at time t = 0, then a formula for the temperature T (t) can be obtained by solving the initial-value problem Natural position

dT = k(T − Te ), dt

Stretched

Released

m Figure 8.1.2

Natural position

m 0

Figure 8.1.3

T (0) = T0

(7)

VIBRATIONS OF SPRINGS We conclude this section with an engineering model that leads to a second-order differential equation. As shown in Figure 8.1.2, consider a block of mass m attached to the end of a horizontal spring. Assume that the block is then set into vibratory motion by pulling the spring beyond its natural position and releasing it at time t = 0. We will be interested in finding a mathematical model that describes the vibratory motion of the block over time. To translate this problem into mathematical form, we introduce a horizontal x-axis whose positive direction is to the right and whose origin is at the right end of the spring when the spring is in its natural position (Figure 8.1.3). Our goal is to find a model for the coordinate x = x(t) of the point of attachment of the block to the spring as a function of time. In developing this model, we will assume that the only force on the mass m is the restoring force of the spring, and we will ignore the influence of other forces such as friction, air resistance, and so forth. Recall from Hooke’s Law (Section 6.6) that when the connection point has coordinate x(t), the restoring force is −kx(t), where k is the spring constant. [The negative sign is due to the fact that the restoring force is to the left when x(t) is positive, and the restoring force is to the right when x(t) is negative.] It follows from Newton’s

566

Chapter 8 / Mathematical Modeling with Differential Equations

Second Law of Motion [Equation (5) of Section 6.6] that this restoring force is equal to the product of the mass m and the acceleration d 2 x /dt 2 of the mass. In other words, we have d 2x = −kx dt 2 which is a second-order differential equation for x. If at time t = 0 the mass is released from rest at position x(0) = x0 , then a formula for x(t) can be found by solving the initial-value problem d 2x m 2 = −kx, x(0) = x0 , x ′ (0) = 0 (8) dt [If at time t = 0 the mass is given an initial velocity v0 = 0, then the condition x ′ (0) = 0 must be replaced by x ′ (0) = v0 .] m

✔QUICK CHECK EXERCISES 8.1

(See page 568 for answers.)

1. Match each differential equation with its family of solutions. dy (a) x (i) y = x 2 + C =y dx (ii) y = C1 sin 2x + C2 cos 2x (b) y ′′ = 4y dy = 2x (c) (iii) y = C1 e2x + C2 e−2x dx d 2y = −4y (d) (iv) y = Cx dx 2 2. If y = C1 e2x + C2 xe2x is the general solution of a differential equation, then the order of the equation is , and a solution to the differential equation that satisﬁes the initial conditions y(0) = 1, y ′ (0) = 4 is given by y = .

3. The graph of a differentiable function y = y(x) passes through the point (0, 1) and at every point P (x, y) on the graph the tangent line is perpendicular to the line through P and the origin. Find an initial-value problem whose solution is y(x). 4. A glass of ice water with a temperature of 36 ◦ F is placed in a room with a constant temperature of 68 ◦ F. Assuming that Newton’s Law of Cooling applies, ﬁnd an initial-value problem whose solution is the temperature of water t minutes after it is placed in the room. [Note: The differential equation will involve a constant of proportionality.]

EXERCISE SET 8.1 3

1. Conﬁrm that y = 3ex is a solution of the initial-value problem y ′ = 3x 2 y, y(0) = 3. 2. Conﬁrm that y = 41 x 4 + 2 cos x + 1 is a solution of the initial-value problem y ′ = x 3 − 2 sin x, y(0) = 3.

3–4 State the order of the differential equation, and conﬁrm that

the functions in the given family are solutions. ■ dy 3. (a) (1 + x) = y; y = c(1 + x) dx ′′ (b) y + y = 0; y = c1 sin t + c2 cos t dy 4. (a) 2 + y = x − 1; y = ce−x /2 + x − 3 dx (b) y ′′ − y = 0; y = c1 et + c2 e−t 5–8 True–False Determine whether the statement is true or false. Explain your answer. ■

5. The equation

dy dx

dy + 2y dx

is an example of a second-order differential equation.

6. The differential equation dy = 2y + 1 dx has a solution that is constant. 7. We expect the general solution of the differential equation d 3y d 2y dy + 3 − + 4y = 0 3 2 dx dx dx to involve three arbitrary constants. 8. If every solution to a differential equation can be expressed in the form y = Aex+b for some choice of constants A and b, then the differential equation must be of second order. 9–14 In each part, verify that the functions are solutions of the

differential equation by substituting the functions into the equation. ■ 9. y ′′ + y ′ − 2y = 0 (a) e−2x and ex (b) c1 e−2x + c2 ex (c1 , c2 constants)

8.1 Modeling with Differential Equations

10. y ′′ − y ′ − 6y = 0 (a) e−2x and e3x (b) c1 e−2x + c2 e3x (c1 , c2 constants)

11. y ′′ − 4y ′ + 4y = 0 (a) e2x and xe2x (b) c1 e2x + c2 xe2x (c1 , c2 constants) 12. y ′′ − 8y ′ + 16y = 0 (a) e4x and xe4x (b) c1 e4x + c2 xe4x (c1 , c2 constants)

13. y ′′ + 4y = 0 (a) sin 2x and cos 2x (b) c1 sin 2x + c2 cos 2x (c1 , c2 constants)

14. y ′′ + 4y ′ + 13y = 0 (a) e−2x sin 3x and e−2x cos 3x (b) e−2x (c1 sin 3x + c2 cos 3x) (c1 , c2 constants) 15–20 Use the results of Exercises 9–14 to ﬁnd a solution to the initial-value problem. ■

15. y ′′ + y ′ − 2y = 0, y(0) = −1, y ′ (0) = −4 16. y ′′ − y ′ − 6y = 0, y(0) = 1, y ′ (0) = 8

17. y ′′ − 4y ′ + 4y = 0, y(0) = 2, y ′ (0) = 2

18. y ′′ − 8y ′ + 16y = 0, y(0) = 1, y ′ (0) = 1 19. y ′′ + 4y = 0, y(0) = 1, y ′ (0) = 2

20. y ′′ + 4y ′ + 13y = 0, y(0) = −1, y ′ (0) = −1 21–26 Find a solution to the initial-value problem. ■

21. y ′ + 4x = 2, y(0) = 3

22. y ′′ + 6x = 0, y(0) = 1, y ′ (0) = 2

23. y ′ − y 2 = 0, y(1) = 2 [Hint: Assume the solution has an inverse function x = x(y). Find, and solve, a differential equation that involves x ′ (y).] 24. y ′ = 1 + y 2 , y(0) = 0

(See Exercise 23.)

25. x 2 y ′ + 2xy = 0, y(1) = 2 [Hint: Interpret the left-hand side of the equation as the derivative of a product of two functions.] 26. xy ′ + y = ex , y(1) = 1 + e

(See Exercise 25.)

F O C U S O N C O N C E P TS

27. (a) Suppose that a quantity y = y(t) increases at a rate that is proportional to the square of the amount present, and suppose that at time t = 0, the amount present is y0 . Find an initial-value problem whose solution is y(t). (b) Suppose that a quantity y = y(t) decreases at a rate that is proportional to the square of the amount present, and suppose that at a time t = 0, the amount present is y0 . Find an initial-value problem whose solution is y(t). 28. (a) Suppose that a quantity y = y(t) changes in such √ a way that dy/dt = k y, where k > 0. Describe how y changes in words.

567

(b) Suppose that a quantity y = y(t) changes in such a way that dy/dt = −ky 3 , where k > 0. Describe how y changes in words. 29. (a) Suppose that a particle moves along an s-axis in such a way that its velocity v(t) is always half of s(t). Find a differential equation whose solution is s(t). (b) Suppose that an object moves along an s-axis in such a way that its acceleration a(t) is always twice the velocity. Find a differential equation whose solution is s(t). 30. Suppose that a body moves along an s-axis through a resistive medium in such a way that the velocity v = v(t) decreases at a rate that is twice the square of the velocity. (a) Find a differential equation whose solution is the velocity v(t). (b) Find a differential equation whose solution is the position s(t). 31. Consider a solution y = y(t) to the uninhibited population growth model. (a) Use Equation (3) to explain why y will be an increasing function of t. (b) Use Equation (3) to explain why the graph y = y(t) will be concave up. 32. Consider the logistic model for population growth. (a) Explain why there are two constant solutions to this model. (b) For what size of the population will the population be growing most rapidly? 33. Consider the model for the spread of disease. (a) Explain why there are two constant solutions to this model. (b) For what size of the infected population is the disease spreading most rapidly? 34. Explain why there is exactly one constant solution to the Newton’s Law of Cooling model. 35. Show that if c1 and c2 are any constants, the function k k t + c2 sin t x = x(t) = c1 cos m m is a solution to the differential equation for the vibrating spring. (The corresponding motion of the spring is referred to as simple harmonic motion.) 36. (a) Use the result of Exercise 35 to solve the initial-value problem in (8). (b) Find the amplitude, period, and frequency of your answer to part (a), and interpret each of these in terms of the motion of the spring. 37. Writing Select one of the models in this section and write a paragraph that discusses conditions under which the model would not be appropriate. How might you modify the model to take those conditions into account?

568

Chapter 8 / Mathematical Modeling with Differential Equations

✔QUICK CHECK ANSWERS 8.1 1. (a) (iv) (b) (iii) (c) (i) (d) (ii)

8.2

2. 2; e2x + 2xe2x

x dy = − , y(0) = 1 dx y

dT = k(T − 68), T (0) = 36 dt

SEPARATION OF VARIABLES In this section we will discuss a method, called “separation of variables,” that can be used to solve a large class of ﬁrst-order differential equations of a particular form. We will use this method to investigate mathematical models for exponential growth and decay, including population models and carbon dating.

Some writers define a separable equation to be one that can be written in the form dy /dx = G(x)H (y). Explain why this is equivalent to our definition.

FIRST-ORDER SEPARABLE EQUATIONS We will now consider a method of solution that can often be applied to ﬁrst-order equations that are expressible in the form dy (1) = g(x) h(y) dx Such ﬁrst-order equations are said to be separable. Some examples of separable equations are given in Table 8.2.1. The name “separable” arises from the fact that Equation (1) can be rewritten in the differential form

(2)

h(y) dy = g(x) dx

in which the expressions involving x and y appear on opposite sides. The process of rewriting (1) in form (2) is called separating variables. Table 8.2.1

form (1)

equation

h(y)

g(x)

dy = x 2 y3 dx

dy =x dx 1 dy = x2 y3 dx

dy =y dx

1 dy =1 y dx

1 y

y dy =y− x dx

1 dy = 1 − 1x y dx

1 y

1 − 1x

dy x = dx y

1 y3

To motivate a method for solving separable equations, assume that h(y) and g(x) are continuous functions of their respective variables, and let H (y) and G(x) denote antiderivatives of h(y) and g(x), respectively. Consider the equation that results if we integrate both sides of (2), the left side with respect to y and the right side with respect to x. We then have h(y) dy = g(x) dx (3) or, equivalently,

H (y) = G(x) + C

(4)

where C denotes a constant. We claim that a differentiable function y = y(x) is a solution to (1) if and only if y satisﬁes Equation (4) for some choice of the constant C.

8.2 Separation of Variables

569

Suppose that y = y(x) is a solution to (1). It then follows from the chain rule that dH dy dy dG d [H (y)] = = h(y) = g(x) = dx dy dx dx dx

(5)

Since the functions H (y) and G(x) have the same derivative with respect to x, they must differ by a constant (Theorem 4.8.3). It then follows that y satisfies (4) for an appropriate choice of C. Conversely, if y = y(x) is defined implicitly by Equation (4), then implicit differentiation shows that (5) is satisfied, and thus y(x) is a solution to (1) (Exercise 67). Because of this, it is common practice to refer to Equation (4) as the “solution” to (1). In summary, we have the following procedure for solving (1), called separation of variables:

Separation of Variables Step 1. Separate the variables in (1) by rewriting the equation in the differential form h(y) dy = g(x) dx Step 2. Integrate both sides of the equation in Step 1 (the left side with respect to y and the right side with respect to x): h(y) dy = g(x) dx

Step 3. If H (y) is any antiderivative of h(y) and G(x) is any antiderivative of g(x), then the equation H (y) = G(x) + C will generally deﬁne a family of solutions implicitly. In some cases it may be possible to solve this equation explicitly for y.

Example 1

Solve the differential equation

dy = −4xy 2 dx and then solve the initial-value problem For an initial-value problem in which the differential equation is separable, you can either use the initial condition to solve for C , as in Example 1, or replace the indefinite integrals in Step 2 by definite integrals (Exercise 68).

dy = −4xy 2 , dx

y(0) = 1

Solution. For y = 0 we can write the differential equation in form (1) as 1 dy = −4x y 2 dx

Separating variables and integrating yields

1 dy = −4x dx y2 1 dy = −4x dx y2 −

1 = −2x 2 + C y

Solving for y as a function of x, we obtain y=

1 2x 2 − C

570

Chapter 8 / Mathematical Modeling with Differential Equations y

The initial condition y(0) = 1 requires that y = 1 when x = 0. Substituting these values into our solution yields C = −1 (verify). Thus, a solution to the initial-value problem is

2 1 x −2

1 (6) 2x 2 + 1 Some integral curves and our solution of the initial-value problem are graphed in Figure 8.2.1. y=

−1

One aspect of our solution to Example 1 deserves special comment. Had the initial condition been y(0) = 0 instead of y(0) = 1, the method we used would have failed to yield a solution to the resulting initial-value problem (Exercise 25). This is due to the fact that we assumed y = 0 in order to rewrite the equation dy/dx = −4xy 2 in the form

dy = −4xy 2 dx

Integral curves for

Figure 8.2.1

1 dy = −4x y 2 dx

It is important to be aware of such assumptions when manipulating a differential equation algebraically.

Example 2

Solve the initial-value problem (4y − cos y)

dy − 3x 2 = 0, dx

y(0) = 0

Solution. We can write the differential equation in form (1) as The solution of an initial-value problem in x and y can sometimes be expressed explicitly as a function of x [as in Formula (6) of Example 1], or explicitly as a function of y [as in Formula (8) of Example 2]. However, sometimes the solution cannot be expressed in either such form, so the only option is to express it implicitly as an equation in x and y .

(4y − cos y)

dy = 3x 2 dx

Separating variables and integrating yields

(4y − cos y) dy = 3x 2 dx (4y − cos y) dy = 3x 2 dx 2y 2 − sin y = x 3 + C

For the initial-value problem, the initial condition y(0) = 0 requires that y = 0 if x = 0. Substituting these values into (7) to determine the constant of integration yields C = 0 (verify). Thus, the solution of the initial-value problem is

y 3 2

2y 2 − sin y = x 3

1 x −2

−1

−1 −2 −3 Integral curves for

(4y − cos y) Figure 8.2.2

(7)

dy − 3x 2 = 0 dx

or x=

2y 2 − sin y

(8)

Some integral curves and the solution of the initial-value problem in Example 2 are graphed in Figure 8.2.2. Initial-value problems often result from geometrical questions, as in the following example. Example 3 Find a curve in the xy-plane that passes through (0, 3) and whose tangent line at a point (x, y) has slope 2x /y 2 .

8.2 Separation of Variables T E C H N O LO GY M A ST E R Y Some computer algebra systems can graph implicit equations. Figure 8.2.2 shows the graphs of (7) for C = 0, ±1, ±2, and ±3. If you have a CAS that can graph implicit equations, try to duplicate this figure.

571

Solution. Since the slope of the tangent line is dy/dx, we have 2x dy = 2 dx y

(9)

and, since the curve passes through (0, 3), we have the initial condition y(0) = 3 Equation (9) is separable and can be written as y 2 dy = 2x dx so

y 2 dy =

2x dx

1 3 y 3

= x2 + C

It follows from the initial condition that y = 3 if x = 0. Substituting these values into the last equation yields C = 9 (verify), so the equation of the desired curve is 1 3 y 3

= x2 + 9

or y = (3x 2 + 27)1/3

EXPONENTIAL GROWTH AND DECAY MODELS The population growth and pharmacology models developed in Section 8.1 are examples of a general class of models called exponential models. In general, exponential models arise in situations where a quantity increases or decreases at a rate that is proportional to the amount of the quantity present. More precisely, we make the following deﬁnition.

8.2.1 definition A quantity y = y(t) is said to have an exponential growth model if it increases at a rate that is proportional to the amount of the quantity present, and it is said to have an exponential decay model if it decreases at a rate that is proportional to the amount of the quantity present. Thus, for an exponential growth model, the quantity y(t) satisﬁes an equation of the form dy (10) = ky (k > 0) dt and for an exponential decay model, the quantity y(t) satisﬁes an equation of the form dy = −ky (k > 0) (11) dt The constant k is called the growth constant or the decay constant, as appropriate.

Equations (10) and (11) are separable since they have the form of (1), but with t rather than x as the independent variable. To illustrate how these equations can be solved, suppose that a positive quantity y = y(t) has an exponential growth model and that we know the amount of the quantity at some point in time, say y = y0 when t = 0. Thus, a formula for y(t) can be obtained by solving the initial-value problem dy = ky, y(0) = y0 dt Separating variables and integrating yields 1 dy = k dt y or (since y > 0)

ln y = kt + C

(12)

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Chapter 8 / Mathematical Modeling with Differential Equations

The initial condition implies that y = y0 when t = 0. Substituting these values in (12) yields C = ln y0 (verify). Thus, ln y = kt + ln y0 from which it follows that

y = eln y = ekt+ln y0

or, equivalently,

y = y0 ekt

(13)

We leave it for you to show that if y = y(t) has an exponential decay model, and if y(0) = y0 , then y = y0 e−kt (14) INTERPRETING THE GROWTH AND DECAY CONSTANTS The signiﬁcance of the constant k in Formulas (13) and (14) can be understood by reexamining the differential equations that gave rise to these formulas. For example, in the case of the exponential growth model, Equation (10) can be rewritten as

k= It is standard practice in applications to call (15) the growth rate, even though it is misleading (the growth rate is dy/dt ). However, the practice is so common that we will follow it here.

dy/dt y

(15)

which states that the growth rate as a fraction of the entire population remains constant over time, and this constant is k. For this reason, k is called the relative growth rate of the population. It is usual to express the relative growth rate as a percentage. Thus, a relative growth rate of 3% per unit of time in an exponential growth model means that k = 0.03. Similarly, the constant k in an exponential decay model is called the relative decay rate. Example 4 According to the U.S. Census Bureau, the world population in 2011 was 6.9 billion and growing at a rate of about 1.10% per year. Assuming an exponential growth model, estimate the world population at the beginning of the year 2030.

Solution. We assume that the population at the beginning of 2011 was 6.9 billion and let t = time elapsed from the beginning of 2011 (in years) y = world population (in billions) Since the beginning of 2011 corresponds to t = 0, it follows from the given data that y0 = y(0) = 6.9 billion

In Example 4 the growth rate was given, so there was no need to calculate it. If the growth rate or decay rate is unknown, then it can be calculated using the initial condition and the value of y at another point in time (Exercise 44).

Since the growth rate is 1.10% (k = 0.011), it follows from (13) that the world population at time t will be y(t) = y0 ekt = 6.9e0.011t (16) Since the beginning of the year 2030 corresponds to an elapsed time of t = 19 years (2030 − 2011 = 19), it follows from (16) that the world population by the year 2030 will be y(19) = 6.9e0.011(19) ≈ 8.5 which is a population of approximately 8.5 billion. DOUBLING TIME AND HALF-LIFE

If a quantity y has an exponential growth model, then the time required for the original size to double is called the doubling time, and if y has an exponential decay model, then the time required for the original size to reduce by half is called the half-life. As it turns out, doubling time and half-life depend only on the growth or decay rate and not on the amount present initially. To see why this is so, suppose that y = y(t) has an exponential growth model y = y0 ekt (17)

8.2 Separation of Variables y

573

and let T denote the amount of time required for y to double in size. Thus, at time t = T the value of y will be 2y0 , and hence from (17)

8y0

2y0 = y0 ekT

or ekT = 2

Taking the natural logarithm of both sides yields kT = ln 2, which implies that the doubling time is 1 T = ln 2 (18) k

4y0 2y0 y0

Exponential growth model with doubling time T

We leave it as an exercise to show that Formula (18) also gives the half-life of an exponential decay model. Observe that this formula does not involve the initial amount y0 , so that in an exponential growth or decay model, the quantity y doubles (or reduces by half ) every T units (Figure 8.2.3). Example 5 It follows from (18) that with a continued growth rate of 1.10% per year, the doubling time for the world population will be

1 ln 2 ≈ 63 0.011 or approximately 63 years. Thus, with a continued 1.10% annual growth rate the population of 6.9 billion in 2011 will double to 13.8 billion by the year 2074 and will double again to 27.6 billion by 2137. T =

y0 /2

RADIOACTIVE DECAY

y0 /4 y0 /8

Exponential decay model with half-life T

Figure 8.2.3

It is a fact of physics that radioactive elements disintegrate spontaneously in a process called radioactive decay. Experimentation has shown that the rate of disintegration is proportional to the amount of the element present, which implies that the amount y = y(t) of a radioactive element present as a function of time has an exponential decay model. Every radioactive element has a speciﬁc half-life; for example, the half-life of radioactive carbon-14 is about 5730 years. Thus, from (18), the decay constant for this element is 1 ln 2 ln 2 = ≈ 0.000121 T 5730 and this implies that if there are y0 units of carbon-14 present at time t = 0, then the number of units present after t years will be approximately k=

y(t) = y0 e−0.000121t

(19)

Example 6 If 100 grams of radioactive carbon-14 are stored in a cave for 1000 years, how many grams will be left at that time?

Solution. From (19) with y0 = 100 and t = 1000, we obtain

y(1000) = 100e−0.000121(1000) = 100e−0.121 ≈ 88.6

Thus, about 88.6 grams will be left. CARBON DATING When the nitrogen in the Earth’s upper atmosphere is bombarded by cosmic radiation, the radioactive element carbon-14 is produced. This carbon-14 combines with oxygen to form carbon dioxide, which is ingested by plants, which in turn are eaten by animals. In this way all living plants and animals absorb quantities of radioactive carbon-14. In 1947 ∗ the American nuclear scientist W. F. Libby proposed the theory that the percentage of

∗

W. F. Libby, “Radiocarbon Dating,” American Scientist, Vol. 44, 1956, pp. 98–112.

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Chapter 8 / Mathematical Modeling with Differential Equations

carbon-14 in the atmosphere and in living tissues of plants is the same. When a plant or animal dies, the carbon-14 in the tissue begins to decay. Thus, the age of an artifact that contains plant or animal material can be estimated by determining what percentage of its original carbon-14 content remains. Various procedures, called carbon dating or carbon-14 dating, have been developed for measuring this percentage.

Example 7 In 1988 the Vatican authorized the British Museum to date a cloth relic known as the Shroud of Turin, possibly the burial shroud of Jesus of Nazareth. This cloth, which ﬁrst surfaced in 1356, contains the negative image of a human body that was widely believed to be that of Jesus. The report of the British Museum showed that the ﬁbers in the cloth contained between 92% and 93% of their original carbon-14. Use this information to estimate the age of the shroud.

Solution. From (19), the fraction of the original carbon-14 that remains after t years is y(t) = e−0.000121t y0 Taking the natural logarithm of both sides and solving for t, we obtain 1 y(t) t =− ln 0.000121 y0 Thus, taking y(t)/y0 to be 0.93 and 0.92, we obtain t =−

Patrick Mesner/Liaison Agency, Inc./Getty Images

The Shroud of Turin

1 ln(0.93) ≈ 600 0.000121

1 ln(0.92) ≈ 689 0.000121 This means that when the test was done in 1988, the shroud was between 600 and 689 years old, thereby placing its origin between 1299 A.D. and 1388 A.D. Thus, if one accepts the validity of carbon-14 dating, the Shroud of Turin cannot be the burial shroud of Jesus of Nazareth. t =−

✔QUICK CHECK EXERCISES 8.2

(See page 579 for answers.)

1. Solve the ﬁrst-order separable equation dy = g(x) dx by completing the following steps: h(y)

Step 1. Separate the variables by writing the equation in the differential form . Step 2. Integrate both sides of the equation in Step 1: . Step 3. If H (y) is any antiderivative of h(y), G(x) is any antiderivative of g(x), and C is an unspeciﬁed constant, then, as suggested by Step 2, the equation will generally deﬁne a family of solutions to h(y) dy/dx = g(x) implicitly.

2. Suppose that a quantity y = y(t) has an exponential growth model with growth constant k > 0. (a) y(t) satisfies a first-order differential equation of the form dy/dt = . (b) In terms of k, the doubling time of the quantity is . (c) If y0 = y(0) is the initial amount of the quantity, then an explicit formula for y(t) is given by y(t) = . 3. Suppose that a quantity y = y(t) has an exponential decay model with decay constant k > 0. (a) y(t) satisfies a first-order differential equation of the form dy/dt = . (b) In terms of k, the half-life of the quantity is . (c) If y0 = y(0) is the initial amount of the quantity, then an explicit formula for y(t) is given by y(t) = .

8.2 Separation of Variables

575

4. The initial-value problem dy x = − , y(0) = 1 dx y has solution y(x) = .

EXERCISE SET 8.2

Graphing Utility

CAS

1–10 Solve the differential equation by separation of variables. Where reasonable, express the family of solutions as explicit functions of x. ■ dy y dy 1. = 2. = 2(1 + y 2 )x dx x dx √ dy x3 1 + x 2 dy = −x 4. (1 + x 4 ) = 3. 1 + y dx dx y 5. (2 + 2y 2 )y ′ = ex y 6. y ′ = −xy −y

′

7. e sin x − y cos x = 0 8. y − (1 + x)(1 + y ) = 0 dy y2 − y dy 9. − =0 10. y − sec x = 0 dx sin x dx

11–14 Solve the initial-value problem by separation of vari-

ables. ■ 3x 2 , y(0) = π 2y + cos y 12. y ′ − xey = 2ey , y(0) = 0 2t + 1 dy 13. = , y(0) = −1 dt 2y − 2 14. y ′ cosh2 x − y cosh 2x = 0, y(0) = 3 11. y ′ =

23. If a radioactive element has a half-life of 1 minute, and if a container holds 32 g of the element at 1:00 p.m., then the amount remaining at 1:05 p.m. will be 1 g. 24. If a population is growing exponentially, then the time it takes the population to quadruple is independent of the size of the population. 25. Suppose that the initial condition in Example 1 had been y(0) = 0. Show that none of the solutions generated in Example 1 satisfy this initial condition, and then solve the initial-value problem dy = −4xy 2 , y(0) = 0 dx Why does the method of Example 1 fail to produce this particular solution? 26. Find all ordered pairs (x0 , y0 ) such that if the initial condition in Example 1 is replaced by y(x0 ) = y0 , the solution of the resulting initial-value problem is deﬁned for all real numbers.

15. (a) Sketch some typical integral curves of the differential equation y ′ = y /2x. (b) Find an equation for the integral curve that passes through the point (2, 1).

27. Find an equation of a curve with x-intercept 2 whose tangent line at any point (x, y) has slope xe−y .

16. (a) Sketch some typical integral curves of the differential equation y ′ = −x /y. (b) Find an equation for the integral curve that passes through the point (3, 4).

28. Use a graphing utility to generate a curve that passes through the point (1, 1) and whose tangent line at (x, y) is perpendicular to the line through (x, y) with slope −2y /(3x 2 ).

17–18 Solve the differential equation and then use a graphing

utility to generate ﬁve integral curves for the equation. ■ dy 17. (x 2 + 4) + xy = 0 18. (cos y)y ′ = cos x dx C

22. A differential equation of the form dy h(x) = g(y) dx is not separable.

19–20 Solve the differential equation. If you have a CAS with implicit plotting capability, use the CAS to generate ﬁve integral curves for the equation. ■ x2 y 19. y ′ = 20. y ′ = 1 − y2 1 + y2 21–24 True–False Determine whether the statement is true or false. Explain your answer. ■

21. Every differential equation of the form y ′ = f(y) is separable.

29. Suppose that an initial population of 10,000 bacteria grows exponentially at a rate of 2% per hour and that y = y(t) is the number of bacteria present t hours later. (a) Find an initial-value problem whose solution is y(t). (b) Find a formula for y(t). (c) How long does it take for the initial population of bacteria to double? (d) How long does it take for the population of bacteria to reach 45,000?

30. A cell of the bacterium E. coli divides into two cells every 20 minutes when placed in a nutrient culture. Let y = y(t) be the number of cells that are present t minutes after a single cell is placed in the culture. Assume that the growth of the bacteria is approximated by an exponential growth model. (a) Find an initial-value problem whose solution is y(t). (b) Find a formula for y(t). (cont.)

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Chapter 8 / Mathematical Modeling with Differential Equations

(c) How many cells are present after 2 hours? (d) How long does it take for the number of cells to reach 1,000,000? 31. Radon-222 is a radioactive gas with a half-life of 3.83 days. This gas is a health hazard because it tends to get trapped in the basements of houses, and many health officials suggest that homeowners seal their basements to prevent entry of the gas. Assume that 5.0 × 107 radon atoms are trapped in a basement at the time it is sealed and that y(t) is the number of atoms present t days later. (a) Find an initial-value problem whose solution is y(t). (b) Find a formula for y(t). (c) How many atoms will be present after 30 days? (d) How long will it take for 90% of the original quantity of gas to decay? 32. Polonium-210 is a radioactive element with a half-life of 140 days. Assume that 10 milligrams of the element are placed in a lead container and that y(t) is the number of milligrams present t days later. (a) Find an initial-value problem whose solution is y(t). (b) Find a formula for y(t). (c) How many milligrams will be present after 10 weeks? (d) How long will it take for 70% of the original sample to decay? 33. Suppose that 100 fruit flies are placed in a breeding container that can support at most 10,000 flies. Assuming that the population grows exponentially at a rate of 2% per day, how long will it take for the container to reach capacity? 34. Suppose that the town of Grayrock had a population of 10,000 in 2006 and a population of 12,000 in 2011. Assuming an exponential growth model, in what year will the population reach 20,000? 35. A scientist wants to determine the half-life of a certain radioactive substance. She determines that in exactly 5 days a 10.0-milligram sample of the substance decays to 3.5 milligrams. Based on these data, what is the half-life? 36. Suppose that 30% of a certain radioactive substance decays in 5 years. (a) What is the half-life of the substance in years? (b) Suppose that a certain quantity of this substance is stored in a cave. What percentage of it will remain after t years? F O C U S O N C O N C E P TS

37. (a) Make a conjecture about the effect on the graphs of y = y0 ekt and y = y0 e−kt of varying k and keeping y0 fixed. Confirm your conjecture with a graphing utility. (b) Make a conjecture about the effect on the graphs of y = y0 ekt and y = y0 e−kt of varying y0 and keeping k fixed. Confirm your conjecture with a graphing utility.

38. (a) What effect does increasing y0 and keeping k ﬁxed have on the doubling time or half-life of an exponential model? Justify your answer. (b) What effect does increasing k and keeping y0 ﬁxed have on the doubling time and half-life of an exponential model? Justify your answer. 39. (a) There is a trick, called the Rule of 70, that can be used to get a quick estimate of the doubling time or half-life of an exponential model. According to this rule, the doubling time or half-life is roughly 70 divided by the percentage growth or decay rate. For example, we showed in Example 5 that with a continued growth rate of 1.10% per year the world population would double every 63 years. This result agrees with the Rule of 70, since 70/1.10 ≈ 63.6. Explain why this rule works. (b) Use the Rule of 70 to estimate the doubling time of a population that grows exponentially at a rate of 1% per year. (c) Use the Rule of 70 to estimate the half-life of a population that decreases exponentially at a rate of 3.5% per hour. (d) Use the Rule of 70 to estimate the growth rate that would be required for a population growing exponentially to double every 10 years. 40. Find a formula for the tripling time of an exponential growth model. 41. In 1950, a research team digging near Folsom, New Mexico, found charred bison bones along with some leaf-shaped projectile points (called the “Folsom points”) that had been made by a Paleo-Indian hunting culture. It was clear from the evidence that the bison had been cooked and eaten by the makers of the points, so that carbon-14 dating of the bones made it possible for the researchers to determine when the hunters roamed North America. Tests showed that the bones contained between 27% and 30% of their original carbon14. Use this information to show that the hunters lived roughly between 9000 b.c. and 8000 b.c. 42. (a) Use a graphing utility to make a graph of prem versus t, where prem is the percentage of carbon-14 that remains in an artifact after t years. (b) Use the graph to estimate the percentage of carbon-14 that would have to have been present in the 1988 test of the Shroud of Turin for it to have been the burial shroud of Jesus of Nazareth (see Example 7). 43. (a) It is currently accepted that the half-life of carbon-14 might vary ±40 years from its nominal value of 5730 years. Does this variation make it possible that the Shroud of Turin dates to the time of Jesus of Nazareth (see Example 7)? (b) Review the subsection of Section 3.5 entitled Error Propagation, and then estimate the percentage error that

8.2 Separation of Variables

577

results in the computed age of an artifact from an r% error in the half-life of carbon-14.

48. What is the effective annual interest rate for an interest rate of r% per year compounded continuously?

44. Suppose that a quantity y has an exponential growth model y = y0 ekt or an exponential decay model y = y0 e−kt , and it is known that y = y1 if t = t1 . In each case ﬁnd a formula for k in terms of y0 , y1 , and t1 , assuming that t1 = 0.

49. Assume that y = y(t) satisﬁes the logistic equation with y0 = y(0) the initial value of y. (a) Use separation of variables to derive the solution y0 L y= y0 + (L − y0 )e−kt (b) Use part (a) to show that lim y(t) = L.

45. (a) Show that if a quantity y = y(t) has an exponential model, and if y(t1 ) = y1 and y(t2 ) = y2 , then the doubling time or the half-life T is (t2 − t1 ) ln 2 T = ln(y2 /y1 )

(b) In a certain 1-hour period the number of bacteria in a colony increases by 25%. Assuming an exponential growth model, what is the doubling time for the colony?

46. Suppose that P dollars is invested at an annual interest rate of r × 100%. If the accumulated interest is credited to the account at the end of the year, then the interest is said to be compounded annually; if it is credited at the end of each 6-month period, then it is said to be compounded semiannually; and if it is credited at the end of each 3-month period, then it is said to be compounded quarterly. The more frequently the interest is compounded, the better it is for the investor since more of the interest is itself earning interest. (a) Show that if interest is compounded n times a year at equally spaced intervals, then the value A of the investment after t years is r nt A=P 1+ n (b) One can imagine interest to be compounded each day, each hour, each minute, and so forth. Carried to the limit one can conceive of interest compounded at each instant of time; this is called continuous compounding. Thus, from part (a), the value A of P dollars after t years when invested at an annual rate of r × 100%, compounded continuously, is r nt A = lim P 1 + n → +⬁ n

Use the fact that limx → 0 (1 + x)1/x = e to prove that A = P ert . (c) Use the result in part (b) to show that money invested at continuous compound interest increases at a rate proportional to the amount present.

47. (a) If $1000 is invested at 8% per year compounded continuously (Exercise 46), what will the investment be worth after 5 years? (b) If it is desired that an investment at 8% per year compounded continuously should have a value of $10,000 after 10 years, how much should be invested now? (c) How long does it take for an investment at 8% per year compounded continuously to double in value?

t → +⬁

50. Use your answer to Exercise 49 to derive a solution to the model for the spread of disease [Equation (6) of Section 8.1]. 51. The graph of a solution to the logistic equation is known as a logistic curve, and if y0 > 0, it has one of four general shapes, depending on the relationship between y0 and L. In each part, assume that k = 1 and use a graphing utility to plot a logistic curve satisfying the given condition. (a) y0 > L (b) y0 = L (c) L/2 ≤ y0 < L (d) 0 < y0 < L/2 52–53 The graph of a logistic model

y0 L y0 + (L − y0 )e−kt is shown. Estimate y0 , L, and k. ■ y=

52.

1000

53.

600

6 4

200

t 200

600

1000

t 2

8 10

54. Plot a solution to the initial-value problem dy y y, y0 = 1 = 0.98 1 − dt 5 55. Suppose that the growth of a population y = y(t) is given by the logistic equation 60 y= 5 + 7e−t (a) What is the population at time t = 0? (b) What is the carrying capacity L? (c) What is the constant k? (d) When does the population reach half of the carrying capacity? (e) Find an initial-value problem whose solution is y(t). 56. Suppose that the growth of a population y = y(t) is given by the logistic equation 1000 y= 1 + 999e−0.9t (a) What is the population at time t = 0? (b) What is the carrying capacity L? (c) What is the constant k?

(cont.)

578

Chapter 8 / Mathematical Modeling with Differential Equations

(d) When does the population reach 75% of the carrying capacity? (e) Find an initial-value problem whose solution is y(t). 57. Suppose that a university residence hall houses 1000 students. Following the semester break, 20 students in the hall return with the flu, and 5 days later 35 students have the flu. (a) Use the result of Exercise 50 to find the number of students who will have the flu t days after returning to school. (b) Make a table that illustrates how the flu spreads day to day over a 2-week period. (c) Use a graphing utility to generate a graph that illustrates how the flu spreads over a 2-week period. 58. Suppose that at time t = 0 an object with temperature T0 is placed in a room with constant temperature Ta . If T0 < Ta , then the temperature of the object will increase, and if T0 > Ta , then the temperature will decrease. Assuming that Newton’s Law of Cooling applies, show that in both cases the temperature T (t) at time t is given by

62. A bullet of mass m, ﬁred straight up with an initial velocity of v0 , is slowed by the force of gravity and a drag force of air resistance kv 2 , where k is a positive constant. As the bullet moves upward, its velocity v satisﬁes the equation dv = −(kv 2 + mg) dt where g is the constant acceleration due to gravity. (a) Show that if x = x(t) is the height of the bullet above the barrel opening at time t, then m

dv = −(kv 2 + mg) dx (b) Express x in terms of v given that x = 0 when v = v0 . (c) Assuming that mv

v0 = 988 m/s, g = 9.8 m/s2

m = 3.56 × 10−3 kg, k = 7.3 × 10−6 kg/m use the result in part (b) to ﬁnd out how high the bullet rises. [Hint: Find the velocity of the bullet at its highest point.]

T (t) = Ta + (T0 − Ta )e−kt where k is a positive constant. ◦

59. A cup of water with a temperature of 95 C is placed in a room with a constant temperature of 21 ◦ C. (a) Assuming that Newton’s Law of Cooling applies, use the result of Exercise 58 to ﬁnd the temperature of the water t minutes after it is placed in the room. [Note: The solution will involve a constant of proportionality.] (b) How many minutes will it take for the water to reach a temperature of 51 ◦ C if it cools to 85 ◦ C in 1 minute? 60. A glass of lemonade with a temperature of 40 ◦ F is placed in a room with a constant temperature of 70 ◦ F, and 1 hour later its temperature is 52 ◦ F. Show that t hours after the lemonade is placed in the room its temperature is approximated by T = 70 − 30e−0.5t . 61. A rocket, ﬁred upward from rest at time t = 0, has an initial mass of m0 (including its fuel). Assuming that the fuel is consumed at a constant rate k, the mass m of the rocket, while fuel is being burned, will be given by m = m0 − kt. It can be shown that if air resistance is neglected and the fuel gases are expelled at a constant speed c relative to the rocket, then the velocity v of the rocket will satisfy the equation m

63–64 Suppose that a tank containing a liquid is vented to the air at the top and has an outlet at the bottom through which the liquid can drain. It follows from Torricelli’s law in physics that if the outlet is opened at time t = 0, then at each instant the depth of the liquid h(t) and the area A(h) of the liquid’s surface are related by √ dh A(h) = −k h dt where k is a positive constant that depends on such factors as the viscosity of the liquid and the cross-sectional area of the outlet. Use this result in these exercises, assuming that h is in feet, A(h) is in square feet, and t is in seconds. ■

63. Suppose that the cylindrical tank in the accompanying figure is filled to a depth of 4 feet at time t = 0 and that the constant in Torricelli’s law is k = 0.025. (a) Find h(t). (b) How many minutes will it take for the tank to drain completely? 64. Follow the directions of Exercise 63 for the cylindrical tank in the accompanying figure, assuming that the tank is filled to a depth of 4 feet at time t = 0 and that the constant in Torricelli’s law is k = 0.025.

dv = ck − mg dt

where g is the acceleration due to gravity. (a) Find v(t) keeping in mind that the mass m is a function of t. (b) Suppose that the fuel accounts for 80% of the initial mass of the rocket and that all of the fuel is consumed in 100 s. Find the velocity of the rocket in meters per second at the instant the fuel is exhausted. [Note: Take g = 9.8 m/s2 and c = 2500 m/s.]

1 ft

6 ft

4 ft

Figure Ex-63

Figure Ex-64

8.3 Slope Fields; Euler’s Method

65. Suppose that a particle moving along the x-axis encounters a resisting force that results in an acceleration of 1 2 a = dv /dt = − 32 v . If x = 0 cm and v = 128 cm/s at time t = 0, ﬁnd the velocity v and position x as a function of t for t ≥ 0.

66. Suppose that a particle moving along the x-axis encounters a resisting force √ that results in an acceleration of a = dv /dt = −0.02 v. Given that x = 0 cm and v = 9 cm/s at time t = 0, ﬁnd the velocity v and position x as a function of t for t ≥ 0. F O C U S O N C O N C E P TS

67. Use implicit differentiation to prove that any differentiable function deﬁned implicitly by Equation (4) will be a solution to (1). 68. Prove that a solution to the initial-value problem dy h(y) = g(x), y(x0 ) = y0 dx

579

is deﬁned implicitly by the equation x y g(s) ds h(r) dr = x0

69. Let L denote a tangent line at (x, y) to a solution of Equation (1), and let (x1 , y1 ), (x2 , y2 ) denote any two points on L. Prove that Equation (2) is satisﬁed by dy = y = y2 − y1 and dx = x = x2 − x1 . 70. Writing A student objects to the method of separation of variables because it often produces an equation in x and y instead of an explicit function y = f(x). Discuss the pros and cons of this student’s position. 71. Writing A student objects to Step 2 in the method of separation of variables because one side of the equation is integrated with respect to x while the other side is integrated with respect to y. Answer this student’s objection. [Hint: Recall the method of integration by substitution.]

✔QUICK CHECK ANSWERS 8.2 1. Step 1: h(y) dy = g(x) dx; Step 2: 3. (a) −ky (b)

8.3

ln 2 (c) y0 e−kt k

h(y) dy = √ 4. y = 1 − x 2

g(x) dx; Step 3: H (y) = G(x) + C

2. (a) ky (b)

ln 2 (c) y0 ekt k

SLOPE FIELDS; EULER’S METHOD In this section we will reexamine the concept of a slope ﬁeld and we will discuss a method for approximating solutions of ﬁrst-order equations numerically. Numerical approximations are important in cases where the differential equation cannot be solved exactly.

FUNCTIONS OF TWO VARIABLES We will be concerned here with ﬁrst-order equations that are expressed with the derivative by itself on one side of the equation. For example,

y′ = x3 In applied problems involving time, it is usual to use t as the independent variable, in which case one would be concerned with equations of the form y ′ = f(t, y), where y ′ = dy/dt .

and y ′ = sin(xy)

The ﬁrst of these equations involves only x on the right side, so it has the form y ′ = f(x). However, the second equation involves both x and y on the right side, so it has the form y ′ = f(x, y), where the symbol f(x, y) stands for a function of the two variables x and y. Later in the text we will study functions of two variables in more depth, but for now it will sufﬁce to think of f(x, y) as a formula that produces a unique output when values of x and y are given as inputs. For example, if f(x, y) = x 2 + 3y

580

Chapter 8 / Mathematical Modeling with Differential Equations

and if the inputs are x = 2 and y = −4, then the output is f(2, −4) = 22 + 3(−4) = 4 − 12 = −8

SLOPE FIELDS y Slope = f (x, y)

(x, y) x

In Section 5.2 we introduced the concept of a slope ﬁeld in the context of differential equations of the form y ′ = f(x); the same principles apply to differential equations of the form y ′ = f(x, y) To see why this is so, let us review the basic idea. If we interpret y ′ as the slope of a tangent line, then the differential equation states that at each point (x, y) on an integral curve, the slope of the tangent line is equal to the value of f at that point (Figure 8.3.1). For example, suppose that f(x, y) = y − x, in which case we have the differential equation y′ = y − x

At each point (x, y) on an integral curve of y′ = f (x, y), the tangent line has slope f (x, y).

(1)

A geometric description of the set of integral curves can be obtained by choosing a rectangular grid of points in the xy-plane, calculating the slopes of the tangent lines to the integral curves at the gridpoints, and drawing small segments of the tangent lines through those points. The resulting picture is called a slope field or a direction field for the differential equation because it shows the “slope” or “direction” of the integral curves at the gridpoints. The more gridpoints that are used, the better the description of the integral curves. For example, Figure 8.3.2 shows two slope fields for (1)—the first was obtained by hand calculation using the 49 gridpoints shown in the accompanying table, and the second, which gives a clearer picture of the integral curves, was obtained using 625 gridpoints and a CAS.

Figure 8.3.1

values of f (x, y) = y − x y = −3 y = −2 y = −1 y = 0

y=1

y=2

y=3

x = −3

x = −2

−1

x = −1

−2

−1

x=0

−3

−2

−1

x=1

−4

−3

−2

−1

x=2

−5

−4

−3

−2

−1

−2

x=3

−6

−5

−4

−3

−2

−1

−3

-3

1 x

−3

−2

−1

x −3

−2

−1

Figure 8.3.2

It so happens that Equation (1) can be solved exactly using a method we will introduce in Section 8.4. We leave it for you to conﬁrm that the general solution of this equation is y = x + 1 + Cex Confirm that the first slope field in Figure 8.3.2 is consistent with the accompanying table in that figure.

(2)

Figure 8.3.3 shows some of the integral curves superimposed on the slope field. Note that it was not necessary to have the general solution to construct the slope field. Indeed, slope fields are important precisely because they can be constructed in cases where the differential equation cannot be solved exactly.

8.3 Slope Fields; Euler’s Method y

581

EULER’S METHOD

Consider an initial-value problem of the form

y ′ = f(x, y),

1 x −3

−2

−1

−1 −2 −3

Figure 8.3.3

y(x0 ) = y0

The slope ﬁeld for the differential equation y ′ = f(x, y) gives us a way to visualize the solution of the initial-value problem, since the graph of the solution is the integral curve that passes through the point (x0 , y0 ). The slope ﬁeld will also help us to develop a method for approximating the solution to the initial-value problem numerically. We will not attempt to approximate y(x) for all values of x; rather, we will choose some small increment x and focus on approximating the values of y(x) at a succession of x-values spaced x units apart, starting from x0 . We will denote these x-values by x1 = x0 + x,

x2 = x1 + x,

x3 = x2 + x,

x4 = x3 + x, . . .

and we will denote the approximations of y(x) at these points by y1 ≈ y(x1 ),

(x4, y4) (x3, y3) (x2, y2) (x1, y1)

(x0, y0)

Δx Figure 8.3.4

(xn+1, yn+1)

Slope = f (xn, yn )

(xn , yn ) Figure 8.3.5

y2 ≈ y(x2 ),

y3 ≈ y(x3 ),

y4 ≈ y(x4 ), . . .

The technique that we will describe for obtaining these approximations is called Euler’s Method. Although there are better approximation methods available, many of them use Euler’s Method as a starting point, so the underlying concepts are important to understand. The basic idea behind Euler’s Method is to start at the known initial point (x0 , y0 ) and draw a line segment in the direction determined by the slope field until we reach the point (x1 , y1 ) with x-coordinate x1 = x0 + x (Figure 8.3.4). If x is small, then it is reasonable to expect that this line segment will not deviate much from the integral curve y = y(x), and thus y1 should closely approximate y(x1 ). To obtain the subsequent approximations, we repeat the process using the slope field as a guide at each step. Starting at the endpoint (x1 , y1 ), we draw a line segment determined by the slope field until we reach the point (x2 , y2 ) with x-coordinate x2 = x1 + x, and from that point we draw a line segment determined by the slope field to the point (x3 , y3 ) with x-coordinate x3 = x2 + x, and so forth. As indicated in Figure 8.3.4, this procedure produces a polygonal path that tends to follow the integral curve closely, so it is reasonable to expect that the y-values y2 , y3 , y4 , . . . will closely approximate y(x2 ), y(x3 ), y(x4 ), . . . . To explain how the approximations y1 , y2 , y3 , . . . can be computed, let us focus on a typical line segment. As indicated in Figure 8.3.5, assume that we have found the point (xn , yn ), and we are trying to determine the next point (xn+1 , yn+1 ), where xn+1 = xn + x. Since the slope of the line segment joining the points is determined by the slope field at the starting point, the slope is f(xn , yn ), and hence

yn+1 − yn

yn+1 − yn yn+1 − yn = = f(xn , yn ) xn+1 − xn x

Δx

which we can rewrite as yn+1 = yn + f(xn , yn ) x This formula, which is the heart of Euler’s Method, tells us how to use each approximation to compute the next approximation.

582

Chapter 8 / Mathematical Modeling with Differential Equations

Euler’s Method To approximate the solution of the initial-value problem y ′ = f(x, y),

y(x0 ) = y0

proceed as follows: Step 1. Choose a nonzero number x to serve as an increment or step size along the x-axis, and let x1 = x0 + x,

x2 = x1 + x,

x3 = x2 + x, . . .

Step 2. Compute successively y1 y2 y3 .. . yn+1

= y0 + f(x0 , y0 ) x = y1 + f(x1 , y1 ) x = y2 + f(x2 , y2 ) x = yn + f(xn , yn ) x

The numbers y1 , y2 , y3 , . . . in these equations are the approximations of y(x1 ), y(x2 ), y(x3 ), . . . . Example 1 Use Euler’s Method with a step size of 0.1 to make a table of approximate values of the solution of the initial-value problem over the interval 0 ≤ x ≤ 1.

y ′ = y − x,

y(0) = 2

(3)

Solution. In this problem we have f(x, y) = y − x, x0 = 0, and y0 = 2. Moreover, since the step size is 0.1, the x-values at which the approximate values will be obtained are x1 = 0.1,

x2 = 0.2,

The ﬁrst three approximations are

x3 = 0.3, . . . ,

x9 = 0.9,

x10 = 1

y1 = y0 + f(x0 , y0 ) x = 2 + (2 − 0)(0.1) = 2.2 y2 = y1 + f(x1 , y1 ) x = 2.2 + (2.2 − 0.1)(0.1) = 2.41 y3 = y2 + f(x2 , y2 ) x = 2.41 + (2.41 − 0.2)(0.1) = 2.631

Here is a way of organizing all 10 approximations rounded to ﬁve decimal places: euler's method for y′ = y − x, y(0) = 2 with Δx = 0.1 n

f (xn, yn )Δx

yn+1 = yn + f (xn, yn )Δx

0 1 2 3 4 5 6 7 8 9 10

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

2.00000 2.20000 2.41000 2.63100 2.86410 3.11051 3.37156 3.64872 3.94359 4.25795 4.59374

0.20000 0.21000 0.22100 0.23310 0.24641 0.26105 0.27716 0.29487 0.31436 0.33579 —

2.20000 2.41000 2.63100 2.86410 3.11051 3.37156 3.64872 3.94359 4.25795 4.59374 —

8.3 Slope Fields; Euler’s Method

583

Observe that each entry in the last column becomes the next entry in the third column. This is reminiscent of Newton’s Method in which each successive approximation is used to ﬁnd the next. ACCURACY OF EULER’S METHOD As a rule of thumb, the absolute error in an approximation produced by Euler’s Method is proportional to the step size. Thus, reducing the step size by half reduces the absolute and percentage errors by roughly half. However, reducing the step size increases the amount of computation, thereby increasing the potential for more roundoff error. Such matters are discussed in courses on differential equations or numerical analysis.

It follows from (3) and the initial condition y(0) = 2 that the exact solution of the initialvalue problem in Example 1 is y = x + 1 + ex Thus, in this case we can compare the approximate values of y(x) produced by Euler’s Method with decimal approximations of the exact values (Table 8.3.1). In Table 8.3.1 the absolute error is calculated as |exact value − approximation| and the percentage error as |exact value − approximation| × 100% |exact value| Table 8.3.1

Notice that the absolute error tends to increase as x moves away from x0 .

✔QUICK CHECK EXERCISES 8.3

exact solution

euler approximation

absolute error

percentage error

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

2.00000 2.20517 2.42140 2.64986 2.89182 3.14872 3.42212 3.71375 4.02554 4.35960 4.71828

2.00000 2.20000 2.41000 2.63100 2.86410 3.11051 3.37156 3.64872 3.94359 4.25795 4.59374

0.00000 0.00517 0.01140 0.01886 0.02772 0.03821 0.05056 0.06503 0.08195 0.10165 0.12454

0.00 0.23 0.47 0.71 0.96 1.21 1.48 1.75 2.04 2.33 2.64

(See page 586 for answers.) y

1. Match each differential equation with its slope ﬁeld. (a) y ′ = 2xy 2 (b) y ′ = e−y ′ (c) y = y (d) y ′ = 2xy y

III Figure Ex-1

2. The slope ﬁeld for y ′ = y /x at the 16 gridpoints (x, y), where x = −2, −1, 1, 2 and y = −2, −1, 1, 2 is shown in

Chapter 8 / Mathematical Modeling with Differential Equations

584

the accompanying figure. Use this slope field and geometric reasoning to find the integral curve that passes through the point (1, 2).

3. When using Euler’s Method on the initial-value problem y ′ = f(x, y), y(x0 ) = y0 , we obtain yn+1 from yn , xn , and x by means of the formula yn+1 = .

4. Consider the initial-value problem y ′ = y, y(0) = 1. (a) Use Euler’s Method with two steps to approximate y(1). (b) What is the exact value of y(1)?

y 2 1 x −2

−1

−1 −2

Figure Ex-2

EXERCISE SET 8.3

Graphing Utility

1. Sketch the slope ﬁeld for y ′ = xy /4 at the 25 gridpoints (x, y), where x = −2, −1, . . . , 2 and y = −2, −1, . . . , 2.

x+y x−y

(e) y ′ =

2. Sketch the slope ﬁeld for y ′ + y = 2 at the 25 gridpoints (x, y), where x = 0, 1, . . . , 4 and y = 0, 1, . . . , 4.

(f ) y ′ = (sin x)(sin y) y

3. A slope field for the differential equation y ′ = 1 − y is shown in the accompanying figure. In each part, sketch the graph of the solution that satisfies the initial condition. (a) y(0) = −1 (b) y(0) = 1 (c) y(0) = 2

y 3 2

1 −3

−2

−1

x 3

−1 −2 −3

Figure Ex-3

4. Solve the initial-value problems in Exercise 3, and use a graphing utility to conﬁrm that the integral curves for these solutions are consistent with the sketches you obtained from the slope ﬁeld.

III

IV y

F O C U S O N C O N C E P TS

5. Use the slope field in Exercise 3 to make a conjecture about the behavior of the solutions of y ′ = 1 − y as x → +⬁, and confirm your conjecture by examining the general solution of the equation. 6. In parts (a)–(f ), match the differential equation with the slope field, and explain your reasoning. (a) y ′ = 1/x (b) y ′ = 1/y ′ −x 2 (c) y = e (d) y ′ = y 2 − 1

V Figure Ex-6

8.3 Slope Fields; Euler’s Method 7–10 Use Euler’s Method with the given step size x or t to

approximate the solution of the initial-value problem over the stated interval. Present your answer as a table and as a graph. ■

7. dy/dx =

√ 3

y, y(0) = 1, 0 ≤ x ≤ 4, x = 0.5

8. dy/dx = x − y 2 , y(0) = 1, 0 ≤ x ≤ 2, x = 0.25 9. dy/dt = cos y, y(0) = 1, 0 ≤ t ≤ 2, t = 0.5

10. dy/dt = e−y , y(0) = 0, 0 ≤ t ≤ 1, t = 0.1

11. Consider the initial-value problem y ′ = sin πt,

y(0) = 0

Use Euler’s Method with ﬁve steps to approximate y(1). 12–15 True–False Determine whether the statement is true or

false. Explain your answer. ■ 12. If the graph of y = f(x) is an integral curve for a slope field, then so is any vertical translation of this graph. 13. Every integral curve for the slope field dy /dx = exy is the graph of an increasing function of x. 14. Every integral curve for the slope field dy /dx = ey is concave up. 15. If p(y) is a cubic polynomial in y, then the slope field dy /dx = p(y) has an integral curve that is a horizontal line. F O C U S O N C O N C E P TS

16. (a) Show that the solution of the initial-value problem 2 y ′ = e−x , y(0) = 0 is x 2 e−t dt y(x) = 0

(b) Use Euler’s Method with x = 0.05 to approximate the value of 1 2 y(1) = e−t dt 0

and compare the answer to that produced by a calculating utility with a numerical integration capability.

17. The accompanying figure shows a slope field for the differential equation y ′ = −x /y. (a) Use the slope field to estimate y 21 for the solution that satisfies the given initial condition y(0) = 1. (b) Compare your estimate to the exact value of y 21 . y

2 1

x −2

−1

−1 −2

Figure Ex-17

585

18. Refer to slope field II in Quick Check Exercise 1. (a) Does the slope field appear to have a horizontal line as an integral curve? (b) Use the differential equation for the slope field to verify your answer to part (a). 19. Refer to the slope field in Exercise 3 and consider the integral curve through (0, −1). (a) Use the slope field to estimate where the integral curve intersects the x-axis. (b) Compare your estimate in part (a) with the exact value of the x-intercept for the integral curve. 20. Consider the initial-value problem √ dy y = , dx 2

y(0) = 1

(a) Use Euler’s Method with step sizes of x = 0.2, 0.1, and 0.05 to obtain three approximations of y(1). (b) Find y(1) exactly. 21. A slope field of the form y ′ = f(y) is said to be autonomous. (a) Explain why the tangent segments along any horizontal line will be parallel for an autonomous slope field. (b) The word autonomous means “independent.” In what sense is an autonomous slope field independent? (c) Suppose that G(y) is an antiderivative of 1/[f(y)] and that C is a constant. Explain why any differentiable function defined implicitly by G(y) − x = C will be a solution to the equation y ′ = f(y). √ 22. (a) Solve the equation y ′ = y and show that every nonconstant solution has a graph that is everywhere concave up. (b) Explain how the conclusion in part (a) may be ob√ tained directly from the equation y ′ = y without solving. 23. (a) Find a slope field whose integral curve through (1, 1) satisfies xy 3 − x 2 y = 0 by differentiating this equation implicitly. (b) Prove that if y(x) is any integral curve of the slope field in part (a), then x[y(x)]3 − x 2 y(x) will be a constant function. (c) Find an equation that implicitly defines the integral curve through (−1, −1) of the slope field in part (a).

24. (a) Find a slope field whose integral curve through (0, 0) satisfies xey + yex = 0 by differentiating this equation implicitly. (b) Prove that if y(x) is any integral curve of the slope field in part (a), then xey(x) + y(x)ex will be a constant function. (c) Find an equation that implicitly defines the integral curve through (1, 1) of the slope field in part (a).

586

Chapter 8 / Mathematical Modeling with Differential Equations

25. Consider the initial-value problem y â&#x20AC;˛ = y, y(0) = 1, and let yn denote the approximation of y(1) using Eulerâ&#x20AC;&#x2122;s Method with n steps. (a) What would you conjecture is the exact value of limn â&#x2020;&#x2019; +âŹ yn ? Explain your reasoning. (b) Find an explicit formula for yn and use it to verify your conjecture in part (a).

26. Writing Explain the connection between Eulerâ&#x20AC;&#x2122;s Method and the local linear approximation discussed in Section 3.5. 27. Writing Given a slope ďŹ eld, what features of an integral curve might be discussed from the slope ďŹ eld? Apply your ideas to the slope ďŹ eld in Exercise 3.

â&#x153;&#x201D;QUICK CHECK ANSWERS 8.3 1. (a) IV (b) III (c) I (d) II

8.4

2. y = 2x, x > 0

4. (a) 2.25 (b) e

3. yn + f(xn , yn ) x

FIRST-ORDER DIFFERENTIAL EQUATIONS AND APPLICATIONS In this section we will discuss a general method that can be used to solve a large class of ďŹ rst-order differential equations. We will use this method to solve differential equations related to the problems of mixing liquids and free fall retarded by air resistance. FIRST-ORDER LINEAR EQUATIONS The simplest ďŹ rst-order equations are those that can be written in the form

dy = q(x) dx Such equations can often be solved by integration. For example, if

(1)

dy = x3 dx

(2)

then y=

x 3 dx =

x4 +C 4

is the general solution of (2) on the interval (â&#x2C6;&#x2019;âŹ , +âŹ ). More generally, a ďŹ rst-order differential equation is called linear if it is expressible in the form dy + p(x)y = q(x) dx

(3)

Equation (1) is the special case of (3) that results when the function p(x) is identically 0. Some other examples of ďŹ rst-order linear differential equations are dy + x 2 y = ex , dx

dy + (sin x)y + x 3 = 0, dx

dy + 5y = 2 dx

p(x) = x 2 , q(x) = ex

p(x) = sin x, q(x) = â&#x2C6;&#x2019;x 3

p(x) = 5, q(x) = 2

We will assume that the functions p(x) and q(x) in (3) are continuous on a common interval, and we will look for a general solution that is valid on that interval. One method for doing this is based on the observation that if we deďŹ ne Îź = Îź(x) by Îź=e

p(x) dx

(4)

8.4 First-Order Differential Equations and Applications

587

then d dÎź = e p(x) dx Âˇ dx dx

Thus,

p(x) dx = Îźp(x)

dy dÎź dy d (Îźy) = Îź + y=Îź + Îźp(x)y dx dx dx dx If (3) is multiplied through by Îź, it becomes Îź

(5)

dy + Îźp(x)y = Îźq(x) dx

Combining this with (5) we have d (Îźy) = Îźq(x) (6) dx This equation can be solved for y by integrating both sides with respect to x and then dividing through by Îź to obtain 1 Îźq(x) dx (7) y= Îź which is a general solution of (3) on the interval. The function Îź in (4) is called an integrating factor for (3), and this method for ďŹ nding a general solution of (3) is called the method of integrating factors. Although one could simply memorize Formula (7), we recommend solving ďŹ rst-order linear equations by actually carrying out the steps used to derive this formula:

The Method of Integrating Factors Step 1. Calculate the integrating factor Îź=e

p(x) dx

Since any Îź will sufďŹ ce, we can take the constant of integration to be zero in this step. Step 2. Multiply both sides of (3) by Îź and express the result as d (Îźy) = Îźq(x) dx Step 3. Integrate both sides of the equation obtained in Step 2 and then solve for y. Be sure to include a constant of integration in this step.

Example 1

Solve the differential equation dy â&#x2C6;&#x2019; y = e2x dx

Solution. Comparing the given equation to (3), we see that we have a ďŹ rst-order linear equation with p(x) = â&#x2C6;&#x2019;1 and q(x) = e2x . These coefďŹ cients are continuous on the interval (â&#x2C6;&#x2019;âŹ , +âŹ ), so the method of integrating factors will produce a general solution on this interval. The ďŹ rst step is to compute the integrating factor. This yields Îź=e

p(x) dx

(â&#x2C6;&#x2019;1) dx

= eâ&#x2C6;&#x2019;x

588

Chapter 8 / Mathematical Modeling with Differential Equations

Next we multiply both sides of the given equation by Îź to obtain eâ&#x2C6;&#x2019;x

dy â&#x2C6;&#x2019; eâ&#x2C6;&#x2019;x y = eâ&#x2C6;&#x2019;x e2x dx

which we can rewrite as d â&#x2C6;&#x2019;x [e y] = ex dx Integrating both sides of this equation with respect to x we obtain Confirm that the solution obtained in Example 1 agrees with that obtained by substituting the integrating factor into Formula (7).

eâ&#x2C6;&#x2019;x y = ex + C Finally, solving for y yields the general solution y = e2x + Cex A differential equation of the form P (x)

dy + Q(x)y = R(x) dx

can be solved by dividing through by P (x) to put the equation in the form of (3) and then applying the method of integrating factors. However, the resulting solution will only be valid on intervals where p(x) = Q(x)/P (x) and q(x) = R(x)/P (x) are both continuous. Example 2

Solve the initial-value problem x

dy â&#x2C6;&#x2019; y = x, dx

y(1) = 2

Solution. This differential equation can be written in the form of (3) by dividing through by x. This yields

1 dy â&#x2C6;&#x2019; y=1 dx x

(8)

where q(x) = 1 is continuous on (â&#x2C6;&#x2019;âŹ , +âŹ ) and p(x) = â&#x2C6;&#x2019;1/x is continuous on (â&#x2C6;&#x2019;âŹ , 0) and (0, +âŹ ). Since we need p(x) and q(x) to be continuous on a common interval, and since our initial condition requires a solution for x = 1, we will ďŹ nd a general solution of (8) on the interval (0, +âŹ ). On this interval we have |x| = x, so that 1 Taking the constant of dx = â&#x2C6;&#x2019; ln |x| = â&#x2C6;&#x2019; ln x p(x) dx = â&#x2C6;&#x2019; integration to be 0 x Thus, an integrating factor that will produce a general solution on the interval (0, +âŹ ) is Îź=e

p(x) dx

= eâ&#x2C6;&#x2019; ln x = eln(1/x) =

1 x

Multiplying both sides of Equation (8) by this integrating factor yields 1 1 1 dy â&#x2C6;&#x2019; 2y = x dx x x or d dx

1 1 y = x x

8.4 First-Order Differential Equations and Applications

It is not accidental that the initial-value problem in Example 2 has a unique solution. If the coefficients of (3) are continuous on an open interval that contains the point x0 , then for any y0 there will be a unique solution of (3) on that interval that satisfies the initial condition y(x0 ) = y0 [Exercise 29(b)].

589

Therefore, on the interval (0, +⬁), 1 y= x

1 dx = ln x + C x

from which it follows that y = x ln x + Cx

(9)

The initial condition y(1) = 2 requires that y = 2 if x = 1. Substituting these values into (9) and solving for C yields C = 2 (verify), so the solution of the initial-value problem is y = x ln x + 2x We conclude this section with some applications of ﬁrst-order differential equations. MIXING PROBLEMS

In a typical mixing problem, a tank is filled to a specified level with a solution that contains a known amount of some soluble substance (say salt). The thoroughly stirred solution is allowed to drain from the tank at a known rate, and at the same time a solution with a known concentration of the soluble substance is added to the tank at a known rate that may or may not differ from the draining rate. As time progresses, the amount of the soluble substance in the tank will generally change, and the usual mixing problem seeks to determine the amount of the substance in the tank at a specified time. This type of problem serves as a model for many kinds of problems: discharge and filtration of pollutants in a river, injection and absorption of medication in the bloodstream, and migrations of species into and out of an ecological system, for example. 5 gal/min

Example 3 At time t = 0, a tank contains 4 lb of salt dissolved in 100 gal of water. Suppose that brine containing 2 lb of salt per gallon of brine is allowed to enter the tank at a rate of 5 gal/min and that the mixed solution is drained from the tank at the same rate (Figure 8.4.1). Find the amount of salt in the tank after 10 minutes.

100 gal

Solution. Let y(t) be the amount of salt (in pounds) after t minutes. We are given that y(0) = 4, and we want to find y(10). We will begin by finding a differential equation that is satisfied by y(t). To do this, observe that dy/dt, which is the rate at which the amount of salt in the tank changes with time, can be expressed as 5 gal/min

dy = rate in − rate out dt

Figure 8.4.1

(10)

where rate in is the rate at which salt enters the tank and rate out is the rate at which salt leaves the tank. But the rate at which salt enters the tank is rate in = (2 lb/gal) · (5 gal/min) = 10 lb/min Since brine enters and drains from the tank at the same rate, the volume of brine in the tank stays constant at 100 gal. Thus, after t minutes have elapsed, the tank contains y(t) lb of salt per 100 gal of brine, and hence the rate at which salt leaves the tank at that instant is y(t) y(t) / rate out = lb gal · (5 gal/min) = lb/min 100 20 Therefore, (10) can be written as dy y = 10 − dt 20

dy y + = 10 dt 20

Chapter 8 / Mathematical Modeling with Differential Equations

590

which is a ďŹ rst-order linear differential equation satisďŹ ed by y(t). Since we are given that y(0) = 4, the function y(t) can be obtained by solving the initial-value problem y dy + = 10, dt 20

y(0) = 4

The integrating factor for the differential equation is Îź=e

(1/20) dt

= et /20

If we multiply the differential equation through by Îź, then we obtain d t /20 (e y) = 10et /20 dt

et /20 y =

10et /20 dt = 200et /20 + C

y(t) = 200 + Ceâ&#x2C6;&#x2019;t /20

(11)

The initial condition states that y = 4 when t = 0. Substituting these values into (11) and solving for C yields C = â&#x2C6;&#x2019;196 (verify), so

Amount of salt y (lb)

y(t) = 200 â&#x2C6;&#x2019; 196eâ&#x2C6;&#x2019;t /20

The graph of (12) is shown in Figure 8.4.2. At time t = 10 the amount of salt in the tank is

200 150

â&#x2030;&#x2C6; 81.1

y(10) = 200 â&#x2C6;&#x2019; 196eâ&#x2C6;&#x2019;0.5 â&#x2030;&#x2C6; 81.1 lb

y = 200 â&#x2C6;&#x2019; 196eâ&#x2C6;&#x2019;t / 20

100 50 0

(12)

Notice that it follows from (11) that 0 10 20 30 40 50 60 70 80 Time t (min)

lim y(t) = 200

tâ&#x2020;&#x2019;+âŹ

Figure 8.4.2

for all values of C, so regardless of the amount of salt that is present in the tank initially, the amount of salt in the tank will eventually stabilize at 200 lb. This can also be seen geometrically from the slope ďŹ eld for the differential equation shown in Figure 8.4.3. This slope ďŹ eld suggests the following: If the amount of salt present in the tank is greater than 200 lb initially, then the amount of salt will decrease steadily over time toward a limiting value of 200 lb; and if the amount of salt is less than 200 lb initially, then it will increase steadily toward a limiting value of 200 lb. The slope ďŹ eld also suggests that if the amount present initially is exactly 200 lb, then the amount of salt in the tank will stay constant at 200 lb. This can also be seen from (11), since C = 0 in this case (verify).

The graph shown in Figure 8.4.2 suggests that y(t) â&#x2020;&#x2019; 200 as t â&#x2020;&#x2019; +âŹ . This means that over an extended period of time the amount of salt in the tank tends toward 200 lb. Give an informal physical argument to explain why this result is to be expected.

A MODEL OF FREE-FALL MOTION RETARDED BY AIR RESISTANCE y

300

In Section 5.7 we considered the free-fall model of an object moving along a vertical axis near the surface of the Earth. It was assumed in that model that there is no air resistance and that the only force acting on the object is the Earthâ&#x20AC;&#x2122;s gravity. Our goal here is to ďŹ nd a model that takes air resistance into account. For this purpose we make the following assumptions:

250 200 150 100

t 0

Figure 8.4.3

100

150

â&#x20AC;˘ The object moves along a vertical s-axis whose origin is at the surface of the Earth and whose positive direction is up (Figure 5.7.7).

â&#x20AC;˘ At time t = 0 the height of the object is s0 and the velocity is v0 . â&#x20AC;˘ The only forces on the object are the force FG = â&#x2C6;&#x2019;mg of the Earthâ&#x20AC;&#x2122;s gravity acting down and the force FR of air resistance acting opposite to the direction of motion. The force FR is called the drag force.

8.4 First-Order Differential Equations and Applications

591

In the case of free-fall motion retarded by air resistance, the net force acting on the object is 2

FG + FR = −mg + FR

and the acceleration is d s /dt , so Newton’s Second Law of Motion [Equation (5) of Section 6.6] implies that d 2s −mg + FR = m 2 (13) dt Experimentation has shown that the force FR of air resistance depends on the shape of the object and its speed—the greater the speed, the greater the drag force. There are many possible models for air resistance, but one of the most basic assumes that the drag force FR is proportional to the velocity of the object, that is, FR = −cv

where c is a positive constant that depends on the object’s shape and properties of the ∗ air. (The minus sign ensures that the drag force is opposite to the direction of motion.) Substituting this in (13) and writing d 2 s /dt 2 as dv /dt, we obtain dv −mg − cv = m dt Dividing by m and rearranging we obtain c dv + v = −g dt m which is a first-order linear differential equation in the unknown function v = v(t) with p(t) = c/m and q(t) = −g [see (3)]. For a specific object, the coefficient c can be determined experimentally, so we will assume that m, g, and c are known constants. Thus, the velocity function v = v(t) can be obtained by solving the initial-value problem c dv + v = −g, v(0) = v0 (14) dt m Once the velocity function is found, the position function s = s(t) can be obtained by solving the initial-value problem ds = v(t), s(0) = s0 (15) dt In Exercise 25 we will ask you to solve (14) and show that mg mg − (16) v(t) = e−ct /m v0 + c c Note that mg (17) lim v(t) = − t → +⬁ c (verify). Thus, the speed |v(t)| does not increase indefinitely, as in free fall; rather, because of the air resistance, it approaches a finite limiting speed vτ given by mg mg (18) vτ = − = c c This is called the terminal speed of the object, and (17) is called its terminal velocity. REMARK

Intuition suggests that near the limiting velocity, the velocity v(t) changes very slowly; that is, dv /dt ≈ 0. Thus, it should not be surprising that the limiting velocity can be obtained informally from (14) by setting dv /dt = 0 in the differential equation and solving for v . This yields

v=−

mg c

which agrees with (17).

∗

Other common models assume that FR = −cv 2 or, more generally, FR = −cv p for some value of p.

Chapter 8 / Mathematical Modeling with Differential Equations

592

✔QUICK CHECK EXERCISES 8.4

(See page 594 for answers.)

1. Solve the ﬁrst-order linear differential equation

2. An integrating factor for

dy + p(x)y = q(x) dx

y dy + = q(x) dx x

by completing the following steps: .

is Step 1. Calculate the integrating factor μ =

Step 2. Multiply both sides of the equation by the integrating factor and express the result as d [ dx

3. At time t = 0, a tank contains 30 oz of salt dissolved in 60 gal of water. Then brine containing 5 oz of salt per gallon of brine is allowed to enter the tank at a rate of 3 gal/min and the mixed solution is drained from the tank at the same rate. Give an initial-value problem satisﬁed by the amount of salt y(t) in the tank at time t. Do not solve the problem.

Step 3. Integrate both sides of the equation obtained in Step 2 and solve for y = .

EXERCISE SET 8.4

Graphing Utility

1–6 Solve the differential equation by the method of integrating

factors. ■ dy 1. + 4y = e−3x dx 3. y ′ + y = cos(ex ) 5. (x 2 + 1)

dy + 2xy = x dx dy + 4y = 1 4. 2 dx dy 1 =0 +y+ 6. dx 1 − ex 2.

dy + xy = 0 dx

14. In our model for free-fall motion retarded by air resistance, the terminal velocity is proportional to the weight of the falling object. 15. A slope field for the differential equation y ′ = 2y − x is shown in the accompanying figure. In each part, sketch the graph of the solution that satisfies the initial condition. (a) y(1) = 1 (b) y(0) = −1 (c) y(−1) = 0 y

7–10 Solve the initial-value problem. ■

dy + y = x, y(1) = 2 dx dy 8. x − y = x 2 , y(1) = −1 dx dy 9. − 2xy = 2x, y(0) = 3 dx dy 10. + y = 2, y(0) = 1 dt

7. x

11–14 True–False Determine whether the statement is true or

false. Explain your answer. ■ 11. If y1 and y2 are two solutions to a ﬁrst-order linear differential equation, then y = y1 + y2 is also a solution.

12. If the ﬁrst-order linear differential equation dy + p(x)y = q(x) dx

has a solution that is a constant function, then q(x) is a constant multiple of p(x). 13. In a mixing problem, we expect the concentration of the dissolved substance within the tank to approach a ﬁnite limit over time.

x −2

−1

−1 −2

Figure Ex-15

16. Solve the initial-value problems in Exercise 15, and use a graphing utility to conﬁrm that the integral curves for these solutions are consistent with the sketches you obtained from the slope ﬁeld. F O C U S O N C O N C E P TS

17. Use the slope ﬁeld in Exercise 15 to make a conjecture about the effect of y0 on the behavior of the solution of the initial-value problem y ′ = 2y − x, y(0) = y0 as x → +⬁, and check your conjecture by examining the solution of the initial-value problem. 18. Consider the slope ﬁeld in Exercise 15.

(a) Use Euler’s Method with x = 0.1 to estimate y 21 for the solution that satisﬁes the initial condition y(0) = 1.

8.4 First-Order Differential Equations and Applications

(b) Would you conjecture your answer in part (a) to

be greater than or less than the actual value of y 21 ? Explain. (c) Check your conjecture in part (b) by ﬁnding the ex act value of y 21 . 19. (a) Use Euler’s Method with a step size of x = 0.2 to approximate the solution of the initial-value problem y ′ = x + y,

y(0) = 1

over the interval 0 ≤ x ≤ 1. (b) Solve the initial-value problem exactly, and calculate the error and the percentage error in each of the approximations in part (a). (c) Sketch the exact solution and the approximate solution together. 20. It was stated at the end of Section 8.3 that reducing the step size in Euler’s Method by half reduces the error in each approximation by about half. Conﬁrm that the error in y(1) is reduced by about half if a step size of x = 0.1 is used in Exercise 19. 21. At time t = 0, a tank contains 25 oz of salt dissolved in 50 gal of water. Then brine containing 4 oz of salt per gallon of brine is allowed to enter the tank at a rate of 2 gal/min and the mixed solution is drained from the tank at the same rate. (a) How much salt is in the tank at an arbitrary time t? (b) How much salt is in the tank after 25 min? 22. A tank initially contains 200 gal of pure water. Then at time t = 0 brine containing 5 lb of salt per gallon of brine is allowed to enter the tank at a rate of 20 gal/min and the mixed solution is drained from the tank at the same rate. (a) How much salt is in the tank at an arbitrary time t? (b) How much salt is in the tank after 30 min? 23. A tank with a 1000 gal capacity initially contains 500 gal of water that is polluted with 50 lb of particulate matter. At time t = 0, pure water is added at a rate of 20 gal/min and the mixed solution is drained off at a rate of 10 gal/min. How much particulate matter is in the tank when it reaches the point of overﬂowing? 24. The water in a polluted lake initially contains 1 lb of mercury salts per 100,000 gal of water. The lake is circular with diameter 30 m and uniform depth 3 m. Polluted water is pumped from the lake at a rate of 1000 gal/h and is replaced with fresh water at the same rate. Construct a table that shows the amount of mercury in the lake (in lb) at the end of each hour over a 12-hour period. Discuss any assumptions you made. [Note: Use 1 m3 = 264 gal.]

25. (a) Use the method of integrating factors to derive solution (16) to the initial-value problem (14). [Note: Keep in mind that c, m, and g are constants.] (b) Show that (16) can be expressed in terms of the terminal speed (18) as v(t) = e−gt /vτ (v0 + vτ ) − vτ

593

(c) Show that if s(0) = s0 , then the position function of the object can be expressed as vτ s(t) = s0 − vτ t + (v0 + vτ )(1 − e−gt /vτ ) g 26. Suppose a fully equipped skydiver weighing 240 lb has a terminal speed of 120 ft/s with a closed parachute and 24 ft/s with an open parachute. Suppose further that this skydiver is dropped from an airplane at an altitude of 10,000 ft, falls for 25 s with a closed parachute, and then falls the rest of the way with an open parachute. (a) Assuming that the skydiver’s initial vertical velocity is zero, use Exercise 25 to find the skydiver’s vertical velocity and height at the time the parachute opens. [Note: Take g = 32 ft/s2 .] (b) Use a calculating utility to find a numerical solution for the total time that the skydiver is in the air. 27. The accompanying figure is a schematic diagram of a basic RL series electrical circuit that contains a power source with a time-dependent voltage of V (t) volts (V), a resistor with a constant resistance of R ohms ( ), and an inductor with a constant inductance of L henrys (H). If you don’t know anything about electrical circuits, don’t worry; all you need to know is that electrical theory states that a current of I (t) amperes (A) flows through the circuit where I (t) satisfies the differential equation L

dI + RI = V (t) dt

(a) Find I (t) if R = 10 , L = 5 H, V is a constant 20 V, and I (0) = 0 A. (b) What happens to the current over a long period of time? R

V(t)

Figure Ex-27

28. Find I (t) for the electrical circuit in Exercise 27 if R = 6 , L = 3 H, V (t) = 3 sin t V, and I (0) = 15 A. F O C U S O N C O N C E P TS

29. (a) Prove that any function y = y(x) deﬁned by Equation (7) will be a solution to (3). (b) Consider the initial-value problem dy + p(x)y = q(x), dx

y(x0 ) = y0

where the functions p(x) and q(x) are both continuous on some open interval. Using the general solution for a ﬁrst-order linear equation, prove that this initial-value problem has a unique solution on the interval.

594

Chapter 8 / Mathematical Modeling with Differential Equations

30. (a) Prove that solutions need not be unique for nonlinear initial-value problems by ďŹ nding two solutions to dy y = x, y(0) = 0 dx (b) Prove that solutions need not exist for nonlinear initial-value problems by showing that there is no solution for dy y = â&#x2C6;&#x2019;x, y(0) = 0 dx

31. Writing Explain why the quantity Îź in the Method of Integrating Factors is called an â&#x20AC;&#x153;integrating factorâ&#x20AC;? and explain its role in this method. 32. Writing Suppose that a given ďŹ rst-order differential equation can be solved both by the method of integrating factors and by separation of variables. Discuss the advantages and disadvantages of each method.

â&#x153;&#x201D;QUICK CHECK ANSWERS 8.4 1. Step 1: e

p(x) dx

; Step 2: Îźy, Îźq(x); Step 3:

CHAPTER 8 REVIEW EXERCISES

1 Îź

Îźq(x) dx

2. x

dy y + = 15, y(0) = 30 dt 20

CAS

1. Classify the following ďŹ rst-order differential equations as separable, linear, both, or neither. dy dy (a) â&#x2C6;&#x2019; 3y = sin x (b) + xy = x dx dx dy dy â&#x2C6;&#x2019;x =1 (d) + xy 2 = sin(xy) (c) y dx dx 2. Which of the given differential equations are separable? dy dy f(x) (a) = f(x)g(y) (b) = dx dx g(y)

dy dy (c) = f(x) + g(y) (d) = f(x)g(y) dx dx 3â&#x20AC;&#x201C;5 Solve the differential equation by the method of separation of variables. â&#x2013; dy dy 3. = (1 + y 2 )x 2 4. 3 tan y â&#x2C6;&#x2019; sec x = 0 dx dx 5. (1 + y 2 )y â&#x20AC;˛ = ex y 6â&#x20AC;&#x201C;8 Solve the initial-value problem by the method of separation of variables. â&#x2013; y5 , y(1) = 1 6. y â&#x20AC;˛ = 1 + y 2 , y(0) = 1 7. y â&#x20AC;˛ = x(1 + y 4 )

8. y â&#x20AC;˛ = 4y 2 sec2 2x, y(Ď&#x20AC;/8) = 1

9. Sketch the integral curve of y â&#x20AC;˛ = â&#x2C6;&#x2019;2xy 2 that passes through the point (0, 1).

10. Sketch the integral curve of 2yy â&#x20AC;˛ = 1 that passes through the point (0, 1) and the integral curve that passes through the point (0, â&#x2C6;&#x2019;1).

11. Sketch the slope ďŹ eld for y â&#x20AC;˛ = xy /8 at the 25 gridpoints (x, y), where x = 0, 1, . . . , 4 and y = 0, 1, . . . , 4.

12. Solve the differential equation y â&#x20AC;˛ = xy /8, and ďŹ nd a family of integral curves for the slope ďŹ eld in Exercise 11.

13â&#x20AC;&#x201C;14 Use Eulerâ&#x20AC;&#x2122;s Method with the given step size x to approximate the solution of the initial-value problem over the stated interval. Present your answer as a table and as a graph. â&#x2013;

13. dy/dx =

â&#x2C6;&#x161;

y, y(0) = 1, 0 â&#x2030;¤ x â&#x2030;¤ 4, x = 0.5

14. dy/dx = sin y, y(0) = 1, 0 â&#x2030;¤ x â&#x2030;¤ 2, x = 0.5 15. Consider the initial-value problem y â&#x20AC;˛ = cos 2Ď&#x20AC;t,

y(0) = 1

Use Eulerâ&#x20AC;&#x2122;s Method with ďŹ ve steps to approximate y(1). 16. Cloth found in an Egyptian pyramid contains 78.5% of its original carbon-14. Estimate the age of the cloth. 17. In each part, ďŹ nd an exponential growth model y = y0 ekt that satisďŹ es the stated conditions. (a) y0 = 2; doubling time T = 5 (b) y(0) = 5; growth rate 1.5% (c) y(1) = 1; y(10) = 100 (d) y(1) = 1; doubling time T = 5

18. Suppose that an initial population of 5000 bacteria grows exponentially at a rate of 1% per hour and that y = y(t) is the number of bacteria present after t hours. (a) Find an initial-value problem whose solution is y(t). (b) Find a formula for y(t). (c) What is the doubling time for the population? (d) How long does it take for the population of bacteria to reach 30,000?

19â&#x20AC;&#x201C;20 Solve the differential equation by the method of integrating factors. â&#x2013; dy dy 1 19. =0 + 3y = eâ&#x2C6;&#x2019;2x 20. +yâ&#x2C6;&#x2019; dx dx 1 + ex 21â&#x20AC;&#x201C;23 Solve the initial-value problem by the method of integrating factors. â&#x2013;

Chapter 8 Making Connections

21. y ′ − xy = x, y(0) = 3

22. xy ′ + 2y = 4x 2 , y(1) = 2

23. y ′ cosh x + y sinh x = cosh2 x, y(0) = 2 24. (a) Solve the initial-value problem y ′ − y = x sin 3x,

y(0) = 1

by the method of integrating factors, using a CAS to perform any difﬁcult integrations. (b) Use the CAS to solve the initial-value problem directly, and conﬁrm that the answer is consistent with that obtained in part (a). (c) Graph the solution. 25. A tank contains 1000 gal of fresh water. At time t = 0 min, brine containing 5 oz of salt per gallon of brine is poured into the tank at a rate of 10 gal/min, and the mixed solution is drained from the tank at the same rate. After 15 min that

595

process is stopped and fresh water is poured into the tank at the rate of 5 gal/min, and the mixed solution is drained from the tank at the same rate. Find the amount of salt in the tank at time t = 30 min. 26. Suppose that a room containing 1200 ft3 of air is free of carbon monoxide. At time t = 0 cigarette smoke containing 4% carbon monoxide is introduced at the rate of 0.1 ft3 /min, and the well-circulated mixture is vented from the room at the same rate. (a) Find a formula for the percentage of carbon monoxide in the room at time t. (b) Extended exposure to air containing 0.012% carbon monoxide is considered dangerous. How long will it take to reach this level? Source: This is based on a problem from William E. Boyce and Richard C. DiPrima, Elementary Differential Equations, 7th ed., John Wiley & Sons, New York, 2001.

CHAPTER 8 MAKING CONNECTIONS 1. Consider the ﬁrst-order differential equation dy + py = q dx where p and q are constants. If y = y(x) is a solution to this equation, deﬁne u = u(x) = q − py(x). (a) Without solving the differential equation, show that u grows exponentially as a function of x if p < 0, and decays exponentially as a function of x if 0 < p. (b) Use the result of part (a) and Equations (13–14) of Section 8.2 to solve the initial-value problem dy + 2y = 4, dx

y(0) = −1

2. Consider a differential equation of the form dy = f (ax + by + c) dx where f is a function of a single variable. If y = y(x) is a solution to this equation, deﬁne u = u(x) = ax + by(x) + c. (a) Find a separable differential equation that is satisﬁed by the function u. (b) Use your answer to part (a) to solve dy 1 = dx x+y

3. A first-order differential equation is homogeneous if it can be written in the form y dy =f for x = 0 dx x where f is a function of a single variable. If y = y(x) is a solution to a first-order homogeneous differential equation, define u = u(x) = y(x)/x. (a) Find a separable differential equation that is satisfied by the function u. (b) Use your answer to part (a) to solve dy x−y = dx x+y

4. A first-order differential equation is called a Bernoulli equation if it can be written in the form dy + p(x)y = q(x)y n for n = 0, 1 dx If y = y(x) is a solution to a Bernoulli equation, define u = u(x) = [y(x)]1−n . (a) Find a first-order linear differential equation that is satisfied by u. (b) Use your answer to part (a) to solve the initial-value problem dy 1 x − y = −2xy 2 , y(1) = dx 2

9 INFINITE SERIES

iStockphoto

Perspective creates the illusion that the sequence of railroad ties continues indeﬁnitely but converges toward a single point inﬁnitely far away.

9.1

In this chapter we will be concerned with infinite series, which are sums that involve infinitely many terms. Infinite series play a fundamental role in both mathematics and science—they are used, for example, to approximate trigonometric functions and logarithms, to solve differential equations, to evaluate difficult integrals, to create new functions, and to construct mathematical models of physical laws. Since it is impossible to add up infinitely many numbers directly, one goal will be to define exactly what we mean by the sum of an infinite series. However, unlike finite sums, it turns out that not all infinite series actually have a sum, so we will need to develop tools for determining which infinite series have sums and which do not. Once the basic ideas have been developed we will begin to apply our work; we will show how infinite series are used to evaluate such quantities as ln 2, e, sin 3 ◦ , and π, how they are used to create functions, and finally, how they are used to model physical laws.

SEQUENCES In everyday language, the term “sequence” means a succession of things in a deﬁnite order—chronological order, size order, or logical order, for example. In mathematics, the term “sequence” is commonly used to denote a succession of numbers whose order is determined by a rule or a function. In this section, we will develop some of the basic ideas concerning sequences of numbers. DEFINITION OF A SEQUENCE

Stated informally, an infinite sequence, or more simply a sequence, is an unending succession of numbers, called terms. It is understood that the terms have a definite order; that is, there is a first term a1 , a second term a2 , a third term a3 , a fourth term a4 , and so forth. Such a sequence would typically be written as a1 , a2 , a3 , a4 , . . . where the dots are used to indicate that the sequence continues indefinitely. Some specific examples are 1, 2, 3, 4, . . . , 1, 21 , 31 , 41 , . . . , 2, 4, 6, 8, . . . ,

1, −1, 1, −1, . . .

Each of these sequences has a deﬁnite pattern that makes it easy to generate additional terms if we assume that those terms follow the same pattern as the displayed terms. However, 596

9.1 Sequences

597

such patterns can be deceiving, so it is better to have a rule or formula for generating the terms. One way of doing this is to look for a function that relates each term in the sequence to its term number. For example, in the sequence 2, 4, 6, 8, . . . each term is twice the term number; that is, the nth term in the sequence is given by the formula 2n. We denote this by writing the sequence as 2, 4, 6, 8, . . . , 2n, . . . We call the function f(n) = 2n the general term of this sequence. Now, if we want to know a speciﬁc term in the sequence, we need only substitute its term number in the formula for the general term. For example, the 37th term in the sequence is 2 · 37 = 74. Example 1

Table 9.1.1

term number term

(a)

1 2 3 4 , , , ,... 2 3 4 5

(b)

(c)

1 , − 23 , 43 , − 45 , . . . 2

(d) 1, 3, 5, 7, . . .

1 1 1 1 , , , ,... 2 4 8 16

Solution (a). In Table 9.1.1, the four known terms have been placed below their term

1 2 3 4 ... n ... 2 3 4 5 n+1

numbers, from which we see that the numerator is the same as the term number and the denominator is one greater than the term number. This suggests that the nth term has numerator n and denominator n + 1, as indicated in the table. Thus, the sequence can be expressed as n 1 2 3 4 , , , ,..., ,... 2 3 4 5 n+1

Table 9.1.2

Solution (b). In Table 9.1.2, the denominators of the four known terms have been ex-

term ... number 1 2 3 4 term

In each part, ﬁnd the general term of the sequence.

...

3 4 ... n ...

1 1 1 1 ... 1 ... 2 22 23 24 2n

pressed as powers of 2 and the ﬁrst four terms have been placed below their term numbers, from which we see that the exponent in the denominator is the same as the term number. This suggests that the denominator of the nth term is 2n , as indicated in the table. Thus, the sequence can be expressed as 1 1 1 1 1 , , , ,..., n,... 2 4 8 16 2

Solution (c). This sequence is identical to that in part (a), except for the alternating signs. Thus, the nth term in the sequence can be obtained by multiplying the nth term in part (a) by (−1)n+1 . This factor produces the correct alternating signs, since its successive values, starting with n = 1, are 1, −1, 1, −1, . . . . Thus, the sequence can be written as 2 3 4 n 1 , − , , − , . . . , (−1)n+1 ,... 2 3 4 5 n+1

Solution (d). In Table 9.1.3, the four known terms have been placed below their term

Table 9.1.3

term ... number 1 2 3 4 term

...

1 3 5 7 . . . 2n − 1 . . .

numbers, from which we see that each term is one less than twice its term number. This suggests that the nth term in the sequence is 2n − 1, as indicated in the table. Thus, the sequence can be expressed as 1, 3, 5, 7, . . . , 2n − 1, . . . When the general term of a sequence a1 , a2 , a3 , . . . , an , . . .

(1)

598

Chapter 9 / Infinite Series

is known, there is no need to write out the initial terms, and it is common to write only the general term enclosed in braces. Thus, (1) might be written as {an }+⬁ n=1

or as

{an }⬁n=1

For example, here are the four sequences in Example 1 expressed in brace notation. A sequence cannot be uniquely determined from a few initial terms. For example, the sequence whose general term is

f(n) = 13 (3 − 5n + 6n2 − n3 ) has 1, 3, and 5 as its first three terms, but its fourth term is also 5.

sequence

brace notation

1 , 2, 3, 4, . . . , n , . . . 2 3 4 5 n+1 1 , 1, 1, 1 , . . . , 1 , . . . 2 4 8 16 2n 1 , − 2 , 3 , − 4 , . . . , (−1) n +1 n , . . . n+1 2 3 4 5 1, 3, 5, 7, . . . , 2n − 1, . . .

n +∞ n + 1 n=1 1 +∞ 2n n=1 (−1) n +1

n n+1

∞ {2n − 1}+n=1

+∞ n=1

The letter n in (1) is called the index for the sequence. It is not essential to use n for the index; any letter not reserved for another purpose can be used. For example, we might view the general term of the sequence a1 , a2 , a3 , . . . to be the kth term, in which case we would denote this sequence as {ak }+⬁ k=1 . Moreover, it is not essential to start the index at 1; sometimes it is more convenient to start it at 0 (or some other integer). For example, consider the sequence 1 1 1 1, , 2 , 3 , . . . 2 2 2 One way to write this sequence is 1 +⬁ 2n−1 n=1 However, the general term will be simpler if we think of the initial term in the sequence as the zeroth term, in which case we can write the sequence as +⬁ 1 2n n=0 We began this section by describing a sequence as an unending succession of numbers. Although this conveys the general idea, it is not a satisfactory mathematical definition because it relies on the term “succession,” which is itself an undefined term. To motivate a precise definition, consider the sequence 2, 4, 6, 8, . . . , 2n, . . . If we denote the general term by f(n) = 2n, then we can write this sequence as f(1), f(2), f(3), . . . , f(n), . . .

which is a “list” of values of the function f(n) = 2n,

n = 1, 2, 3, . . .

whose domain is the set of positive integers. This suggests the following deﬁnition. 9.1.1 definition A sequence is a function whose domain is a set of integers. Typically, the domain of a sequence is the set of positive integers or the set of nonnegative integers. We will regard the expression {an }+⬁ n=1 to be an alternative notation for the function f(n) = an , n = 1, 2, 3, . . . , and we will regard {an }+⬁ n=0 to be an alternative notation for the function f(n) = an , n = 0, 1, 2, 3, . . . .

9.1 Sequences

599

GRAPHS OF SEQUENCES

Since sequences are functions, it makes sense to talk about the graph of a sequence. For example, the graph of the sequence {1/n}+⬁ n=1 is the graph of the equation

When the starting value for the index of a sequence is not relevant to the discussion, it is common to use a notation such as {an } in which there is no reference to the starting value of n. We can distinguish between different sequences by using different letters for their general terms; thus, {an }, {bn }, and {cn } denote three different sequences.

1 , n

n = 1, 2, 3, . . .

Because the right side of this equation is deﬁned only for positive integer values of n, the graph consists of a succession of isolated points (Figure 9.1.1a). This is different from the graph of 1 y= , x≥1 x which is a continuous curve (Figure 9.1.1b). y

1 n 1

1 y = n , n = 1, 2, 3, ...

1 y = x, x≥1

(a)

(b)

Figure 9.1.1

LIMIT OF A SEQUENCE

Since sequences are functions, we can inquire about their limits. However, because a sequence {an } is only deﬁned for integer values of n, the only limit that makes sense is the limit of an as n → +⬁. In Figure 9.1.2 we have shown the graphs of four sequences, each of which behaves differently as n → +⬁:

• • • •

The terms in the sequence {n + 1} increase without bound.

The terms in the sequence {(−1)n+1 } oscillate between −1 and 1. The terms in the sequence {n/(n + 1)} increase toward a “limiting value” of 1. n The terms in the sequence 1 + − 21 also tend toward a “limiting value” of 1, but do so in an oscillatory fashion.

7 6 5 4 3 2 1

1 n 1 2 3 4 5 6 7 8 9 n

−1

+∞ n=1

1 2

1 2 3 4 5 6 7 8 9

n+1

1 2 3 4 5 6 7 8 9

(−1)n+1

+∞ n =1

n n+1

+∞ n=1

n 1 2 3 4 5 6 7 8 9

1 + 冸− 12 冹n

+∞ n =1

Figure 9.1.2

Informally speaking, the limit of a sequence {an } is intended to describe how an behaves as n → +⬁. To be more speciﬁc, we will say that a sequence {an } approaches a limit L if the terms in the sequence eventually become arbitrarily close to L. Geometrically, this

600

Chapter 9 / Infinite Series y

y= L+e y= L−e

From this point on, the terms in the sequence are all within e units of L.

n 1

Figure 9.1.3

...

means that for any positive number ǫ there is a point in the sequence after which all terms lie between the lines y = L − ǫ and y = L + ǫ (Figure 9.1.3). The following deﬁnition makes these ideas precise.

How would you define these limits?

lim an = +⬁

n → +⬁

lim an = −⬁

n → +⬁

9.1.2 definition A sequence {an } is said to converge to the limit L if given any ǫ > 0, there is a positive integer N such that |an − L| < ǫ for n ≥ N. In this case we write lim an = L n → +⬁

A sequence that does not converge to some ﬁnite limit is said to diverge.

Example 2 The ﬁrst two sequences in Figure 9.1.2 diverge, and the second two converge to 1; that is, n n lim =1 = 1 and lim 1 + − 21 n → +⬁ n → +⬁ n + 1 The following theorem, which we state without proof, shows that the familiar properties of limits apply to sequences. This theorem ensures that the algebraic techniques used to ﬁnd limits of the form lim can also be used for limits of the form lim . n → +⬁

x → +⬁

9.1.3 theorem Suppose that the sequences {an } and {bn } converge to limits L1 and L2 , respectively, and c is a constant. Then: (a) (b) (c) (d ) Additional limit properties follow from those in Theorem 9.1.3. For example, use part (e) to show that if an → L and m is a positive integer, then m

lim (an ) = L

n → +⬁

(e) (f)

lim c = c

n → +⬁

lim can = c lim an = cL1 n → +⬁

n → +⬁

lim (an + bn ) = lim an + lim bn = L1 + L2 n → +⬁

n → +⬁

lim (an − bn ) = lim an − lim bn = L1 − L2 n → +⬁

n → +⬁

lim (an bn ) = lim an · lim bn = L1 L2 n → +⬁

n → +⬁

lim

n → +⬁

an bn

n → +⬁

lim an

n → +⬁

lim bn

n → +⬁

L1 L2

(if L2 = 0)

If the general term of a sequence is f(n), where f(x) is a function deﬁned on the entire interval [1, +⬁), then the values of f(n) can be viewed as “sample values” of f(x) taken

9.1 Sequences y

601

at the positive integers. Thus,

if f(x) → L as x → +⬁,

f (3) f (2) f (1)

x 1

If f (x) → L as x → + ∞, then f (n) → L as n → + ∞.

(a) y

L f (x)

then

f(n) → L as n → +⬁

(Figure 9.1.4a). However, the converse is not true; that is, one cannot infer that f(x) → L as x → +⬁ from the fact that f(n) → L as n → +⬁ (Figure 9.1.4b). Example 3 In each part, determine whether the sequence converges or diverges by examining the limit as n → +⬁. +⬁ +⬁ n n n+1 (b) (−1) (a) 2n + 1 n=1 2n + 1 n=1 +⬁ 1 (d) {8 − 2n}+⬁ (c) (−1)n+1 n=1 n n=1

Solution (a). Dividing numerator and denominator by n and using Theorem 9.1.3 yields x 1

f (n) → L as n → + ∞, but f (x) diverges by oscillation as x → + ∞.

(b) Figure 9.1.4

lim 1 lim 1 n 1 n → +⬁ n → +⬁ = = lim = n → +⬁ 2n + 1 n → +⬁ 2 + 1/n lim (2 + 1/n) lim 2 + lim 1/n lim

n → +⬁

1 1 = = 2+0 2

n → +⬁

Thus, the sequence converges to 21 .

Solution (b). This sequence is the same as that in part (a), except for the factor of (−1)n+1 , which oscillates between +1 and −1. Thus, the terms in this sequence oscillate between positive and negative values, with the odd-numbered terms being identical to those in part (a) and the even-numbered terms being the negatives of those in part (a). Since the sequence in part (a) has a limit of 21 , it follows that the odd-numbered terms in this sequence approach 21 , and the even-numbered terms approach − 21 . Therefore, this sequence has no limit—it diverges.

Solution (c). Since 1/n → 0, the product (−1)n+1 (1/n) oscillates between positive and

negative values, with the odd-numbered terms approaching 0 through positive values and the even-numbered terms approaching 0 through negative values. Thus, lim (−1)n+1

n → +⬁

1 =0 n

so the sequence converges to 0.

Solution (d).

Example 4 its limit.

lim (8 − 2n) = −⬁, so the sequence {8 − 2n}+⬁ n=1 diverges.

n → +⬁

In each part, determine whether the sequence converges, and if so, ﬁnd

(a) 1,

1 1 1 1 , , ,..., n,... 2 22 23 2

(b) 1, 2, 22 , 23 , . . . , 2n , . . .

Solution. Replacing n by x in the ﬁrst sequence produces the power function (1/2)x , and replacing n by x in the second sequence produces the power function 2x . Now recall that if 0 < b < 1, then bx → 0 as x → +⬁, and if b > 1, then bx → +⬁ as x → +⬁ (Figure 0.5.1).

602

Chapter 9 / Infinite Series

Thus,

1 = 0 and lim 2n = +⬁ n → +⬁ 2n So, the sequence {1/2n } converges to 0, but the sequence {2n } diverges. lim

n → +⬁

Example 5

Find the limit of the sequence

Solution. The expression

n +⬁ en

n=1

n en is an indeterminate form of type ⬁/⬁, so L’Hôpital’s rule is indicated. However, we cannot apply this rule directly to n/en because the functions n and en have been deﬁned here only at the positive integers, and hence are not differentiable functions. To circumvent this problem we extend the domains of these functions to all real numbers, here implied by replacing n by x, and apply L’Hôpital’s rule to the limit of the quotient x /ex . This yields x 1 lim x = lim x = 0 x → +⬁ e x → +⬁ e from which we can conclude that n lim n = 0 n → +⬁ e lim

n → +⬁

Example 6

Solution. lim

n → +⬁

Show that lim

n → +⬁

√ n n = 1.

√ n n = lim n1/n = lim e(1/n) ln n = e0 = 1 n → +⬁

n → +⬁

By L’Hôpital’s rule applied to (1/x) ln x

Sometimes the even-numbered and odd-numbered terms of a sequence behave sufﬁciently differently that it is desirable to investigate their convergence separately. The following theorem, whose proof is omitted, is helpful for that purpose. 9.1.4 theorem A sequence converges to a limit L if and only if the sequences of even-numbered terms and odd-numbered terms both converge to L.

Example 7 The sequence

{cn}

1 1 1 1 1 1 , , , , , ,... 2 3 22 32 23 33 converges to 0, since the even-numbered terms and the odd-numbered terms both converge to 0, and the sequence 1, 1 , 1, 1 , 1, 1 , . . . 2

diverges, since the odd-numbered terms converge to 1 and the even-numbered terms converge to 0.

{bn}

{an} If an → L and cn → L, then

bn → L. Figure 9.1.5

THE SQUEEZING THEOREM FOR SEQUENCES The following theorem, illustrated in Figure 9.1.5, is an adaptation of the Squeezing Theorem (1.6.4) to sequences. This theorem will be useful for ﬁnding limits of sequences that cannot be obtained directly. The proof is omitted.

9.1 Sequences

603

9.1.5 theorem (The Squeezing Theorem for Sequences) Let {an }, {bn }, and {cn } be sequences such that an ≤ bn ≤ cn ( for all values of n beyond some index N ) If the sequences {an } and {cn } have a common limit L as n → +⬁, then {bn } also has the limit L as n → +⬁.

Recall that if n is a positive integer, then n! (read “n factorial”) is the product of the first n positive integers. In addition, it is convenient to define 0! = 1.

Example 8 quence

Use numerical evidence to make a conjecture about the limit of the se +⬁ n! nn n=1

and then conﬁrm that your conjecture is correct. Table 9.1.4

n 1 2 3 4 5 6 7 8 9 10 11 12

n! nn 1.0000000000 0.5000000000 0.2222222222 0.0937500000 0.0384000000 0.0154320988 0.0061198990 0.0024032593 0.0009366567 0.0003628800 0.0001399059 0.0000537232

Solution. Table 9.1.4, which was obtained with a calculating utility, suggests that the limit of the sequence may be 0. To conﬁrm this we need to examine the limit of an =

n! nn

as n → +⬁. Although this is an indeterminate form of type ⬁/⬁, L’Hôpital’s rule is not helpful because we have no deﬁnition of x! for values of x that are not integers. However, let us write out some of the initial terms and the general term in the sequence: 1·2 1 1·2·3 2 1 1·2·3·4 3 1 = , a3 = = < , a4 = = < ,... 2·2 2 3·3·3 9 3 4·4·4·4 32 4 If n > 1, the general term of the sequence can be rewritten as

1 · 2 · 3···n 1 2 · 3···n = an = n · n · n···n n n · n···n

a1 = 1,

a2 =

from which it follows that an ≤ 1/n (why?). It is now evident that

1 n However, the two outside expressions have a limit of 0 as n → +⬁; thus, the Squeezing Theorem for Sequences implies that an → 0 as n → +⬁, which conﬁrms our conjecture. 0 ≤ an ≤

The following theorem is often useful for ﬁnding the limit of a sequence with both positive and negative terms—it states that if the sequence {|an |} that is obtained by taking the absolute value of each term in the sequence {an } converges to 0, then {an } also converges to 0.

9.1.6

theorem If lim |an | = 0, then lim an = 0. n → +⬁

n → +⬁

proof Depending on the sign of an , either an = |an | or an = −|an |. Thus, in all cases we have −|an | ≤ an ≤ |an | However, the limit of the two outside terms is 0, and hence the limit of an is 0 by the Squeezing Theorem for Sequences. ■

604

Chapter 9 / Infinite Series

Example 9

Consider the sequence

1 1 1 1 1, â&#x2C6;&#x2019; , 2 , â&#x2C6;&#x2019; 3 , . . . , (â&#x2C6;&#x2019;1)n n , . . . 2 2 2 2 If we take the absolute value of each term, we obtain the sequence 1 1 1 1 , , ,..., n,... 2 22 23 2 which, as shown in Example 4, converges to 0. Thus, from Theorem 9.1.6 we have 1 lim (â&#x2C6;&#x2019;1)n n = 0 n â&#x2020;&#x2019; +âŹ 2 1,

SEQUENCES DEFINED RECURSIVELY

Some sequences do not arise from a formula for the general term, but rather from a formula or set of formulas that specify how to generate each term in the sequence from terms that precede it; such sequences are said to be deďŹ ned recursively, and the deďŹ ning formulas are called recursion formulas. A good example is the mechanicâ&#x20AC;&#x2122;s rule for approximating square roots. In Exercise 25 of Section 4.7 you were asked to show that

a 1 xn + (2) x1 = 1, xn+1 = 2 xn â&#x2C6;&#x161; describes the sequence produced by Newtonâ&#x20AC;&#x2122;s Method to approximate a as a zero of the function f(x) = x 2 â&#x2C6;&#x2019; a. Tableâ&#x2C6;&#x161;9.1.5 shows the ďŹ rst ďŹ ve terms in an application of the mechanicâ&#x20AC;&#x2122;s rule to approximate 2. Table 9.1.5

1 2

ĺ&#x2020;¸xn + x2n ĺ&#x2020;š

x1 = 1,

xn +1 =

x1 = 1

(Starting value)

decimal approximation 1.00000000000

ĺ&#x2020;&#x2039;1 + 21 ĺ&#x2020;&#x152; = 32

1.50000000000

ĺ&#x2020;&#x2039;

1.41666666667

x2 =

1 2

x3 =

1 3 2 2

x4 =

1 17 2 12

x5 =

1 577 2 408

x6 =

1 665,857 2 470,832

ĺ&#x2020;&#x2039;

ĺ&#x2020;&#x2039; ĺ&#x2020;&#x2039;

2 3/2

ĺ&#x2020;&#x152; = 1712 ĺ&#x2020;&#x152;

2 + 17/12 =

2 577/408

577 408

1.41421568627

ĺ&#x2020;&#x152; = 665,857 470,832

1.41421356237

2 665,857/470,832

ĺ&#x2020;&#x152; = 886,731,088,897 627,013,566,048

1.41421356237

It would take us too far aďŹ eld to investigate the convergence of sequences deďŹ ned recursively, but we will conclude this section with a useful technique that can sometimes be used to compute limits of such sequences.

Example 10 Assuming that the sequence in Table 9.1.5 converges, show that the limit â&#x2C6;&#x161; 2.

9.1 Sequences

605

Solution. Assume that xn → L, where L is to be determined. Since n + 1 → +⬁ as n → +⬁, it is also true that xn+1 → L as n → +⬁. Thus, if we take the limit of the expression

2 1 xn + xn+1 = 2 xn

as n → +⬁, we obtain 1 2 L= L+ 2 L

which can be rewritten as L2 = √ 2. The negative solution of this equation is extraneous because xn > 0 for all n, so L = 2.

✔QUICK CHECK EXERCISES 9.1

(See page 607 for answers.)

1. Consider the sequence 4, 6, 8, 10, 12, . . . . (a) If {an }+⬁ , n=1 denotes this sequence, then a1 = a4 = , and a7 = . The general term is an = . (b) If {bn }+⬁ , denotes this sequence, then b0 = n=0 b4 = , and b8 = . The general term is bn = . 2. What does it mean to say that a sequence {an } converges?

3. Consider sequences {an } and {bn }, where an → 2 as n → +⬁ and bn = (−1)n . Determine which of the following se-

EXERCISE SET 9.1

quences converge and which diverge. If a sequence converges, indicate its limit. 2 (a) {bn } (b) {3a (c) {b n − 1} n} 1 bn (e) (f ) (d) {an + bn } an2 + 3 1000 4. Suppose that {an }, {bn }, and {cn } are sequences such that an ≤ bn ≤ cn for all n ≥ 10, and that {an } and {cn } both converge to 12. Then the Theorem for Sequences implies that {bn } converges to .

Graphing Utility

1. In each part, ﬁnd a formula for the general term of the sequence, starting with n = 1. 1 1 1 1 1 1 (a) 1, , , ,... (b) 1, − , , − , . . . 3 9 27 3 9 27 1 3 5 7 4 9 16 1 (c) , , , , . . . , √ , √ (d) √ , √ ,... 3 4 2 4 6 8 π π π 5π 2. In each part, ﬁnd two formulas for the general term of the sequence, one starting with n = 1 and the other with n = 0. (a) 1, −r, r 2 , −r 3 , . . . (b) r, −r 2 , r 3 , −r 4 , . . .

3. (a) Write out the first four terms of the sequence {1 + (−1)n }, starting with n = 0. (b) Write out the first four terms of the sequence {cos nπ}, starting with n = 0. (c) Use the results in parts (a) and (b) to express the general term of the sequence 4, 0, 4, 0, . . . in two different ways, starting with n = 0. 4. In each part, find a formula for the general term using factorials and starting with n = 1. (a) 1 · 2, 1 · 2 · 3 · 4, 1 · 2 · 3 · 4 · 5 · 6, 1 · 2 · 3 · 4 · 5 · 6 · 7 · 8, . . . (b) 1, 1 · 2 · 3, 1 · 2 · 3 · 4 · 5, 1 · 2 · 3 · 4 · 5 · 6 · 7, . . .

π x and deﬁne se2 quences {an } and {bn } by an = f(2n) and bn = f(2n + 1). ■ 5–6 Let f be the function f(x) = cos

5. (a) Does limx → +⬁ f(x) exist? Explain. (b) Evaluate a1 , a2 , a3 , a4 , and a5 . (c) Does {an } converge? If so, find its limit. 6. (a) Evaluate b1 , b2 , b3 , b4 , and b5 . (b) Does {bn } converge? If so, find its limit. (c) Does {f(n)} converge? If so, find its limit. 7–22 Write out the first five terms of the sequence, determine

whether the sequence converges, and if so ﬁnd its limit. ■ +⬁ +⬁ n2 n 8. 9. {2}+⬁ 7. n=1 n + 2 n=1 2n + 1 n=1 +⬁

ln n +⬁ π +⬁ 1 11. 12. n sin 10. ln n n=1 n n=1 n n=1 +⬁ (−1)n+1 13. {1 + (−1)n }+⬁ 14. n=1 n2 n=1 +⬁

3 n +⬁ 2n 16. 15. (−1)n 3 n + 1 n=1 2n n=1 n +⬁ π (n + 1)(n + 2) +⬁ 17. 18. 2n2 4n n=1 n=1 √ 19. {n2 e−n }+⬁ 20. { n2 + 3n − n}+⬁ n=1 n=1

606

21.

Chapter 9 / Infinite Series

n+3 n+1

n +⬁

n=1

22.

+⬁

2 n 1− n n=1

23–30 Find the general term of the sequence, starting with n = 1, determine whether the sequence converges, and if so ﬁnd its limit. ■ 1 3 5 7 1 2 3 23. , , , , . . . 24. 0, 2 , 2 , 2 , . . . 2 4 6 8 2 3 4 1 1 1 1 25. , − , , − ,... 26. −1, 2, −3, 4, −5, . . . 3 9 27 81

1 1 1 1 1 1 1 27. 1 − − − − , , , ,... 2 3 2 3 4 5 4 3 3 3 28. 3, , 2 , 3 , . . . 2 2 2 √ √ √ √ √ √ 29. ( 2 − 3 ), ( 3 − 4 ), ( 4 − 5 ), . . . 1 1 1 1 30. 5 , − 6 , 7 , − 8 , . . . 3 3 3 3 31–34 True–False Determine whether the statement is true or false. Explain your answer. ■

31. Sequences are functions. 32. If {an } and {bn } are sequences such that {an + bn } converges, then {an } and {bn } converge.

33. If {an } diverges, then an → +⬁ or an → −⬁. 34. If the graph of y = f(x) has a horizontal asymptote as x → +⬁, then the sequence {f(n)} converges.

35–36 Use numerical evidence to make a conjecture about the limit of the sequence, and then use the Squeezing Theorem for Sequences (Theorem 9.1.5) to conﬁrm that your conjecture is correct. ■

sin2 n 1+n n 35. lim 36. lim n → +⬁ n → +⬁ n 2n F O C U S O N C O N C E P TS

37. Give two examples of sequences, all of whose terms are between −10 and 10, that do not converge. Use graphs of your sequences to explain their properties. 38. (a) Suppose that f satisﬁes limx → 0+ f(x) = +⬁. Is it possible that the sequence {f(1/n)} converges? Explain. (b) Find a function f such that limx → 0+ f(x) does not exist but the sequence {f(1/n)} converges.

39. (a) Starting with n = 1, write out the ﬁrst six terms of the sequence {an }, where 1, if n is odd an = n, if n is even

(b) Starting with n = 1, and considering the even and odd terms separately, ﬁnd a formula for the general term of the sequence 1 1 1 1, 2 , 3, 4 , 5, 6 , . . . 2 2 2

(c) Starting with n = 1, and considering the even and odd terms separately, ﬁnd a formula for the general term of the sequence 1 1 1 1 1 1 1 1 1, , , , , , , , , . . . 3 3 5 5 7 7 9 9 (d) Determine whether the sequences in parts (a), (b), and (c) converge. For those that do, ﬁnd the limit. 40. For what positive values of b does the sequence b, 0, b2 , 0, b3 , 0, b4 , . . . converge? Justify your answer. 41. Assuming that the sequence given in Formula (2) of this section converges, use the method √ of Example 10 to show that the limit of this sequence is a. 42. Consider the sequence √ a1 = 6 √ a2 = 6 + 6 a3 =

a4 =

6+ 6+ 6+

√

.. . (a) Find a recursion formula for an+1 . (b) Assuming that the sequence converges, use the method of Example 10 to find the limit. 43. (a) A bored student enters the number 0.5 in a calculator display and then repeatedly computes the square of the number in the display. Taking a0 = 0.5, find a formula for the general term of the sequence {an } of numbers that appear in the display. (b) Try this with a calculator and make a conjecture about the limit of an . (c) Confirm your conjecture by finding the limit of an . (d) For what values of a0 will this procedure produce a convergent sequence? 44. Let 2x, 0 ≤ x < 0.5 f(x) = 2x − 1, 0.5 ≤ x < 1 Does the sequence f(0.2), f(f(0.2)), f(f(f(0.2))), . . . converge? Justify your reasoning. 45. (a) Use a graphing utility to generate the graph of the equation y = (2x + 3x )1/x , and then use the graph to make a conjecture about the limit of the sequence {(2n + 3n )1/n }+⬁ n=1 (b) Confirm your conjecture by calculating the limit. 46. Consider the sequence {an }+⬁ n=1 whose nth term is n 1 1 an = n k=1 1 + (k /n)

Show that limn → +⬁ an = ln 2 by interpreting an as the Riemann sum of a deﬁnite integral.

9.2 Monotone Sequences

47. The sequence whose terms are 1, 1, 2, 3, 5, 8, 13, 21, . . . is called the Fibonacci sequence in honor of the Italian mathematician Leonardo (“Fibonacci”) da Pisa (c. 1170–1250). This sequence has the property that after starting with two 1’s, each term is the sum of the preceding two. (a) Denoting the sequence by {an } and starting with a1 = 1 and a2 = 1, show that an+2 an =1+ if n ≥ 1 an+1 an+1 (b) Give a reasonable informal argument to show that if the sequence {an+1 /an } converges to some limit L, then the sequence {an+2 /an+1 } must also converge to L. (c) Assuming that the sequence {an+1 /an } converges, show √ that its limit is (1 + 5 )/2. 48. If we accept the fact that the sequence {1/n}+⬁ n=1 converges to the limit L = 0, then according to Deﬁnition 9.1.2, for every ǫ > 0 there exists a positive integer N such that |an − L| = |(1/n) − 0| < ǫ when n ≥ N . In each part, ﬁnd the smallest possible value of N for the given value of ǫ. (a) ǫ = 0.5 (b) ǫ = 0.1 (c) ǫ = 0.001

607

49. If we accept the fact that the sequence +⬁ n n + 1 n=1 converges to the limit L = 1, then according to Deﬁnition 9.1.2, for every ǫ > 0 there exists an integer N such that n |an − L| = − 1 < ǫ n+1 when n ≥ N. In each part, ﬁnd the smallest value of N for the given value of ǫ. (a) ǫ = 0.25 (b) ǫ = 0.1 (c) ǫ = 0.001

50. Use Deﬁnition 9.1.2 to prove that (a) the sequence {1/n}+⬁ converges to 0 n=1 +⬁ n (b) the sequence converges to 1. n + 1 n=1 51. Writing Discuss, with examples, various ways that a sequence could diverge.

52. Writing Discuss the convergence of the sequence {r n } considering the cases |r| < 1, |r| > 1, r = 1, and r = −1 separately.

✔QUICK CHECK ANSWERS 9.1 1. (a) 4; 10; 16; 2n + 2 (b) 4; 12; 20; 2n + 4 (d) diverges (e) converges to

9.2

1 7

2. lim an exists n → +⬁

(f ) diverges

3. (a) diverges (b) converges to 5 (c) converges to 1

4. Squeezing; 12

MONOTONE SEQUENCES There are many situations in which it is important to know whether a sequence converges, but the value of the limit is not relevant to the problem at hand. In this section we will study several techniques that can be used to determine whether a sequence converges. TERMINOLOGY We begin with some terminology.

9.2.1

Note that an increasing sequence need not be strictly increasing, and a decreasing sequence need not be strictly decreasing.

definition A sequence {an }+⬁ n=1 is called strictly increasing if increasing if strictly decreasing if decreasing if

a1 < a2 < a3 < · · · < an < · · · a1 ≤ a2 ≤ a3 ≤ · · · ≤ an ≤ · · · a1 > a2 > a3 > · · · > an > · · · a1 ≥ a2 ≥ a3 ≥ · · · ≥ an ≥ · · ·

A sequence that is either increasing or decreasing is said to be monotone, and a sequence that is either strictly increasing or strictly decreasing is said to be strictly monotone.

Some examples are given in Table 9.2.1 and their corresponding graphs are shown in Figure 9.2.1. The ﬁrst and second sequences in Table 9.2.1 are strictly monotone; the third

608

Chapter 9 / Infinite Series

and fourth sequences are monotone but not strictly monotone; and the ﬁfth sequence is neither strictly monotone nor monotone. Table 9.2.1

sequence

description

1 , 2, 3 , . . . , n , . . . 2 3 4 n+1 1 1 1, , ,..., 1 ,... n 2 3 1, 1, 2, 2, 3, 3, . . . 1, 1, 1 , 1 , 1 , 1 , . . . 2 2 3 3 1 1, – , 1 , – 1 , . . . , (−1) n+1 1 , . . . n 2 3 4

Strictly increasing Strictly decreasing Increasing; not strictly increasing Decreasing; not strictly decreasing Neither increasing nor decreasing

0.8

0.6

0.4

0.2

n 2

n n+1

10 12

6 5 4 3 2 1

n 2

+∞ n =1

1 n

10 12

+∞ n =1

10 12

1, 1, 2, 2, 3, 3, ...

0.8 0.5

0.6

0.4 1

0.2

n 2

Can a sequence be both increasing and decreasing? Explain.

1 (−1)n+1 n

+∞ n=1

−0.5

10 12

1, 1, 12 , 12 , 13 , 13 , ... Figure 9.2.1

TESTING FOR MONOTONICITY

Frequently, one can guess whether a sequence is monotone or strictly monotone by writing out some of the initial terms. However, to be certain that the guess is correct, one must give a precise mathematical argument. Table 9.2.2 provides two ways of doing this, one based Table 9.2.2

difference between successive terms

ratio of successive terms

an+1 − an > 0 an+1 − an < 0 an+1 − an ≥ 0 an+1 − an ≤ 0

an+1/an > 1 an+1/an < 1 an+1/an ≥ 1 an+1/an ≤ 1

conclusion Strictly increasing Strictly decreasing Increasing Decreasing

9.2 Monotone Sequences

609

on differences of successive terms and the other on ratios of successive terms. It is assumed in the latter case that the terms are positive. One must show that the speciﬁed conditions hold for all pairs of successive terms. y

Example 1

Use differences of successive terms to show that

1 2 3 n , , ,..., ,... 2 3 4 n+1 (Figure 9.2.2) is a strictly increasing sequence.

0.8 0.6 0.4 0.2

n 2

n n+1 Figure 9.2.2

10 12

+∞ n=1

Solution. The pattern of the initial terms suggests that the sequence is strictly increasing. To prove that this is so, let n an = n+1 We can obtain an+1 by replacing n by n + 1 in this formula. This yields an+1 = Thus, for n ≥ 1 an+1 − an =

n+1 n+1 = (n + 1) + 1 n+2

n+1 n n2 + 2n + 1 − n2 − 2n 1 − = = >0 n+2 n+1 (n + 1)(n + 2) (n + 1)(n + 2)

which proves that the sequence is strictly increasing.

Example 2 Use ratios of successive terms to show that the sequence in Example 1 is strictly increasing.

Solution. As shown in the solution of Example 1, n n+1 and an+1 = n+1 n+2 Forming the ratio of successive terms we obtain an =

(n + 1)/(n + 2) n+1 n+1 n2 + 2n + 1 an+1 = = · = an n/(n + 1) n+2 n n2 + 2n

(1)

from which we see that an+1 /an > 1 for n ≥ 1. This proves that the sequence is strictly increasing. The following example illustrates still a third technique for determining whether a sequence is strictly monotone. Example 3

In Examples 1 and 2 we proved that the sequence

1 2 3 n , , ,..., ,... 2 3 4 n+1 is strictly increasing by considering the difference and ratio of successive terms. Alternatively, we can proceed as follows. Let x f(x) = x+1 so that the nth term in the given sequence is an = f(n). The function f is increasing for x ≥ 1 since (x + 1)(1) − x(1) 1 f ′(x) = = >0 (x + 1)2 (x + 1)2

610

Chapter 9 / Infinite Series

Thus,

Table 9.2.3

conclusion for derivative of the sequence f for x ≥ 1 with an = f (n) f ′(x) > 0 f ′(x) < 0 f ′(x) ≥ 0 f ′(x) ≤ 0

an = f(n) < f(n + 1) = an+1

which proves that the given sequence is strictly increasing.

Strictly increasing Strictly decreasing Increasing Decreasing

In general, if f(n) = an is the nth term of a sequence, and if f is differentiable for x ≥ 1, then the results in Table 9.2.3 can be used to investigate the monotonicity of the sequence. PROPERTIES THAT HOLD EVENTUALLY

Sometimes a sequence will behave erratically at ﬁrst and then settle down into a deﬁnite pattern. For example, the sequence 9, −8, −17, 12, 1, 2, 3, 4, . . .

(2)

is strictly increasing from the fifth term on, but the sequence as a whole cannot be classified as strictly increasing because of the erratic behavior of the first four terms. To describe such sequences, we introduce the following terminology.

9.2.2 definition If discarding ﬁnitely many terms from the beginning of a sequence produces a sequence with a certain property, then the original sequence is said to have that property eventually.

For example, although we cannot say that sequence (2) is strictly increasing, we can say that it is eventually strictly increasing. y 3000

Example 4

10 n Show that the sequence n!

+⬁

is eventually strictly decreasing.

n=1

2000

Solution. We have 1000

an = n 5

10 n/n! Figure 9.2.3

20 +∞ n=1

10 n n!

and an+1 =

10 n+1 (n + 1)!

10 n+1 /(n + 1)! 10 n+1 n! n! 10 an+1 (3) = = = 10 = n n / an 10 n! 10 (n + 1)! (n + 1)n! n+1 From (3), an+1 /an < 1 for all n ≥ 10, so the sequence is eventually strictly decreasing, as conﬁrmed by the graph in Figure 9.2.3. AN INTUITIVE VIEW OF CONVERGENCE

Informally stated, the convergence or divergence of a sequence does not depend on the behavior of its initial terms, but rather on how the terms behave eventually. For example, the sequence 1 1 1 3, −9, −13, 17, 1, , , , . . . 2 3 4 eventually behaves like the sequence 1,

1 1 1 , ,..., ,... 2 3 n

and hence has a limit of 0. CONVERGENCE OF MONOTONE SEQUENCES The following two theorems, whose proofs are discussed at the end of this section, show that a monotone sequence either converges or becomes inﬁnite—divergence by oscillation cannot occur.

9.2 Monotone Sequences

611

9.2.3 theorem If a sequence {an } is eventually increasing, then there are two possibilities: (a) There is a constant M, called an upper bound for the sequence, such that an ≤ M for all n, in which case the sequence converges to a limit L satisfying L ≤ M. (b) No upper bound exists, in which case lim an = +⬁. n → +⬁

9.2.4 theorem If a sequence {an } is eventually decreasing, then there are two possibilities: Theorems 9.2.3 and 9.2.4 are examples of existence theorems; they tell us whether a limit exists, but they do not provide a method for finding it.

(a) There is a constant M, called a lower bound for the sequence, such that an ≥ M for all n, in which case the sequence converges to a limit L satisfying L ≥ M. (b) No lower bound exists, in which case lim an = −⬁. n → +⬁

Example 5

10 n Show that the sequence n!

+⬁

converges and ﬁnd its limit.

n=1

Solution. We showed in Example 4 that the sequence is eventually strictly decreasing. Since all terms in the sequence are positive, it is bounded below by M = 0, and hence Theorem 9.2.4 guarantees that it converges to a nonnegative limit L. However, the limit is not evident directly from the formula 10 n /n! for the nth term, so we will need some ingenuity to obtain it. It follows from Formula (3) of Example 4 that successive terms in the given sequence are related by the recursion formula 10 (4) an an+1 = n+1 where an = 10 n /n!. We will take the limit as n → +⬁ of both sides of (4) and use the fact that = lim a = L lim a n → +⬁

We obtain L = lim an+1 = lim n → +⬁

so that

n → +⬁

n+1

10 an n+1

n → +⬁

= lim

n → +⬁

10 n =0 n → +⬁ n!

10 lim an = 0 · L = 0 n + 1 n → +⬁

L = lim

In the exercises we will show that the technique illustrated in the last example can be adapted to obtain xn =0 lim (5) n → +⬁ n! for any real value of x (Exercise 31). This result will be useful in our later work.

THE COMPLETENESS AXIOM In this text we have accepted the familiar properties of real numbers without proof, and indeed, we have not even attempted to deﬁne the term real number. Although this is sufﬁcient for many purposes, it was recognized by the late nineteenth century that the study of limits

612

Chapter 9 / Infinite Series

and functions in calculus requires a precise axiomatic formulation of the real numbers analogous to the axiomatic development of Euclidean geometry. Although we will not attempt to pursue this development, we will need to discuss one of the axioms about real numbers in order to prove Theorems 9.2.3 and 9.2.4. But ﬁrst we will introduce some terminology. If S is a nonempty set of real numbers, then we call u an upper bound for S if u is greater than or equal to every number in S, and we call l a lower bound for S if l is smaller than or equal to every number in S. For example, if S is the set of numbers in the interval (1, 3), then u = 10, 4, 3.2, and 3 are upper bounds for S and l = −10, 0, 0.5, and 1 are lower bounds for S. Observe also that u = 3 is the smallest of all upper bounds and l = 1 is the largest of all lower bounds. The existence of a smallest upper bound and a largest lower bound for S is not accidental; it is a consequence of the following axiom.

9.2.5 axiom (The Completeness Axiom) If a nonempty set S of real numbers has an upper bound, then it has a smallest upper bound (called the least upper bound ), and if a nonempty set S of real numbers has a lower bound, then it has a largest lower bound (called the greatest lower bound ).

proof of theorem 9.2.3 (a) We will prove the result for increasing sequences, and leave it for the reader to adapt the argument to sequences that are eventually increasing. Assume there exists a number M such that an ≤ M for n = 1, 2, . . . . Then M is an upper bound for the set of terms in the sequence. By the Completeness Axiom there is a least upper bound for the terms; call it L. Now let ǫ be any positive number. Since L is the least upper bound for the terms, L − ǫ is not an upper bound for the terms, which means that there is at least one term aN such that a >L−ǫ N

Moreover, since {an } is an increasing sequence, we must have an ≥ aN > L − ǫ

(6)

when n ≥ N . But an cannot exceed L since L is an upper bound for the terms. This observation together with (6) tells us that L ≥ an > L − ǫ for n ≥ N, so all terms from the Nth on are within ǫ units of L. This is exactly the requirement to have lim an = L

n → +⬁

Finally, L ≤ M since M is an upper bound for the terms and L is the least upper bound. This proves part (a). (b) If there is no number M such that an ≤ M for n = 1, 2, . . . , then no matter how large we choose M, there is a term aN such that aN > M and, since the sequence is increasing, an ≥ aN > M when n ≥ N. Thus, the terms in the sequence become arbitrarily large as n increases. That is, lim an = +⬁ ■ n → +⬁

We omit the proof of Theorem 9.2.4 since it is similar to that of 9.2.3.

9.2 Monotone Sequences

✔QUICK CHECK EXERCISES 9.2

613

(See page 614 for answers.)

1. Classify each sequence as (I) increasing, (D) decreasing, or (N) neither increasing nor decreasing. {2n} {2−n } 5−n −1 n2 n2 n (−1) n2 2. Classify each sequence as (M) monotonic, (S) strictly monotonic, or (N) not monotonic.

3. Since

{n + (−1)n } {3n + (−1)n }

{2n + (−1)n }

n/[2(n + 1)] n2 = 2 > (n − 1)/(2n) n −1

the sequence {(n − 1)/(2n)} is strictly

4. Since

d [(x − 8)2 ] > 0 for x > dx

the sequence {(n − 8)2 } is

strictly

EXERCISE SET 9.2 1–6 Use the difference an+1 − an to show that the given sequence {an } is strictly increasing or strictly decreasing. ■ +⬁ +⬁ 1 n 1 +⬁ 1. 2. 1 − 3. n n=1 n n=1 2n + 1 n=1 +⬁ n 4. 5. {n − 2n }+⬁ 6. {n − n2 }+⬁ n=1 n=1 4n − 1 n=1 7–12 Use the ratio an+1 /an to show that the given sequence {an }

is strictly increasing or strictly decreasing. ■ n +⬁ +⬁ n 2 9. {ne−n }+⬁ 7. 8. n=1 2n + 1 n=1 1 + 2n n=1 n +⬁ n +⬁ n +⬁ 10 n 5 10. 11. 12. (2n)! n=1 n! n=1 2(n2 ) n=1

13–16 True–False Determine whether the statement is true or

false. Explain your answer. ■ 13. If an+1 − an > 0 for all n ≥ 1, then the sequence {an } is strictly increasing. 14. A sequence {an } is monotone if an+1 − an = 0 for all n ≥ 1. 15. Any bounded sequence converges.

16. If {an } is eventually increasing, then a100 < a200 . 17–20 Use differentiation to show that the given sequence is

strictly increasing or strictly decreasing. ■ +⬁ n ln(n + 2) +⬁ 17. 18. 2n + 1 n=1 n+2 n=1 19. {tan

−1

n}+⬁ n=1

20.

{ne−2n }+⬁ n=1

21–24 Show that the given sequence is eventually strictly increasing or eventually strictly decreasing. ■ +⬁ n 21. {2n2 − 7n}+⬁ 22. n=1 n2 + 10 n=1 +⬁ n! 23. 24. {n5 e−n }+⬁ n=1 3n n=1

F O C U S O N C O N C E P TS

25. Suppose that {an } is a monotone sequence such that 1 ≤ an ≤ 2 for all n. Must the sequence converge? If so, what can you say about the limit? 26. Suppose that {an } is a monotone sequence such that an ≤ 2 for all n. Must the sequence converge? If so, what can you say about the limit? √ the sequence defined recursively by a1 = 2 27. Let {an } be √ and an+1 = 2 + an for n ≥ 1. (a) List the first three terms of the sequence. (b) Show that an < 2 for n ≥ 1. 2 (c) Show that an+1 − an2 = (2 − an )(1 + an ) for n ≥ 1. (d) Use the results in parts (b) and (c) to show that {an } is a strictly increasing sequence. [Hint: If x and y are positive real numbers such that x 2 − y 2 > 0, then it follows by factoring that x − y > 0.] (e) Show that {an } converges and find its limit L.

28. Let {an } be the sequence deﬁned recursively by a1 = 1 and an+1 = 21 [an + (3/an )] for n ≥ 1. √ (a) Show that an ≥ 3 for n ≥ 2. [Hint: What is the minimum value of 21 [x + (3/x)] for x > 0?] (b) Show that {an } is eventually decreasing. [Hint: Examine an+1 − an or an+1 /an and use the result in part (a).] (c) Show that {an } converges and ﬁnd its limit L.

29–30 The Beverton–Holt model is used to describe changes in a population from one generation to the next under certain assumptions. If the population in generation n is given by xn , the Beverton–Holt model predicts that the population in the next generation satisﬁes RKxn xn+1 = K + (R − 1)xn

for some positive constants R and K with R > 1. These exercises explore some properties of this population model. ■

614

Chapter 9 / Infinite Series

29. Let {xn } be the sequence of population values defined recursively by x1 = 60, and for n ≥ 1, xn+1 is given by the Beverton–Holt model with R = 10 and K = 300. (a) List the first four terms of the sequence {xn }. (b) If 0 < xn < 300, show that 0 < xn+1 < 300. Conclude that 0 < xn < 300 for n ≥ 1. (c) Show that {xn } is increasing. (d) Show that {xn } converges and find its limit L.

(b) Use the result in part (a) to show that (n + 1)n+1 nn < n! < , en−1 en

(c) Use the Squeezing Theorem for Sequences (Theorem 9.1.5) and the result in part (b) to show that √ n 1 n! lim = n → +⬁ n e

30. Let {xn } be a sequence of population values deﬁned recursively by the Beverton–Holt model for which x1 > K. Assume that the constants R and K satisfy R > 1 and K > 0. (a) If xn > K, show that xn+1 > K. Conclude that xn > K for all n ≥ 1. (b) Show that {xn } is decreasing. (c) Show that {xn } converges and ﬁnd its limit L.

y = ln x

31. The goal of this exercise is to establish Formula (5), namely, xn =0 lim n → +⬁ n! Let an = |x|n /n! and observe that the case where x = 0 is obvious, so we will focus on the case where x = 0. (a) Show that |x| an+1 = an n+1 (b) Show that the sequence {an } is eventually strictly decreasing. (c) Show that the sequence {an } converges.

1 2 3

...

x 1 2 3

...

n n+1

Figure Ex-32

33. Use the left inequality in Exercise 32(b) to show that √ n n! = +⬁ lim n → +⬁

34. Writing Give an example of an increasing sequence that is not eventually strictly increasing. What can you conclude about the terms of any such sequence? Explain. 35. Writing Discuss the appropriate use of “eventually” for various properties of sequences. For example, which is a useful expression: “eventually bounded” or “eventually monotone”?

32. (a) Compare appropriate areas in the accompanying ﬁgure to deduce the following inequalities for n ≥ 2: n n+1 ln x dx < ln n! < ln x dx 1

n>1

✔QUICK CHECK ANSWERS 9.2 1. I; D; N; I; N

9.3

2. N; M; S

3. 1; increasing

4. 8; eventually; increasing

INFINITE SERIES The purpose of this section is to discuss sums that contain inﬁnitely many terms. The most familiar examples of such sums occur in the decimal representations of real numbers. For example, when we write 31 in the decimal form 13 = 0.3333 . . . , we mean 1 = 0.3 + 0.03 + 0.003 + 0.0003 + · · · 3 which suggests that the decimal representation of many real numbers.

1 3

can be viewed as a sum of inﬁnitely

SUMS OF INFINITE SERIES

Our first objective is to define what is meant by the “sum” of infinitely many real numbers. We begin with some terminology.

9.3 Infinite Series

9.3.1

615

definition An inﬁnite series is an expression that can be written in the form ⬁ k=1

uk = u1 + u2 + u3 + · · · + uk + · · ·

The numbers u1 , u2 , u3 , . . . are called the terms of the series.

Since it is impossible to add infinitely many numbers together directly, sums of infinite series are defined and computed by an indirect limiting process. To motivate the basic idea, consider the decimal (1) 0.3333 . . . This can be viewed as the infinite series 0.3 + 0.03 + 0.003 + 0.0003 + · · · or, equivalently, 3 3 3 3 + + 3 + 4 + ··· 10 102 10 10

(2)

Since (1) is the decimal expansion of 31 , any reasonable definition for the sum of an infinite series should yield 31 for the sum of (2). To obtain such a definition, consider the following sequence of (finite) sums: 3 10 3 s2 = 10 3 s3 = 10 3 s4 = 10 .. . s1 =

= 0.3 3 = 0.33 102 3 3 + 2 + 3 = 0.333 10 10 3 3 3 + 2 + 3 + 4 = 0.3333 10 10 10 +

The sequence of numbers s1 , s2 , s3 , s4 , . . . (Figure 9.3.1) can be viewed as a succession of approximations to the “sum” of the inﬁnite series, which we want to be 31 . As we progress through the sequence, more and more terms of the inﬁnite series are used, and the approximations get better and better, suggesting that the desired sum of 13 might be the limit of this sequence of approximations. To see that this is so, we must calculate the limit of the general term in the sequence of approximations, namely,

0.4

1/3 0.3

sn = n 1

(3)

The problem of calculating

0.3, 0.33, 0.333, ... Figure 9.3.1

3 3 3 + 2 + ··· + n 10 10 10

lim sn = lim

n → +⬁

3 3 3 + + ··· + n 10 102 10

is complicated by the fact that both the last term and the number of terms in the sum change with n. It is best to rewrite such limits in a closed form in which the number of terms does not vary, if possible. (See the discussion of closed form and open form following Example 1 2 in Section 5.4.) To do this, we multiply both sides of (3) by 10 to obtain 1 3 3 3 3 sn = 2 + 3 + · · · + n + n+1 10 10 10 10 10

(4)

616

Chapter 9 / Infinite Series

and then subtract (4) from (3) to obtain 1 3 3 sn = − 10 10 10 n+1

9 3 1 sn = 1− n 10 10 10

1 1 1− n sn = 3 10 sn −

Since 1/10 n → 0 as n → +⬁, it follows that

1 1 1 lim sn = lim 1− n = n → +⬁ 3 n → +⬁ 10 3

which we denote by writing 1 3 3 3 3 = + 2 + 3 + ··· + n + ··· 3 10 10 10 10 Motivated by the preceding example, we are now ready to deﬁne the general concept of the “sum” of an inﬁnite series u 1 + u 2 + u 3 + · · · + uk + · · · We begin with some terminology: Let sn denote the sum of the initial terms of the series, up to and including the term with index n. Thus, s1 = u1 s2 = u1 + u2 s3 = u1 + u2 + u3 .. . sn = u1 + u2 + u3 + · · · + un =

k=1

The number sn is called the nth partial sum of the series and the sequence {sn }+⬁ n=1 is called the sequence of partial sums. As n increases, the partial sum sn = u1 + u2 + · · · + un includes more and more terms of the series. Thus, if sn tends toward a limit as n → +⬁, it is reasonable to view this limit as the sum of all the terms in the series. This suggests the following deﬁnition.

WARNING In everyday language the words “sequence” and “series” are often used interchangeably. However, in mathematics there is a distinction between these two words—a sequence is a succession whereas a series is a sum. It is essential that you keep this distinction in mind.

9.3.2 definition Let {sn } be the sequence of partial sums of the series u1 + u2 + u3 + · · · + uk + · · · If the sequence {sn } converges to a limit S, then the series is said to converge to S, and S is called the sum of the series. We denote this by writing S=

⬁

k=1

If the sequence of partial sums diverges, then the series is said to diverge. A divergent series has no sum.

Example 1

Determine whether the series 1 − 1 + 1 − 1 + 1 − 1 + ···

converges or diverges. If it converges, ﬁnd the sum.

9.3 Infinite Series

617

Solution. It is tempting to conclude that the sum of the series is zero by arguing that the positive and negative terms cancel one another. However, this is not correct; the problem is that algebraic operations that hold for finite sums do not carry over to infinite series in all cases. Later, we will discuss conditions under which familiar algebraic operations can be applied to infinite series, but for this example we turn directly to Definition 9.3.2. The partial sums are s1 = 1 s2 = 1 − 1 = 0 s3 = 1 − 1 + 1 = 1

s4 = 1 − 1 + 1 − 1 = 0 y

and so forth. Thus, the sequence of partial sums is

1.5

1, 0, 1, 0, 1, 0, . . .

(Figure 9.3.2). Since this is a divergent sequence, the given series diverges and consequently has no sum.

0.5 n 2

1, 0, 1, 0, 1, 0, ... Figure 9.3.2

GEOMETRIC SERIES

In many important series, each term is obtained by multiplying the preceding term by some ﬁxed constant. Thus, if the initial term of the series is a and each term is obtained by multiplying the preceding term by r, then the series has the form ⬁ k=0

ar k = a + ar + ar 2 + ar 3 + · · · + ar k + · · ·

(a = 0)

(5)

Such series are called geometric series, and the number r is called the ratio for the series. Here are some examples: 1 + 2 + 4 + 8 + · · · + 2k + · · ·

a = 1, r = 2

3 3 3 3 + + 3 + ··· + k + ··· 10 102 10 10 1 1 1 1 1 − + − + · · · + (−1)k+1 k + · · · 2 4 8 16 2 1 + 1 + 1 + ··· + 1 + ···

Sometimes it is desirable to start the index of summation of an infinite series at k = 0 rather than k = 1, in which case we would call u0 the zeroth term and s0 = u0 the zeroth partial sum. One can prove that changing the starting value for the index of summation of an infinite series has no effect on the convergence, the divergence, or the sum. If we had started the index at k = 1 in (5), then the series would be expressed as ⬁

ar k−1

3 10 , r

1 10

a = 21 , r = − 21 a = 1, r = 1

1 − 1 + 1 − 1 + · · · + (−1)k+1 + · · ·

a = 1, r = −1

1 + x + x2 + x3 + · · · + xk + · · ·

a = 1, r = x

The following theorem is the fundamental result on convergence of geometric series. 9.3.3

theorem A geometric series ⬁ k=0

ar k = a + ar + ar 2 + · · · + ar k + · · ·

(a = 0)

converges if |r| < 1 and diverges if |r| ≥ 1. If the series converges, then the sum is ⬁ k=0

ar k =

a 1−r

k=1

Since this expression is more complicated than (5), we started the index at k = 0.

proof Let us treat the case |r| = 1 ﬁrst. If r = 1, then the series is a + a + a + a + ···

Chapter 9 / Infinite Series

618

so the nth partial sum is sn = (n + 1)a and lim sn = lim (n + 1)a = ±⬁ n → +⬁

n → +⬁

(the sign depending on whether a is positive or negative). This proves divergence. If r = −1, the series is a − a + a − a + ··· so the sequence of partial sums is a, 0, a, 0, a, 0, . . . which diverges. Now let us consider the case where |r| = 1. The nth partial sum of the series is sn = a + ar + ar 2 + · · · + ar n

(6)

Multiplying both sides of (6) by r yields rsn = ar + ar 2 + · · · + ar n + ar n+1

(7)

and subtracting (7) from (6) gives sn − rsn = a − ar n+1 or (1 − r)sn = a − ar n+1

(8)

Since r = 1 in the case we are considering, this can be rewritten as sn =

If |r| > 1, then either r > 1 or r < −1. In the case r > 1, r n+1 increases without bound as n → +⬁, and in the case r < −1, r n+1 oscillates between positive and negative values that grow in magnitude, so {sn } diverges in both cases. ■ Example 2 sum.

y 20/ 3 6 5

n 2

6 ∞

Partial sums for k=0

Figure 9.3.3

(9)

If |r| < 1, then r n+1 goes to 0 as n → +⬁ (can you see why?), so {sn } converges. From (9) a lim sn = n → +⬁ 1−r

Note that (6) is an open form for sn , while (9) is a closed form for sn . In general, one needs a closed form to calculate the limit.

a a − ar n+1 = (1 − r n+1 ) 1−r 1−r

In each part, determine whether the series converges, and if so ﬁnd its ⬁ ⬁ 5 (a) (b) 32k 51−k k 4 k=0 k=1

Solution (a). This is a geometric series with a = 5 and r = 41 . Since |r| = series converges and the sum is

5 a = 1−r 1−

5 4k

1 4

< 1, the

20 3

(Figure 9.3.3).

Solution (b). This is a geometric series in concealed form, since we can rewrite it as ⬁ k=1

Since r =

9 5

2k 1−k

3 5

k−1 ⬁ ⬁ 9 9k = 9 = k−1 5 5 k=1 k=1

> 1, the series diverges.

9.3 Infinite Series

Example 3

619

Find the rational number represented by the repeating decimal 0.784784784 . . .

T E C H N O LO GY M A ST E R Y Computer algebra systems have commands for finding sums of convergent series. If you have a CAS, use it to compute the sums in Examples 2 and 3.

Solution. We can write 0.784784784 . . . = 0.784 + 0.000784 + 0.000000784 + · · · so the given decimal is the sum of a geometric series with a = 0.784 and r = 0.001. Thus, 0.784784784 . . . =

a 0.784 0.784 784 = = = 1−r 1 − 0.001 0.999 999

Example 4 In each part, ﬁnd all values of x for which the series converges, and ﬁnd the sum of the series for those values of x. ⬁ 3x 2 3x 3 3(−1)k k 3x + − + ··· + x + ··· xk (b) 3 − (a) 2 4 8 2k k=0

Solution (a). The expanded form of the series is ⬁ k=0

xk = 1 + x + x2 + · · · + xk + · · ·

The series is a geometric series with a = 1 and r = x, so it converges if |x| < 1 and diverges otherwise. When the series converges its sum is ⬁ k=0

xk =

1 1−x

Solution (b). This is a geometric series with a = 3 and r = −x /2. It converges if | −x /2| < 1, or equivalently, when |x| < 2. When the series converges its sum is ⬁ x k 6 3 x = = 3 − 2 2+x 1− − k=0 2

TELESCOPING SUMS

Example 5

Determine whether the series ⬁ k=1

1 1 1 1 1 = + + + + ··· k(k + 1) 1·2 2·3 3·4 4·5

converges or diverges. If it converges, ﬁnd the sum.

Solution. The nth partial sum of the series is sn =

n k=1

1 1 1 1 1 = + + + ··· + k(k + 1) 1·2 2·3 3·4 n(n + 1)

We will begin by rewriting sn in closed form. This can be accomplished by using the method of partial fractions to obtain (verify) 1 1 1 = − k(k + 1) k k+1

620

Chapter 9 / Infinite Series

The sum in (10) is an example of a telescoping sum. The name is derived from the fact that in simplifying the sum, one term in each parenthetical expression cancels one term in the next parenthetical expression, until the entire sum collapses (like a folding telescope) into just two terms.

from which we obtain the sum n

1 1 sn = − k k+1 k=1

1 1 1 1 1 1 1 = 1− − − − + + + ··· + 2 2 3 3 4 n n+1

1 1 1 1 1 1 1 + − + + ··· + − + − =1+ − + 2 2 3 3 n n n+1 1 =1− n+1

⬁ Thus, 1 1 1− = lim sn = lim =1 n → +⬁ k(k + 1) n → +⬁ n+1 k=1

(10)

HARMONIC SERIES

One of the most important of all diverging series is the harmonic series,

⬁ 1

3 2

{sn}

k=1

1 n 2

(a) y 6 5 4 3

{s2 n}

2 1

2n 20 21 22 23 24 25 26 27

(b) Partial sums for the harmonic series Figure 9.3.4

s2 = 1 + s4 = s2 + s8 = s4 +

.. .

1 1 1 1 + + + + ··· 2 3 4 5

1 2 1 3 1 5 1 9

> + +

1 2 1 4 1 6

1 2

2 2

> s2 + +

1 7

= s2 + 21 > 23 > s4 + 18 + 81 + 18 + 81 = s4 +

1 4

1 8

1 4

1 1 1 1 1 1 1 + 10 + 11 + 12 + 13 + 14 + 15 + 16 1 1 1 1 1 1 1 + 16 + 16 + 16 + 16 + 16 + 16 + > s8 + 16

1 16

1 2

4 2

= s8 +

1 2

5 2

n+1 2 If M is any constant, we can ﬁnd a positive integer n such that (n + 1)/2 > M. But for this n n+1 >M s2 n > 2 so that no constant M is greater than or equal to every partial sum of the harmonic series. This proves divergence. This divergence proof, which predates the discovery of calculus, is due to a French bishop and teacher, Nicole Oresme (1323–1382). This series eventually attracted the interest of Johann and Jakob Bernoulli (p. 700) and led them to begin thinking about the general concept of convergence, which was a new idea at that time. s2 n >

This is a proof of the divergence of the harmonic series, as it appeared in an appendix of Jakob Bernoulli’s posthumous publication, Ars Conjectandi, which appeared in 1713.

=1+

which arises in connection with the overtones produced by a vibrating musical string. It is not immediately evident that this series diverges. However, the divergence will become apparent when we examine the partial sums in detail. Because the terms in the series are all positive, the partial sums 1 1 1 1 1 1 s1 = 1, s2 = 1 + , s3 = 1 + + , s4 = 1 + + + , . . . 2 2 3 2 3 4 form a strictly increasing sequence s1 < s2 < s3 < · · · < sn < · · · (Figure 9.3.4a). Thus, by Theorem 9.2.3 we can prove divergence by demonstrating that there is no constant M that is greater than or equal to every partial sum. To this end, we will consider some selected partial sums, namely, s2 , s4 , s8 , s16 , s32 , . . . . Note that the subscripts are successive powers of 2, so that these are the partial sums of the form s2n (Figure 9.3.4b). These partial sums satisfy the inequalities

s16 = s8 +

Courtesy Lilly Library, Indiana University

9.3 Infinite Series

✔QUICK CHECK EXERCISES 9.3

(See page 623 for answers.)

1. In mathematics, the terms “sequence” and “series” have different meanings: a is a succession, whereas a is a sum.

4. A geometric series is a series of the form ⬁

2. Consider the series

k=0

⬁ k=1

This series converges to diverges if .

1 2k

If {sn } is the sequence of partial sums for this series, then s1 = , s2 = , s3 = , s4 = , and sn = . 3. What does it mean to say that a series uk converges?

EXERCISE SET 9.3

sums, ﬁnd a closed form for the nth partial sum, and determine whether the series converges by calculating the limit of the nth partial sum. If the series converges, then state its sum. ■

(b) (c)

2. (a)

2 2 2 2 + + 2 + · · · + k−1 + · · · 5 5 5 1 2 22 2k−1 + + + ··· + + ··· 4 4 4 4 1 1 1 1 + + + ··· + + ··· 2·3 3·4 4·5 (k + 1)(k + 2) ⬁

⬁ ⬁ k 1 1 1 k−1 4 (c) − (b) 4 k+3 k+4 k=1 k=1 k=1

11.

⬁ 7 (−1)k−1 k−1 6 k=1 ⬁ k=1

1 (k + 2)(k + 3)

⬁ k=1

1 9k 2 + 3k − 2

⬁

1 k−2

⬁

k+2

k=3

13.

k=1

4 7k−1

. This series

5. The harmonic series has the form ⬁ k=1

Does the harmonic series converge or diverge?

15. Match a series from one of Exercises 3, 5, 7, or 9 with the graph of its sequence of partial sums. y y (a) (b)

⬁ k+2 2 k=1

⬁

3 k+1 − 2 k=1

6 4 2

n 2

10.

k=2

12.

⬁ k−1 e k=5

14.

1 k2 − 1

⬁ k=1

53k 71−k

8 10

(c)

8 10

(d)

0.3

0.2

0.2 0.1 0.1 n

n 2

8 10

16. Match a series from one of Exercises 4, 6, 8, or 10 with the graph of its sequence of partial sums. y y (a) (b) 60

0.8

0.6

⬁

1 1 − 8. 2k 2k+1 k=1 ⬁

1 0.8 0.6 0.4 0.2

3–14 Determine whether the series converges, and if so ﬁnd its

sum. ■ ⬁

3 k−1 3. − 4 k=1

CAS

1–2 In each part, ﬁnd exact values for the ﬁrst four partial

1. (a)

621

0.4

0.2

n 2

–30 –60

8 10

(c)

(d) 0.6

1 0.8 0.6 0.4 0.2

0.4 0.2 n 2

8 10

n 2

8 10

622

Chapter 9 / Infinite Series

17–20 True–False Determine whether the statement is true or

false. Explain your answer. ■ 17. An inﬁnite series converges if its sequence of terms converges. 18. The geometric series a + ar + ar 2 + · · · + ar n + · · · converges provided |r| < 1. 19. The harmonic series diverges.

20. An inﬁnite series converges if its sequence of partial sums is bounded and monotone. 21–24 Express the repeating decimal as a fraction. ■

21. 0.9999 . . .

22. 0.4444 . . .

23. 5.373737 . . .

24. 0.451141414 . . .

25. Recall that a terminating decimal is a decimal whose digits are all 0 from some point on (0.5 = 0.50000 . . . , for example). Show that a decimal of the form 0.a1 a2 . . . an 9999 . . . , where an = 9, can be expressed as a terminating decimal. F O C U S O N C O N C E P TS

26. The great Swiss mathematician Leonhard Euler (biography on p. 3) sometimes reached incorrect conclusions in his pioneering work on inﬁnite series. For example, Euler deduced that 1 2

and

= 1 − 1 + 1 − 1 + ···

−1 = 1 + 2 + 4 + 8 + · · ·

1 2 3 k + ln + ln + · · · + ln + ··· 2 3 4 k+1

1 1 1 (b) ln 1 − + ln 1 − + ln 1 − + ··· 4 9 16

1 + ··· + ln 1 − (k + 1)2 30. Use geometric series to show that ⬁ 1 (−1)k x k = if −1 < x < 1 (a) 1+x k=0 ⬁ 1 (x − 3)k = if 2 < x < 4 (b) 4 − x k=0 ⬁ 1 (−1)k x 2k = if −1 < x < 1. (c) 1 + x2 k=0 (a) ln

31. In each part, ﬁnd all values of x for which the series converges, and ﬁnd the sum of the series for those values of x. (a) x − x 3 + x 5 − x 7 + x 9 − · · · 2 4 8 16 1 (b) 2 + 3 + 4 + 5 + 6 + · · · x x x x x (c) e−x + e−2x + e−3x + e−4x + e−5x + · · ·

32. Show that for all real values of x sin x −

33. Let a1 be any real number, and let {an } be the sequence deﬁned recursively by

by substituting x = −1 and x = 2 in the formula

1 = 1 + x + x2 + x3 + · · · 1−x What was the problem with his reasoning?

27. A ball is dropped from a height of 10 m. Each time it strikes the ground it bounces vertically to a height that is 43 of the preceding height. Find the total distance the ball will travel if it is assumed to bounce infinitely often. 28. The accompanying figure shows an “infinite staircase” constructed from cubes. Find the total volume of the staircase, given that the largest cube has a side of length 1 and each successive cube has a side whose length is half that of the preceding cube.

...

Figure Ex-28

29. In each part, ﬁnd a closed form for the nth partial sum of the series, and determine whether the series converges. If so, ﬁnd its sum.

1 2 1 1 2 sin x sin x + sin3 x − sin4 x + · · · = 2 4 8 2 + sin x

an+1 = 21 (an + 1) Make a conjecture about the limit of the sequence, and conﬁrm your conjecture by expressing an in terms of a1 and taking the limit. √ ⬁ √ k+1− k 34. Show: = 1. √ k2 + k k=1 ⬁

1 3 1 − = . 35. Show: k k + 2 2 k=1

1 1 1 3 + + + ··· = . 1·3 2·4 3·5 4 1 1 1 1 37. Show: + + + ··· = . 1·3 3·5 5·7 2 38. In his Treatise on the Conﬁgurations of Qualities and Motions (written in the 1350s), the French Bishop of Lisieux, Nicole Oresme, used a geometric method to ﬁnd the sum of the series

36. Show:

⬁ 1 2 3 k 4 = + + + + ··· k 2 2 4 8 16 k=1

In part (a) of the accompanying ﬁgure, each term in the series is represented by the area of a rectangle, and in

9.4 Convergence Tests

623

Source: This problem is based on “Trisection of an Angle in an Inﬁnite Number of Steps” by Eric Kincannon, which appeared in The College Mathematics Journal, Vol. 21, No. 5, November 1990.

part (b) the conﬁguration in part (a) has been divided into rectangles with areas A1 , A2 , A3 , . . . . Find the sum A1 + A2 + A3 + · · ·.

R1 R3

1 1 1 1 1

1 A1

1 1 2

1 4

R4 R2

1 1 1 A3 A2

1 1 8 16

u Initial side C

(a)

(b) Not to scale

Figure Ex-38

39. As shown in the accompanying ﬁgure, suppose that an angle θ is bisected using a straightedge and compass to produce ray R1 , then the angle between R1 and the initial side is bisected to produce ray R2 . Thereafter, rays R3 , R4 , R5 , . . . are constructed in succession by bisecting the angle between the preceding two rays. Show that the sequence of angles that these rays make with the initial side has a limit of θ /3.

Figure Ex-39

40. In each part, use a CAS to ﬁnd the sum of the series if it converges, and then conﬁrm the result by hand calculation. ⬁ ⬁ ⬁ 33k 1 (a) (−1)k+1 2k 32−k (b) (c) k−1 2−1 5 4k k=1 k=1 k=1 41. Writing Discuss the similarities and differences between what it means for a sequence to converge and what it means for a series to converge. 42. Writing Read about Zeno’s dichotomy paradox in an appropriate reference work and relate the paradox in a setting that is familiar to you. Discuss a connection between the paradox and geometric series.

✔QUICK CHECK ANSWERS 9.3 1 3 7 15 1 ; ; ; ; 1 − n 3. The sequence of partial sums converges. 2 4 8 16 2 1 a 4. ar k (a = 0); ; |r| < 1; |r| ≥ 1 5. ; diverge 1−r k 1. sequence; series

9.4

CONVERGENCE TESTS In the last section we showed how to find the sum of a series by finding a closed form for the nth partial sum and taking its limit. However, it is relatively rare that one can find a closed form for the nth partial sum of a series, so alternative methods are needed for finding the sum of a series. One possibility is to prove that the series converges, and then to approximate the sum by a partial sum with sufficiently many terms to achieve the desired degree of accuracy. In this section we will develop various tests that can be used to determine whether a given series converges or diverges.

THE DIVERGENCE TEST

In stating general results about convergence or divergence of series, it is convenient to use the notation uk as a generic notation for a series, thus avoiding the issue of whether the sum begins with k = 0 or k = 1 or some other value. Indeed, we will see shortly that the starting index value is irrelevant to the issue of convergence. The kth term in an inﬁnite series uk is called the general term of the series. The following theorem establishes

624

Chapter 9 / Infinite Series

a relationship between the limit of the general term and the convergence properties of a series.

9.4.1 theorem (The Divergence Test) (a) If lim uk = 0, then the series k → +⬁

(b) If lim uk = 0, then the series k → +⬁

uk diverges. uk may either converge or diverge.

proof (a) To prove this result, it sufﬁces to show that if the series converges, then limk → +⬁ uk = 0 (why?). We will prove this alternative form of (a). Let us assume that the series converges. The general term uk can be written as (1)

uk = sk − sk−1

where sk is the sum of the terms through uk and sk−1 is the sum of the terms through uk−1 . If S denotes the sum of the series, then limk → +⬁ sk = S, and since (k − 1) → +⬁ as k → +⬁, we also have limk → +⬁ sk−1 = S. Thus, from (1) lim uk = lim (sk − sk−1 ) = S − S = 0

k → +⬁

proof (b) To prove this result, it sufﬁces to produce both a convergent series and a divergent series for which limk → +⬁ uk = 0. The following series both have this property: 1 1 1 + ··· + k + ··· + 2 2 22

and 1 +

1 1 1 + + ··· + + ··· 2 3 k

The ﬁrst is a convergent geometric series and the second is the divergent harmonic series. ■

The alternative form of part (a) given in the preceding proof is sufﬁciently important that we state it separately for future reference.

WARNING The converse of Theorem 9.4.2 is false; that is, showing that

lim uk = 0

k → +⬁

9.4.2 theorem If the series

does not prove that uk converges, since this property may hold for divergent as well as convergent series. This is illustrated in the proof of part (b) of Theorem 9.4.1.

uk converges, then lim uk = 0. k → +⬁

Example 1 The series ⬁ k=1

k 1 2 3 k = + + + ··· + + ··· k+1 2 3 4 k+1

diverges since lim

k → +⬁

k 1 = lim = 1 = 0 k + 1 k → +⬁ 1 + 1/k

ALGEBRAIC PROPERTIES OF INFINITE SERIES

For brevity, the proof of the following result is omitted.

9.4 Convergence Tests

625

9.4.3 See Exercises 27 and 28 for an explo uk or ration of what happens when vk diverge.

theorem (a) If uk and vk are convergent series, then (uk + vk ) and (uk − vk ) are convergent series and the sums of these series are related by ⬁ ⬁ ⬁ (uk + vk ) = uk + vk k=1

k=1

⬁ ⬁ ⬁ (uk − vk ) = uk − vk k=1

k=1

(b) If c is a nonzero constant, then the series uk and cuk both converge or both diverge. In the case of convergence, the sums are related by ⬁ k=1

cuk = c

⬁

k=1

(c) Convergence or divergence is unaffected by deleting a ﬁnite number of terms from a series; in particular, for any positive integer K, the series WARNING Do not read too much into part (c) of Theorem 9.4.3. Although convergence is not affected when finitely many terms are deleted from the beginning of a convergent series, the sum of the series is changed by the removal of those terms.

⬁

k=1 ⬁

k=K

uk = u1 + u2 + u3 + · · · uk = uK + uK+1 + uK+2 + · · ·

both converge or both diverge.

Example 2

Find the sum of the series ⬁

2 3 − k−1 4k 5 k=1

Solution. The series ⬁ 3 3 3 3 = + 2 + 3 + ··· k 4 4 4 4 k=1

is a convergent geometric series a = 43 , r = 41 , and the series

⬁ 2 2 2 2 = 2 + + 2 + 3 + ··· k−1 5 5 5 5 k=1

is also a convergent geometric series a = 2, r = 15 . Thus, from Theorems 9.4.3(a) and 9.3.3 the given series converges and ⬁ ⬁ ⬁

2 3 2 3 − − = k k−1 k k−1 4 5 4 5 k=1 k=1 k=1 =

3 4

1−

1 4

−

2 1−

1 5

=−

3 2

Chapter 9 / Infinite Series

626

Example 3 (a)

⬁ 5 k=1

Determine whether the following series converge or diverge.

=5+

5 5 5 + + ··· + + ··· 2 3 k

(b)

⬁ 1 1 1 1 = + + + ··· k 10 11 12 k=10

Solution. The ﬁrst series is a constant times the divergent harmonic series, and hence diverges by part (b) of Theorem 9.4.3. The second series results by deleting the ﬁrst nine terms from the divergent harmonic series, and hence diverges by part (c) of Theorem 9.4.3.

THE INTEGRAL TEST The expressions y

u2 u 3 u4

y = f (x) un

1 2 3 4

...

⬁ 1 k2 k=1

and

are related in that the integrand in the improper integral results when the index k in the general term of the series is replaced by x and the limits of summation in the series are replaced by the corresponding limits of integration. The following theorem shows that there is a relationship between the convergence of the series and the integral.

k=1

1 2 3 4

y = f (x)

...

(b) Figure 9.4.1

1 dx x2

9.4.4 theorem (The Integral Test) Let uk be a series with positive terms. If f is a function that is decreasing and continuous on an interval [a, +⬁) and such that uk = f(k) for all k ≥ a, then +⬁ ⬁ f(x) dx uk and

(a)

+⬁

n+1

u2 u 3 u 4

both converge or both diverge.

The proof of the integral test is deferred to the end of this section. However, the gist of the proof is captured in Figure 9.4.1: if the integral diverges, then so does the series (Figure 9.4.1a), and if the integral converges, then so does the series (Figure 9.4.1b). Example 4 Show that the integral test applies, and use the integral test to determine whether the following series converge or diverge. (a)

⬁ 1 k=1

(b)

⬁ 1 2 k k=1

Solution (a). We already know that this is the divergent harmonic series, so the integral test will simply illustrate another way of establishing the divergence. Note ﬁrst that the series has positive terms, so the integral test is applicable. If we replace k by x in the general term 1/k, we obtain the function f(x) = 1/x, which is decreasing and continuous for x ≥ 1 (as required to apply the integral test with a = 1). Since +⬁ b 1 1 dx = lim dx = lim [ln b − ln 1] = +⬁ b → +⬁ 1 x b → +⬁ x 1

the integral diverges and consequently so does the series.

9.4 Convergence Tests WARNING In part (b) of Example 4, do not erroneously conclude that the sum of the series is 1 because the value of the corresponding integral is 1. You can see that this is not so since the sum of the first two terms alone exceeds 1. Later, we will see that the sum of the series is actually Ď&#x20AC;2 /6.

627

Solution (b). Note ďŹ rst that the series has positive terms, so the integral test is applicable. If we replace k by x in the general term 1/k 2 , we obtain the function f(x) = 1/x 2 , which is decreasing and continuous for x â&#x2030;Ľ 1. Since b +âŹ dx 1 b 1 1 â&#x2C6;&#x2019; =1 1 â&#x2C6;&#x2019; dx = lim = lim = lim b â&#x2020;&#x2019; +âŹ 1 x 2 b â&#x2020;&#x2019; +âŹ x2 x 1 b â&#x2020;&#x2019; +âŹ b 1 the integral converges and consequently the series converges by the integral test with a = 1.

p-SERIES The series in Example 4 are special cases of a class of series called p-series or hyperharmonic series. A p-series is an inďŹ nite series of the form âŹ 1 1 1 1 = 1 + p + p + ÂˇÂˇÂˇ + p + ÂˇÂˇÂˇ p k 2 3 k k=1

where p > 0. Examples of p-series are âŹ 1 k=1 âŹ k=1 âŹ k=1

=1+

1 1 1 + + ÂˇÂˇÂˇ + + ÂˇÂˇÂˇ 2 3 k

p=1

1 1 1 1 = 1 + 2 + 2 + ÂˇÂˇÂˇ + 2 + ÂˇÂˇÂˇ 2 k 2 3 k

p=2

1 1 1 1 â&#x2C6;&#x161; = 1 + â&#x2C6;&#x161; + â&#x2C6;&#x161; + ÂˇÂˇÂˇ + â&#x2C6;&#x161; + ÂˇÂˇÂˇ k k 2 3

1 2

The following theorem tells when a p-series converges.

9.4.5

theorem (Convergence of p-Series) âŹ 1 1 1 1 = 1 + p + p + ÂˇÂˇÂˇ + p + ÂˇÂˇÂˇ p k 2 3 k k=1

converges if p > 1 and diverges if 0 < p â&#x2030;¤ 1.

proof To establish this result when p = 1, we will use the integral test. b 1â&#x2C6;&#x2019;p b +âŹ x 1â&#x2C6;&#x2019;p b 1 1 â&#x2C6;&#x2019;p dx = lim â&#x2C6;&#x2019; = lim x dx = lim b â&#x2020;&#x2019; +âŹ 1 b â&#x2020;&#x2019; +âŹ 1 â&#x2C6;&#x2019; p b â&#x2020;&#x2019; +âŹ 1 â&#x2C6;&#x2019; p 1 xp 1â&#x2C6;&#x2019;p 1

Assume ďŹ rst that p > 1. Then 1 â&#x2C6;&#x2019; p < 0, so b1â&#x2C6;&#x2019;p â&#x2020;&#x2019; 0 as b â&#x2020;&#x2019; +âŹ . Thus, the integral converges [its value is â&#x2C6;&#x2019;1/(1 â&#x2C6;&#x2019; p)] and consequently the series also converges. Now assume that 0 < p < 1. It follows that 1 â&#x2C6;&#x2019; p > 0 and b1â&#x2C6;&#x2019;p â&#x2020;&#x2019; +âŹ as b â&#x2020;&#x2019; +âŹ , so the integral and the series diverge. The case p = 1 is the harmonic series, which was previously shown to diverge. â&#x2013; Example 5

1 1 1 1+ â&#x2C6;&#x161; + ÂˇÂˇÂˇ + â&#x2C6;&#x161; + ÂˇÂˇÂˇ + â&#x2C6;&#x161; 3 3 3 k 2 3

diverges since it is a p-series with p =

1 3

< 1.

Chapter 9 / Infinite Series

628

PROOF OF THE INTEGRAL TEST

Before we can prove the integral test, we need a basic result about convergence of series with nonnegative terms. If u1 + u2 + u3 + · · · + uk + · · · is such a series, then its sequence of partial sums is increasing, that is, s1 ≤ s2 ≤ s3 ≤ · · · ≤ sn ≤ · · · Thus, from Theorem 9.2.3 the sequence of partial sums converges to a limit S if and only if it has some upper bound M, in which case S ≤ M. If no upper bound exists, then the sequence of partial sums diverges. Since convergence of the sequence of partial sums corresponds to convergence of the series, we have the following theorem.

9.4.6 theorem If constant M such that

uk is a series with nonnegative terms, and if there is a sn = u1 + u2 + · · · + un ≤ M

for every n, then the series converges and the sum S satisﬁes S ≤ M. If no such M exists, then the series diverges.

In words, this theorem implies that a series with nonnegative terms converges if and only if its sequence of partial sums is bounded above. proof of theorem 9.4.4 We need only show that the series converges when the integral converges and that the series diverges when the integral diverges. For simplicity, we will limit the proof to the case where a = 1. Assume that f(x) satisﬁes the hypotheses of the theorem for x ≥ 1. Since f(1) = u1 , f(2) = u2 , . . . , f(n) = un , . . . y

y = f (x)

u3 u 4 un

1 2 3 4

...

n+1

the values of u1 , u2 , . . . , un , . . . can be interpreted as the areas of the rectangles shown in Figure 9.4.2. The following inequalities result by comparing the areas under the curve y = f(x) to the areas of the rectangles in Figure 9.4.2 for n > 1: n+1 Figure 9.4.2a f(x) dx < u1 + u2 + · · · + un = sn 1 n Figure 9.4.2b f(x) dx sn − u1 = u2 + u3 + · · · + un < 1

These inequalities can be combined as n+1 f(x) dx < sn < u1 +

(a) y

If the integral in (2) u2 u 3 u 4

y = f (x) un

1 2 3 4

...

(b) Figure 9.4.2

f(x) dx

(2)

+⬁ 1

f(x) dx converges to a ﬁnite value L, then from the right-hand inequality +⬁ n sn < u1 + f(x) dx = u1 + L f(x) dx < u1 + 1

Thus, each partial sum is less than the ﬁnite constant +⬁ u1 + L, and the series converges by Theorem 9.4.6. On the other hand, if the integral 1 f(x) dx diverges, then n+1 f(x) dx = +⬁ lim n → +⬁ 1

so that from the left-hand inequality in (2), sn → +⬁ as n → +⬁. This implies that the series also diverges. ■

9.4 Convergence Tests

â&#x153;&#x201D;QUICK CHECK EXERCISES 9.4

(See page 631 for answers.)

1. The divergence test says that if uk diverges.

= 0, then the series

a1 = 3,

k=1

it follows that âŹ ak =

ak = 1,

and

k=2

âŹ

and

k=1

bk = 5

âŹ k=1

âŹ (2ak + bk ) =

This series converges if .

k=1

EXERCISE SET 9.4

Graphing Utility

2. Use Theorem 9.4.3 to ďŹ nd the sum of each series. âŹ âŹ 1 7 2k+1 â&#x2C6;&#x2019;k k+1 (a) (b) â&#x2C6;&#x2019; 7 3 â&#x2C6;&#x2019; k k 2 â&#x2C6;&#x2019; 1 10kâ&#x2C6;&#x2019;1 5 k=2 k=1 3â&#x20AC;&#x201C;4 For each given p-series, identify p and determine whether

the series converges. â&#x2013; âŹ âŹ 1 1 3. (a) (b) â&#x2C6;&#x161; 3 k k k=1 k=1 4. (a)

k â&#x2C6;&#x2019;4/3

(b)

k=1

âŹ k=1

1 â&#x2C6;&#x161; 4 k

(c)

âŹ

k â&#x2C6;&#x2019;1

(d)

(c)

âŹ

1 â&#x2C6;&#x161; 3 5 k

k=1

(d)

âŹ

k=1

âŹ k k e k=1 âŹ 1 (c) â&#x2C6;&#x161; k k=1

6. (a)

8. (a)

k=1

(b)

k=1

âŹ 1 (d) k! k=1

(b) (d)

âŹ k=1 âŹ k=1

1 k

ln k â&#x2C6;&#x161;

â&#x2C6;&#x161;

(b) (b)

âŹ k=1

1 1 + 9k 2

âŹ

1 (4 + 2k)3/2

k=1

âŹ k=1

1 k+6

âŹ

k=1 âŹ k=1

k ln(k + 1)

10.

13.

16.

k +1 k2 + 3

19.

âŹ

2 2 1 k sin 21. k k=1 âŹ

7k â&#x2C6;&#x2019;1.01

k=5

âŹ 3 5k k=1

âŹ

k=1 âŹ k=1 âŹ k=1

k=1

1 â&#x2C6;&#x161; 3 2k â&#x2C6;&#x2019; 1

âŹ

11.

â&#x2C6;&#x2019;k 2

14.

k=3 âŹ

k=1

tan k 1 + k2 22.

âŹ

20.

k=1

k 2 eâ&#x2C6;&#x2019;k

1 k+5

âŹ ln k

17.

â&#x2C6;&#x2019;1

â&#x2C6;&#x161;

1 1+ k

â&#x2C6;&#x161;

â&#x2C6;&#x2019;k

1 k2

k=1

24.

âŹ

sech2 k

k=1

25â&#x20AC;&#x201C;26 Use the integral test to investigate the relationship between the value of p and the convergence of the series. â&#x2013; âŹ âŹ 1 1 25. 26. p k(ln k) k(ln k)[ln(ln k)]p k=2 k=3 F O C U S O N C O N C E P TS

k+3

7â&#x20AC;&#x201C;8 ConďŹ rm that the integral test is applicable and use it to

determine whether the series converges. â&#x2013;

k=1

k 1 + k2

âŹ

âŹ 1 12. â&#x2C6;&#x161; k e k=1

23. âŹ

1 5k + 2

9â&#x20AC;&#x201C;24 Determine whether the series converges. â&#x2013;

18. 1 kĎ&#x20AC;

âŹ k=1

15.

5â&#x20AC;&#x201C;6 Apply the divergence test and state what it tells you about

the series. â&#x2013; âŹ k2 + k + 3 5. (a) 2k 2 + 1 k=1 âŹ cos kĎ&#x20AC; (c)

7. (a)

k â&#x2C6;&#x2019;2/3

k=1

âŹ

. This series diverges if

CAS

1. Use

Theorem 9.4.3 sum of each

series.

to ďŹ nd the 1 1 1 1 1 1 (a) + + + 2 + ÂˇÂˇÂˇ + + k + ÂˇÂˇÂˇ 2 k 2 4 2 4 2 4 âŹ

1 1 (b) â&#x2C6;&#x2019; k 5 k(k + 1) k=1

âŹ

+âŹ â&#x2C6;&#x161; (1/ x) dx = +âŹ , the 3. Since 1 test applied to the series âŹ k=1 shows that this series . 4. A p-series is a series of the form

2. Given that

âŹ

629

27. Suppose uk converges and the se that the series ries vk diverges. Show that the series (uk + vk ) and (uk â&#x2C6;&#x2019; vk ) both diverge. [Hint: Assume that (uk + vk ) converges and use Theorem 9.4.3 to obtain a contradiction.]

630

Chapter 9 / Infinite Series

28. Find examples to show that if the series uk and both diverge, then the series (u + v v k k k ) and (uk â&#x2C6;&#x2019; vk ) may either converge or diverge. 29â&#x20AC;&#x201C;30 Use the results of Exercises 27 and 28, if needed, to determine whether each series converges or diverges. â&#x2013; âŹ âŹ 2 kâ&#x2C6;&#x2019;1 1 1 1 29. (a) (b) â&#x2C6;&#x2019; 3/2 + 3 k 3k + 2 k k=1 k=1

30. (a)

âŹ k=2

1 1 â&#x2C6;&#x2019; 2 k(ln k)2 k

(b)

âŹ

keâ&#x2C6;&#x2019;k

k=2

1 + k ln k

y = f (x) un+1 un+2 ... x n n+1

1 y

y = f (x) un+1

31â&#x20AC;&#x201C;34 Trueâ&#x20AC;&#x201C;False Determine whether the statement is true or

false. Explain your answer. â&#x2013; 31. If uk converges to L, then (1/uk ) converges to 1/L. 32. If cuk diverges for some constant c, then uk must diverge.

33. The integral test can be used to prove that a series diverges. âŹ 1 34. The series is a p-series. pk k=1 35. Use a CAS to conďŹ rm that âŹ 1 Ď&#x20AC;2 = k2 6 k=1

and

âŹ 1 Ď&#x20AC;4 = k4 90 k=1

f(x) dx <

n+1

âŹ

k=n+1

uk <

+âŹ

+âŹ n+1

f(x) dx < S < sn +

f(x) dx

(b) Show that if S is the sum of the series is the nth partial sum, then sn +

... x 1

n n+1

Figure Ex-36

37. (a) It was stated in Exercise 35 that âŹ Ď&#x20AC;2 1 = k2 6 k=1

sn +

36â&#x20AC;&#x201C;40 Exercise 36 will show how a partial sum can be used to obtain upper and lower bounds on the sum of a series when the hypotheses of the integral test are satisďŹ ed. This result will be needed in Exercises 37â&#x20AC;&#x201C;40. â&#x2013; 36. (a) Let âŹ k=1 uk be a convergent series with positive terms, and let f be a function that is decreasing and continuous on [n, +âŹ ) and such that uk = f(k) for k â&#x2030;Ľ n. Use an area argument and the accompanying ďŹ gure to show that +âŹ

un+2 un+3

Show that if sn is the nth partial sum of this series, then

and then use these results in each part to ďŹ nd the sum of the series. âŹ âŹ âŹ 1 1 3k 2 â&#x2C6;&#x2019; 1 (b) (c) (a) 4 2 k k (k â&#x2C6;&#x2019; 1)4 k=3 k=2 k=1

...

âŹ

k=1

+âŹ n

uk and sn

f(x) dx

Ď&#x20AC;2 1 1 < < sn + n+1 6 n

(b) Calculate s3 exactly, and then use the result in part (a) to show that 29 Ď&#x20AC;2 61 < < 18 6 36 (c) Use a calculating utility to conďŹ rm that the inequalities in part (b) are correct. (d) Find upper and lower bounds on the error that results if the sum of the series is approximated by the 10th partial sum. 38. In each part, ďŹ nd upper and lower bounds on the error that results if the sum of the series is approximated by the 10th partial sum. âŹ âŹ âŹ 1 k 1 (a) (b) (c) 2 2+1 k (2k + 1) k e k=1 k=1 k=1

39. It was stated in Exercise 35 that

âŹ Ď&#x20AC;4 1 = k4 90 k=1

(a) Let sn be the nth partial sum of the series above. Show that Ď&#x20AC;4 1 1 sn + < < sn + 3 3(n + 1)3 90 3n (b) We can use a partial sum of the series to approximate Ď&#x20AC;4 /90 to three decimal-place accuracy by capturing the

9.5 The Comparison, Ratio, and Root Tests

sum of the series in an interval of length 0.001 (or less). Find the smallest value of n such that the interval containing π4 /90 in part (a) has a length of 0.001 or less. (c) Approximate π4 /90 to three decimal places using the midpoint of an interval of width at most 0.001 that contains the sum of the series. Use a calculating utility to confirm that your answer is within 0.0005 of π4 /90. 40. We showed in Section 9.3 that the harmonic series ⬁k=1 1/k diverges. Our objective in this problem is to demonstrate that although the partial sums of this series approach +⬁, they increase extremely slowly. (a) Use inequality (2) to show that for n ≥ 2 ln(n + 1) < sn < 1 + ln n (b) Use the inequalities in part (a) to find upper and lower bounds on the sum of the first million terms in the series.

631

(c) Show that the sum of the ﬁrst billion terms in the series is less than 22. (d) Find a value of n so that the sum of the ﬁrst n terms is greater than 100.

41. Use a graphing utility to conﬁrm that the integral test applies to the series ⬁k=1 k 2 e−k , and then determine whether the series converges.

42. (a) Show that the hypotheses of the integral test are satisﬁed by the series ⬁k=1 1/(k 3 + 1). (b) Use a CAS and the integral test to conﬁrm that the series converges. (c) Construct a table of partial sums for n = 10, 20, 30, . . . , 100, showing at least six decimal places. (d) Based on your table, make a conjecture about the sum of the series to three decimal-place accuracy. (e) Use part (b) of Exercise 36 to check your conjecture.

✔QUICK CHECK ANSWERS 9.4 1. lim uk k → +⬁

9.5

2. −2; 7

1 3. integral; √ ; diverges k

1 ; p > 1; 0 < p ≤ 1 kp

THE COMPARISON, RATIO, AND ROOT TESTS In this section we will develop some more basic convergence tests for series with nonnegative terms. Later, we will use some of these tests to study the convergence of Taylor series.

THE COMPARISON TEST We will begin with a test that is useful in its own right and is also the building block for other important convergence tests. The underlying idea of this test is to use the known convergence or divergence of a series to deduce the convergence or divergence of another series.

9.5.1 theorem (The Comparison Test) Let negative terms and suppose that It is not essential in Theorem 9.5.1 that the condition ak ≤ bk hold for all k , as stated; the conclusions of the theorem remain true if this condition is eventually true.

⬁

k=1

ak and

⬁

k=1

bk be series with non-

a1 ≤ b1 , a2 ≤ b2 , a3 ≤ b3 , . . . , ak ≤ bk , . . . (a) If the “bigger series” bk converges, then the “smaller series” ak also converges. (b) If the “smaller series” ak diverges, then the “bigger series” bk also diverges.

We have left the proof of this theorem for the exercises; however, it is easy to visualize why the theorem is true by interpreting the terms in the series as areas of rectangles

632

Chapter 9 / Infinite Series

(Figure total 9.5.1). The comparison test states that if the total area bk is finite, then the area ak must also be finite; and if the total area ak is infinite, then the total area bk must also be infinite.

b3 b4 b5

a1 a 2 a 3

a4 a5

... 5 ...

...

USING THE COMPARISON TEST There are two steps required for using the comparison test to determine whether a series uk with positive terms converges:

k ...

Step 1. Guess at whether the series For each rectangle, ak denotes the area of the blue portion and bk denotes the combined area of the white and blue portions.

Figure 9.5.1

uk converges or diverges.

Step 2. Find a series that proves the guess to be correct. That is, if we guess that uk diverges, we must find a divergent series whose terms are “smaller” than the corresponding terms of uk , and if we guess that uk converges, we must find a convergent series whose terms are “bigger” than the corresponding terms of uk . In most cases, the series uk being considered will have its general term uk expressed as a fraction. To help with the guessing process in the first step, we have formulated two principles that are based on the form of the denominator for uk . These principles sometimes suggest whether a series is likely to converge or diverge. We have called these “informal principles” because they are not intended as formal theorems. In fact, we will not guarantee that they always work. However, they work often enough to be useful.

9.5.2 informal principle Constant terms in the denominator of uk can usually be deleted without affecting the convergence or divergence of the series.

9.5.3 informal principle If a polynomial in k appears as a factor in the numerator or denominator of uk , all but the leading term in the polynomial can usually be discarded without affecting the convergence or divergence of the series.

Example 1 Use the comparison test to determine whether the following series converge or diverge. ⬁ ⬁ 1 1 (a) (b) √ 2 1 2k + k k− 2 k=1 k=1

Solution (a). According to Principle 9.5.2, we should be able to drop the constant in the denominator without affecting the convergence or divergence. Thus, the given series is likely to behave like ⬁ 1 (1) √ k k=1 which is a divergent p-series p = 21 . Thus, we will guess that the given series diverges and try to prove this by ﬁnding a divergent series that is “smaller” than the given series. However, series (1) does the trick since 1 1 for k = 1, 2, . . . >√ √ 1 k− 2 k

Thus, we have proved that the given series diverges.

9.5 The Comparison, Ratio, and Root Tests

633

Solution (b). According to Principle 9.5.3, we should be able to discard all but the leading term in the polynomial without affecting the convergence or divergence. Thus, the given series is likely to behave like ⬁ ⬁ 1 1 1 (2) = 2k 2 2 k=1 k 2 k=1

which converges since it is a constant times a convergent p-series (p = 2). Thus, we will guess that the given series converges and try to prove this by ﬁnding a convergent series that is “bigger” than the given series. However, series (2) does the trick since

1 1 < 2 for k = 1, 2, . . . 2k 2 + k 2k Thus, we have proved that the given series converges.

THE LIMIT COMPARISON TEST

In the last example, Principles 9.5.2 and 9.5.3 provided the guess about convergence or divergence as well as the series needed to apply the comparison test. Unfortunately, it is not always so straightforward to ﬁnd the series required for comparison, so we will now consider an alternative to the comparison test that is usually easier to apply. The proof is given in Appendix D.

9.5.4 theorem (The Limit Comparison Test) Let ak and bk be series with positive terms and suppose that ak ρ = lim k → +⬁ bk If ρ is ﬁnite and ρ > 0, then the series both converge or both diverge.

The cases where ρ = 0 or ρ = +⬁ are discussed in the exercises (Exercise 56). To we must again ﬁrst guess at the convergence or divergence use the limit comparison test of ak and then ﬁnd a series bk that supports our guess. The following example illustrates this principle. Example 2 Use the limit comparison test to determine whether the following series converge or diverge. (a)

⬁ k=1

1 √ k+1

(b)

⬁ k=1

1 2 2k + k

(c)

⬁ 3k 3 − 2k 2 + 4

k7 − k3 + 2

k=1

Solution (a). As in Example 1, Principle 9.5.2 suggests that the series is likely to behave like the divergent p-series (1). To prove that the given series diverges, we will apply the limit comparison test with

We obtain

1 ak = √ k+1

1 and bk = √ k

√ ak k = lim ρ = lim = lim √ k → +⬁ bk k → +⬁ k + 1 k → +⬁

=1 1 1+ √ k Since ρ is ﬁnite and positive, it follows from Theorem 9.5.4 that the given series diverges.

634

Chapter 9 / Infinite Series

Solution (b). As in Example 1, Principle 9.5.3 suggests that the series is likely to behave like the convergent series (2). To prove that the given series converges, we will apply the limit comparison test with ak = We obtain

1 2k 2 + k

and

bk =

1 2k 2

ak 2k 2 = lim = lim k → +⬁ bk k → +⬁ 2k 2 + k k → +⬁

=1 1 k Since ρ is ﬁnite and positive, it follows from Theorem 9.5.4 that the given series converges, which agrees with the conclusion reached in Example 1 using the comparison test. ρ = lim

Solution (c). From Principle 9.5.3, the series is likely to behave like ⬁ 3k 3 k=1

⬁ 3 4 k k=1

(3)

which converges since it is a constant times a convergent p-series. Thus, the given series is likely to converge. To prove this, we will apply the limit comparison test to series (3) and the given series. We obtain 3k 3 − 2k 2 + 4 7 6 4 k 7 − k 3 + 2 = lim 3k − 2k + 4k = 1 ρ = lim 7 3 3 k → +⬁ 3k − 3k + 6 k → +⬁ k4 Since ρ is ﬁnite and nonzero, it follows from Theorem 9.5.4 that the given series converges, since (3) converges.

THE RATIO TEST The comparison test and the limit comparison test hinge on first making a guess about convergence and then finding an appropriate series for comparison, both of which can be difficult tasks in cases where Principles 9.5.2 and 9.5.3 cannot be applied. In such cases the next test can often be used, since it works exclusively with the terms of the given series—it requires neither an initial guess about convergence nor the discovery of a series for comparison. Its proof is given in Appendix D.

9.5.5 theorem (The Ratio Test) that

Let

uk be a series with positive terms and suppose uk+1 ρ = lim k → +⬁ uk

(a) If ρ < 1, the series converges. (b) If ρ > 1 or ρ = +⬁, the series diverges. (c) If ρ = 1, the series may converge or diverge, so that another test must be tried.

Example 3 Each of the following series has positive terms, so the ratio test applies. In each part, use the ratio test to determine whether the following series converge or diverge. (a)

⬁ 1 k! k=1

(b)

⬁ k k 2 k=1

(c)

⬁ kk k=1

(d)

⬁ (2k)! k=3

(e)

⬁ k=1

1 2k − 1

9.5 The Comparison, Ratio, and Root Tests

635

Solution (a). The series converges, since Ď = lim

k â&#x2020;&#x2019; +âŹ

1/(k + 1)! k! 1 uk+1 = lim = lim = lim =0<1 k â&#x2020;&#x2019; +âŹ k â&#x2020;&#x2019; +âŹ (k + 1)! k â&#x2020;&#x2019; +âŹ k + 1 uk 1/k!

Solution (b). The series converges, since uk+1 k + 1 2k k+1 1 1 = lim k+1 Âˇ = lim = <1 k â&#x2020;&#x2019; +âŹ uk k â&#x2020;&#x2019; +âŹ 2 k 2 k â&#x2020;&#x2019; +âŹ k 2

Ď = lim

Solution (c). The series diverges, since

(k + 1)k+1 k! (k + 1)k 1 k uk+1 = lim = lim =e>1 1 + Âˇ k = lim k â&#x2020;&#x2019; +âŹ (k + 1)! k â&#x2020;&#x2019; +âŹ k â&#x2020;&#x2019; +âŹ k â&#x2020;&#x2019; +âŹ uk k kk k

Ď = lim

See Formula (7) of Section 1.3

Solution (d). The series diverges, since [2(k + 1)]! uk+1 Ď = lim = lim k â&#x2020;&#x2019; +âŹ k â&#x2020;&#x2019; +âŹ uk 4k+1

(2k + 2)(2k + 1)(2k)! = lim Âˇ k â&#x2020;&#x2019; +âŹ (2k)!

4k (2k + 2)! 1 = lim Âˇ Âˇ (2k)! k â&#x2020;&#x2019; +âŹ (2k)! 4 1 1 lim (2k + 2)(2k + 1) = +âŹ = 4 4 k â&#x2020;&#x2019; +âŹ

Solution (e). The ratio test is of no help since Ď = lim

k â&#x2020;&#x2019; +âŹ

uk+1 2k â&#x2C6;&#x2019; 1 1 2k â&#x2C6;&#x2019; 1 = lim Âˇ = lim =1 k â&#x2020;&#x2019; +âŹ 2(k + 1) â&#x2C6;&#x2019; 1 k â&#x2020;&#x2019; +âŹ 2k + 1 uk 1

However, the integral test proves that the series diverges since b +âŹ b dx dx 1 = lim = lim ln(2x â&#x2C6;&#x2019; 1) = +âŹ 2x â&#x2C6;&#x2019; 1 b â&#x2020;&#x2019; +âŹ 1 2x â&#x2C6;&#x2019; 1 b â&#x2020;&#x2019; +âŹ 2 1 1

Both the comparison test and the limit comparison test would also have worked here (verify).

THE ROOT TEST

In cases where it is difďŹ cult or inconvenient to ďŹ nd the limit required for the ratio test, the next test is sometimes useful. Since its proof is similar to the proof of the ratio test, we will omit it.

9.5.6 that

theorem (The Root Test) Let

uk be a series with positive terms and suppose â&#x2C6;&#x161; k Ď = lim uk = lim (uk )1/k k â&#x2020;&#x2019; +âŹ

k â&#x2020;&#x2019; +âŹ

(a) If Ď < 1, the series converges. (b) If Ď > 1 or Ď = +âŹ , the series diverges. (c) If Ď = 1, the series may converge or diverge, so that another test must be tried. Example 4 diverge.

Use the root test to determine whether the following series converge or âŹ

âŹ 4k â&#x2C6;&#x2019; 5 k 1 (b) (a) 2k + 1 (ln(k + 1))k k=2 k=1

636

Chapter 9 / Infinite Series

Solution (a). The series diverges, since ρ = lim (uk )1/k = lim k → +⬁

k → +⬁

4k − 5 =2>1 2k + 1

Solution (b). The series converges, since ρ = lim (uk )1/k = lim k → +⬁

✔QUICK CHECK EXERCISES 9.5

k → +⬁

(See page 637 for answers.)

1–4 Select between converges or diverges to ﬁll the ﬁrst blank.

3. Since (k + 1)!/3k+1 k+1 = lim = +⬁ k → +⬁ k → +⬁ k!/3k 3 by the test. the series ⬁k=1 k!/3k

■

1. The series

⬁ 2k 2 + 1 2k 8/3 − 1 k=1

lim

4. Since

by comparison with the p-series 2. Since

1 =0<1 ln(k + 1)

(k + 1)3 /3k+1 lim = lim k → +⬁ k → +⬁ k 3 /3k ⬁ 3/ k the series k=1 k 3 by the

⬁

1 1+ k 3

k=1

lim

k → +⬁

the series

1 = 3 test.

⬁

k=1

k k /2

1/k

= lim

1/k k/2

k → +⬁

1 =0 k 1/2

by the

test.

EXERCISE SET 9.5 1–2 Make a guess about the convergence or divergence of the

series, and conﬁrm your guess using the comparison test. ■ ⬁ ⬁ 1 3 1. (a) (b) 2 5k − k k − 41 k=1 k=1 2. (a)

⬁ k+1 k2 − k k=2

(b)

⬁ k=1

2 +k

3. In each part, use the comparison test to show that the series converges. ⬁ ⬁ 1 5 sin2 k (a) (b) k 3 +5 k! k=1 k=1

4. In each part, use the comparison test to show that the series diverges. ⬁ ⬁ k ln k (a) (b) /2 3 k k − 21 k=1 k=1

5–10 Use the limit comparison test to determine whether the

series converges. ■ ⬁ 4k 2 − 2k + 6 5. 8k 7 + k − 8 k=1 7.

⬁ k=1

5 3k + 1

⬁ k=1

1 9k + 6

⬁

k(k + 3) (k + 1)(k + 2)(k + 5)

k=1

⬁ k=1

1 √ 3 2 8k − 3k

10.

⬁ k=1

1 (2k + 3)17

11–16 Use the ratio test to determine whether the series converges. If the test is inconclusive, then say so. ■ ⬁ ⬁ ⬁ 1 4k 3k 13. 12. 11. 2 k! k 5k k=1 k=1 k=1

k ⬁ ⬁ ⬁ k k! 1 16. 15. k 14. 3 2+1 2 k k k=1 k=1 k=1 17–20 Use the root test to determine whether the series converges. If the test is inconclusive, then say so. ■ ⬁

⬁

3k + 2 k k k 17. 18. 2k − 1 100 k=1 k=1

19.

⬁ k k 5 k=1

20.

⬁ (1 − e−k )k k=1

21–24 True–False Determine whether the statement is true or false. Explain your answer. ■

21. The limit comparison test decides convergence based on a limit of the quotient of consecutive terms in a series. 22. If limk → +⬁ (uk+1 /uk ) = 5, then uk diverges.

9.5 The Comparison, Ratio, and Root Tests

23. If limk â&#x2020;&#x2019; +âŹ (k 2 uk ) = 5, then

uk converges.

24. The root test decides convergence based on a limit of kth roots of terms in the sequence of partial sums for a series. 25â&#x20AC;&#x201C;49 Use any method to determine whether the series converges. â&#x2013;

25.

âŹ 7k k=0

28.

âŹ k!10k

3k â&#x2C6;&#x161; âŹ k 31. 3+1 k k=1

26.

k=1

29.

k=1

âŹ

âŹ âŹ

37.

âŹ k=1

âŹ k! 40. k2 e k=1

43.

âŹ k=1

46.

49.

âŹ

k=1 âŹ k=1

1 4 + 2â&#x2C6;&#x2019;k 5k + k k! + 3 ln k 3k

k 50 eâ&#x2C6;&#x2019;k

30. âŹ k=1

34.

k=1

36.

âŹ (k + 4)!

4!k!4k âŹ â&#x2C6;&#x161; k ln k 44. 3+1 k k=1 k=0

47.

k=0

âŹ 4 + | cos x|

âŹ k! kk k=1

âŹ

54. (a) Make a conjecture about the convergence of the series âŹ k=1 sin(Ď&#x20AC;/k) by considering the local linear approximation of sin x at x = 0. (b) Try to conďŹ rm your conjecture using the limit comparison test. 55. (a) We will see later that the polynomial 1 â&#x2C6;&#x2019; x 2 /2 is the â&#x20AC;&#x153;local quadraticâ&#x20AC;? approximation for cos x at x = 0. Make a conjecture about the convergence of the series

âŹ 1 1 â&#x2C6;&#x2019; cos k k=1

k2 k3 + 1

âŹ 2 + (â&#x2C6;&#x2019;1)k

k=1

41.

âŹ

(k!)2 (2k)!

39.

âŹ ln k k=1

2 âŹ

k k 42. k+1 k=1 45.

âŹ tanâ&#x2C6;&#x2019;1 k

k=1

48.

âŹ [Ď&#x20AC;(k + 1)]k

k k+1

k=1

50. For what positive values of Îą does the series converge?

âŹ

1Âˇ3 1Âˇ3Âˇ5 1Âˇ3Âˇ5Âˇ7 + + + ÂˇÂˇÂˇ 3! 5! 7! â&#x2C6;&#x161; 53. Show that ln x < x if x > 0, and use this result to investigate the convergence of âŹ âŹ ln k 1 (a) (b) 2 k (ln k)2 k=1 k=2

52. 1 +

F O C U S O N C O N C E P TS

4 2 + 3k k

k=1

38.

âŹ k2

k=1

32.

â&#x2C6;&#x161;

1 â&#x2C6;&#x161; 1+ k

27.

k=1

1 k(k + 1) k=1 â&#x2C6;&#x161; âŹ 2+ k 35. (k + 1)3 â&#x2C6;&#x2019; 1 k=1 33.

1 2k + 1

k=1 (Îą

k/ Îą

k )

51â&#x20AC;&#x201C;52 Find the general term of the series and use the ratio test

to show that the series converges. â&#x2013; 1Âˇ2 1Âˇ2Âˇ3 1Âˇ2Âˇ3Âˇ4 + + + ÂˇÂˇÂˇ 51. 1 + 1Âˇ3 1Âˇ3Âˇ5 1Âˇ3Âˇ5Âˇ7

by considering this approximation. (b) Try to conďŹ rm your conjecture using the limit comparison test.

56. Let ak and bk be series with positive terms. Prove: (a) If limk â&#x2020;&#x2019; +âŹ (ak /bk ) = 0 and bk converges, then ak converges. (b) If limk â&#x2020;&#x2019; +âŹ (ak /bk ) = +âŹ and bk diverges, then ak diverges. 57. Use Theorem 9.4.6 to prove the comparison test (Theorem 9.5.1). 58. Writing What does the ratio test tell you about the convergence of a geometric series? Discuss similarities between geometric series and series to which the ratio test applies. 59. Writing Given an inďŹ nite series, discuss a strategy for deciding what convergence test to use.

â&#x153;&#x201D;QUICK CHECK ANSWERS 9.5 1. diverges; 1/k 2/3

2. converges; ratio

637

3. diverges; ratio

4. converges; root

638

9.6

Chapter 9 / Infinite Series

ALTERNATING SERIES; ABSOLUTE AND CONDITIONAL CONVERGENCE Up to now we have focused exclusively on series with nonnegative terms. In this section we will discuss series that contain both positive and negative terms.

ALTERNATING SERIES

Series whose terms alternate between positive and negative, called alternating series, are of special importance. Some examples are ⬁

(−1)k+1

k=1

⬁

(−1)k

k=1

1 1 1 1 1 = 1 − + − + − ··· k 2 3 4 5

1 1 1 1 1 = −1 + − + − + · · · k 2 3 4 5

In general, an alternating series has one of the following two forms: ⬁ (−1)k+1 ak = a1 − a2 + a3 − a4 + · · ·

(1)

k=1

⬁ (−1)k ak = −a1 + a2 − a3 + a4 − · · ·

(2)

k=1

where the ak ’s are assumed to be positive in both cases. The following theorem is the key result on convergence of alternating series.

9.6.1 theorem (Alternating Series Test) An alternating series of either form (1) or form (2) converges if the following two conditions are satisﬁed: (a) a1 ≥ a2 ≥ a3 ≥ · · · ≥ ak ≥ · · · (b)

lim ak = 0

k → +⬁

a1 a2

proof We will consider only alternating series of form (1). The idea of the proof is to show that if conditions (a) and (b) hold, then the sequences of even-numbered and oddnumbered partial sums converge to a common limit S. It will then follow from Theorem 9.1.4 that the entire sequence of partial sums converges to S. Figure 9.6.1 shows how successive partial sums satisfying conditions (a) and (b) appear when plotted on a horizontal axis. The even-numbered partial sums

a3 a4 a5

s1 = a1

Figure 9.6.1

s2 , s4 , s6 , s8 , . . . , s2n , . . . form an increasing sequence bounded above by a1 , and the odd-numbered partial sums

It is not essential for condition (a) in Theorem 9.6.1 to hold for all terms; an alternating series will converge if condition (b) is true and condition (a) holds eventually.

s1 , s3 , s5 , . . . , s2n−1 , . . . form a decreasing sequence bounded below by 0. Thus, by Theorems 9.2.3 and 9.2.4, the even-numbered partial sums converge to some limit SE and the odd-numbered partial sums converge to some limit SO . To complete the proof we must show that SE = SO . But the

9.6 Alternating Series; Absolute and Conditional Convergence

If an alternating series violates condition (b) of the alternating series test, then the series must diverge by the divergence test (Theorem 9.4.1).

639

(2n)-th term in the series is −a2n , so that s2n − s2n−1 = −a2n , which can be written as s2n−1 = s2n + a2n However, 2n → +⬁ and 2n − 1 → +⬁ as n → +⬁, so that SO = lim s2n−1 = lim (s2n + a2n ) = SE + 0 = SE n → +⬁

n → +⬁

which completes the proof. ■

Example 1

Use the alternating series test to show that the following series converge. (a)

⬁ 1 (−1)k+1 k k=1

(b)

⬁

(−1)k+1

k=1

k+3 k(k + 1)

Solution (a). The two conditions in the alternating series test are satisﬁed since The series in part (a) of Example 1 is called the alternating harmonic series. Note that this series converges, whereas the harmonic series diverges.

ak =

1 1 > = ak+1 k k+1

and

1 =0 k → +⬁ k

lim ak = lim

k → +⬁

Solution (b). The two conditions in the alternating series test are satisﬁed since ak+1 k+4 k(k + 1) k 2 + 4k k 2 + 4k = · = 2 = 2 <1 ak (k + 1)(k + 2) k+3 k + 5k + 6 (k + 4k) + (k + 6) so ak > ak+1 and 1 3 + 2 k+3 k k =0 = lim lim ak = lim 1 k → +⬁ k → +⬁ k → +⬁ k(k + 1) 1+ k APPROXIMATING SUMS OF ALTERNATING SERIES The following theorem is concerned with the error that results when the sum of an alternating series is approximated by a partial sum.

9.6.2 theorem If an alternating series satisﬁes the hypotheses of the alternating series test, and if S is the sum of the series, then: (a) S lies between any two successive partial sums; that is, either sn ≤ S ≤ sn+1 or sn+1 ≤ S ≤ sn

(3)

depending on which partial sum is larger. (b) If S is approximated by sn , then the absolute error |S − sn | satisﬁes |S − sn | ≤ an+1

(4)

Moreover, the sign of the error S − sn is the same as that of the coefﬁcient of an+1 .

Chapter 9 / Infinite Series

640

proof We will prove the theorem for series of form (1). Referring to Figure 9.6.2 and keeping in mind our observation in the proof of Theorem 9.6.1 that the odd-numbered partial sums form a decreasing sequence converging to S and the even-numbered partial sums form an increasing sequence converging to S, we see that successive partial sums oscillate from one side of S to the other in smaller and smaller steps with the odd-numbered partial sums being larger than S and the even-numbered partial sums being smaller than S. Thus, depending on whether n is even or odd, we have

a3 a4 a5

sn ≤ S ≤ sn+1

Figure 9.6.2

sn+1 ≤ S ≤ sn

which proves (3). Moreover, in either case we have |S − sn | ≤ |sn+1 − sn |

(5)

But sn+1 − sn = ±an+1 (the sign depending on whether n is even or odd). Thus, it follows from (5) that |S − sn | ≤ an+1 , which proves (4). Finally, since the odd-numbered partial sums are larger than S and the even-numbered partial sums are smaller than S, it follows that S − sn has the same sign as the coefﬁcient of an+1 (verify). ■ REMARK

Example 2 series is

Later in this chapter we will show that the sum of the alternating harmonic 1 1 1 1 ln 2 = 1 − + − + · · · + (−1)k+1 + · · · 2 3 4 k This is illustrated in Figure 9.6.3.

{sn}

0.8 ln 2 0.6 0.4 0.2

n −0.2 −0.4

In words, inequality (4) states that for a series satisfying the hypotheses of the alternating series test, the magnitude of the error that results from approximating S by sn is at most that of the first term that is not included in the partial sum. Also, note that if a1 > a2 > · · · > ak > · · ·, then inequality (4) can be strengthened to |S − sn | < an+1 .

兵(−1)k+1 1k 其

(a) Accepting this to be so, ﬁnd an upper bound on the magnitude of the error that results if ln 2 is approximated by the sum of the ﬁrst eight terms in the series. (b) Find a partial sum that approximates ln 2 to one decimal-place accuracy (the nearest tenth).

−0.6 Graph of the sequences of terms and nth partial sums for the alternating harmonic series

Figure 9.6.3

Solution (a). It follows from the strengthened form of (4) that | ln 2 − s8 | < a9 =

1 < 0.12 9

(6)

As a check, let us compute s8 exactly. We obtain 533 1 1 1 1 1 1 1 + − + − + − = 2 3 4 5 6 7 8 840 Thus, with the help of a calculator 533 | ln 2 − s8 | = ln 2 − ≈ 0.059 840 s8 = 1 −

This shows that the error is well under the estimate provided by upper bound (6).

Solution (b). For one decimal-place accuracy, we must choose a value of n for which | ln 2 − sn | ≤ 0.05. However, it follows from the strengthened form of (4) that | ln 2 − sn | < an+1 so it sufﬁces to choose n so that an+1 ≤ 0.05.

9.6 Alternating Series; Absolute and Conditional Convergence

641

One way to find n is to use a calculating utility to obtain numerical values for a1 , a2 , a3 , . . . until you encounter the first value that is less than or equal to 0.05. If you do this, you will find that it is a20 = 0.05; this tells us that partial sum s19 will provide the desired accuracy. Another way to find n is to solve the inequality As Example 2 illustrates, the alternating harmonic series does not provide an efficient way to approximate ln 2, since too many terms and hence too much computation is required to achieve reasonable accuracy. Later, we will develop better ways to approximate logarithms.

1 ≤ 0.05 n+1

algebraically. We can do this by taking reciprocals, reversing the sense of the inequality, and then simplifying to obtain n ≥ 19. Thus, s19 will provide the required accuracy, which is consistent with the previous result. With the help of a calculating utility, the value of s19 is approximately s19 ≈ 0.7 and the value of ln 2 obtained directly is approximately ln 2 ≈ 0.69, which agrees with s19 when rounded to one decimal place.

ABSOLUTE CONVERGENCE The series 1 1 1 1 1 1 1 − − 2 + 3 + 4 − 5 − 6 + ··· 2 2 2 2 2 2

does not ﬁt in any of the categories studied so far—it has mixed signs but is not alternating. We will now develop some convergence tests that can be applied to such series.

9.6.3

definition A series ⬁ k=1

uk = u1 + u2 + · · · + uk + · · ·

is said to converge absolutely if the series of absolute values ⬁ k=1

|uk | = |u1 | + |u2 | + · · · + |uk | + · · ·

converges and is said to diverge absolutely if the series of absolute values diverges.

Example 3 (a) 1 −

Determine whether the following series converge absolutely. 1 1 1 1 1 − + 3 + 4 − 5 − ··· 2 22 2 2 2

(b) 1 −

1 1 1 1 + − + − ··· 2 3 4 5

Solution (a). The series of absolute values is the convergent geometric series 1+

1 1 1 1 1 + + 3 + 4 + 5 + ··· 2 22 2 2 2

so the given series converges absolutely.

Solution (b). The series of absolute values is the divergent harmonic series 1+

1 1 1 1 + + + + ··· 2 3 4 5

so the given series diverges absolutely.

Chapter 9 / Infinite Series

642

It is important to distinguish between the notions of convergence and absolute convergence. For example, the series in part (b) of Example 3 converges, since it is the alternating harmonic series, yet we demonstrated that it does not converge absolutely. However, the following theorem shows that if a series converges absolutely, then it converges.

9.6.4 theorem If the series Theorem 9.6.4 provides a way of inferring convergence of a series with positive and negative terms from a related series with nonnegative terms (the series of absolute values). This is important because most of the convergence tests that we have developed apply only to series with nonnegative terms.

⬁ k=1

|uk | = |u1 | + |u2 | + · · · + |uk | + · · ·

converges, then so does the series ⬁ k=1

uk = u1 + u2 + · · · + uk + · · ·

proof We will write the series ⬁ k=1

uk as

uk =

⬁ k=1

(7)

[(uk + |uk |) − |uk |]

We are assuming that |uk | converges, so that if we can show that (uk + |uk |) converges, then it will follow from (7) and Theorem 9.4.3(a) that uk converges. However, the value of uk + |uk | is either 0 or 2|uk |, depending on the sign of uk . Thus, in all cases it is true that

0 ≤ uk + |uk | ≤ 2|uk |

1 0.8 0.6

{sn}

0.4 0.2

n 2

−0.2 −0.4

|uk |; hence

Example 4

{uk}

−0.6

But 2|uk | converges, since it is a constant times the convergent series (uk + |uk |) converges by the comparison test. ■ Show that the following series converge.

(a) 1 −

(a)

1 1 1 1 1 1 − 2 + 3 + 4 − 5 − 6 + ··· 2 2 2 2 2 2

(b)

⬁ cos k k=1

Solution (a). Observe that this is not an alternating series because the signs alternate in

0.8 0.6 0.4

{sn}

0.2

−0.2

−0.4

{uk}

−0.6

(b) Graphs of the sequences of terms and n th partial sums for the series in Example 4

Figure 9.6.4

pairs after the ﬁrst term. Thus, we have no convergence test that can be applied directly. However, we showed in Example 3(a) that the series converges absolutely, so Theorem 9.6.4 implies that it converges (Figure 9.6.4a).

Solution (b). With the help of a calculating utility, you will be able to verify that the signs of the terms in this series vary irregularly. Thus, we will test for absolute convergence. The series of absolute values is ⬁ cos k k2 k=1 However,

cos k 1 k2 ≤ k2

9.6 Alternating Series; Absolute and Conditional Convergence

643

But 1/k 2 is a convergent p-series (p = 2), so the series of absolute values converges by the comparison test. Thus, the given series converges absolutely and hence converges (Figure 9.6.4b).

CONDITIONAL CONVERGENCE Although Theorem 9.6.4 is a useful tool for series that converge absolutely, it provides no information about the convergence or divergence of a series that diverges absolutely. For example, consider the two series

1 1 1 1 + − + · · · + (−1)k+1 + · · · 2 3 4 k 1 1 1 1 −1 − − − − · · · − − · · · 2 3 4 k 1−

(8) (9)

Both of these series diverge absolutely, since in each case the series of absolute values is the divergent harmonic series 1 1 1 + + ··· + + ··· 2 3 k However, series (8) converges, since it is the alternating harmonic series, and series (9) diverges, since it is a constant times the divergent harmonic series. As a matter of terminology, a series that converges but diverges absolutely is said to converge conditionally (or to be conditionally convergent). Thus, (8) is a conditionally convergent series. 1+

Example 5

In Example 1(b) we used the alternating series test to show that the series ⬁ k=1

(−1)k+1

k+3 k(k + 1)

converges. Determine whether this series converges absolutely or converges conditionally.

Solution. We test the series for absolute convergence by examining the series of absolute values:

⬁ ⬁ k+3 (−1)k+1 k + 3 = k(k + 1) k(k + 1) k=1

k=1

Principle 9.5.3 suggests that the series of absolute values should behave like the divergent p-series with p = 1. To prove that the series of absolute values diverges, we will apply the limit comparison test with ak = We obtain

k+3 1 and bk = k(k + 1) k

k(k + 3) k+3 ak = lim =1 = lim k→+⬁ k(k + 1) k→+⬁ k + 1 k→+⬁ bk

ρ = lim

Since ρ is ﬁnite and positive, it follows from the limit comparison test that the series of absolute values diverges. Thus, the original series converges and also diverges absolutely, and so converges conditionally.

THE RATIO TEST FOR ABSOLUTE CONVERGENCE Although one cannot generally infer convergence or divergence of a series from absolute divergence, the following variation of the ratio test provides a way of deducing divergence from absolute divergence in certain situations. We omit the proof.

644

Chapter 9 / Infinite Series

9.6.5 theorem (Ratio Test for Absolute Convergence) Let uk be a series with nonzero terms and suppose that |uk+1 | ρ = lim k → +⬁ |uk | (a) If ρ < 1, then the series uk converges absolutely and therefore converges. (b) If ρ > 1 or if ρ = +⬁, then the series uk diverges. (c) If ρ = 1, no conclusion about convergence or absolute convergence can be drawn from this test.

Example 6 converges.

Use the ratio test for absolute convergence to determine whether the series (a)

⬁ k=1

(−1)k

2k k!

(b)

⬁

(−1)k

k=1

(2k − 1)! 3k

Solution (a). Taking the absolute value of the general term uk we obtain

Thus,

k 2k k2 = |uk | = (−1) k! k!

|uk+1 | k! 2k+1 2 = lim · = lim =0<1 k → +⬁ |uk | k → +⬁ (k + 1)! 2k k → +⬁ k + 1

ρ = lim

which implies that the series converges absolutely and therefore converges.

Solution (b). Taking the absolute value of the general term uk we obtain

Thus,

(2k − 1)! (2k − 1)! |uk | = (−1)k = 3k 3k

|uk+1 | 3k [2(k + 1) − 1]! = lim · k → +⬁ |uk | k → +⬁ 3k+1 (2k − 1)!

ρ = lim = lim

k → +⬁

1 (2k + 1)! 1 · = lim (2k)(2k + 1) = +⬁ 3 (2k − 1)! 3 k → +⬁

which implies that the series diverges.

SUMMARY OF CONVERGENCE TESTS We conclude this section with a summary of convergence tests that can be used for reference. The skill of selecting a good test is developed through lots of practice. In some instances a test may be inconclusive, so another test must be tried.

Summary of Convergence Tests

name Divergence Test (9.4.1)

Integral Test (9.4.4)

statement If lim uk ≠ 0, then

comments If lim uk = 0, then k→ + ∞ may not converge.

uk diverges.

k→ + ∞

Let uk be a series with positive terms. If f is a function that is decreasing and continuous on an interval [a, + ∞) and such that uk = f (k) for all k ≥ a, then ∞ k =1

uk may or

This test only applies to series that have positive terms.

+∞

and

Try this test when f (x) is easy to integrate.

f (x) dx a

both converge or both diverge. ∞

Let a and k =1 k terms such that Comparison Test (9.5.1)

∞

b k =1 k

be series with nonnegative

a 1 ≤ b 1, a 2 ≤ b 2, . . . , a k ≤ b k , . . . If bk converges, then ak converges, and if diverges, then bk diverges. Let

ak and

bk be series with positive terms and let a r = lim k k→ + ∞ bk

Limit Comparison Test (9.5.4)

If 0 < r < + ∞, then both series converge or both diverge. Let Ratio Test (9.5.5)

Root Test (9.5.6)

Try this test as a last resort; other tests are often easier to apply.

This is easier to apply than the comparison test, but still requires some skill in choosing the series bk for comparison.

uk be a series with positive terms and suppose that r=

uk+1 k→ + ∞ uk lim

(a) Series converges if r < 1. (b) Series diverges if r > 1 or r = + ∞. (c) The test is inconclusive if r = 1. Let

This test only applies to series with nonnegative terms.

Try this test when uk involves factorials or kth powers.

uk be a series with positive terms and suppose that r=

lim

k→ + ∞

(a) The series converges if r < 1. (b) The series diverges if r > 1 or r = + ∞. (c) The test is inconclusive if r = 1.

Try this test when uk involves kth powers.

If ak > 0 for k = 1, 2, 3, . . . , then the series Alternating Series Test (9.6.1)

a1 − a2 + a3 − a4 + . . . −a1 + a2 – a3 + a4 − . . . converge if the following conditions hold: (a) a1 ≥ a2 ≥ a3 ≥ . . . (b) lim ak = 0

This test applies only to alternating series.

k→ + ∞

Let Ratio Test for Absolute Convergence (9.6.5)

uk be a series with nonzero terms and suppose that

| uk+1 | k→ + ∞ | uk |

r = lim

(a) The series converges absolutely if r < 1. (b) The series diverges if r > 1 or r = + ∞. (c) The test is inconclusive if r = 1.

The series need not have positive terms and need not be alternating to use this test.

646

Chapter 9 / Infinite Series

✔QUICK CHECK EXERCISES 9.6

(See page 648 for answers.) ⬁ 3k − 1 (−1)k : 9k + 15 k=1 ⬁ 1 (−1)k (c) : k(k + 2) k=1 ⬁ 1 (−1)k+1 √ : (d) 4 3 k k=1

1. What characterizes an alternating series? 2. (a) The series

(b)

⬁ (−1)k+1

k=1

converges by the alternating series test since and . (b) If S=

⬁ (−1)k+1 k=1

and

then |S − s9 | <

s9 =

4. Given that

9 (−1)k+1 k=1

16.

ﬁes the hypotheses of the alternating series test (Theorem 9.6.1). ⬁ k (−1)k+1 k 2. 3 k=1

2k + 1

k=1

3–6 Determine whether the alternating series converges; justify your answer. ■ ⬁ ⬁ k+1 k+1 (−1)k+1 3. (−1)k+1 √ 4. 3k + 1 k+1 k=1 k=1 ⬁ 5. (−1)k+1 e−k k=1

⬁ ln k (−1)k 6. k k=3

7–12 Use the ratio test for absolute convergence (Theorem

9.6.5) to determine whether the series converges or diverges. If the test is inconclusive, say so. ■ ⬁

⬁ 3 k 2k 7. (−1)k+1 − 8. 5 k! k=1 k=1 9.

⬁

(−1)k+1

k=1

⬁ k3 (−1)k k 11. e k=1

3 k2

10.

⬁ k=1

12.

⬁ k=1

(−1)k

k 5k

(−1)k+1

kk k!

13–28 Classify each series as absolutely convergent, conditionally convergent, or divergent. ■ ⬁ ⬁ ⬁ (−1)k+1 (−1)k+1 (−4)k 14. 13. 15. 3k k2 k 4/3 k=1 k=1 k=1

⬁ (−1)k+1

k=1

■

1 k

CAS

1–2 Show that the series converges by conﬁrming that it satis-

⬁ (−1)k+1

1 (k + 1)4 /4k+1 = lim = k → +⬁ k → +⬁ k 4 /4k 4 4 ⬁ k 4/ k is the series k=1 (−1) k 4 conditionally convergent, absolutely convergent, or divergent? lim

3. Classify each sequence as conditionally convergent, absolutely convergent, or divergent. ⬁ 1 (a) (−1)k+1 : k k=1

EXERCISE SET 9.6

19.

21.

⬁

sin

kπ 2

−

1 ln k

20.

k3 + 1

⬁ sin k

⬁ k=1

(2k − 1)!

26.

√

(−1)k k(k + 1)

⬁ k cos kπ k=1

28.

⬁ (−1)k+1 k 2 k=1

24.

⬁ (−1)k ln k k=3

k=1

⬁ (−1)k+1 k! k=1

18.

22.

k ln k

⬁

k=2

27.

k=1

⬁ (−1)k k=2

25.

⬁ cos kπ

⬁ k+2 (−1)k+1 k(k + 3) k=1

k=1

23.

17.

⬁ k=1

k2 + 1

(−1)k+1

32k−1 k2 + 1

29–32 True–False Determine whether the statement is true or

false. Explain your answer. ■ 29. An alternating series is one whose terms alternate between even and odd. 30. If a series satisﬁes the hypothesis of the alternating series test, then the sequence of partial sums of the series oscillates between overestimates and underestimates for the sum of the series. 31. If a series converges, then either it converges absolutely or it converges conditionally. 32. If (uk )2 converges, then uk converges absolutely.

9.6 Alternating Series; Absolute and Conditional Convergence 33–36 Each series satisﬁes the hypotheses of the alternating

series test. For the stated value of n, ﬁnd an upper bound on the absolute error that results if the sum of the series is approximated by the nth partial sum. ■ ⬁ ⬁ (−1)k+1 (−1)k+1 33. ; n=7 34. ; n=5 k k! k=1 k=1 35.

36.

⬁ (−1)k+1 ; n = 99 √ k k=1

⬁ k=1

(−1)k+1 ; n=3 (k + 1) ln(k + 1)

37–40 Each series satisﬁes the hypotheses of the alternating

series test. Find a value of n for which the nth partial sum is ensured to approximate the sum of the series to the stated accuracy. ■

37.

38.

⬁ (−1)k+1

; |error| < 0.0001

k=1 ⬁

(−1)k+1 ; |error| < 0.00001 k!

⬁

(−1)k+1 ; one decimal place (k + 1) ln(k + 1)

k=1

⬁ (−1)k+1 ; two decimal places 39. √ k k=1

40.

k=1

41–42 Find an upper bound on the absolute error that results if s10 is used to approximate the sum of the given geometric series. Compute s10 rounded to four decimal places and compare this value with the exact sum of the series. ■ 3 3 3 3 2 4 8 41. − + − + · · · 42. 1 − + − + ··· 4 8 16 32 3 9 27 43–46 Each series satisﬁes the hypotheses of the alternating se-

ries test. Approximate the sum of the series to two decimal-place accuracy. ■ 1 1 1 1 1 1 43. 1 − + − + · · · 44. 1 − + − + · · · 3! 5! 7! 2! 4! 6! 1 1 1 1 45. + − + ··· − 1 · 2 2 · 22 3 · 23 4 · 24 1 1 1 1 46. 5 − 5 + 5 − 5 + ··· 1 +4·1 3 +4·3 5 +4·5 7 +4·7 F O C U S O N C O N C E P TS C

47. The purpose of this exercise is to show that the error bound in part (b) of Theorem 9.6.2 can be overly conservative in certain cases. (a) Use a CAS to conﬁrm that π 1 1 1 = 1 − + − + ··· 4 3 5 7 (b) Use the CAS to show that |(π/4) − s25 | < 10−2 .

647

(c) According to the error bound in part (b) of Theorem 9.6.2, what value of n is required to ensure that |(π/4) − sn | < 10−2 ? 48. Prove: If a series ak converges absolutely, then the series ak2 converges. 49. (a) Find 2examples to show that if ak converges, then ak may diverge or converge. 2 (b) Find examples to show that if ak converges, then ak may diverge or converge. 50. Let that

uk be a series and deﬁne series

pk and

qk so

uk > 0 0, uk ≥ 0 and qk = uk ≤ 0 −uk , uk < 0 (a) Show that uk converges absolutely if and only if converge. pk and qk both (b) Show that if one of pk or qk converges and the other diverges, then u diverges. k (c) Show that if uk converges conditionally, then both pk and qk diverge. pk =

uk , 0,

51. It can be proved that the terms of any conditionally convergent series can be rearranged to give either a divergent series or a conditionally convergent series whose sum is any given number S. For example, we stated in Example 2 that 1 1 1 1 1 ln 2 = 1 − + − + − + · · · 2 3 4 5 6 Show that we can rearrange this series so that its sum is 1 ln 2 by rewriting it as 2

1 1 1 1 1 1 1 1 − − − − + + + ··· 1− − 2 4 3 6 8 5 10 12 [Hint: Add the ﬁrst two terms in each grouping.]

52–54 Exercise 51 illustrates that one of the nuances of “conditional” convergence is that the sum of a series that converges conditionally depends on the order that the terms of the series are summed. Absolutely convergent series are more dependable, however. It can be proved that any series that is constructed from an absolutely convergent series by rearranging the terms will also be absolutely convergent and has the same sum as the original series. Use this fact together with parts (a) and (b) of Theorem 9.4.3 in these exercises. ■

52. It was stated in Exercise 35 of Section 9.4 that 1 1 1 π2 = 1 + 2 + 2 + 2 + ··· 6 2 3 4 Use this to show that π2 1 1 1 = 1 + 2 + 2 + 2 + ··· 8 3 5 7 53. Use the series for π2 /6 given in the preceding exercise to show that π2 1 1 1 = 1 − 2 + 2 − 2 + ··· 12 2 3 4

648

Chapter 9 / Infinite Series

54. It was stated in Exercise 35 of Section 9.4 that π4 1 1 1 = 1 + 4 + 4 + 4 + ··· 90 2 3 4 Use this to show that 1 1 1 π4 = 1 + 4 + 4 + 4 + ··· 96 3 5 7 55. Writing Consider the series 1 2 1 2 1 2 1 1 − + − + − + − + ··· 2 3 3 4 4 5 5

Determine whether this series converges and use this series as an example in a discussion of the importance of hypotheses (a) and (b) of the alternating series test (Theorem 9.6.1). 56. Writing Discuss the ways that conditional convergence is “conditional.” In particular, describe how one could rearrange the terms of a conditionally convergent series uk so that the resulting series diverges, either to +⬁ or to −⬁. [Hint: See Exercise 50.]

✔QUICK CHECK ANSWERS 9.6 1 1 1 1 1 1 ≥ · · ·; lim 2 = 0 (b) ≥ ≥ ··· ≥ 2 ≥ 2 k → +⬁ 4 9 k (k + 1) k 100 3. (a) conditionally convergent (b) divergent (c) absolutely convergent (d) conditionally convergent 4. absolutely convergent 1. Terms alternate between positive and negative.

9.7

2. (a) 1 ≥

MACLAURIN AND TAYLOR POLYNOMIALS In a local linear approximation the tangent line to the graph of a function is used to obtain a linear approximation of the function near the point of tangency. In this section we will consider how one might improve on the accuracy of local linear approximations by using higher-order polynomials as approximating functions. We will also investigate the error associated with such approximations. LOCAL QUADRATIC APPROXIMATIONS

Recall from Formula (1) in Section 3.5 that the local linear approximation of a function f at x0 is (1) f(x) ≈ f(x0 ) + f ′(x0 )(x − x0 ) In this formula, the approximating function p(x) = f(x0 ) + f ′(x0 )(x − x0 )

Local linear approximation

x0 Figure 9.7.1

is a first-degree polynomial satisfying p(x0 ) = f(x0 ) and p ′ (x0 ) = f ′(x0 ) (verify). Thus, the local linear approximation of f at x0 has the property that its value and the value of its first derivative match those of f at x0 . If the graph of a function f has a pronounced “bend” at x0 , then we can expect that the accuracy of the local linear approximation of f at x0 will decrease rapidly as we progress away from x0 (Figure 9.7.1). One way to deal with this problem is to approximate the function f at x0 by a polynomial p of degree 2 with the property that the value of p and the values of its first two derivatives match those of f at x0 . This ensures that the graphs of f and p not only have the same tangent line at x0 , but they also bend in the same direction at x0 (both concave up or concave down). As a result, we can expect that the graph of p will remain close to the graph of f over a larger interval around x0 than the graph of the local linear approximation. The polynomial p is called the local quadratic approximation of f at x = x0 . To illustrate this idea, let us try to find a formula for the local quadratic approximation of a function f at x = 0. This approximation has the form f(x) ≈ c0 + c1 x + c2 x 2

where c0 , c1 , and c2 must be chosen so that the values of p(x) = c0 + c1 x + c2 x 2

(2)

9.7 Maclaurin and Taylor Polynomials

649

and its ﬁrst two derivatives match those of f at 0. Thus, we want p(0) = f(0),

p′ (0) = f ′(0),

p′′ (0) = f ′′(0)

(3)

But the values of p(0), p ′ (0), and p ′′ (0) are as follows: p(x) = c0 + c1 x + c2 x 2

p(0) = c0

p ′′ (x) = 2c2

p ′′ (0) = 2c2

′

p (x) = c1 + 2c2 x

p ′ (0) = c1

Thus, it follows from (3) that c0 = f(0),

c1 = f ′(0),

c2 =

f ′′(0) 2

and substituting these in (2) yields the following formula for the local quadratic approximation of f at x = 0: f(x) ≈ f(0) + f ′(0)x +

f ′′(0) 2 x 2

(4)

Example 1 Find the local linear and quadratic approximations of ex at x = 0, and graph ex and the two approximations together.

Solution. If we let f(x) = ex , then f ′(x) = f ′′(x) = ex ; and hence y = 1+x+

f(0) = f ′(0) = f ′′(0) = e0 = 1

x 2

y = 1+x 2

Thus, from (4) the local quadratic approximation of ex at x = 0 is

x2 2 and the local linear approximation (which is the linear part of the local quadratic approximation) is ex ≈ 1 + x ex ≈ 1 + x +

y = ex −2

Figure 9.7.2

x 2

The graphs of ex and the two approximations are shown in Figure 9.7.2. As expected, the local quadratic approximation is more accurate than the local linear approximation near x = 0. MACLAURIN POLYNOMIALS

It is natural to ask whether one can improve on the accuracy of a local quadratic approximation by using a polynomial of degree 3. Speciﬁcally, one might look for a polynomial of degree 3 with the property that its value and the values of its ﬁrst three derivatives match

Colin Maclaurin (1698–1746) Scottish mathematician. Maclaurin’s father, a minister, died when the boy was only six months old, and his mother when he was nine years old. He was then raised by an uncle who was also a minister. Maclaurin entered Glasgow University as a divinity student but switched to mathematics after one year. He received his Master’s degree at age 17 and, in spite of his youth, began teaching at Marischal College in Aberdeen, Scotland. He met Isaac Newton during a visit to London in 1719 and from that time on became Newton’s disciple. During that era, some of Newton’s analytic methods were bitterly attacked by major

mathematicians and much of Maclaurin’s important mathematical work resulted from his efforts to defend Newton’s ideas geometrically. Maclaurin’s work, A Treatise of Fluxions (1742), was the ﬁrst systematic formulation of Newton’s methods. The treatise was so carefully done that it was a standard of mathematical rigor in calculus until the work of Cauchy in 1821. Maclaurin was also an outstanding experimentalist; he devised numerous ingenious mechanical devices, made important astronomical observations, performed actuarial computations for insurance societies, and helped to improve maps of the islands around Scotland. [Image: © Bettmann/Corbis Images]

650

Chapter 9 / Infinite Series

those of f at a point; and if this provides an improvement in accuracy, why not go on to polynomials of even higher degree? Thus, we are led to consider the following general problem.

9.7.1 problem Given a function f that can be differentiated n times at x = x0 , ﬁnd a polynomial p of degree n with the property that the value of p and the values of its ﬁrst n derivatives match those of f at x0 .

We will begin by solving this problem in the case where x0 = 0. Thus, we want a polynomial p(x) = c0 + c1 x + c2 x 2 + c3 x 3 + · · · + cn x n (5) such that f(0) = p(0), But

f ′(0) = p′ (0),

f ′′(0) = p′′ (0), . . . ,

f (n) (0) = p(n) (0)

(6)

p(x) = c0 + c1 x + c2 x 2 + c3 x 3 + · · · + cn x n

p ′ (x) = c1 + 2c2 x + 3c3 x 2 + · · · + ncn x n−1 p ′′ (x) = 2c2 + 3 · 2c3 x + · · · + n(n − 1)cn x n−2

p ′′′ (x) = 3 · 2c3 + · · · + n(n − 1)(n − 2)cn x n−3 .. . p(n) (x) = n(n − 1)(n − 2) · · · (1)cn Thus, to satisfy (6) we must have f(0) = p(0) ′

′

= c0

f (0) = p (0) = c1

f ′′(0) = p′′ (0) = 2c2 = 2!c2 f ′′′(0) = p′′′ (0) = 3 · 2c3 = 3!c3 .. . f (n) (0) = p(n) (0) = n(n − 1)(n − 2) · · · (1)cn = n!cn which yields the following values for the coefﬁcients of p(x): c0 = f(0),

c1 = f ′(0),

c2 =

f ′′(0) , 2!

c3 =

f ′′′(0) ,..., 3!

cn =

f (n) (0) n!

The polynomial that results by using these coefﬁcients in (5) is called the nth Maclaurin polynomial for f .

Local linear approximations and local quadratic approximations at x = 0 of a function f are special cases of the MacLaurin polynomials for f . Verify that f(x) ≈ p1 (x) is the local linear approximation of f at x = 0, and f(x) ≈ p2 (x) is the local quadratic approximation at x = 0.

9.7.2 definition If f can be differentiated n times at 0, then we deﬁne the nth Maclaurin polynomial for f to be pn (x) = f(0) + f ′(0)x +

f ′′(0) 2 f ′′′(0) 3 f (n) (0) n x + x + ··· + x 2! 3! n!

(7)

Note that the polynomial in (7) has the property that its value and the values of its ﬁrst n derivatives match the values of f and its ﬁrst n derivatives at x = 0.

9.7 Maclaurin and Taylor Polynomials

Example 2

651

Find the Maclaurin polynomials p0 , p1 , p2 , p3 , and pn for ex .

Solution. Let f(x) = ex . Thus, and

f ′(x) = f ′′(x) = f ′′′(x) = · · · = f (n) (x) = ex

f(0) = f ′(0) = f ′′(0) = f ′′′(0) = · · · = f (n) (0) = e0 = 1 Therefore, p0 (x) = f(0) = 1

p1 (x) = f(0) + f ′(0)x = 1 + x

y=e

p1(x)

p0 (x)

x −2

−1

Figure 9.7.3

f ′′(0) 2 f ′′′(0) 3 x + x 2! 3!

x3 1 1 x2 + = 1 + x + x2 + x3 2! 3! 2 6

f ′′(0) 2 f (n) (0) n x + ··· + x 2! n! x2 xn =1+x+ + ··· + 2! n!

p2(x)

p3 (x) = f(0) + f ′(0)x +

pn (x) = f(0) + f ′(0)x +

f ′′(0) 2 x2 1 x =1+x+ = 1 + x + x2 2! 2! 2

=1+x+

p3(x)

p2 (x) = f(0) + f ′(0)x +

Figure 9.7.3 shows the graph of ex (in blue) and the graph of the ﬁrst four Maclaurin polynomials. Note that the graphs of p1 (x), p2 (x), and p3 (x) are virtually indistinguishable from the graph of ex near x = 0, so these polynomials are good approximations of ex for x near 0. However, the farther x is from 0, the poorer these approximations become. This is typical of the Maclaurin polynomials for a function f(x); they provide good approximations of f(x) near 0, but the accuracy diminishes as x progresses away from 0. It is usually the case that the higher the degree of the polynomial, the larger the interval on which it provides a speciﬁed accuracy. Accuracy issues will be investigated later.

Augustin Louis Cauchy (1789–1857) French mathematician. Cauchy’s early education was acquired from his father, a barrister and master of the classics. Cauchy entered L’Ecole Polytechnique in 1805 to study engineering, but because of poor health, was advised to concentrate on mathematics. His major mathematical work began in 1811 with a series of brilliant solutions to some difﬁcult outstanding problems. In 1814 he wrote a treatise on integrals that was to become the basis for modern complex variable theory; in 1816 there followed a classic paper on wave propagation in liquids that won a prize from the French Academy; and in 1822 he wrote a paper that formed the basis of modern elasticity theory. Cauchy’s mathematical contributions for the next 35 years were brilliant and staggering in quantity, over 700 papers ﬁlling 26 modern volumes. Cauchy’s work initiated the era of modern analysis. He brought to mathematics standards of precision and rigor undreamed of by Leibniz and Newton.

Cauchy’s life was inextricably tied to the political upheavals of the time. A strong partisan of the Bourbons, he left his wife and children in 1830 to follow the Bourbon king Charles X into exile. For his loyalty he was made a baron by the ex-king. Cauchy eventually returned to France, but refused to accept a university position until the government waived its requirement that he take a loyalty oath. It is difﬁcult to get a clear picture of the man. Devoutly Catholic, he sponsored charitable work for unwed mothers, criminals, and relief for Ireland. Yet other aspects of his life cast him in an unfavorable light. The Norwegian mathematician Abel described him as, “mad, inﬁnitely Catholic, and bigoted.” Some writers praise his teaching, yet others say he rambled incoherently and, according to a report of the day, he once devoted an entire lecture to extracting the square root of seventeen to ten decimal places by a method well known to his students. In any event, Cauchy is undeniably one of the greatest minds in the history of science. [Image: http://en.wikipedia.org/wiki/File:Augustin-Louis_Cauchy_1901.jpg]

652

Chapter 9 / Infinite Series

Example 3

Find the nth Maclaurin polynomials for (a) sin x

(b) cos x

Solution (a). In the Maclaurin polynomials for sin x, only the odd powers of x appear explicitly. To see this, let f(x) = sin x; thus, f(x) = sin x

f(0) = 0

f ′(x) = cos x

f ′(0) = 1

f ′′(x) = − sin x

f ′′(0) = 0

f ′′′(x) = − cos x

f ′′′(0) = −1

Since f (4) (x) = sin x = f(x), the pattern 0, 1, 0, −1 will repeat as we evaluate successive derivatives at 0. Therefore, the successive Maclaurin polynomials for sin x are p0 (x) = 0

p1 (x) = 0 + x

p2 (x) = 0 + x + 0

x3 3! x3 p4 (x) = 0 + x + 0 − +0 3! x3 x5 p5 (x) = 0 + x + 0 − +0+ 3! 5! 3 x5 x p6 (x) = 0 + x + 0 − +0+ +0 3! 5! x3 x5 x7 p7 (x) = 0 + x + 0 − +0+ +0− 3! 5! 7! Because of the zero terms, each even-order Maclaurin polynomial [after p0 (x)] is the same as the preceding odd-order Maclaurin polynomial. That is, p3 (x) = 0 + x + 0 −

y 2

p1(x) p5(x) c

C −2

y = sin x p3(x) p7(x)

Figure 9.7.4

x3 x5 x7 x 2k+1 + − + · · · + (−1)k (k = 0, 1, 2, . . .) 3! 5! 7! (2k + 1)! The graphs of sin x, p1 (x), p3 (x), p5 (x), and p7 (x) are shown in Figure 9.7.4. p2k+1 (x) = p2k+2 (x) = x −

Solution (b). In the Maclaurin polynomials for cos x, only the even powers of x appear explicitly; the computations are similar to those in part (a). The reader should be able to show that p0 (x) = p1 (x) = 1

y 2

p4(x) p0(x) c

C −2

p2(x) y = cos x

Figure 9.7.5

p6(x)

p2 (x) = p3 (x) = 1 −

x2 2!

p4 (x) = p5 (x) = 1 −

x2 x4 + 2! 4!

x4 x6 x2 + − 2! 4! 6! In general, the Maclaurin polynomials for cos x are given by p6 (x) = p7 (x) = 1 −

x2 x4 x6 x 2k + − + · · · + (−1)k (k = 0, 1, 2, . . .) 2! 4! 6! (2k)! The graphs of cos x, p0 (x), p2 (x), p4 (x), and p6 (x) are shown in Figure 9.7.5. p2k (x) = p2k+1 (x) = 1 −

9.7 Maclaurin and Taylor Polynomials

653

TAYLOR POLYNOMIALS

Up to now we have focused on approximating a function f in the vicinity of x = 0. Now we will consider the more general case of approximating f in the vicinity of an arbitrary domain value x0 . The basic idea is the same as before; we want to ﬁnd an nth-degree polynomial p with the property that its value and the values of its ﬁrst n derivatives match those of f at x0 . However, rather than expressing p(x) in powers of x, it will simplify the computations if we express it in powers of x − x0 ; that is, p(x) = c0 + c1 (x − x0 ) + c2 (x − x0 )2 + · · · + cn (x − x0 )n

(8)

We will leave it as an exercise for you to imitate the computations used in the case where x0 = 0 to show that c0 = f(x0 ),

c1 = f ′(x0 ),

c2 =

f ′′(x0 ) , 2!

c3 =

f ′′′(x0 ) ,..., 3!

cn =

f (n) (x0 ) n!

Substituting these values in (8) we obtain a polynomial called the nth Taylor polynomial about x = x0 for f .

Local linear approximations and local quadratic approximations at x = x0 of a function f are special cases of the Taylor polynomials for f . Verify that f(x) ≈ p1 (x) is the local linear approximation of f at x = x0 , and f(x) ≈ p2 (x) is the local quadratic approximation at x = x0 .

9.7.3 definition If f can be differentiated n times at x0 , then we deﬁne the nth Taylor polynomial for f about x = x0 to be f ′′(x0 ) (x − x0 )2 2! f (n) (x0 ) f ′′′(x0 ) (x − x0 )3 + · · · + (x − x0 )n + 3! n!

pn (x) = f(x0 ) + f ′(x0 )(x − x0 ) +

Example 4 The Maclaurin polynomials are the special cases of the Taylor polynomials in which x0 = 0. Thus, theorems about Taylor polynomials also apply to Maclaurin polynomials.

(9)

Find the ﬁrst four Taylor polynomials for ln x about x = 2.

Solution. Let f(x) = ln x. Thus, f(x) = ln x f ′(x) = 1/x

f ′′(x) = −1/x 2 f ′′′(x) = 2/x 3

Brook Taylor (1685–1731) English mathematician. Taylor was born of well-to-do parents. Musicians and artists were entertained frequently in the Taylor home, which undoubtedly had a lasting influence on him. In later years, Taylor published a definitive work on the mathematical theory of perspective and obtained major mathematical results about the vibrations of strings. There also exists an unpublished work, On Musick, that was intended to be part of a joint paper with Isaac Newton. Taylor’s life was scarred with unhappiness, illness, and tragedy. Because his first wife was not rich enough to suit his father, the two men argued bitterly and parted ways. Subsequently, his wife died in childbirth. Then, after he remarried, his

f(2) = ln 2 f ′(2) = 1/2

f ′′(2) = −1/4 f ′′′(2) = 1/4

second wife also died in childbirth, though his daughter survived. Taylor’s most productive period was from 1714 to 1719, during which time he wrote on a wide range of subjects—magnetism, capillary action, thermometers, perspective, and calculus. In his ﬁnal years, Taylor devoted his writing efforts to religion and philosophy. According to Taylor, the results that bear his name were motivated by coffeehouse conversations about works of Newton on planetary motion and works of Halley (“Halley’s comet”) on roots of polynomials. Unfortunately, Taylor’s writing style was so terse and hard to understand that he never received credit for many of his innovations. [Image: http://en.wikipedia.org/wiki/File:BTaylor.jpg]

Chapter 9 / Infinite Series

654

Substituting in (9) with x0 = 2 yields y 2 1

p0 (x) = f(2) = ln 2

y = ln x p3(x) p2(x) p1(x) p0 (x)

p1 (x) = f(2) + f ′(2)(x − 2) = ln 2 + 21 (x − 2)

x 1

p2 (x) = f(2) + f ′(2)(x − 2) +

f ′′(2) (x − 2)2 = ln 2 + 21 (x − 2) − 18 (x − 2)2 2!

p3 (x) = f(2) + f ′(2)(x − 2) +

f ′′(2) f ′′′(2) (x − 2)2 + (x − 2)3 2! 3!

−1 −2

Figure 9.7.6

= ln 2 + 21 (x − 2) − 18 (x − 2)2 +

1 (x 24

− 2)3

The graph of ln x (in blue) and its ﬁrst four Taylor polynomials about x = 2 are shown in Figure 9.7.6. As expected, these polynomials produce their best approximations of ln x near 2. SIGMA NOTATION FOR TAYLOR AND MACLAURIN POLYNOMIALS

Frequently, we will want to express Formula (9) in sigma notation. To do this, we use the notation f (k) (x0 ) to denote the kth derivative of f at x = x0 , and we make the convention that f (0) (x0 ) denotes f(x0 ). This enables us to write n f (k) (x0 ) k=0

(x − x0 )k = f(x0 ) + f ′(x0 )(x − x0 )

f ′′(x0 ) f (n) (x0 ) (x − x0 )2 + · · · + (x − x0 )n 2! n! In particular, we can write the nth Maclaurin polynomial for f(x) as +

n f (k) (0) k=0

Example 5

x k = f(0) + f ′(0)x +

f ′′(0) 2 f (n) (0) n x + ··· + x 2! n!

(10)

(11)

Find the nth Maclaurin polynomial for 1 1−x

and express it in sigma notation.

Solution. Let f(x) = 1/(1 − x). The values of f and its ﬁrst k derivatives at x = 0 are as follows:

f(x)

f ′(x)

f ′′(x) = T E C H N O LO GY M A ST E R Y Computer algebra systems have commands for generating Taylor polynomials of any specified degree. If you have a CAS, use it to find some of the Maclaurin and Taylor polynomials in Examples 3, 4, and 5.

f ′′′(x) =

1 1−x

1 (1 − x)2 2 (1 − x)3 3·2 (1 − x)4

f(0)

= 1 = 0!

f ′(0)

= 1 = 1!

f ′′(0) = 2 = 2! f ′′′(0) = 3!

f (4) (x) =

4·3·2 (1 − x)5

f (4) (0) = 4!

f (k) (x) =

k! (1 − x)k+1

f (k) (0) = k!

.. .

9.7 Maclaurin and Taylor Polynomials

655

Thus, substituting f (k) (0) = k! into Formula (11) yields the nth Maclaurin polynomial for 1/(1 − x): pn (x) =

Example 6 sigma notation.

n k=0

xk = 1 + x + x2 + · · · + xn

(n = 0, 1, 2, . . .)

Find the nth Taylor polynomial for 1/x about x = 1 and express it in

Solution. Let f(x) = 1/x. The computations are similar to those in Example 5. We leave it for you to show that f(1) = 1,

f ′(1) = −1,

f (4) (1) = 4!, . . . ,

f ′′(1) = 2!,

f (k) (1) = (−1)k k!

f ′′′(1) = −3!,

Thus, substituting f (k) (1) = (−1)k k! into Formula (10) with x0 = 1 yields the nth Taylor polynomial for 1/x: n (−1)k (x − 1)k = 1 − (x − 1) + (x − 1)2 − (x − 1)3 + · · · + (−1)n (x − 1)n k=0

THE nTH REMAINDER

It will be convenient to have a notation for the error in the approximation f(x) ≈ pn (x). Accordingly, we will let Rn (x) denote the difference between f(x) and its nth Taylor polynomial; that is, Rn (x) = f(x) − pn (x) = f(x) −

n f (k) (x0 ) k=0

(x − x0 )k

(12)

This can also be written as f(x) = pn (x) + Rn (x) =

n f (k) (x0 ) k=0

(x − x0 )k + Rn (x)

(13)

The function Rn (x) is called the nth remainder for the Taylor series of f , and Formula (13) is called Taylor’s formula with remainder. Finding a bound for Rn (x) gives an indication of the accuracy of the approximation pn (x) ≈ f(x). The following theorem, which is proved in Appendix D, provides such a bound.

The bound for |Rn (x)| in (14) is called the Lagrange error bound .

9.7.4 theorem (The Remainder Estimation Theorem) If the function f can be differentiated n + 1 times on an interval containing the number x0 , and if M is an upper bound for |f (n+1) (x)| on the interval, that is, |f (n+1) (x)| ≤ M for all x in the interval, then |Rn (x)| ≤ for all x in the interval.

M |x − x0 |n+1 (n + 1)!

(14)

656

Chapter 9 / Infinite Series

Example 7 Use an nth Maclaurin polynomial for ex to approximate e to ﬁve decimalplace accuracy.

Solution. We note ﬁrst that the exponential function ex has derivatives of all orders for every real number x. From Example 2, the nth Maclaurin polynomial for ex is n xk k=0

from which we have

e = e1 ≈

=1+x+

n 1k k=0

x2 xn + ··· + 2! n!

=1+1+

1 1 + ··· + 2! n!

Thus, our problem is to determine how many terms to include in a Maclaurin polynomial for ex to achieve ﬁve decimal-place accuracy; that is, we want to choose n so that the absolute value of the nth remainder at x = 1 satisﬁes |Rn (1)| ≤ 0.000005

To determine n we use the Remainder Estimation Theorem with f(x) = ex , x = 1, x0 = 0, and the interval [0, 1]. In this case it follows from (14) that |Rn (1)| ≤

M M · |1 − 0|n+1 = (n + 1)! (n + 1)!

(15)

where M is an upper bound on the value of f (n+1) (x) = ex for x in the interval [0, 1]. However, ex is an increasing function, so its maximum value on the interval [0, 1] occurs at x = 1; that is, ex ≤ e on this interval. Thus, we can take M = e in (15) to obtain e (16) |Rn (1)| ≤ (n + 1)!

Unfortunately, this inequality is not very useful because it involves e, which is the very quantity we are trying to approximate. However, if we accept that e < 3, then we can replace (16) with the following less precise, but more easily applied, inequality: |Rn (1)| ≤

3 (n + 1)!

Thus, we can achieve ﬁve decimal-place accuracy by choosing n so that 3 ≤ 0.000005 (n + 1)!

or (n + 1)! ≥ 600,000

Since 9! = 362,880 and 10! = 3,628,800, the smallest value of n that meets this criterion is n = 9. Thus, to ﬁve decimal-place accuracy

1 1 1 1 1 1 1 1 + + + + + + + ≈ 2.71828 2! 3! 4! 5! 6! 7! 8! 9! As a check, a calculator’s 12-digit representation of e is e ≈ 2.71828182846, which agrees with the preceding approximation when rounded to ﬁve decimal places. e ≈1+1+

Example 8 Use the Remainder Estimation Theorem to ﬁnd an interval containing x = 0 throughout which f(x) = cos x can be approximated by p(x) = 1 − (x 2 /2!) to three decimal-place accuracy.

Solution. We note first that f(x) = cos x has derivatives of all orders for every real number x, so the first hypothesis of the Remainder Estimation Theorem is satisfied over any interval that we choose. The given polynomial p(x) is both the second and the third

9.7 Maclaurin and Taylor Polynomials

657

Maclaurin polynomial for cos x; we will choose the degree n of the polynomial to be as large as possible, so we will take n = 3. Our problem is to determine an interval on which the absolute value of the third remainder at x satisﬁes |R3 (x)| ≤ 0.0005 We will use the Remainder Estimation Theorem with f(x) = cos x, n = 3, and x0 = 0. It follows from (14) that |R3 (x)| ≤

y y = | f (x) − p(x)|

M|x|4 M |x − 0|3+1 = (3 + 1)! 24

(17)

|x|4 (18) 24 Thus we can achieve three decimal-place accuracy by choosing values of x for which |R3 (x)| ≤

0.0005 0.0004 0.0003 0.0002 0.0001

|x|4 ≤ 0.0005 or |x| ≤ 0.3309 24 so the interval [−0.3309, 0.3309] is one option. We can check this answer by graphing |f (x) − p(x)| over the interval [−0.3309, 0.3309] (Figure 9.7.7).

−0.3309

0.3309

Figure 9.7.7

✔QUICK CHECK EXERCISES 9.7

(See page 659 for answers.)

1. If f can be differentiated three times at 0, then the third Maclaurin polynomial for f is p3 (x) = . 2. The third Maclaurin polynomial for f(x) = e2x is p3 (x) =

x +

3. If f(2) = 3, f ′(2) = −4, and f ′′(2) = 10, then the second Taylor polynomial for f about x = 2 is p2 (x) = .

5. (a) If a function f has nth Taylor polynomial pn (x) about x = x0 , then the nth remainder Rn (x) is defined by Rn (x) = . (b) Suppose that a function f can be differentiated five times on an interval containing x0 = 2 and that |f (5) (x)| ≤ 20 for all x in the interval. Then the fourth remainder satisfies |R4 (x)| ≤ for all x in the interval.

4. The third Taylor polynomial for f(x) = x 5 about x = −1 is p3 (x) = + (x + 1) +

EXERCISE SET 9.7

(x + 1)2 +

(x + 1)3

Graphing Utility

1–2 In each part, ﬁnd the local quadratic approximation of f at x = x0 , and use that approximation to ﬁnd the local linear approximation of f at x0 . Use a graphing utility to graph f and the two approximations on the same screen. ■

1. (a) f(x) = e−x ; x0 = 0

(b) f(x) = cos x; x0 = 0 √ x; x0 = 1 √ 3. (a) Find the local quadratic approximation of x at x0√= 1. (b) Use the result obtained in part (a) to approximate 1.1, and compare your approximation to that produced directly by your calculating utility. [Note: See Example 1 of Section 3.5.] 2. (a) f(x) = sin x; x0 = π/2

(b) f(x) =

4. (a) Find the local quadratic approximation of cos x at x0 = 0. (b) Use the result obtained in part (a) to approximate cos 2 ◦ , and compare the approximation to that produced directly by your calculating utility. 5. Use an appropriate local quadratic approximation to approximate tan 61 ◦ , and compare the result to that produced directly by your calculating utility. 6. Use an√ appropriate local quadratic approximation to approximate 36.03, and compare the result to that produced directly by your calculating utility.

658

Chapter 9 / Infinite Series

7–16 Find the Maclaurin polynomials of orders n = 0, 1, 2, 3,

and 4, and then ﬁnd the nth Maclaurin polynomials for the function in sigma notation. ■ 7. e−x 8. eax 9. cos πx 1 10. sin πx 11. ln(1 + x) 12. 1+x 13. cosh x 14. sinh x 15. x sin x

36. 1/e; three decimal-place accuracy F O C U S O N C O N C E P TS

37. Which of the functions graphed in the following ﬁgure is most likely to have p(x) = 1 − x + 2x 2 as its secondorder Maclaurin polynomial? Explain your reasoning.

16. xex

17–24 Find the Taylor polynomials of orders n = 0, 1, 2, 3, and 4 about x = x0 , and then ﬁnd the nth Taylor polynomial for the function in sigma notation. ■

17. ex ; x0 = 1 1 19. ; x0 = −1 x

18. e−x ; x0 = ln 2 1 20. ; x0 = 3 x+2 π 22. cos x; x0 = 2 24. ln x; x0 = e

1 21. sin πx; x0 = 2 23. ln x; x0 = 1

25. (a) Find the third Maclaurin polynomial for

f(x) = 1 + 2x − x 2 + x 3

(b) Find the third Taylor polynomial about x = 1 for

f(x) = 1 + 2(x − 1) − (x − 1)2 + (x − 1)3

26. (a) Find the nth Maclaurin polynomial for

f(x) = c0 + c1 x + c2 x 2 + · · · + cn x n

(b) Find the nth Taylor polynomial about x = 1 for

f(x) = c0 + c1 (x − 1) + c2 (x − 1)2 + · · · + cn (x − 1)n

27–30 Find the ﬁrst four distinct Taylor polynomials about x = x0 , and use a graphing utility to graph the given function and the Taylor polynomials on the same screen. ■ 27. f(x) = e−2x ; x0 = 0 28. f(x) = sin x; x0 = π/2

29. f(x) = cos x; x0 = π

30. ln(x + 1); x0 = 0

31–34 True–False Determine whether the statement is true or false. Explain your answer. ■

31. The equation of a tangent line to a differentiable function is a ﬁrst-degree Taylor polynomial for that function. 32. The graph of a function f and the graph of its Maclaurin polynomial have a common y-intercept. 33. If p6 (x) is the sixth-degree Taylor polynomial for a function f about x = x0 , then p6(4) (x0 ) = 4!f (4) (x0 ).

34. If p4 (x) is the fourth-degree Maclaurin polynomial for ex , then 9 |e2 − p4 (2)| ≤ 5!

y x

III

38. Suppose that the values of a function f and its ﬁrst three derivatives at x = 1 are f ′(1) = −3,

f(1) = 2,

f ′′(1) = 0,

f ′′′(1) = 6

Find as many Taylor polynomials for f as you can about x = 1.

39. Let p1 (x) and p2 (x) be the local linear and local quadratic approximations of f(x) = esin x at x = 0. (a) Use a graphing utility to generate the graphs of f(x), p1 (x), and p2 (x) on the same screen for −1 ≤ x ≤ 1. (b) Construct a table of values of f(x), p1 (x), and p2 (x) for x = −1.00, −0.75, −0.50, −0.25, 0, 0.25, 0.50, 0.75, 1.00. Round the values to three decimal places. (c) Generate the graph of |f(x) − p1 (x)|, and use the graph to determine an interval on which p1 (x) approximates f(x) with an error of at most ±0.01. [Suggestion: Review the discussion relating to Figure 3.5.4.] (d) Generate the graph of |f(x) − p2 (x)|, and use the graph to determine an interval on which p2 (x) approximates f(x) with an error of at most ±0.01.

40. (a) The accompanying ﬁgure shows a sector of radius r and central angle 2α. Assuming that the angle α is small, use the local quadratic approximation of cos α at α = 0 to show that x ≈ rα 2 /2. (b) Assuming that the Earth is a sphere of radius 4000 mi, use the result in part (a) to approximate the maximum amount by which a 100 mi arc along the equator will diverge from its chord. x

35–36 Use the method of Example 7 to approximate the given expression to the speciﬁed accuracy. Check your answer to that produced directly by your calculating utility. ■ √ 35. e; four decimal-place accuracy

r Figure Ex-40

9.8 Maclaurin and Taylor Series; Power Series

41. (a) Find an interval [0, b] over which ex can be approximated by 1 + x + (x 2 /2!) to three decimal-place accuracy throughout the interval. (b) Check your answer in part (a) by graphing

2 x e − 1 + x + x 2! over the interval you obtained.

42. Show that the nth Taylor polynomial for sinh x about x = ln 4 is n 16 − (−1)k (x − ln 4)k 8k! k=0

659

43–46 Use the Remainder Estimation Theorem to ﬁnd an interval containing x = 0 over which f(x) can be approximated by p(x) to three decimal-place accuracy throughout the interval. Check your answer by graphing |f(x) − p(x)| over the interval you obtained. ■ x3 43. f(x) = sin x; p(x) = x − 3! x2 x4 44. f(x) = cos x; p(x) = 1 − + 2! 4! 1 45. f(x) = ; p(x) = 1 − x 2 + x 4 1 + x2 x2 x3 46. f(x) = ln(1 + x); p(x) = x − + 2 3

✔QUICK CHECK ANSWERS 9.7 f ′′(0) 2 f ′′′(0) 3 x + x 2! 3! 5. (a) f(x) − pn (x) (b) 61 |x − 2|5

1. f(0) + f ′(0)x +

9.8

2. 1; 2; 2; 43

3. 3 − 4(x − 2) + 5(x − 2)2

4. −1; 5; −10; 10

MACLAURIN AND TAYLOR SERIES; POWER SERIES Recall from the last section that the nth Taylor polynomial pn (x) at x = x0 for a function f was deﬁned so its value and the values of its ﬁrst n derivatives match those of f at x0 . This being the case, it is reasonable to expect that for values of x near x0 the values of pn (x) will become better and better approximations of f(x) as n increases, and may possibly converge to f(x) as n → +⬁. We will explore this idea in this section.

MACLAURIN AND TAYLOR SERIES

In Section 9.7 we deﬁned the nth Maclaurin polynomial for a function f as n f (k) (0) k=0

x k = f(0) + f ′(0)x +

f ′′(0) 2 f (n) (0) n x + ··· + x 2! n!

and the nth Taylor polynomial for f about x = x0 as n f (k) (x0 ) k=0

(x − x0 )k = f(x0 ) + f ′(x0 )(x − x0 ) +

f ′′(x0 ) f (n) (x0 ) (x − x0 )2 + · · · + (x − x0 )n 2! n!

It is not a big step to extend the notions of Maclaurin and Taylor polynomials to series by not stopping the summation index at n. Thus, we have the following deﬁnition.

660

Chapter 9 / Infinite Series

9.8.1 definition If f has derivatives of all orders at x0 , then we call the series ⬁ f (k) (x0 )

k=0

f ′′(x0 ) (x − x0 )2 2! f (k) (x0 ) +··· + (x − x0 )k + · · · k!

(x − x0 )k = f(x0 ) + f ′(x0 )(x − x0 ) +

(1)

the Taylor series for f about x = x0 . In the special case where x0 = 0, this series becomes ⬁ f (k) (0) k=0

x k = f(0) + f ′(0)x +

f ′′(0) 2 f (k) (0) k x + ··· + x + ··· 2! k!

(2)

in which case we call it the Maclaurin series for f .

Note that the nth Maclaurin and Taylor polynomials are the nth partial sums for the corresponding Maclaurin and Taylor series. Example 1

Find the Maclaurin series for (a) ex

(b) sin x

(d)

1 1−x

Solution (a). In Example 2 of Section 9.7 we found that the nth Maclaurin polynomial for ex is pn (x) =

n xk k=0

=1+x+

x2 xn + ··· + 2! n!

Thus, the Maclaurin series for ex is ⬁ x2 xk xk =1+x+ + ··· + + ··· k! 2! k! k=0

Solution (b). In Example 3(a) of Section 9.7 we found that the Maclaurin polynomials for sin x are given by x3 x5 x7 x 2k+1 + − + · · · + (−1)k (k = 0, 1, 2, . . .) 3! 5! 7! (2k + 1)! Thus, the Maclaurin series for sin x is ⬁ x 2k+1 x3 x5 x7 x 2k+1 (−1)k =x− + − + · · · + (−1)k + ··· (2k + 1)! 3! 5! 7! (2k + 1)! k=0 p2k+1 (x) = p2k+2 (x) = x −

Solution (c). In Example 3(b) of Section 9.7 we found that the Maclaurin polynomials for cos x are given by x2 x4 x6 x 2k + − + · · · + (−1)k (k = 0, 1, 2, . . .) 2! 4! 6! (2k)! Thus, the Maclaurin series for cos x is ⬁ x2 x4 x6 x 2k x 2k =1− + − + · · · + (−1)k + ··· (−1)k (2k)! 2! 4! 6! (2k)! k=0 p2k (x) = p2k+1 (x) = 1 −

9.8 Maclaurin and Taylor Series; Power Series

661

Solution (d). In Example 5 of Section 9.7 we found that the nth Maclaurin polynomial for 1/(1 − x) is pn (x) =

n k=0

xk = 1 + x + x2 + · · · + xn

(n = 0, 1, 2, . . .)

Thus, the Maclaurin series for 1/(1 − x) is ⬁ k=0

Example 2

xk = 1 + x + x2 + · · · + xk + · · ·

Find the Taylor series for 1/x about x = 1.

Solution. In Example 6 of Section 9.7 we found that the nth Taylor polynomial for 1/x about x = 1 is n k=0

(−1)k (x − 1)k = 1 − (x − 1) + (x − 1)2 − (x − 1)3 + · · · + (−1)n (x − 1)n

Thus, the Taylor series for 1/x about x = 1 is ⬁ k=0

(−1)k (x − 1)k = 1 − (x − 1) + (x − 1)2 − (x − 1)3 + · · · + (−1)k (x − 1)k + · · ·

POWER SERIES IN x

Maclaurin and Taylor series differ from the series that we have considered in Sections 9.3 to 9.6 in that their terms are not merely constants, but instead involve a variable. These are examples of power series, which we now deﬁne. If c0 , c1 , c2 , . . . are constants and x is a variable, then a series of the form ⬁ k=0

ck x k = c0 + c1 x + c2 x 2 + · · · + ck x k + · · ·

(3)

is called a power series in x. Some examples are ⬁ k=0

xk = 1 + x + x2 + x3 + · · ·

⬁ xk

k=0 ⬁

=1+x+

(−1)k

k=0

x2 x3 + + ··· 2! 3!

x 2k x2 x4 x6 =1− + − + ··· (2k)! 2! 4! 6!

From Example 1, these are the Maclaurin series for the functions 1/(1 − x), ex , and cos x, respectively. Indeed, every Maclaurin series ⬁ f (k) (0) k=0

is a power series in x.

x k = f(0) + f ′(0)x +

f ′′(0) 2 f (k) (0) k x + ··· + x + ··· 2! k!

662

Chapter 9 / Infinite Series

RADIUS AND INTERVAL OF CONVERGENCE

If a numerical value is substituted for x in a power series ck x k , then the resulting series of numbers may either converge or diverge. This leads to the problem of determining the set of x-values for which a given power series converges; this is called its convergence set. Observe that every power series in x converges at x = 0, since substituting this value in (3) produces the series c0 + 0 + 0 + 0 + · · · + 0 + · · · whose sum is c0 . In some cases x = 0 may be the only number in the convergence set; in other cases the convergence set is some ﬁnite or inﬁnite interval containing x = 0. This is the content of the following theorem, whose proof will be omitted.

9.8.2 theorem For any power series in x, exactly one of the following is true: (a) The series converges only for x = 0. (b) The series converges absolutely (and hence converges) for all real values of x. (c) The series converges absolutely (and hence converges) for all x in some ﬁnite open interval (−R, R) and diverges if x < −R or x > R. At either of the values x = R or x = −R, the series may converge absolutely, converge conditionally, or diverge, depending on the particular series.

This theorem states that the convergence set for a power series in x is always an interval centered at x = 0 (possibly just the value x = 0 itself or possibly inﬁnite). For this reason, the convergence set of a power series in x is called the interval of convergence. In the case where the convergence set is the single value x = 0 we say that the series has radius of convergence 0, in the case where the convergence set is (−⬁, +⬁) we say that the series has radius of convergence +ⴥ, and in the case where the convergence set extends between −R and R we say that the series has radius of convergence R (Figure 9.8.1). Diverges

Diverges Radius of convergence R = 0

0 Converges

Radius of convergence R = + ∞

0 Converges

Diverges

Figure 9.8.1

−R

Diverges

Radius of convergence R

FINDING THE INTERVAL OF CONVERGENCE The usual procedure for ﬁnding the interval of convergence of a power series is to apply the ratio test for absolute convergence (Theorem 9.6.5). The following example illustrates how this works.

Example 3 Find the interval of convergence and radius of convergence of the following power series. (a)

⬁ k=0

(b)

⬁ xk k=0

(c)

⬁ k=0

k!x k

(d)

⬁ (−1)k x k 3k (k + 1) k=0

9.8 Maclaurin and Taylor Series; Power Series

663

Solution (a). Applying the ratio test for absolute convergence to the given series, we obtain

uk+1 = lim Ď = lim k â&#x2020;&#x2019; +âŹ uk k â&#x2020;&#x2019; +âŹ

k+1 x lim |x| = |x| xk = k â&#x2020;&#x2019; +âŹ

so the series converges absolutely if Ď = |x| < 1 and diverges if Ď = |x| > 1. The test is inconclusive if |x| = 1 (i.e., if x = 1 or x = â&#x2C6;&#x2019;1), which means that we will have to investigate convergence at these values separately. At these values the series becomes âŹ

1k = 1 + 1 + 1 + 1 + Âˇ Âˇ Âˇ

x=1

(â&#x2C6;&#x2019;1)k = 1 â&#x2C6;&#x2019; 1 + 1 â&#x2C6;&#x2019; 1 + Âˇ Âˇ Âˇ

x = â&#x2C6;&#x2019;1

k=0 âŹ k=0

both of which diverge; thus, the interval of convergence for the given power series is (â&#x2C6;&#x2019;1, 1), and the radius of convergence is R = 1.

Solution (b). Applying the ratio test for absolute convergence to the given series, we obtain

k+1 x x uk+1 k! =0 Ď = lim = lim = lim Âˇ k â&#x2020;&#x2019; +âŹ uk k â&#x2020;&#x2019; +âŹ (k + 1)! x k k â&#x2020;&#x2019; +âŹ k + 1

Since Ď < 1 for all x, the series converges absolutely for all x. Thus, the interval of convergence is (â&#x2C6;&#x2019;âŹ , +âŹ ) and the radius of convergence is R = +âŹ .

Solution (c). If x = 0, then the ratio test for absolute convergence yields

k+1 uk+1 = lim (k + 1)!x = lim |(k + 1)x| = +âŹ Ď = lim k â&#x2020;&#x2019; +âŹ k â&#x2020;&#x2019; +âŹ uk k â&#x2020;&#x2019; +âŹ k!x k

Therefore, the series diverges for all nonzero values of x. Thus, the interval of convergence is the single value x = 0 and the radius of convergence is R = 0.

Solution (d). Since |(â&#x2C6;&#x2019;1)k | = |(â&#x2C6;&#x2019;1)k+1 | = 1, we obtain

uk+1 x k+1 3k (k + 1) Ď = lim = lim Âˇ k â&#x2020;&#x2019; +âŹ uk k â&#x2020;&#x2019; +âŹ 3k+1 (k + 2) xk

|x| k+1 = lim Âˇ k â&#x2020;&#x2019; +âŹ 3 k+2

|x| 1 + (1/k) |x| lim = = 3 k â&#x2020;&#x2019; +âŹ 1 + (2/k) 3

The ratio test for absolute convergence implies that the series converges absolutely if |x| < 3 and diverges if |x| > 3. The ratio test fails to provide any information when |x| = 3, so the cases x = â&#x2C6;&#x2019;3 and x = 3 need separate analyses. Substituting x = â&#x2C6;&#x2019;3 in the given series yields âŹ âŹ âŹ (â&#x2C6;&#x2019;1)k (â&#x2C6;&#x2019;3)k (â&#x2C6;&#x2019;1)k (â&#x2C6;&#x2019;1)k 3k 1 = = k (k + 1) k (k + 1) 3 3 k + 1 k=0 k=0 k=0 which is the divergent harmonic series 1 + given series yields

1 2

1 3

1 4

+ Âˇ Âˇ Âˇ. Substituting x = 3 in the

âŹ âŹ 1 1 1 (â&#x2C6;&#x2019;1)k (â&#x2C6;&#x2019;1)k 3k = = 1 â&#x2C6;&#x2019; + â&#x2C6;&#x2019; + ÂˇÂˇÂˇ k 3 (k + 1) k+1 2 3 4 k=0 k=0

which is the conditionally convergent alternating harmonic series. Thus, the interval of convergence for the given series is (â&#x2C6;&#x2019;3, 3] and the radius of convergence is R = 3.

664

Chapter 9 / Infinite Series

POWER SERIES IN x – x 0

If x0 is a constant, and if x is replaced by x − x0 in (3), then the resulting series has the form ⬁ k=0

ck (x − x0 )k = c0 + c1 (x − x0 ) + c2 (x − x0 )2 + · · · + ck (x − x0 )k + · · ·

This is called a power series in x − x0 . Some examples are ⬁ (x − 1)k k=0

k+1

=1+

⬁ (−1)k (x + 3)k

k=0

(x − 1) (x − 1)2 (x − 1)3 + + + ··· 2 3 4

= 1 − (x + 3) +

(x + 3)2 (x + 3)3 − + ··· 2! 3!

x0 = 1

x0 = −3

The ﬁrst of these is a power series in x − 1 and the second is a power series in x + 3. Note that a power series in x is a power series in x − x0 in which x0 = 0. More generally, the Taylor series ⬁ f (k) (x0 ) (x − x0 )k k! k=0 is a power series in x − x0 . The main result on convergence of a power series in x − x0 can be obtained by substituting x − x0 for x in Theorem 9.8.2. This leads to the following theorem. 9.8.3 theorem For a power series statements is true:

ck (x − x0 )k , exactly one of the following

(a) The series converges only for x = x0 . (b) The series converges absolutely (and hence converges) for all real values of x. (c) The series converges absolutely (and hence converges) for all x in some ﬁnite open interval (x0 − R, x0 + R) and diverges if x < x0 − R or x > x0 + R. At either of the values x = x0 − R or x = x0 + R, the series may converge absolutely, converge conditionally, or diverge, depending on the particular series.

It follows from this theorem that the set of values for which a power series in x − x0 converges is always an interval centered at x = x0 ; we call this the interval of convergence (Figure 9.8.2). In part (a) of Theorem 9.8.3 the interval of convergence reduces to the single value x = x0 , in which case we say that the series has radius of convergence R = 0; in part Diverges

Diverges Radius of convergence R = 0

x0 Converges

Radius of convergence R = + ∞

x0 Diverges

Figure 9.8.2

x0 − R

Converges

Diverges

x0 + R

Radius of convergence R

9.8 Maclaurin and Taylor Series; Power Series

665

(b) the interval of convergence is inďŹ nite (the entire real line), in which case we say that the series has radius of convergence R = +â´Ľ; and in part (c) the interval extends between x0 â&#x2C6;&#x2019; R and x0 + R, in which case we say that the series has radius of convergence R. Example 4

Find the interval of convergence and radius of convergence of the series âŹ (x â&#x2C6;&#x2019; 5)k

k=1

Solution. We apply the ratio test for absolute convergence. k+1 uk+1 k 2 = lim (x â&#x2C6;&#x2019; 5) Âˇ Ď = lim k â&#x2020;&#x2019; +âŹ uk k â&#x2020;&#x2019; +âŹ (k + 1)2 (x â&#x2C6;&#x2019; 5)k 2

k = lim |x â&#x2C6;&#x2019; 5| k â&#x2020;&#x2019; +âŹ k+1 2

1 = |x â&#x2C6;&#x2019; 5| lim = |x â&#x2C6;&#x2019; 5| k â&#x2020;&#x2019; +âŹ 1 + (1/k) Thus, the series converges absolutely if |x â&#x2C6;&#x2019; 5| < 1, or â&#x2C6;&#x2019;1 < x â&#x2C6;&#x2019; 5 < 1, or 4 < x < 6. The series diverges if x < 4 or x > 6. To determine the convergence behavior at the endpoints x = 4 and x = 6, we substitute these values in the given series. If x = 6, the series becomes âŹ 1k k=1

âŹ 1 1 1 1 = 1 + 2 + 2 + 2 + ÂˇÂˇÂˇ 2 k 2 3 4 k=1

which is a convergent p-series (p = 2). If x = 4, the series becomes It will always be a waste of time to test for convergence at the endpoints of the interval of convergence using the ratio test, since Ď will always be 1 at those points if

uk+1 lim k â&#x2020;&#x2019; +âŹ uk

âŹ (â&#x2C6;&#x2019;1)k k=1

= â&#x2C6;&#x2019;1 +

1 1 1 â&#x2C6;&#x2019; 2 + 2 â&#x2C6;&#x2019; ÂˇÂˇÂˇ 22 3 4

Since this series converges absolutely, the interval of convergence for the given series is [4, 6]. The radius of convergence is R = 1 (Figure 9.8.3).

exists. Explain why this must be so.

Series diverges

Series converges absolutely

4 Figure 9.8.3

R=1

x0 = 5

R=1

Series diverges

FUNCTIONS DEFINED BY POWER SERIES

If a function f is expressed as a power series on some interval, then we say that the power series represents f on that interval. For example, we saw in Example 4(a) of Section 9.3 that âŹ 1 xk = 1â&#x2C6;&#x2019;x k=0 if |x| < 1, so this power series represents the function 1/(1 â&#x2C6;&#x2019; x) on the interval â&#x2C6;&#x2019;1 < x < 1.

666

Chapter 9 / Infinite Series 1 0.8 0.6 0.4 0.2

-15 -10 -5 -0.2 -0.4

Sometimes new functions actually originate as power series, and the properties of the functions are developed by working with their power series representations. For example, the functions ⬁ (−1)k x 2k x2 x4 x6 J0 (x) = = 1 − + − + ··· (4) 22k (k!)2 22 (1!)2 24 (2!)2 26 (3!)2 k=0 and

y = J0 (x)

J1 (x) =

0.6 0.4 0.2 -15 -10 -5 -0.2 -0.4 -0.6

y = J1(x) Generated by Mathematica Figure 9.8.4 T E C H N O LO GY M A ST E R Y Many computer algebra systems have the Bessel functions as part of their libraries. If you have a CAS with Bessel functions, use it to generate the graphs in Figure 9.8.4.

⬁ k=0

(5)

which are called Bessel functions in honor of the German mathematician and astronomer Friedrich Wilhelm Bessel (1784–1846), arise naturally in the study of planetary motion and in various problems that involve heat flow. To find the domains of these functions, we must determine where their defining power series converge. For example, in the case of J0 (x) we have uk+1 x 2(k+1) 22k (k!)2 = lim · ρ = lim k → +⬁ uk k → +⬁ 22(k+1) [(k + 1)!]2 x 2k x2 =0<1 = lim k → +⬁ 4(k + 1)2

so the series converges for all x; that is, the domain of J0 (x) is (−⬁, +⬁). We leave it as an exercise (Exercise 61) to show that the power series for J1 (x) also converges for all x. Computer-generated graphs of J0 (x) and J1 (x) are shown in Figure 9.8.4.

✔QUICK CHECK EXERCISES 9.8

(See page 668 for answers.)

1. If f has derivatives of all orders at x0 , then the Taylor series for f about x = x0 is deﬁned to be ⬁ k=0

2. Since

(−1)k x 2k+1 x x3 x5 = − 3 + 5 − ··· + 1)! 2 2 (1!)(2!) 2 (2!)(3!)

22k+1 (k!)(k

k+1 k+1 2 x = 2|x| lim k → +⬁ 2k x k

the radius of convergence for the inﬁnite series is . 3. Since

(b) When x = 3,

⬁ ⬁ 1 1 √ (x − 4)k = √ (−1)k k k k=1 k=1

Does this series converge or diverge? ⬁

k=0

2k x k

k+1 k+1 3x (3 x )/(k + 1)! =0 lim lim = k→ +⬁ k + 1 k → +⬁ (3k x k )/k! the interval of convergence for the series ⬁k=0 (3k /k!)x k is .

4. (a) Since

(x − 4)k+1 /√k + 1 k lim = lim (x − 4) √ k → +⬁ (x − 4)k / k k → +⬁ k + 1 = |x − 4|

the for the inﬁnite series √ of convergence ⬁ radius k / (1 k)(x − 4) is . k=1

Does this series converge or diverge? (d) The for the inﬁnite series √ of convergence ⬁ interval k / k)(x − 4) is . (1 k=1

9.8 Maclaurin and Taylor Series; Power Series

EXERCISE SET 9.8

Graphing Utility

CAS

1–10 Use sigma notation to write the Maclaurin series for the

29.

function. ■ 2. eax

1. e−x

3. cos πx

1 1+x 9. x sin x

5. ln(1 + x) 8. sinh x

4. sin πx

32.

7. cosh x

10. xe

35.

11–18 Use sigma notation to write the Taylor series about

x = x0 for the function. ■

11. ex ; x0 = 1 1 13. ; x0 = −1 x

15. sin πx; x0 =

1 2

17. ln x; x0 = 1

37. 12. e−x ; x0 = ln 2 1 14. ; x0 = 3 x+2 π 16. cos x; x0 = 2 18. ln x; x0 = e

19–22 Find the interval of convergence of the power series, and

ﬁnd a familiar function that is represented by the power series on that interval. ■ 19. 1 − x + x 2 − x 3 + · · · + (−1)k x k + · · ·

20. 1 + x 2 + x 4 + · · · + x 2k + · · · 2

x x2 x3 xk + − + · · · + (−1)k k + · · · 2 4 8 2 (b) Find f(0) and f(1).

24. Suppose that the function f is represented by the power series x − 5 (x − 5)2 (x − 5)3 − + ··· + 2 3 3 33 (a) Find the domain of f . (b) Find f(3) and f(6). f(x) = 1 −

25–28 True–False Determine whether the statement is true or

false. Explain your answer. ■ 25. If a power series in x converges conditionally at x = 3, then the series converges if |x| < 3 and diverges if |x| > 3.

26. The ratio test is often useful to determine convergence at the endpoints of the interval of convergence of a power series. 27. The Maclaurin series for a polynomial function has radius of convergence +⬁. ⬁ xk 28. The series converges if |x| < 1. k! k=0

29–50 Find the radius of convergence and the interval of convergence. ■

30.

⬁ k! k x k 2 k=0 ⬁

k=1 ⬁

⬁

k=0 ⬁ k=1

x (−1)k−1 √

38.

47.

⬁ k=1

49.

(x + 1)k k

k=0 ⬁

(x + 1) k2 + 4

(−1)k x 2k (2k)!

xk k(ln k)2

⬁ x 2k+1 (−1)k 42. (2k + 1)! k=0

44.

46.

⬁ (x − 3)k

k=0 ⬁

48.

(−1)k

k=0

(x − 4)k (k + 1)2

⬁ (2k + 1)! k=1

50.

(2k + 1)!

xk ln k

k+1

⬁ (−1)k+1 40.

2k+1

⬁ πk (x − 1)2k k=0

k=2

⬁ (−2)k x k+1

k=2

(−1)k+1 (−1)k

k=0 ⬁

34.

k=0

3k k x k!

⬁ (−1)k x k

31.

5k k x k2 36.

xk 41. 1 + k2 k=0 ⬁ k 3 (x + 5)k 43. 4 k=0 ⬁

3k x k

k=1 ⬁ k=0

33.

xk k(k + 1)

k=1

23. Suppose that the function f is represented by the power series

(a) Find the domain of f .

⬁ xk k+1 k=0

⬁

22. 1 − (x + 3) + (x + 3) − (x + 3) + · · · + (−1)k (x + 3)k | . . .

f(x) = 1 −

39.

45.

21. 1 + (x − 2) + (x − 2) + · · · + (x − 2) + · · · 2

667

(x − 2)k

⬁ (2x − 3)k k=0

42k

51. Use the root test to ﬁnd the interval of convergence of ⬁ xk (ln k)k k=2

52. Find the domain of the function ⬁ 1 · 3 · 5 · · · (2k − 1) k x f(x) = (2k − 2)! k=1

53. Show that the series

x x2 x3 + − + ··· 2! 4! 6! is the Maclaurin series for the function √ x≥0 cos x, f(x) = √ cosh −x, x < 0 [Hint: Use the Maclaurin series for cos x and cosh √ x to ob√ tain series for cos x, where x ≥ 0, and cosh −x, where x ≤ 0.] 1−

F O C U S O N C O N C E P TS

54. If a function f is represented by a power series on an interval, then the graphs of the partial sums can be used as approximations to the graph of f . (a) Use a graphing utility to generate the graph of 1/(1 − x) together with the graphs of the ﬁrst four partial sums of its Maclaurin series over the interval (−1, 1). (cont.)

Chapter 9 / Infinite Series

668

(b) In general terms, where are the graphs of the partial sums the most accurate?

61. Show that the power series representation of the Bessel function J1 (x) converges for all x [Formula (5)].

55. Prove: (a) If f is an even function, then all odd powers of x in its Maclaurin series have coefficient 0. (b) If f is an odd function, then all even powers of x in its Maclaurin series have coefficient 0. 56. Suppose that the power series ck (x − x0 )k has radius of convergence R and p is a nonzero constant. What can you say about the radius of convergence of the power series pck (x − x0 )k ? Explain your reasoning. [Hint: See Theorem 9.4.3.] 57. Suppose that the power series ck (x − x0 )k has a finite radius of convergence R, and the power series dk (x − x0 )k has a radius of convergence of +⬁. What the radius of convergence of can you say about (ck + dk )(x − x0 )k ? Explain your reasoning. 58. Suppose that the power series ck (x − x0 )k has a finite radius of convergence R and the power series 1 dk (x − x0 )k has a finite radius of convergence R2 . What the radius of convergence of can you say about (ck + dk )(x − x0 )k ? Explain your reasoning. [Hint: The case R1 = R2 requires special attention.]

62. Approximate the values of the Bessel functions J0 (x) and J1 (x) at x = 1, each to four decimal-place accuracy.

59. Show that if p is a positive integer, then the power series ⬁ (pk)! k x (k!)p k=0 has a radius of convergence of 1/p p . 60. Show that if p and q are positive integers, then the power series ⬁ (k + p)! k x k!(k + q)! k=0

63. If the constant p in the general p-series is replaced by a variable x for x > 1, then the resulting function is called the Riemann zeta function and is denoted by ζ (x) =

⬁ 1 x k k=1

(a) Let sn be the nth partial sum of the series for ζ (3.7). Find n such that sn approximates ζ (3.7) to two decimalplace accuracy, and calculate sn using this value of n. [Hint: Use the right inequality in Exercise 36(b) of Section 9.4 with f(x) = 1/x 3.7 .] (b) Determine whether your CAS can evaluate the Riemann zeta function directly. If so, compare the value produced by the CAS to the value of sn obtained in part (a). 64. Prove: If limk → +⬁ |ck |1/k = L, where L = 0, then 1/L is the radius of convergence of the power series ⬁k=0 ck x k . 65. Prove: If the power series ⬁k=0 ck x k has radius of convergence √ R, then the series ⬁k=0 ck x 2k has radius of convergence R. 66. Prove: If the interval of convergence of the series ⬁ k k=0 ck (x − x0 ) is (x0 − R, x0 + R], then the series converges conditionally at x0 + R.

67. Writing The sine function can be deﬁned geometrically from the unit circle or analytically from its Maclaurin series. Discuss the advantages of each representation with regard to providing information about the sine function.

has a radius of convergence of +⬁.

✔QUICK CHECK ANSWERS 9.8 1.

f (k) (x0 ) (x − x0 )k k!

9.9

1 2

3. (−⬁, +⬁)

4. (a) 1 (b) converges (c) diverges (d) [3, 5)

CONVERGENCE OF TAYLOR SERIES In this section we will investigate when a Taylor series for a function converges to that function on some interval, and we will consider how Taylor series can be used to approximate values of trigonometric, exponential, and logarithmic functions. THE CONVERGENCE PROBLEM FOR TAYLOR SERIES

Recall that the nth Taylor polynomial for a function f about x = x0 has the property that its value and the values of its ﬁrst n derivatives match those of f at x0 . As n increases,

9.9 Convergence of Taylor Series

669

more and more derivatives match up, so it is reasonable to hope that for values of x near x0 the values of the Taylor polynomials might converge to the value of f(x); that is, f(x) = lim

n â&#x2020;&#x2019; +âŹ

n f (k) (x0 )

k=0

(x â&#x2C6;&#x2019; x0 )k

(1)

However, the nth Taylor polynomial for f is the nth partial sum of the Taylor series for f , so (1) is equivalent to stating that the Taylor series for f converges at x, and its sum is f(x). Thus, we are led to consider the following problem.

Problem 9.9.1 is concerned not only with whether the Taylor series of a function f converges, but also whether it converges to the function f itself. Indeed, it is possible for a Taylor series of a function f to converge to values different from f(x) for certain values of x (Exercise 14).

9.9.1 problem Given a function f that has derivatives of all orders at x = x0 , determine whether there is an open interval containing x0 such that f(x) is the sum of its Taylor series about x = x0 at each point in the interval; that is, f(x) = for all values of x in the interval.

âŹ f (k) (x0 ) k=0

(x â&#x2C6;&#x2019; x0 )k

(2)

One way to show that (1) holds is to show that n f (k) (x0 ) k lim f(x) â&#x2C6;&#x2019; (x â&#x2C6;&#x2019; x0 ) = 0 n â&#x2020;&#x2019; +âŹ k! k=0

However, the difference appearing on the left side of this equation is the nth remainder for the Taylor series [Formula (12) of Section 9.7]. Thus, we have the following result.

9.9.2

theorem The equality f(x) =

âŹ f (k) (x0 ) k=0

(x â&#x2C6;&#x2019; x0 )k

holds at a point x if and only if lim Rn (x) = 0. n â&#x2020;&#x2019; +âŹ

ESTIMATING THE nTH REMAINDER

It is relatively rare that one can prove directly that Rn (x) â&#x2020;&#x2019; 0 as n â&#x2020;&#x2019; +âŹ . Usually, this is proved indirectly by ďŹ nding appropriate bounds on |Rn (x)| and applying the Squeezing Theorem for Sequences. The Remainder Estimation Theorem (Theorem 9.7.4) provides a useful bound for this purpose. Recall that this theorem asserts that if M is an upper bound for |f (n+1) (x)| on an interval containing x0 , then |Rn (x)| â&#x2030;¤

M |x â&#x2C6;&#x2019; x0 |n+1 (n + 1)!

(3)

for all x in that interval. The following example illustrates how the Remainder Estimation Theorem is applied. Example 1 is, cos x =

Show that the Maclaurin series for cos x converges to cos x for all x; that âŹ k=0

(â&#x2C6;&#x2019;1)k

x2 x4 x6 x 2k =1â&#x2C6;&#x2019; + â&#x2C6;&#x2019; + ÂˇÂˇÂˇ (2k)! 2! 4! 6!

(â&#x2C6;&#x2019;âŹ < x < +âŹ )

670

Chapter 9 / Infinite Series

Solution. From Theorem 9.9.2 we must show that Rn (x) → 0 for all x as n → +⬁. For this purpose let f(x) = cos x, so that for all x we have f (n+1) (x) = ± cos x

or f (n+1) (x) = ± sin x

In all cases we have |f (n+1) (x)| ≤ 1, so we can apply (3) with M = 1 and x0 = 0 to conclude that |x|n+1 0 ≤ |Rn (x)| ≤ (4) (n + 1)!

The method of Example 1 can be easily modified to prove that the Taylor series for sin x and cos x about any point x = x0 converge to sin x and cos x , respectively, for all x (Exercises 21 and 22). For reference, some of the most important Maclaurin series are listed in Table 9.9.1 at the end of this section.

However, it follows from Formula (5) of Section 9.2 with n + 1 in place of n and |x| in place of x that |x|n+1 =0 (5) lim n → +⬁ (n + 1)! Using this result and the Squeezing Theorem for Sequences (Theorem 9.1.5), it follows from (4) that |Rn (x)| → 0 and hence that Rn (x) → 0 as n → +⬁ (Theorem 9.1.6). Since this is true for all x, we have proved that the Maclaurin series for cos x converges to cos x for all x. This is illustrated in Figure 9.9.1, where we can see how successive partial sums approximate the cosine curve more and more closely. y

p12

p16

2 1

y = cos x 1

p2n =

k =0

2k (−1) k x (2k)!

−1 −2

Figure 9.9.1

p10

p14

p18

APPROXIMATING TRIGONOMETRIC FUNCTIONS

In general, to approximate the value of a function f at a point x using a Taylor series, there are two basic questions that must be answered:

• About what point x0 should the Taylor series be expanded? • How many terms in the series should be used to achieve the desired accuracy? In response to the first question, x0 needs to be a point at which the derivatives of f can be evaluated easily, since these values are needed for the coefficients in the Taylor series. Furthermore, if the function f is being evaluated at x, then x0 should be chosen as close as possible to x, since Taylor series tend to converge more rapidly near x0 . For example, to approximate sin 3 ◦ (= π/60 radians), it would be reasonable to take x0 = 0, since π/60 is close to 0 and the derivatives of sin x are easy to evaluate at 0. On the other hand, to approximate sin 85 ◦ (= 17π/36 radians), it would be more natural to take x0 = π/2, since 17π/36 is close to π/2 and the derivatives of sin x are easy to evaluate at π/2. In response to the second question posed above, the number of terms required to achieve a specific accuracy needs to be determined on a problem-by-problem basis. The next example gives two methods for doing this. Example 2 Use the Maclaurin series for sin x to approximate sin 3 ◦ to five decimalplace accuracy.

9.9 Convergence of Taylor Series

671

Solution. In the Maclaurin series sin x =

⬁ k=0

(−1)k

x3 x5 x7 x 2k+1 =x− + − + ··· (2k + 1)! 3! 5! 7!

(6)

the angle x is assumed to be in radians (because the differentiation formulas for the trigonometric functions were derived with this assumption). Since 3 ◦ = π/60 radians, it follows from (6) that π (π/60)3 (π/60)5 (π/60)7 π − = + − + ··· (7) sin 3 ◦ = sin 60 60 3! 5! 7! We must now determine how many terms in the series are required to achieve five decimalplace accuracy. We will consider two possible approaches, one using the Remainder Estimation Theorem (Theorem 9.7.4) and the other using the fact that (7) satisfies the hypotheses of the alternating series test (Theorem 9.6.1). Method 1. (The Remainder Estimation Theorem) Since we want to achieve five decimal-place accuracy, our goal is to choose n so that the absolute value of the nth remainder at x = π/60 does not exceed 0.000005 = 5 × 10−6 ; π that is, (8) Rn ≤ 0.000005 60

However, if we let f(x) = sin x, then f (n+1) (x) is either ± sin x or ± cos x, and in either case |f (n+1) (x)| ≤ 1 for all x. Thus, it follows from the Remainder Estimation Theorem with M = 1, x0 = 0, and x = π/60 that π (π/60)n+1 ≤ Rn 60 (n + 1)!

Thus, we can satisfy (8) by choosing n so that

(π/60)n+1 ≤ 0.000005 (n + 1)!

With the help of a calculating utility you can verify that the smallest value of n that meets this criterion is n = 3. Thus, to achieve five decimal-place accuracy we need only keep terms up to the third power in (7). This yields π (π/60)3 sin 3 ◦ ≈ − ≈ 0.05234 (9) 60 3! (verify). As a check, a calculator gives sin 3 ◦ ≈ 0.05233595624, which agrees with (9) when rounded to five decimal places. Method 2. (The Alternating Series Test) We leave it for you to check that (7) satisfies the hypotheses of the alternating series test (Theorem 9.6.1). Let sn denote the sum of the terms in (7) up to and including the nth power of π/60. Since the exponents in the series are odd integers, the integer n must be odd, and the exponent of the first term not included in the sum sn must be n + 2. Thus, it follows from part (b) of Theorem 9.6.2 that (π/60)n+2 |sin 3 ◦ − sn | < (n + 2)!

This means that for ﬁve decimal-place accuracy we must look for the ﬁrst positive odd integer n such that (π/60)n+2 ≤ 0.000005 (n + 2)!

672

Chapter 9 / Infinite Series

With the help of a calculating utility you can verify that the smallest value of n that meets this criterion is n = 3. This agrees with the result obtained above using the Remainder Estimation Theorem and hence leads to approximation (9) as before. ROUNDOFF AND TRUNCATION ERROR There are two types of errors that occur when computing with series. The first, called truncation error, is the error that results when a series is approximated by a partial sum; and the second, called roundoff error, is the error that arises from approximations in numerical computations. For example, in our derivation of (9) we took n = 3 to keep the truncation error below 0.000005. However, to evaluate the partial sum we had to approximate π, thereby introducing roundoff error. Had we not exercised some care in choosing this approximation, the roundoff error could easily have degraded the final result. Methods for estimating and controlling roundoff error are studied in a branch of mathematics called numerical analysis. However, as a rule of thumb, to achieve n decimal-place accuracy in a final result, all intermediate calculations must be accurate to at least n + 1 decimal places. Thus, in (9) at least six decimal-place accuracy in π is required to achieve the five decimal-place accuracy in the final numerical result. As a practical matter, a good working procedure is to perform all intermediate computations with the maximum number of digits that your calculating utility can handle and then round at the end. APPROXIMATING EXPONENTIAL FUNCTIONS

Example 3 ex =

Show that the Maclaurin series for ex converges to ex for all x; that is,

⬁ xk k=0

=1+x+

x2 x3 xk + + ··· + + ··· 2! 3! k!

(−⬁ < x < +⬁)

Solution. Let f(x) = ex , so that f (n+1) (x) = ex We want to show that Rn (x) → 0 as n → +⬁ for all x in the interval −⬁ < x < +⬁. However, it will be helpful here to consider the cases x ≤ 0 and x > 0 separately. If x ≤ 0, then we will take the interval in the Remainder Estimation Theorem (Theorem 9.7.4) to be [x, 0], and if x > 0, then we will take it to be [0, x]. Since f (n+1) (x) = ex is an increasing function, it follows that if c is in the interval [x, 0], then |f (n+1) (c)| ≤ |f (n+1) (0)| = e0 = 1 and if c is in the interval [0, x], then |f (n+1) (c)| ≤ |f (n+1) (x)| = ex Thus, we can apply Theorem 9.7.4 with M = 1 in the case where x ≤ 0 and with M = ex in the case where x > 0. This yields 0 ≤ |Rn (x)| ≤

|x|n+1 (n + 1)!

0 ≤ |Rn (x)| ≤ ex

|x|n+1 (n + 1)!

if x ≤ 0 if x > 0

Thus, in both cases it follows from (5) and the Squeezing Theorem for Sequences that |Rn (x)| → 0 as n → +⬁, which in turn implies that Rn (x) → 0 as n → +⬁. Since this is true for all x, we have proved that the Maclaurin series for ex converges to ex for all x.

9.9 Convergence of Taylor Series

673

Since the Maclaurin series for ex converges to ex for all x, we can use partial sums of the Maclaurin series to approximate powers of e to arbitrary precision. Recall that in Example 7 of Section 9.7 we were able to use the Remainder Estimation Theorem to determine that evaluating the ninth Maclaurin polynomial for ex at x = 1 yields an approximation for e with ďŹ ve decimal-place accuracy: e â&#x2030;&#x2C6;1+1+

In Example 2 of Section 9.6, we stated without proof that

ln 2 = 1 â&#x2C6;&#x2019;

1 1 1 1 + â&#x2C6;&#x2019; + â&#x2C6;&#x2019; ÂˇÂˇÂˇ 2 3 4 5

This result can be obtained by letting x = 1 in (10), but as indicated in the text discussion, this series converges too slowly to be of practical use.

James Gregory (1638 â&#x20AC;&#x201C; 1675) Scottish mathematician and astronomer. Gregory, the son of a minister, was famous in his time as the inventor of the Gregorian reďŹ&#x201A;ecting telescope, so named in his honor. Although he is not generally ranked with the great mathematicians, much of his work relating to calculus was studied by Leibniz and Newton and undoubtedly inďŹ&#x201A;uenced some of their discoveries. There is a manuscript, discovered posthumously, which shows that Gregory had anticipated Taylor series well before Taylor. [Image: http://en.wikipedia.org/wiki/ File:James_Gregory.jpeg]

1 1 1 1 1 1 1 1 + + + + + + + â&#x2030;&#x2C6; 2.71828 2! 3! 4! 5! 6! 7! 8! 9!

APPROXIMATING LOGARITHMS The Maclaurin series x2 x3 x4 ln(1 + x) = x â&#x2C6;&#x2019; (10) + â&#x2C6;&#x2019; + ÂˇÂˇÂˇ (â&#x2C6;&#x2019;1 < x â&#x2030;¤ 1) 2 3 4 is the starting point for the approximation of natural logarithms. Unfortunately, the usefulness of this series is limited because of its slow convergence and the restriction â&#x2C6;&#x2019;1 < x â&#x2030;¤ 1. However, if we replace x by â&#x2C6;&#x2019;x in this series, we obtain

x3 x4 x2 â&#x2C6;&#x2019; â&#x2C6;&#x2019; â&#x2C6;&#x2019; ÂˇÂˇÂˇ 2 3 4 and on subtracting (11) from (10) we obtain

1+x x3 x5 x7 ln =2 x+ + + + ÂˇÂˇÂˇ 1â&#x2C6;&#x2019;x 3 5 7 ln(1 â&#x2C6;&#x2019; x) = â&#x2C6;&#x2019;x â&#x2C6;&#x2019;

(â&#x2C6;&#x2019;1 â&#x2030;¤ x < 1)

(â&#x2C6;&#x2019;1 < x < 1)

(11)

(12)

Series (12), ďŹ rst obtained by James Gregory in 1668, can be used to compute the natural logarithm of any positive number y by letting y=

1+x 1â&#x2C6;&#x2019;x

yâ&#x2C6;&#x2019;1 y+1

or, equivalently,

(13)

and noting that â&#x2C6;&#x2019;1 < x < 1. For example, to compute ln 2 we let y = 2 in (13), which yields x = 31 . Substituting this value in (12) gives 1 3 1 5 1 7 1 3 3 3 ln 2 = 2 + + + + ÂˇÂˇÂˇ (14) 3 3 5 7 In Exercise 19 we will ask you to show that ďŹ ve decimal-place accuracy can be achieved using the partial sum with terms up to and including the 13th power of 13 . Thus, to ďŹ ve decimal-place accuracy 1 3 1 5 1 7 1 13 1 3 3 3 ln 2 â&#x2030;&#x2C6; 2 + + + + ÂˇÂˇÂˇ + 3 â&#x2030;&#x2C6; 0.69315 3 3 5 7 13 (verify). As a check, a calculator gives ln 2 â&#x2030;&#x2C6; 0.69314718056, which agrees with the preceding approximation when rounded to ďŹ ve decimal places. APPROXIMATING Ď&#x20AC;

In the next section we will show that tanâ&#x2C6;&#x2019;1 x = x â&#x2C6;&#x2019; Letting x = 1, we obtain

x5 x7 x3 + â&#x2C6;&#x2019; + ÂˇÂˇÂˇ 3 5 7

(â&#x2C6;&#x2019;1 â&#x2030;¤ x â&#x2030;¤ 1)

Ď&#x20AC; 1 1 1 = tanâ&#x2C6;&#x2019;1 1 = 1 â&#x2C6;&#x2019; + â&#x2C6;&#x2019; + Âˇ Âˇ Âˇ 4 3 5 7

(15)

674

Chapter 9 / Infinite Series

1 1 1 Ď&#x20AC; = 4 1 â&#x2C6;&#x2019; + â&#x2C6;&#x2019; + ÂˇÂˇÂˇ 3 5 7 This famous series, obtained by Leibniz in 1674, converges too slowly to be of computational value. A more practical procedure for approximating Ď&#x20AC; uses the identity 1 1 Ď&#x20AC; = tanâ&#x2C6;&#x2019;1 + tanâ&#x2C6;&#x2019;1 (16) 4 2 3 which was derived in Exercise 60 of Section 0.4. By using this identity and series (15) to approximate tanâ&#x2C6;&#x2019;1 21 and tanâ&#x2C6;&#x2019;1 13 , the value of Ď&#x20AC; can be approximated efďŹ ciently to any degree of accuracy.

BINOMIAL SERIES

Let f(x) = (1 + x)m . Verify that

f(0) = 1

f â&#x20AC;˛(0) = m

f â&#x20AC;˛â&#x20AC;˛(0) = m(m â&#x2C6;&#x2019; 1)

f â&#x20AC;˛â&#x20AC;˛â&#x20AC;˛(0) = m(m â&#x2C6;&#x2019; 1)(m â&#x2C6;&#x2019; 2) .. . f (k) (0) = m(m â&#x2C6;&#x2019; 1) Âˇ Âˇ Âˇ (m â&#x2C6;&#x2019; k + 1)

If m is a real number, then the Maclaurin series for (1 + x)m is called the binomial series; it is given by m(m â&#x2C6;&#x2019; 1) Âˇ Âˇ Âˇ (m â&#x2C6;&#x2019; k + 1) k m(m â&#x2C6;&#x2019; 1) 2 m(m â&#x2C6;&#x2019; 1)(m â&#x2C6;&#x2019; 2) 3 x + x + ÂˇÂˇÂˇ + x + ÂˇÂˇÂˇ 1 + mx + 2! 3! k! In the case where m is a nonnegative integer, the function f(x) = (1 + x)m is a polynomial of degree m, so f (m+1) (0) = f (m+2) (0) = f (m+3) (0) = Âˇ Âˇ Âˇ = 0 and the binomial series reduces to the familiar binomial expansion m(m â&#x2C6;&#x2019; 1) 2 m(m â&#x2C6;&#x2019; 1)(m â&#x2C6;&#x2019; 2) 3 x + x + Âˇ Âˇ Âˇ + xm (1 + x)m = 1 + mx + 2! 3! which is valid for â&#x2C6;&#x2019;âŹ < x < +âŹ . It can be proved that if m is not a nonnegative integer, then the binomial series converges to (1 + x)m if |x| < 1. Thus, for such values of x m(m â&#x2C6;&#x2019; 1) Âˇ Âˇ Âˇ (m â&#x2C6;&#x2019; k + 1) k m(m â&#x2C6;&#x2019; 1) 2 (17) x + ÂˇÂˇÂˇ + x + ÂˇÂˇÂˇ (1 + x)m = 1 + mx + 2! k! or in sigma notation, (1 + x)m = 1 +

Example 4

âŹ m(m â&#x2C6;&#x2019; 1) Âˇ Âˇ Âˇ (m â&#x2C6;&#x2019; k + 1)

k=1

Find binomial series for 1 (a) (1 + x)2

(b) â&#x2C6;&#x161;

if |x| < 1

(18)

1 1+x

Solution (a). Since the general term of the binomial series is complicated, you may ďŹ nd it helpful to write out some of the beginning terms of the series, as in Formula (17), to see developing patterns. Substituting m = â&#x2C6;&#x2019;2 in this formula yields (â&#x2C6;&#x2019;2)(â&#x2C6;&#x2019;3) 2 1 = (1 + x)â&#x2C6;&#x2019;2 = 1 + (â&#x2C6;&#x2019;2)x + x (1 + x)2 2! +

(â&#x2C6;&#x2019;2)(â&#x2C6;&#x2019;3)(â&#x2C6;&#x2019;4) 3 (â&#x2C6;&#x2019;2)(â&#x2C6;&#x2019;3)(â&#x2C6;&#x2019;4)(â&#x2C6;&#x2019;5) 4 x + x + ÂˇÂˇÂˇ 3! 4!

3! 2 4! 3 5! 4 x â&#x2C6;&#x2019; x + x â&#x2C6;&#x2019; ÂˇÂˇÂˇ 2! 3! 4! 2 3 4 = 1 â&#x2C6;&#x2019; 2x + 3x â&#x2C6;&#x2019; 4x + 5x â&#x2C6;&#x2019; Âˇ Âˇ Âˇ âŹ = (â&#x2C6;&#x2019;1)k (k + 1)x k = 1 â&#x2C6;&#x2019; 2x +

k=0

9.9 Convergence of Taylor Series

Solution (b). Substituting m = − 21 in (17) yields

y 4 p3 (x)

(1 + x)2 x

−1

1 −1

p3(x) = 1 − 2x + 3x2 − 4x3

y 4

1 1 1 1 − 2 − 2 − 1 − 21 − 2 3 −2 −2 − 1 2 1 =1− x+ x + x + ··· √ 2 2! 3! 1+x 1·3 2 1·3·5 3 1 x − 3 x + ··· =1− x+ 2 2 2 · 2! 2 · 3! ⬁ 1 · 3 · 5 · · · (2k − 1) k =1+ x (−1)k 2k k! k=1 1

3 2

1 √1 + x

Figure 9.9.2 shows the graphs of the functions in Example 4 compared to their thirddegree Maclaurin polynomials. SOME IMPORTANT MACLAURIN SERIES

For reference, Table 9.9.1 lists the Maclaurin series for some of the most important functions, together with a speciﬁcation of the intervals over which the Maclaurin series converge to those functions. Some of these results are derived in the exercises and others will be derived in the next section using some special techniques that we will develop.

1 x −1

1 −1

p3(x) = 1 − 12 x + Figure 9.9.2

675

Table 9.9.1 some important maclaurin series

p3(x)

3 2 x 8

−

5 16

maclaurin series 1 = ∞ x k = 1 + x + x 2 + x 3 + ... 1−x

interval of convergence −1 < x < 1

k =0

∞ 1 (−1) k x 2k = 1 − x 2 + x 4 − x 6 + ... = 2 1+x k =0 ∞ k x = 1 + x + x 2 + x 3 + x 4 + ... ex = k! 2! 3! 4! k =0 ∞

sin x =

k =0

2k+1 3 5 7 (−1) k x = x − x + x − x + ... (2k + 1)! 3! 5! 7!

∞

2 4 6 2k (−1)k x = 1 − x + x − x + ... 2! 4! 6! (2k)! k =0 ∞ 2 3 4 k ln (1 + x) = (−1) k+1 x = x − x + x − x + ... k 2 3 4 k =1

cos x =

∞

3 5 7 2k+1 (−1) k x = x − x + x − x + ... 3 5 7 2k + 1 k =0 ∞ 3 5 7 2k+1 x x x x sinh x = =x+ + + + ... 3! 5! 7! (2k + 1)! k =0

tan−1 x =

∞

x 2k = 1 + x 2 + x 4 + x 6 + ... 2! 4! 6! k =0 (2k)! ∞ m(m − 1) ... ( m − k + 1) (1 + x)m = 1 + xk k! k =1 cosh x =

−1 < x < 1 −∞ < x < +∞ −∞ < x < +∞ −∞ < x < +∞ −1 < x ≤ 1 −1 ≤ x ≤ 1 −∞ < x < +∞ −∞ < x < +∞ −1 < x < 1* (m ≠ 0, 1, 2, ...)

*The behavior at the endpoints depends on m: For m > 0 the series converges absolutely at both endpoints; for m ≤ −1 the series diverges at both endpoints; and for −1 < m < 0 the series converges conditionally at x = 1 and diverges at x = −1.

676

Chapter 9 / Infinite Series

✔QUICK CHECK EXERCISES 9.9 1. cos x = 2. ex =

(See page 677 for answers.)

⬁

4. If m is a real number but not a nonnegative integer, the ⬁ binomial series 1+

k=0

⬁

k=1

k=0

3. ln(1 + x) =

⬁

converges to (1 + x) if |x| < for x in the interval

k=1

EXERCISE SET 9.9

Graphing Utility

CAS

1. Use the Remainder Estimation Theorem and the method of Example 1 to prove that the Taylor series for sin x about x = π/4 converges to sin x for all x.

2. Use the Remainder Estimation Theorem and the method of Example 3 to prove that the Taylor series for ex about x = 1 converges to ex for all x.

3–10 Approximate the speciﬁed function value as indicated and

check your work by comparing your answer to the function value produced directly by your calculating utility. ■ 3. Approximate sin 4 ◦ to five decimal-place accuracy using both of the methods given in Example 2. 4. Approximate cos 3 ◦ to three decimal-place accuracy using both of the methods given in Example 2. 5. Approximate cos 0.1 to five decimal-place accuracy using the Maclaurin series for cos x. 6. Approximate tan−1 0.1 to three decimal-place accuracy using the Maclaurin series for tan−1 x. 7. Approximate sin 85 ◦ to four decimal-place accuracy using an appropriate Taylor series. 8. Approximate cos(−175 ◦ ) to four decimal-place accuracy using a Taylor series. 9. Approximate sinh 0.5 to three decimal-place accuracy using the Maclaurin series for sinh x. 10. Approximate cosh 0.1 to three decimal-place accuracy using the Maclaurin series for cosh x. 11. (a) Use Formula (12) in the text to find a series that converges to ln 1.25. (b) Approximate ln 1.25 using the first two terms of the series. Round your answer to three decimal places, and compare the result to that produced directly by your calculating utility. 12. (a) Use Formula (12) to find a series that converges to ln 3. (b) Approximate ln 3 using the first two terms of the series. Round your answer to three decimal places, and compare the result to that produced directly by your calculating utility.

F O C U S O N C O N C E P TS

13. (a) Use the Maclaurin series for tan−1 x to approximate tan−1 21 and tan−1 31 to three decimal-place accuracy. (b) Use the results in part (a) and Formula (16) to approximate π. (c) Would you be willing to guarantee that your answer in part (b) is accurate to three decimal places? Explain your reasoning. (d) Compare your answer in part (b) to that produced by your calculating utility. 14. The purpose of this exercise is to show that the Taylor series of a function f may possibly converge to a value different from f(x) for certain values of x. Let 2 e−1/x , x = 0 f(x) = 0, x=0 (a) Use the deﬁnition of a derivative to show that f ′(0) = 0. (b) With some difﬁculty it can be shown that if n ≥ 2, then f (n) (0) = 0. Accepting this fact, show that the Maclaurin series of f converges for all x but converges to f(x) only at x = 0. 15. (a) Find an upper bound on the error that can result if cos x is approximated by 1 − (x 2 /2!) + (x 4 /4!) over the interval [−0.2, 0.2]. (b) Check your answer in part (a) by graphing

2 4 cos x − 1 − x + x 2! 4! over the interval.

16. (a) Find an upper bound on the error that can result if ln(1 + x) is approximated by x over the interval [−0.01, 0.01]. (b) Check your answer in part (a) by graphing | ln(1 + x) − x| over the interval.

9.9 Convergence of Taylor Series

17. Use Formula (17) for the binomial series to obtain the Maclaurin series for â&#x2C6;&#x161; 1 1 3 (a) (c) . (b) 1 + x 1+x (1 + x)3 18. If m is any real number, and k is a nonnegative integer, then we deďŹ ne the binomial coefďŹ cient

m m = 1 and by the formulas 0 k

m(m â&#x2C6;&#x2019; 1)(m â&#x2C6;&#x2019; 2) Âˇ Âˇ Âˇ (m â&#x2C6;&#x2019; k + 1) m = k k! for k â&#x2030;Ľ 1. Express Formula (17) in the text in terms of binomial coefďŹ cients.

(e) Use the results in parts (c) and (d) to show that ďŹ ve decimal-place accuracy will be achieved if n satisďŹ es

n+1 1 1 n+1 1 â&#x2030;¤ 0.000005 + n+1 3 2 and then show that the smallest value of n that satisďŹ es this condition is n = 13.

20. Use Formula (12) and the method of Exercise 19 to approximate ln 35 to ďŹ ve decimal-place accuracy. Then check your work by comparing your answer to that produced directly by your calculating utility. 21. Prove: The Taylor series for cos x about any value x = x0 converges to cos x for all x.

19. In this exercise we will use the Remainder Estimation Theorem to determine the number of terms that are required in Formula (14) to approximate ln 2 to ďŹ ve decimal-place accuracy. For this purpose let

22. Prove: The Taylor series for sin x about any value x = x0 converges to sin x for all x. 23. Research has shown that the proportion p of the population with IQs (intelligence quotients) between Îą and Î˛ is approximately Î˛ 1 1 xâ&#x2C6;&#x2019;100 2 p= â&#x2C6;&#x161; eâ&#x2C6;&#x2019; 2 ( 16 ) dx 16 2Ď&#x20AC; Îą Use the ďŹ rst three terms of an appropriate Maclaurin series to estimate the proportion of the population that has IQs between 100 and 110.

1+x = ln(1 + x) â&#x2C6;&#x2019; ln(1 â&#x2C6;&#x2019; x) (â&#x2C6;&#x2019;1 < x < 1) f(x) = ln 1â&#x2C6;&#x2019;x

(a) Show that

f (n+1) (x) = n!

(â&#x2C6;&#x2019;1)n 1 + (1 + x)n+1 (1 â&#x2C6;&#x2019; x)n+1

(b) Use the triangle inequality [Theorem 0.1.4(d )] to show that 1 1 |f (n+1) (x)| â&#x2030;¤ n! + (1 + x)n+1 (1 â&#x2C6;&#x2019; x)n+1 (c) Since we want to achieve ďŹ ve decimal-place accuracy, our goal is to choose n so that the absolute value of the nth remainder at x = 13 does not exceed the value 0.000005 = 0.5 Ă&#x2014; 10â&#x2C6;&#x2019;5 ; that is, Rn 31 â&#x2030;¤ 0.000005. Use the Remainder Estimation Theorem to show that this condition will be satisďŹ ed if n is chosen so that

n+1 M 1 â&#x2030;¤ 0.000005 (n + 1)! 3 where |f (n+1) (x)| â&#x2030;¤ M on the interval 0, 13 . (d) Use the result in part (b) to show that M can be taken as 1 M = n! 1 + n+1 2 3

24. (a) In 1706 the British astronomer and mathematician John Machin discovered the following formula for Ď&#x20AC;/4, called Machinâ&#x20AC;&#x2122;s formula: Ď&#x20AC; 1 1 = 4 tanâ&#x2C6;&#x2019;1 â&#x2C6;&#x2019; tanâ&#x2C6;&#x2019;1 4 5 239 Use a CAS to approximate Ď&#x20AC;/4 using Machinâ&#x20AC;&#x2122;s formula to 25 decimal places. (b) In 1914 the brilliant Indian mathematician Srinivasa Ramanujan (1887â&#x20AC;&#x201C;1920) showed that â&#x2C6;&#x161; âŹ 1 8 (4k)!(1103 + 26,390k) = Ď&#x20AC; 9801 k=0 (k!)4 3964k Use a CAS to compute the ďŹ rst four partial sums in Ramanujanâ&#x20AC;&#x2122;s formula.

â&#x153;&#x201D;QUICK CHECK ANSWERS 9.9 1. (â&#x2C6;&#x2019;1)k

x 2k (2k)!

xk k!

3. (â&#x2C6;&#x2019;1)k+1

xk ; (â&#x2C6;&#x2019;1, 1] k

677

m(m â&#x2C6;&#x2019; 1) Âˇ Âˇ Âˇ (m â&#x2C6;&#x2019; k + 1) k x ;1 k!

678

Chapter 9 / Infinite Series

9.10

DIFFERENTIATING AND INTEGRATING POWER SERIES; MODELING WITH TAYLOR SERIES In this section we will discuss methods for ďŹ nding power series for derivatives and integrals of functions, and we will discuss some practical methods for ďŹ nding Taylor series that can be used in situations where it is difďŹ cult or impossible to ďŹ nd the series directly. DIFFERENTIATING POWER SERIES We begin by considering the following problem.

9.10.1 problem Suppose that a function f is represented by a power series on an open interval. How can we use the power series to ďŹ nd the derivative of f on that interval? The solution to this problem can be motivated by considering the Maclaurin series for sin x: x5 x7 x3 + â&#x2C6;&#x2019; + ÂˇÂˇÂˇ (â&#x2C6;&#x2019;âŹ < x < +âŹ ) sin x = x â&#x2C6;&#x2019; 3! 5! 7! Of course, we already know that the derivative of sin x is cos x; however, we are concerned here with using the Maclaurin series to deduce this. The solution is easyâ&#x20AC;&#x201D;all we need to do is differentiate the Maclaurin series term by term and observe that the resulting series is the Maclaurin series for cos x: d x5 x7 x2 x4 x6 x3 + â&#x2C6;&#x2019; + ÂˇÂˇÂˇ = 1 â&#x2C6;&#x2019; 3 + 5 â&#x2C6;&#x2019; 7 + ÂˇÂˇÂˇ xâ&#x2C6;&#x2019; dx 3! 5! 7! 3! 5! 7! x2 x4 x6 =1â&#x2C6;&#x2019; + â&#x2C6;&#x2019; + Âˇ Âˇ Âˇ = cos x 2! 4! 6! Here is another example. d x d x3 x4 x2 [e ] = + + + ÂˇÂˇÂˇ 1+x+ dx dx 2! 3! 4! x x2 x3 x2 x3 = 1 + 2 + 3 + 4 + ÂˇÂˇÂˇ = 1 + x + + + Âˇ Âˇ Âˇ = ex 2! 3! 4! 2! 3! The preceding computations suggest that if a function f is represented by a power series on an open interval, then a power series representation of f â&#x20AC;˛ on that interval can be obtained by differentiating the power series for f term by term. This is stated more precisely in the following theorem, which we give without proof. 9.10.2 theorem (Differentiation of Power Series) Suppose that a function f is represented by a power series in x â&#x2C6;&#x2019; x0 that has a nonzero radius of convergence R; that âŹ is, ck (x â&#x2C6;&#x2019; x0 )k (x0 â&#x2C6;&#x2019; R < x < x0 + R) f(x) = k=0

Then:

(a) The function f is differentiable on the interval (x0 â&#x2C6;&#x2019; R, x0 + R). (b) If the power series representation for f is differentiated term by term, then the resulting series has radius of convergence R and converges to f â&#x20AC;˛ on the interval (x0 â&#x2C6;&#x2019; R, x0 + R); that is, âŹ d [ck (x â&#x2C6;&#x2019; x0 )k ] (x0 â&#x2C6;&#x2019; R < x < x0 + R) f â&#x20AC;˛(x) = dx k=0

9.10 Differentiating and Integrating Power Series; Modeling with Taylor Series

679

This theorem has an important implication about the differentiability of functions that are represented by power series. According to the theorem, the power series for f ′ has the same radius of convergence as the power series for f , and this means that the theorem can be applied to f ′ as well as f . However, if we do this, then we conclude that f ′ is differentiable on the interval (x0 − R, x0 + R), and the power series for f ′′ has the same radius of convergence as the power series for f and f ′ . We can now repeat this process ad inﬁnitum, applying the theorem successively to f ′′ , f ′′′ , . . . , f (n) , . . . to conclude that f has derivatives of all orders on the interval (x0 − R, x0 + R). Thus, we have established the following result.

9.10.3 theorem If a function f can be represented by a power series in x − x0 with a nonzero radius of convergence R, then f has derivatives of all orders on the interval (x0 − R, x0 + R).

In short, it is only the most “well-behaved” functions that can be represented by power series; that is, if a function f does not possess derivatives of all orders on an interval (x0 − R, x0 + R), then it cannot be represented by a power series in x − x0 on that interval. Example 1 In Section 9.8, we showed that the Bessel function J0 (x), represented by the power series ⬁ (−1)k x 2k J0 (x) = (1) 22k (k!)2 k=0 has radius of convergence +⬁ [see Formula (4) of that section and the related discussion]. Thus, J0 (x) has derivatives of all orders on the interval (−⬁, +⬁), and these can be obtained by differentiating the series term by term. For example, if we write (1) as J0 (x) = 1 +

⬁ (−1)k x 2k k=1

22k (k!)2

and differentiate term by term, we obtain See Exercise 45 for a relationship between J0′ (x) and J1 (x).

REMARK

J0′ (x) =

⬁ (−1)k (2k)x 2k−1 k=1

22k (k!)2

⬁ k=1

(−1)k x 2k−1 22k−1 k!(k − 1)!

The computations in this example use some techniques that are worth noting. First, when a power series is expressed in sigma notation, the formula for the general term of the series will often not be of a form that can be used for differentiating the constant term. Thus, if the series has a nonzero constant term, as here, it is usually a good idea to split it off from the summation before differentiating. Second, observe how we simplified the final formula by canceling the factor k from one of the factorials in the denominator. This is a standard simplification technique.

INTEGRATING POWER SERIES

Since the derivative of a function that is represented by a power series can be obtained by differentiating the series term by term, it should not be surprising that an antiderivative of a function represented by a power series can be obtained by integrating the series term by term. For example, we know that sin x is an antiderivative of cos x. Here is how this result

680

Chapter 9 / Infinite Series

can be obtained by integrating the Maclaurin series for cos x term by term: x2 x4 x6 1â&#x2C6;&#x2019; cos x dx = + â&#x2C6;&#x2019; + Âˇ Âˇ Âˇ dx 2! 4! 6! 5 3 x x x7 = xâ&#x2C6;&#x2019; + â&#x2C6;&#x2019; + ÂˇÂˇÂˇ + C 3(2!) 5(4!) 7(6!) x3 x5 x7 = xâ&#x2C6;&#x2019; + â&#x2C6;&#x2019; + Âˇ Âˇ Âˇ + C = sin x + C 3! 5! 7! The same idea applies to deďŹ nite integrals. For example, by direct integration we have 1 1 dx Ď&#x20AC; Ď&#x20AC; â&#x2C6;&#x2019;1 = tan x = tanâ&#x2C6;&#x2019;1 1 â&#x2C6;&#x2019; tan 0 = â&#x2C6;&#x2019; 0 = 2 1 + x 4 4 0 0 and we will show later in this section that 1 1 1 Ď&#x20AC; = 1 â&#x2C6;&#x2019; + â&#x2C6;&#x2019; + ÂˇÂˇÂˇ 4 3 5 7 Thus, 1 dx 1 1 1 = 1 â&#x2C6;&#x2019; + â&#x2C6;&#x2019; + ÂˇÂˇÂˇ 2 3 5 7 0 1+x

(2)

Here is how this result can be obtained by integrating the Maclaurin series for 1/(1 + x 2 ) term by term (see Table 9.9.1): 1 1 dx [1 â&#x2C6;&#x2019; x 2 + x 4 â&#x2C6;&#x2019; x 6 + Âˇ Âˇ Âˇ] dx = 2 0 1+x 0 1 x3 x5 x7 1 1 1 = xâ&#x2C6;&#x2019; + â&#x2C6;&#x2019; + ÂˇÂˇÂˇ = 1 â&#x2C6;&#x2019; + â&#x2C6;&#x2019; + ÂˇÂˇÂˇ 3 5 7 3 5 7 0

The preceding computations are justiďŹ ed by the following theorem, which we give without proof.

Theorems 9.10.2 and 9.10.4 tell us how to use a power series representation of a function f to produce power series representations of f â&#x20AC;˛ (x) and f(x) dx that have the same radius of convergence as f . However, the intervals of convergence for these series may not be the same because their convergence behavior may differ at the endpoints of the interval. (See Exercises 25 and 26.)

9.10.4 theorem (Integration of Power Series) Suppose that a function f is represented by a power series in x â&#x2C6;&#x2019; x0 that has a nonzero radius of convergence R; that is, f(x) =

âŹ k=0

ck (x â&#x2C6;&#x2019; x0 )k

(x0 â&#x2C6;&#x2019; R < x < x0 + R)

(a) If the power series representation of f is integrated term by term, then the resulting series has radius of convergence R and converges to an antiderivative for f(x) on the interval (x0 â&#x2C6;&#x2019; R, x0 + R); that is, âŹ ck k+1 f(x) dx = (x â&#x2C6;&#x2019; x0 ) +C (x0 â&#x2C6;&#x2019; R < x < x0 + R) k+1 k=0 (b) If Îą and Î˛ are points in the interval (x0 â&#x2C6;&#x2019; R, x0 + R), and if the power series representation of f is integrated term by term from Îą to Î˛, then the resulting series converges absolutely on the interval (x0 â&#x2C6;&#x2019; R, x0 + R) and Î˛ âŹ Î˛ k f(x) dx = ck (x â&#x2C6;&#x2019; x0 ) dx Îą

k=0

Îą

9.10 Differentiating and Integrating Power Series; Modeling with Taylor Series

681

POWER SERIES REPRESENTATIONS MUST BE TAYLOR SERIES

For many functions it is difficult or impossible to find the derivatives that are required to obtain a Taylor series. For example, to find the Maclaurin series for 1/(1 + x 2 ) directly would require some tedious derivative computations (try it). A more practical approach is to substitute −x 2 for x in the geometric series 1 = 1 + x + x2 + x3 + x4 + · · · 1−x to obtain

(−1 < x < 1)

1 = 1 − x2 + x4 − x6 + x8 − · · · 1 + x2

However, there are two questions of concern with this procedure:

• Where does the power series that we obtained for 1/(1 + x 2 ) actually converge to •

1/(1 + x 2 )? How do we know that the power series we have obtained is actually the Maclaurin series for 1/(1 + x 2 )?

The ﬁrst question is easy to resolve. Since the geometric series converges to 1/(1 − x) if |x| < 1, the second series will converge to 1/(1 + x 2 ) if |−x 2 | < 1 or |x 2 | < 1. However, this is true if and only if |x| < 1, so the power series we obtained for the function 1/(1 + x 2 ) converges to this function if −1 < x < 1. The second question is more difﬁcult to answer and leads us to the following general problem.

9.10.5 problem Suppose that a function f is represented by a power series in x − x0 that has a nonzero radius of convergence. What relationship exists between the given power series and the Taylor series for f about x = x0 ?

The answer is that they are the same; and here is the theorem that proves it.

Theorem 9.10.6 tells us that no matter how we arrive at a power series representation of a function f , be it by substitution, by differentiation, by integration, or by some algebraic process, that series will be the Taylor series for f about x = x0 , provided the series converges to f on some open interval containing x0 .

9.10.6 theorem If a function f is represented by a power series in x − x0 on some open interval containing x0 , then that power series is the Taylor series for f about x = x0 .

proof Suppose that f(x) = c0 + c1 (x − x0 ) + c2 (x − x0 )2 + · · · + ck (x − x0 )k + · · · for all x in some open interval containing x0 . To prove that this is the Taylor series for f about x = x0 , we must show that ck =

f (k) (x0 ) k!

for k = 0, 1, 2, 3, . . .

However, the assumption that the series converges to f(x) on an open interval containing x0 ensures that it has a nonzero radius of convergence R; hence we can differentiate term

682

Chapter 9 / Infinite Series

by term in accordance with Theorem 9.10.2. Thus, f(x) = c0 + c1 (x − x0 ) + c2 (x − x0 )2 + c3 (x − x0 )3 + c4 (x − x0 )4 + · · · f ′(x) = c1 + 2c2 (x − x0 ) + 3c3 (x − x0 )2 + 4c4 (x − x0 )3 + · · · f ′′(x) = 2!c2 + (3 · 2)c3 (x − x0 ) + (4 · 3)c4 (x − x0 )2 + · · ·

f ′′′(x) = 3!c3 + (4 · 3 · 2)c4 (x − x0 ) + · · · .. . On substituting x = x0 , all the powers of x − x0 drop out, leaving f(x0 ) = c0 ,

f ′(x0 ) = c1 ,

f ′′(x0 ) = 2!c2 ,

f ′′′(x0 ) = 3!c3 , . . .

from which we obtain f ′′(x0 ) f ′′′(x0 ) , c3 = ,... 2! 3! which shows that the coefﬁcients c0 , c1 , c2 , c3 , . . . are precisely the coefﬁcients in the Taylor series about x0 for f(x). ■ c0 = f(x0 ),

c1 = f ′(x0 ),

c2 =

SOME PRACTICAL WAYS TO FIND TAYLOR SERIES

Example 2

Find Taylor series for the given functions about the given x0 . 2

(a) e−x ,

x0 = 0

(b) ln x,

x0 = 1

1 , x

(c)

x0 = 1 2

Solution (a). The simplest way to ﬁnd the Maclaurin series for e−x is to substitute −x 2

for x in the Maclaurin series

ex = 1 + x + to obtain

x3 x4 x2 + + + ··· 2! 3! 4!

e−x = 1 − x 2 +

(3)

x6 x8 x4 − + − ··· 2! 3! 4! 2

Since (3) converges for all values of x, so will the series for e−x .

Solution (b). We begin with the Maclaurin series for ln(1 + x), which can be found in

Table 9.9.1:

x3 x4 x2 + − + ··· 2 3 4 Substituting x − 1 for x in this series gives ln(1 + x) = x −

(−1 < x ≤ 1)

(x − 1)2 (x − 1)3 (x − 1)4 (4) + − + ··· 2 3 4 Since the original series converges when −1 < x ≤ 1, the interval of convergence for (4) will be −1 < x − 1 ≤ 1 or, equivalently, 0 < x ≤ 2. ln(1 + [x − 1]) = ln x = (x − 1) −

Solution (c). Since 1/x is the derivative of ln x, we can differentiate the series for ln x found in (b) to obtain 2(x − 1) 3(x − 1)2 4(x − 1)3 1 =1− + − + ··· x 2 3 4 = 1 − (x − 1) + (x − 1)2 − (x − 1)3 + · · ·

(5)

9.10 Differentiating and Integrating Power Series; Modeling with Taylor Series

683

By Theorem 9.10.2, we know that the radius of convergence for (5) is the same as that for (4), which is R = 1. Thus the interval of convergence for (5) must be at least 0 < x < 2. Since the behaviors of (4) and (5) may differ at the endpoints x = 0 and x = 2, those must be checked separately. When x = 0, (5) becomes 1 â&#x2C6;&#x2019; (â&#x2C6;&#x2019;1) + (â&#x2C6;&#x2019;1)2 â&#x2C6;&#x2019; (â&#x2C6;&#x2019;1)3 + Âˇ Âˇ Âˇ = 1 + 1 + 1 + 1 + Âˇ Âˇ Âˇ which diverges by the divergence test. Similarly, when x = 2, (5) becomes 1 â&#x2C6;&#x2019; 1 + 1 2 â&#x2C6;&#x2019; 13 + Âˇ Âˇ Âˇ = 1 â&#x2C6;&#x2019; 1 + 1 â&#x2C6;&#x2019; 1 + Âˇ Âˇ Âˇ which also diverges by the divergence test. Thus the interval of convergence for (5) is 0 < x < 2.

Find the Maclaurin series for tanâ&#x2C6;&#x2019;1 x.

Example 3

Solution. It would be tedious to ďŹ nd the Maclaurin series directly. A better approach is to start with the formula

1 dx = tanâ&#x2C6;&#x2019;1 x + C 1 + x2

and integrate the Maclaurin series

1 = 1 â&#x2C6;&#x2019; x2 + x4 â&#x2C6;&#x2019; x6 + x8 â&#x2C6;&#x2019; Âˇ Âˇ Âˇ 1 + x2

(â&#x2C6;&#x2019;1 < x < 1)

term by term. This yields tan or

â&#x2C6;&#x2019;1

1 x+C = dx = [1 â&#x2C6;&#x2019; x 2 + x 4 â&#x2C6;&#x2019; x 6 + x 8 â&#x2C6;&#x2019; Âˇ Âˇ Âˇ] dx 1 + x2 x5 x7 x9 x3 â&#x2C6;&#x2019;1 tan x = x â&#x2C6;&#x2019; + â&#x2C6;&#x2019; + â&#x2C6;&#x2019; ÂˇÂˇÂˇ â&#x2C6;&#x2019; C 3 5 7 9

The constant of integration can be evaluated by substituting x = 0 and using the condition tanâ&#x2C6;&#x2019;1 0 = 0. This gives C = 0, so that tanâ&#x2C6;&#x2019;1 x = x â&#x2C6;&#x2019;

REMARK

x5 x7 x9 x3 + â&#x2C6;&#x2019; + â&#x2C6;&#x2019; ÂˇÂˇÂˇ 3 5 7 9

(â&#x2C6;&#x2019;1 < x < 1)

(6)

Observe that neither Theorem 9.10.2 nor Theorem 9.10.3 addresses what happens at the endpoints of the interval of convergence. However, it can be proved that if the Taylor series for f about x = x0 converges to f(x) for all x in the interval (x0 â&#x2C6;&#x2019; R, x0 + R), and if the Taylor series converges at the right endpoint x0 + R , then the value that it converges to at that point is the limit of f(x) as x â&#x2020;&#x2019; x0 + R from the left; and if the Taylor series converges at the left endpoint x0 â&#x2C6;&#x2019; R , then the value that it converges to at that point is the limit of f(x) as x â&#x2020;&#x2019; x0 â&#x2C6;&#x2019; R from the right. For example, the Maclaurin series for tanâ&#x2C6;&#x2019;1 x given in (6) converges at both x = â&#x2C6;&#x2019;1 and x = 1, since the hypotheses of the alternating series test (Theorem 9.6.1) are satisfied at those points. Thus, the continuity of tanâ&#x2C6;&#x2019;1 x on the interval [â&#x2C6;&#x2019;1, 1] implies that at x = 1 the Maclaurin series converges to

lim tanâ&#x2C6;&#x2019;1 x = tanâ&#x2C6;&#x2019;1 1 =

x â&#x2020;&#x2019; 1â&#x2C6;&#x2019;

Ď&#x20AC; 4

and at x = â&#x2C6;&#x2019;1 it converges to

lim tanâ&#x2C6;&#x2019;1 x = tanâ&#x2C6;&#x2019;1 (â&#x2C6;&#x2019;1) = â&#x2C6;&#x2019;

x â&#x2020;&#x2019; â&#x2C6;&#x2019;1+

Ď&#x20AC; 4

This shows that the Maclaurin series for tanâ&#x2C6;&#x2019;1 x actually converges to tanâ&#x2C6;&#x2019;1 x on the closed interval â&#x2C6;&#x2019;1 â&#x2030;¤ x â&#x2030;¤ 1. Moreover, the convergence at x = 1 establishes Formula (2).

684

Chapter 9 / Infinite Series

APPROXIMATING DEFINITE INTEGRALS USING TAYLOR SERIES Taylor series provide an alternative to Simpsonâ&#x20AC;&#x2122;s rule and other numerical methods for approximating deďŹ nite integrals.

Example 4 Approximate the integral 1

eâ&#x2C6;&#x2019;x dx

to three decimal-place accuracy by expanding the integrand in a Maclaurin series and integrating term by term. 2

Solution. We found in Example 2(a) that the Maclaurin series for eâ&#x2C6;&#x2019;x is 2

eâ&#x2C6;&#x2019;x = 1 â&#x2C6;&#x2019; x 2 +

x4 x6 x8 â&#x2C6;&#x2019; + â&#x2C6;&#x2019; ÂˇÂˇÂˇ 2! 3! 4!

Therefore,

e 0

â&#x2C6;&#x2019;x 2

x4 x6 x8 1â&#x2C6;&#x2019;x + dx = â&#x2C6;&#x2019; + â&#x2C6;&#x2019; Âˇ Âˇ Âˇ dx 2! 3! 4! 0 1 x3 x5 x7 x9 = xâ&#x2C6;&#x2019; + â&#x2C6;&#x2019; + â&#x2C6;&#x2019; ÂˇÂˇÂˇ 3 5(2!) 7(3!) 9(4!) 0

=1â&#x2C6;&#x2019; =

âŹ k=0

1 1 1 1 + â&#x2C6;&#x2019; + â&#x2C6;&#x2019; ÂˇÂˇÂˇ 3 5 Âˇ 2! 7 Âˇ 3! 9 Âˇ 4!

(â&#x2C6;&#x2019;1)k (2k + 1)k!

Since this series clearly satisďŹ es the hypotheses of the alternating series test (Theorem 9.6.1), it follows from Theorem 9.6.2 that if we approximate the integral by sn (the nth partial sum of the series), then 1 1 1 â&#x2C6;&#x2019;x 2 = e dx â&#x2C6;&#x2019; s n < [2(n + 1) + 1](n + 1)! (2n + 3)(n + 1)! 0

Thus, for three decimal-place accuracy we must choose n such that 1 â&#x2030;¤ 0.0005 = 5 Ă&#x2014; 10â&#x2C6;&#x2019;4 (2n + 3)(n + 1)!

What advantages does the method of Example 4 have over Simpsonâ&#x20AC;&#x2122;s rule? What are its disadvantages?

With the help of a calculating utility you can show that the smallest value of n that satisďŹ es this condition is n = 5. Thus, the value of the integral to three decimal-place accuracy is 1 1 1 1 1 1 2 eâ&#x2C6;&#x2019;x dx â&#x2030;&#x2C6; 1 â&#x2C6;&#x2019; + â&#x2C6;&#x2019; + â&#x2C6;&#x2019; â&#x2030;&#x2C6; 0.747 3 5 Âˇ 2! 7 Âˇ 3! 9 Âˇ 4! 11 Âˇ 5! 0 As a check, a calculator with a built-in numerical integration capability produced the approximation 0.746824, which agrees with our result when rounded to three decimal places.

FINDING TAYLOR SERIES BY MULTIPLICATION AND DIVISION The following examples illustrate some algebraic techniques that are sometimes useful for ďŹ nding Taylor series.

9.10 Differentiating and Integrating Power Series; Modeling with Taylor Series x4 . . . – 2 × 3 5 x x x− + –... 3 5 1 − x2 +

685

Example 5 Find the ﬁrst three nonzero terms in the Maclaurin series for the function 2 f(x) = e−x tan−1 x. 2

x5 . . . – 2 3 5 x x7 . . . x − + − + 3 3 6 5 7 x x − +... 5 5

x − x3 +

4 3

x− x + 3

31 5 x 30

−...

x 3 2x 5 . . . + + 3 15

2 4 3 5 1− x + x − . . . x − x + x − . . . 2 24 6 120 3 5 x− x + x −... 2 24

x3 − 3 x3 − 3

x5 + . . . 30 x5 + . . . 6 2x 5 + . . . 15

T E C H N O LO GY M A ST E R Y If you have a CAS, use its capability for multiplying and dividing polynomials to perform the computations in Examples 5 and 6.

Solution. Using the series for e−x and tan−1 x obtained in Examples 2 and 3 gives e

−x 2

tan

−1

x4 − ··· x = 1−x + 2

x3 x5 x− + − ··· 3 5

Multiplying, as shown in the margin, we obtain 31 4 2 e−x tan−1 x = x − x 3 + x 5 − · · · 3 30 More terms in the series can be obtained by including more terms in the factors. Moreover, one can prove that a series obtained by this method converges at each point in the intersection of the intervals of convergence of the factors (and possibly on a larger interval). Thus, we can be certain that the series we have obtained converges for all x in the interval −1 ≤ x ≤ 1 (why?).

Example 6

Find the ﬁrst three nonzero terms in the Maclaurin series for tan x.

Solution. Using the ﬁrst three terms in the Maclaurin series for sin x and cos x, we can express tan x as

x3 x5 + − ··· sin x 3! 5! = tan x = cos x x4 x2 + − ··· 1− 2! 4! Dividing, as shown in the margin, we obtain x−

tan x = x +

2x 5 x3 + + ··· 3 15

MODELING PHYSICAL LAWS WITH TAYLOR SERIES Taylor series provide an important way of modeling physical laws. To illustrate the idea we will consider the problem of modeling the period of a simple pendulum (Figure 9.10.1). As explained in Chapter 7 Making Connections Exercise 5, the period T of such a pendulum is given by L π/2 1 T =4 (7) dφ g 0 1 − k 2 sin2 φ

where

Figure 9.10.1

L = length of the supporting rod g = acceleration due to gravity k = sin(θ0 /2), where θ0 is the initial angle of displacement from the vertical The integral, which is called a complete elliptic integral of the first kind, cannot be expressed in terms of elementary functions and is often approximated by numerical methods. Unfortunately, numerical values are so specific that they often give little insight into general physical principles. However, if we expand the integrand of (7) in a series and integrate term by term, then we can generate an infinite series that can be used to construct various mathematical models for the period T that give a deeper understanding of the behavior of the pendulum.

686

Chapter 9 / Infinite Series

To obtainâ&#x2C6;&#x161;a series for the integrand, we will substitute â&#x2C6;&#x2019;k 2 sin2 Ď&#x2020; for x in the binomial series for 1/ 1 + x that we derived in Example 4(b) of Section 9.9. If we do this, then we can rewrite (7) as 1Âˇ3 4 4 1Âˇ3Âˇ5 6 6 L Ď&#x20AC;/2 1 2 2 k sin Ď&#x2020; + Âˇ Âˇ Âˇ dĎ&#x2020; (8) 1 + k sin Ď&#x2020; + 2 k sin Ď&#x2020; + 3 T =4 g 0 2 2 2! 2 3! If we integrate term by term, then we can produce a series that converges to the period T . However, one of the most important cases of pendulum motion occurs when the initial displacement is small, in which case all subsequent displacements are small, and we can assume that k = sin(Î¸0 /2) â&#x2030;&#x2C6; 0. In this case we expect the convergence of the series for T to be rapid, and we can approximate the sum of the series by dropping all but the constant term in (8). This yields L T = 2Ď&#x20AC; (9) g

ÂŠ ACE STOCK LIMITED/Alamy

Understanding the motion of a pendulum played a critical role in the advance of accurate timekeeping with the development of the pendulum clock in the seventeenth century.

which is called the ďŹ rst-order model of T or the model for small vibrations. This model can be improved on by using more terms in the series. For example, if we use the ďŹ rst two terms in the series, we obtain the second-order model

L k2 1+ (10) T = 2Ď&#x20AC; g 4 (verify).

â&#x153;&#x201D;QUICK CHECK EXERCISES 9.10 1. The Maclaurin series for e for x in the series

â&#x2C6;&#x2019;x 2

e =

obtained by substituting â&#x2C6;&#x2019;x

k=0

âŹ xk k=0

âŹ xk k+1 k=0

x2 x x2 = 1+x+ + ÂˇÂˇÂˇ 1+ + + ÂˇÂˇÂˇ 2! 2 3

x +

x + ÂˇÂˇÂˇ

âŹ

(a) f â&#x20AC;˛â&#x20AC;˛(1) = (b) f(x) =

CAS

1. In each part, obtain the Maclaurin series for the function by making an appropriate substitution in the Maclaurin series for 1/(1 â&#x2C6;&#x2019; x). Include the general term in your answer, and state the radius of convergence of the series. 1 1 1 1 (a) (c) (b) (d) 2 1+x 1â&#x2C6;&#x2019;x 1 â&#x2C6;&#x2019; 2x 2â&#x2C6;&#x2019;x

2. In each part, obtain the Maclaurin series for the function by making an appropriate substitution in the Maclaurin series for ln(1 + x). Include the general term in your answer, and

(x â&#x2C6;&#x2019; 1)2 + +

âŹ (â&#x2C6;&#x2019;1)k (x â&#x2C6;&#x2019; 1)k (k + 1)! k=0

(x â&#x2C6;&#x2019; 1)

+ +

k=0

x2 + Âˇ Âˇ Âˇ

4. Suppose that f(1) = 4 and f â&#x20AC;˛(x) =

EXERCISE SET 9.10

âŹ xk

2 is eâ&#x2C6;&#x2019;x = âŹ k=0 . âŹ d xk 2. = (â&#x2C6;&#x2019;1)k+1 dx k=1 k =

(See page 689 for answers.)

(x â&#x2C6;&#x2019; 1)3 + Âˇ Âˇ Âˇ

âŹ k=1

state the radius of convergence of the series. (a) ln(1 â&#x2C6;&#x2019; x) (b) ln(1 + x 2 ) (c) ln(1 + 2x) (d) ln(2 + x) 3. In each part, obtain the ďŹ rst four nonzero terms of the Maclaurin series for the function by making an appropriate substitution in one of the binomial series obtained in Example 4 of Section 9.9. (a) (2 + x)â&#x2C6;&#x2019;1/2 (b) (1 â&#x2C6;&#x2019; x 2 )â&#x2C6;&#x2019;2

9.10 Differentiating and Integrating Power Series; Modeling with Taylor Series

4. (a) Use the Maclaurin series for 1/(1 − x) to find the Maclaurin series for 1/(a − x), where a = 0, and state the radius of convergence of the series. (b) Use the binomial series for 1/(1 + x)2 obtained in Example 4 of Section 9.9 to find the first four nonzero terms in the Maclaurin series for 1/(a + x)2 , where a = 0, and state the radius of convergence of the series. 5–8 Find the first four nonzero terms of the Maclaurin series for the function by making an appropriate substitution in a known Maclaurin series and performing any algebraic operations that are required. State the radius of convergence of the series. ■ 2

5. (a) sin 2x

(b) e−2x

6. (a) cos 2x x2 7. (a) 1 + 3x x 8. (a) x−1

(b) x 2 ex

(d) x 2 cos πx (d) sin(x 2 ) 2 3/2

(b) x sinh 2x

(b) 3 cosh(x 2 )

(c)

x (1 + 2x)3

9–10 Find the ﬁrst four nonzero terms of the Maclaurin series

for the function by using an appropriate trigonometric identity or property of logarithms and then substituting in a known Maclaurin series. ■ 9. (a) sin2 x

(b) ln[(1 + x 3 )12 ]

1−x 10. (a) cos2 x (b) ln 1+x 11. (a) Use a known Maclaurin series to ﬁnd the Taylor series of 1/x about x = 1 by expressing this function as 1 1 = x 1 − (1 − x)

(b) Find the interval of convergence of the Taylor series. 12. Use the method of Exercise 11 to ﬁnd the Taylor series of 1/x about x = x0 , and state the interval of convergence of the Taylor series. 13–14 Find the ﬁrst four nonzero terms of the Maclaurin series for the function by multiplying the Maclaurin series of the factors. ■ √ 13. (a) ex sin x (b) 1 + x ln(1 + x) 2

14. (a) e−x cos x

(b) (1 + x 2 )4/3 (1 + x)1/3

15–16 Find the ﬁrst four nonzero terms of the Maclaurin series for the function by dividing appropriate Maclaurin series. ■

1 sin x 15. (a) sec x = (b) cos x ex −1 ln(1 + x) tan x (b) 16. (a) 1+x 1−x 17. Use the Maclaurin series for ex and e−x to derive the Maclaurin series for sinh x and cosh x. Include the general terms in your answers and state the radius of convergence of each series.

18. Use the Maclaurin series for sinh x and cosh x to obtain the ﬁrst four nonzero terms in the Maclaurin series for tanh x.

687

19–20 Find the ﬁrst ﬁve nonzero terms of the Maclaurin series for the function by using partial fractions and a known Maclaurin series. ■

19.

4x − 2 x2 − 1

20.

x 3 + x 2 + 2x − 2 x2 − 1

21–22 Conﬁrm the derivative formula by differentiating the appropriate Maclaurin series term by term. ■ d 1 d [cos x] = − sin x (b) [ln(1 + x)] = 21. (a) dx dx 1+x d d 1 22. (a) [sinh x] = cosh x (b) [tan−1 x] = dx dx 1 + x2 23–24 Conﬁrm the integration formula by integrating the appropriate Maclaurin series term by term. ■ 23. (a) ex dx = ex + C (b) sinh x dx = cosh x + C 24. (a) sin x dx = − cos x + C 1 (b) dx = ln(1 + x) + C 1+x 25. Consider the series ⬁ x k+1 (k + 1)(k + 2) k=0

Determine the intervals of convergence for this series and for the series obtained by differentiating this series term by term.

26. Consider the series ⬁ (−3)k k=1

Determine the intervals of convergence for this series and for the series obtained by integrating this series term by term. 27. (a) Use the Maclaurin series for 1/(1 − x) to find the Maclaurin series for x f(x) = 1 − x2 (b) Use the Maclaurin series obtained in part (a) to find f (5) (0) and f (6) (0). (c) What can you say about the value of f (n) (0)? 28. Let f(x) = x 2 cos 2x. Use the method of Exercise 27 to find f (99) (0). 29–30 The limit of an indeterminate form as x → x0 can sometimes be found by expanding the functions involved in Taylor series about x = x0 and taking the limit of the series term by term. Use this method to find the limits in these exercises. ■ sin x tan−1 x − x 29. (a) lim (b) lim x →0 x x →0 x3 √ ln 1 + x − sin 2x 1 − cos x (b) lim 30. (a) lim x → 0 x →0 sin x x

Chapter 9 / Infinite Series

688

31–34 Use Maclaurin series to approximate the integral to three decimal-place accuracy. ■ 1/2 1 2 tan−1 (2x 2 ) dx sin(x ) dx 32. 31. 0

33.

0.2

40. (a) Use the relationship 1 dx = sin−1 x + C √ 1 − x2 to ﬁnd the ﬁrst four nonzero terms in the Maclaurin series for sin−1 x. (b) Express the series in sigma notation. (c) What is the radius of convergence?

3 1 + x 4 dx

34.

1/2

dx √ 4 2 x +1

F O C U S O N C O N C E P TS

41. We showed by Formula (19) of Section 8.2 that if there are y0 units of radioactive carbon-14 present at time t = 0, then the number of units present t years later is

35. (a) Find the Maclaurin series for e . What is the radius of convergence? (b) Explain two different ways to use the Maclaurin se4 4 ries for ex to ﬁnd a series for x 3 ex . Conﬁrm that both methods produce the same series. 36. (a) Differentiate the Maclaurin series for 1/(1 − x), and use the result to show that ⬁ x for −1 < x < 1 kx k = (1 − x)2 k=1 (b) Integrate the Maclaurin series for 1/(1 − x), and use the result to show that ⬁ xk = − ln(1 − x) for −1 < x < 1 k k=1 (c) Use the result in part (b) to show that ⬁ xk = ln(1 + x) (−1)k+1 k k=1

for −1 < x < 1

(d) Show that the series in part (c) converges if x = 1. (e) Use the remark following Example 3 to show that ⬁ xk = ln(1 + x) (−1)k+1 k k=1

for −1 < x ≤ 1

37. Use the results in Exercise 36 to ﬁnd the sum of the series. ⬁ 1 3 4 2 k = + 2 + 3 + 4 + ··· (a) k 3 3 3 3 3 k=1 ⬁ 1 1 1 1 1 = + + + + ··· (b) k 2 3 k(4 ) 4 2(4 ) 3(4 ) 4(44 ) k=1

38. Use the results in Exercise 36 to find the sum of each series. ⬁ 1 1 1 1 (a) (−1)k+1 = 1 − + − + · · · k 2 3 4 k=1 ⬁ k e − 1 (e − 1)2 (e − 1)3 (e − 1) = + − + ··· (b) k 2 ke e 2(e ) 3(e3 ) k=1 39. (a) Use the relationship 1 dx = sinh−1 x + C √ 1 + x2 to find the first four nonzero terms in the Maclaurin series for sinh−1 x. (b) Express the series in sigma notation. (c) What is the radius of convergence?

y(t) = y0 e−0.000121t

(a) Express y(t) as a Maclaurin series. (b) Use the ﬁrst two terms in the series to show that the number of units present after 1 year is approximately (0.999879)y0 . (c) Compare this to the value produced by the formula for y(t). C

42. Suppose that a simple pendulum with a length of L = 1 meter is given an initial displacement of θ0 = 5 ◦ from the vertical. (a) Approximate the period T of the pendulum using Formula (9) for the ﬁrst-order model of T . [Note: Take g = 9.8 m/s2 .] (b) Approximate the period of the pendulum using Formula (10) for the second-order model. (c) Use the numerical integration capability of a CAS to approximate the period of the pendulum from Formula (7), and compare it to the values obtained in parts (a) and (b). 43. Use the ﬁrst three nonzero terms in Formula (8) and the Wallis sine formula in the Endpaper Integral Table (Formula 122) to obtain a model for the period of a simple pendulum. 44. Recall that the gravitational force exerted by the Earth on an object is called the object’s weight (or more precisely, its Earth weight). If an object of mass m is on the surface of the Earth (mean sea level), then the magnitude of its weight is mg, where g is the acceleration due to gravity at the Earth’s surface. A more general formula for the magnitude of the gravitational force that the Earth exerts on an object of mass m is mgR 2 F = (R + h)2 where R is the radius of the Earth and h is the height of the object above the Earth’s surface. (a) Use the binomial series for 1/(1 + x)2 obtained in Example 4 of Section 9.9 to express F as a Maclaurin series in powers of h/R. (b) Show that if h = 0, then F = mg. (c) Show that if h/R ≈ 0, then F ≈ mg − (2mgh/R). [Note: The quantity 2mgh/R can be thought of as a “correction term” for the weight that takes the object’s height above the Earth’s surface into account.] (d) If we assume that the Earth is a sphere of radius R = 4000 mi at mean sea level, by approximately what

Chapter 9 Review Exercises

percentage does a person’s weight change in going from mean sea level to the top of Mt. Everest (29,028 ft)? 45. (a) Show that the Bessel function J0 (x) given by Formula (4) of Section 9.8 satisﬁes the differential equation xy ′′ + y ′ + xy = 0. (This is called the Bessel equation of order zero.) (b) Show that the Bessel function J1 (x) given by Formula (5) of Section 9.8 satisﬁes the differential equation x 2 y ′′ + xy ′ + (x 2 − 1)y = 0. (This is called the Bessel equation of order one.) (c) Show that J0′ (x) = −J1 (x).

689

46. Prove: If the power series ⬁k=0 ak x k and ⬁k=0 bk x k have the same sum on an interval (−r, r), then ak = bk for all values of k. 47. Writing Evaluate the limit x − sin x lim x →0 x3 in two ways: using L’Hôpital’s rule and by replacing sin x by its Maclaurin series. Discuss how the use of a series can give qualitative information about how the value of an indeterminate limit is approached.

✔QUICK CHECK ANSWERS 9.10 1. (−1)k

x 2k k!

2. 1; −1; 1; −1; (−1)k x k

3. 1;

3 4 ; 2 3

4. (a) −

(x − 1)k 1 1 1 (b) 4; 1; − ; ; 4; (−1)k+1 2 4 18 k · (k!)

CHAPTER 9 REVIEW EXERCISES 1. What is the difference between an infinite sequence and an infinite series? 2. What is meant by the sum of an infinite series? 3. (a) What is a geometric series? Give some examples of convergent and divergent geometric series. (b) What is a p-series? Give some examples of convergent and divergent p-series. 4. State conditions under which an alternating series is guaranteed to converge. 5. (a) What does it mean to say that an infinite series converges absolutely? (b) What relationship exists between convergence and absolute convergence of an infinite series? 6. State the Remainder Estimation Theorem, and describe some of its uses. 7. If a power series in x − x0 has radius of convergence R, what can you say about the set of x-values at which the series converges? 8. (a) Write down the formula for the Maclaurin series for f in sigma notation. (b) Write down the formula for the Taylor series for f about x = x0 in sigma notation.

9. Are the following statements true or false? If true, state a theorem to justify your conclusion; if false, then give a counterexample. (a) If uk converges, then uk → 0 as k → +⬁. (b) If uk → 0 as k → +⬁, then uk converges. (c) If f(n) = an for n = 1, 2, 3, . . . , and if an → L as n → +⬁, then f(x) → L as x → +⬁. (d) If f(n) = an for n = 1, 2, 3, . . . , and if f(x) → L as x → +⬁, then an → L as n → +⬁. (e) If 0 < an < 1, then {a n } converges. (f ) If 0 < uk < 1, then uk converges.

(g) If uk and vk converge, then (uk + vk ) diverges. (h) If uk and vk diverge, then (uk − vk ) converges. (i) If 0 ≤ uk ≤ vk and vk converges, then uk converges. ( j) If 0 ≤ uk ≤ vk and uk diverges, then vk diverges. (k) If an inﬁnite series converges, then it converges absolutely. (l) If an inﬁnite series diverges absolutely, then it diverges. 10. State whether each of the following is true or false. Justify your answers. (a) The function f(x) = x 1/3 has a Maclaurin series. (b) 1 + 21 − 21 + 31 − 13 + 41 − 41 + · · · = 1 (c) 1 + 21 − 21 + 21 − 21 + 21 − 21 + · · · = 1

11. Find the general term of the sequence, starting with n = 1, determine whether the sequence converges, and if so find its limit. 3 4 5 (a) 2 , , ,... 2 − 1 2 32 − 2 2 4 2 − 3 2 1 2 3 4 (b) , − , , − , . . . 3 5 7 9 12. Suppose that the sequence {ak } is defined recursively by √ a0 = c, ak+1 = ak Assuming that the sequence converges, find its limit if (a) c = 21 (b) c = 23 .

13. Show that the sequence is eventually strictly monotone. +⬁ +⬁ 100n (a) (n − 10)4 n=0 (b) (2n)!(n!) n=1 14. (a) Give an example of a bounded sequence that diverges. (b) Give an example of a monotonic sequence that diverges. 15–20 Use any method to determine whether the series converge. ■

690

Chapter 9 / Infinite Series

15. (a) 16. (a) 17. (a)

⬁ 1 5k k=1

(b)

⬁ k+4 (−1)k 2 k +k k=1 ⬁ k=1

(b)

1 k 3 + 2k + 1

(b)

⬁ ln k 18. (a) √ k k k=1

19. (a) 20. (a)

⬁

k=1 ⬁ k=1

√

(b)

⬁

k=1 ⬁

(b)

k+1

k −1/2 2 + sin2 k

(b)

1 +1

k=1

⬁ k=1

1 (3 + k)2/5

⬁

k 4 /3 8k 2 + 5k + 1

(−1)k+1

k=1

k+2 3k − 1

⬁ cos(1/k)

k=1 ⬁ k=1

(−1)k+1 k2 + 1

21. Find the exact error that results when the sum of the geometric series ⬁k=0 (1/5)k is approximated by the sum of the ﬁrst 100 terms in the series. n 1 22. Suppose that uk = 2 − . Find n k=1 ⬁ uk . (c) (b) lim uk (a) u100 k → +⬁

k=1

23. In each part, determine whether the series converges; if so, ﬁnd its sum. ⬁

⬁ 3 2 [ln(k + 1) − ln k] − (b) (a) 2k 3k k=1 k=1 ⬁ ⬁ 1 [tan−1 (k + 1) − tan−1 k] (d) (c) k(k + 2) k=1 k=1

24. It can be proved that

√ n n! = +⬁ and lim

n → +⬁

√ n 1 n! lim = n → +⬁ n e

In each part, use these limits and the root test to determine whether the series converges. ⬁ ⬁ 2k kk (a) (b) k! k! k=0 k=0

25. Let a, b, and p be positive constants. For which values of ⬁ 1 converge? p does the series (a + bk)p k=1

26. Find the interval of convergence of ⬁ (x − x0 )k k=0

29. (a) Find the ﬁrst ﬁve Maclaurin polynomials of the function p(x) = 1 − 7x + 5x 2 + 4x 3 . (b) Make a general statement about the Maclaurin polynomials of a polynomial of degree n. 30. Show that the approximation sin x ≈ x −

is accurate to four decimal places if 0 ≤ x ≤ π/4. 31. Use a Maclaurin series and properties of alternating series to show that | ln(1 + x) − x| ≤ x 2 /2 if 0 < x < 1. 32. Use Maclaurin series to approximate the integral 1 1 − cos x dx x 0 to three decimal-place accuracy.

33. In parts (a)–(d), find the sum of the series by associating it with some Maclaurin series. 4 8 16 (a) 2 + + + + ··· 2! 3! 4! π5 π7 π3 + − + ··· (b) π − 3! 5! 7! e2 e4 e6 (c) 1 − + − + ··· 2! 4! 6! (ln 3)2 (ln 3)3 (d) 1 − ln 3 + − + ··· 2! 3! 34. In each part, write out the first four terms of the series, and then find the radius of convergence. ⬁ 1 · 2 · 3···k (a) xk 1 · 4 · 7 · · · (3k − 2) k=1 ⬁ 1 · 2 · 3···k (−1)k x 2k+1 (b) 1 · 3 · 5 · · · (2k − 1) k=1 √ √ 35. Use an appropriate Taylor series for 3 x to approximate 3 28 to three decimal-place accuracy, and check your answer by comparing it to that produced directly by your calculating utility. 36. Differentiate the Maclaurin series for xex and use the result ⬁ to show that k+1 = 2e k! k=0

37. Use the supplied Maclaurin series for sin x and cos x to ﬁnd the ﬁrst four nonzero terms of the Maclaurin series for the given functions.

(b > 0)

27. (a) Show that k k ≥ k!. (b) Use the comparison test to show that

⬁

sin x = cos x =

k −k converges.

k=1

(c) Use the root test to show that the series converges. 28. Does the series 1 − 23 + 53 − 47 + 59 + · · · converge? Justify your answer.

x3 x5 + 3! 5!

(a) sin x cos x

⬁ x 2k+1 (−1)k (2k + 1)! k=0

⬁ x 2k (−1)k (2k)! k=0

(b)

1 2

sin 2x

Chapter 9 Making Connections

691

CHAPTER 9 MAKING CONNECTIONS 1. As shown in the accompanying ďŹ gure, suppose that lines L1 and L2 form an angle Î¸ , 0 < Î¸ < Ď&#x20AC;/2, at their point of intersection P . A point P0 is chosen that is on L1 and a units from P . Starting from P0 a zig-zag path is constructed by successively going back and forth between L1 and L2 along a perpendicular from one line to the other. Find the following sums in terms of Î¸ and a. (a) P0 P1 + P1 P2 + P2 P3 + Âˇ Âˇ Âˇ (b) P0 P1 + P2 P3 + P4 P5 + Âˇ Âˇ Âˇ (c) P1 P2 + P3 P4 + P5 P6 + Âˇ Âˇ Âˇ L2

P0 P2 P4 P6 a

Figure Ex-1

2. (a) Find A and B such that 2k B 2k A + (3k+1 â&#x2C6;&#x2019; 2k+1 )(3k â&#x2C6;&#x2019; 2k ) 3k â&#x2C6;&#x2019; 2 k 3k+1 â&#x2C6;&#x2019; 2k+1 (b) Use the result in part (a) to ďŹ nd a closed form for the nth partial sum of the series 6k

âŹ k=1

6k (3k+1 â&#x2C6;&#x2019; 2k+1 )(3k â&#x2C6;&#x2019; 2k )

and then ďŹ nd the sum of the series. Source: This exercise is adapted from a problem that appeared in the Forty-Fifth Annual William Lowell Putnam Competition.

3. Show that the alternating p-series 1 1 1 1 + p â&#x2C6;&#x2019; p + Âˇ Âˇ Âˇ + (â&#x2C6;&#x2019;1)k+1 p + Âˇ Âˇ Âˇ p 2 3 4 k converges absolutely if p > 1, converges conditionally if 0 < p â&#x2030;¤ 1, and diverges if p â&#x2030;¤ 0. 1â&#x2C6;&#x2019;

4. As illustrated in the accompanying ďŹ gure, a bug, starting at point A on a 180 cm wire, walks the length of the wire, stops and walks in the opposite direction for half the length of the wire, stops again and walks in the opposite direction for onethird the length of the wire, stops again and walks in the opposite direction for one-fourth the length of the wire, and so forth until it stops for the 1000th time.

(a) Give upper and lower bounds on the distance between the bug and point A when it ďŹ nally stops. [Hint: As stated in Example 2 of Section 9.6, assume that the sum of the alternating harmonic series is ln 2.] (b) Give upper and lower bounds on the total distance that the bug has traveled when it ďŹ nally stops. [Hint: Use inequality (2) of Section 9.4.] A

180 cm Figure Ex-4

5. In Section 6.6 we deďŹ ned the kinetic energy K of a particle with mass m and velocity v to be K = 21 mv 2 [see Formula (7) of that section]. In this formula the mass m is assumed to be constant, and K is called the Newtonian kinetic energy. However, in Albert Einsteinâ&#x20AC;&#x2122;s relativity theory the mass m increases with the velocity and the kinetic energy K is given by the formula 1 2 K = m0 c â&#x2C6;&#x2019;1 1 â&#x2C6;&#x2019; (v /c)2 in which m0 is the mass of the particle when its velocity is zero, and c is the speed of light. This is called the relativistic kinetic energy. Use an appropriate binomial series to show that if the velocity is small compared to the speed of light (i.e., v /c â&#x2030;&#x2C6; 0), then the Newtonian and relativistic kinetic energies are in close agreement.

6. In Section 8.4 we studied the motion of a falling object that has mass m and is retarded by air resistance. We showed that if the initial velocity is v0 and the drag force FR is proportional to the velocity, that is, FR = â&#x2C6;&#x2019;cv, then the velocity of the object at time t is mg mg v(t) = eâ&#x2C6;&#x2019;ct /m v0 + â&#x2C6;&#x2019; c c where g is the acceleration due to gravity [see Formula (16) of Section 8.4]. (a) Use a Maclaurin series to show that if ct /m â&#x2030;&#x2C6; 0, then the velocity can be approximated as cv 0 v(t) â&#x2030;&#x2C6; v0 â&#x2C6;&#x2019; +g t m (b) Improve on the approximation in part (a).

EXPANDING THE CALCULUS HORIZON To learn how ecologists use mathematical models based on the process of iteration to study the growth and decline of animal populations, see the module entitled Iteration and Dynamical Systems at: www.wiley.com/college/anton

10 PARAMETRIC AND POLAR CURVES; CONIC SECTIONS Gilbert S. Grant/Photo Researchers, Inc.

Mathematical curves, such as the spirals in the center of a sunﬂower, can be described conveniently using ideas developed in this chapter.

10.1

In this chapter we will study alternative ways of expressing curves in the plane. We will begin by studying parametric curves: curves described in terms of component functions. This study will include methods for finding tangent lines to parametric curves. We will then introduce polar coordinate systems and discuss methods for finding tangent lines to polar curves, arc length of polar curves, and areas enclosed by polar curves. Our attention will then turn to a review of the basic properties of conic sections: parabolas, ellipses, and hyperbolas. Finally, we will consider conic sections in the context of polar coordinates and discuss some applications in astronomy.

PARAMETRIC EQUATIONS; TANGENT LINES AND ARC LENGTH FOR PARAMETRIC CURVES Graphs of functions must pass the vertical line test, a limitation that excludes curves with self-intersections or even such basic curves as circles. In this section we will study an alternative method for describing curves algebraically that is not subject to the severe restriction of the vertical line test. We will then derive formulas required to ﬁnd slopes, tangent lines, and arc lengths of these parametric curves. We will conclude with an investigation of a classic parametric curve known as the cycloid. PARAMETRIC EQUATIONS

Suppose that a particle moves along a curve C in the xy-plane in such a way that its x- and y-coordinates, as functions of time, are x = f (t),

(x, y) x

A moving particle with trajectory C

Figure 10.1.1

692

y = g(t)

We call these the parametric equations of motion for the particle and refer to C as the trajectory of the particle or the graph of the equations (Figure 10.1.1). The variable t is called the parameter for the equations. Example 1 Sketch the trajectory over the time interval 0 ≤ t ≤ 10 of the particle whose parametric equations of motion are x = t − 3 sin t,

y = 4 − 3 cos t

(1)

10.1 Parametric Equations; Tangent Lines and Arc Length for Parametric Curves

693

Solution. One way to sketch the trajectory is to choose a representative succession of times, plot the (x, y) coordinates of points on the trajectory at those times, and connect the points with a smooth curve. The trajectory in Figure 10.1.2 was obtained in this way from the data in Table 10.1.1 in which the approximate coordinates of the particle are given at time increments of 1 unit. Observe that there is no t-axis in the picture; the values of t appear only as labels on the plotted points, and even these are usually omitted unless it is important to emphasize the locations of the particle at speciﬁc times. Table 10.1.1

y 8

t=2 T E C H N O LO GY M A ST E R Y

t=3

t=1

t = 10

t=8

Read the documentation for your graphing utility to learn how to graph parametric equations, and then generate the trajectory in Example 1. Explore the behavior of the particle beyond time t = 10.

t=9 t=4

t=5

t=0 −2

t=7 4

t=6 6

x 10

Figure 10.1.2

0 1 2 3 4 5 6 7 8 9 10

x 0.0 −1.5 −0.7 2.6 6.3 7.9 6.8 5.0 5.0 7.8 11.6

y 1.0 2.4 5.2 7.0 6.0 3.1 1.1 1.7 4.4 6.7 6.5

Although parametric equations commonly arise in problems of motion with time as the parameter, they arise in other contexts as well. Thus, unless the problem dictates that the parameter t in the equations x = f (t), y = g(t) represents time, it should be viewed simply as an independent variable that varies over some interval of real numbers. (In fact, there is no need to use the letter t for the parameter; any letter not reserved for another purpose can be used.) If no restrictions on the parameter are stated explicitly or implied by the equations, then it is understood that it varies from −⬁ to +⬁. To indicate that a parameter t is restricted to an interval [a, b], we will write x = f (t), Example 2

x = cos t, y = sin t (0 ≤ t ≤ 2c) Figure 10.1.3

y = sin t

(0 ≤ t ≤ 2π)

(2)

Solution. One way to ﬁnd the graph is to eliminate the parameter t by noting that

(x, y)

t x

(a ≤ t ≤ b)

Find the graph of the parametric equations x = cos t,

y = g(t)

(1, 0)

x 2 + y 2 = sin2 t + cos2 t = 1

Thus, the graph is contained in the unit circle x 2 + y 2 = 1. Geometrically, the parameter t can be interpreted as the angle swept out by the radial line from the origin to the point (x, y) = (cos t, sin t) on the unit circle (Figure 10.1.3). As t increases from 0 to 2π, the point traces the circle counterclockwise, starting at (1, 0) when t = 0 and completing one full revolution when t = 2π. One can obtain different portions of the circle by varying the interval over which the parameter varies. For example, x = cos t,

y = sin t

represents just the upper semicircle in Figure 10.1.3.

(0 ≤ t ≤ π)

(3)

Chapter 10 / Parametric and Polar Curves; Conic Sections

694

(1, 0)

x −t 1

ORIENTATION The direction in which the graph of a pair of parametric equations is traced as the parameter increases is called the direction of increasing parameter or sometimes the orientation imposed on the curve by the equations. Thus, we make a distinction between a curve, which is a set of points, and a parametric curve, which is a curve with an orientation imposed on it by a set of parametric equations. For example, we saw in Example 2 that the circle represented parametrically by (2) is traced counterclockwise as t increases and hence has counterclockwise orientation. As shown in Figures 10.1.2 and 10.1.3, the orientation of a parametric curve can be indicated by arrowheads. To obtain parametric equations for the unit circle with clockwise orientation, we can replace t by −t in (2) and use the identities cos(−t) = cos t and sin(−t) = − sin t. This yields x = cos t, y = − sin t (0 ≤ t ≤ 2π)

y (x, y)

x = cos(−t), y = sin(−t) (0 ≤ t ≤ 2c)

Here, the circle is traced clockwise by a point that starts at (1, 0) when t = 0 and completes one full revolution when t = 2π (Figure 10.1.4).

Figure 10.1.4

T E C H N O L O GY M A ST E R Y

When parametric equations are graphed using a calculator, the orientation can often be determined by watching the direction in which the graph is traced on the screen. However, many computers graph so fast that it is often hard to discern the orientation. See if you can use your graphing utility to confirm that (3) has a counterclockwise orientation.

Example 3

Graph the parametric curve

x = 2t − 3,

y = 6t − 7

by eliminating the parameter, and indicate the orientation on the graph.

Solution. To eliminate the parameter we will solve the ﬁrst equation for t as a function

of x, and then substitute this expression for t into the second equation: t = 21 (x + 3) y = 6 21 (x + 3) − 7 y = 3x + 2

Thus, the graph is a line of slope 3 and y-intercept 2. To ﬁnd the orientation we must look to the original equations; the direction of increasing t can be deduced by observing that x increases as t increases or by observing that y increases as t increases. Either piece of information tells us that the line is traced left to right as shown in Figure 10.1.5.

x = 2t − 3, y = 6t − 7 Figure 10.1.5

REMARK

(−1, 1)

(1, 1)

Not all parametric equations produce curves with definite orientations; if the equations are badly behaved, then the point tracing the curve may leap around sporadically or move back and forth, failing to determine a definite direction. For example, if

x = sin t, y = sin2 t

Figure 10.1.6

then the point (x, y) moves along the parabola y = x 2 . However, the value of x varies periodically between −1 and 1, so the point (x, y) moves periodically back and forth along the parabola between the points (−1, 1) and (1, 1) (as shown in Figure 10.1.6). Later in the text we will discuss restrictions that eliminate such erratic behavior, but for now we will just avoid such complications.

EXPRESSING ORDINARY FUNCTIONS PARAMETRICALLY An equation y = f(x) can be expressed in parametric form by introducing the parameter t = x; this yields the parametric equations

x = t,

y = f (t)

10.1 Parametric Equations; Tangent Lines and Arc Length for Parametric Curves

695

For example, the portion of the curve y = cos x over the interval [−2π, 2π] can be expressed parametrically as x = t, y = cos t (−2π ≤ t ≤ 2π) (Figure 10.1.7). y

t = −2c

t = 2c

t=0

x −7

t = −c

Figure 10.1.7

−1

t=c

f 8

f −1

If a function f is one-to-one, then it has an inverse function f −1 . In this case the equation y = f −1 (x) is equivalent to x = f(y). We can express the graph of f −1 in parametric form by introducing the parameter y = t; this yields the parametric equations x = f(t),

x −6

y=t

For example, Figure 10.1.8 shows the graph of f(x) = x 5 + x + 1 and its inverse. The graph of f can be repesented parametrically as

x = t,

−6

y = t5 + t + 1

and the graph of f −1 can be represented parametrically as x = t 5 + t + 1,

Figure 10.1.8

y=t

TANGENT LINES TO PARAMETRIC CURVES We will be concerned with curves that are given by parametric equations

x = f(t),

y = g(t)

in which f(t) and g(t) have continuous ﬁrst derivatives with respect to t. It can be proved that if dx /dt = 0, then y is a differentiable function of x, in which case the chain rule implies that dy dy/dt (4) = dx dx/dt This formula makes it possible to ﬁnd dy/dx directly from the parametric equations without eliminating the parameter. Example 4

Find the slope of the tangent line to the unit circle x = cos t,

y = sin t

(0 ≤ t ≤ 2π)

at the point where t = π/6 (Figure 10.1.9). 1 c/6

Solution. From (4), the slope at a general point on the circle is dy dy/dt cos t = = = − cot t dx dx/dt − sin t

Thus, the slope at t = π/6 is Figure 10.1.9

√ dy π = − cot = − 3 dx t=π/6 6

(5)

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Chapter 10 / Parametric and Polar Curves; Conic Sections

Note that Formula (5) makes sense geometrically because the radius from the origin to the point P (cos t, sin t) has slope m = tan t . Thus the tangent line at P , being perpendicular to the radius, has slope

−

1 1 =− = − cot t m tan t

It follows from Formula (4) that the tangent line to a parametric curve will be horizontal at those points where dy/dt = 0 and dx/dt = 0, since dy/dx = 0 at such points. Two different situations occur when dx/dt = 0. At points where dx/dt = 0 and dy/dt = 0, the right side of (4) has a nonzero numerator and a zero denominator; we will agree that the curve has inﬁnite slope and a vertical tangent line at such points. At points where dx/dt and dy/dt are both zero, the right side of (4) becomes an indeterminate form; we call such points singular points. No general statement can be made about the behavior of parametric curves at singular points; they must be analyzed case by case.

(Figure 10.1.10).

Example 5 In a disastrous ﬁrst ﬂight, an experimental paper airplane follows the trajectory of the particle in Example 1:

x = t − 3 sin t, 1 t

P(cos t, sin t) x

Radius OP has slope m = tan t.

Figure 10.1.10

y = 4 − 3 cos t

(t ≥ 0)

but crashes into a wall at time t = 10 (Figure 10.1.11). (a) At what times was the airplane ﬂying horizontally? (b) At what times was it ﬂying vertically?

Solution (a). The airplane was ﬂying horizontally at those times when dy/dt = 0 and dx/dt = 0. From the given trajectory we have

dy dx = 1 − 3 cos t (6) = 3 sin t and dt dt Setting dy/dt = 0 yields the equation 3 sin t = 0, or, more simply, sin t = 0. This equation has four solutions in the time interval 0 ≤ t ≤ 10: t = 0,

t = π,

t = 2π,

t = 3π

Since dx/dt = 1 − 3 cos t = 0 for these values of t (verify), the airplane was ﬂying horizontally at times t = 0,

t = π ≈ 3.14,

t = 2π ≈ 6.28,

and t = 3π ≈ 9.42

which is consistent with Figure 10.1.11. Stanislovas Kairys/iStockphoto

The complicated motion of a paper airplane is best described mathematically using parametric equations.

Solution (b). The airplane was ﬂying vertically at those times when dx/dt = 0 and dy/dt = 0. Setting dx/dt = 0 in (6) yields the equation 1 − 3 cos t = 0

cos t =

1 3

This equation has three solutions in the time interval 0 ≤ t ≤ 10 (Figure 10.1.12): t = 2π − cos−1 13 ,

t = cos−1 31 ,

t = 2π + cos−1

1 3

y 8

t=3

t=9

t=2

t = 10

t=4 t=8

1 3

t=5

t=1 t=0 −2

Figure 10.1.11

t=7

y = cos t

t=6

x 12

cos−1 13 −1

Figure 10.1.12

x o

10.1 Parametric Equations; Tangent Lines and Arc Length for Parametric Curves

697

Since dy/dt = 3 sin t is not zero at these points (why?), it follows that the airplane was ﬂying vertically at times t = cos−1

1 3

≈ 1.23,

t ≈ 2π − 1.23 ≈ 5.05,

t ≈ 2π + 1.23 ≈ 7.51

which again is consistent with Figure 10.1.11. y 6

Example 6 The curve represented by the parametric equations x = t 2, x 5

y = t3

(−⬁ < t < +⬁)

is called a semicubical parabola. The parameter t can be eliminated by cubing x and squaring y, from which it follows that y 2 = x 3 . The graph of this equation, shown in Figure 10.1.13, consists of two branches: an upper branch obtained by graphing y = x 3/2 and a lower branch obtained by graphing y = −x 3/2 . The two branches meet at the origin, which corresponds to t = 0 in the parametric equations. This is a singular point because the derivatives dx/dt = 2t and dy/dt = 3t 2 are both zero there.

−6

x = t 2, y = t 3 (− ∞ < t < + ∞) Figure 10.1.13

Example 7 Without eliminating the parameter, ﬁnd dy/dx and d 2 y/dx 2 at (1, 1) and (1, −1) on the semicubical parabola given by the parametric equations in Example 6.

Solution. From (4) we have dy/dt 3t 2 3 dy = = = t dx dx/dt 2t 2

(t = 0)

(7)

and from (4) applied to y ′ = dy/dx we have

WARNING Although it is true that

dy dy/dt = dx/dt dx you cannot conclude that d 2 y/dx 2 is the quotient of d 2 y/dt 2 and d 2 x/dt 2 . To illustrate that this conclusion is erroneous, show that for the parametric curve in Example 7,

d 2 y d 2 y/dt 2 = dx 2 t=1 d 2 x/dt 2 t=1

dy ′ dy ′ /dt 3/2 3 d 2y = = = = 2 / dx dx dx dt 2t 4t

(8)

Since the point (1, 1) on the curve corresponds to t = 1 in the parametric equations, it follows from (7) and (8) that 3 d 2 y 3 dy = = and 2 dx t=1 2 dx t=1 4

Similarly, the point (1, −1) corresponds to t = −1 in the parametric equations, so applying (7) and (8) again yields dy d 2 y 3 3 and =− =− 2 dx t=−1 2 dx t=−1 4

Note that the values we obtained for the ﬁrst and second derivatives are consistent with the graph in Figure 10.1.13, since at (1, 1) on the upper branch the tangent line has positive slope and the curve is concave up, and at (1, −1) on the lower branch the tangent line has negative slope and the curve is concave down. Finally, observe that we were able to apply Formulas (7) and (8) for both t = 1 and t = −1, even though the points (1, 1) and (1, −1) lie on different branches. In contrast, had we chosen to perform the same computations by eliminating the parameter, we would have had to obtain separate derivative formulas for y = x 3/2 and y = −x 3/2 . ARC LENGTH OF PARAMETRIC CURVES The following result provides a formula for ﬁnding the arc length of a curve from parametric equations for the curve. Its derivation is similar to that of Formula (3) in Section 6.4 and will be omitted.

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Chapter 10 / Parametric and Polar Curves; Conic Sections

Formulas (4) and (5) in Section 6.4 can be viewed as special cases of (9). For example, Formula (4) in Section 6.4 can be obtained from (9) by writing y = f(x) parametrically as

x = t, y = f(t) and Formula (5) in Section 6.4 can be obtained by writing x = g(y) parametrically as

x = g(t), y = t

10.1.1 arc length formula for parametric curves If no segment of the curve represented by the parametric equations x = x(t),

y = y(t)

(a ≤ t ≤ b)

is traced more than once as t increases from a to b, and if dx /dt and dy /dt are continuous functions for a ≤ t ≤ b, then the arc length L of the curve is given by b 2 2 dx dy L= (9) + dt dt dt a Example 8 equations

Use (9) to ﬁnd the circumference of a circle of radius a from the parametric x = a cos t,

y = a sin t

(0 ≤ t ≤ 2π)

Solution. L=

2π

dx dt

dy dt

dt =

2π

(−a sin t)2 + (a cos t)2 dt

2π

a dt = at

2π 0

= 2πa

THE CYCLOID (THE APPLE OF DISCORD) The results of this section can be used to investigate a curve known as a cycloid. This curve, which is one of the most signiﬁcant in the history of mathematics, can be generated by a point on a circle that rolls along a straight line (Figure 10.1.14). This curve has a fascinating history, which we will discuss shortly; but ﬁrst we will show how to obtain parametric equations for it. For this purpose, let us assume that the circle has radius a and rolls along the positive x-axis of a rectangular coordinate system. Let P (x, y) be the point on the circle that traces the cycloid, and assume that P is initially at the origin. We will take as our parameter the angle θ that is swept out by the radial line to P as the circle rolls (Figure 10.1.14). It is standard here to regard θ as positive, even though it is generated by a clockwise rotation. The motion of P is a combination of the movement of the circle’s center parallel to the x-axis and the rotation of P about the center. As the radial line sweeps out an angle θ, the point P traverses an arc of length aθ, and the circle moves a distance aθ along the x-axis. Thus, as suggested by Figure 10.1.15, the center moves to the point (aθ, a), and the coordinates of P are

x = aθ − a sin θ,

y = a − a cos θ

(10)

These are the equations of the cycloid in terms of the parameter θ. One of the reasons the cycloid is important in the history of mathematics is that the study of its properties helped to spur the development of early versions of differentiation and integration. Work on the cycloid was carried out by some of the most famous names in seventeenth century mathematics, including Johann and Jakob Bernoulli, Descartes, L’Hôpital, Newton, and Leibniz. The curve was named the “cycloid” by the Italian mathematician and astronomer, Galileo, who spent over 40 years investigating its properties. An early problem of interest was that of constructing tangent lines to the cycloid. This problem was ﬁrst solved by Descartes, and then by Fermat, whom Descartes had challenged with the question. A modern solution to this problem follows directly from the parametric equations (10) and Formula (4). For example, using Formula (4), it is straightforward to show that the x-intercepts of the cycloid are cusps and that there is a horizontal tangent line to the cycloid halfway between adjacent x-intercepts (Exercise 60).

10.1 Parametric Equations; Tangent Lines and Arc Length for Parametric Curves

699

P(x, y) a

y = a − a cos u

a x

x = au − a sin u au Figure 10.1.15

Q Figure 10.1.16

a sin u

åa

A cycloid

Figure 10.1.14

u a cos u

Another early problem was determining the arc length of an arch of the cycloid. This was solved in 1658 by the famous British architect and mathematician, Sir Christopher Wren. He showed that the arc length of one arch of the cycloid is exactly eight times the radius of the generating circle. [For a solution to this problem using Formula (9), see Exercise 71.] The cycloid is also important historically because it provides the solution to two famous mathematical problems—the brachistochrone problem (from Greek words meaning “shortest time”) and the tautochrone problem (from Greek words meaning “equal time”). The brachistochrone problem is to determine the shape of a wire along which a bead might slide from a point P to another point Q, not directly below, in the shortest time. The tautochrone problem is to find the shape of a wire from P to Q such that two beads started at any points on the wire between P and Q reach Q in the same amount of time. The solution to both problems turns out to be an inverted cycloid (Figure 10.1.16). In June of 1696, Johann Bernoulli posed the brachistochrone problem in the form of a challenge to other mathematicians. At first, one might conjecture that the wire should form a straight line, since that shape results in the shortest distance from P to Q. However, the inverted cycloid allows the bead to fall more rapidly at first, building up sufficient speed to reach Q in the shortest time, even though it travels a longer distance. The problem was solved by Newton, Leibniz, and L’Hôpital, as well as by Johann Bernoulli and his older brother Jakob; it was formulated and solved incorrectly years earlier by Galileo, who thought the answer was a circular arc. In fact, Johann was so impressed with his brother Jakob’s solution that he claimed it to be his own. (This was just one of many disputes about the cycloid that eventually led to the curve being known as the “apple of discord.”) One solution of the brachistochrone problem leads to the differential equation 2

dy 1+ y = 2a (11) dx where a is a positive constant. We leave it as an exercise (Exercise 72) to show that the cycloid provides a solution to this differential equation.

Newton’s solution of the brachistochrone problem in his own handwriting

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Chapter 10 / Parametric and Polar Curves; Conic Sections

✔QUICK CHECK EXERCISES 10.1

(See page 705 for answers.)

1. Find parametric equations for a circle of radius 2, centered at (3, 5). 2. The graph of the curve described by the parametric equations x = 4t − 1, y = 3t + 2 is a straight line with slope and y-intercept . 3. Suppose that a parametric curve C is given by the equations x = f(t), y = g(t) for 0 ≤ t ≤ 1. Find parametric equations for C that reverse the direction the curve is traced as the parameter increases from 0 to 1.

EXERCISE SET 10.1

Graphing Utility

x = f(t),

y = g(t)

we can use the formula dy/dx =

5. Let L be the length of the curve x = ln t,

y = sin t

(1 ≤ t ≤ π)

An integral expression for L is

CAS

1. (a) By eliminating the parameter, sketch the trajectory over the time interval 0 ≤ t ≤ 5 of the particle whose parametric equations of motion are x = t − 1,

4. To ﬁnd dy /dx directly from the parametric equations

y =t +1

Johann (left) and Jakob (right) Bernoulli Members of an amazing Swiss family that included several generations of outstanding mathematicians and scientists. Nikolaus Bernoulli (1623–1708), a druggist, ﬂed from Antwerp to escape religious persecution and ultimately settled in Basel, Switzerland. There he had three sons, Jakob I (also called Jacques or James), Nikolaus, and Johann I (also called Jean or John). The Roman numerals are used to distinguish family members with identical names (see the family tree below). Following Newton and Leibniz, the Bernoulli brothers, Jakob I and Johann I, are considered by some to be the two most important founders of calculus. Jakob I was self-taught in mathematics. His father wanted him to study for the ministry, but he turned to mathematics and in 1686 became a professor at the University of Basel. When he started working in mathematics, he knew nothing of Newton’s and Leibniz’ work. He eventually became familiar with Newton’s results, but because so little of Leibniz’ work was published, Jakob duplicated many of Leibniz’ results. Jakob’s younger brother Johann I was urged to enter into business by his father. Instead, he turned to medicine and studied mathematics under the guidance of his older brother. He eventually became a mathematics professor at Gröningen in Holland, and then, when Jakob died in 1705, Johann succeeded him as mathematics professor at Basel. Throughout their lives, Jakob I and Johann I had a mutual passion for criticizing each other’s work, which frequently erupted into ugly confrontations. Leibniz tried to mediate the disputes, but Jakob, who resented Leibniz’ superior intellect, accused him of siding with Johann, and thus Leibniz became entangled in the arguments. The brothers often worked on common problems that they posed as challenges to one another. Johann, interested in gaining fame, often used unscrupulous means to make himself appear the originator of his brother’s results; Jakob occasionally retaliated. Thus, it is often difﬁcult to determine who deserves credit for many results. However, both men made major contributions

(b) Indicate the direction of motion on your sketch. (c) Make a table of x- and y-coordinates of the particle at times t = 0, 1, 2, 3, 4, 5. (d) Mark the position of the particle on the curve at the times in part (c), and label those positions with the values of t. to the development of calculus. In addition to his work on calculus, Jakob helped establish fundamental principles in probability, including the Law of Large Numbers, which is a cornerstone of modern probability theory. Among the other members of the Bernoulli family, Daniel, son of Johann I, is the most famous. He was a professor of mathematics at St. Petersburg Academy in Russia and subsequently a professor of anatomy and then physics at Basel. He did work in calculus and probability, but is best known for his work in physics. A basic law of ﬂuid ﬂow, called Bernoulli’s principle, is named in his honor. He won the annual prize of the French Academy 10 times for work on vibrating strings, tides of the sea, and kinetic theory of gases. Johann II succeeded his father as professor of mathematics at Basel. His research was on the theory of heat and sound. Nikolaus I was a mathematician and law scholar who worked on probability and series. On the recommendation of Leibniz, he was appointed professor of mathematics at Padua and then went to Basel as a professor of logic and then law. Nikolaus II was professor of jurisprudence in Switzerland and then professor of mathematics at St. Petersburg Academy. Johann III was a professor of mathematics and astronomy in Berlin and Jakob II succeeded his uncle Daniel as professor of mathematics at St. Petersburg Academy in Russia. Truly an incredible family! Nikolaus Bernoulli (1623−1708) Jakob I Nikolaus (1654−1705) (Jacques, James) Nikolaus I (1687−1759)

Johann I (1667−1748) (Jean, John) Nikolaus II Daniel (1695−1726) (1700−1782)

Johann II (1710−1790)

Johann III Jakob II (1744−1807) (1759−1789)

10.1 Parametric Equations; Tangent Lines and Arc Length for Parametric Curves

2. (a) By eliminating the parameter, sketch the trajectory over the time interval 0 ≤ t ≤ 1 of the particle whose parametric equations of motion are x = cos(πt),

y = sin(πt)

(b) Indicate the direction of motion on your sketch. (c) Make a table of x- and y-coordinates of the particle at times t = 0, 0.25, 0.5, 0.75, 1. (d) Mark the position of the particle on the curve at the times in part (c), and label those positions with the values of t. 3–12 Sketch the curve by eliminating the parameter, and indi-

cate the direction of increasing t. ■ 3. x = 3t − 4, y = 6t + 2

(0 ≤ t ≤ 2π)

7. x = 3 + 2 cos t, y = 2 + 4 sin t

(0 ≤ t ≤ 2π) 8. x = sec t, y = tan t (π ≤ t < 3π/2) 9. x = cos 2t, y = sin t (−π/2 ≤ t ≤ π/2)

10. x = 4t + 3, y = 16t 2 − 9

11. x = 2 sin2 t, y = 3 cos2 t 12. x = sec2 t, y = tan2 t

(b) Assuming that the plane flies in a room in which the floor is at y = 0, explain why the plane will not crash into the floor. [For simplicity, ignore the physical size of the plane by treating it as a particle.] (c) How high must the ceiling be to ensure that the plane does not touch or crash into it? 21–22 Graph the equation using a graphing utility. ■

21. (a) x = y 2 + 2y + 1 (b) x = sin y, −2π ≤ y ≤ 2π

22. (a) x = y + 2y 3 − y 5 (b) x = tan y, −π/2 < y < π/2 F O C U S O N C O N C E P TS

4. x = t − 3, y = 3t − 7 (0 ≤ t ≤ 3) 5. x = 2 cos t, y = 5 sin t √ 6. x = t, y = 2t + 4

(0 ≤ t ≤ π/2)

(0 ≤ t < π/2)

23. In each part, match the parametric equation with one of the curves√labeled (I) – (VI), and explain your reasoning. (a) x = t, y = sin 3t (b) x = 2 cos t, y = 3 sin t (c) x = t cos t, y = t sin t 3t 3t 2 (d) x = , y= 3 1+t 1 + t3 3 t 2t 2 (e) x = , y = 1 + t2 1 + t2 1 (f ) x = 2 cos t, y = sin 2t y

13–18 Find parametric equations for the curve, and check your

work by generating the curve with a graphing utility. ■

y x

13. A circle of radius 5, centered at the origin, oriented clockwise. 14. The portion of the circle x 2 + y 2 = 1 that lies in the third quadrant, oriented counterclockwise. 15. A vertical line intersecting the x-axis at x = 2, oriented upward. 16. The ellipse x 2 /4 + y 2 /9 = 1, oriented counterclockwise. 17. The portion of the parabola x = y 2 joining (1, −1) and (1, 1), oriented down to up.

18. The circle of radius 4, centered at (1, −3), oriented counterclockwise. 19. (a) Use a graphing utility to generate the trajectory of a particle whose equations of motion over the time interval 0 ≤ t ≤ 5 are x = 6t −

1 3 t , 2

y =1+

1 2 t 2

(b) Make a table of x- and y-coordinates of the particle at times t = 0, 1, 2, 3, 4, 5. (c) At what times is the particle on the y-axis? (d) During what time interval is y < 5? (e) At what time does the x-coordinate of the particle reach a maximum? 20. (a) Use a graphing utility to generate the trajectory of a paper airplane whose equations of motion for t ≥ 0 are x = t − 2 sin t,

701

y = 3 − 2 cos t

I y

III

y x

IV Figure Ex-23

24. (a) Identify the orientation of the curves in Exercise 23. (b) Explain why the parametric curve x = t 2,

y = t4

(−1 ≤ t ≤ 1)

does not have a deﬁnite orientation.

25. (a) Suppose that the line segment from the point P (x0 , y0 ) to Q(x1 , y1 ) is represented parametrically by x = x0 + (x1 − x0 )t, y = y0 + (y1 − y0 )t

(0 ≤ t ≤ 1)

and that R(x, y) is the point on the line segment corresponding to a speciﬁed value of t (see the accompanying ﬁgure on the next page). Show that t = r /q, where r is the distance from P to R and q is the distance from P to Q. (cont.)

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Chapter 10 / Parametric and Polar Curves; Conic Sections

(b) What value of t produces the midpoint between points P and Q? (c) What value of t produces the point that is three-fourths of the way from P to Q? t=1

x = t 3,

Q(x 1, y1)

t t=0

y = t + t6

(−⬁ < t < +⬁)

is concave down for t < 0.

R(x, y)

P(x 0, y0)

Figure Ex-25

26. Find parametric equations for the line segment joining P (2, −1) and Q(3, 1), and use the result in Exercise 25 to ﬁnd (a) the midpoint between P and Q (b) the point that is one-fourth of the way from P to Q (c) the point that is three-fourths of the way from P to Q. 27. (a) Show that the line segment joining the points (x0 , y0 ) and (x1 , y1 ) can be represented parametrically as t − t0 , x = x0 + (x1 − x0 ) t1 − t 0 (t0 ≤ t ≤ t1 ) t − t0 y = y0 + (y1 − y0 ) t1 − t 0 (b) Which way is the line segment oriented? (c) Find parametric equations for the line segment traced from (3, −1) to (1, 4) as t varies from 1 to 2, and check your result with a graphing utility. 28. (a) By eliminating the parameter, show that if a and c are not both zero, then the graph of the parametric equations x = at + b, y = ct + d (t0 ≤ t ≤ t1 ) is a line segment. (b) Sketch the parametric curve x = 2t − 1, y = t + 1 (1 ≤ t ≤ 2) and indicate its orientation. (c) What can you say about the line in part (a) if a or c (but not both) is zero? (d) What do the equations represent if a and c are both zero? 29–32 Use a graphing utility and parametric equations to display the graphs of f and f −1 on the same screen. ■

29. f(x) = x 3 + 0.2x − 1, −1 ≤ x ≤ 2 √ 30. f(x) = x 2 + 2 + x, −5 ≤ x ≤ 5 31. f(x) = cos(cos 0.5x), 0 ≤ x ≤ 3 32. f(x) = x + sin x,

35. For the parametric curve x = x(t), y = 3t 4 − 2t 3 , the derivative of y with respect to x is computed by dy 12t 3 − 6t 2 = dx x ′ (t) 36. The curve represented by the parametric equations

0≤x≤6

33–36 True–False Determine whether the statement is true or

false. Explain your answer. ■ 33. The equation y = 1 − x 2 can be described parametrically by x = sin t, y = cos2 t.

34. The graph of the parametric equations x = f(t), y = t is the reﬂection of the graph of y = f(x) about the x-axis.

37. Parametric curves can be defined piecewise by using different formulas for different values of the parameter. Sketch the curve that is represented piecewise by the parametric equations x = 2t, y = 4t 2 0 ≤ t ≤ 21 1 ≤t ≤1 x = 2 − 2t, y = 2t 2 38. Find parametric equations for the rectangle in the accompanying figure, assuming that the rectangle is traced counterclockwise as t varies from 0 to 1, starting at 21 , 21 when t = 0. [Hint: Represent the rectangle piecewise, letting t vary from 0 to 41 for the first edge, from 41 to 21 for the second edge, and so forth.] y

冸− 12 ,

1 2

冹

冸 12 , 12 冹

冸− 12 ,

− 12 冹

冸 12 , − 12 冹

Figure Ex-38

39. (a) Find parametric equations for the ellipse that is centered at the origin and has intercepts (4, 0), (−4, 0), (0, 3), and (0, −3). (b) Find parametric equations for the ellipse that results by translating the ellipse in part (a) so that its center is at (−1, 2). (c) Confirm your results in parts (a) and (b) using a graphing utility. 40. We will show later in the text that if a projectile is fired from ground level with an initial speed of v0 meters per second at an angle α with the horizontal, and if air resistance is neglected, then its position after t seconds, relative to the coordinate system in the accompanying figure on the next page is x = (v0 cos α)t, y = (v0 sin α)t − 21 gt 2 where g ≈ 9.8 m/s2 . (a) By eliminating the parameter, show that the trajectory lies on the graph of a quadratic polynomial. (b) Use a graphing utility to sketch the trajectory if α = 30 ◦ and v0 = 1000 m/s. (c) Using the trajectory in part (b), how high does the shell rise? (cont.)

10.1 Parametric Equations; Tangent Lines and Arc Length for Parametric Curves

(d) Using the trajectory in part (b), how far does the shell travel horizontally? y

703

53–54 Find all values of t at which the parametric curve has (a) a horizontal tangent line and (b) a vertical tangent line. ■ 53. x = 2 sin t, y = 4 cos t (0 ≤ t ≤ 2π)

54. x = 2t 3 − 15t 2 + 24t + 7, y = t 2 + t + 1

55. In the mid-1850s the French physicist Jules Antoine Lissajous (1822–1880) became interested in parametric equations of the form

Figure Ex-40

F O C U S O N C O N C E P TS

41. (a) Find the slope of the tangent line to the parametric curve x = t /2, y = t 2 + 1 at t = −1 and at t = 1 without eliminating the parameter. (b) Check your answers in part (a) by eliminating the parameter and differentiating an appropriate function of x. 42. (a) Find the slope of the tangent line to the parametric curve x = 3 cos t, y = 4 sin t at t = π/4 and at t = 7π/4 without eliminating the parameter. (b) Check your answers in part (a) by eliminating the parameter and differentiating an appropriate function of x. 43. For the parametric curve in Exercise 41, make a conjecture about the sign of d 2 y/dx 2 at t = −1 and at t = 1, and conﬁrm your conjecture without eliminating the parameter. 44. For the parametric curve in Exercise 42, make a conjecture about the sign of d 2 y/dx 2 at t = π/4 and at t = 7π/4, and conﬁrm your conjecture without eliminating the parameter. 45–50 Find dy/dx and d 2 y/dx 2 at the given point without eliminating the parameter. ■ √ 45. x = t, y = 2t + 4; t = 1

46. x = 21 t 2 + 1, y = 13 t 3 − t; t = 2 47. x = sec t, y = tan t; t = π/3 48. x = sinh t, y = cosh t; t = 0

49. x = θ + cos θ, y = 1 + sin θ; θ = π/6 50. x = cos φ, y = 3 sin φ; φ = 5π/6

51. (a) Find the equation of the tangent line to the curve x = et ,

y = e−t

at t = 1 without eliminating the parameter. (b) Find the equation of the tangent line in part (a) by eliminating the parameter. 52. (a) Find the equation of the tangent line to the curve x = 2t + 4,

y = 8t 2 − 2t + 4

at t = 1 without eliminating the parameter. (b) Find the equation of the tangent line in part (a) by eliminating the parameter.

x = sin at,

y = sin bt

in the course of studying vibrations that combine two perpendicular sinusoidal motions. If a /b is a rational number, then the combined effect of the oscillations is a periodic motion along a path called a Lissajous curve. (a) Use a graphing utility to generate the complete graph of the Lissajous curves corresponding to a = 1, b = 2; a = 2, b = 3; a = 3, b = 4; and a = 4, b = 5. (b) The Lissajous curve x = sin t,

y = sin 2t

(0 ≤ t ≤ 2π)

crosses itself at the origin (see Figure Ex-55). Find equations for the two tangent lines at the origin. 56. The prolate cycloid x = 2 − π cos t,

y = 2t − π sin t

(−π ≤ t ≤ π)

crosses itself at a point on the x-axis (see the accompanying ﬁgure). Find equations for the two tangent lines at that point. y

Figure Ex-55

Figure Ex-56 2

57. Show that the curve x = t , y = t 3 − 4t intersects itself at the point (4, 0), and ﬁnd equations for the two tangent lines to the curve at the point of intersection. 58. Show that the curve with parametric equations x = t 2 − 3t + 5,

y = t 3 + t 2 − 10t + 9

intersects itself at the point (3, 1), and ﬁnd equations for the two tangent lines to the curve at the point of intersection. 59. (a) Use a graphing utility to generate the graph of the parametric curve x = cos3 t,

y = sin3 t

(0 ≤ t ≤ 2π)

and make a conjecture about the values of t at which singular points occur. (b) Conﬁrm your conjecture in part (a) by calculating appropriate derivatives. 60. Verify that the cycloid described by Formula (10) has cusps at its x-intercepts and horizontal tangent lines at midpoints between adjacent x-intercepts (see Figure 10.1.14).

Chapter 10 / Parametric and Polar Curves; Conic Sections

704

61. (a) What is the slope of the tangent line at time t to the trajectory of the paper airplane in Example 5? (b) What was the airplane’s approximate angle of inclination when it crashed into the wall? 62. Suppose that a bee follows the trajectory x = t − 2 cos t,

y = 2 − 2 sin t

(0 ≤ t ≤ 10)

(a) At what times was the bee ﬂying horizontally? (b) At what times was the bee ﬂying vertically?

63. Consider the family of curves described by the parametric equations x = a cos t + h,

y = b sin t + k

(0 ≤ t < 2π)

where a = 0 and b = 0. Describe the curves in this family if (a) h and k are ﬁxed but a and b can vary (b) a and b are ﬁxed but h and k can vary (c) a = 1 and b = 1, but h and k vary so that h = k + 1.

64. (a) Use a graphing utility to study how the curves in the family x = 2a cos2 t,

y = 2a cos t sin t

(−2π < t < 2π)

change as a varies from 0 to 5. (b) Conﬁrm your conclusion algebraically. (c) Write a brief paragraph that describes your ﬁndings.

F O C U S O N C O N C E P TS

73. The amusement park rides illustrated in the accompanying ﬁgure consist of two connected rotating arms of length 1—an inner arm that rotates counterclockwise at 1 radian per second and an outer arm that can be programmed to rotate either clockwise at 2 radians per second (the Scrambler ride) or counterclockwise at 2 radians per second (the Calypso ride). The center of the rider cage is at the end of the outer arm. (a) Show that in the Scrambler ride the center of the cage has parametric equations x = cos t + cos 2t,

(b) Find parametric equations for the center of the cage in the Calypso ride, and use a graphing utility to conﬁrm that the center traces the curve shown in the accompanying ﬁgure. (c) Do you think that a rider travels the same distance in one revolution of the Scrambler ride as in one revolution of the Calypso ride? Justify your conclusion.

65–70 Find the exact arc length of the curve over the stated interval. ■

65. x = t 2 , y = 13 t 3 (0 ≤ t ≤ 1) √ 66. x = t − 2, y = 2t 3/4 (1 ≤ t ≤ 16) 67. x = cos 3t, y = sin 3t

(0 ≤ t ≤ π)

68. x = sin t + cos t, y = sin t − cos t

(0 ≤ t ≤ π)

69. x = e2t (sin t + cos t), y = e2t (sin t − cos t) (−1 ≤ t ≤ 1) 70. x = 2 sin−1 t, y = ln(1 − t 2 ) 0 ≤ t ≤ 21

71. (a) Use Formula (9) to show that the length L of one arch of a cycloid is given by 2π L=a 2(1 − cos θ ) dθ 0

(b) Use a CAS to show that L is eight times the radius of the wheel that generates the cycloid (see the accompanying ﬁgure). L = 8a a x 0

Scrambler ride

Figure Ex-71

72. Use the parametric equations in Formula (10) to verify that the cycloid provides one solution to the differential equation 2

dy 1+ y = 2a dx where a is a positive constant.

Calypso ride

Figure Ex-73

74. (a) If a thread is unwound from a ﬁxed circle while being held taut (i.e., tangent to the circle), then the end of the thread traces a curve called an involute of a circle. Show that if the circle is centered at the origin, has radius a, and the end of the thread is initially at the point (a, 0), then the involute can be expressed parametrically as x = a(cos θ + θ sin θ ),

y = sin t − sin 2t

y = a(sin θ − θ cos θ)

where θ is the angle shown in part (a) of the accompanying ﬁgure on the next page. (b) Assuming that the dog in part (b) of the accompanying ﬁgure on the next page unwinds its leash while keeping it taut, for what values of θ in the interval 0 ≤ θ ≤ 2π will the dog be walking North? South? East? West? (c) Use a graphing utility to generate the curve traced by the dog, and show that it is consistent with your answer in part (b).

10.2 Polar Coordinates

75. Find the area of the surface generated by revolving x = t 2 , y = 3t (0 ≤ t ≤ 2) about the x-axis.

1 u

(a, 0)

x = r cos t,

75–80 If f ′(t) and g ′ (t) are continuous functions, and if no

segment of the curve y = g(t)

77. Find the area of the surface generated by revolving the curve x = cos2 t, y = sin2 t (0 ≤ t ≤ π/2) about the y-axis.

79. By revolving the semicircle

(b)

Figure Ex-74

x = f(t),

76. Find the area of the surface generated by revolving the curve x = et cos t, y = et sin t (0 ≤ t ≤ π/2) about the x-axis.

78. Find the area of the surface generated by revolving x = 6t, y = 4t 2 (0 ≤ t ≤ 1) about the y-axis.

(a)

705

is traced more than once, then it can be shown that the area of the surface generated by revolving this curve about the x-axis is 2 2 b dx dy S= 2πy + dt dt dt a and the area of the surface generated by revolving the curve about the y-axis is 2 b dx 2 dy 2πx + dt S= dt dt a [The derivations are similar to those used to obtain Formulas (4) and (5) in Section 6.5.] Use the formulas above in these exercises. ■

(0 ≤ t ≤ π)

80. The equations x = aφ − a sin φ,

(a ≤ t ≤ b)

y = r sin t

about the x-axis, show that the surface area of a sphere of radius r is 4πr 2 . y = a − a cos φ

(0 ≤ φ ≤ 2π)

represent one arch of a cycloid. Show that the surface area generated by revolving this curve about the x-axis is given by S = 64πa 2 /3.

81. Writing Consult appropriate reference works and write an essay on American mathematician Nathaniel Bowditch (1773–1838) and his investigation of Bowditch curves (better known as Lissajous curves; see Exercise 55). 82. Writing What are some of the advantages of expressing a curve parametrically rather than in the form y = f(x)?

✔QUICK CHECK ANSWERS 10.1 1. x = 3 + 2 cos t, y = 5 + 2 sin t π 5. (1/t)2 + cos2 t dt

(0 ≤ t ≤ 2π)

3 ; 4

2.75

3. x = f(1 − t), y = g(1 − t)

dy /dt g ′ (t) = ′ dx /dt f (t)

10.2

POLAR COORDINATES Up to now we have specified the location of a point in the plane by means of coordinates relative to two perpendicular coordinate axes. However, sometimes a moving point has a special affinity for some fixed point, such as a planet moving in an orbit under the central attraction of the Sun. In such cases, the path of the particle is best described by its angular direction and its distance from the fixed point. In this section we will discuss a new kind of coordinate system that is based on this idea. POLAR COORDINATE SYSTEMS A polar coordinate system in a plane consists of a fixed point O, called the pole (or origin), and a ray emanating from the pole, called the polar axis. In such a coordinate system

Chapter 10 / Parametric and Polar Curves; Conic Sections

706

P(r, u)

we can associate with each point P in the plane a pair of polar coordinates (r, θ), where r is the distance from P to the pole and θ is an angle from the polar axis to the ray OP (Figure 10.2.1). The number r is called the radial coordinate of P and the number θ the angular coordinate (or polar angle) of P . In Figure 10.2.2, the points (6, π/4), (5, 2π/3), (3, 5π/4), and (4, 11π/6) are plotted in polar coordinate systems. If P is the pole, then r = 0, but there is no clearly deﬁned polar angle. We will agree that an arbitrary angle can be used in this case; that is, (0, θ) are polar coordinates of the pole for all choices of θ.

r u

O Pole

Polar axis

Figure 10.2.1

(6, c/4)

(5, 2c/3)

5c/4

11c/6

2c/3

c /4

(4, 11c/6)

(3, 5c/4)

Figure 10.2.2

The polar coordinates of a point are not unique. For example, the polar coordinates (1, 7π/4),

(1, −π/4),

and (1, 15π/4)

all represent the same point (Figure 10.2.3). 15c/4

7c/4 −c/4 1 Figure 10.2.3

1 (1, −c/4)

(1, 7c/4)

(1, 15c/4)

In general, if a point P has polar coordinates (r, θ), then (r, θ + 2nπ) and

5c/4

Polar axis

e id

r Te

P(3, 5c/4)

l na

P(−3, c/4) Figure 10.2.4

c /4

Polar axis

(r, θ − 2nπ)

are also polar coordinates of P for any nonnegative integer n. Thus, every point has infinitely many pairs of polar coordinates. As defined above, the radial coordinate r of a point P is nonnegative, since it represents the distance from P to the pole. However, it will be convenient to allow for negative values of r as well. To motivate an appropriate definition, consider the point P with polar coordinates (3, 5π/4). As shown in Figure 10.2.4, we can reach this point by rotating the polar axis through an angle of 5π/4 and then moving 3 units from the pole along the terminal side of the angle, or we can reach the point P by rotating the polar axis through an angle of π/4 and then moving 3 units from the pole along the extension of the terminal side. This suggests that the point (3, 5π/4) might also be denoted by (−3, π/4), with the minus sign serving to indicate that the point is on the extension of the angle’s terminal side rather than on the terminal side itself. In general, the terminal side of the angle θ + π is the extension of the terminal side of θ, so we define negative radial coordinates by agreeing that (−r, θ)

and (r, θ + π)

are polar coordinates of the same point. RELATIONSHIP BETWEEN POLAR AND RECTANGULAR COORDINATES

Frequently, it will be useful to superimpose a rectangular xy-coordinate system on top of a polar coordinate system, making the positive x-axis coincide with the polar axis. If this is done, then every point P will have both rectangular coordinates (x, y) and polar coordinates

10.2 Polar Coordinates c/ 2

707

(r, Î¸). As suggested by Figure 10.2.5, these coordinates are related by the equations

(x, y) P (r, u) r

x = r cos Î¸,

y = r sin u

x = r cos u

r 2 = x2 + y2,

c/ 2

(1)

These equations are well suited for ďŹ nding x and y when r and Î¸ are known. However, to ďŹ nd r and Î¸ when x and y are known, it is preferable to use the identities sin2 Î¸ + cos2 Î¸ = 1 and tan Î¸ = sin Î¸ / cos Î¸ to rewrite (1) as

Figure 10.2.5

tan Î¸ =

y x

(2)

Example 1 Find the rectangular coordinates of the point P whose polar coordinates are (r, Î¸) = (6, 2Ď&#x20AC;/3) (Figure 10.2.6).

y = r sin Î¸

Solution. Substituting the polar coordinates r = 6 and Î¸ = 2Ď&#x20AC;/3 in (1) yields

r=6

2c/ 3 x

Figure 10.2.6

c/ 2

2Ď&#x20AC; 1 =6 â&#x2C6;&#x2019; = â&#x2C6;&#x2019;3 x = 6 cos 3 2 â&#x2C6;&#x161;

â&#x2C6;&#x161; 2Ď&#x20AC; 3 y = 6 sin =6 =3 3 3 2 â&#x2C6;&#x161; Thus, the rectangular coordinates of P are (x, y) = (â&#x2C6;&#x2019;3, 3 3).

Example â&#x2C6;&#x161; 2 Find polar coordinates of the point P whose rectangular coordinates are (â&#x2C6;&#x2019;2, â&#x2C6;&#x2019;2 3 ) (Figure 10.2.7).

Solution. We will ďŹ nd the polar coordinates (r, Î¸) of P that satisfy the conditions r > 0 and 0 â&#x2030;¤ Î¸ < 2Ď&#x20AC;. From the ďŹ rst equation in (2),

â&#x2C6;&#x161; r 2 = x 2 + y 2 = (â&#x2C6;&#x2019;2)2 + (â&#x2C6;&#x2019;2 3 )2 = 4 + 12 = 16

P(â&#x2C6;&#x2019;2, â&#x2C6;&#x2019;2â&#x2C6;&#x161;3) Figure 10.2.7

so r = 4. From the second equation in (2),

â&#x2C6;&#x161; â&#x2C6;&#x2019;2 3 â&#x2C6;&#x161; y = 3 tan Î¸ = = x â&#x2C6;&#x2019;2 â&#x2C6;&#x161; From this and the fact that (â&#x2C6;&#x2019;2, â&#x2C6;&#x2019;2 3 ) lies in the third quadrant, it follows that the angle satisfying the requirement 0 â&#x2030;¤ Î¸ < 2Ď&#x20AC; is Î¸ = 4Ď&#x20AC;/3. Thus, (r, Î¸) = (4, 4Ď&#x20AC;/3) are polar coordinates of P . All other polar coordinates of P are expressible in the form 4Ď&#x20AC; Ď&#x20AC; 4, + 2nĎ&#x20AC; or â&#x2C6;&#x2019;4, + 2nĎ&#x20AC; 3 3 where n is an integer.

GRAPHS IN POLAR COORDINATES We will now consider the problem of graphing equations in r and Î¸ , where Î¸ is assumed to be measured in radians. Some examples of such equations are

r = 1,

Î¸ = Ď&#x20AC;/4,

r = Î¸,

r = sin Î¸,

r = cos 2Î¸

In a rectangular coordinate system the graph of an equation in x and y consists of all points whose coordinates (x, y) satisfy the equation. However, in a polar coordinate system, points have inďŹ nitely many different pairs of polar coordinates, so that a given point may have some polar coordinates that satisfy an equation and others that do not. Given an equation

708

Chapter 10 / Parametric and Polar Curves; Conic Sections

in r and θ, we deﬁne its graph in polar coordinates to consist of all points with at least one pair of coordinates (r, θ) that satisfy the equation. Example 3

Sketch the graphs of (a) r = 1

(b) θ =

π 4

in polar coordinates.

Solution (a). For all values of θ, the point (1, θ) is 1 unit away from the pole. Since θ is arbitrary, the graph is the circle of radius 1 centered at the pole (Figure 10.2.8a).

Solution (b). For all values of r, the point (r, π/4) lies on a line that makes an angle of π/4 with the polar axis (Figure 10.2.8b). Positive values of r correspond to points on the line in the ﬁrst quadrant and negative values of r to points on the line in the third quadrant. Thus, in absence of any restriction on r, the graph is the entire line. Observe, however, that had we imposed the restriction r ≥ 0, the graph would have been just the ray in the ﬁrst quadrant. c/ 2

c/ 2

Figure 10.2.8

5c/ 2 c/ 2 c

r=1

u = c/4

(a)

(b)

Equations r = f(θ) that express r as a function of θ are especially important. One way to graph such an equation is to choose some typical values of θ , calculate the corresponding values of r, and then plot the resulting pairs (r, θ) in a polar coordinate system. The next two examples illustrate this process.

9c/ 2

c/ 4

3c/ 2

Example 4 Sketch the graph of r = θ (θ ≥ 0) in polar coordinates by plotting points.

7c/ 2

Solution. Observe that as θ increases, so does r; thus, the graph is a curve that spirals out

r = u (u ≥ 0) Figure 10.2.9

Graph the spiral r = θ (θ ≤ 0). Compare your graph to that in Figure 10.2.9.

from the pole as θ increases. A reasonably accurate sketch of the spiral can be obtained by plotting the points that correspond to values of θ that are integer multiples of π/2, keeping in mind that the value of r is always equal to the value of θ (Figure 10.2.9). Example 5 points.

Sketch the graph of the equation r = sin θ in polar coordinates by plotting

Solution. Table 10.2.1 shows the coordinates of points on the graph at increments of π/6. These points are plotted in Figure 10.2.10. Note, however, that there are 13 points listed in the table but only 6 distinct plotted points. This is because the pairs from θ = π on yield

10.2 Polar Coordinates

709

duplicates of the preceding points. For example, (−1/2, 7π/6) and (1/2, π/6) represent the same point. Table 10.2.1 u (radians)

r = sin u

1 2

√3

(r, u)

(0, 0)

1 π

√3 , π

冸2 , 6冹冸 2

冸 √2 , 4 冹

√3

1 2

−

1 2

3 −√ 2

−1

3 −√ 2

√3 , 2π

冹冸1, 2 冹冸 2

冹冸2,

5π 6

冹

(0, c)

冸− 2 ,

7π 6

√3 4π

冹冸− 2 ,

冹冸−1,

3π 2

1 2

−

√3 5π

冹冸− 2 ,

1 11π 6

冹冸− 2 ,

冹

(0, 2π )

Observe that the points in Figure 10.2.10 appear to lie on a circle. We can conﬁrm that this is so by expressing the polar equation r = sin θ in terms of x and y. To do this, we multiply the equation through by r to obtain

冸1, 6 冹冸√2 , 8 冹

r 2 = r sin θ

冸12 , a 冹

which now allows us to apply Formulas (1) and (2) to rewrite the equation as

冸12 , 2 冹

x2 + y2 = y

(0, 0)

Rewriting this equation as x 2 + y 2 − y = 0 and then completing the square yields 2 x 2 + y − 21 = 41 which is a circle of radius 21 centered at the point 0, 21 in the xy-plane. It is often useful to view the equation r = f(θ) as an equation in rectangular coordinates (rather than polar coordinates) and graphed in a rectangular θ r-coordinate system. For example, Figure 10.2.11 shows the graph of r = sin θ displayed using rectangular θ rcoordinates. This graph can actually help to visualize how the polar graph in Figure 10.2.10 is generated:

r = sin u Figure 10.2.10

r 1

r = sin u u c

• At θ = 0 we have r = 0, which corresponds to the pole (0, 0) on the polar graph. • As θ varies from 0 to π/2, the value of r increases from 0 to 1, so the point (r, θ)

−1

•

Figure 10.2.11

•

u c

−1

moves along the circle from the pole to the high point at (1, π/2). As θ varies from π/2 to π, the value of r decreases from 1 back to 0, so the point (r, θ) moves along the circle from the high point back to the pole. As θ varies from π to 3π/2, the values of r are negative, varying from 0 to −1. Thus, the point (r, θ) moves along the circle from the pole to the high point at (1, π/2), which is the same as the point (−1, 3π/2). This duplicates the motion that occurred for 0 ≤ θ ≤ π/2. As θ varies from 3π/2 to 2π, the value of r varies from −1 to 0. Thus, the point (r, θ) moves along the circle from the high point back to the pole, duplicating the motion that occurred for π/2 ≤ θ ≤ π.

Example 6

Sketch the graph of r = cos 2θ in polar coordinates.

Solution. Instead of plotting points, we will use the graph of r = cos 2θ in rectangular

r = cos 2u Figure 10.2.12

coordinates (Figure 10.2.12) to visualize how the polar graph of this equation is generated. The analysis and the resulting polar graph are shown in Figure 10.2.13. This curve is called a four-petal rose.

710

Chapter 10 / Parametric and Polar Curves; Conic Sections

r varies from −1 to 0 as u

r varies from 0 to 1 as u

r varies from 1 to 0 as u

r varies from 0 to −1 as u

r varies from −1 to 0 as u

r varies from 0 to 1 as u

r varies from 1 to 0 as u

r varies from 0 to −1 as u

varies from 0 to c/ 4.

varies from c/ 4 to c/ 2.

c/ 2 to 3c/ 4.

varies from

varies from 3c/ 4 to c.

varies from c to 5c/ 4.

varies from

5c/ 4 to 3c/ 2.

3c/ 2 to 7c/ 4.

7c/ 4 to 2c.

Figure 10.2.13

SYMMETRY TESTS

Observe that the polar graph of r = cos 2θ in Figure 10.2.13 is symmetric about the x-axis and the y-axis. This symmetry could have been predicted from the following theorem, which is suggested by Figure 10.2.14 (we omit the proof).

10.2.1

theorem (Symmetry Tests)

(a) A curve in polar coordinates is symmetric about the x-axis if replacing θ by −θ in its equation produces an equivalent equation (Figure 10.2.14a). The converse of each part of Theorem 10.2.1 is false. See Exercise 79.

(b) A curve in polar coordinates is symmetric about the y-axis if replacing θ by π − θ in its equation produces an equivalent equation (Figure 10.2.14b). (c) A curve in polar coordinates is symmetric about the origin if replacing θ by θ + π, or replacing r by −r in its equation produces an equivalent equation (Figure 10.2.14c).

c/ 2

c/ 2 (r, c − u)

(r, u)

c/ 2 (r, u)

(r, u) 0 (r, u + c) or (−r, u)

(r, −u)

Figure 10.2.14

(a)

(b)

(c)

Example 7 Use Theorem 10.2.1 to conﬁrm that the graph of r = cos 2θ in Figure 10.2.13 is symmetric about the x-axis and y-axis.

A graph that is symmetric about both the x -axis and the y -axis is also symmetric about the origin. Use Theorem 10.2.1(c) to verify that the curve in Example 7 is symmetric about the origin.

Solution. To test for symmetry about the x-axis, we replace θ by −θ. This yields r = cos(−2θ ) = cos 2θ Thus, replacing θ by −θ does not alter the equation. To test for symmetry about the y-axis, we replace θ by π − θ. This yields r = cos 2(π − θ ) = cos(2π − 2θ ) = cos(−2θ) = cos 2θ Thus, replacing θ by π − θ does not alter the equation.

10.2 Polar Coordinates

711

Example 8 Sketch the graph of r = a(1 − cos θ) in polar coordinates, assuming a to be a positive constant.

Solution. Observe ﬁrst that replacing θ by −θ does not alter the equation, so we know

in advance that the graph is symmetric about the polar axis. Thus, if we graph the upper half of the curve, then we can obtain the lower half by reflection about the polar axis. As in our previous examples, we will first graph the equation in rectangular θ r-coordinates. This graph, which is shown in Figure 10.2.15a, can be obtained by rewriting the given equation as r = a − a cos θ, from which we see that the graph in rectangular θ r-coordinates can be obtained by first reflecting the graph of r = a cos θ about the x-axis to obtain the graph of r = −a cos θ, and then translating that graph up a units to obtain the graph of r = a − a cos θ. Now we can see the following:

• • • •

As θ varies from 0 to π/3, r increases from 0 to a /2. As θ varies from π/3 to π/2, r increases from a /2 to a. As θ varies from π/2 to 2π/3, r increases from a to 3a /2. As θ varies from 2π/3 to π, r increases from 3a /2 to 2a.

This produces the polar curve shown in Figure 10.2.15b. The rest of the curve can be obtained by continuing the preceding analysis from π to 2π or, as noted above, by reﬂecting the portion already graphed about the x-axis (Figure 10.2.15c). This heart-shaped curve is called a cardioid (from the Greek word kardia meaning “heart” ). r

2a 3a 2 a a 2

冸 3a2 ,

2c 3

冹

冸

c 2

冹

a c

冸2 , 3 冹

u 4 68

(2a, c) r = a(1 − cos u)

(a)

(b)

(c)

Figure 10.2.15

Example 9

Sketch the graph of r 2 = 4 cos 2θ in polar coordinates.

Solution. This equation does not express r as a function of θ, since solving for r in terms of θ yields two functions: √ √ r = 2 cos 2θ and r = −2 cos 2θ Thus, to graph the equation r 2 = 4 cos 2θ we will have to graph the two functions separately and then combine those graphs. √ We will start with the graph of r = 2 cos 2θ . Observe first that this equation is not changed if we replace θ by −θ or if we replace θ by π − θ . Thus, the graph is symmetric about the x-axis and the y-axis. This means that the entire graph can be obtained by graphing the portion in the first quadrant, reflecting that portion about the y-axis to obtain the portion in the second quadrant, and then reflecting those two portions about the x-axis to obtain the portions in the third and fourth quadrants.

712

Chapter 10 / Parametric and Polar Curves; Conic Sections

√ To begin the analysis, we will graph the equation r = 2 cos 2θ in rectangular θ rcoordinates (see Figure 10.2.16a). Note that there are gaps in that graph over the intervals π/4 < θ < 3π/4 and 5π/4 < θ < 7π/4 because cos 2θ is negative for those values of θ. From this graph we can see the following:

• As θ varies from 0 to π/4, r decreases from 2 to 0. • As θ varies from π/4 to π/2, no points are generated on the polar graph. This produces the portion of the graph shown in Figure 10.2.16b. As noted above, we can complete the graph by a reﬂection about the y-axis followed by a reﬂection about the x-axis (Figure 10.2.16c). The resulting propeller-shaped graph is called a lemniscate (from the Greek word lemniscos for a looped We leave it for you √ ribbon resembling the number 8). √ to verify that the equation r = 2 cos 2θ has the same graph as r = −2 cos 2θ , but traced in a diagonally opposite manner. Thus, the graph of the equation r 2 = 4 cos 2θ consists of two identical superimposed lemniscates.

c/ 2 r

c/ 2

u = c/ 4

r = 2√cos 2u

2 0

2 u 3

−2

(a)

(b)

(c)

Figure 10.2.16

c/ 2

FAMILIES OF LINES AND RAYS THROUGH THE POLE

(a)

If θ0 is a ﬁxed angle, then for all values of r the point (r, θ0 ) lies on the line that makes an angle of θ = θ0 with the polar axis; and, conversely, every point on this line has a pair of polar coordinates of the form (r, θ0 ). Thus, the equation θ = θ0 represents the line that passes through the pole and makes an angle of θ0 with the polar axis (Figure 10.2.17a). If r is restricted to be nonnegative, then the graph of the equation θ = θ0 is the ray that emanates from the pole and makes an angle of θ0 with the polar axis (Figure 10.2.17b). Thus, as θ0 varies, the equation θ = θ0 produces either a family of lines through the pole or a family of rays through the pole, depending on the restrictions on r.

u = u0

c/ 2

FAMILIES OF CIRCLES We will consider three families of circles in which a is assumed to be a positive constant: u0

u = u0 (r ≥ 0)

(b) Figure 10.2.17

r=a

r = 2a cos θ

r = 2a sin θ

(3–5)

The equation r = a represents a circle of radius a centered at the pole (Figure 10.2.18a). Thus, as a varies, this equation produces a family of circles centered at the pole. For families (4) and (5), recall from plane geometry that a triangle that is inscribed in a circle with a diameter of the circle for a side must be a right triangle. Thus, as indicated in Figures 10.2.18b and 10.2.18c, the equation r = 2a cos θ represents a circle of radius a, centered on the x-axis and tangent to the y-axis at the origin; similarly, the equation r = 2a sin θ represents a circle of radius a, centered on the y-axis and tangent to the x-axis at the origin. Thus, as a varies, Equations (4) and (5) produce the families illustrated in Figures 10.2.18d and 10.2.18e.

10.2 Polar Coordinates c/ 2

c/ 2

713

c/ 2

P(r, u) a

r u 2a

P(a, u) 0

P(r, u)

u 0

r u 0

r=a

r = 2a cos u

r = 2a sin u

r = 2a cos u

r = 2a sin u

(a)

(b)

(c)

(d )

(e)

Figure 10.2.18

FAMILIES OF ROSE CURVES

In polar coordinates, equations of the form Observe that replacing θ by −θ does not change the equation r = 2a cos θ and that replacing θ by π − θ does not change the equation r = 2a sin θ. This explains why the circles in Figure 10.2.18d are symmetric about the x -axis and those in Figure 10.2.18e are symmetric about the y -axis.

What do the graphs of one-petal roses look like?

r = a sin nθ

r = a cos nθ

(6–7)

in which a > 0 and n is a positive integer represent families of ﬂower-shaped curves called roses (Figure 10.2.19). The rose consists of n equally spaced petals of radius a if n is odd and 2n equally spaced petals of radius a if n is even. It can be shown that a rose with an even number of petals is traced out exactly once as θ varies over the interval 0 ≤ θ < 2π and a rose with an odd number of petals is traced out exactly once as θ varies over the interval 0 ≤ θ < π (Exercise 78). A four-petal rose of radius 1 was graphed in Example 6. rose curves n=2

n=3

n=4

n=5

n=6

r = a sin nu

r = a cos nu

Figure 10.2.19

FAMILIES OF CARDIOIDS AND LIMAÇONS

Equations with any of the four forms r = a ± b sin θ

r = a ± b cos θ

(8–9)

in which a > 0 and b > 0 represent polar curves called limaçons (from the Latin word limax for a snail-like creature that is commonly called a “slug”). There are four possible shapes for a limaçon that are determined by the ratio a /b (Figure 10.2.20). If a = b (the case a /b = 1), then the limaçon is called a cardioid because of its heart-shaped appearance, as noted in Example 8.

714

Chapter 10 / Parametric and Polar Curves; Conic Sections

Figure 10.2.20

a/b < 1

a/ b = 1

1 < a/ b < 2

a/b ≥ 2

Limaçon with inner loop

Cardioid

Dimpled limaçon

Convex limaçon

Example 10 Figure 10.2.21 shows the family of limaçons r = a + cos θ with the constant a varying from 0.25 to 2.50 in steps of 0.25. In keeping with Figure 10.2.20, the limaçons evolve from the loop type to the convex type. As a increases from the starting value of 0.25, the loops get smaller and smaller until the cardioid is reached at a = 1. As a increases further, the limaçons evolve through the dimpled type into the convex type.

a = 0.25

a = 0.5

a = 0.75

a=1

a = 1.25

a = 1.5

a = 1.75

a=2

a = 2.25

a = 2.5

r = a + cos u Figure 10.2.21

FAMILIES OF SPIRALS A spiral is a curve that coils around a central point. Spirals generally have “left-hand” and “right-hand” versions that coil in opposite directions, depending on the restrictions on the polar angle and the signs of constants that appear in their equations. Some of the more common types of spirals are shown in Figure 10.2.22 for nonnegative values of θ, a, and b.

c/ 2

0 0

Archimedean spiral

Parabolic spiral

Logarithmic spiral

Lituus spiral

Hyperbolic spiral

r = au

r = a √u

r = ae bu

r = a /√u

r = a /u

Figure 10.2.22

SPIRALS IN NATURE

Spirals of many kinds occur in nature. For example, the shell of the chambered nautilus (below) forms a logarithmic spiral, and a coiled sailor’s rope forms an Archimedean spiral. Spirals also occur in ﬂowers, the tusks of certain animals, and in the shapes of galaxies.

10.2 Polar Coordinates

715

Courtesy NASA & The Hubble Heritage Team

The shell of the chambered nautilus reveals a logarithmic spiral. The animal lives in the outermost chamber.

A sailor’s coiled rope forms an Archimedean spiral.

A spiral galaxy.

GENERATING POLAR CURVES WITH GRAPHING UTILITIES

For polar curves that are too complicated for hand computation, graphing utilities can be used. Although many graphing utilities are capable of graphing polar curves directly, some are not. However, if a graphing utility is capable of graphing parametric equations, then it can be used to graph a polar curve r = f(θ) by converting this equation to parametric form. This can be done by substituting f(θ) for r in (1). This yields x = f(θ) cos θ,

y = f(θ) sin θ

(10)

which is a pair of parametric equations for the polar curve in terms of the parameter θ.

Example 11 Express the polar equation 5θ 2 parametrically, and generate the polar graph from the parametric equations using a graphing utility. r = 2 + cos

Solution. Substituting the given expression for r in x = r cos θ and y = r sin θ yields the parametric equations

5θ x = 2 + cos cos θ, 2

T E C H N O LO GY M A ST E R Y Use a graphing utility to duplicate the curve in Figure 10.2.23. If your graphing utility requires that t be used as the parameter, then you will have to replace θ by t in (10) to generate the graph.

5θ y = 2 + cos sin θ 2

Next, we need to find an interval over which to vary θ to produce the entire graph. To find such an interval, we will look for the smallest number of complete revolutions that must occur until the value of r begins to repeat. Algebraically, this amounts to finding the smallest positive integer n such that 5θ 5(θ + 2nπ) = 2 + cos 2 + cos 2 2 or cos

5θ 5θ + 5nπ = cos 2 2

716

Chapter 10 / Parametric and Polar Curves; Conic Sections c/ 2

r = 2 + cos

For this equality to hold, the quantity 5nπ must be an even multiple of π; the smallest n for which this occurs is n = 2. Thus, the entire graph will be traced in two revolutions, which means it can be generated from the parametric equations

5θ 5θ cos θ, y = 2 + cos sin θ (0 ≤ θ ≤ 4π) x = 2 + cos 2 2 This yields the graph in Figure 10.2.23.

5u 2

Figure 10.2.23

✔QUICK CHECK EXERCISES 10.2

(See page 719 for answers.)

1. (a) Rectangular coordinates of a point (x, y) may be recovered from its polar coordinates (r, θ ) by means of the equations x = and y = . (b) Polar coordinates (r, θ ) may be recovered from rectangular coordinates (x, y) by means of the equations r2 = and tan θ = . 2. Find the rectangular coordinates of the points whose polar coordinates are given. (a) (4, π/3) (b) (2, −π/6) (c) (6, −2π/3)

(d) (4, 5π/4)

EXERCISE SET 10.2

4. In each part, state the name that describes the polar curve most precisely: a rose, a line, a circle, a limaçon, a cardioid, a spiral, a lemniscate, or none of these. (a) r = 1 − θ (b) r = 1 + 2 sin θ (c) r = sin 2θ (d) r = cos2 θ (e) r = csc θ (f ) r = 2 + 2 cos θ (g) r = −2 sin θ

Graphing Utility

1–2 Plot the points in polar coordinates. ■

1. (a) (3, π/4) (d) (4, 7π/6) 2. (a) (2, −π/3) (d) (−5, −π/6)

3. In each part, ﬁnd polar coordinates satisfying the stated conditions for the point whose rectangular coordinates are √ (1, 3 ). (a) r ≥ 0 and 0 ≤ θ < 2π (b) r ≤ 0 and 0 ≤ θ < 2π

(b) (5, 2π/3) (e) (−6, −π) (b) (3/2, −7π/4) (e) (2, 4π/3)

3–4 Find the rectangular coordinates of the points whose polar

coordinates are given. ■ 3. (a) (6, π/6) (b) (d) (0, −π) (e) 4. (a) (−2, π/4) (b) (d) (3, 0) (e)

(7, 2π/3) (7, 17π/6) (6, −π/4) (−4, −3π/2)

5. In each part, a point is given in rectangular coordinates. Find two pairs of polar coordinates for the point, one pair satisfying r ≥ 0 and 0 ≤ θ < 2π, and the second pair satisfying r ≥ 0 and −2π < θ ≤ 0. √ (a) (−5, 0) (b) (2 3, √ −2) (c) (0, −2) (d) (−8, −8) (e) (−3, 3 3 ) (f ) (1, 1)

6. In each part, ﬁnd polar coordinates satisfying the stated conditions for the point whose rectangular coordinates are √ (− 3, 1).

(a) (b) (c) (d)

r r r r

≥0 ≤0 ≥0 ≤0

and and and and

0 ≤ θ < 2π 0 ≤ θ < 2π −2π < θ ≤ 0 −π < θ ≤ π

7–8 Use a calculating utility, where needed, to approximate the

polar coordinates of the points whose rectangular coordinates are given. ■ 7. (a) (3, 4) 8. (a) (−3, 4)

(b) (6, −8)

(b) (−3, 1.7)

9–10 Identify the curve by transforming the given polar equa-

tion to rectangular coordinates. ■ 9. (a) r = 2

10. (a) r = 5 sec θ (c) r = 4 cos θ + 4 sin θ

(b) r sin θ = 4

6 3 cos θ + 2 sin θ (b) r = 2 sin θ (d) r = sec θ tan θ (d) r =

11–12 Express the given equations in polar coordinates. ■

11. (a) x = 3 (c) x 2 + y 2 + 6y = 0

(b) x 2 + y 2 = 7 (d) 9xy = 4

10.2 Polar Coordinates

(b) x 2 + y 2 = 5 (d) x 2 (x 2 + y 2 ) = y 2

12. (a) y = −3 (c) x 2 + y 2 + 4x = 0

27. r = 3(1 + sin θ )

F O C U S O N C O N C E P TS

13–16 A graph is given in a rectangular θ r-coordinate system. Sketch the corresponding graph in polar coordinates. r

13.

14.

u c

u 6

17–20 Find an equation for the given polar graph. [Note: Nu-

meric labels on these graphs represent distances to the origin.] ■

17. (a)

39. r 2 = 16 sin 2θ

40. r = 4θ

(θ ≤ 0)

38. r 2 = cos 2θ 42. r = 4θ

(θ ≥ 0)

44. r = 3 sin 2θ

46. r = 2 cos 3θ

48. If the graph of r = f(θ ) drawn in rectangular θrcoordinates is symmetric about the r-axis, then the graph of r = f(θ ) drawn in polar coordinates is symmetric about the x-axis.

−1

36. r = 5 − 2 cos θ

47–50 True–False Determine whether the statement is true or false. Explain your answer. ■ 47. The polar coordinate pairs (−1, π/3) and (1, −2π/3) describe the same point.

32. r = 4 + 3 cos θ

35. r − 5 = 3 sin θ

45. r = 9 sin 4θ

16.

34. r = 3 + 4 cos θ

43. r = −2 cos 2θ

−2

33. r = 3 − sin θ

41. r = 4θ

−3

15.

30. r = 1 + 2 sin θ

37. r = −3 − 4 sin θ

■

28. r = 5 − 5 sin θ

29. r = 4 − 4 cos θ

31. r = −1 − cos θ

717

(b)

(c)

49. The portion of the polar graph of r = sin 2θ for values of θ between π/2 and π is contained in the second quadrant. 50. The graph of a dimpled limaçon passes through the polar origin.

1 5

Circle

18. (a)

Cardioid

(b)

1 3

Limaçon

19. (a)

Circle 3

Three-petal rose

(b)

Four-petal rose

20. (a)

Limaçon

Lemniscate

(b)

(c)

51–55 Determine a shortest parameter interval on which a complete graph of the polar equation can be generated, and then use a graphing utility to generate the polar graph. ■ θ θ 52. r = sin 51. r = cos 2 2 θ θ 53. r = 1 − 2 sin 54. r = 0.5 + cos 4 3 θ 55. r = cos 5 56. The accompanying figure shows the graph of the “butterfly curve” θ r = ecos θ − 2 cos 4θ + sin3 4 Determine a shortest parameter interval on which the complete butterfly can be generated, and then check your answer using a graphing utility. c/ 2

1 6

Cardioid

Five-petal rose

Circle

21–46 Sketch the curve in polar coordinates. ■

π 3 24. r = 4 cos θ

21. θ =

3π 4 25. r = 6 sin θ

22. θ = −

23. r = 3 26. r − 2 = 2 cos θ

Figure Ex-56

718

Chapter 10 / Parametric and Polar Curves; Conic Sections

57. The accompanying ďŹ gure shows the Archimedean spiral r = Î¸ /2 produced with a graphing calculator. (a) What interval of values for Î¸ do you think was used to generate the graph? (b) Duplicate the graph with your own graphing utility.

61â&#x20AC;&#x201C;62 A polar graph of r = f(Î¸ ) is given over the stated interval. Sketch the graph of Ď&#x20AC; (a) r = f(â&#x2C6;&#x2019;Î¸ ) (b) r = f Î¸ â&#x2C6;&#x2019; 2 Ď&#x20AC; (c) r = f Î¸ + (d) r = â&#x2C6;&#x2019;f(Î¸ ). â&#x2013; 2 / 61. 0 â&#x2030;¤ Î¸ â&#x2030;¤ Ď&#x20AC; 2 62. Ď&#x20AC;/2 â&#x2030;¤ Î¸ â&#x2030;¤ Ď&#x20AC; c/ 2

[â&#x2C6;&#x2019;9, 9] Ă&#x2014; [â&#x2C6;&#x2019;6, 6] xScl = 1, yScl = 1

c/ 2

(1, c/ 4)

(1, 3c/ 4)

Figure Ex-57

Figure Ex-61

58. Find equations for the two families of circles in the accompanying ďŹ gure. c/ 2

c/ 2

Figure Ex-62

63â&#x20AC;&#x201C;64 Use the polar graph from the indicated exercise to sketch the graph of (a) r = f(Î¸ ) + 1 (b) r = 2f (Î¸ ) â&#x2C6;&#x2019; 1. â&#x2013;

63. Exercise 61

64. Exercise 62

65. Show that if the polar graph of r = f(Î¸ ) is rotated counterclockwise around the origin through an angle Îą, then r = f(Î¸ â&#x2C6;&#x2019; Îą) is an equation for the rotated curve. [Hint: If (r0 , Î¸0 ) is any point on the original graph, then (r0 , Î¸0 + Îą) is a point on the rotated graph.]

I Figure Ex-58

66. Use the result in Exercise 65 to ďŹ nd an equation for the lemniscate that results when the lemniscate in Example 9 is rotated counterclockwise through an angle of Ď&#x20AC;/2.

59. (a) Show that if a varies, then the polar equation r = a sec Î¸ (â&#x2C6;&#x2019;Ď&#x20AC;/2 < Î¸ < Ď&#x20AC;/2)

describes a family of lines perpendicular to the polar axis. (b) Show that if b varies, then the polar equation r = b csc Î¸

(0 < Î¸ < Ď&#x20AC;)

describes a family of lines parallel to the polar axis. F O C U S O N C O N C E P TS

60. The accompanying ďŹ gure shows graphs of the Archiâ&#x2C6;&#x161; medean spiral r = Î¸ and the parabolic spiral r = Î¸ . Which is which? Explain your reasoning.

I Figure Ex-60

r = A sin Î¸ + B cos Î¸ is a circle. Find its radius. (b) Derive Formulas (4) and (5) from the formula given in part (a). 69. Find the highest point on the cardioid r = 1 + cos Î¸.

70. Find the leftmost point on the upper half of the cardioid r = 1 + cos Î¸.

c/ 2

67. Use the result in Exercise 65 to ďŹ nd an equation for the cardioid r = 1 + cos Î¸ after it has been rotated through the given angle, and check your answer with a graphing utility. Ď&#x20AC; Ď&#x20AC; 5Ď&#x20AC; (a) (b) (c) Ď&#x20AC; (d) 4 2 4 68. (a) Show that if A and B are not both zero, then the graph of the polar equation

71. Show that in a polar coordinate system the distance d between the points (r1 , Î¸1 ) and (r2 , Î¸2 ) is d = r12 + r22 â&#x2C6;&#x2019; 2r1 r2 cos(Î¸1 â&#x2C6;&#x2019; Î¸2 )

72â&#x20AC;&#x201C;74 Use the formula obtained in Exercise 71 to ďŹ nd the distance between the two points indicated in polar coordinates. â&#x2013; 72. (3, Ď&#x20AC;/6) and (2, Ď&#x20AC;/3)

10.3 Tangent Lines, Arc Length, and Area for Polar Curves

73. Successive tips of the four-petal rose r = cos 2θ . Check your answer using geometry. 74. Successive tips of the three-petal rose r = sin 3θ. Check your answer using trigonometry. 75. In the late seventeenth century the Italian astronomer Giovanni Domenico Cassini (1625–1712) introduced the family of curves 2

2 2

(x + y + a ) − b − 4a x = 0

(a > 0, b > 0)

in his studies of the relative motions of the Earth and the Sun. These curves, which are called Cassini ovals, have one of the three basic shapes shown in the accompanying ﬁgure. (a) Show that if a = b, then the polar equation of the Cassini oval is r 2 = 2a 2 cos 2θ , which is a lemniscate. (b) Use the formula in Exercise 71 to show that the lemniscate in part (a) is the curve traced by a point that moves in such a way that the product of its distances from the polar points (a, 0) and (a, π) is a 2 . y

a=b

719

76. Show that the hyperbolic spiral r = 1/θ (θ > 0) has a horizontal asymptote at y = 1 by showing that y → 1 and x → +⬁ as θ → 0+ . Conﬁrm this result by generating the spiral with a graphing utility. 77. Show that the spiral r = 1/θ 2 does not have any horizontal asymptotes. 78. Prove that a rose with an even number of petals is traced out exactly once as θ varies over the interval 0 ≤ θ < 2π and a rose with an odd number of petals is traced out exactly once as θ varies over the interval 0 ≤ θ < π.

79. (a) Use a graphing utility to conﬁrm that the graph of r = 2 − sin(θ /2) (0 ≤ θ ≤ 4π) is symmetric about the x-axis. (b) Show that replacing θ by −θ in the polar equation r = 2 − sin(θ /2) does not produce an equivalent equation. Why does this not contradict the symmetry demonstrated in part (a)? 80. Writing Use a graphing utility to investigate how the family of polar curves r = 1 + a cos nθ is affected by changing the values of a and n, where a is a positive real number and n is a positive integer. Write a brief paragraph to explain your conclusions. 81. Writing Why do you think the adjective “polar” was chosen in the name “polar coordinates”?

a>b

a<b Figure Ex-75

76–77 Vertical and horizontal asymptotes of polar curves can sometimes be detected by investigating the behavior of x = r cos θ and y = r sin θ as θ varies. This idea is used in these exercises. ■

✔QUICK CHECK ANSWERS 10.2 √ √ √ √ √ 1. (a) r cos θ ; r sin θ (b) x 2 + y 2 ; y /x 2. (a) (2, 2 3) (b) ( 3, −1) (c) (−3, −3 3) (d) (−2 2, −2 2) 3. (a) (2, π/3) (b) (−2, 4π/3) 4. (a) spiral (b) limaçon (c) rose (d) none of these (e) line (f ) cardioid (g) circle

10.3

TANGENT LINES, ARC LENGTH, AND AREA FOR POLAR CURVES In this section we will derive the formulas required to ﬁnd slopes, tangent lines, and arc lengths of polar curves. We will then show how to ﬁnd areas of regions that are bounded by polar curves. TANGENT LINES TO POLAR CURVES

Our ﬁrst objective in this section is to ﬁnd a method for obtaining slopes of tangent lines to polar curves of the form r = f(θ) in which r is a differentiable function of θ. We showed in the last section that a curve of this form can be expressed parametrically in terms of the parameter θ by substituting f(θ) for r in the equations x = r cos θ and y = r sin θ. This yields x = f(θ) cos θ, y = f(θ) sin θ

720

Chapter 10 / Parametric and Polar Curves; Conic Sections

from which we obtain dx dr cos θ = −f (θ) sin θ + f ′(θ) cos θ = −r sin θ + dθ dθ (1) dr dy ′ = f (θ) cos θ + f (θ) sin θ = r cos θ + sin θ dθ dθ Thus, if dx/dθ and dy/dθ are continuous and if dx/dθ = 0, then y is a differentiable function of x, and Formula (4) in Section 10.1 with θ in place of t yields dr r cos θ + sin θ dy/dθ dy dθ = = dr dx dx/dθ −r sin θ + cos θ dθ

(2)

Example 1 Find the slope of the tangent line to the circle r = 4 cos θ at the point where θ = π/4.

Solution. From (2) with r = 4 cos θ, so that dr /dθ = −4 sin θ, we obtain dy 4 cos2 θ − 4 sin2 θ cos2 θ − sin2 θ = =− dx −8 sin θ cos θ 2 sin θ cos θ

c/ 2

Using the double-angle formulas for sine and cosine,

Tangent

c/ 4 4

r = 4 cos u Figure 10.3.1

cos 2θ dy =− = − cot 2θ dx sin 2θ Thus, at the point where θ = π/4 the slope of the tangent line is dy π m= = − cot = 0 dx θ=π/4 2

which implies that the circle has a horizontal tangent line at the point where θ = π/4 (Figure 10.3.1). Example 2 Find the points on the cardioid r = 1 − cos θ at which there is a horizontal tangent line, a vertical tangent line, or a singular point.

Solution. A horizontal tangent line will occur where dy/dθ = 0 and dx/dθ = 0, a ver-

tical tangent line where dy/dθ = 0 and dx/dθ = 0, and a singular point where dy/dθ = 0 and dx/dθ = 0. We could ﬁnd these derivatives from the formulas in (1). However, an alternative approach is to go back to basic principles and express the cardioid parametrically by substituting r = 1 − cos θ in the conversion formulas x = r cos θ and y = r sin θ. This yields x = (1 − cos θ) cos θ, y = (1 − cos θ ) sin θ (0 ≤ θ ≤ 2π) Differentiating these equations with respect to θ and then simplifying yields (verify) dx = sin θ(2 cos θ − 1), dθ

dy = (1 − cos θ)(1 + 2 cos θ ) dθ

Thus, dx/dθ = 0 if sin θ = 0 or cos θ = 21 , and dy/dθ = 0 if cos θ = 1 or cos θ = − 21 . We leave it for you to solve these equations and show that the solutions of dx/dθ = 0 on the interval 0 ≤ θ ≤ 2π are dx = 0: dθ

θ = 0,

π 5π , π, , 2π 3 3

10.3 Tangent Lines, Arc Length, and Area for Polar Curves c/ 2

and the solutions of dy/dθ = 0 on the interval 0 ≤ θ ≤ 2π are

冸 12 , c3 冹 0

冸

1 , 5c 2 3

2π 4π , , 2π 3 3 Thus, horizontal tangent lines occur at θ = 2π/3 and θ = 4π/3; vertical tangent lines occur at θ = π/3, π, and 5π/3; and singular points occur at θ = 0 and θ = 2π (Figure 10.3.2). Note, however, that r = 0 at both singular points, so there is really only one singular point on the cardioid—the pole. dy = 0: dθ

冸 32 , 2c3 冹 (2, c)

721

冹

冸 32 , 4c3 冹

θ = 0,

TANGENT LINES TO POLAR CURVES AT THE ORIGIN

r = 1 − cos u

Formula (2) reveals some useful information about the behavior of a polar curve r = f(θ) that passes through the origin. If we assume that r = 0 and dr /dθ = 0 when θ = θ0 , then it follows from Formula (2) that the slope of the tangent line to the curve at θ = θ0 is

Figure 10.3.2

dr 0 + sin θ0 dy dθ = sin θ0 = tan θ = 0 dr dx cos θ0 0 + cos θ0 dθ c/ 2

(Figure 10.3.3). However, tan θ0 is also the slope of the line θ = θ0 , so we can conclude that this line is tangent to the curve at the origin. Thus, we have established the following result.

r = f (u)

u = u0

10.3.1 theorem If the polar curve r = f(θ) passes through the origin at θ = θ0 , and if dr /dθ = 0 at θ = θ0 , then the line θ = θ0 is tangent to the curve at the origin.

Slope = tan u0

This theorem tells us that equations of the tangent lines at the origin to the curve r = f(θ) can be obtained by solving the equation f(θ) = 0. It is important to keep in mind, however, that r = f(θ) may be zero for more than one value of θ, so there may be more than one tangent line at the origin. This is illustrated in the next example.

Figure 10.3.3

c/ 2

u = 2c/ 3

Example 3 The three-petal rose r = sin 3θ in Figure 10.3.4 has three tangent lines at the origin, which can be found by solving the equation

u = c/ 3

sin 3θ = 0 0

It was shown in Exercise 78 of Section 10.2 that the complete rose is traced once as θ varies over the interval 0 ≤ θ < π, so we need only look for solutions in this interval. We leave it for you to conﬁrm that these solutions are θ = 0,

r = sin 3u

θ=

π , 3

and θ =

2π 3

Since dr /dθ = 3 cos 3θ = 0 for these values of θ, these three lines are tangent to the rose at the origin, which is consistent with the ﬁgure.

Figure 10.3.4

ARC LENGTH OF A POLAR CURVE A formula for the arc length of a polar curve r = f(θ) can be derived by expressing the curve in parametric form and applying Formula (9) of Section 10.1 for the arc length of a parametric curve. We leave it as an exercise to show the following.

722

Chapter 10 / Parametric and Polar Curves; Conic Sections

10.3.2 arc length formula for polar curves If no segment of the polar curve r = f(θ) is traced more than once as θ increases from α to β, and if dr /dθ is continuous for α ≤ θ ≤ β, then the arc length L from θ = α to θ = β is L= c/ 2 r=e

Solution. (c, e c )

(1, 0)

Figure 10.3.5

[f(θ)]2 + [f ′(θ)]2 dθ =

r2 +

dr dθ

dθ

(3)

Find the arc length of the spiral r = eθ in Figure 10.3.5 between θ = 0

Example 4 and θ = π.

π√

Example 5

r2 +

dr dθ

2 eθ dθ =

dθ =

√ θ 2e

(eθ )2 + (eθ )2 dθ

√ π 2(e − 1) ≈ 31.3

Find the total arc length of the cardioid r = 1 + cos θ.

Solution. The cardioid is traced out once as θ varies from θ = 0 to θ = 2π. Thus, L=

dr dθ

dθ = =

c/ 2

√ 2

=2 r = 1 + cos u

=2 0

Figure 10.3.6

2π

(1 + cos θ )2 + (− sin θ )2 dθ

2π √

0 2π

1 + cos θ dθ

cos2 21 θ dθ

2π

Identity (45) of Appendix B

cos 1 θ dθ 2

Since cos 21 θ changes sign at π, we must split the last integral into the sum of two integrals: the integral from 0 to π plus the integral from π to 2π. However, the integral from π to 2π is equal to the integral from 0 to π, since the cardioid is symmetric about the polar axis (Figure 10.3.6). Thus,

π π 2π 1 1 1 cos 2 θ dθ = 8 sin 2 θ = 8 L=2 cos 2 θ dθ = 4 0

AREA IN POLAR COORDINATES We begin our investigation of area in polar coordinates with a simple case. r = f (u) R u=b

Figure 10.3.7

u=a

10.3.3 area problem in polar coordinates Suppose that α and β are angles that satisfy the condition α < β ≤ α + 2π and suppose that f(θ) is continuous and nonnegative for α ≤ θ ≤ β. Find the area of the region R enclosed by the polar curve r = f(θ) and the rays θ = α and θ = β (Figure 10.3.7).

10.3 Tangent Lines, Arc Length, and Area for Polar Curves

u = u2

u = un−1

Δun

Δu2

... ..

u=b

u = u1

u=a

Δu1 Figure 10.3.8

u = u *k

723

In rectangular coordinates we obtained areas under curves by dividing the region into an increasing number of vertical strips, approximating the strips by rectangles, and taking a limit. In polar coordinates rectangles are clumsy to work with, and it is better to partition the region into wedges by using rays θ = θ1 , θ = θ2 , . . . , θ = θn−1 such that

α < θ1 < θ2 < · · · < θn−1 < β

(Figure 10.3.8). As shown in that ﬁgure, the rays divide the region R into n wedges with areas A1 , A2 , . . . , An and central angles θ1 , θ2 , . . . , θn . The area of the entire region n can be written as A = A1 + A2 + · · · + An = Ak (4) k=1

r = f (u) Δuk u *k Figure 10.3.9

u=b Figure 10.3.10

If θk is small, then we can approximate the area Ak of the kth wedge by the area of a sector with central angle θk and radius f(θk∗ ), where θ = θk∗ is any ray that lies in the kth wedge (Figure 10.3.9). Thus, from (4) and Formula (5) of Appendix B for the area of a sector, we obtain n n 1 A= Ak ≈ (5) [f(θk∗ )]2 θk 2 k=1

u=a

k=1

If we now increase n in such a way that max θk → 0, then the sectors will become better and better approximations of the wedges and it is reasonable to expect that (5) will approach the exact value of the area A (Figure 10.3.10); that is, β n 1 1 ∗ 2 A = lim [f(θk )] θk = [f(θ)]2 dθ 2 2 max θk → 0

k=1

Note that the discussion above can easily be adapted to the case where f(θ) is nonpositive for α ≤ θ ≤ β. We summarize this result below. 10.3.4

area in polar coordinates

If α and β are angles that satisfy the condition

α < β ≤ α + 2π and if f(θ) is continuous and either nonnegative or nonpositive for α ≤ θ ≤ β, then the area A of the region R enclosed by the polar curve r = f(θ) (α ≤ θ ≤ β) and the lines θ = α and θ = β is A=

β 1 [f(θ)]2 2

dθ =

β 1 2 r 2

dθ

(6)

The hardest part of applying (6) is determining the limits of integration. This can be done as follows: Area in Polar Coordinates: Limits of Integration Step 1. Sketch the region R whose area is to be determined. Step 2. Draw an arbitrary “radial line” from the pole to the boundary curve r = f(θ). Step 3. Ask, “Over what interval of values must θ vary in order for the radial line to sweep out the region R?” Step 4. Your answer in Step 3 will determine the lower and upper limits of integration.

724

Chapter 10 / Parametric and Polar Curves; Conic Sections

Example 6 r = 1 − cos θ. r = 1 − cos u

c/ 2

Find the area of the region in the ﬁrst quadrant that is within the cardioid

Solution. The region and a typical radial line are shown in Figure 10.3.11. For the radial

line to sweep out the region, θ must vary from 0 to π/2. Thus, from (6) with α = 0 and β = π/2, we obtain π/2 1 π/2 1 π/2 1 2 A= r dθ = (1 − cos θ)2 dθ = (1 − 2 cos θ + cos2 θ ) dθ 2 2 2 0 0 0

With the help of the identity cos2 θ = 21 (1 + cos 2θ ), this can be rewritten as The shaded region is swept out by the radial line as u varies from 0 to c/ 2.

Figure 10.3.11

1 2

π/2 0

Example 7

π/2 1 1 3 1 3 3 = π−1 − 2 cos θ + cos 2θ dθ = θ − 2 sin θ + sin 2θ 2 2 2 2 4 8 0

Find the entire area within the cardioid of Example 6.

Solution. For the radial line to sweep out the entire cardioid, θ must vary from 0 to 2π. Thus, from (6) with α = 0 and β = 2π, 2π 1 2 1 2π r dθ = (1 − cos θ )2 dθ A= 2 2 0 0 If we proceed as in Example 6, this reduces to 1 3π 1 2π 3 − 2 cos θ + cos 2θ dθ = A= 2 0 2 2 2

Alternative Solution. Since the cardioid is symmetric about the x-axis, we can calculate the portion of the area above the x-axis and double the result. In the portion of the cardioid above the x-axis, θ ranges from 0 to π, so that π π 1 2 3π (1 − cos θ)2 dθ = r dθ = A=2 2 0 2 0 USING SYMMETRY Although Formula (6) is applicable if r = f(θ) is negative, area computations can sometimes be simpliﬁed by using symmetry to restrict the limits of integration to intervals where r ≥ 0. This is illustrated in the next example.

Example 8

Find the area of the region enclosed by the rose curve r = cos 2θ.

Solution. Referring to Figure 10.2.13 and using symmetry, the area in the ﬁrst quadrant that is swept out for 0 ≤ θ ≤ π/4 is one-eighth of the total area inside the rose. Thus, from Formula (6) π/4 π/4 1 2 A=8 cos2 2θ dθ r dθ = 4 2 0 0 π/4 π/4 1 =4 (1 + cos 4θ ) dθ (1 + cos 4θ ) dθ = 2 2 0 0

π/4 1 π = 2θ + sin 4θ = 2 2 0

10.3 Tangent Lines, Arc Length, and Area for Polar Curves

725

Sometimes the most natural way to satisfy the restriction α < β ≤ α + 2π required by Formula (6) is to use a negative value for α. For example, suppose that we are interested in finding the area of the shaded region in Figure 10.3.12a. The first step would be to determine the intersections of the cardioid r = 4 + 4 cos θ and the circle r = 6, since this information is needed for the limits of integration. To find the points of intersection, we can equate the two expressions for r. This yields 4 + 4 cos θ = 6 or

cos θ =

1 2

which is satisﬁed by the positive angles π 5π and θ = 3 3 However, there is a problem here because the radial lines to the circle and cardioid do not sweep through the shaded region shown in Figure 10.3.12b as θ varies over the interval π/3 ≤ θ ≤ 5π/3. There are two ways to circumvent this problem—one is to take advantage of the symmetry by integrating over the interval 0 ≤ θ ≤ π/3 and doubling the result, and the second is to use a negative lower limit of integration and integrate over the interval −π/3 ≤ θ ≤ π/3 (Figure 10.3.12c). The two methods are illustrated in the next example. θ=

c/ 2

c/ 2 u=4

r=6

(b)

c/ 2 u=4

u=k

r = 4 + 4 cos u

(a)

c/ 2 u=4

u=$

(c)

u=4

u=$

(d)

u=$

(e)

Figure 10.3.12

Example 9 Find the area of the region that is inside of the cardioid r = 4 + 4 cos θ and outside of the circle r = 6.

Solution Using a Negative Angle. The area of the region can be obtained by subtracting the areas in Figures 10.3.12d and 10.3.12e: π/3 π/3 1 2 1 Area inside cardioid A= (4 + 4 cos θ)2 dθ − (6) dθ minus area inside circle. −π/3 2 −π/3 2 π/3 π/3 1 (16 cos θ + 8 cos2 θ − 10) dθ [(4 + 4 cos θ)2 − 36] dθ = = 2 / / −π 3 −π 3 √ π/3 = 16 sin θ + (4θ + 2 sin 2θ) − 10 θ −π/3 = 18 3 − 4π

Solution Using Symmetry. Using symmetry, we can calculate the area above the polar axis and double it. This yields (verify) π/3 √ √ 1 A=2 [(4 + 4 cos θ )2 − 36] dθ = 2(9 3 − 2π) = 18 3 − 4π 2 0 which agrees with the preceding result.

726

Chapter 10 / Parametric and Polar Curves; Conic Sections c/ 2

r = 1 − cos u

INTERSECTIONS OF POLAR GRAPHS

In the last example we found the intersections of the cardioid and circle by equating their expressions for r and solving for θ. However, because a point can be represented in different ways in polar coordinates, this procedure will not always produce all of the intersections. For example, the cardioids

r = 1 + cos u

冸1, 6冹 0

冸1, i冹

Figure 10.3.13

The orbits intersect, but the satellites do not collide.

r = 1 − cos θ

and r = 1 + cos θ

(7)

intersect at three points: the pole, the point (1, π/2), and the point (1, 3π/2) (Figure 10.3.13). Equating the right-hand sides of the equations in (7) yields 1 − cos θ = 1 + cos θ or cos θ = 0, so π θ = + kπ, k = 0, ±1, ±2, . . . 2 Substituting any of these values in (7) yields r = 1, so that we have found only two distinct points of intersection, (1, π/2) and (1, 3π/2); the pole has been missed. This problem occurs because the two cardioids pass through the pole at different values of θ—the cardioid r = 1 − cos θ passes through the pole at θ = 0, and the cardioid r = 1 + cos θ passes through the pole at θ = π. The situation with the cardioids is analogous to two satellites circling the Earth in intersecting orbits (Figure 10.3.14). The satellites will not collide unless they reach the same point at the same time. In general, when looking for intersections of polar curves, it is a good idea to graph the curves to determine how many intersections there should be.

Figure 10.3.14

✔QUICK CHECK EXERCISES 10.3

(See page 729 for answers.)

1. (a) To obtain dy /dx directly from the polar equation r = f(θ ), we can use the formula dy dy /dθ = = dx dx /dθ (b) Use the formula in part (a) to ﬁnd dy /dx directly from the polar equation r = csc θ. ′

2. (a) What conditions on f(θ0 ) and f (θ0 ) guarantee that the line θ = θ0 is tangent to the polar curve r = f(θ ) at the origin?

EXERCISE SET 10.3

Graphing Utility

(b) What are the values of θ0 in [0, 2π] at which the lines θ = θ0 are tangent at the origin to the four-petal rose r = cos 2θ ?

3. (a) To ﬁnd the arc length L of the polar curve r = f(θ ) (α ≤ θ ≤ β), we can use the formula L = . (b) The polar curve r = sec θ (0 ≤ θ ≤ π/4) has arc length L= .

4. The area of the region enclosed by a nonnegative polar curve r = f(θ ) (α ≤ θ ≤ β) and the lines θ = α and θ = β is given by the deﬁnite integral . 5. Find the area of the circle r = a by integration.

CAS

1–6 Find the slope of the tangent line to the polar curve for the

c/ 2

given value of θ. ■ 1. r = 2 sin θ ; θ = π/6 3. r = 1/θ ; θ = 2 5. r = sin 3θ; θ = π/4

7. r = 2 + 2 sin θ 8. r = 1 − 2 sin θ

6. r = 4 − 3 sin θ; θ = π

7–8 Calculate the slopes of the tangent lines indicated in the

accompanying ﬁgures. ■

2. r = 1 + cos θ ; θ = π/2 4. r = a sec 2θ; θ = π/6 Figure Ex-7

Figure Ex-8

9–10 Find polar coordinates of all points at which the polar

curve has a horizontal or a vertical tangent line. ■ 9. r = a(1 + cos θ )

10. r = a sin θ

10.3 Tangent Lines, Arc Length, and Area for Polar Curves 11–12 Use a graphing utility to make a conjecture about the

number of points on the polar curve at which there is a horizontal tangent line, and conﬁrm your conjecture by ﬁnding appropriate derivatives. ■ 11. r = sin θ cos2 θ

12. r = 1 − 2 sin θ

13–18 Sketch the polar curve and ﬁnd polar equations of the tangent lines to the curve at the pole. ■ √ 13. r = 2 cos 3θ 14. r = 4 sin θ 15. r = 4 cos 2θ

16. r = sin 2θ

17. r = 1 − 2 cos θ

18. r = 2θ

19–22 Use Formula (3) to calculate the arc length of the polar curve. ■

19. The entire circle r = a

727

27. In each part, ﬁnd the area of the circle by integration. (a) r = 2a sin θ (b) r = 2a cos θ

28. (a) Show that r = 2 sin θ + 2 cos θ is a circle. (b) Find the area of the circle using a geometric formula and then by integration.

29–34 Find the area of the region described. ■

29. The region that is enclosed by the cardioid r = 2 + 2 sin θ.

30. The region in the ﬁrst quadrant within the cardioid r = 1 + cos θ. 31. The region enclosed by the rose r = 4 cos 3θ. 32. The region enclosed by the rose r = 2 sin 2θ.

20. The entire circle r = 2a cos θ

33. The region enclosed by the inner loop of the limaçon r = 1 + 2 cos θ. [Hint: r ≤ 0 over the interval of integration.]

22. r = e3θ from θ = 0 to θ = 2

34. The region swept out by a radial line from the pole to the curve r = 2/θ as θ varies over the interval 1 ≤ θ ≤ 3.

21. The entire cardioid r = a(1 − cos θ ) 23. (a) Show that the arc length of one petal of the rose r = cos nθ is given by 2

π/(2n)

1 + (n2 − 1) sin2 nθ dθ

(b) Use the numerical integration capability of a calculating utility to approximate the arc length of one petal of the four-petal rose r = cos 2θ. (c) Use the numerical integration capability of a calculating utility to approximate the arc length of one petal of the n-petal rose r = cos nθ for n = 2, 3, 4, . . . , 20; then make a conjecture about the limit of these arc lengths as n → +⬁.

35–38 Find the area of the shaded region. ■

35.

36.

r = 1 + cos u r = cos u

r = √cos 2u r = 2 cos u

37.

38.

24. (a) Sketch the spiral r = e−θ /8 (0 ≤ θ < +⬁). (b) Find an improper integral for the total arc length of the spiral. (c) Show that the integral converges and ﬁnd the total arc length of the spiral.

39–46 Find the area of the region described. ■

25. Write down, but do not evaluate, an integral for the area of each shaded region.

39. The region inside the circle r = 3 sin θ and outside the cardioid r = 1 + sin θ.

(a)

(b)

r = 4 cos u r = 4√3 sin u

r = 1 + cos u r = 3 cos u

40. The region outside the cardioid r = 2 − 2 cos θ and inside the circle r = 4.

(c)

41. The region inside the cardioid r = 2 + 2 cos θ and outside the circle r = 3.

r = 1 − cos u

(d)

r = 2 cos u

(e)

r = sin 2u

42. The region that is common to the circles r = 2 cos θ and r = 2 sin θ.

43. The region between the loops of the limaçon r =

(f)

1 2

+ cos θ.

44. The region inside the cardioid r = 2 + 2 cos θ and to the right of the line r cos θ = 23 .

r=u

r = 1 − sin u

r = cos 2u

26. Find the area of the shaded region in Exercise 25(d).

45. The region inside the circle r = 2 and to the right of the line √ r = 2 sec θ .

46. The region √ inside the rose r = 2a cos 2θ and outside the circle r = a 2.

728

Chapter 10 / Parametric and Polar Curves; Conic Sections

47–50 True–False Determine whether the statement is true or false. Explain your answer. ■ 47. The x-axis is tangent to the polar curve r = cos(θ /2) at θ = 3π. √ 48. The arc length of the polar curve r = θ for 0 ≤ θ ≤ π/2 is given by π/2 1 L= 1+ dθ 4θ 0

Bug

t= 5s

49. The area of a sector with central angle θ taken from a circle of radius r is θr 2 . 50. The expression 1 2

π/4 −π/4

(1 −

√

F O C U S O N C O N C E P TS

51. (a) Find the error: The area that is inside the lemniscate r 2 = a 2 cos 2θ is A= =

2π

1 2 a 4

1 2 r 2

dθ =

sin 2θ

2π 0

2π

1 2 a 2

cos 2θ dθ

(b) Find the correct area. (c) Find the area inside the lemniscate r 2 = 4 cos 2θ √ and outside the circle r = 2.

52. Find the area inside the curve r 2 = sin 2θ.

53. A radial line is drawn from the origin to the spiral r = aθ (a > 0 and θ ≥ 0). Find the area swept out during the second revolution of the radial line that was not swept out during the first revolution. 54. As illustrated in the accompanying figure, suppose that a rod with one end fixed at the pole of a polar coordinate system rotates counterclockwise at the constant rate of 1 rad/s. At time t = 0 a bug on the rod is 10 mm from the pole and is moving outward along the rod at the constant speed of 2 mm/s. (a) Find an equation of the form r = f(θ ) for the path of motion of the bug, assuming that θ = 0 when t = 0. (b) Find the distance the bug travels along the path in part (a) during the first 5 s. Round your answer to the nearest tenth of a millimeter.

Figure Ex-54

55. (a) Show that the Folium of Descartes x 3 − 3xy + y 3 = 0 can be expressed in polar coordinates as 3 sin θ cos θ r= cos3 θ + sin3 θ (b) Use a CAS to show that the area inside of the loop is 23 (Figure 3.1.3a).

56. (a) What is the area that is enclosed by one petal of the rose r = a cos nθ if n is an even integer? (b) What is the area that is enclosed by one petal of the rose r = a cos nθ if n is an odd integer? (c) Use a CAS to show that the total area enclosed by the rose r = a cos nθ is πa 2 /2 if the number of petals is even. [Hint: See Exercise 78 of Section 10.2.] (d) Use a CAS to show that the total area enclosed by the rose r = a cos nθ is πa 2 /4 if the number of petals is odd.

2 cos θ )2 dθ

computes√the area enclosed by the inner loop of the limaçon r = 1 − 2 cos θ.

t=0s

57. One of the most famous problems in Greek antiquity was “squaring the circle,” that is, using a straightedge and compass to construct a square whose area is equal to that of a given circle. It was proved in the nineteenth century that no such construction is possible. However, show that the shaded areas in the accompanying ﬁgure are equal, thereby “squaring the crescent.” c/ 2

Figure Ex-57

58. Use a graphing utility to generate the polar graph of the equation r = cos 3θ + 2, and ﬁnd the area that it encloses.

59. Use a graphing utility to generate the graph of the bifolium r = 2 cos θ sin2 θ , and ﬁnd the area of the upper loop. 60. Use Formula (9) of Section 10.1 to derive the arc length formula for polar curves, Formula (3).

61. As illustrated in the accompanying ﬁgure, let P (r, θ ) be a point on the polar curve r = f(θ ), let ψ be the smallest counterclockwise angle from the extended radius OP to the

10.3 Tangent Lines, Arc Length, and Area for Polar Curves

tangent line at P , and let φ be the angle of inclination of the tangent line. Derive the formula r tan ψ = / dr dθ by substituting tan φ for dy/dx in Formula (2) and applying the trigonometric identity tan φ − tan θ tan(φ − θ) = 1 + tan φ tan θ c

r = f (u)

P(r, u) Tangent line

u O

surface of revolution that is generated by revolving a parametric curve about the x-axis or y-axis. Use those formulas to derive the following formulas for the areas of the surfaces of revolution that are generated by revolving the portion of the polar curve r = f(θ ) from θ = α to θ = β about the polar axis and about the line θ = π/2:

dr 2 2πr sin θ S= dθ + dθ α 2 β dr dθ 2πr cos θ r 2 + S= dθ α

About θ = 0

About θ = π/2

(b) State conditions under which these formulas hold.

f Figure Ex-61

62–63 Use the formula for ψ obtained in Exercise 61. ■

62. (a) Use the trigonometric identity θ 1 − cos θ tan = 2 sin θ to show that if (r, θ ) is a point on the cardioid r = 1 − cos θ (0 ≤ θ < 2π) then ψ = θ /2. (b) Sketch the cardioid and show the angle ψ at the points where the cardioid crosses the y-axis. (c) Find the angle ψ at the points where the cardioid crosses the y-axis. 63. Show that for a logarithmic spiral r = aebθ , the angle from the radial line to the tangent line is constant along the spiral (see the accompanying ﬁgure). [Note: For this reason, logarithmic spirals are sometimes called equiangular spirals.]

Figure Ex-63

64. (a) In the discussion associated with Exercises 75–80 of Section 10.1, formulas were given for the area of the

65–68 Sketch the surface, and use the formulas in Exercise 64 to ﬁnd the surface area. ■

65. The surface generated by revolving the circle r = cos θ about the line θ = π/2.

66. The surface generated by revolving the spiral r = eθ (0 ≤ θ ≤ π/2) about the line θ = π/2.

67. The “apple” generated by revolving the upper half of the cardioid r = 1 − cos θ (0 ≤ θ ≤ π) about the polar axis. 68. The sphere of radius a generated by revolving the semicircle r = a in the upper half-plane about the polar axis.

69. Writing (a) Show that if 0 ≤ θ1 < θ2 ≤ π and if r1 and r2 are positive, then the area A of a triangle with vertices (0, 0), (r1 , θ1 ), and (r2 , θ2 ) is A = 21 r1 r2 sin(θ2 − θ1 )

(b) Use the formula obtained in part (a) to describe an approach to answer Area Problem 10.3.3 that uses an approximation of the region R by triangles instead of circular wedges. Reconcile your approach with Formula (6). 70. Writing In order to ﬁnd the area of a region bounded by two polar curves it is often necessary to determine their points of intersection. Give an example to illustrate that the points of intersection of curves r = f(θ ) and r = g(θ ) may not coincide with solutions to f (θ ) = g(θ ). Discuss some strategies for determining intersection points of polar curves and provide examples to illustrate your strategies.

✔QUICK CHECK ANSWERS 10.3 dr dy π 3π 5π 7π dθ = 0 2. (a) f(θ0 ) = 0, f ′(θ0 ) = 0 (b) θ0 = , , , 1. (a) (b) dr dx 4 4 4 4 −r sin θ + cos θ dθ 2 β β β 2π dr 1 2 1 1 2 2 2 3. (a) [f(θ )] dθ = r dθ 5. a dθ = πa 2 r + dθ (b) 1 4. 2 2 2 dθ α α α 0 r cos θ + sin θ

729

730

Chapter 10 / Parametric and Polar Curves; Conic Sections

10.4

CONIC SECTIONS ∗

In this section we will discuss some of the basic geometric properties of parabolas, ellipses, and hyperbolas. These curves play an important role in calculus and also arise naturally in a broad range of applications in such ﬁelds as planetary motion, design of telescopes and antennas, geodetic positioning, and medicine, to name a few.

CONIC SECTIONS

Circles, ellipses, parabolas, and hyperbolas are called conic sections or conics because they can be obtained as intersections of a plane with a double-napped circular cone (Figure 10.4.1). If the plane passes through the vertex of the double-napped cone, then the intersection is a point, a pair of intersecting lines, or a single line. These are called degenerate conic sections.

Circle

A point

Ellipse

Parabola

A pair of intersecting lines

Hyperbola

A single line

Figure 10.4.1

∗

Some students may already be familiar with the material in this section, in which case it can be treated as a review. Instructors who want to spend some additional time on precalculus review may want to allocate more than one lecture on this material.

10.4 Conic Sections

731

DEFINITIONS OF THE CONIC SECTIONS Although we could derive properties of parabolas, ellipses, and hyperbolas by deﬁning them as intersections with a double-napped cone, it will be better suited to calculus if we begin with equivalent deﬁnitions that are based on their geometric properties.

10.4.1 definition A parabola is the set of all points in the plane that are equidistant from a ﬁxed line and a ﬁxed point not on the line.

All points on the parabola are equidistant from the focus and directrix.

Axis Focus

The line is called the directrix of the parabola, and the point is called the focus (Figure 10.4.2). A parabola is symmetric about the line that passes through the focus at right angles to the directrix. This line, called the axis or the axis of symmetry of the parabola, intersects the parabola at a point called the vertex.

10.4.2 definition An ellipse is the set of all points in the plane, the sum of whose distances from two ﬁxed points is a given positive constant that is greater than the distance between the ﬁxed points.

Vertex Directrix

Figure 10.4.2

The two ﬁxed points are called the foci (plural of “focus”) of the ellipse, and the midpoint of the line segment joining the foci is called the center (Figure 10.4.3a). To help visualize Deﬁnition 10.4.2, imagine that two ends of a string are tacked to the foci and a pencil traces a curve as it is held tight against the string (Figure 10.4.3b). The resulting curve will be an ellipse since the sum of the distances to the foci is a constant, namely, the total length of the string. Note that if the foci coincide, the ellipse reduces to a circle. For ellipses other than circles, the line segment through the foci and across the ellipse is called the major axis (Figure 10.4.3c), and the line segment across the ellipse, through the center, and perpendicular to the major axis is called the minor axis. The endpoints of the major axis are called vertices. The sum of the distances to the foci is constant.

Minor axis

Major axis Vertex

Focus

Figure 10.4.3

Center

(a)

Vertex

Focus

(b)

(c)

10.4.3 definition A hyperbola is the set of all points in the plane, the difference of whose distances from two ﬁxed distinct points is a given positive constant that is less than the distance between the ﬁxed points.

The two ﬁxed points are called the foci of the hyperbola, and the term “difference” that is used in the deﬁnition is understood to mean the distance to the farther focus minus the distance to the closer focus. As a result, the points on the hyperbola form two branches, each

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Chapter 10 / Parametric and Polar Curves; Conic Sections

“wrapping around” the closer focus (Figure 10.4.4a). The midpoint of the line segment joining the foci is called the center of the hyperbola, the line through the foci is called the focal axis, and the line through the center that is perpendicular to the focal axis is called the conjugate axis. The hyperbola intersects the focal axis at two points called the vertices. Associated with every hyperbola is a pair of lines, called the asymptotes of the hyperbola. These lines intersect at the center of the hyperbola and have the property that as a point P moves along the hyperbola away from the center, the vertical distance between P and one of the asymptotes approaches zero (Figure 10.4.4b).

Conjugate axis

The distance from the farther focus minus the distance to the closer focus is constant.

Center

Focal axis

Focus

Focus Vertex

Vertex

These distances approach zero as the point moves away from the center.

x These distances approach zero as the point moves away from the center.

(a)

(b)

Figure 10.4.4

EQUATIONS OF PARABOLAS IN STANDARD POSITION

It is traditional in the study of parabolas to denote the distance between the focus and the vertex by p. The vertex is equidistant from the focus and the directrix, so the distance between the vertex and the directrix is also p; consequently, the distance between the focus and the directrix is 2p (Figure 10.4.5). As illustrated in that ﬁgure, the parabola passes through two of the corners of a box that extends from the vertex to the focus along the axis of symmetry and extends 2p units above and 2p units below the axis of symmetry. The equation of a parabola is simplest if the vertex is the origin and the axis of symmetry is along the x-axis or y-axis. The four possible such orientations are shown in Figure 10.4.6. These are called the standard positions of a parabola, and the resulting equations are called the standard equations of a parabola.

2p p

Axis

Directrix

Figure 10.4.5

parabolas in standard position y

y y=p

x ( p, 0)

x (−p, 0)

x = −p

x x=p

y 2 = 4px

Figure 10.4.6

y 2 = −4px

(0, p)

(0, −p)

y = −p x 2 = 4py

x 2 = −4py

10.4 Conic Sections

733

To illustrate how the equations in Figure 10.4.6 are obtained, we will derive the equation for the parabola with focus (p, 0) and directrix x = −p. Let P (x, y) be any point on the parabola. Since P is equidistant from the focus and directrix, the distances PF and PD in Figure 10.4.7 are equal; that is, PF = PD (1)

D(−p, y)

P(x, y) x

F(p, 0)

where D(−p, y) is the foot of the perpendicular from P to the directrix. From the distance formula, the distances PF and PD are (2) PF = (x − p)2 + y 2 and PD = (x + p)2

Substituting in (1) and squaring yields x = −p

and after simplifying

Figure 10.4.7

(x − p)2 + y 2 = (x + p)2

(3)

y 2 = 4px

(4)

The derivations of the other equations in Figure 10.4.6 are similar. A TECHNIQUE FOR SKETCHING PARABOLAS

Parabolas can be sketched from their standard equations using four basic steps:

Sketching a Parabola from Its Standard Equation Step 1. Determine whether the axis of symmetry is along the x-axis or the y-axis. Referring to Figure 10.4.6, the axis of symmetry is along the x-axis if the equation has a y 2 -term, and it is along the y-axis if it has an x 2 -term. Step 2. Determine which way the parabola opens. If the axis of symmetry is along the x-axis, then the parabola opens to the right if the coefficient of x is positive, and it opens to the left if the coefficient is negative. If the axis of symmetry is along the y-axis, then the parabola opens up if the coefficient of y is positive, and it opens down if the coefficient is negative. Step 3. Determine the value of p and draw a box extending p units from the origin along the axis of symmetry in the direction in which the parabola opens and extending 2p units on each side of the axis of symmetry. Step 4. Using the box as a guide, sketch the parabola so that its vertex is at the origin and it passes through the corners of the box (Figure 10.4.8).

Rough sketch

Figure 10.4.8

Example 1

Sketch the graphs of the parabolas (a) x 2 = 12y

and show the focus and directrix of each.

(0, 3)

Solution (a). This equation involves x 2 , so the axis of symmetry is along the y-axis, and x

−6

y = −3

x 2 = 12y Figure 10.4.9

(b) y 2 + 8x = 0

the coefﬁcient of y is positive, so the parabola opens upward. From the coefﬁcient of y, we obtain 4p = 12 or p = 3. Drawing a box extending p = 3 units up from the origin and 2p = 6 units to the left and 2p = 6 units to the right of the y-axis, then using corners of the box as a guide, yields the graph in Figure 10.4.9. The focus is p = 3 units from the vertex along the axis of symmetry in the direction in which the parabola opens, so its coordinates are (0, 3). The directrix is perpendicular to the axis of symmetry at a distance of p = 3 units from the vertex on the opposite side from the focus, so its equation is y = −3.

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Chapter 10 / Parametric and Polar Curves; Conic Sections y

Solution (b). We ﬁrst rewrite the equation in the standard form y 2 = −8x

(−2, 0)

−4

This equation involves y 2 , so the axis of symmetry is along the x-axis, and the coefﬁcient of x is negative, so the parabola opens to the left. From the coefﬁcient of x we obtain 4p = 8, so p = 2. Drawing a box extending p = 2 units left from the origin and 2p = 4 units above and 2p = 4 units below the x-axis, then using corners of the box as a guide, yields the graph in Figure 10.4.10. Example 2 Find an equation of the parabola that is symmetric about the y-axis, has its vertex at the origin, and passes through the point (5, 2).

x=2

Solution. Since the parabola is symmetric about the y-axis and has its vertex at the origin, y 2 = −8x

the equation is of the form x 2 = 4py

Figure 10.4.10

or x 2 = −4py

where the sign depends on whether the parabola opens up or down. But the parabola must open up since it passes through the point (5, 2), which lies in the ﬁrst quadrant. Thus, the equation is of the form (5) x 2 = 4py Since the parabola passes through (5, 2), we must have 52 = 4p · 2 or 4p = (5) becomes y x 2 = 25 2

25 . 2

Therefore,

EQUATIONS OF ELLIPSES IN STANDARD POSITION c

c b

b a

Figure 10.4.11

√b 2 + c 2

It is traditional in the study of ellipses to denote the length of the major axis by 2a, the length of the minor axis by 2b, and the distance between the foci by 2c (Figure 10.4.11). The number a is called the semimajor axis and the number b the semiminor axis (standard but odd terminology, since a and b are numbers, not geometric axes). There is a basic relationship between the numbers a, b, and c that can be obtained by examining the sum of the distances to the foci from a point P at the end of the major axis and from a point Q at the end of the minor axis (Figure 10.4.12). From Deﬁnition 10.4.2, these sums must be equal, so we obtain 2 b2 + c2 = (a − c) + (a + c) from which it follows that

b c

P a−c a

Figure 10.4.12

b c

Figure 10.4.13

√ b2 + c2

(6)

√ a 2 − b2

(7)

or, equivalently,

From (6), the distance from a focus to an end of the minor axis is a (Figure 10.4.13), which implies that for all points on the ellipse the sum of the distances to the foci is 2a. It also follows from (6) that a ≥ b with the equality holding only when c = 0. Geometrically, this means that the major axis of an ellipse is at least as large as the minor axis and that the two axes have equal length only when the foci coincide, in which case the ellipse is a circle. The equation of an ellipse is simplest if the center of the ellipse is at the origin and the foci are on the x-axis or y-axis. The two possible such orientations are shown in Figure 10.4.14.

10.4 Conic Sections

735

These are called the standard positions of an ellipse, and the resulting equations are called the standard equations of an ellipse.

ellipses in standard position y

a (0, c) b x

−a (−c, 0)

(c, 0)

−b

−b −a

(0, −c)

x 2 y2 + =1 b2 a2

x 2 y2 + =1 a 2 b2 Figure 10.4.14

To illustrate how the equations in Figure 10.4.14 are obtained, we will derive the equation for the ellipse with foci on the x-axis. Let P (x, y) be any point on that ellipse. Since the sum of the distances from P to the foci is 2a, it follows (Figure 10.4.15) that

P(x, y)

PF ′ + PF = 2a x

F ′(−c, 0)

F(c, 0)

so (x + c)2 + y 2 + (x − c)2 + y 2 = 2a

Transposing the second radical to the right side of the equation and squaring yields Figure 10.4.15

(x + c)2 + y 2 = 4a 2 − 4a (x − c)2 + y 2 + (x − c)2 + y 2

and, on simplifying,

c (x − c)2 + y 2 = a − x a

(8)

Squaring again and simplifying yields y2 x2 + 2 =1 2 a a − c2 which, by virtue of (6), can be written as y2 x2 + =1 a2 b2

(9)

Conversely, it can be shown that any point whose coordinates satisfy (9) has 2a as the sum of its distances from the foci, so that such a point is on the ellipse.

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Chapter 10 / Parametric and Polar Curves; Conic Sections

A TECHNIQUE FOR SKETCHING ELLIPSES

Ellipses can be sketched from their standard equations using three basic steps:

Sketching an Ellipse from Its Standard Equation Step 1. Determine whether the major axis is on the x-axis or the y-axis. This can be ascertained from the sizes of the denominators in the equation. Referring to Figure 10.4.14, and keeping in mind that a 2 > b2 (since a > b), the major axis is along the x-axis if x 2 has the larger denominator, and it is along the y-axis if y 2 has the larger denominator. If the denominators are equal, the ellipse is a circle. Step 2. Determine the values of a and b and draw a box extending a units on each side of the center along the major axis and b units on each side of the center along the minor axis. Step 3. Using the box as a guide, sketch the ellipse so that its center is at the origin and it touches the sides of the box where the sides intersect the coordinate axes (Figure 10.4.16).

Rough sketch

Figure 10.4.16

Example 3

Sketch the graphs of the ellipses

y 4

(a)

(0, √7)

x2 y2 + =1 9 16

(b) x 2 + 2y 2 = 4

showing the foci of each. x −3

Solution (a). Since y 2 has the larger denominator, the major axis is along the y-axis. Moreover, since a 2 > b2 , we must have a 2 = 16 and b2 = 9, so

a=4 −4

Figure 10.4.17 y

√2

Solution (b). We ﬁrst rewrite the equation in the standard form (√2, 0) x

−2

2 −√2

x 2 y2 + =1 4 2 Figure 10.4.18

b=3

Drawing a box extending 4 units on each side of the origin along the y-axis and 3 units on each side of the origin along the x-axis as a guide yields the graph in Figure 10.4.17. The foci lie c units on each side of the center along the major axis, where c is given by (7). From the values of a 2 and b2 above, we obtain √ √ c = a 2 − b2 = 16 − 9 = 7 ≈ 2.6 √ √ Thus, the coordinates of the foci are (0, 7 ) and (0, − 7 ), since they lie on the y-axis.

(0, −√7)

x2 y2 + =1 9 16

(−√2, 0)

and

x2 y2 + =1 4 2 Since x 2 has the larger denominator, the major axis lies along the x-axis, and we have a 2 = 4 and b2 = 2. Drawing a√box extending a = 2 units on each side of the origin along the x-axis and extending b = 2 ≈ 1.4 units on each side of the origin along the y-axis as a guide yields the graph in Figure 10.4.18. From (7), we obtain √ c = a 2 − b2 = 2 ≈ 1.4 √ √ Thus, the coordinates of the foci are ( 2, 0) and (− 2, 0), since they lie on the x-axis.

10.4 Conic Sections

Example 4 points (0, ±4).

737

Find an equation for the ellipse with foci (0, ±2) and major axis with end-

Solution. From Figure 10.4.14, the equation has the form y2 x2 + 2 =1 2 b a and from the given information, a = 4 and c = 2. It follows from (6) that b2 = a 2 − c2 = 16 − 4 = 12

so the equation of the ellipse is

x2 y2 + =1 12 16

EQUATIONS OF HYPERBOLAS IN STANDARD POSITION

It is traditional in the study of hyperbolas to denote the distance between the vertices by 2a, the distance between the foci by 2c (Figure 10.4.19), and to deﬁne the quantity b as √ b = c2 − a 2 (10)

c a

This relationship, which can also be expressed as √ c = a 2 + b2

is pictured geometrically in Figure 10.4.20. As illustrated in that ﬁgure, and as we will show later in this section, the asymptotes pass through the corners of a box extending b units on each side of the center along the conjugate axis and a units on each side of the center along the focal axis. The number a is called the semifocal axis of the hyperbola and the number b the semiconjugate axis. (As with the semimajor and semiminor axes of an ellipse, these are numbers, not geometric axes.) If V is one vertex of a hyperbola, then, as illustrated in Figure 10.4.21, the distance from V to the farther focus minus the distance from V to the closer focus is

Figure 10.4.19

a b

(11)

[(c − a) + 2a] − (c − a) = 2a

Figure 10.4.20

V a c−a Figure 10.4.21

Thus, for all points on a hyperbola, the distance to the farther focus minus the distance to the closer focus is 2a. The equation of a hyperbola has an especially convenient form if the center of the hyperbola is at the origin and the foci are on the x-axis or y-axis. The two possible such orientations are shown in Figure 10.4.22. These are called the standard positions of a hyperbola, and the resulting equations are called the standard equations of a hyperbola. The derivations of these equations are similar to those already given for parabolas and ellipses, so we will leave them as exercises. However, to illustrate how the equations of the asymptotes are derived, we will derive those equations for the hyperbola y2 x2 − 2 =1 2 a b

a c−a

We can rewrite this equation as b2 2 (x − a 2 ) a2 which is equivalent to the pair of equations b 2 b 2 y= x − a 2 and y = − x − a2 a a y2 =

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Chapter 10 / Parametric and Polar Curves; Conic Sections

hyperbolas in standard position y

b y = ax

(0, c)

a x

(â&#x2C6;&#x2019;c, 0)

â&#x2C6;&#x2019;a

â&#x2C6;&#x2019;b

(c, 0)

b â&#x2C6;&#x2019;a

â&#x2C6;&#x2019;b b y =â&#x2C6;&#x2019;ax

x2 a

â&#x2C6;&#x2019;

a x b

y2 b

(0, â&#x2C6;&#x2019;c)

â&#x2C6;&#x2019;

x2 b2

a y=â&#x2C6;&#x2019; x b

Figure 10.4.22

b y = a â&#x2C6;&#x161;x 2 â&#x2C6;&#x2019; a 2

b y = ax

Thus, in the ďŹ rst quadrant, the vertical distance between the line y = (b/a)x and the hyperbola can be written as b 2 b xâ&#x2C6;&#x2019; x â&#x2C6;&#x2019; a2 a a (Figure 10.4.23). But this distance tends to zero as x â&#x2020;&#x2019; +âŹ since lim

x â&#x2020;&#x2019; +âŹ

b b b 2 2 x â&#x2C6;&#x2019;a = lim x â&#x2C6;&#x2019; x 2 â&#x2C6;&#x2019; a2 xâ&#x2C6;&#x2019; x â&#x2020;&#x2019; +âŹ a a a 2 â&#x2C6;&#x2019; a2 2 â&#x2C6;&#x2019; a2 x + x x x â&#x2C6;&#x2019; b = lim x â&#x2020;&#x2019; +âŹ a x + x 2 â&#x2C6;&#x2019; a2 = lim

x â&#x2020;&#x2019; +âŹ

Figure 10.4.23

ab =0 x + x 2 â&#x2C6;&#x2019; a2

The analysis in the remaining quadrants is similar.

A QUICK WAY TO FIND ASYMPTOTES There is a trick that can be used to avoid memorizing the equations of the asymptotes of a hyperbola. They can be obtained, when needed, by replacing 1 by 0 on the right side of the hyperbola equation, and then solving for y in terms of x. For example, for the hyperbola

y2 x2 â&#x2C6;&#x2019; =1 a2 b2 we would write y2 x2 â&#x2C6;&#x2019; =0 a2 b2

or y 2 =

which are the equations for the asymptotes.

b2 2 x a2

b or y = Âą x a

10.4 Conic Sections

739

A TECHNIQUE FOR SKETCHING HYPERBOLAS

Hyperbolas can be sketched from their standard equations using four basic steps:

Sketching a Hyperbola from Its Standard Equation Step 1. Determine whether the focal axis is on the x-axis or the y-axis. This can be ascertained from the location of the minus sign in the equation. Referring to Figure 10.4.22, the focal axis is along the x-axis when the minus sign precedes the y 2 -term, and it is along the y-axis when the minus sign precedes the x 2 -term. Step 2. Determine the values of a and b and draw a box extending a units on either side of the center along the focal axis and b units on either side of the center along the conjugate axis. (The squares of a and b can be read directly from the equation.) Step 3. Draw the asymptotes along the diagonals of the box. Step 4. Using the box and the asymptotes as a guide, sketch the graph of the hyperbola (Figure 10.4.24).

Rough sketch

Figure 10.4.24

Example 5 3 y=− x 2

Sketch the graphs of the hyperbolas

y2 x2 − =1 4 9 showing their vertices, foci, and asymptotes. (a)

3 y= x 2

(b) y 2 − x 2 = 1

Solution (a). The minus sign precedes the y 2 -term, so the focal axis is along the x-axis. From the denominators in the equation we obtain x −√13

√13 5

x2 y2 − =1 4 9 Figure 10.4.25

y 4

y=x

√2 x −4

−√2 −4

Figure 10.4.26

and

b2 = 9

Since a and b are positive, we must have a = 2 and b = 3. Recalling that the vertices lie a units on each side of the center on the focal axis, it follows that their coordinates in this case are (2, 0) and (−2, 0). Drawing a box extending a = 2 units along the x-axis on each side of the origin and b = 3 units on each side of the origin along the y-axis, then drawing the asymptotes along the diagonals of the box as a guide, yields the graph in Figure 10.4.25. To obtain equations for the asymptotes, we replace 1 by 0 in the given equation; this yields y2 3 x2 − = 0 or y = ± x 4 9 2 The foci lie c units on each side of the center along the focal axis, where c is given by (11). From the values of a 2 and b2 above we obtain √ √ c = a 2 + b2 = 4 + 9 = 13 ≈ 3.6 √ √ Since the foci lie on the x-axis in this case, their coordinates are ( 13, 0) and (− 13, 0).

−5

y = −x

a2 = 4

Solution (b). The minus sign precedes the x 2 -term, so the focal axis is along the y-axis. From the denominators in the equation we obtain a 2 = 1 and b2 = 1, from which it follows that a = 1 and b = 1 Thus, the vertices are at (0, −1) and (0, 1). Drawing a box extending a = 1 unit on either side of the origin along the y-axis and b = 1 unit on either side of the origin along the x-axis, then drawing the asymptotes, yields the graph in Figure 10.4.26. Since the box is actually

740

Chapter 10 / Parametric and Polar Curves; Conic Sections

A hyperbola in which a = b, as in part (b) of Example 5, is called an equilateral hyperbola. Such hyperbolas always have perpendicular asymptotes.

a square, the asymptotes are perpendicular and have equations y = ±x. This can also be seen by replacing 1 by 0 in the given equation, which yields y 2 − x 2 = 0 or y = ±x. Also, √ √ c = a 2 + b2 = 1 + 1 = 2 √ √ so the foci, which lie on the y-axis, are (0, − 2 ) and (0, 2 ). Example 6 Find the equation of the hyperbola with vertices (0, ±8) and asymptotes y = ± 43 x.

Solution. Since the vertices are on the y-axis, the equation of the hyperbola has the form (y 2 /a 2 ) − (x 2 /b2 ) = 1 and the asymptotes are

a y=± x b From the locations of the vertices we have a = 8, so the given equations of the asymptotes yield a 8 4 y=± x=± x=± x b b 3 from which it follows that b = 6. Thus, the hyperbola has the equation y2 x2 − =1 64 36

TRANSLATED CONICS

Equations of conics that are translated from their standard positions can be obtained by replacing x by x − h and y by y − k in their standard equations. For a parabola, this translates the vertex from the origin to the point (h, k); and for ellipses and hyperbolas, this translates the center from the origin to the point (h, k). Parabolas with vertex (h, k) and axis parallel to x-axis (y − k)2 = 2

4p(x − h)

(y − k) = −4p(x − h)

[Opens right]

(12)

[Opens left]

(13)

Parabolas with vertex (h, k) and axis parallel to y-axis (x − h)2 = 2

4p(y − k)

(x − h) = −4p(y − k)

[Opens up]

(14)

[Opens down]

(15)

Ellipse with center (h, k) and major axis parallel to x-axis (x − h)2 (y − k)2 + = 1 [b < a] a2 b2

(16)

Ellipse with center (h, k) and major axis parallel to y-axis (x − h)2 (y − k)2 + = 1 [b < a] b2 a2

(17)

Hyperbola with center (h, k) and focal axis parallel to x-axis (x − h)2 (y − k)2 − =1 a2 b2

(18)

Hyperbola with center (h, k) and focal axis parallel to y-axis (y − k)2 (x − h)2 − =1 2 a b2

(19)

10.4 Conic Sections

741

Example 7 Find an equation for the parabola that has its vertex at (1, 2) and its focus at (4, 2).

Solution. Since the focus and vertex are on a horizontal line, and since the focus is to the right of the vertex, the parabola opens to the right and its equation has the form (y − k)2 = 4p(x − h) Since the vertex and focus are 3 units apart, we have p = 3, and since the vertex is at (h, k) = (1, 2), we obtain (y − 2)2 = 12(x − 1) Sometimes the equations of translated conics occur in expanded form, in which case we are faced with the problem of identifying the graph of a quadratic equation in x and y: Ax 2 + Cy 2 + Dx + Ey + F = 0

(20)

The basic procedure for determining the nature of such a graph is to complete the squares of the quadratic terms and then try to match up the resulting equation with one of the forms of a translated conic. Example 8

Describe the graph of the equation y 2 − 8x − 6y − 23 = 0

Solution. The equation involves quadratic terms in y but none in x, so we ﬁrst take all of the y-terms to one side: y 2 − 6y = 8x + 23

(−4, 3)

Next, we complete the square on the y-terms by adding 9 to both sides:

(−2, 3) x

(y − 3)2 = 8x + 32 Finally, we factor out the coefﬁcient of the x-term to obtain

Directrix

x = −6

y − 8x − 6y − 23 = 0 2

Figure 10.4.27

(y − 3)2 = 8(x + 4) This equation is of form (12) with h = −4, k = 3, and p = 2, so the graph is a parabola with vertex (−4, 3) opening to the right. Since p = 2, the focus is 2 units to the right of the vertex, which places it at the point (−2, 3); and the directrix is 2 units to the left of the vertex, which means that its equation is x = −6. The parabola is shown in Figure 10.4.27. Example 9

Describe the graph of the equation 16x 2 + 9y 2 − 64x − 54y + 1 = 0

Solution. This equation involves quadratic terms in both x and y, so we will group the x-terms and the y-terms on one side and put the constant on the other: (16x 2 − 64x) + (9y 2 − 54y) = −1 Next, factor out the coefﬁcients of x 2 and y 2 and complete the squares: 16(x 2 − 4x + 4) + 9(y 2 − 6y + 9) = −1 + 64 + 81 or

16(x − 2)2 + 9(y − 3)2 = 144

742

Chapter 10 / Parametric and Polar Curves; Conic Sections y

Finally, divide through by 144 to introduce a 1 on the right side: (2, 7)

(2, 3)

(−1, 3)

(x − 2)2 (y − 3)2 + =1 9 16 This is an equation of form (17), with h = 2, k = 3, a 2 = 16, and b2 = 9. Thus, the graph of the equation is an ellipse with center (2, 3) and major axis parallel to the y-axis. Since a = 4, the major axis extends 4 units above and 4 units below the center, so its endpoints are (2, 7) and (2, −1) (Figure 10.4.28). Since b = 3, the minor axis extends 3 units to the left and 3 units to the right of the center, so its endpoints are (−1, 3) and (5, 3). Since √ √ c = a 2 − b2 = 16 − 9 = 7 √ √ the foci lie √7 units above and below the center, placing them at the points (2, 3 + 7 ) and (2, 3 − 7 ).

(2, 3 + √7 )

(5, 3)

(2, −1)

(2, 3 − √7 )

16x 2 + 9y 2 − 64x − 54y + 1 = 0 Figure 10.4.28

Example 10 Describe the graph of the equation x 2 − y 2 − 4x + 8y − 21 = 0

Solution. This equation involves quadratic terms in both x and y, so we will group the x-terms and the y-terms on one side and put the constant on the other: (x 2 − 4x) − (y 2 − 8y) = 21 We leave it for you to verify by completing the squares that this equation can be written as y = −x + 6

(2 − 3√2, 4)

y= x+2

(2 + 3√2, 4)

x 2 − y 2 − 4x + 8y − 21 = 0 Figure 10.4.29

(x − 2)2 (y − 4)2 − =1 (21) 9 9 This is an equation of form (18) with h = 2, k = 4, a 2 = 9, and b2 = 9. Thus, the equation represents a hyperbola with center (2, 4) and focal axis parallel to the x-axis. Since a = 3, the vertices are located 3 units to the √left and 3 units √ of the center, or at the points √ √ to the right 2, so the foci are located 3 2 (−1, 4) and (5, 4). From (11), c = a 2 + b2 = 9 + 9 = 3 √ √ units to the left and right of the center, or at the points (2 − 3 2, 4) and (2 + 3 2, 4). The equations of the asymptotes may be found using the trick of replacing 1 by 0 in (21) to obtain (x − 2)2 (y − 4)2 − =0 9 9 This can be written as y − 4 = ±(x − 2), which yields the asymptotes y =x+2

and y = −x + 6

With the aid of a box extending a = 3 units left and right of the center and b = 3 units above and below the center, we obtain the sketch in Figure 10.4.29.

REFLECTION PROPERTIES OF THE CONIC SECTIONS

Parabolas, ellipses, and hyperbolas have certain reﬂection properties that make them extremely valuable in various applications. In the exercises we will ask you to prove the following results. 10.4.4 theorem (Reﬂection Property of Parabolas) The tangent line at a point P on a parabola makes equal angles with the line through P parallel to the axis of symmetry and the line through P and the focus (Figure 10.4.30a).

10.4 Conic Sections

743

10.4.5 theorem (Reﬂection Property of Ellipses) A line tangent to an ellipse at a point P makes equal angles with the lines joining P to the foci (Figure 10.4.30b).

10.4.6 theorem (Reﬂection Property of Hyperbolas) A line tangent to a hyperbola at a point P makes equal angles with the lines joining P to the foci (Figure 10.4.30c).

Tangent line at P

Axis of symmetry

P a

a P

a a

Focus

P Tangent line at P

Tangent line at P

(a)

Figure 10.4.30

(b)

(c)

APPLICATIONS OF THE CONIC SECTIONS

John Mead/Science Photo Library/Photo Researchers

Incoming signals are reflected by the parabolic antenna to the receiver at the focus.

Fermat’s principle in optics implies that light reflects off of a surface at an angle equal to its angle of incidence. (See Exercise 64 in Section 4.5.) In particular, if a reflecting surface is generated by revolving a parabola about its axis of symmetry, it follows from Theorem 10.4.4 that all light rays entering parallel to the axis will be reflected to the focus (Figure 10.4.31a); conversely, if a light source is located at the focus, then the reflected rays will all be parallel to the axis (Figure 10.4.31b). This principle is used in certain telescopes to reflect the approximately parallel rays of light from the stars and planets off of a parabolic mirror to an eyepiece at the focus; and the parabolic reflectors in flashlights and automobile headlights utilize this principle to form a parallel beam of light rays from a bulb placed at the focus. The same optical principles apply to radar signals and sound waves, which explains the parabolic shape of many antennas.

Figure 10.4.31

(a)

(b)

Visitors to various rooms in the United States Capitol Building and in St. Paul’s Cathedral in London are often astonished by the “whispering gallery” effect in which two people at opposite ends of the room can hear one another’s whispers very clearly. Such rooms have ceilings with elliptical cross sections and common foci. Thus, when the two people stand at the foci, their whispers are reﬂected directly to one another off of the elliptical ceiling. Hyperbolic navigation systems, which were developed in World War II as navigational aids to ships, are based on the deﬁnition of a hyperbola. With these systems the ship receives

744

Chapter 10 / Parametric and Polar Curves; Conic Sections

Ship

synchronized radio signals from two widely spaced transmitters with known positions. The ship’s electronic receiver measures the difference in reception times between the signals and then uses that difference to compute the difference 2a between its distances from the two transmitters. This information places the ship somewhere on the hyperbola whose foci are at the transmitters and whose points have 2a as the difference in their distances from the foci. By repeating the process with a second set of transmitters, the position of the ship can be approximated as the intersection of two hyperbolas (Figure 10.4.32). [The modern global positioning system (GPS) is based on the same principle.]

Atlantic Ocean

Figure 10.4.32

✔QUICK CHECK EXERCISES 10.4

(See page 748 for answers.)

1. Identify the conic. (a) The set of points in the plane, the sum of whose distances to two fixed points is a positive constant greater than the distance between the fixed points is . (b) The set of points in the plane, the difference of whose distances to two fixed points is a positive constant less than the distance between the fixed points is . (c) The set of points in the plane that are equidistant from a fixed line and a fixed point not on the line is . 2. (a) The equation of the parabola with focus (p, 0) and directrix x = −p is . (b) The equation of the parabola with focus (0, p) and directrix y = −p is .

3. (a) Suppose that an ellipse has semimajor axis a and semiminor axis b. Then for all points on the ellipse, the sum of the distances to the foci is equal to . (b) The two standard equations of an ellipse with semimajor axis a and semiminor axis b are and .

EXERCISE SET 10.4

Graphing Utility

(c) Suppose that an ellipse has semimajor axis a, semiminor axis b, and foci (±c, 0). Then c may be obtained from a and b by the equation c = .

4. (a) Suppose that a hyperbola has semifocal axis a and semiconjugate axis b. Then for all points on the hyperbola, the difference of the distance to the farther focus minus the distance to the closer focus is equal to . (b) The two standard equations of a hyperbola with semifocal axis a and semiconjugate axis b are and . (c) Suppose that a hyperbola in standard position has semifocal axis a, semiconjugate axis b, and foci (±c, 0). Then c may be obtained from a and b by the equation c= . The equations of the asymptotes of this hyperbola are y = ± .

CAS

F O C U S O N C O N C E P TS

1. In parts (a)–(f), ﬁnd the equation of the conic. y (a) (b) 2

x 0 −1 −2

1 x

−1

−2

−3 −3 −2 −1

(d) 3

1 1

−3 −3 −2 −1

10.4 Conic Sections y

(e) 3 2

18. Vertex (5, −3); axis parallel to the y-axis; passes through (9, 5).

(f ) 3 2

−1

−1 −2

−3 −3 −2 −1

−2

−3 −3 −2 −1

2. (a) Find the focus and directrix for each parabola in Exercise 1. (b) Find the foci of the ellipses in Exercise 1. (c) Find the foci and the equations of the asymptotes of the hyperbolas in Exercise 1. 3–6 Sketch the parabola, and label the focus, vertex, and directrix. ■

3. (a) y 2 = 4x

4. (a) y 2 = −10x

(b) x 2 = −8y (b) x 2 = 4y

5. (a) (y − 1) = −12(x + 4) (b) (x − 1) = 2 y − 21 2

6. (a) y 2 − 6y − 2x + 1 = 0

745

(b) y = 4x 2 + 8x + 5

7–10 Sketch the ellipse, and label the foci, vertices, and ends of the minor axis. ■ x2 y2 7. (a) + =1 (b) 9x 2 + y 2 = 9 16 9 x2 y2 8. (a) + =1 (b) 4x 2 + y 2 = 36 25 4 9. (a) (x + 3)2 + 4(y − 5)2 = 16 (b) 41 x 2 + 19 (y + 2)2 − 1 = 0

19–22 Find an equation for the ellipse that satisﬁes the given conditions. ■

19. (a) Ends of major axis (±3, 0); ends of minor axis (0, ±2). (b) Length of minor axis 8; foci (0, ±3). √ 20. (a) Foci (±1, √ 0); b = 2. (b) c = 2 3; a = 4; center at the origin; foci on a coordinate axis (two answers). 21. (a) Ends of major axis (0, ±6); passes through (−3, 2). (b) Foci (−1, 1) and (−1, 3); minor axis of length 4. 22. (a) Center at (0, 0); major and minor axes along the coordinate axes; passes through (3, 2) and (1, 6). (b) Foci (2, 1) and (2, −3); major axis of length 6. 23–26 Find an equation for a hyperbola that satisﬁes the given conditions. [Note: In some cases there may be more than one hyperbola.] ■

23. (a) Vertices (±2, 0); foci (±3, 0). (b) Vertices (0, ±2); asymptotes y = ± 23 x.

24. (a) Asymptotes y = ± 23 x; b = 4. (b) Foci (0, ±5); asymptotes y = ±2x. 25. (a) Asymptotes y = ± 43 x; c = 5. (b) Foci (±3, 0); asymptotes y = ±2x.

26. (a) Vertices (0, 6) and (6, 6); foci 10 units apart. (b) Asymptotes y = x − 2 and y = −x + 4; passes through the origin. 27–30 True–False Determine whether the statement is true or false. Explain your answer. ■

10. (a) 9x 2 + 4y 2 − 18x + 24y + 9 = 0 (b) 5x 2 + 9y 2 + 20x − 54y = −56

27. A hyperbola is the set of all points in the plane that are equidistant from a ﬁxed line and a ﬁxed point not on the line.

11–14 Sketch the hyperbola, and label the vertices, foci, and

28. If an ellipse is not a circle, then the foci of an ellipse lie on the major axis of the ellipse.

asymptotes. ■ x2 y2 11. (a) − =1 (b) 9y 2 − x 2 = 36 16 9 y2 x2 12. (a) − =1 (b) 16x 2 − 25y 2 = 400 9 25 (y + 4)2 (x − 2)2 13. (a) − =1 3 5 2 (b) 16(x + 1) − 8(y − 3)2 = 16 14. (a) x 2 − 4y 2 + 2x + 8y − 7 = 0 (b) 16x 2 − y 2 − 32x − 6y = 57

15–18 Find an equation for the parabola that satisﬁes the given conditions. ■

15. (a) Vertex (0, 0); focus (3, 0). (b) Vertex (0, 0); directrix y = 41 .

16. (a) Focus (6, 0); directrix x = −6. (b) Focus (1, 1); directrix y = −2.

√ 17. Axis y = 0; passes through (3, 2) and (2, − 2).

29. If a parabola has equation y 2 = 4px, where p is a positive constant, then the perpendicular distance from the parabola’s focus to its directrix is p. 30. The hyperbola (y 2 /a 2 ) − x 2 = 1 has asymptotes the lines y = ±x /a.

31. (a) As illustrated in the accompanying ﬁgure, a parabolic arch spans a road 40 ft wide. How high is the arch if a center section of the road 20 ft wide has a minimum clearance of 12 ft? (b) How high would the center be if the arch were the upper half of an ellipse?

12 ft

12 ft 20 ft 40 ft

Figure Ex-31

746

Chapter 10 / Parametric and Polar Curves; Conic Sections

32. (a) Find an equation for the parabolic arch with base b and height h, shown in the accompanying ﬁgure. (b) Find the area under the arch. y

冸 12 b, h冹

x (b, 0)

Figure Ex-32

33. Show that the vertex is the closest point on a parabola to the focus. [Suggestion: Introduce a convenient coordinate system and use Definition 10.4.1.] 34. As illustrated in the accompanying figure, suppose that a comet moves in a parabolic orbit with the Sun at its focus and that the line from the Sun to the comet makes an angle of 60 ◦ with the axis of the parabola when the comet is 40 million miles from the center of the Sun. Use the result in Exercise 33 to determine how close the comet will come to the center of the Sun. 35. For the parabolic reflector in the accompanying figure, how far from the vertex should the light source be placed to produce a beam of parallel rays?

60°

1 ft

39. Find an equation of the ellipse traced by a point that moves so that the sum of its distances to (4, 1) and (4, 5) is 12. 40. Find the equation of the hyperbola traced by a point that moves so that the difference between its distances to (0, 0) and (1, 1) is 1. 41. Suppose that the base of a solid is elliptical with a major axis of length 9 and a minor axis of length 4. Find the volume of the solid if the cross sections perpendicular to the major axis are squares (see the accompanying ﬁgure). 42. Suppose that the base of a solid is elliptical with a major axis of length 9 and a minor axis of length 4. Find the volume of the solid if the cross sections perpendicular to the minor axis are equilateral triangles (see the accompanying ﬁgure).

Figure Ex-41

Figure Ex-42

43. Show that an ellipse with semimajor axis a and semiminor axis b has area A = πab. F O C U S O N C O N C E P TS

44. Show that if a plane is not parallel to the axis of a right circular cylinder, then the intersection of the plane and cylinder is an ellipse (possibly a circle). [Hint: Let θ be the angle shown in the accompanying ﬁgure, introduce coordinate axes as shown, and express x ′ and y ′ in terms of x and y.]

1 ft Figure Ex-34

Figure Ex-35

36. (a) Show that the right and left branches of the hyperbola y2 x2 − 2 =1 2 a b can be represented parametrically as x = a cosh t, x = −a cosh t,

y = b sinh t y = b sinh t

(−⬁ < t < +⬁) (−⬁ < t < +⬁)

(b) Use a graphing utility to generate both branches of the hyperbola x 2 − y 2 = 1 on the same screen.

37. (a) Show that the right and left branches of the hyperbola y2 x2 − 2 =1 2 a b can be represented parametrically as x = a sec t, x = −a sec t,

y = b tan t y = b tan t

(−π/2 < t < π/2) (−π/2 < t < π/2)

(b) Use a graphing utility to generate both branches of the hyperbola x 2 − y 2 = 1 on the same screen.

38. Find an equation of the parabola traced by a point that moves so that its distance from (2, 4) is the same as its distance to the x-axis.

y′

x′

Figure Ex-44

45. As illustrated in the accompanying ﬁgure on the next page, a carpenter needs to cut an elliptical hole in a sloped roof through which a circular vent pipe of diameter D is to be inserted vertically. The carpenter wants to draw the outline of the hole on the roof using a pencil, two tacks, and a piece of string (as in Figure 10.4.3b). The center point of the ellipse is known, and common sense suggests that its major axis must be perpendicular to the drip line of the roof. The carpenter needs to determine the length L of the string and the distance T between a tack and the center point. The architect’s plans show that the pitch of the roof is p (pitch = rise over run; see the accompanying ﬁgure). Find T and L in terms of D and p. Source: This exercise is based on an article by William H. Enos, which appeared in the Mathematics Teacher, Feb. 1991, p. 148.

10.4 Conic Sections

Vent pipe

(a) Sketch the solid generated by revolving R about the x-axis, and ﬁnd its volume. (b) Sketch the solid generated by revolving R about the y-axis, and ﬁnd its volume.

Rise

Drip line

747

Run

Figure Ex-45

46. As illustrated in the accompanying ﬁgure, suppose that two observers are stationed at the points F1 (c, 0) and F2 (−c, 0) in an xy-coordinate system. Suppose also that the sound of an explosion in the xy-plane is heard by the F1 observer t seconds before it is heard by the F2 observer. Assuming that the speed of sound is a constant v, show that the explosion occurred somewhere on the hyperbola x2 y2 − 2 =1 2 2 v t /4 c − (v 2 t 2 /4)

50. As illustrated in the accompanying ﬁgure, the tank of an oil truck is 18 ft long and has elliptical cross sections that are 6 ft wide and 4 ft high. (a) Show that the volume V of oil in the tank (in cubic feet) when it is ﬁlled to a depth of h feet is

−1 h − 2 2 V = 27 4 sin + (h − 2) 4h − h + 2π 2 (b) Use the numerical root-ﬁnding capability of a CAS to determine how many inches from the bottom of a dipstick the calibration marks should be placed to indicate when the tank is 41 , 21 , and 43 full. Dipstick 18′

4′

x F2 (−c, 0 )

F1(c, 0 )

Figure Ex-46

47. As illustrated in the accompanying ﬁgure, suppose that two transmitting stations are positioned 100 km apart at points F1 (50, 0) and F2 (−50, 0) on a straight shoreline in an xy-coordinate system. Suppose also that a ship is traveling parallel to the shoreline but 200 km at sea. Find the coordinates of the ship if the stations transmit a pulse simultaneously, but the pulse from station F1 is received by the ship 100 microseconds sooner than the pulse from station F2 . [Assume that the pulses travel at the speed of light (299,792,458 m/s).] y

200 km x F2 (−50, 0 )

F1(50, 0 )

Figure Ex-47

48. A nuclear cooling tower is to have a height of h feet and the shape of the solid that is generated by revolving the region R enclosed by the right branch of the hyperbola 1521x 2 − 225y 2 = 342,225 and the lines x = 0, y = −h/2, and y = h/2 about the y-axis. (a) Find the volume of the tower. (b) Find the lateral surface area of the tower. 49. Let R be the region that is above the x-axis and en2 2 2 2 2 2 closed √ between the curve b x − a y = a b and the line x = a 2 + b2 .

6′

Figure Ex-50

51. Prove: The line tangent to the parabola x 2 = 4py at the point (x0 , y0 ) is x0 x = 2p(y + y0 ).

52. Prove: The line tangent to the ellipse x2 y2 + =1 a2 b2 at the point (x0 , y0 ) has the equation yy0 xx0 + 2 =1 2 a b 53. Prove: The line tangent to the hyperbola x2 y2 − =1 a2 b2 at the point (x0 , y0 ) has the equation yy0 xx0 − 2 =1 a2 b 54. Use the results in Exercises 52 and 53 to show that if an ellipse and a hyperbola have the same foci, then at each point of intersection their tangent lines are perpendicular. 55. Find two values of k such that the line x + 2y = k is tangent to the ellipse x 2 + 4y 2 = 8. Find the points of tangency.

56. Find the coordinates of all points on the hyperbola 4x 2 − y 2 = 4 where the two lines that pass through the point and the foci are perpendicular.

57. A line tangent to the hyperbola 4x 2 − y 2 = 36 intersects the y-axis at the point (0, 4). Find the point(s) of tangency. 58. Consider the second-degree equation Ax 2 + Cy 2 + Dx + Ey + F = 0

(cont.)

748

59.

60. 61. 62. 63.

64.

Chapter 10 / Parametric and Polar Curves; Conic Sections

where A and C are not both 0. Show by completing the square: (a) If AC > 0, then the equation represents an ellipse, a circle, a point, or has no graph. (b) If AC < 0, then the equation represents a hyperbola or a pair of intersecting lines. (c) If AC = 0, then the equation represents a parabola, a pair of parallel lines, or has no graph. In each part, use the result in Exercise 58 to make a statement about the graph of the equation, and then check your conclusion by completing the square and identifying the graph. (a) x 2 − 5y 2 − 2x − 10y − 9 = 0 (b) x 2 − 3y 2 − 6y − 3 = 0 (c) 4x 2 + 8y 2 + 16x + 16y + 20 = 0 (d) 3x 2 + y 2 + 12x + 2y + 13 = 0 (e) x 2 + 8x + 2y + 14 = 0 (f ) 5x 2 + 40x + 2y + 94 = 0 Derive the equation x 2 = 4py in Figure 10.4.6. Derive the equation (x 2 /b2 ) + (y 2 /a 2 ) = 1 given in Figure 10.4.14. Derive the equation (x 2 /a 2 ) − (y 2 /b2 ) = 1 given in Figure 10.4.22. Prove Theorem 10.4.4. [Hint: Choose coordinate axes so that the parabola has the equation x 2 = 4py. Show that the tangent line at P (x0 , y0 ) intersects the y-axis at Q(0, −y0 ) and that the triangle whose three vertices are at P , Q, and the focus is isosceles.] Given two intersecting lines, let L2 be the line with the larger angle of inclination φ2 , and let L1 be the line with the smaller

angle of inclination φ1 . We deﬁne the angle θ between L1 and L2 by θ = φ2 − φ1 . (See the accompanying ﬁgure.) (a) Prove: If L1 and L2 are not perpendicular, then tan θ =

m2 − m1 1 + m 1 m2

where L1 and L2 have slopes m1 and m2 . (b) Prove Theorem 10.4.5. [Hint: Introduce coordinates so that the equation (x 2 /a 2 ) + (y 2 /b2 ) = 1 describes the ellipse, and use part (a).] (c) Prove Theorem 10.4.6. [Hint: Introduce coordinates so that the equation (x 2 /a 2 ) − (y 2 /b2 ) = 1 describes the hyperbola, and use part (a).] y

L1 L2

Figure Ex-64

65. Writing Suppose that you want to draw an ellipse that has given values for the lengths of the major and minor axes by using the method shown in Figure 10.4.3b. Assuming that the axes are drawn, explain how a compass can be used to locate the positions for the tacks. 66. Writing List the forms for standard equations of parabolas, ellipses, and hyperbolas, and write a summary of techniques for sketching conic sections from their standard equations.

✔QUICK CHECK ANSWERS 10.4 1. (a) an ellipse (b) a hyperbola (c) a parabola 2. (a) y 2 = 4px (b) x 2 = 4py √ √ x2 b y2 x2 y2 y2 y2 x2 x2 3. (a) 2a (b) 2 + 2 = 1; 2 + 2 = 1 (c) a 2 − b2 4. (a) 2a (b) 2 − 2 = 1; 2 − 2 = 1 (c) a 2 + b2 ; x a b b a a b a b a

10.5

ROTATION OF AXES; SECOND-DEGREE EQUATIONS In the preceding section we obtained equations of conic sections with axes parallel to the coordinate axes. In this section we will study the equations of conics that are “tilted” relative to the coordinate axes. This will lead us to investigate rotations of coordinate axes. QUADRATIC EQUATIONS IN x AND y We saw in Examples 8 to 10 of the preceding section that equations of the form

Ax 2 + Cy 2 + Dx + Ey + F = 0

(1)

can represent conic sections. Equation (1) is a special case of the more general equation Ax 2 + Bxy + Cy 2 + Dx + Ey + F = 0

(2)

which, if A, B, and C are not all zero, is called a quadratic equation in x and y. It is usually the case that the graph of any second-degree equation is a conic section. If B = 0,

10.5 Rotation of Axes; Second-Degree Equations y 3

P(x, y)

(1, 2)

1 x −3

−2

−1

1 −1

(−1, −2)

−2 −3

Figure 10.5.1

749

then (2) reduces to (1) and the conic section has its axis or axes parallel to the coordinate axes. However, if B = 0, then (2) contains a cross-product term Bxy, and the graph of the conic section represented by the equation has its axis or axes “tilted” relative to the coordinate axes. As an illustration, consider the ellipse with foci F1 (1, 2) and F2 (−1, −2) and such that the sum of the distances from each point P (x, y) on the ellipse to the foci is 6 units. Expressing this condition as an equation, we obtain (Figure 10.5.1) (x − 1)2 + (y − 2)2 + (x + 1)2 + (y + 2)2 = 6

Squaring both sides, then isolating the remaining radical, then squaring again ultimately yields 8x 2 − 4xy + 5y 2 = 36

as the equation of the ellipse. This is of form (2) with A = 8, B = −4, C = 5, D = 0, E = 0, and F = −36.

ROTATION OF AXES To study conics that are tilted relative to the coordinate axes it is frequently helpful to rotate the coordinate axes, so that the rotated coordinate axes are parallel to the axes of the conic. Before we can discuss the details, we need to develop some ideas about rotation of coordinate axes. In Figure 10.5.2a the axes of an xy-coordinate system have been rotated about the origin through an angle θ to produce a new x ′ y ′ -coordinate system. As shown in the ﬁgure, each point P in the plane has coordinates (x ′ , y ′ ) as well as coordinates (x, y). To see how the two are related, let r be the distance from the common origin to the point P , and let α be the angle shown in Figure 10.5.2b. It follows that

x = r cos(θ + α),

and

x ′ = r cos α,

y = r sin(θ + α)

(3)

y ′ = r sin α

(4)

Using familiar trigonometric identities, the relationships in (3) can be written as x = r cos θ cos α − r sin θ sin α y = r sin θ cos α + r cos θ sin α

and on substituting (4) in these equations we obtain the following relationships called the rotation equations: x = x ′ cos θ − y ′ sin θ (5) y = x ′ sin θ + y ′ cos θ y

(x, y) P (x', y' )

y' x'

a u

Figure 10.5.2

(a)

u x x'

(b)

Example 1 Suppose that the axes of an xy-coordinate system are rotated through an angle of θ = 45 ◦ to obtain an x ′ y ′ -coordinate system. Find the equation of the curve in x ′ y ′ -coordinates.

x 2 − xy + y 2 − 6 = 0

750

Chapter 10 / Parametric and Polar Curves; Conic Sections

√

Solution. Substituting sin θ = sin 45 ◦ = 1/ 2 and cos θ = cos 45 ◦ = 1/ 2 in (5)

y y'

yields the rotation equations

45°

x 2 − xy + y 2 − 6 = 0 Figure 10.5.3

x′ y′ x′ y′ x=√ −√ and y = √ + √ 2 2 2 2 Substituting these into the given equation yields ′ ′ ′ ′ x x x y′ 2 y′ y′ y′ 2 x − √ −√ −6=0 + √ +√ √ −√ √ +√ 2 2 2 2 2 2 2 2 or x ′ 2 − 2x ′ y ′ + y ′ 2 − x ′ 2 + y ′ 2 + x ′ 2 + 2x ′ y ′ + y ′ 2 =6 2 or

y′2 x′2 + =1 12 4 which is the equation of an ellipse (Figure 10.5.3).

If the rotation equations (5) are solved for x ′ and y ′ in terms of x and y, one obtains (Exercise 16): x ′ = x cos θ + y sin θ (6) y ′ = −x sin θ + y cos θ Example 2 Find the new coordinates of the point (2, 4) if the coordinate axes are rotated through an angle of θ = 30 ◦ . √

Solution. Using the rotation equations in (6) with x = 2, y = 4, cos θ = cos 30 ◦ = 3/2,

and sin θ = sin 30 ◦ = 1/2, we obtain √ √ x ′ = 2( 3/2) + 4(1/2) = 3 + 2 √ √ y ′ = −2(1/2) + 4( 3/2) = −1 + 2 3 √ √ Thus, the new coordinates are ( 3 + 2, −1 + 2 3 ). ELIMINATING THE CROSS-PRODUCT TERM

In Example 1 we were able to identify the curve x 2 − xy + y 2 − 6 = 0 as an ellipse because the rotation of axes eliminated the xy-term, thereby reducing the equation to a familiar form. This occurred because the new x ′ y ′ -axes were aligned with the axes of the ellipse. The following theorem tells how to determine an appropriate rotation of axes to eliminate the cross-product term of a second-degree equation in x and y. 10.5.1

theorem If the equation Ax 2 + Bxy + Cy 2 + Dx + Ey + F = 0 ′ ′

It is always possible to satisfy (8) with an angle θ in the interval

0 < θ < π/2 We will always choose θ in this way.

(7)

is such that B = 0, and if an x y -coordinate system is obtained by rotating the xy-axes through an angle θ satisfying A−C (8) cot 2θ = B then, in x ′ y ′ -coordinates, Equation (7 ) will have the form A′ x ′ 2 + C ′ y ′ 2 + D ′ x ′ + E ′ y ′ + F ′ = 0

proof Substituting (5) into (7) and simplifying yields A′ x ′ 2 + B ′ x ′ y ′ + C ′ y ′ 2 + D ′ x ′ + E ′ y ′ + F ′ = 0

10.5 Rotation of Axes; Second-Degree Equations

where

751

A′ = A cos2 θ + B cos θ sin θ + C sin2 θ

B ′ = B(cos2 θ − sin2 θ ) + 2(C − A) sin θ cos θ C ′ = A sin2 θ − B sin θ cos θ + C cos2 θ

D ′ = D cos θ + E sin θ

(9)

E ′ = −D sin θ + E cos θ

F′ = F

(Verify.) To complete the proof we must show that B ′ = 0 if cot 2θ = or, equivalently,

A−C B

A−C cos 2θ = sin 2θ B

(10)

However, by using the trigonometric double-angle formulas, we can rewrite B ′ in the form B ′ = B cos 2θ − (A − C) sin 2θ Thus, B ′ = 0 if θ satisﬁes (10). ■ Example 3

Identify and sketch the curve xy = 1.

Solution. As a ﬁrst step, we will rotate the coordinate axes to eliminate the cross-product term. Comparing the given equation to (7), we have A = 0,

B = 1,

C=0

Thus, the desired angle of rotation must satisfy 0−0 A−C = =0 B 1 ◦ This condition can be met √ by taking 2θ = π/2◦ or θ = √π/4 = 45 . Making the substitutions ◦ cos θ = cos 45 = 1/ 2 and sin θ = sin 45 = 1/ 2 in (5) yields cot 2θ =

y y'

xy = 1 x'

x′ y′ x=√ −√ 2 2

and

x′ y′ y=√ +√ 2 2

Substituting these in the equation xy = 1 yields ′ ′ x y′ y′ x =1 √ −√ √ +√ 2 2 2 2 Figure 10.5.4

y′ 2 x′ 2 − =1 2 2

′ ′ which is√ the equation in √the x y -coordinate system of an equilateral hyperbola with vertices at ( 2, 0) and (− 2, 0) in that coordinate system (Figure 10.5.4).

In problems where it is inconvenient to solve A−C B for θ, the values of sin θ and cos θ needed for the rotation equations can be obtained by ﬁrst calculating cos 2θ and then computing sin θ and cos θ from the identities 1 − cos 2θ 1 + cos 2θ and cos θ = sin θ = 2 2 cot 2θ =

752

Chapter 10 / Parametric and Polar Curves; Conic Sections y

Example 4

Identify and sketch the curve 153x 2 − 192xy + 97y 2 − 30x − 40y − 200 = 0

25 24

Solution. We have A = 153, B = −192, and C = 97, so

56 7 A−C =− =− B 192 24 Since θ is to be chosen in the range 0 < θ < π/2, this relationship is represented by the 7 , which implies that triangle in Figure 10.5.5. From that triangle we obtain cos 2θ = − 25 7 1 − 25 3 1 + cos 2θ cos θ = = = 2 2 5 7 1 + 25 1 − cos 2θ 4 = = sin θ = 2 2 5

cot 2θ =

−7 Figure 10.5.5

y y'

Substituting these values in (5) yields the rotation equations x

x = 35 x ′ − 45 y ′

and y = 45 x ′ + 35 y ′

and substituting these in turn in the given equation yields 153 (3x ′ 25

− 4y ′ )2 −

192 (3x ′ 25

− 4y ′ )(4x ′ + 3y ′ ) +

97 (4x ′ 25

+ 3y ′ )2

(3x ′ − 4y ′ ) − − 30 5 (x' − 1) 2 + y' 2 = 1 9 Figure 10.5.6

which simpliﬁes to

40 (4x ′ 5

+ 3y ′ ) − 200 = 0

25x ′ 2 + 225y ′ 2 − 50x ′ − 200 = 0

x ′ 2 + 9y ′ 2 − 2x ′ − 8 = 0

Completing the square yields There is a method for deducing the kind of curve represented by a seconddegree equation directly from the equation itself without rotating coordinate axes. For a discussion of this topic, see the section on the discriminant that appears in Web Appendix K.

(x ′ − 1)2 + y′2 = 1 9 which is the equation in the x ′ y ′ -coordinate system of an ellipse with center (1, 0) in that coordinate system and semiaxes a = 3 and b = 1 (Figure 10.5.6).

✔QUICK CHECK EXERCISES 10.5

(See page 754 for answers.)

1. Suppose that an xy-coordinate system is rotated θ radians to produce a new x ′ y ′ -coordinate system. (a) x and y may be obtained from x ′ , y ′ , and θ using the rotation equations x = and y = . (b) x ′ and y ′ may be obtained from x, y, and θ using the equations x ′ = and y ′ = .

2. If the equation

Ax 2 + Bxy + Cy 2 + Dx + Ey + F = 0 is such that B = 0, then the xy-term in this equation can be

eliminated by a rotation of axes through an angle θ satisfying cot 2θ = .

3. In each part, determine a rotation angle θ that will eliminate the xy-term. 2 (a) 2x 2 + xy √ + 2y +2x − y = 0 2 (b) x + 2√3xy + 3y − 2x + y = 1 (c) 3x 2 + 3xy + 2y 2 + y = 0

4. Express 2x 2 + xy + 2y 2 = 1 in the x ′ y ′ -coordinate system obtained by rotating the xy-coordinate system through the angle θ = π/4.

10.5 Rotation of Axes; Second-Degree Equations

753

EXERCISE SET 10.5 1. Let an x ′ y ′ -coordinate system be obtained by rotating an xy-coordinate system through an angle of θ = 60 ◦ . (a) Find the x ′ y ′ -coordinates of the point whose xy-coordinates are (−2, 6). √ (b) Find an equation of the curve 3xy + y 2 = 6 in x ′ y ′ -coordinates. (c) Sketch the curve in part (b), showing both xy-axes and x ′ y ′ -axes. 2. Let an x ′ y ′ -coordinate system be obtained by rotating an xy-coordinate system through an angle of θ = 30 ◦ . (a) Find the x ′ y ′ -coordinates of the point whose xy-coor√ dinates are (1, − 3). √ (b) Find an equation of the curve 2x 2 + 2 3xy = 3 in x ′ y ′ -coordinates. (c) Sketch the curve in part (b), showing both xy-axes and x ′ y ′ -axes. 3–12 Rotate the coordinate axes to remove the xy-term. Then

identify the type of conic and sketch its graph. ■ 4. x 2 − xy + y 2 − 2 = 0

3. xy = −9

5. x 2 + 4xy − 2y 2 − 6 = 0 √ 6. 31x 2 + 10 3xy + 21y 2 − 144 = 0 √ √ 7. x 2 + 2 3xy + 3y 2 + 2 3x − 2y = 0

8. 34x 2 − 24xy + 41y 2 − 25 = 0

9. 9x 2 − 24xy + 16y 2 − 80x − 60y + 100 = 0 √ √ 10. 5x 2 − 6xy + 5y 2 − 8 2x + 8 2y = 8

11. 52x 2 − 72xy + 73y 2 + 40x + 30y − 75 = 0

12. 6x 2 + 24xy − y 2 − 12x + 26y + 11 = 0

13. Let an x ′ y ′ -coordinate system be obtained by rotating an xycoordinate system through an angle of 45 ◦ . Use (6) to ﬁnd an equation of the curve 3x ′ 2 + y ′ 2 = 6 in xy-coordinates. 14. Let an x ′ y ′ -coordinate system be obtained by rotating an xy-coordinate system through an angle of 30 ◦ . Use (5) to ﬁnd an equation in x′ y′ -coordinates of the curve y = x 2 . F O C U S O N C O N C E P TS ′ ′

15. Let an x y -coordinate system be obtained by rotating an xy-coordinate system through an angle θ . Prove: For every value of θ , the equation x 2 + y 2 = r 2 becomes the equation x ′ 2 + y ′ 2 = r 2 . Give a geometric explanation.

18. Let an x ′ y ′ -coordinate system be obtained by rotating an xy-coordinate system through an angle θ. Explain how to ﬁnd the xy-equation of a line whose x ′ y ′ -equation is known. 19–22 Show that the graph of the given equation is a parabola. Find its vertex, focus, and directrix. ■ √ √ 19. x 2 + 2xy + y 2 + 4 2x − 4 2y = 0 √ √ 20. x 2 − 2 3xy + 3y 2 − 8 3x − 8y = 0

21. 9x 2 − 24xy + 16y 2 − 80x − 60y + 100 = 0 √ √ 22. x 2 + 2 3xy + 3y 2 + 16 3x − 16y − 96 = 0

23–26 Show that the graph of the given equation is an ellipse. Find its foci, vertices, and the ends of its minor axis. ■

23. 288x 2 − 168xy + 337y 2 − 3600 = 0

24. 25x 2 − 14xy + 25y 2 − 288 = 0 √ √ 25. 31x 2 + 10 3xy + 21y 2 − 32x + 32 3y − 80 = 0 √ √ 26. 43x 2 − 14 3xy + 57y 2 − 36 3x − 36y − 540 = 0

27–30 Show that the graph of the given equation is a hyperbola. Find its foci, vertices, and asymptotes. ■ √ 27. x 2 − 10 3xy + 11y 2 + 64 = 0

28. 17x 2 − 312xy + 108y 2 − 900 = 0 √ √ 29. 32y 2 − 52xy − 7x 2 + 72 5x − 144 5y + 900 = 0 √ 2 √ √ 2 √ 30. 2 2y + 5 2xy + 2 2x + 18x + 18y + 36 2 = 0 31. Show that the graph of the equation

√ √ x+ y=1 is a portion of a parabola. [Hint: First rationalize the equation and then perform a rotation of axes.]

F O C U S O N C O N C E P TS

32. Derive the expression for B ′ in (9). 33. Use (9) to prove that B 2 − 4AC = B ′ 2 − 4A′ C ′ for all values of θ . 34. Use (9) to prove that A + C = A′ + C ′ for all values of θ.

16. Derive (6) by solving the rotation equations in (5) for x ′ and y ′ in terms of x and y.

35. Prove: If A = C in (7), then the cross-product term can be eliminated by rotating through 45 ◦ .

17. Let an x ′ y ′ -coordinate system be obtained by rotating an xy-coordinate system through an angle θ . Explain how to ﬁnd the xy-coordinates of a point whose x ′ y ′ coordinates are known.

36. Prove: If B = 0, then the graph of x 2 + Bxy + F = 0 is a hyperbola if F = 0 and two intersecting lines if F = 0.

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Chapter 10 / Parametric and Polar Curves; Conic Sections

✔QUICK CHECK ANSWERS 10.5 1. (a) x ′ cos θ − y ′ sin θ ; x ′ sin θ + y ′ cos θ (b) x cos θ + y sin θ; −x sin θ + y cos θ

4. 5x ′2 + 3y ′2 = 2

10.6

A−C B

3. (a)

π π π (b) (c) 4 3 6

CONIC SECTIONS IN POLAR COORDINATES It will be shown later in the text that if an object moves in a gravitational field that is directed toward a fixed point (such as the center of the Sun), then the path of that object must be a conic section with the fixed point at a focus. For example, planets in our solar system move along elliptical paths with the Sun at a focus, and the comets move along parabolic, elliptical, or hyperbolic paths with the Sun at a focus, depending on the conditions under which they were born. For applications of this type it is usually desirable to express the equations of the conic sections in polar coordinates with the pole at a focus. In this section we will show how to do this. THE FOCUS–DIRECTRIX CHARACTERIZATION OF CONICS To obtain polar equations for the conic sections we will need the following theorem.

10.6.1 theorem (Focus–Directrix Property of Conics) Suppose that a point P moves in the plane determined by a ﬁxed point (called the focus) and a ﬁxed line (called the directrix), where the focus does not lie on the directrix. If the point moves in such a way that its distance to the focus divided by its distance to the directrix is some constant e (called the eccentricity), then the curve traced by the point is a conic section. Moreover, the conic is

It is an unfortunate historical accident that the letter e is used for the base of the natural logarithm as well as for the eccentricity of conic sections. However, as a practical matter the appropriate interpretation will usually be clear from the context in which the letter is used.

(a) a parabola if e = 1

(b) an ellipse if 0 < e < 1

We will not give a formal proof of this theorem; rather, we will use the speciﬁc cases in Figure 10.6.1 to illustrate the basic ideas. For the parabola, we will take the directrix to be x = −p, as usual; and for the ellipse and the hyperbola we will take the directrix to be x = a 2 /c. We want to show in all three cases that if P is a point on the graph, F is the focus, and D is the directrix, then the ratio PF /PD is some constant e, where e = 1 for the parabola, 0 < e < 1 for the ellipse, and e > 1 for the hyperbola. We will give the arguments for the parabola and ellipse and leave the argument for the hyperbola as an exercise. y

y y

D P(x, y)

P(x, y)

D x

F(p, 0)

F(c, 0)

x = −p

x = a2/c x = a2/c e=1

Figure 10.6.1

0<e <1

e>1

10.6 Conic Sections in Polar Coordinates

755

For the parabola, the distance PF to the focus is equal to the distance PD to the directrix, so that PF /PD = 1, which is what we wanted to show. For the ellipse, we rewrite Equation (8) of Section 10.4 as c a2 c −x (x − c)2 + y 2 = a − x = a a c

But the expression on the left side is the distance PF, and the expression in the parentheses on the right side is the distance PD, so we have shown that PF =

c PD a

Thus, PF /PD is constant, and the eccentricity is e=

c a

(1)

If we rule out the degenerate case where a = 0 or c = 0, then it follows from Formula (7) of Section 10.4 that 0 < c < a, so 0 < e < 1, which is what we wanted to show. We will leave it as an exercise to show that the eccentricity of the hyperbola in Figure 10.6.1 is also given by Formula (1), but in this case it follows from Formula (11) of Section 10.4 that c > a, so e > 1. ECCENTRICITY OF AN ELLIPSE AS A MEASURE OF FLATNESS The eccentricity of an ellipse can be viewed as a measure of its ﬂatness—as e approaches 0 the ellipses become more and more circular, and as e approaches 1 they become more and more ﬂat (Figure 10.6.2). Table 10.6.1 shows the orbital eccentricities of various celestial objects. Note that most of the planets actually have fairly circular orbits.

e=0

Table 10.6.1 e = 0.5

celestial body e = 0.8

Ellipses with a common focus and equal semimajor axes

Figure 10.6.2

e = 0.9

Mercury Venus Earth Mars Jupiter Saturn Uranus Neptune Pluto Halley's comet

eccentricity 0.206 0.007 0.017 0.093 0.048 0.056 0.046 0.010 0.249 0.970

POLAR EQUATIONS OF CONICS P(r, u)

r F Pole

u r cos u d Directrix

Figure 10.6.3

Our next objective is to derive polar equations for the conic sections from their focus– directrix characterizations. We will assume that the focus is at the pole and the directrix is either parallel or perpendicular to the polar axis. If the directrix is parallel to the polar axis, then it can be above or below the pole; and if the directrix is perpendicular to the polar axis, then it can be to the left or right of the pole. Thus, there are four cases to consider. We will derive the formulas for the case in which the directrix is perpendicular to the polar axis and to the right of the pole. As illustrated in Figure 10.6.3, let us assume that the directrix is perpendicular to the polar axis and d units to the right of the pole, where the constant d is known. If P is a point

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Chapter 10 / Parametric and Polar Curves; Conic Sections

on the conic and if the eccentricity of the conic is e, then it follows from Theorem 10.6.1 that PF /PD = e or, equivalently, that (2)

PF = ePD

However, it is evident from Figure 10.6.3 that PF = r and PD = d − r cos θ. Thus, (2) can be written as r = e(d − r cos θ) which can be solved for r and expressed as r=

ed 1 + e cos θ

(verify). Observe that this single polar equation can represent a parabola, an ellipse, or a hyperbola, depending on the value of e. In contrast, the rectangular equations for these conics all have different forms. The derivations in the other three cases are similar.

10.6.2 theorem If a conic section with eccentricity e is positioned in a polar coordinate system so that its focus is at the pole and the corresponding directrix is d units from the pole and is either parallel or perpendicular to the polar axis, then the equation of the conic has one of four possible forms, depending on its orientation: Directrix

Directrix Focus

Focus

ed 1 + e cos θ

Directrix right of pole

ed 1 − e cos θ

(3–4)

Directrix left of pole

Directrix Focus Focus Directrix

ed 1 + e sin θ

Directrix above pole

ed 1 − e sin θ

(5–6)

Directrix below pole

SKETCHING CONICS IN POLAR COORDINATES

Precise graphs of conic sections in polar coordinates can be generated with graphing utilities. However, it is often useful to be able to make quick sketches of these graphs that show their orientations and give some sense of their dimensions. The orientation of a conic relative to the polar axis can be deduced by matching its equation with one of the four forms in Theorem 10.6.2. The key dimensions of a parabola are determined by the constant p (Figure 10.4.5) and those of ellipses and hyperbolas by the constants a, b, and c (Figures 10.4.11 and 10.4.20). Thus, we need to show how these constants can be obtained from the polar equations.

10.6 Conic Sections in Polar Coordinates

Example 1

Sketch the graph of r =

757

2 in polar coordinates. 1 − cos θ

Solution. The equation is an exact match to (4) with d = 2 and e = 1. Thus, the graph is a parabola with the focus at the pole and the directrix 2 units to the left of the pole. This tells us that the parabola opens to the right along the polar axis and p = 1. Thus, the parabola looks roughly like that sketched in Figure 10.6.4.

All of the important geometric information about an ellipse can be obtained from the values of a, b, and c in Figure 10.6.5. One way to find these values from the polar equation of an ellipse is based on finding the distances from the focus to the vertices. As shown in the figure, let r0 be the distance from the focus to the closest vertex and r1 the distance to the farthest vertex. Thus,

Rough sketch

Figure 10.6.4

a a

r0 = a − c

(7)

and r1 = a + c

from which it follows that a = 21 (r1 + r0 )

c = 21 (r1 − r0 )

(8–9)

Moreover, it also follows from (7) that

Figure 10.6.5

r 0 r1 = a 2 − c 2 = b 2 In words, Formula (8) states that a is the arithmetic average (also called the arithmetic mean) of r0 and r1 , and Formula (10) states that b is the geometric mean of r0 and r1 .

Thus, b=

Example 2

√ r0 r1

(10)

Find the constants a, b, and c for the ellipse r =

6 . 2 + cos θ

Solution. This equation does not match any of the forms in Theorem 10.6.2 because they all require a constant term of 1 in the denominator. However, we can put the equation into one of these forms by dividing the numerator and denominator by 2 to obtain r= Rough sketch

This is an exact match to (3) with d = 6 and e = 21 , so the graph is an ellipse with the directrix 6 units to the right of the pole. The distance r0 from the focus to the closest vertex can be obtained by setting θ = 0 in this equation, and the distance r1 to the farthest vertex can be obtained by setting θ = π. This yields

Figure 10.6.6

r0 = a b

3 1 + cos θ 1 2

3 3 = 3 = 2, 1 + cos 0 2 1 2

r1 =

3 3 = 1 =6 1 + cos π 2 1 2

Thus, from Formulas (8), (10), and (9), respectively, we obtain √ √ a = 21 (r1 + r0 ) = 4, b = r0 r1 = 2 3, c = 21 (r1 − r0 ) = 2

The ellipse looks roughly like that sketched in Figure 10.6.6. r1 r0 Figure 10.6.7

All of the important information about a hyperbola can be obtained from the values of a, b, and c in Figure 10.6.7. As with the ellipse, one way to ﬁnd these values from the polar equation of a hyperbola is based on ﬁnding the distances from the focus to the vertices. As

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Chapter 10 / Parametric and Polar Curves; Conic Sections

shown in the ﬁgure, let r0 be the distance from the focus to the closest vertex and r1 the distance to the farthest vertex. Thus, r0 = c − a

and

(11)

r1 = c + a

from which it follows that a = 21 (r1 − r0 )

c = 21 (r1 + r0 )

(12–13)

Moreover, it also follows from (11) that In words, Formula (13) states that c is the arithmetic mean of r0 and r1 , and Formula (14) states that b is the geometric mean of r0 and r1 .

r0 r1 = c2 − a 2 = b2 from which it follows that

Example 3

Sketch the graph of r =

√ r0 r1

(14)

2 in polar coordinates. 1 + 2 sin θ

Solution. This equation is an exact match to (5) with d = 1 and e = 2. Thus, the graph

is a hyperbola with its directrix 1 unit above the pole. However, it is not so straightforward to compute the values of r0 and r1 , since hyperbolas in polar coordinates are generated in a strange way as θ varies from 0 to 2π. This can be seen from Figure 10.6.8a, which is the graph of the given equation in rectangular θr-coordinates. It follows from this graph that the corresponding polar graph is generated in pieces (see Figure 10.6.8b):

• As θ varies over the interval 0 ≤ θ < 7π/6, the value of r is positive and varies from • • •

2 down to 2/3 and then to +⬁, which generates part of the lower branch. As θ varies over the interval 7π/6 < θ ≤ 3π/2, the value of r is negative and varies from −⬁ to −2, which generates the right part of the upper branch. As θ varies over the interval 3π/2 ≤ θ < 11π/6, the value of r is negative and varies from −2 to −⬁, which generates the left part of the upper branch. As θ varies over the interval 11π/6 < θ ≤ 2π, the value of r is positive and varies from +⬁ to 2, which ﬁlls in the missing piece of the lower right branch.

r 2 1

−1

e i m

−2 −3

2 1 + 2 sin u

(a)

Rough sketch

(b)

(c)

Figure 10.6.8

To obtain a rough sketch of a hyperbola, it is generally sufficient to locate the center, the asymptotes, and the points where θ = 0, θ = π/2, θ = π, and θ = 3π/2.

It is now clear that we can obtain r0 by setting θ = π/2 and r1 by setting θ = 3π/2. Keeping in mind that r0 and r1 are positive, this yields 2 2 2 2 = =2 = , r1 = r0 = 1 + 2 sin(π/2) 3 1 + 2 sin(3π/2) −1

10.6 Conic Sections in Polar Coordinates

759

Thus, from Formulas (12), (14), and (13), respectively, we obtain √ √ 1 2 3 2 1 4 a = (r1 − r0 ) = , b = r0 r1 = , c = (r1 + r0 ) = 2 3 3 2 3 Thus, the hyperbola looks roughly like that sketched in Figure 10.6.8c.

APPLICATIONS IN ASTRONOMY

In 1609 Johannes Kepler published a book known as Astronomia Nova (or sometimes Commentaries on the Motions of Mars) in which he succeeded in distilling thousands of years of observational astronomy into three beautiful laws of planetary motion (Figure 10.6.9).

10.6.3

kepler’s laws

• First law (Law of Orbits). Each planet moves in an elliptical orbit with the Sun a

at a focus. Sun

• Second law (Law of Areas). The radial line from the center of the Sun to the •

Equal areas are swept out in equal times, and the square of the period T is proportional to a 3.

center of a planet sweeps out equal areas in equal times. Third law (Law of Periods). The square of a planet’s period (the time it takes the planet to complete one orbit about the Sun) is proportional to the cube of the semimajor axis of its orbit.

Figure 10.6.9

Apogee

Figure 10.6.10

Perigee

Kepler’s laws, although stated for planetary motion around the Sun, apply to all orbiting celestial bodies that are subjected to a single central gravitational force—artiﬁcial satellites subjected only to the central force of Earth’s gravity and moons subjected only to the central gravitational force of a planet, for example. Later in the text we will derive Kepler’s laws from basic principles, but for now we will show how they can be used in basic astronomical computations. In an elliptical orbit, the closest point to the focus is called the perigee and the farthest point the apogee (Figure 10.6.10). The distances from the focus to the perigee and apogee

Johannes Kepler (1571–1630) German astronomer and physicist. Kepler, whose work provided our contemporary view of planetary motion, led a fascinating but ill-starred life. His alcoholic father made him work in a family-owned tavern as a child, later withdrawing him from elementary school and hiring him out as a ﬁeld laborer, where the boy contracted smallpox, permanently crippling his hands and impairing his eyesight. In later years, Kepler’s ﬁrst wife and several children died, his mother was accused of witchcraft, and being a Protestant he was often subjected to persecution by Catholic authorities. He was often impoverished, eking out a living as an astrologer and prognosticator. Looking back on his unhappy childhood, Kepler described his father as “criminally inclined” and “quarrelsome” and his mother as “garrulous” and “bad-tempered.” However, it was his mother who left an indelible mark on the sixyear-old Kepler by showing him the comet of 1577; and in later life he personally prepared her defense against the witchcraft charges. Kepler became acquainted with the work of Copernicus as a student at the University of Tübingen, where he received his master’s degree in 1591. He continued on as a theological student, but at the

urging of the university officials he abandoned his clerical studies and accepted a position as a mathematician and teacher in Graz, Austria. However, he was expelled from the city when it came under Catholic control, and in 1600 he finally moved on to Prague, where he became an assistant at the observatory of the famous Danish astronomer Tycho Brahe. Brahe was a brilliant and meticulous astronomical observer who amassed the most accurate astronomical data known at that time; and when Brahe died in 1601 Kepler inherited the treasure-trove of data. After eight years of intense labor, Kepler deciphered the underlying principles buried in the data and in 1609 published his monumental work, Astronomia Nova, in which he stated his first two laws of planetary motion. Commenting on his discovery of elliptical orbits, Kepler wrote, “I was almost driven to madness in considering and calculating this matter. I could not find out why the planet would rather go on an elliptical orbit (rather than a circle). Oh ridiculous me!” It ultimately remained for Isaac Newton to discover the laws of gravitation that explained the reason for elliptical orbits. [Image: http://commons.wikimedia.org/wiki/File:Kepler.png]

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Chapter 10 / Parametric and Polar Curves; Conic Sections

are called the perigee distance and apogee distance, respectively. For orbits around the Sun, it is more common to use the terms perihelion and aphelion, rather than perigee and apogee, and to measure time in Earth years and distances in astronomical units (AU), where 1 AU is the semimajor axis a of the Earth’s orbit (approximately 150 × 106 km or 92.9 × 106 mi). With this choice of units, the constant of proportionality in Kepler’s third law is 1, since a = 1 AU produces a period of T = 1 Earth year. In this case Kepler’s third law can be expressed as T = a 3/2 (15) Shapes of elliptical orbits are often speciﬁed by giving the eccentricity e and the semimajor axis a, so it is useful to express the polar equations of an ellipse in terms of these constants. Figure 10.6.11, which can be obtained from the ellipse in Figure 10.6.1 and the relationship c = ea, implies that the distance d between the focus and the directrix is

Directrix

Center

Focus

a a(1 − e2 ) a − c = − ea = e e e

(16)

from which it follows that ed = a(1 − e2 ). Thus, depending on the orientation of the ellipse, the formulas in Theorem 10.6.2 can be expressed in terms of a and e as ae a

a e

Figure 10.6.11

a(1 − e2 ) 1 ± e cos θ

+: Directrix right of pole −: Directrix left of pole

a(1 − e2 ) 1 ± e sin θ

(17–18)

+: Directrix above pole −: Directrix below pole

Moreover, it is evident from Figure 10.6.11 that the distances from the focus to the closest and farthest vertices can be expressed in terms of a and e as r0 = a − ea = a(1 − e) and

Halley's comet

c/ 2

(19–20)

Example 4 Halley’s comet (last seen in 1986) has an eccentricity of 0.97 and a semimajor axis of a = 18.1 AU. 0

Figure 10.6.12

r1 = a + ea = a(1 + e)

(a) Find the equation of its orbit in the polar coordinate system shown in Figure 10.6.12. (b) Find the period of its orbit. (c) Find its perihelion and aphelion distances.

Solution (a). From (17), the polar equation of the orbit has the form r=

a(1 − e2 ) 1 + e cos θ

But a(1 − e2 ) = 18.1[1 − (0.97)2 ] ≈ 1.07. Thus, the equation of the orbit is r= Science Photo Library/Photo Researchers

Halley’s comet photographed April 21, 1910 in Peru.

1.07 1 + 0.97 cos θ

Solution (b). From (15), with a = 18.1, the period of the orbit is T = (18.1)3/2 ≈ 77 years

Solution (c). Since the perihelion and aphelion distances are the distances to the closest and farthest vertices, respectively, it follows from (19) and (20) that r0 = a − ea = a(1 − e) = 18.1(1 − 0.97) ≈ 0.543 AU r1 = a + ea = a(1 + e) = 18.1(1 + 0.97) ≈ 35.7 AU

10.6 Conic Sections in Polar Coordinates

761

or since 1 AU ≈ 150 × 106 km, the perihelion and aphelion distances in kilometers are r0 = 18.1(1 − 0.97)(150 × 106 ) ≈ 81,500,000 km r1 = 18.1(1 + 0.97)(150 × 106 ) ≈ 5,350,000,000 km

Minimum distance

Example 5 An Apollo lunar lander orbits the Moon in an elliptic orbit with eccentricity e = 0.12 and semimajor axis a = 2015 km. Assuming the Moon to be a sphere of radius 1740 km, ﬁnd the minimum and maximum heights of the lander above the lunar surface (Figure 10.6.13).

Solution. If we let r0 and r1 denote the minimum and maximum distances from the center of the Moon, then the minimum and maximum distances from the surface of the Moon will be dmin = r0 − 1740 dmax = r1 − 1740

or from Formulas (19) and (20) Maximum distance

dmin = r0 − 1740 = a(1 − e) − 1740 = 2015(0.88) − 1740 = 33.2 km dmax = r1 − 1740 = a(1 + e) − 1740 = 2015(1.12) − 1740 = 516.8 km

Figure 10.6.13 [Image: NASA]

✔QUICK CHECK EXERCISES 10.6

(See page 763 for answers.)

1. In each part, name the conic section described. (a) The set of points whose distance to the point (2, 3) is half the distance to the line x + y = 1 is . (b) The set of points whose distance to the point (2, 3) is equal to the distance to the line x + y = 1 is . (c) The set of points whose distance to the point (2, 3) is twice the distance to the line x + y = 1 is .

2. In each part: (i) Identify the polar graph as a parabola, an ellipse, or a hyperbola; (ii) state whether the directrix is above, below, to the left, or to the right of the pole; and (iii) ﬁnd the distance from the pole to the directrix. 1 1 (a) r = (b) r = 4 + cos θ 1 − 4 cos θ

EXERCISE SET 10.6

1 4 (d) r = 4 + 4 sin θ 1 − sin θ 3. If the distance from a vertex of an ellipse to the nearest focus is r0 , and if the distance from that vertex to the farthest focus is r1 , then the semimajor axis is a = and the semiminor axis is b = . (c) r =

4. If the distance from a vertex of a hyperbola to the nearest focus is r0 , and if the distance from that vertex to the farthest focus is r1 , then the semifocal axis is a = and the semiconjugate axis is b = .

Graphing Utility

1–2 Find the eccentricity and the distance from the pole to the

directrix, and sketch the graph in polar coordinates. ■ 3 3 1. (a) r = (b) r = 2 − 2 cos θ 2 + sin θ 5 4 (b) r = 2. (a) r = 2 + 3 cos θ 3 + 3 sin θ

3–4 Use Formulas (3)–(6) to identify the type of conic and its orientation. Check your answer by generating the graph with a graphing utility. ■ 8 16 3. (a) r = (b) r = 1 − sin θ 4 + 3 sin θ

4. (a) r =

4 2 − 3 sin θ

(b) r =

12 4 + cos θ

5–6 Find a polar equation for the conic that has its focus at the pole and satisﬁes the stated conditions. Points are in polar coordinates and directrices in rectangular coordinates for simplicity. (In some cases there may be more than one conic that satisﬁes the conditions.) ■

5. (a) Ellipse; e = 43 ; directrix x = 2. (b) Parabola; directrix x = 1. (c) Hyperbola; e = 43 ; directrix y = 3.

762

Chapter 10 / Parametric and Polar Curves; Conic Sections

6. (a) Ellipse; ends of major axis (2, π/2) and (6, 3π/2). (b) Parabola; vertex √ (2, π). (c) Hyperbola; e = 2; vertex (2, 0). 7–8 Find the distances from the pole to the vertices, and then

apply Formulas (8)–(10) to ﬁnd the equation of the ellipse in rectangular coordinates. ■ 6 1 7. (a) r = (b) r = 2 + sin θ 2 − cos θ 6 8 8. (a) r = (b) r = 5 + 2 cos θ 4 − 3 sin θ 9–10 Find the distances from the pole to the vertices, and then

apply Formulas (12)–(14) to ﬁnd the equation of the hyperbola in rectangular coordinates. ■ 3 5 9. (a) r = (b) r = 1 + 2 sin θ 2 − 3 cos θ 4 15 10. (a) r = (b) r = 1 − 2 sin θ 2 + 8 cos θ 11–12 Find a polar equation for the ellipse that has its focus at

the pole and satisﬁes the stated conditions. ■ 11. (a) Directrix to the right of the pole; a = 8; e = 21 . (b) Directrix below the pole; a = 4; e = 35 . 12. (a) Directrix to the left of the pole; b = 4; e = 53 . (b) Directrix above the pole; c = 5; e = 15 .

13. Find the polar equation of an equilateral hyperbola with a focus at the pole and vertex (5, 0). F O C U S O N C O N C E P TS

14. Prove that a√ hyperbola is an equilateral hyperbola if and only if e = 2.

15. (a) Show that the coordinates of the point P on the hyperbola in Figure 10.6.1 satisfy the equation c (x − c)2 + y 2 = x − a a (b) Use the result obtained in part (a) to show that PF /PD = c/a.

16. (a) Show that the eccentricity of an ellipse can be expressed in terms of r0 and r1 as r1 − r0 e= r1 + r 0 (b) Show that 1+e r1 = r0 1−e 17. (a) Show that the eccentricity of a hyperbola can be expressed in terms of r0 and r1 as r1 + r0 e= r1 − r 0 (b) Show that e+1 r1 = r0 e−1

18. (a) Sketch the curves 1 r= 1 + cos θ

and

1 1 − cos θ

(b) Find polar coordinates of the intersections of the curves in part (a). (c) Show that the curves are orthogonal, that is, their tangent lines are perpendicular at the points of intersection. 19–22 True–False Determine whether the statement is true or

false. Explain your answer. ■ 19. If an ellipse is not a circle, then the eccentricity of the ellipse is less than one. 20. A parabola has eccentricity greater than one. 21. If one ellipse has foci that are farther apart than those of a second ellipse, then the eccentricity of the ﬁrst is greater than that of the second. 22. If d is a positive constant, then the conic section with polar equation d r= 1 + cos θ is a parabola. 23–28 Use the following values, where needed:

radius of the Earth = 4000 mi = 6440 km 1 year (Earth year) = 365 days (Earth days) 1 AU = 92.9 × 106 mi = 150 × 106 km ■

23. The dwarf planet Pluto has eccentricity e = 0.249 and semimajor axis a = 39.5 AU. (a) Find the period T in years. (b) Find the perihelion and aphelion distances. (c) Choose a polar coordinate system with the center of the Sun at the pole, and find a polar equation of Pluto’s orbit in that coordinate system. (d) Make a sketch of the orbit with reasonably accurate proportions. 24. (a) Let a be the semimajor axis of a planet’s orbit around the Sun, and let T be its period. Show that if T is measured in days and a is measured in kilometers, then T = (365 × 10−9 )(a /150)3/2 . (b) Use the result in part (a) to find the period of the planet Mercury in days, given that its semimajor axis is a = 57.95 × 106 km. (c) Choose a polar coordinate system with the Sun at the pole, and find an equation for the orbit of Mercury in that coordinate system given that the eccentricity of the orbit is e = 0.206. (d) Use a graphing utility to generate the orbit of Mercury from the equation obtained in part (c). 25. The Hale–Bopp comet, discovered independently on July 23, 1995 by Alan Hale and Thomas Bopp, has an orbital eccentricity of e = 0.9951 and a period of 2380 years. (a) Find its semimajor axis in astronomical units (AU). (b) Find its perihelion and aphelion distances. (cont.)

Chapter 10 Review Exercises

(c) Choose a polar coordinate system with the center of the Sun at the pole, and ﬁnd an equation for the Hale–Bopp orbit in that coordinate system. (d) Make a sketch of the Hale–Bopp orbit with reasonably accurate proportions.

On November 6, 1997 the spacecraft Galileo was placed in a Jovian orbit to study the moon Europa. The orbit had eccentricity 0.814580 and semimajor axis 3,514,918.9 km. Find Galileo’s minimum and maximum heights above the molecular hydrogen layer (see the accompanying ﬁgure).

26. Mars has a perihelion distance of 204,520,000 km and an aphelion distance of 246,280,000 km. (a) Use these data to calculate the eccentricity, and compare your answer to the value given in Table 10.6.1. (b) Find the period of Mars. (c) Choose a polar coordinate system with the center of the Sun at the pole, and ﬁnd an equation for the orbit of Mars in that coordinate system. (d) Use a graphing utility to generate the orbit of Mars from the equation obtained in part (c). 27. Vanguard 1 was launched in March 1958 into an orbit around the Earth with eccentricity e = 0.21 and semimajor axis 8864.5 km. Find the minimum and maximum heights of Vanguard 1 above the surface of the Earth. 28. The planet Jupiter is believed to have a rocky core of radius 10,000 km surrounded by two layers of hydrogen— a 40,000 km thick layer of compressed metallic-like hydrogen and a 20,000 km thick layer of ordinary molecular hydrogen. The visible features, such as the Great Red Spot, are at the outer surface of the molecular hydrogen layer.

763

Not to scale Figure Ex-28 [Image: NASA]

29. Writing Discuss how a hyperbola’s eccentricity e affects the shape of the hyperbola. How is the shape affected as e approaches 1? As e approaches +⬁? Draw some pictures to illustrate your conclusions. 30. Writing Discuss the relationship between the eccentricity e of an ellipse and the distance z between the directrix and center of the ellipse. For example, if the foci remain ﬁxed, what happens to z as e approaches 0?

✔QUICK CHECK ANSWERS 10.6 1. (a) an ellipse (b) a parabola (c) a hyperbola 2. (a) (i) ellipse (ii) to the right of the pole (iii) distance = 1 (b) (i) hyperbola (ii) to the left of the pole (iii) distance = 41 (c) (i) parabola (ii) above the pole (iii) distance = √ √ (d) (i) parabola (ii) below the pole (iii) distance = 4 3. 21 (r1 + r0 ); r0 r1 4. 21 (r1 − r0 ); r0 r1

CHAPTER 10 REVIEW EXERCISES

1 4

Graphing Utility

1. Find parametric equations for the portion of the circle x 2 + y 2 = 2 that lies outside the ﬁrst quadrant, oriented clockwise. Check your work by generating the curve with a graphing utility. 2. (a) Suppose that the equations x = f (t), y = g(t) describe a curve C as t increases from 0 to 1. Find parametric equations that describe the same curve C but traced in the opposite direction as t increases from 0 to 1. (b) Check your work using the parametric graphing feature of a graphing utility by generating the line segment between (1, 2) and (4, 0) in both possible directions as t increases from 0 to 1. 3. (a) Find the slope of the tangent line to the parametric curve x = t 2 + 1, y = t /2 at t = −1 and t = 1 without eliminating the parameter. (b) Check your answers in part (a) by eliminating the parameter and differentiating a function of x.

4. Find dy /dx and d 2 y /dx 2 at t = 2 for the parametric curve x = 21 t 2 , y = 31 t 3 .

5. Find all values of t at which a tangent line to the parametric curve x = 2 cos t, y = 4 sin t is (a) horizontal (b) vertical. 6. Find the exact arc length of the curve x = 1 − 5t 4 , y = 4t 5 − 1

(0 ≤ t ≤ 1)

7. In each part, ﬁnd the rectangular coordinates of the point whose polar coordinates are given. (a) (−8, π/4) (b) (7, −π/4) (c) (8, 9π/4) (d) (5, 0) (e) (−2, −3π/2) (f ) (0, π)

8. Express the point whose xy-coordinates are (−1, 1) in polar coordinates with (a) r > 0, 0 ≤ θ < 2π (b) r < 0, 0 ≤ θ < 2π (c) r > 0, −π < θ ≤ π (d) r < 0, −π < θ ≤ π.

764

Chapter 10 / Parametric and Polar Curves; Conic Sections

9. In each part, use a calculating utility to approximate the polar coordinates of the point whose rectangular coordinates are given. (a) (4, 3) (b) (2, â&#x2C6;&#x2019;5) (c) (1, tanâ&#x2C6;&#x2019;1 1)

10. In each part, state the name that describes the polar curve most precisely: a rose, a line, a circle, a limaĂ§on, a cardioid, a spiral, a lemniscate, or none of these. (a) r = 3 cos Î¸ (b) r = cos 3Î¸ 3 (c) r = (d) r = 3 â&#x2C6;&#x2019; cos Î¸ cos Î¸ (e) r = 1 â&#x2C6;&#x2019; 3 cos Î¸ (f ) r 2 = 3 cos Î¸ (g) r = (3 cos Î¸)2 (h) r = 1 + 3Î¸

11. In each part, identify the curve by converting the polar equation to rectangular coordinates. Assume that a > 0. Î¸ (a) r = a sec2 (b) r 2 cos 2Î¸ = a 2 2 Ď&#x20AC; (c) r = 4 csc Î¸ â&#x2C6;&#x2019; (d) r = 4 cos Î¸ + 8 sin Î¸ 4 12. In each part, express the given equation in polar coordinates. (a) x = 7 (b) x 2 + y 2 = 9 (c) x 2 + y 2 â&#x2C6;&#x2019; 6y = 0 (d) 4xy = 9 13â&#x20AC;&#x201C;17 Sketch the curve in polar coordinates. â&#x2013;

Ď&#x20AC; 6 15. r = 3(1 â&#x2C6;&#x2019; sin Î¸)

13. Î¸ =

17. r = 3 â&#x2C6;&#x2019; cos Î¸

14. r = 6 cos Î¸ 16. r = sin 2Î¸

20. Determine the slope of the tangent line to the polar curve r = 1 + sin Î¸ at Î¸ = Ď&#x20AC;/4. 21. A parametric curve of the form (0 < t < 2Ď&#x20AC;)

is called a conchoid of Nicomedes (see the accompanying ďŹ gure for the case 0 < a < b). (a) Describe how the conchoid x = cot t + 4 cos t,

Figure Ex-21

22. (a) Find the arc length of the polar curve r = 1/Î¸ for Ď&#x20AC;/4 â&#x2030;¤ Î¸ â&#x2030;¤ Ď&#x20AC;/2. (b) What can you say about the arc length of the portion of the curve that lies inside the circle r = 1? 23. Find the area of the region that is enclosed by the cardioid r = 2 + 2 cos Î¸ . 24. Find the area of the region in the ďŹ rst quadrant within the cardioid r = 1 + sin Î¸ . 25. Find the area of the region that is common to the circles r = 1, r = 2 cos Î¸ , and r = 2 sin Î¸.

26. Find the area of the region that is inside the cardioid r = a(1 + sin Î¸ ) and outside the circle r = a sin Î¸ .

27. y 2 = 6x

19. (a) Find the minimum and maximum x-coordinates of points on the cardioid r = 1 â&#x2C6;&#x2019; cos Î¸ . (b) Find the minimum and maximum y-coordinates of points on the cardioid in part (a).

y = a + b sin t

27â&#x20AC;&#x201C;30 Sketch the parabola, and label the focus, vertex, and directrix. â&#x2013;

18. (a) Show that the maximum â&#x2C6;&#x161; value of the y-coordinate of points on the curve r = 1/ Î¸ for Î¸ in the interval (0, Ď&#x20AC;] occurs when tan Î¸ = 2Î¸ . (b) Use a calculating utility to solve the equation in part (a) to at least four decimal-place accuracy. (c) Use the result of part (b) to approximate the maximum value of y for 0 < Î¸ â&#x2030;¤ Ď&#x20AC;.

x = a cot t + b cos t,

(d) Find a polar equation r = f(Î¸ ) for the conchoid in part (a), and then ďŹ nd polar equations for the tangent lines to the conchoid at the pole.

y = 1 + 4 sin t

is generated as t varies over the interval 0 < t < 2Ď&#x20AC;. (b) Find the horizontal asymptote of the conchoid given in part (a). (c) For what values of t does the conchoid in part (a) have a horizontal tangent line? A vertical tangent line?

29. (y + 1)2 = â&#x2C6;&#x2019;7(x â&#x2C6;&#x2019; 4)

28. x 2 = â&#x2C6;&#x2019;9y 2 30. x â&#x2C6;&#x2019; 21 = 2(y â&#x2C6;&#x2019; 1)

31â&#x20AC;&#x201C;34 Sketch the ellipse, and label the foci, the vertices, and the ends of the minor axis. â&#x2013; x2 y2 31. + =1 32. 4x 2 + 9y 2 = 36 4 25 33. 9(x â&#x2C6;&#x2019; 1)2 + 16(y â&#x2C6;&#x2019; 3)2 = 144

34. 3(x + 2)2 + 4(y + 1)2 = 12

35â&#x20AC;&#x201C;37 Sketch the hyperbola, and label the vertices, foci, and asymptotes. â&#x2013; x2 y2 35. â&#x2C6;&#x2019; =1 36. 9y 2 â&#x2C6;&#x2019; 4x 2 = 36 16 4 (x â&#x2C6;&#x2019; 2)2 (y â&#x2C6;&#x2019; 4)2 37. â&#x2C6;&#x2019; =1 9 4 38. In each part, sketch the graph of the conic section with reasonably accurate proportions. (a) x 2 â&#x2C6;&#x2019; 4x + 8y + 36 = 0 (b) 3x 2 + 4y 2 â&#x2C6;&#x2019; 30x â&#x2C6;&#x2019; 8y + 67 = 0 (c) 4x 2 â&#x2C6;&#x2019; 5y 2 â&#x2C6;&#x2019; 8x â&#x2C6;&#x2019; 30y â&#x2C6;&#x2019; 21 = 0 39â&#x20AC;&#x201C;41 Find an equation for the conic described. â&#x2013;

39. A parabola with vertex (0, 0) and focus (0, â&#x2C6;&#x2019;4). â&#x2C6;&#x161; 40. An ellipse with the ends of the major axis (0, Âą 5) and the ends of the minor axis (Âą1, 0). 41. A hyperbola with vertices (0, Âą3) and asymptotes y = Âąx.

Chapter 10 Review Exercises

42. It can be shown in the accompanying figure that hanging cables form parabolic arcs rather than catenaries if they are subjected to uniformly distributed downward forces along their length. For example, if the weight of the roadway in a suspension bridge is assumed to be uniformly distributed along the supporting cables, then the cables can be modeled by parabolas. (a) Assuming a parabolic model, find an equation for the cable in the accompanying figure, taking the y-axis to be vertical and the origin at the low point of the cable. (b) Find the length of the cable between the supports. 470 ft

Figure Ex-42

43. It will be shown later in this text that if a projectile is launched with speed v0 at an angle α with the horizontal and at a height y0 above ground level, then the resulting trajectory relative to the coordinate system in the accompanying ﬁgure will have parametric equations y = y0 + (v0 sin α)t − 21 gt 2

where g is the acceleration due to gravity. (a) Show that the trajectory is a parabola. (b) Find the coordinates of the vertex. y

a x

44. Mickey Mantle is recognized as baseball’s unofficial king of long home runs. On April 17, 1953 Mantle blasted a pitch by Chuck Stobbs of the hapless Washington Senators out of Griffith Stadium, just clearing the 50 ft wall at the 391 ft marker in left center. Assuming that the ball left the bat at a height of 3 ft above the ground and at an angle of 45 ◦ , use the parametric equations in Exercise 43 with g = 32 ft/s2 to find (a) the speed of the ball as it left the bat (b) the maximum height of the ball (c) the distance along the ground from home plate to where the ball struck the ground. 45–47 Rotate the coordinate axes to remove the xy-term, and

then name the conic. ■

48. Rotate the coordinate axes to show that the graph of 17x 2 − 312xy + 108y 2 + 1080x − 1440y + 4500 = 0 is a hyperbola. Then ﬁnd its vertices, foci, and asymptotes. 49. In each part: (i) Identify the polar graph as a parabola, an ellipse, or a hyperbola; (ii) state whether the directrix is above, below, to the left, or to the right of the pole; and (iii) ﬁnd the distance from the pole to the directrix. 1 1 (a) r = (b) r = 3 + cos θ 1 − 3 cos θ 1 3 (c) r = (d) r = 3(1 + sin θ ) 1 − sin θ

50. (a) Ellipse with eccentricity e = 27 and ends of the minor axis at the points (0, ±3). (b) Parabola with vertex at the origin, focus on the y-axis, and directrix passing through the point (7, 4). (c) Hyperbola that has the same foci as the ellipse 3x 2 + 16y 2 = 48 and asymptotes y = ±2x /3.

51. (a) Ellipse with center (−3, 2), vertex (2, 2), and eccentricity e = 45 . (b) Parabola with focus (−2, −2) and vertex (−2, 0). (c) Hyperbola with vertex (−1, 7) and asymptotes y − 5 = ±8(x + 1).

52. Use the parametric equations x = a cos t, y = b sin t to show that the circumference C of an ellipse with semimajor axis a and eccentricity e is π/2 C = 4a 1 − e2 sin2 u du 0

Figure Ex-43

45. x 2 + y 2 − 3xy − 3 = 0

√ 46. 7x 2 + 2 3xy + 5y 2 − 4 = 0 √ 2 √ √ 47. 4 5x + 4 5xy + 5y 2 + 5x − 10y = 0

50–51 Find an equation in xy-coordinates for the conic section that satisﬁes the given conditions. ■

4200 ft

x = (v0 cos α)t,

765

53. Use Simpson’s rule or the numerical integration capability of a graphing utility to approximate the circumference of the ellipse 4x 2 + 9y 2 = 36 from the integral obtained in Exercise 52.

54. (a) Calculate the eccentricity of the Earth’s orbit, given that the ratio of the distance between the center of the Earth and the center of the Sun at perihelion to the distance 59 between the centers at aphelion is 61 . (b) Find the distance between the center of the Earth and the center of the Sun at perihelion, given that the average value of the perihelion and aphelion distances between the centers is 93 million miles. (c) Use the result in Exercise 52 and Simpson’s rule or the numerical integration capability of a graphing utility to approximate the distance that the Earth travels in 1 year (one revolution around the Sun).

766

Chapter 10 / Parametric and Polar Curves; Conic Sections

CHAPTER 10 MAKING CONNECTIONS C

CAS

1. Recall from Section 5.10 that the Fresnel sine and cosine functions are deﬁned as 2 x 2 x πt πt S(x) = cos dt and C(x) = dt sin 2 2 0 0 The following parametric curve, which is used to study amplitudes of light waves in optics, is called a clothoid or Cornu spiral in honor of the French scientist Marie Alfred Cornu (1841–1902): 2 t πu x = C(t) = du cos 2 0 (−⬁ < t < +⬁) 2 t πu y = S(t) = sin du 2 0 (a) Use a CAS to graph the Cornu spiral. (b) Describe the behavior of the spiral as t → +⬁ and as t → −⬁. (c) Find the arc length of the spiral for −1 ≤ t ≤ 1.

2. (a) The accompanying ﬁgure shows an ellipse with semimajor axis a and semiminor axis b. Express the coordinates of the points P , Q, and R in terms of t. (b) How does the geometric interpretation of the parameter t differ between a circle x = a cos t,

y = a sin t

x = a cos t,

y = b sin t?

and an ellipse

triangle and the other end to a ﬁxed point F. A pencil holds the string taut against the base of the triangle as the edge opposite Q slides along a horizontal line L below F. Show that the pencil traces an arc of a parabola with focus F and directrix L.

Q F

L Figure Ex-3

4. The accompanying ﬁgure shows a method for constructing a hyperbola. A corner of a ruler is pinned to a ﬁxed point F1 and the ruler is free to rotate about that point. A piece of string whose length is less than that of the ruler is tacked to a point F2 and to the free corner Q of the ruler on the same edge as F1 . A pencil holds the string taut against the top edge of the ruler as the ruler rotates about the point F1 . Show that the pencil traces an arc of a hyperbola with foci F1 and F2 . Q

b P

5. Consider an ellipse √ E with semimajor axis a and semiminor axis b, and set c = a 2 − b2 . (a) Show that the ellipsoid that results when E is revolved about its major axis has volume V = 43 πab2 and surface area b a −1 c S = 2πab + sin a c a

Figure Ex-4

Figure Ex-2

3. The accompanying ﬁgure shows Kepler’s method for constructing a parabola. A piece of string the length of the left edge of the drafting triangle is tacked to the vertex Q of the

(b) Show that the ellipsoid that results when E is revolved about its minor axis has volume V = 43 πa 2 b and surface area a b a+c S = 2πab + ln b c b

EXPANDING THE CALCULUS HORIZON To learn how polar coordinates and conic sections can be used to analyze the possibility of a collision between a comet and Earth, see the module entitled Comet Collision at: www.wiley.com/college/anton

11 THREE-DIMENSIONAL SPACE; VECTORS Craig Aurness/© Corbis Images

To describe fully the motion of a boat, one must specify its speed and direction of motion at each instant. Speed and direction together describe a “vector” quantity. We will study vectors in this chapter.

11.1

In this chapter we will discuss rectangular coordinate systems in three dimensions, and we will study the analytic geometry of lines, planes, and other basic surfaces. The second theme of this chapter is the study of vectors. These are the mathematical objects that physicists and engineers use to study forces, displacements, and velocities of objects moving on curved paths. More generally, vectors are used to represent all physical entities that involve both a magnitude and a direction for their complete description. We will introduce various algebraic operations on vectors, and we will apply these operations to problems involving force, work, and rotational tendencies in two and three dimensions. Finally, we will discuss cylindrical and spherical coordinate systems, which are appropriate in problems that involve various kinds of symmetries and also have specific applications in navigation and celestial mechanics.

RECTANGULAR COORDINATES IN 3-SPACE; SPHERES; CYLINDRICAL SURFACES In this section we will discuss coordinate systems in three-dimensional space and some basic facts about surfaces in three dimensions.

RECTANGULAR COORDINATE SYSTEMS

Figure 11.1.1

In the remainder of this text we will call three-dimensional space 3-space, two-dimensional space (a plane) 2-space, and one-dimensional space (a line) 1-space. Just as points in 2space can be placed in one-to-one correspondence with pairs of real numbers using two perpendicular coordinate lines, so points in 3-space can be placed in one-to-one correspondence with triples of real numbers by using three mutually perpendicular coordinate lines, called the x-axis, the y-axis, and the z-axis, positioned so that their origins coincide (Figure 11.1.1). The three coordinate axes form a three-dimensional rectangular coordinate system (or Cartesian coordinate system). The point of intersection of the coordinate axes is called the origin of the coordinate system. Rectangular coordinate systems in 3-space fall into two categories: left-handed and right-handed. A right-handed system has the property that when the ﬁngers of the right hand are cupped so that they curve from the positive x-axis toward the positive y-axis, the thumb points (roughly) in the direction of the positive z-axis (Figure 11.1.2). A similar 767

Chapter 11 / Three-Dimensional Space; Vectors

768

Left-handed y

Right-handed x

Figure 11.1.2

property holds for a left-handed coordinate system (Figure 11.1.2). We will use only righthanded coordinate systems in this text. The coordinate axes, taken in pairs, determine three coordinate planes: the xy-plane, the xz-plane, and the yz-plane (Figure 11.1.3). To each point P in 3-space we can assign a triple of real numbers by passing three planes through P parallel to the coordinate planes and letting a, b, and c be the coordinates of the intersections of those planes with the xaxis, y-axis, and z-axis, respectively (Figure 11.1.4). We call a, b, and c the x-coordinate, y-coordinate, and z-coordinate of P , respectively, and we denote the point P by (a, b, c) or by P (a, b, c). Figure 11.1.5 shows the points (4, 5, 6) and (−3, 2, −4). z

z c

(4, 5, 6) P xy -plane

(a, b, c) y

y O

(−3, 2, −4)

yz -plane xz -plane

Figure 11.1.3

Figure 11.1.4

Figure 11.1.5

Just as the coordinate axes in a two-dimensional coordinate system divide 2-space into four quadrants, so the coordinate planes of a three-dimensional coordinate system divide 3-space into eight parts, called octants. The set of points with three positive coordinates forms the ﬁrst octant; the remaining octants have no standard numbering. You should be able to visualize the following facts about three-dimensional rectangular coordinate systems:

region

description

xy-plane xz-plane yz-plane x-axis y-axis z-axis

Consists of all points of the form (x, y, 0) Consists of all points of the form (x, 0, z) Consists of all points of the form (0, y, z) Consists of all points of the form (x, 0, 0) Consists of all points of the form (0, y, 0) Consists of all points of the form (0, 0, z)

DISTANCE IN 3-SPACE; SPHERES

Recall that in 2-space the distance d between the points P1 (x1 , y1 ) and P2 (x2 , y2 ) is d = (x2 − x1 )2 + (y2 − y1 )2

(1)

The distance formula in 3-space has the same form, but it has a third term to account for the added dimension. (We will see that this is a common occurrence in extending formulas from 2-space to 3-space.) The distance between the points P1 (x1 , y1 , z1 ) and P2 (x2 , y2 , z2 ) is d = (x2 − x1 )2 + (y2 − y1 )2 + (z2 − z1 )2 (2)

We leave the proof of (2) as an exercise (Exercise 7).

11.1 Rectangular Coordinates in 3-Space; Spheres; Cylindrical Surfaces

Example 1

769

Find the distance d between the points (2, 3, −1) and (4, −1, 3).

Solution. From Formula (2) d= In an xy -coordinate system, the set of points (x, y) whose coordinates satisfy an equation in x and y is called the graph of the equation. Analogously, in an xyz-coordinate system, the set of points (x, y, z) whose coordinates satisfy an equation in x , y , and z is called the graph of the equation.

√ (4 − 2)2 + (−1 − 3)2 + (3 + 1)2 = 36 = 6

Recall that the standard equation of the circle in 2-space that has center (x0 , y0 ) and radius r is (x − x0 )2 + (y − y0 )2 = r 2 (3) This follows from distance formula (1) and the fact that the circle consists of all points in 2-space whose distance from (x0 , y0 ) is r. Analogously, the standard equation of the sphere in 3-space that has center (x0 , y0 , z0 ) and radius r is (x − x0 )2 + (y − y0 )2 + (z − z0 )2 = r 2

(4)

This follows from distance formula (2) and the fact that the sphere consists of all points in 3-space whose distance from (x0 , y0 , z0 ) is r. Note that (4) has the same form as the standard equation for the circle in 2-space, but with an additional term to account for the third coordinate. Some examples of the standard equation of the sphere are given in the following table: equation

graph

(x − 3)2 + (y − 2)2 + (z − 1)2 = 9 (x + 1)2 + y 2 + (z + 4)2 = 5 x2 + y2 + z2 = 1

Sphere with center (3, 2, 1) and radius 3 Sphere with center (−1, 0, −4) and radius √5 Sphere with center (0, 0, 0) and radius 1

If the terms in (4) are expanded and like terms are collected, then the resulting equation has the form x 2 + y 2 + z2 + Gx + Hy + I z + J = 0 (5) The following example shows how the center and radius of a sphere that is expressed in this form can be obtained by completing the squares. Example 2

Find the center and radius of the sphere x 2 + y 2 + z2 − 2x − 4y + 8z + 17 = 0

Solution. We can put the equation in the form of (4) by completing the squares: (x 2 − 2x) + (y 2 − 4y) + (z2 + 8z) = −17 (x 2 − 2x + 1) + (y 2 − 4y + 4) + (z2 + 8z + 16) = −17 + 21 (x − 1)2 + (y − 2)2 + (z + 4)2 = 4 which is the equation of the sphere with center (1, 2, −4) and radius 2. In general, completing the squares in (5) produces an equation of the form (x − x0 )2 + (y − y0 )2 + (z − z0 )2 = k

√ If k > 0, then the graph of this equation is a sphere with center (x0 , y0 , z0 ) and radius k. If k = 0, then the sphere has radius zero, so the graph is the single point (x0 , y0 , z0 ). If k < 0, the equation is not satisﬁed by any values of x, y, and z (why?), so it has no graph.

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Chapter 11 / Three-Dimensional Space; Vectors

11.1.1

theorem An equation of the form x 2 + y 2 + z2 + Gx + Hy + I z + J = 0

represents a sphere, a point, or has no graph.

(x, y, z) y

(x, y, 0 ) x

Figure 11.1.6

2-space

x 2 + z2 = 1 x

11.1.2 theorem An equation that contains only two of the variables x, y, and z represents a cylindrical surface in an xyz-coordinate system. The surface can be obtained by graphing the equation in the coordinate plane of the two variables that appear in the equation and then translating that graph parallel to the axis of the missing variable.

3-space

CYLINDRICAL SURFACES Although it is natural to graph equations in two variables in 2-space and equations in three variables in 3-space, it is also possible to graph equations in two variables in 3-space. For example, the graph of the equation y = x 2 in an xy-coordinate system is a parabola; however, there is nothing to prevent us from inquiring about its graph in an xyz-coordinate system. To obtain this graph we need only observe that the equation y = x 2 does not impose any restrictions on z. Thus, if we ďŹ nd values of x and y that satisfy this equation, then the coordinates of the point (x, y, z) will also satisfy the equation for arbitrary values of z. Geometrically, the point (x, y, z) lies on the vertical line through the point (x, y, 0) in the xy-plane, which means that we can obtain the graph of y = x 2 in an xyz-coordinate system by ďŹ rst graphing the equation in the xy-plane and then translating that graph parallel to the z-axis to generate the entire graph (Figure 11.1.6). The process of generating a surface by translating a plane curve parallel to some line is called extrusion, and surfaces that are generated by extrusion are called cylindrical surfaces. A familiar example is the surface of a right circular cylinder, which can be generated by translating a circle parallel to the axis of the cylinder. The following theorem provides basic information about graphing equations in two variables in 3-space:

x2 + z2 = 1 y

Example 3

Sketch the graph of x 2 + z2 = 1 in 3-space.

Solution. Since y does not appear in this equation, the graph is a cylindrical surface generated by extrusion parallel to the y-axis. In the xz-plane the graph of the equation x 2 + z2 = 1 is a circle. Thus, in 3-space the graph is a right circular cylinder along the y-axis (Figure 11.1.7).

Figure 11.1.7

Example 4 In an xy -coordinate system, the graph of the equation x = 1 is a line parallel to the y -axis. What is the graph of this equation in an xyz-coordinate system?

Sketch the graph of z = sin y in 3-space.

Solution. (See Figure 11.1.8.) z z

z = sin y

z = sin y y

Figure 11.1.8

2-space

3-space

11.1 Rectangular Coordinates in 3-Space; Spheres; Cylindrical Surfaces

✔QUICK CHECK EXERCISES 11.1

(See page 773 for answers.)

1. The distance between the points (1, −2, 0) and (4, 0, 5) is . 2. The graph of (x − 3)2 + (y − 2)2 + (z + 1)2 = 16 is a of radius centered at . 3. The shortest distance from the point (4, 0, 5) to the sphere (x − 1)2 + (y + 2)2 + z2 = 36 is .

EXERCISE SET 11.1

(b)

z y

2. A cube of side 4 has its geometric center at the origin and its faces parallel to the coordinate planes. Sketch the cube and give the coordinates of the corners.

F O C U S O N C O N C E P TS

3. Suppose that a box has its faces parallel to the coordinate planes and the points (4, 2, −2) and (−6, 1, 1) are endpoints of a diagonal. Sketch the box and give the coordinates of the remaining six corners. 4. Suppose that a box has its faces parallel to the coordinate planes and the points (x1 , y1 , z1 ) and (x2 , y2 , z2 ) are endpoints of a diagonal. (a) Find the coordinates of the remaining six corners. (b) Show that the midpoint of the line segment joining (x1 , y1 , z1 ) and (x2 , y2 , z2 ) is 1 2

4. Let S be the graph of x 2 + z2 + 6z = 16 in 3-space. (a) The intersection of S with the xz-plane is a circle with center and radius . (b) The intersection of S with the xy-plane is two lines, x= and x = . (c) The intersection of S with the yz-plane is two lines, z= and z = .

Graphing Utility

1. In each part, ﬁnd the coordinates of the eight corners of the box. (a)

771

(x1 + x2 ), 21 (y1 + y2 ), 21 (z1 + z2 )

[Suggestion: Apply Theorem H.2 in Web Appendix H to three appropriate edges of the box.] 5. Interpret the graph of x = 1 in the contexts of (a) a number line (b) 2-space (c) 3-space. 6. Consider the points P (3, 1, 0) and Q(1, 4, 4). (a) Sketch the triangle with vertices P , Q, and (1, 4, 0). Without computing distances, explain why this triangle is a right triangle, and then apply the Theorem of Pythagoras twice to ﬁnd the distance from P to Q. (b) Repeat part (a) using the points P , Q, and (3, 4, 0). (c) Repeat part (a) using the points P , Q, and (1, 1, 4).

7. (a) Consider a box whose sides have lengths a, b, and c. Use the Theorem of Pythagoras√to show that a diagonal of the box has length d = a 2 + b2 + c2 . [Hint: Use the Theorem of Pythagoras to find the length of a diagonal of the base and then again to find the length of a diagonal of the entire box.] (b) Use the result of part (a) to derive formula (2). 8. (a) Make a conjecture about the set of points in 3-space that are equidistant from the origin and the point (1, 0, 0). (b) Confirm your conjecture in part (a) by using distance formula (2). 9. Find the center and radius of the sphere that has (1, −2, 4) and (3, 4, −12) as endpoints of a diameter. [See Exercise 4.] 10. Show that (4, 5, 2), (1, 7, 3), and (2, 4, 5) are vertices of an equilateral triangle. 11. (a) Show that (2, 1, 6), (4, 7, 9), and (8, 5, −6) are the vertices of a right triangle. (b) Which vertex is at the 90 ◦ angle? (c) Find the area of the triangle. 12. Find the distance from the point (−5, 2, −3) to the (a) xy-plane (b) xz-plane (c) yz-plane (d) x-axis (e) y-axis (f ) z-axis. 13. In each part, find the standard equation of the sphere that satisfies the stated conditions. (a) Center (7, 1, 1); radius = 4. (b) Center (1, 0, −1); diameter = 8. (c) Center (−1, 3, 2) and passing through the origin. (d) A diameter has endpoints (−1, 2, 1) and (0, 2, 3). 14. Find equations of two spheres that are centered at the origin and are tangent to the sphere of radius 1 centered at (3, −2, 4). 15. In each part, find an equation of the sphere with center (2, −1, −3) and satisfying the given condition. (a) Tangent to the xy-plane (b) Tangent to the xz-plane (c) Tangent to the yz-plane

16. (a) Find an equation of the sphere that is inscribed in the cube that is centered at the point (−2, 1, 3) and has sides of length 1 that are parallel to the coordinate planes. (b) Find an equation of the sphere that is circumscribed about the cube in part (a). (cont.)

Chapter 11 / Three-Dimensional Space; Vectors

772

(c) Find an equation of the sphere that is inscribed in the cube determined by the planes x = 6, x = 2, y = 5, y = 9, z = 0, and z = 4. (d) Find an equation of the sphere that is circumscribed about the cube in part (c). 17. A sphere has center in the ﬁrst octant and is tangent to each of the three coordinate planes. Show that the center of the sphere is at a point of the form (r, r, r), where r is the radius of the sphere. 18. A sphere has center in the ﬁrst octant and is tangent to each of the three coordinate planes. The distance from the ori√ gin to the sphere is 3 − 3 units. Find an equation for the sphere. 19–22 True–False Determine whether the statement is true or

false. Explain your answer. ■ 19. By deﬁnition, a “cylindrical surface” is a right circular cylinder whose axis is parallel to one of the coordinate axes. 20. The graph of x 2 + y 2 = 1 in 3-space is a circle of radius 1 centered at the origin. 21. If a point belongs to both the xy-plane and the xz-plane, then the point lies on the x-axis. 22. A sphere with center P (x0 , y0 , z0 ) and radius r consists of all points (x, y, z) that satisfy the inequality (x − x0 )2 + (y − y0 )2 + (z − z0 )2 ≤ r 2 23–28 Describe the surface whose equation is given. ■

23. x 2 + y 2 + z2 + 10x + 4y + 2z − 19 = 0

24. x 2 + y 2 + z2 − y = 0

25. 2x 2 + 2y 2 + 2z2 − 2x − 3y + 5z − 2 = 0

26. x 2 + y 2 + z2 + 2x − 2y + 2z + 3 = 0 2

27. x + y + z − 3x + 4y − 8z + 25 = 0 28. x 2 + y 2 + z2 − 2x − 6y − 8z + 1 = 0

29. In each part, sketch the portion of the surface that lies in the ﬁrst octant. (a) y = x (b) y = z (c) x = z 30. In each part, sketch the graph of the equation in 3-space. (a) x = 1 (b) y = 1 (c) z = 1

31. In each part, sketch the graph of the equation in 3-space. (a) x 2 + y 2 = 25 (b) y 2 + z2 = 25 (c) x 2 + z2 = 25

34. Find equations for the following right circular cylinders. Each cylinder has radius a and is tangent to two coordinate planes. z z z (a) (b) (c) (a, 0, a)

(0, a, a) y

(a, a, 0) x

35–44 Sketch the surface in 3-space. ■

35. y = sin x

37. z = 1 − y

39. 2x + z = 3

41. 4x 2 + 9z2 = 36

43. y 2 − 4z2 = 4

36. y = ex

38. z = cos x

40. 2x + 3y = 6 √ 3−x

42. z =

44. yz = 1

45. Use a graphing utility to generate the curve y = x 3 /(1 + x 2 ) in the xy-plane, and then use the graph to help sketch the surface z = y 3 /(1 + y 2 ) in 3-space. 46. Use a graphing utility to generate the curve y = x /(1 + x 4 ) in the xy-plane, and then use the graph to help sketch the surface z = y /(1 + y 4 ) in 3-space. 47. If a bug walks on the sphere

x 2 + y 2 + z2 + 2x − 2y − 4z − 3 = 0

how close and how far can it get from the origin? 48. Describe the set of all points in 3-space whose coordinates satisfy the inequality x 2 + y 2 + z2 − 2x + 8z ≤ 8.

49. Describe the set of all points in 3-space whose coordinates satisfy the inequality y 2 + z2 + 6y − 4z > 3.

50. The distance between a point P (x, y, z) and the point A(1, −2, 0) is twice the distance between P and the point B(0, 1, 1). Show that the set of all such points is a sphere, and ﬁnd the center and radius of the sphere. 51. As shown in the accompanying ﬁgure, a bowling ball of radius R is placed inside a box just large enough to hold it, and it is secured for shipping by packing a Styrofoam sphere into each corner of the box. Find the radius of the largest Styrofoam sphere that can be used. [Hint: Take the origin of a Cartesian coordinate system at a corner of the box with the coordinate axes along the edges.]

32. In each part, sketch the graph of the equation in 3-space. (a) x = y 2 (b) z = x 2 (c) y = z2

33. In each part, write an equation for the surface. (a) The plane that contains the x-axis and the point (0, 1, 2). (b) The plane that contains the y-axis and the point (1, 0, 2). (c) The right circular cylinder that has radius 1 and is centered on the line parallel to the z-axis that passes through the point (1, 1, 0). (d) The right circular cylinder that has radius 1 and is centered on the line parallel to the y-axis that passes through the point (1, 0, 1).

Figure Ex-51

52. Consider the equation x 2 + y 2 + z2 + Gx + Hy + I z + J = 0

and let K = G2 + H 2 + I 2 − 4J .

(cont.)

11.2 Vectors

(a) Prove that the equation represents a sphere if K > 0, a point if K = 0, and has no graph if K < 0. (b) In the case where K > 0, find the center and radius of the sphere. 53. (a) The accompanying figure shows a surface of revolution that is generated by revolving the curve y = f (x) in the xy-plane about the x-axis. Show that the equation of this surface is y 2 + z2 = [f (x)]2 . [Hint: Each point on the curve traces a circle as it revolves about the x-axis.] (b) Find an equation of the surface of revolution that is generated by revolving the curve y = ex in the xy-plane about the x-axis. (c) Show that the ellipsoid 3x 2 + 4y 2 + 4z2 = 16 is a surface of revolution about the x-axis by finding a curve y = f (x) in the xy-plane that generates it.

773

54. In each part, use the idea in Exercise 53(a) to derive a formula for the stated surface of revolution. (a) The surface generated by revolving the curve x = f (y) in the xy-plane about the y-axis. (b) The surface generated by revolving the curve y = f (z) in the yz-plane about the z-axis. (c) The surface generated by revolving the curve z = f (x) in the xz-plane about the x-axis. 55. Show that for all values of θ and φ, the point (a sin φ cos θ, a sin φ sin θ, a cos φ) lies on the sphere x 2 + y 2 + z2 = a 2 . 56. Writing Explain how you might determine whether a set of points in 3-space is the graph of an equation involving at most two of the variables x, y, and z.

57. Writing Discuss what happens geometrically when equations in x, y, and z are replaced by inequalities. For example, compare the graph of x 2 + y 2 + z2 = 1 with the set of points that satisfy the inequality x 2 + y 2 + z2 ≤ 1.

y = f (x) x

Figure Ex-53

✔QUICK CHECK ANSWERS 11.1 1.

√

11.2

2. sphere; 4; (3, 2, −1)

√ 38 − 6

4. (a) (0, 0, −3); 5 (b) 4; −4 (c) 2; −8

VECTORS Many physical quantities such as area, length, mass, and temperature are completely described once the magnitude of the quantity is given. Such quantities are called “scalars.” Other physical quantities, called “vectors,” are not completely determined until both a magnitude and a direction are speciﬁed. For example, winds are usually described by giving their speed and direction, say 20 mi/h northeast. The wind speed and wind direction together form a vector quantity called the wind velocity. Other examples of vectors are force and displacement. In this section we will develop the basic mathematical properties of vectors. VECTORS IN PHYSICS AND ENGINEERING A particle that moves along a line can move in only two directions, so its direction of motion can be described by taking one direction to be positive and the other negative. Thus, the displacement or change in position of the point can be described by a signed real number. For example, a displacement of 3 (= +3) describes a position change of 3 units in the positive direction, and a displacement of −3 describes a position change of 3 units in the negative direction. However, for a particle that moves in two dimensions or three dimensions, a plus or minus sign is no longer sufﬁcient to specify the direction of motion—other methods are required. One method is to use an arrow, called a vector, that points in the direction of motion and whose length represents the distance from the starting point to the ending point; this is called the displacement vector for the motion. For example, Figure 11.2.1a shows the displacement vector of a particle that moves from point A to point B along a circuitous path. Note that the length of the arrow describes the

774

Chapter 11 / Three-Dimensional Space; Vectors B

A A displacement vector

(a) Rope

10 lb

A force vector acting on a block

(b) 3 mi/h 45°

2 mi/h

Two velocity vectors that affect the motion of the boat

(a)

(b)

Figure 11.2.2

w v

and that the sum (gray arrow) coincides with the diagonal of the parallelogram determined by v and w when these vectors are positioned so they have the same initial point. Since the initial and terminal points of 0 coincide, it follows that

w+v w

(b) Figure 11.2.3

11.2.1 definition If v and w are vectors, then the sum v + w is the vector from the initial point of v to the terminal point of w when the vectors are positioned so the initial point of w is at the terminal point of v (Figure 11.2.3a).

v+w =w+v

v+w

VECTORS VIEWED GEOMETRICALLY Vectors can be represented geometrically by arrows in 2-space or 3-space; the direction of the arrow specifies the direction of the vector, and the length of the arrow describes its magnitude. The tail of the arrow is called the initial point of the vector, and the tip of the arrow the terminal point. We will denote vectors with lowercase boldface type such as a, k, v, w, and x. When discussing vectors, we will refer to real numbers as scalars. Scalars will be denoted by lowercase italic type such as a, k, v, w, and x. Two vectors, v and w, are considered to be equal (also called equivalent) if they have the same length and same direction, in which case we write v = w. Geometrically, two vectors are equal if they are translations of one another; thus, the three vectors in Figure 11.2.2a are equal, even though they are in different positions. Because vectors are not affected by translation, the initial point of a vector v can be moved to any convenient point A by making an appropriate translation. If the initial point −→ of v is A and the terminal point is B, then we write v = AB when we want to emphasize the initial and terminal points (Figure 11.2.2b). If the initial and terminal points of a vector coincide, then the vector has length zero; we call this the zero vector and denote it by 0. The zero vector does not have a specific direction, so we will agree that it can be assigned any convenient direction in a specific problem. There are various algebraic operations that are performed on vectors, all of whose definitions originated in physics. We begin with vector addition.

In Figure 11.2.3b we have constructed two sums, v + w (from purple arrows) and w + v (from green arrows). It is evident that

(a)

distance between the starting and ending points and not the actual distance traveled by the particle. Arrows are not limited to describing displacements—they can be used to describe any physical quantity that involves both a magnitude and a direction. Two important examples are forces and velocities. For example, the arrow in Figure 11.2.1b represents a force vector of 10 lb acting in a specific direction on a block, and the arrows in Figure 11.2.1c show the velocity vector of a boat whose motor propels it parallel to the shore at 2 mi/h and the velocity vector of a 3 mi/h wind acting at an angle of 45 ◦ with the shoreline. Intuition suggests that the two velocity vectors will combine to produce some net velocity for the boat at an angle to the shoreline. Thus, our first objective in this section is to define mathematical operations on vectors that can be used to determine the combined effect of vectors.

0+v =v+0=v 11.2.2 definition If v is a nonzero vector and k is a nonzero real number (a scalar), then the scalar multiple kv is deﬁned to be the vector whose length is |k| times the length of v and whose direction is the same as that of v if k > 0 and opposite to that of v if k < 0. We deﬁne kv = 0 if k = 0 or v = 0.

11.2 Vectors

2v v 1 v 2

(−1)v

775

Figure 11.2.4 shows the geometric relationship between a vector v and various scalar multiples of it. Observe that if k and v are nonzero, then the vectors v and kv lie on the same line if their initial points coincide and lie on parallel or coincident lines if they do not. Thus, we say that v and kv are parallel vectors. Observe also that the vector (−1)v has the same length as v but is oppositely directed. We call (−1)v the negative of v and denote it by −v (Figure 11.2.5). In particular, −0 = (−1)0 = 0. Vector subtraction is deﬁned in terms of addition and scalar multiplication by

冸− 23 冹v

v − w = v + (−w) The difference v − w can be obtained geometrically by ﬁrst constructing the vector −w and then adding v and −w, say by the parallelogram method (Figure 11.2.6a). However, if v and w are positioned so their initial points coincide, then v − w can be formed more directly, as shown in Figure 11.2.6b, by drawing the vector from the terminal point of w (the second term) to the terminal point of v (the ﬁrst term). In the special case where v = w the terminal points of the vectors coincide, so their difference is 0; that is,

Figure 11.2.4

v + (−v) = v − v = 0

−w

−v

v−w

(a)

(v1, v2 ) Figure 11.2.5

v−w

(b)

Figure 11.2.6

VECTORS IN COORDINATE SYSTEMS

(v1, v2, v3) v y

Figure 11.2.7

Problems involving vectors are often best solved by introducing a rectangular coordinate system. If a vector v is positioned with its initial point at the origin of a rectangular coordinate system, then its terminal point will have coordinates of the form (v1 , v2 ) or (v1 , v2 , v3 ), depending on whether the vector is in 2-space or 3-space (Figure 11.2.7). We call these coordinates the components of v, and we write v in component form using the bracket notation v = v1 , v2 or v = v1 , v2 , v3 2-space

3-space

In particular, the zero vectors in 2-space and 3-space are Note the difference in notation between a point (v1 , v2 ) and a vector v1 , v2 .

0 = 0, 0

and 0 = 0, 0, 0

respectively. Components provide a simple way of identifying equivalent vectors. For example, consider the vectors v = v1 , v2 and w = w1 , w2 in 2-space. If v = w, then the vectors have the same length and same direction, and this means that their terminal points coincide when their initial points are placed at the origin. It follows that v1 = w1 and v2 = w2 , so we have shown that equivalent vectors have the same components. Conversely, if v1 = w1 and v2 = w2 , then the terminal points of the vectors coincide when their initial points are placed at the origin. It follows that the vectors have the same length and same direction, so we have shown that vectors with the same components are equivalent. A similar argument holds for vectors in 3-space, so we have the following result.

Chapter 11 / Three-Dimensional Space; Vectors

776

11.2.3 theorem Two vectors are equivalent if and only if their corresponding components are equal.

For example, a, b, c = 1, −4, 2 if and only if a = 1, b = −4, and c = 2. ARITHMETIC OPERATIONS ON VECTORS The next theorem shows how to perform arithmetic operations on vectors using components.

11.2.4 theorem If v = v1 , v2 and w = w1 , w2 are vectors in 2-space and k is any scalar, then

(v1 + w1, v2 + w2)

(w1, w2 )

v + w = v1 + w1 , v2 + w2 v − w = v1 − w1 , v2 − w2 kv = kv1 , kv2

(v1, v2 )

v v1

Similarly, if v = v1 , v2 , v3 and w = w1 , w2 , w3 are vectors in 3-space and k is any scalar, then

v + w = v1 + w1 , v2 + w2 , v3 + w3 v − w = v1 − w1 , v2 − w2 , v3 − w3 kv = kv1 , kv2 , kv3

(k v1, k v2 ) kv k v2 v2

(v1, v2 )

(1) (2) (3)

(4) (5) (6)

We will not prove this theorem. However, results (1) and (3) should be evident from Figure 11.2.8. Similar ﬁgures in 3-space can be used to motivate (4) and (6). Formulas (2) and (5) can be obtained by writing v + w = v + (−1)w.

k v1 Figure 11.2.8

Example 1

If v = −2, 0, 1 and w = 3, 5, −4 , then v + w = −2, 0, 1 + 3, 5, −4 = 1, 5, −3 3v = −6, 0, 3

−w = −3, −5, 4

w − 2v = 3, 5, −4 − −4, 0, 2 = 7, 5, −6

P1(x1, y1)

P1 P2

P2 (x2, y2 )

VECTORS WITH INITIAL POINT NOT AT THE ORIGIN OP1

OP2 x

O Figure 11.2.9

Recall that we defined the components of a vector to be the coordinates of its terminal point when its initial point is at the origin. We will now consider the problem of finding the components of a vector whose initial point is not at the origin. To be specific, suppose that P1 (x1 , y1 ) and P2 (x2 , y2 ) are points in 2-space and we are interested in finding the −−→ components of the vector P1P2 . As illustrated in Figure 11.2.9, we can write this vector as −−→ −→ −→ P1P2 = OP2 − OP1 = x2 , y2 − x1 , y1 = x2 − x1 , y2 − y1

11.2 Vectors

777

−−→ Thus, we have shown that the components of the vector P1P2 can be obtained by subtracting the coordinates of its initial point from the coordinates of its terminal point. Similar computations hold in 3-space, so we have established the following result. −−→ 11.2.5 theorem If P1P2 is a vector in 2-space with initial point P1 (x1 , y1 ) and terminal point P2 (x2 , y2 ), then −−→ P1P2 = x2 − x1 , y2 − y1

(7)

−−→ Similarly, if P1P2 is a vector in 3-space with initial point P1 (x1 , y1 , z1 ) and terminal point P2 (x2 , y2 , z2 ), then −−→ P1P2 = x2 − x1 , y2 − y1 , z2 − z1

Example 2

(8)

In 2-space the vector from P1 (1, 3) to P2 (4, −2) is −−→ P1P2 = 4 − 1, −2 − 3 = 3, −5

and in 3-space the vector from A(0, −2, 5) to B(3, 4, −1) is −→ AB = 3 − 0, 4 − (−2), −1 − 5 = 3, 6, −6 RULES OF VECTOR ARITHMETIC The following theorem shows that many of the familiar rules of ordinary arithmetic also hold for vector arithmetic.

It follows from part (b) of Theorem 11.2.6 that the expression

u+v+w is unambiguous since the same vector results no matter how the terms are grouped.

11.2.6 theorem For any vectors u, v, and w and any scalars k and l, the following relationships hold: (a) u + v = v + u (e) k(lu) = (kl)u (b) (u + v) + w = u + (v + w) ( f ) k(u + v) = ku + kv (c) u + 0 = 0 + u = u (g) (k + l)u = ku + lu (d ) u + (−u) = 0 (h) 1u = u The results in this theorem can be proved either algebraically by using components or geometrically by treating the vectors as arrows. We will prove part (b) both ways and leave some of the remaining proofs as exercises.

Observe that in Figure 11.2.10 the vectors u, v, and w are positioned “tip to tail” and that

u+v+w is the vector from the initial point of u (the first term in the sum) to the terminal point of w (the last term in the sum). This “tip to tail” method of vector addition also works for four or more vectors (Figure 11.2.11).

proof (b) (algebraic in 2-space) Let u = u1 , u2 , v = v1 , v2 , and w = w1 , w2 . Then (u + v) + w = ( u1 , u2 + v1 , v2 ) + w1 , w2 = u1 + v1 , u2 + v2 + w1 , w2 = (u1 + v1 ) + w1 , (u2 + v2 ) + w2 = u1 + (v1 + w1 ), u2 + (v2 + w2 ) = u1 , u2 + v1 + w1 , v2 + w2 = u + (v + w)

778

Chapter 11 / Three-Dimensional Space; Vectors Q v u

u+v

u + (v + w) (u + v )+w

proof (b) (geometric) in Figure 11.2.10. Then

− → − → v + w = QS and u + (v + w) = PS − → − → u + v = PR and (u + v) + w = PS

−→ −→ − → Let u, v, and w be represented by PQ, QR, and RS as shown

Therefore, (u + v) + w = u + (v + w) ■

Figure 11.2.10

u u+

NORM OF A VECTOR The distance between the initial and terminal points of a vector v is called the length, the norm, or the magnitude of v and is denoted by v . This distance does not change if the vector is translated, so for purposes of calculating the norm we can assume that the vector is positioned with its initial point at the origin (Figure 11.2.12). This makes it evident that the norm of a vector v = v1 , v2 in 2-space is given by

v w

Figure 11.2.11 y

v12 + v22

(9)

v12 + v22 + v32

(10)

v =

and the norm of a vector v = v1 , v2 , v3 in 3-space is given by

(v1, v2 ) ||v||

v =

v2 x

Example 3

(v1, v2, v3)

Find the norms of v = −2, 3 , 10v = −20, 30 , and w = 2, 3, 6 .

Solution. From (9) and (10) √ 13 √ √ 10v = (−20)2 + 302 = 1300 = 10 13 √ √ w = 22 + 32 + 62 = 49 = 7 v =

||v|| v3 y v1 v2

(−2)2 + 32 =

Note that 10v = 10 v in Example 3. This is consistent with Deﬁnition 11.2.2, which stipulated that for any vector v and scalar k, the length of kv must be |k| times the length of v; that is, kv = |k| v (11)

Figure 11.2.12

Thus, for example, 3v = |3| v = 3 v

(0, 1)

−2v = |−2| v = 2 v −1v = |−1| v = v

j x

i z

(0, 0, 1) k j

i (1, 0, 0)

(0, 1, 0)

Figure 11.2.13

This applies to vectors in 2-space and 3-space.

(1, 0)

UNIT VECTORS A vector of length 1 is called a unit vector. In an xy-coordinate system the unit vectors along the x- and y-axes are denoted by i and j, respectively; and in an xyz-coordinate system the unit vectors along the x-, y-, and z-axes are denoted by i, j, and k, respectively (Figure 11.2.13). Thus,

i = 1, 0 ,

j = 0, 1

i = 1, 0, 0 ,

j = 0, 1, 0 ,

In 2-space

k = 0, 0, 1

In 3-space

11.2 Vectors

779

Every vector in 2-space is expressible uniquely in terms of i and j, and every vector in 3-space is expressible uniquely in terms of i, j, and k as follows: v = v1 , v2 = v1 , 0 + 0, v2 = v1 1, 0 + v2 0, 1 = v1 i + v2 j v = v1 , v2 , v3 = v1 1, 0, 0 + v2 0, 1, 0 + v3 0, 0, 1 = v1 i + v2 j + v3 k Example 4 The following table provides some examples of vector notation in 2-space and 3-space. 2-space

The two notations for vectors illustrated in Example 4 are completely interchangeable, the choice being a matter of convenience or personal preference.

3-space

〈2, 3〉 = 2i + 3j

〈2, −3, 4〉 = 2i − 3j + 4k

〈−4, 0〉 = −4i + 0j = −4i

〈0, 3, 0〉 = 3j

〈0, 0〉 = 0i + 0j = 0

〈0, 0, 0〉 = 0i + 0j + 0k = 0

(3i + 2j) + (4i + j) = 7i + 3j

(3i + 2j − k) − (4i − j + 2k) = −i + 3j − 3k

5(6i − 2j) = 30i − 10j

2(i + j − k) + 4(i − j) = 6i − 2j − 2k

|| 2i − 3j || = √ + ||v1i + v2 j || = √v12 + v22 22

(−3)2

= √13

||i + 2j − 3k|| = √12 + 22 + (−3)2 = √14 ||〈v1, v2, v3〉|| = √v12 + v22 + v32

NORMALIZING A VECTOR A common problem in applications is to ﬁnd a unit vector u that has the same direction as some given nonzero vector v. This can be done by multiplying v by the reciprocal of its length; that is, 1 v u= v= v v

is a unit vector with the same direction as v—the direction is the same because k = 1/ v is a positive scalar, and the length is 1 because u = kv = |k| v = k v =

1 v = 1 v

The process of multiplying a vector v by the reciprocal of its length to obtain a unit vector with the same direction is called normalizing v. T E C H N O LO GY M A ST E R Y Many calculating utilities can perform vector operations, and some have builtin norm and normalization operations. If your calculator has these capabilities, use it to check the computations in Examples 1, 3, and 5.

Example 5

Find the unit vector that has the same direction as v = 2i + 2 j − k.

Solution. The vector v has length v =

22 + 22 + (−1)2 = 3

so the unit vector u in the same direction as v is u = 31 v = 23 i + 23 j − 31 k VECTORS DETERMINED BY LENGTH AND ANGLE

If v is a nonzero vector with its initial point at the origin of an xy-coordinate system, and if θ is the angle from the positive x-axis to the radial line through v, then the x-component of v can be written as v cos θ and the y-component as v sin θ (Figure 11.2.14); and hence v can be expressed in trigonometric form as v = v cos θ, sin θ

or v = v cos θi + v sin θ j

(12)

780

Chapter 11 / Three-Dimensional Space; Vectors

In the special case of a unit vector u this simpliﬁes to

u = cos θ, sin θ or

||v||

||v||sin u x u ||v||cos u

u = cos θi + sin θ j

(13)

Example 6 (a) Find the vector of length 2 that makes an angle of π/4 with the positive x-axis. √ (b) Find the angle that the vector v = − 3i + j makes with the positive x-axis.

Solution (a). From (12)

Figure 11.2.14

√ √ π π v = 2 cos i + 2 sin j = 2 i + 2 j 4 4

Solution (b). We will normalize v, then use (13) to ﬁnd sin θ and cos θ , and then use these values to ﬁnd θ. Normalizing v yields √ √ v 3 − 3i + j 1 =− = √ i+ j 2 v 2 2 − 3 + 12 √ Thus, cos θ = − 3/2 and sin θ = 1/2, from which we conclude that θ = 5π/6. VECTORS DETERMINED BY LENGTH AND A VECTOR IN THE SAME DIRECTION

It is a common problem in many applications that a direction in 2-space or 3-space is determined by some known unit vector u, and it is of interest to ﬁnd the components of a vector v that has the same direction as u and some speciﬁed length v . This can be done by expressing v as v = v u

v is equal to its length times a unit vector in the same direction.

and then reading off the components of v u. z

Example 7 Figure 11.2.15 shows a vector v of length through A and B. Find the components of v.

A(0, 0, 4) ||v|| = √5 v

−→

Solution. First we will ﬁnd the components of the vector AB, then we will normalize y

Figure 11.2.15

B(2, 5, 0)

√ 5 that extends along the line

this vector to obtain a unit vector in the direction of v, and then we will multiply this unit vector by v to obtain the vector v. The computations are as follows: −→ AB = 2, 5, 0 − 0, 0, 4 = 2, 5, −4 √ √ −→ AB = 22 + 52 + (−4)2 = 45 = 3 5 −→ 5 4 AB 2 −→ = √ , √ , − √ 3 5 3 5 3 5 AB −→ √ AB 2 5 5 4 4 2 v = v −→ = 5 √ , √ ,− √ = , ,− 3 3 3 3 5 3 5 3 5 AB RESULTANT OF TWO CONCURRENT FORCES The effect that a force has on an object depends on the magnitude and direction of the force and the point at which it is applied. Thus, forces are regarded to be vector quantities and, indeed, the algebraic operations on vectors that we have deﬁned in this section have their origin in the study of forces. For example, it is a fact of physics that if two forces F1 and

11.2 Vectors F1 + F2 F2

F1 The single force F1 + F2 has the same effect as the two forces F1 and F2.

Figure 11.2.16

781

F2 are applied at the same point on an object, then the two forces have the same effect on the object as the single force F1 + F2 applied at the point (Figure 11.2.16). Physicists and engineers call F1 + F2 the resultant of F1 and F2 , and they say that the forces F1 and F2 are concurrent to indicate that they are applied at the same point. In many applications, the magnitudes of two concurrent forces and the angle between them are known, and the problem is to find the magnitude and direction of the resultant. One approach to solving this problem is to use (12) to find the components of the concurrent forces, and then use (1) to find the components of the resultant. The next example illustrates this method. Example 8 Suppose that two forces are applied to an eye bracket, as shown in Figure 11.2.17. Find the magnitude of the resultant and the angle θ that it makes with the positive x-axis.

Solution. Note that F1 makes an angle of 30 ◦ with the positive x-axis and F2 makes an angle of 30 ◦ + 40 ◦ = 70 ◦ with the positive x-axis. Since we are given that F1 = 200 N and F2 = 300 N, (12) yields √ F1 = 200 cos 30 ◦ , sin 30 ◦ = 100 3, 100 and

F2 = 300 cos 70 ◦ , sin 70 ◦ = 300 cos 70 ◦ , 300 sin 70 ◦

Therefore, the resultant F = F1 + F2 has component form √ F = F1 + F2 = 100 3 + 300 cos 70 ◦ , 100 + 300 sin 70 ◦ √ = 100 3 + 3 cos 70 ◦ , 1 + 3 sin 70 ◦ ≈ 275.8, 381.9 The magnitude of the resultant is then

√

2 F = 100 3 + 3 cos 70 ◦ + 1 + 3 sin 70 ◦ ≈ 471 N

Let θ denote the angle F makes with the positive x-axis when the initial point of F is at the origin. Using (12) and equating the x-components of F yield √ √ 100 3 + 300 cos 70 ◦ ◦ F cos θ = 100 3 + 300 cos 70 or cos θ = F

The resultant of three or more concurrent forces can be found by working in pairs. For example, the resultant of three forces can be found by finding the resultant of any two of the forces and then finding the resultant of that resultant with the third force.

Since the terminal point of F is in the ﬁrst quadrant, we have √ ◦ −1 100 3 + 300 cos 70 θ = cos ≈ 54.2 ◦ F (Figure 11.2.18).

||F1 + F2|| ≈ 471 N

|| F2 || = 300 N

|| F1|| = 200 N

40° 30°

|| F1|| = 200 N

54.2° Figure 11.2.17

Figure 11.2.18

782

Chapter 11 / Three-Dimensional Space; Vectors

✔QUICK CHECK EXERCISES 11.2

(See page 785 for answers.)

1. If v = 3, −1, 7 and w = 4, 10, −5 , then (a) v = (b) v + w = (c) v − w = (d) 2v =

2. The unit vector in the direction of v = 3, −1, 7 is . 3. The unit vector in 2-space that makes an angle of π/3 with the positive x-axis is .

EXERCISE SET 11.2

Graphing Utility

1–4 Sketch the vectors with their initial points at the origin. ■

1. (a) 2, 5 (d) −5i + 3 j

(b) −5, −4 (e) 3i − 2 j

2. (a) −3, 7 (d) 4i + 2 j

3. (a) 1, −2, 2 (c) −i + 2 j + 3k

(b) 6, −2 (e) −2i − j

(b) 3, 4, 2 (d) i − j + 2k

5–6 Find the components of the vector, and sketch an equivalent vector with its initial point at the origin. ■ y

(b) (1, 5)

10. (a) Find the terminal point of v = 7, 6 if the initial point is (2, −1). (b) Find the terminal point of v = i + 2 j − 3k if the initial point is (−2, 1, 4). 11–12 Perform the stated operations on the given vectors u, v, and w. ■

(b) 2, 2, −1 (d) 2i + 3 j − k

4. (a) −1, 3, 2 (c) 2 j − k

5. (a)

4. Consider points A(3, 4, 0) and B(0, 0, 5). −→ (a) AB = −→ (b) If v is a vector √ in the same direction as AB and the length of v is 2, then v = .

(0, 0, 4)

11. u = 3i − k, v = i − j + 2k, (a) w − v (c) −v − 2w (e) −8(v + w) + 2u

w = 3j (b) 6u + 4w (d) 4(3u + v) (f ) 3w − (v − w)

12. u = 2, −1, 3 , v = 4, 0, −2 , w = 1, 1, 3 (a) u − w (b) 7v + 3w (c) −w + v (d) 3(u − 7v) (e) −3v − 8w (f ) 2v − (u + w) 13–14 Find the norm of v. ■

13. (a) v = 1, −1 (c) v = −1, 2, 4

(4, 1) x

6. (a) (−3, 3)

(b) (2, 3)

14. (a) v = 3, 4 (c) v = 0, −3, 0

(2, 3, 0)

(0, 4, 4)

(3, 0, 4) y x x

−−→

(b) v = −i + 7 j (d) v = −3i + 2 j + k √ √ (b) v = 2i − 7 j (d) v = i + j + k

15. Let u = i − 3 j + 2k, v = i + j, and w = 2i + 2 j − 4k. Find (a) u + v (b) u + v (c) −2u + 2 v (d) 3u − 5v + w 1 1 (e) w (f ) w . w w 16. Is it possible to have u − v = u + v if u and v are nonzero vectors? Justify your conclusion geometrically.

7–8 Find the components of the vector P1P2 . ■

7. (a) P1 (3, 5), P2 (2, 8) (b) P1 (7, −2), P2 (0, 0) (c) P1 (5, −2, 1), P2 (2, 4, 2) 8. (a) P1 (−6, −2), P2 (−4, −1) (b) P1 (0, 0, 0), P2 (−1, 6, 1) (c) P1 (4, 1, −3), P2 (9, 1, −3)

9. (a) Find the terminal point of v = 3i − 2 j if the initial point is (1, −2). (b) Find the initial point of v = −3, 1, 2 if the terminal point is (5, 0, −1).

17–20 True–False Determine whether the statement is true or false. Explain your answer. ■

17. The norm of the sum of two vectors is equal to the sum of the norms of the two vectors. 18. If two distinct vectors v and w are drawn with the same initial point, then a vector drawn between the terminal points of v and w will be either v − w or w − v.

19. There are exactly two unit vectors that are parallel to a given nonzero vector.

11.2 Vectors

20. Given a nonzero scalar c and vectors b and d, the vector equation ca + b = d has a unique solution a. 21–22 Find unit vectors that satisfy the stated conditions. ■

21. (a) Same direction as −i + 4 j. (b) Oppositely directed to 6i − 4 j + 2k. (c) Same direction as the vector from the point A(−1, 0, 2) to the point B(3, 1, 1). 22. (a) Oppositely directed to 3i − 4 j. (b) Same direction as 2i − j − 2k. (c) Same direction as the vector from the point A(−3, 2) to the point B(1, −1). 23–24 Find the vectors that satisfy the stated conditions. ■

23. (a) Oppositely directed to v = 3, −4 and half the length of v. √ (b) Length 17 and same direction as v = 7, 0, −6 . 24. (a) Same direction as v = −2i + 3 j and three times the length of v. (b) Length 2 and oppositely directed to v = −3i + 4 j + k.

25. In each part, ﬁnd the component form of the vector v in 2space that has the stated length and makes the stated angle θ with the positive x-axis. (a) v = 3; θ = π/4 (b) v = 2; θ = 90 ◦ ◦ (c) v = 5; θ = 120 (d) v = 1; θ = π

26. Find the component forms of v + w and v − w in 2-space, given that v = 1, w = 1, v makes an angle of π/6 with the positive x-axis, and w makes an angle of 3π/4 with the positive x-axis.

27–28 Find the component form of v + w, given that v and w are unit vectors. ■

27.

28.

135° v

120°

30°

29. In each part, sketch the vector u + v + w and express it in component form. y

(a)

(b)

w x

31. Let u = 1, 3 , v = 2, 1 , w = 4, −1 . Find the vector x that satisﬁes 2u − v + x = 7x + w.

32. Let u = −1, 1 , v = 0, 1 , and w = 3, 4 . Find the vector x that satisﬁes u − 2x = x − w + 3v. 33. Find u and v if u + 2v = 3i − k and 3u − v = i + j + k.

34. Find u and v if u + v = 2, −3 and 3u + 2v = −1, 2 .

35. Use vectors to ﬁnd the lengths of the diagonals of the parallelogram that has i + j and i − 2 j as adjacent sides.

36. Use vectors to find the fourth vertex of a parallelogram, three of whose vertices are (0, 0), (1, 3), and (2, 4). [Note: There is more than one answer.] 37. (a) Given that v = 3, find all values of k such that kv = 5. (b) Given that k = −2 and kv = 6, find v . 38. What do you know about k and v if kv = 0?

39. In each part, ﬁnd two unit vectors in 2-space that satisfy the stated condition. (a) Parallel to the line y = 3x + 2 (b) Parallel to the line x + y = 4 (c) Perpendicular to the line y = −5x + 1 40. In each part, ﬁnd two unit vectors in 3-space that satisfy the stated condition. (a) Perpendicular to the xy-plane (b) Perpendicular to the xz-plane (c) Perpendicular to the yz-plane F O C U S O N C O N C E P TS

41. Let r = x, y be an arbitrary vector. In each part, describe the set of all points (x, y) in 2-space that satisfy the stated condition. (a) r = 1 (b) r ≤ 1 (c) r > 1 42. Let r = x, y and r0 = x0 , y0 . In each part, describe the set of all points (x, y) in 2-space that satisfy the stated condition. (a) r − r0 = 1 (b) r − r0 ≤ 1 (c) r − r0 > 1

w w

783

43. Let r = x, y, z be an arbitrary vector. In each part, describe the set of all points (x, y, z) in 3-space that satisfy the stated condition. (a) r = 1 (b) r ≤ 1 (c) r > 1

44. Let r1 = x1 , y1 , r2 = x2 , y2 , and r = x, y . Assuming that k > r2 − r1 , describe the set of all points (x, y) for which r − r1 + r − r2 = k. 45–50 Find the magnitude of the resultant force and the angle that it makes with the positive x-axis. ■

45.

46.

100 N

30. In each part of Exercise 29, sketch the vector u − v + w and express it in component form.

30 lb 60 lb x

60°

120 N x

784

Chapter 11 / Three-Dimensional Space; Vectors

47.

48.

400 N

2 lb

120°

50°

10 ft

27°

30° 400 N

49.

300 lb 60°

30°

100 N

51–52 A particle is said to be in static equilibrium if the resultant of all forces applied to it is zero. In these exercises, ﬁnd the force F that must be applied to the point to produce static equilibrium. Describe F by specifying its magnitude and the angle that it makes with the positive x-axis. ■

51.

52.

10 lb 150 N

120 N 75°

8 lb

60°

100 N

45°

53. The accompanying figure shows a 250 lb traffic light supported by two flexible cables. The magnitudes of the forces that the cables apply to the eye ring are called the cable tensions. Find the tensions in the cables if the traffic light is in static equilibrium (defined above Exercise 51). 54. Find the tensions in the cables shown in the accompanying figure if the block is in static equilibrium (see Exercise 53).

30° 30°

200 N

75°

50 N 75 N x

30° B

20 ft

150 N 40 N 60°

45°

4 lb

50.

(b) Does increasing the sag increase or decrease the forces on the cables? (c) How much sag is required if the cables cannot tolerate forces in excess of 150 N?

60°

Figure Ex-55

100 N Figure Ex-56

57. A vector w is said to be a linear combination of the vectors v1 and v2 if w can be expressed as w = c1 v1 + c2 v2 , where c1 and c2 are scalars. (a) Find scalars c1 and c2 to express the vector 4 j as a linear combination of the vectors v1 = 2i − j and v2 = 4i + 2 j. (b) Show that the vector 3, 5 cannot be expressed as a linear combination of the vectors v1 = 1, −3 and v2 = −2, 6 . 58. A vector w is a linear combination of the vectors v1 , v2 , and v3 if w can be expressed as w = c1 v1 + c2 v2 + c3 v3 , where c1 , c2 , and c3 are scalars. (a) Find scalars c1 , c2 , and c3 to express −1, 1, 5 as a linear combination of v1 = 1, 0, 1 , v2 = 3, 2, 0 , and v3 = 0, 1, 1 . (b) Show that the vector 2i + j − k cannot be expressed as a linear combination of v1 = i − j, v2 = 3i + k, and v3 = 4i − j + k.

59. Use a theorem from plane geometry to show that if u and v are vectors in 2-space or 3-space, then u + v ≤ u + v which is called the triangle inequality for vectors. Give some examples to illustrate this inequality.

60. Prove parts (a), (c), and (e) of Theorem 11.2.6 algebraically in 2-space. 61. Prove parts (d ), (g), and (h) of Theorem 11.2.6 algebraically in 2-space. 62. Prove part ( f ) of Theorem 11.2.6 geometrically.

45°

F O C U S O N C O N C E P TS

200 N Figure Ex-53

Figure Ex-54

55. A block weighing 300 lb is suspended by cables A and B, as shown in the accompanying ﬁgure. Determine the forces that the block exerts along the cables. 56. A block weighing 100 N is suspended by cables A and B, as shown in the accompanying ﬁgure. (a) Use a graphing utility to graph the forces that the block exerts along cables A and B as functions of the “sag” d.

63. Use vectors to prove that the line segment joining the midpoints of two sides of a triangle is parallel to the third side and half as long. 64. Use vectors to prove that the midpoints of the sides of a quadrilateral are the vertices of a parallelogram. 65. Writing Do some research and then write a few paragraphs on the early history of the use of vectors in mathematics. 66. Writing Write a paragraph that discusses some of the similarities and differences between the rules of “vector arithmetic” and the rules of arithmetic of real numbers.

11.3 Dot Product; Projections

785

â&#x153;&#x201D;QUICK CHECK ANSWERS 11.2 1. (a)

â&#x2C6;&#x161;

59 (b) 7, 9, 2 (c) â&#x2C6;&#x2019;1, â&#x2C6;&#x2019;11, 12 (d) 6, â&#x2C6;&#x2019;2, 14

4. (a) â&#x2C6;&#x2019;3, â&#x2C6;&#x2019;4, 5 (b)

11.3

â&#x2020;&#x2019; 1â&#x2C6;&#x2019; AB 5

= â&#x2C6;&#x2019; 35 , â&#x2C6;&#x2019; 45 , 1

3 1 7 1 2. â&#x2C6;&#x161; v = â&#x2C6;&#x161; , â&#x2C6;&#x2019; â&#x2C6;&#x161; , â&#x2C6;&#x161; 59 59 59 59

â&#x2C6;&#x161; â&#x2C6;&#x161; 1 1 3 3 = i+ , j 2 2 2 2

DOT PRODUCT; PROJECTIONS In the last section we deďŹ ned three operations on vectorsâ&#x20AC;&#x201D;addition, subtraction, and scalar multiplication. In scalar multiplication a vector is multiplied by a scalar and the result is a vector. In this section we will deďŹ ne a new kind of multiplication in which two vectors are multiplied to produce a scalar. This multiplication operation has many uses, some of which we will also discuss in this section. DEFINITION OF THE DOT PRODUCT

11.3.1 definition If u = u1 , u2 and v = v1 , v2 are vectors in 2-space, then the dot product of u and v is written as u â´˘ v and is deďŹ ned as In words, the dot product of two vectors is formed by multiplying their corresponding components and adding the resulting products. Note that the dot product of two vectors is a scalar.

u â´˘ v = u1 v1 + u2 v2 Similarly, if u = u1 , u2 , u3 and v = v1 , v2 , v3 are vectors in 3-space, then their dot product is deďŹ ned as u â´˘ v = u1 v1 + u2 v2 + u3 v3 Example 1

3, 5 â´˘ â&#x2C6;&#x2019;1, 2 = 3(â&#x2C6;&#x2019;1) + 5(2) = 7

2, 3 â´˘ â&#x2C6;&#x2019;3, 2 = 2(â&#x2C6;&#x2019;3) + 3(2) = 0

1, â&#x2C6;&#x2019;3, 4 â´˘ 1, 5, 2 = 1(1) + (â&#x2C6;&#x2019;3)(5) + 4(2) = â&#x2C6;&#x2019;6

T E C H N O LO GY M A ST E R Y

Here are the same computations expressed another way: (3i + 5j) â´˘ (â&#x2C6;&#x2019;i + 2 j) = 3(â&#x2C6;&#x2019;1) + 5(2) = 7

Many calculating utilities have a built-in dot product operation. If your calculating utility has this capability, use it to check the computations in Example 1.

(2i + 3j) â´˘ (â&#x2C6;&#x2019;3i + 2 j) = 2(â&#x2C6;&#x2019;3) + 3(2) = 0

(i â&#x2C6;&#x2019; 3j + 4k) â´˘ (i + 5 j + 2k) = 1(1) + (â&#x2C6;&#x2019;3)(5) + 4(2) = â&#x2C6;&#x2019;6 ALGEBRAIC PROPERTIES OF THE DOT PRODUCT The following theorem provides some of the basic algebraic properties of the dot product.

11.3.2 Note the difference between the two zeros that appear in part (e) of Theorem 11.3.2â&#x20AC;&#x201D;the zero on the left side is the zero vector (boldface), whereas the zero on the right side is the zero scalar (lightface).

theorem

If u, v, and w are vectors in 2- or 3-space and k is a scalar, then:

(a) u â´˘ v = v â´˘ u (b) u â´˘ (v + w) = u â´˘ v + u â´˘ w (c) k(u â´˘ v) = (ku) â´˘ v = u â´˘ (kv)

(d ) v â´˘ v = v 2 (e) 0 â´˘ v = 0

We will prove parts (c) and (d) for vectors in 3-space and leave some of the others as exercises.

Chapter 11 / Three-Dimensional Space; Vectors

786

proof (c) Let u = u1 , u2 , u3 and v = v1 , v2 , v3 . Then k(u ⴢ v) = k(u1 v1 + u2 v2 + u3 v3 ) = (ku1 )v1 + (ku2 )v2 + (ku3 )v3 = (ku) ⴢ v Similarly, k(u ⴢ v) = u ⴢ (kv). proof (d) v ⴢ v = v1 v1 + v2 v2 + v3 v3 = v12 + v22 + v32 = v 2 . ■ The following alternative form of the formula in part (d) of Theorem 11.3.2 provides a useful way of expressing the norm of a vector in terms of a dot product: v =

√ vⴢv

(1)

ANGLE BETWEEN VECTORS

u u

u v

Suppose that u and v are nonzero vectors in 2-space or 3-space that are positioned so their initial points coincide. We deﬁne the angle between u and v to be the angle θ determined by the vectors that satisﬁes the condition 0 ≤ θ ≤ π (Figure 11.3.1). In 2-space, θ is the smallest counterclockwise angle through which one of the vectors can be rotated until it aligns with the other. The next theorem provides a way of calculating the angle between two vectors from their components.

u u

v v u

u is the angle between u and v.

11.3.3 theorem If u and v are nonzero vectors in 2-space or 3-space, and if θ is the angle between them, then uⴢv cos θ = (2) u v proof Suppose that the vectors u, v, and v − u are positioned to form three sides of a triangle, as shown in Figure 11.3.2. It follows from the law of cosines that v − u 2 = u 2 + v 2 − 2 u v cos θ

Figure 11.3.1

v−u

Using the properties of the dot product in Theorem 11.3.2, we can rewrite the left side of this equation as v − u 2 = (v − u) ⴢ (v − u) = (v − u) ⴢ v − (v − u) ⴢ u =vⴢv−uⴢv−vⴢu+uⴢu = v 2 − 2u ⴢ v + u 2

u u Figure 11.3.2

(3)

Substituting this back into (3) yields v 2 − 2u ⴢ v + u 2 = u 2 + v 2 − 2 u v cos θ which we can simplify and rewrite as u ⴢ v = u v cos θ Finally, dividing both sides of this equation by u v yields (2). ■ Example 2

Find the angle between the vector u = i − 2 j + 2k and

(a) v = −3i + 6j + 2k

(b) w = 2i + 7 j + 6k

Solution (a). cos θ =

uⴢv −11 11 = =− u v (3)(7) 21

11.3 Dot Product; Projections

787

Thus,

Solution (b). u

≈ 2.12 radians ≈ 121.6 ◦ θ = cos−1 − 11 21 cos θ =

0 uⴢw = =0 u w u w

Thus, θ = π/2, which means that the vectors are perpendicular.

u.v>0

Solution (c).

cos θ =

uⴢz −27 = = −1 u z (3)(9)

Thus, θ = π, which means that the vectors are oppositely directed. (In retrospect, we could have seen this without computing θ, since z = −3u.)

u u u.v<0

INTERPRETING THE SIGN OF THE DOT PRODUCT

It will often be convenient to express Formula (2) as

u ⴢ v = u v cos θ

(4)

which expresses the dot product of u and v in terms of the lengths of these vectors and the angle between them. Since u and v are assumed to be nonzero vectors, this version of the formula makes it clear that the sign of u ⴢ v is the same as the sign of cos θ. Thus, we can tell from the dot product whether the angle between two vectors is acute or obtuse or whether the vectors are perpendicular (Figure 11.3.3).

u u.v= 0 Figure 11.3.3

REMARK

The terms “perpendicular,” “orthogonal,” and “normal” are all commonly used to describe geometric objects that meet at right angles. For consistency, we will say that two vectors are orthogonal, a vector is normal to a plane, and two planes are perpendicular. Moreover, although the zero vector does not make a well-defined angle with other vectors, we will consider 0 to be orthogonal to all vectors. This convention allows us to say that u and v are orthogonal vectors if and only if u ⴢ v = 0, and makes Formula (4) valid if u or v (or both) is zero.

DIRECTION ANGLES

In an xy-coordinate system, the direction of a nonzero vector v is completely determined by the angles α and β between v and the unit vectors i and j (Figure 11.3.4), and in an xyz-coordinate system the direction is completely determined by the angles α, β, and γ between v and the unit vectors i, j, and k (Figure 11.3.5). In both 2-space and 3-space the angles between a nonzero vector v and the vectors i, j, and k are called the direction angles of v, and the cosines of those angles are called the direction cosines of v. Formulas for the direction cosines of a vector can be obtained from Formula (2). For example, if v = v1 i + v2 j + v3 k, then

b a

i Figure 11.3.4 z

cos α =

v1 vⴢi = , v i v

cos β =

vⴢj v2 = , v j v

cos γ =

vⴢk v3 = v k v

Thus, we have the following theorem.

g b a i x

Figure 11.3.5

11.3.4

theorem The direction cosines of a nonzero vector v = v1 i + v2 j + v3 k are cos α =

v1 , v

cos β =

v2 , v

cos γ =

v3 v

788

Chapter 11 / Three-Dimensional Space; Vectors

The direction cosines of a vector v = v1 i + v2 j + v3 k can be computed by normalizing v and reading off the components of v/ v , since v v1 v2 v3 = i+ j+ k = (cos α)i + (cos β)j + (cos γ )k v v v v

We leave it as an exercise for you to show that the direction cosines of a vector satisfy the equation cos2 α + cos2 β + cos2 γ = 1 (5) Example 3 Find the direction cosines of the vector v = 2i − 4 j + 4k, and approximate the direction angles to the nearest degree.

Solution. √ First we will normalize the vector v and then read off the components. We

have v =

4 + 16 + 16 = 6, so that v/ v = 13 i − 23 j + 23 k. Thus, cos α = 31 ,

cos β = − 23 ,

cos γ =

With the help of a calculating utility we obtain α = cos−1 31 ≈ 71 ◦ , β = cos−1 − 23 ≈ 132 ◦ , Example 4

2 3

γ = cos−1

2 3

≈ 48 ◦

Find the angle between a diagonal of a cube and one of its edges.

Solution. Assume that the cube has side a, and introduce a coordinate system as shown

in Figure 11.3.6. In this coordinate system the vector

(0, 0, a)

d = ai + a j + ak

(a, a, a) d y

a i

is a diagonal of the cube and the unit vectors i, j, and k run along the edges. By symmetry, the diagonal makes the same angle with each edge, so it is sufﬁcient to ﬁnd the angle between d and i (the direction angle α). Thus,

(0, a, 0)

(a, 0, 0)

cos α = and hence

dⴢi 1 a a =√ = =√ 2 d i d 3 3a

1 α = cos−1 √ ≈ 0.955 radian ≈ 54.7 ◦ 3

Figure 11.3.6

DECOMPOSING VECTORS INTO ORTHOGONAL COMPONENTS

In many applications it is desirable to “decompose” a vector into a sum of two orthogonal vectors with convenient speciﬁed directions. For example, Figure 11.3.7 shows a block on an inclined plane. The downward force F that gravity exerts on the block can be decomposed into the sum F = F1 + F2

F2 F The force of gravity pulls the block against the ramp and down the ramp.

Figure 11.3.7

where the force F1 is parallel to the ramp and the force F2 is perpendicular to the ramp. The forces F1 and F2 are useful because F1 is the force that pulls the block along the ramp, and F2 is the force that the block exerts against the ramp. Thus, our next objective is to develop a computational procedure for decomposing a vector into a sum of orthogonal vectors. For this purpose, suppose that e1 and e2 are two orthogonal unit vectors in 2-space, and suppose that we want to express a given vector v as a sum v = w1 + w2

11.3 Dot Product; Projections

so that w1 is a scalar multiple of e1 and w2 is a scalar multiple of e2 (Figure 11.3.8a). That is, we want to ﬁnd scalars k1 and k2 such that

w2 e2

v = k1 e1 + k2 e2 e1

v ⴢ e1 = (k1 e1 + k2 e2 ) ⴢ e1

= k1 (e1 ⴢ e1 ) + k2 (e2 ⴢ e1 )

||v|| cos u ||v||

(6)

We can ﬁnd k1 by taking the dot product of v with e1 . This yields

(a) (||v|| sin u)e2

789

= k1 e1 2 + 0 = k1

||v|| sin u

u e1 (||v|| cos u)e1

(b)

Similarly, v ⴢ e2 = (k1 e1 + k2 e2 ) ⴢ e2 = k1 (e1 ⴢ e2 ) + k2 (e2 ⴢ e2 ) = 0 + k2 e2 2 = k2 Substituting these expressions for k1 and k2 in (6) yields

Figure 11.3.8

v = (v ⴢ e1 )e1 + (v ⴢ e2 )e2

(7)

In this formula we call (v ⴢ e1 )e1 and (v ⴢ e2 )e2 the vector components of v along e1 and e2 , respectively; and we call v ⴢ e1 and v ⴢ e2 the scalar components of v along e1 and e2 , respectively. If θ denotes the angle between v and e1 , and the angle between v and e2 is π/2 or less, then the scalar components of v can be written in trigonometric form as v ⴢ e1 = v cos θ

and

v ⴢ e2 = v sin θ

(8)

(Figure 11.3.8b). Moreover, the vector components of v can be expressed as Note that the vector components of v along e1 and e2 are vectors, whereas the scalar components of v along e1 and e2 are numbers.

(v ⴢ e1 )e1 = ( v cos θ)e1

and

(v ⴢ e2 )e2 = ( v sin θ )e2

(9)

and the decomposition (6) can be expressed as v = ( v cos θ )e1 + ( v sin θ)e2

(10)

provided the angle between v and e2 is at most π/2. Example 5

Let

v = 2, 3 ,

1 1 e1 = √ , √ , 2 2

1 1 and e2 = − √ , √ 2 2

Find the scalar components of v along e1 and e2 and the vector components of v along e1 and e2 .

Solution. The scalar components of v along e1 and e2 are

1 1 5 v ⴢ e1 = 2 √ +3 √ =√ 2 2 2 1 1 1 v ⴢ e2 = 2 − √ +3 √ =√ 2 2 2 so the vector components are Notice that in Example 5

(v ⴢ e1 )e1 + (v ⴢ e2 )e2 = 2, 3 = v as guaranteed by (7).

5 1 1 5 5 (v ⴢ e1 )e1 = √ √ , √ = , 2 2 2 2 2 1 1 1 1 1 (v ⴢ e2 )e2 = √ − √ , √ = − , 2 2 2 2 2

790

Chapter 11 / Three-Dimensional Space; Vectors

Example 6 A rope is attached to a 100 lb block on a ramp that is inclined at an angle of 30 ◦ with the ground (Figure 11.3.9a). How much force does the block exert against the ramp, and how much force must be applied to the rope in a direction parallel to the ramp to prevent the block from sliding down the ramp? (Assume that the ramp is smooth, that is, exerts no frictional forces.)

30°

Solution. Let F denote the downward force of gravity on the block (so F = 100 lb),

(a)

F1 60° 30° F2 F

(b) Figure 11.3.9

and let F1 and F2 be the vector components of F parallel and perpendicular to the ramp (as shown in Figure 11.3.9b). The lengths of F1 and F2 are 1 = 50 lb F1 = F cos 60 ◦ = 100 2 √ 3 ◦ F2 = F sin 60 = 100 ≈ 86.6 lb 2

Thus, the block exerts a force of approximately 86.6 lb against the ramp, and it requires a force of 50 lb to prevent the block from sliding down the ramp. ORTHOGONAL PROJECTIONS The vector components of v along e1 and e2 in (7) are also called the orthogonal projections of v on e1 and e2 and are commonly denoted by

proje1 v = (v ⴢ e1 )e1

and proje2 v = (v ⴢ e2 )e2

In general, if e is a unit vector, then we deﬁne the orthogonal projection of v on e to be (11)

proje v = (v ⴢ e)e

The orthogonal projection of v on an arbitrary nonzero vector b can be obtained by normalizing b and then applying Formula (11); that is, b b projb v = v ⴢ b b which can be rewritten as

projb v =

vⴢb b b 2

(12)

Geometrically, if b and v have a common initial point, then projb v is the vector that is determined when a perpendicular is dropped from the terminal point of v to the line through b (illustrated in Figure 11.3.10 in two cases).

Figure 11.3.10

v − proj bv

proj bv

Acute angle between v and b

v − proj bv b

proj bv Obtuse angle between v and b

Moreover, it is evident from Figure 11.3.10 that if we subtract projb v from v, then the resulting vector v − projb v will be orthogonal to b; we call this the vector component of v orthogonal to b.

11.3 Dot Product; Projections

Stated informally, the orthogonal projection projb v is the “shadow” that v casts on the line through b.

Example 7 Find the orthogonal projection of v = i + j + k on b = 2i + 2 j, and then ﬁnd the vector component of v orthogonal to b.

Solution. We have v ⴢ b = (i + j + k) ⴢ (2i + 2 j) = 2 + 2 + 0 = 4

b 2 = 22 + 22 = 8

v = i+j+k k

i+j

791

b = 2i + 2j

Thus, the orthogonal projection of v on b is 4 vⴢb projb v = b = (2i + 2 j) = i + j b 2 8 and the vector component of v orthogonal to b is v − projb v = (i + j + k) − (i + j) = k

These results are consistent with Figure 11.3.11.

Figure 11.3.11

WORK

In Section 6.6 we discussed the work done by a constant force acting on an object that moves along a line. We deﬁned the work W done on the object by a constant force of magnitude F acting in the direction of motion over a distance d to be (13)

W = Fd = force × distance

If we let F denote a force vector of magnitude F = F acting in the direction of motion, then we can write (13) as W = F d

Note that in Formula (14) the quantity F cos θ is the scalar component of force along the displacement vector. Thus, in the case where cos θ > 0, a force of magnitude F acting at an angle θ does the same work as a force of magnitude F cos θ acting in the direction of motion.

Furthermore, if we assume that the object moves along a line from point P to point Q, then −→ d = PQ , so that the work can be expressed entirely in vector form as −→ W = F PQ −→ (Figure 11.3.12a). The vector PQ is called the displacement vector for the object. In the case where a constant force F is not in the direction of motion, but rather makes an angle θ with the displacement vector, then we deﬁne the work W done by F to be −→ −→ W = ( F cos θ) PQ = F ⴢ PQ

(14)

(Figure 11.3.12b). || F ||

|| F ||

u || F || cos u

|| PQ || Work = || F || || PQ ||

(a)

|| PQ || Work = (|| F || cos u) || PQ ||

(b)

Figure 11.3.12

−→ It follows from Formula (14) that the work W can be expressed as W = ( F cos θ ) PQ −→ or as W = F ⴢ PQ. Although these two expressions are mathematically equivalent, in practice it may be more convenient to use one of them instead of the other. Example 8 (a) A wagon is pulled horizontally by exerting a constant force of 10 lb on the handle at an angle of 60 ◦ with the horizontal. How much work is done in moving the wagon 50 ft?

792

Chapter 11 / Three-Dimensional Space; Vectors

(b) A force of F = 3i − j + 2k lb is applied to a point that moves on a line from P (−1, 1, 2) to Q(3, 0, −2). If distance is measured in feet, how much work is done? −→

Solution (a). With F = 10, θ = 60 ◦ , and PQ = 50, it follows that the work done is

Solution (b). done is

1 −→ W = ( F cos θ) PQ = 10 · · 50 = 250 ft·lb 2 −→ Since PQ = (3 − (−1))i + (0 − 1)j + (−2 − 2)k = 4i − j − 4k, the work

−→ W = F ⴢ PQ = (3i − j + 2k) ⴢ (4i − j − 4k) = 12 + 1 − 8 = 5 ft·lb

✔QUICK CHECK EXERCISES 11.3

(See page 794 for answers.)

1. 3, 1, −2 ⴢ 6, 0, 5 =

2. Suppose that u, v, and w are vectors in 3-space such that u = 5, u ⴢ v = 7, and u ⴢ w = −3. (a) u ⴢ u = (b) v ⴢ u = (c) u ⴢ (v − w) = (d) u ⴢ (2w) =

4. The direction cosines of 2, −1, 3 are cos α = cos β = , and cos γ = .

5. The orthogonal projection of v = 10i on b = −3i + j is .

3. For the vectors u and v in the preceding exercise, if the angle between u and v is π/3, then v = .

EXERCISE SET 11.3

CAS

1. In each part, ﬁnd the dot product of the vectors and the cosine of the angle between them. (a) u = i + 2 j, v = 6i − 8 j (b) u = −7, −3 , v = 0, 1 (c) u = i − 3 j + 7k, v = 8i − 2 j − 2k (d) u = −3, 1, 2 , v = 4, 2, −5

2. In each part use the given information to ﬁnd u ⴢ v. (a) u = 1, v = 2, the angle between u and v is π/6. (b) u = 2, v = 3, the angle between u and v is 135 ◦ .

3. In each part, determine whether u and v make an acute angle, an obtuse angle, or are orthogonal. (a) u = 7i + 3 j + 5k, v = −8i + 4 j + 2k (b) u = 6i + j + 3k, v = 4i − 6k (c) u = 1, 1, 1 , v = −1, 0, 0 (d) u = 4, 1, 6 , v = −3, 0, 2

F O C U S O N C O N C E P TS

4. Does the triangle in 3-space with vertices (−1, 2, 3), (2, −2, 0), and (3, 1, −4) have an obtuse angle? Justify your answer. 5. The accompanying figure shows eight vectors that are equally spaced around a circle of radius 1. Find the dot product of v0 with each of the other seven vectors. 6. The accompanying figure shows six vectors that are equally spaced around a circle of radius 5. Find the dot product of v0 with each of the other five vectors.

v2 v3

v0 v5

v7 v6

Figure Ex-5

Figure Ex-6

7. (a) Use vectors to show that A(2, −1, 1), B(3, 2, −1), and C(7, 0, −2) are vertices of a right triangle. At which vertex is the right angle? (b) Use vectors to ﬁnd the interior angles of the triangle with vertices (−1, 0), (2, −1), and (1, 4). Express your answers to the nearest degree. 8. (a) Show that if v = ai + b j is a vector in 2-space, then the vectors v1 = −bi + a j and v2 = bi − a j are both orthogonal to v. (b) Use the result in part (a) to ﬁnd two unit vectors that are orthogonal to the vector v = 3i − 2 j. Sketch the vectors v, v1 , and v2 . 9. Explain why each of the following expressions makes no sense. (a) u ⴢ (v ⴢ w) (b) (u ⴢ v) + w (c) u ⴢ v (d) k ⴢ (u + v)

11.3 Dot Product; Projections

10. Explain why each of the following expressions makes sense. (a) (u ⴢ v)w (b) (u ⴢ v)(v ⴢ w) (c) u ⴢ v + k (d) (ku) ⴢ v

21. Use the result in Exercise 18 to ﬁnd the direction angles of the vector shown in the accompanying ﬁgure to the nearest degree. z

11. Verify parts (b) and (c) of Theorem 11.3.2 for the vectors u = 6i − j + 2k, v = 2i + 7 j + 4k, w = i + j − 3k and k = −5.

12. Let u = 1, 2 , v = 4, −2 , and w = 6, 0 . Find (a) u ⴢ (7v + w) (b) (u ⴢ w)w (c) u (v ⴢ w) (d) ( u v) ⴢ w.

15–16 Find the direction cosines of v and conﬁrm that they satisfy Equation (5). Then use the direction cosines to approximate the direction angles to the nearest degree. ■

(b) v = 2i − 2 j + k

(b) v = 3i − 4k

F O C U S O N C O N C E P TS

17. Show that the direction cosines of a vector satisfy cos2 α + cos2 β + cos2 γ = 1

18. Let θ and λ be the angles shown in the accompanying ﬁgure. Show that the direction cosines of v can be expressed as cos α = cos λ cos θ cos β = cos λ sin θ cos γ = sin λ [Hint: Express v in component form and normalize.]

19. The accompanying ﬁgure shows a cube. (a) Find the angle between the vectors d and u to the nearest degree. (b) Make a conjecture about the angle between the vectors d and v, and conﬁrm your conjecture by computing the angle.

u x

Figure Ex-18

23. Find, to the nearest degree, the angles that a diagonal of a box with dimensions 10 cm by 15 cm by 25 cm makes with the edges of the box. 24. In each part, ﬁnd the vector component of v along b and the vector component of v orthogonal to b. Then sketch the vectors v, projb v, and v − projb v. (a) v = 2i − j, b = 3i + 4 j (b) v = 4, 5 , b = 1, −2 (c) v = −3i − 2 j, b = 2i + j 25. In each part, ﬁnd the vector component of v along b and the vector component of v orthogonal to b. (a) v = 2i − j + 3k, b = i + 2 j + 2k (b) v = 4, −1, 7 , b = 2, 3, −6 26–27 Express the vector v as the sum of a vector parallel to b and a vector orthogonal to b. ■

26. (a) v = 2i − 4 j, b = i + j (b) v = 3i + j − 2k, b = 2i − k (c) v = 4i − 2j + 6k, b = −2i + j − 3k 27. (a) v = −3, 5 , b = 1, 1 (b) v = −2, 1, 6 , b = 0, −2, 1 (c) v = 1, 4, 1 , b = 3, −2, 5 28–31 True–False Determine whether the statement is true or

29. If v and w are nonzero orthogonal vectors, then v + w = 0.

22. Find, to the nearest degree, the acute angle formed by two diagonals of a cube.

28. If a ⴢ b = a ⴢ c and a = 0, then b = c.

Figure Ex-21

false. Explain your answer. ■

14. Find two unit vectors in 2-space that make an angle of 45 ◦ with 4i + 3 j.

15. (a) v = i + j − k

30° 60°

13. Find r so that the vector from the point A(1, −1, 3) to the point B(3, 0, 5) is orthogonal to the vector from A to the point P (r, r, r).

16. (a) v = 3i − 2 j − 6k

793

Figure Ex-19

20. Show that two nonzero vectors v1 and v2 are orthogonal if and only if their direction cosines satisfy cos α1 cos α2 + cos β1 cos β2 + cos γ1 cos γ2 = 0

30. If u is a unit vector that is parallel to a nonzero vector v, then u ⴢ v = ± v . 31. If v and b are nonzero vectors, then the orthogonal projection of v on b is a vector that is parallel to b. 32. If L is a line in 2-space or 3-space that passes through the points A and B, then the distance from a point P to the line −→ L is equal to the length of the component of the vector AP − → that is orthogonal to the vector AB (see the accompanying

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Chapter 11 / Three-Dimensional Space; Vectors

43. Prove that

ﬁgure). Use this result to ﬁnd the distance from the point P (1, 0) to the line through A(2, −3) and B(5, 1).

u + v 2 + u − v 2 = 2 u 2 + 2 v 2

and interpret the result geometrically by translating it into a theorem about parallelograms.

L A

44. Prove: u ⴢ v = 41 u + v 2 − 41 u − v 2 .

Figure Ex-32

33. Use the method of Exercise 32 to ﬁnd the distance from the point P (−3, 1, 2) to the line through A(1, 1, 0) and B(−2, 3, −4).

45. Show that if v1 , v2 , and v3 are mutually orthogonal nonzero vectors in 3-space, and if a vector v in 3-space is expressed as

34. As shown in the accompanying ﬁgure, a child with mass 34 kg is seated on a smooth (frictionless) playground slide that is inclined at an angle of 27 ◦ with the horizontal. Estimate the force that the child exerts on the slide, and estimate how much force must be applied in the direction of P to prevent the child from sliding down the slide. Take the acceleration due to gravity to be 9.8 m/s2 .

v = c1 v1 + c2 v2 + c3 v3 then the scalars c1 , c2 , and c3 are given by the formulas ci = (v ⴢ vi )/ vi 2 , 46. Show that the three vectors v1 = 3i − j + 2k,

35. For the child in Exercise 34, estimate how much force must be applied in the direction of Q (shown in the accompanying ﬁgure) to prevent the child from sliding down the slide?

27° Figure Ex-34

v3 = i − 5 j − 4k

c1 v1 + c2 v2 + c3 v3 = i − j + k C

47. For each x in (−⬁, +⬁), let u(x) be the vector from the origin to the point P (x, y) on the curve y = x 2 + 1, and v(x) the vector from the origin to the point Q(x, y) on the line y = −x − 1. (a) Use a CAS to ﬁnd, to the nearest degree, the minimum angle between u(x) and v(x) for x in (−⬁, +⬁). (b) Determine whether there are any real values of x for which u(x) and v(x) are orthogonal.

48. Let u be a unit vector in the xy-plane of an xyz-coordinate system, and let v be a unit vector in the yz-plane. Let θ1 be the angle between u and i, let θ2 be the angle between v and k, and let θ be the angle between u and v. (a) Show that cos θ = ± sin θ1 sin θ2 . (b) Find θ if θ is acute and θ1 = θ2 = 45 ◦ . (c) Use a CAS to ﬁnd, to the nearest degree, the maximum and minimum values of θ if θ is acute and θ2 = 2θ1 .

27° Figure Ex-35

36. Suppose that the slide in Exercise 34 is 4 m long. Estimate the work done by gravity if the child slides from the top of the slide to the bottom. 37. A box is dragged along the ﬂoor by a rope that applies a force of 50 lb at an angle of 60 ◦ with the ﬂoor. How much work is done in moving the box 15 ft? 38. Find the work done by a force F = −3 j pounds applied to a point that moves on a line from (1, 3) to (4, 7). Assume that distance is measured in feet. 39. A force of F = 4i − 6 j + k newtons is applied to a point that moves a distance of 15 meters in the direction of the vector i + j + k. How much work is done? 40. A boat travels 100 meters due north while the wind exerts a force of 500 newtons toward the northeast. How much work does the wind do? F O C U S O N C O N C E P TS

41. Let u and v be adjacent sides of a parallelogram. Use vectors to prove that the diagonals of the parallelogram are perpendicular if the sides are equal in length. 42. Let u and v be adjacent sides of a parallelogram. Use vectors to prove that the parallelogram is a rectangle if the diagonals are equal in length.

49. Prove parts (b) and (e) of Theorem 11.3.2 for vectors in 3-space.

50. Writing Discuss some of the similarities and differences between the multiplication properties of real numbers and those of the dot product of vectors. 51. Writing Discuss the merits of the following claim: “Suppose an algebraic identity involves only the addition, subtraction, and multiplication of real numbers. If the numbers are replaced by vectors, and the multiplication is replaced by the dot product, then an identity involving vectors will result.”

✔QUICK CHECK ANSWERS 11.3 1. 8

v2 = i + j − k,

are mutually orthogonal, and then use the result of Exercise 45 to ﬁnd scalars c1 , c2 , and c3 so that

i = 1, 2, 3

2. (a) 25 (b) 7 (c) 10 (d) −6

14 5

2 1 3 4. √ ; − √ ; √ 14 14 14

5. 9i − 3j

11.4 Cross Product

11.4

795

CROSS PRODUCT In many applications of vectors in mathematics, physics, and engineering, there is a need to ﬁnd a vector that is orthogonal to two given vectors. In this section we will discuss a new type of vector multiplication that can be used for this purpose. DETERMINANTS

Some of the concepts that we will develop in this section require basic ideas about determinants, which are functions that assign numerical values to square arrays of numbers. For example, if a1 , a2 , b1 , and b2 are real numbers, then we deﬁne a 2 × 2 determinant by

a1 a 2 b1 b 2 = a1b 2 − a 2b1

(1)

The purpose of the arrows is to help you remember the formula—the determinant is the product of the entries on the rightward arrow minus the product of the entries on the leftward arrow. For example,

3 −2 = (3)(5) − (−2)(4) = 15 + 8 = 23 4 5 A 3 × 3 determinant is deﬁned in terms of 2 × 2 determinants by a1 a2 a3 b1 b2 b3 = a1 b2 b3 − a2 b1 b3 + a3 b1 c 2 c 3 c1 c 3 c1 c1 c2 c3

b2 c2

(2)

The right side of this formula is easily remembered by noting that a1 , a2 , and a3 are the entries in the ﬁrst “row” of the left side, and the 2 × 2 determinants on the right side arise by deleting the ﬁrst row and an appropriate column from the left side. The pattern is as follows:

a1 a 2 a 3 a1 a 2 a 3 a1 a 2 a 3 a1 a 2 a 3 b1 b 2 b 3 = a1 b1 b 2 b 3 − a 2 b1 b 2 b 3 + a 3 b1 b 2 b 3 c1 c 2 c 3 c1 c 2 c 3 c1 c 2 c 3 c1 c 2 c 3 For example, 3 1 0

−2 4 3

−5 4 −4 = 3 3 2

1 −4 − (−2) 2 0

1 −4 + (−5) 2 0

= 3(20) + 2(2) − 5(3) = 49

4 3

There are also deﬁnitions of 4 × 4 determinants, 5 × 5 determinants, and higher, but we will not need them in this text. Properties of determinants are studied in a branch of mathematics called linear algebra, but we will only need the two properties stated in the following theorem.

11.4.1

theorem

(a) If two rows in the array of a determinant are the same, then the value of the determinant is 0. (b) Interchanging two rows in the array of a determinant multiplies its value by −1.

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Chapter 11 / Three-Dimensional Space; Vectors

We will give the proofs of parts (a) and (b) for 2 × 2 determinants and leave the proofs for 3 × 3 determinants as exercises. proof (a)

proof (b)

b1 a1

a 1 a1

a2 = a1 a2 − a2 a1 = 0 a2

a b2 = b1 a2 − b2 a1 = −(a1 b2 − a2 b1 ) = − 1 b1 a2

a2 ■ b2

CROSS PRODUCT We now turn to the main concept in this section.

11.4.2 definition If u = u1 , u2 , u3 and v = v1 , v2 , v3 are vectors in 3-space, then the cross product u × v is the vector deﬁned by u 1 u 2 u 1 u 3 u 2 u 3 k j+ i− u×v= (3) v1 v2 v1 v3 v2 v3 or, equivalently,

u × v = (u2 v3 − u3 v2 )i − (u1 v3 − u3 v1 )j + (u1 v2 − u2 v1 )k

(4)

Observe that the right side of Formula (3) has the same form as the right side of Formula (2), the difference being notation and the order of the factors in the three terms. Thus, we can rewrite (3) as i j k u × v = u1 u2 u3 (5) v v v 1 2 3 However, this is just a mnemonic device and not a true determinant since the entries in a determinant are numbers, not vectors.

Example 1

Let u = 1, 2, −2 and v = 3, 0, 1 . Find (a) u × v

(b) v × u

Solution (a). i u × v = 1 3 2 = 0

k −2 1 1 −2 i − 3 1 j 2 0

1 −2 j + 3 1

2 k = 2i − 7j − 6k 0

Solution (b). We could use the method of part (a), but it is really not necessary to perform any computations. We need only observe that reversing u and v interchanges the second and

11.4 Cross Product

797

third rows in (5), which in turn interchanges the rows in the arrays for the 2 × 2 determinants in (3). But interchanging the rows in the array of a 2 × 2 determinant reverses its sign, so the net effect of reversing the factors in a cross product is to reverse the signs of the components. Thus, by inspection v × u = −(u × v) = −2i + 7j + 6k

Example 2

Show that u × u = 0 for any vector u in 3-space.

Solution. We could let u = u1 i + u2 j + u3 k and apply the method in part (a) of Example 1 to show that i j k u × u = u1 u2 u3 = 0 u u u 1 2 3

However, the actual computations are unnecessary. We need only observe that if the two factors in a cross product are the same, then each 2 × 2 determinant in (3) is zero because its array has identical rows. Thus, u × u = 0 by inspection. ALGEBRAIC PROPERTIES OF THE CROSS PRODUCT

Our next goal is to establish some of the basic algebraic properties of the cross product. As you read the discussion, keep in mind the essential differences between the cross product and the dot product:

• The cross product is deﬁned only for vectors in 3-space, whereas the dot product is •

deﬁned for vectors in 2-space and 3-space. The cross product of two vectors is a vector, whereas the dot product of two vectors is a scalar.

The main algebraic properties of the cross product are listed in the next theorem.

11.4.3 theorem Whereas the order of the factors does not matter for ordinary multiplication, or for dot products, it does matter for cross products. Specifically, part (a ) of Theorem 11.4.3 shows that reversing the order of the factors in a cross product reverses the direction of the resulting vector.

If u, v, and w are any vectors in 3-space and k is any scalar, then:

(a) u × v = −(v × u) (b) u × (v + w) = (u × v) + (u × w) (c) (u + v) × w = (u × w) + (v × w) (d ) k(u × v) = (ku) × v = u × (kv) (e) u × 0 = 0 × u = 0 (f) u ×u = 0 Parts (a) and ( f ) were addressed in Examples 1 and 2. The other proofs are left as exercises. The following cross products occur so frequently that it is helpful to be familiar with them: i×j=k j×k=i k×i=j (6) j × i = −k k × j = −i i × k = −j

Chapter 11 / Three-Dimensional Space; Vectors

798

These results are easy to obtain; for example, i j k 1 0 0 i − i × j = 1 0 0 = 0 1 0 0 1 0

Figure 11.4.1 WARNING

1 0 j + 0 0

0 k=k 1

However, rather than computing these cross products each time you need them, you can use the diagram in Figure 11.4.1. In this diagram, the cross product of two consecutive vectors in the counterclockwise direction is the next vector around, and the cross product of two consecutive vectors in the clockwise direction is the negative of the next vector around.

We can write a product of three real numbers as uvw since the associative law u(vw) = (uv)w ensures that the same value for the product results no matter how the factors are grouped. However, the associative law does not hold for cross products. For example,

i × ( j × j) = i × 0 = 0 and (i × j) × j = k × j = −i so that i × ( j × j) = (i × j) × j. Thus, we cannot write a cross product with three vectors as u × v × w, since this expression is ambiguous without parentheses.

GEOMETRIC PROPERTIES OF THE CROSS PRODUCT The following theorem shows that the cross product of two vectors is orthogonal to both factors. This property of the cross product will be used many times in the following sections.

11.4.4

theorem If u and v are vectors in 3-space, then:

(a) u ⴢ (u × v) = 0

(u × v is orthogonal to u)

(b) v ⴢ (u × v) = 0

(u × v is orthogonal to v)

We will prove part (a). The proof of part (b) is similar. proof (a) Let u = u1 , u2 , u3 and v = v1 , v2 , v3 . Then from (4) u × v = u2 v3 − u3 v2 , u3 v1 − u1 v3 , u1 v2 − u2 v1

(7)

so that u ⴢ (u × v) = u1 (u2 v3 − u3 v2 ) + u2 (u3 v1 − u1 v3 ) + u3 (u1 v2 − u2 v1 ) = 0 ■ Example 3 Find a vector that is orthogonal to both of the vectors u = 2, −1, 3 and v = −7, 2, −1 .

Solution. By Theorem 11.4.4, the vector u × v will be orthogonal to both u and v. We compute that

Confirm that u × v in Example 3 is orthogonal to both u and v by computing u ⴢ (u × v) and v ⴢ (u × v).

i j u × v = 2 −1 −7 2

k 3 −1

2 2 −1 3 3 j + i − = −7 −7 −1 2 −1

−1 k = −5i − 19j − 3k 2

11.4 Cross Product

It can be proved that if u and v are nonzero and nonparallel vectors, then the direction ∗ of u × v relative to u and v is determined by a right-hand rule; that is, if the ﬁngers of the right hand are cupped so they curl from u toward v in the direction of rotation that takes u into v in less than 180 ◦ , then the thumb will point (roughly) in the direction of u × v (Figure 11.4.2). For example, we stated in (6) that

u×v

799

i × j = k,

j × k = i,

k×i=j

all of which are consistent with the right-hand rule (verify). The next theorem lists some more important geometric properties of the cross product.

v Figure 11.4.2

11.4.5 theorem Let u and v be nonzero vectors in 3-space, and let θ be the angle between these vectors when they are positioned so their initial points coincide. (a) u × v = u v sin θ (b) The area A of the parallelogram that has u and v as adjacent sides is A = u × v

(8)

(c) u × v = 0 if and only if u and v are parallel vectors, that is, if and only if they are scalar multiples of one another.

proof (a) u v sin θ = u v 1 − cos2 θ

(u ⴢ v)2 Theorem 11.3.3 = u v 1 − u 2 v 2 = u 2 v 2 − (u ⴢ v)2 = (u21 + u22 + u23 )(v12 + v22 + v32 ) − (u1 v1 + u2 v2 + u3 v3 )2 = (u2 v3 − u3 v2 )2 + (u1 v3 − u3 v1 )2 + (u1 v2 − u2 v1 )2 = u × v

See Formula (4).

proof (b) Referring to Figure 11.4.3, the parallelogram that has u and v as adjacent sides can be viewed as having base u and altitude v sin θ. Thus, its area A is

A = (base)(altitude) = u v sin θ = u × v ||v|| sin u

||v|| u

u ||u|| Figure 11.4.3

proof (c) Since u and v are assumed to be nonzero vectors, it follows from part (a) that u × v = 0 if and only if sin θ = 0; this is true if and only if θ = 0 or θ = π (since 0 ≤ θ ≤ π). Geometrically, this means that u × v = 0 if and only if u and v are parallel vectors. ■ Example 4 Find the area of the triangle that is determined by the points P1 (2, 2, 0), P2 (−1, 0, 2), and P3 (0, 4, 3).

∗

Recall that we agreed to consider only right-handed coordinate systems in this text. Had we used left-handed systems instead, a “left-hand rule” would apply here.

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Chapter 11 / Three-Dimensional Space; Vectors

Solution. The area A of the triangle is half the area of the parallelogram determined by −−→ −−→ −−→ −−→ the vectors P1P2 and P1P3 (Figure 11.4.4). But P1P2 = −3, −2, 2 and P1P3 = −2, 2, 3 , −−→ −−→ so P1P2 × P1P3 = −10, 5, −10

P3(0, 4, 3)

P2(⫺1, 0, 2)

(verify), and consequently −−→ −−→ A = 21 P1P2 × P1P3 =

15 2

SCALAR TRIPLE PRODUCTS P1(2, 2, 0)

Figure 11.4.4

If u = u1 , u2 , u3 , v = v1 , v2 , v3 , and w = w1 , w2 , w3 are vectors in 3-space, then the number u ⴢ (v × w) is called the scalar triple product of u, v, and w. It is not necessary to compute the dot product and cross product to evaluate a scalar triple product—the value can be obtained directly from the formula u 1 u 2 u 3 u ⴢ (v × w) = v1 v2 v3 (9) w1 w2 w3

T E C H N O LO GY M A ST E R Y Many calculating utilities have built-in cross product and determinant operations. If your calculating utility has these capabilities, use it to check the computations in Examples 1 and 5.

the validity of which can be seen by writing v 1 v 3 v 1 v 2 v2 v3 k i− j+ u ⴢ (v × w) = u ⴢ w2 w3 w1 w3 w 1 w2 v v1 v3 v1 v2 v3 = u1 2 − u + u 2 3 w2 w3 w1 w3 w1 w2 u1 u2 u3 = v1 v2 v3 w1 w2 w3 Example 5

Calculate the scalar triple product u ⴢ (v × w) of the vectors

u = 3i − 2 j − 5k,

Solution.

v = i + 4 j − 4k,

3 u ⴢ (v × w) = 1 0

−2 4 3

w = 3 j + 2k

−5 −4 = 49 2

GEOMETRIC PROPERTIES OF THE SCALAR TRIPLE PRODUCT

If u, v, and w are nonzero vectors in 3-space that are positioned so their initial points coincide, then these vectors form the adjacent sides of a parallelepiped (Figure 11.4.5). The following theorem establishes a relationship between the volume of this parallelepiped and the scalar triple product of the sides.

v Figure 11.4.5

11.4.6 It follows from Formula (10) that

u ⴢ (v × w) = ±V The + occurs when u makes an acute angle with v × w and the − occurs when it makes an obtuse angle.

theorem Let u, v, and w be nonzero vectors in 3-space.

(a) The volume V of the parallelepiped that has u, v, and w as adjacent edges is V = |u ⴢ (v × w)| (b) u ⴢ (v × w) = 0 if and only if u, v, and w lie in the same plane.

(10)

11.4 Cross Product v×w u

v h = || proj v× w u || Figure 11.4.6

801

proof (a) Referring to Figure 11.4.6, let us regard the base of the parallelepiped with u, v, and w as adjacent sides to be the parallelogram determined by v and w. Thus, the area of the base is v × w , and the altitude h of the parallelepiped (shown in the ﬁgure) is the length of the orthogonal projection of u on the vector v × w. Therefore, from Formula (12) of Section 11.3 we have h = projv × w u =

|u ⴢ (v × w)| |u ⴢ (v × w)| v × w = v × w 2 v × w

It now follows that the volume of the parallelepiped is

V = (area of base)(height) = v × w h = |u ⴢ (v × w)| proof (b) The vectors u, v, and w lie in the same plane if and only if the parallelepiped with these vectors as adjacent sides has volume zero (why?). Thus, from part (a) the vectors lie in the same plane if and only if u ⴢ (v × w) = 0. ■

A good way to remember Formula (11) is to observe that the second expression in the formula can be obtained from the first by leaving the dot, cross, and parentheses fixed, moving the first two vectors to the right, and bringing the third vector to the first position. The same procedure produces the third expression from the second and the first expression from the third (verify).

ALGEBRAIC PROPERTIES OF THE SCALAR TRIPLE PRODUCT We observed earlier in this section that the expression u × v × w must be avoided because it is ambiguous without parentheses. However, the expression u ⴢ v × w is not ambiguous—it has to mean u ⴢ (v × w) and not (u ⴢ v) × w because we cannot form the cross product of a scalar and a vector. Similarly, the expression u × v ⴢ w must mean (u × v) ⴢ w and not u × (v ⴢ w). Thus, when you see an expression of the form u ⴢ v × w or u × v ⴢ w, the cross product is formed ﬁrst and the dot product second. Since interchanging two rows of a determinant multiplies its value by −1, making two row interchanges in a determinant has no effect on its value. This being the case, it follows that u ⴢ (v × w) = w ⴢ (u × v) = v ⴢ (w × u) (11)

since the 3 × 3 determinants that are used to compute these scalar triple products can be obtained from one another by two row interchanges (verify). Another useful formula can be obtained by rewriting the ﬁrst equality in (11) as u ⴢ (v × w) = (u × v) ⴢ w and then omitting the superﬂuous parentheses to obtain uⴢv×w=u×vⴢw

(12)

In words, this formula states that the dot and cross in a scalar triple product can be interchanged (provided the factors are grouped appropriately). DOT AND CROSS PRODUCTS ARE COORDINATE INDEPENDENT

In Deﬁnitions 11.3.1 and 11.4.2 we deﬁned the dot product and the cross product of two vectors in terms of the components of those vectors in a coordinate system. Thus, it is theoretically possible that changing the coordinate system might change u ⴢ v or u × v, since the components of a vector depend on the coordinate system that is chosen. However, the relationships

This independence of a coordinate system is important in applications because it allows us to choose any convenient coordinate system for solving a problem with full confidence that the choice will not affect computations that involve dot products or cross products.

u ⴢ v = u v cos θ u × v = u v sin θ

(13) (14)

that were obtained in Theorems 11.3.3 and 11.4.5 show that this is not the case. Formula (13) shows that the value of u ⴢ v depends only on the lengths of the vectors and the angle between them—not on the coordinate system. Similarly, Formula (14), in combination with the right-hand rule and Theorem 11.4.4, shows that u × v does not depend on the coordinate system (as long as it is right-handed).

802

Chapter 11 / Three-Dimensional Space; Vectors

MOMENTS AND ROTATIONAL MOTION IN 3-SPACE

World Perspectives/Getty Images

Astronauts use tools that are designed to limit forces that would impart unintended rotational motion to a satellite.

Cross products play an important role in describing rotational motion in 3-space. For example, suppose that an astronaut on a satellite repair mission in space applies a force F at a point Q on the surface of a spherical satellite. If the force is directed along a line that passes through the center P of the satellite, then Newton’s Second Law of Motion implies that the force will accelerate the satellite in the direction of F. However, if the astronaut −→ applies the same force at an angle θ with the vector PQ, then F will tend to cause a rotation, as well as an acceleration in the direction of F. To see why this is so, let us resolve F into a sum of orthogonal components F = F1 + F2 , where F1 is the orthogonal projection −→ −→ of F on the vector PQ and F2 is the component of F orthogonal to PQ (Figure 11.4.7). Since the force F1 acts along the line through the center of the satellite, it contributes to the linear acceleration of the satellite but does not cause any rotation. However, the force F2 is −→ tangent to the circle around the satellite in the plane of F and PQ, so it causes the satellite to rotate about an axis that is perpendicular to that plane.

F2 P

u F1

Figure 11.4.7

You know from your own experience that the “tendency” for rotation about an axis depends both on the amount of force and how far from the axis it is applied. For example, it is easier to close a door by pushing on its outer edge than applying the same force close to the hinges. In fact, the tendency of rotation of the satellite can be measured by z

F = 100k

distance from the center × magnitude of the force

(15)

However, F2 = F sin θ, so we can rewrite (15) as

Q(1, 1, 1) y

−→ −→ PQ F sin θ = PQ × F

This is called the scalar moment or torque of F about the point P . Scalar moments have units of force times distance—pound-feet or newton-meters, for example. The vector −→ PQ × F is called the vector moment or torque vector of F about P . −→ Recalling that the direction of PQ × F is determined by the right-hand rule, it follows that the direction of rotation about P that results by applying the force F at the point Q −→ is counterclockwise looking down the axis of PQ × F (Figure 11.4.7). Thus, the vector −→ moment PQ × F captures the essential information about the rotational effect of the force— the magnitude of the cross product provides the scalar moment of the force, and the cross product vector itself provides the axis and direction of rotation.

(a) z

−→ PQ F2

100(i − j)

(b) Figure 11.4.8

Example 6 Figure 11.4.8a shows a force F of 100 N applied in the positive z-direction at the point Q(1, 1, 1) of a cube whose sides have a length of 1 m. Assuming that the cube is free to rotate about the point P (0, 0, 0) (the origin), ﬁnd the scalar moment of the force about P , and describe the direction of rotation.

11.4 Cross Product

803

−→

Solution. The force vector is F = 100k, and the vector from P to Q is PQ = i + j + k, so the vector moment of F about P is i −→ PQ × F = 1 0

k 1 = 100i − 100 j 100 √ Thus, the scalar moment of F about P is 100i − 100 j = 100 2 ≈ 141 N·m, and the direction of rotation is counterclockwise looking along the vector 100i − 100 j = 100(i − j) toward its initial point (Figure 11.4.8b).

✔QUICK CHECK EXERCISES 11.4 3 1. (a) 4

3 (b) 3 5

2 = 5

2. 1, 2, 0 × 3, 0, 4 =

(See page 805 for answers.)

2 2 5

1 1 = 5

3. Suppose that u, v, and w are vectors in 3-space such that u × v = 2, 7, 3 and u × w = −5, 4, 0 . (a) u × u = (b) v × u =

EXERCISE SET 11.4

4. Let u = i − 5k, v = 2i − 4j + k, and w = 3i − 2j + 5k. (a) u ⴢ (v × w) = (b) The volume of the parallelepiped that has u, v, and w as adjacent edges is V = .

CAS z

1. (a) Use a determinant to ﬁnd the cross product

(1, 1, 1)

i × (i + j + k)

v y

(b) Check your answer in part (a) by rewriting the cross product as i × (i + j + k) = (i × i) + (i × j) + (i × k)

u Figure Ex-9

and evaluating each term.

10. Find two unit vectors that are orthogonal to both

2. In each part, use the two methods in Exercise 1 to ﬁnd (a) j × (i + j + k) (b) k × (i + j + k).

u = −7i + 3 j + k,

3–6 Find u × v and check that it is orthogonal to both u and v. ■

v = 2i + 4k

11. Find two unit vectors that are normal to the plane determined by the points A(0, −2, 1), B(1, −1, −2), and C(−1, 1, 0).

3. u = 1, 2, −3 , v = −4, 1, 2

12. Find two unit vectors that are parallel to the yz-plane and are orthogonal to the vector 3i − j + 2k.

5. u = 0, 1, −2 , v = 3, 0, −4

13–16 True–False Determine whether the statement is true or

7. Let u = 2, −1, 3 , v = 0, 1, 7 , and w = 1, 4, 5 . Find (a) u × (v × w) (b) (u × v) × w (c) (u × v) × (v × w) (d) (v × w) × (u × v).

13. If the cross product of two nonzero vectors is the zero vector, then each of the two vectors is a scalar multiple of the other.

8. Use a CAS or a calculating utility that can compute determinants or cross products to solve Exercise 7.

14. For any three vectors a, b, and c, we have a × (b × c) = (a × b) × c.

4. u = 3i + 2 j − k, v = −i − 3 j + k

6. u = 4i + k, v = 2i − j

j 1 0

9. Find the direction cosines of u × v for the vectors u and v in the accompanying ﬁgure.

false. Explain your answer. ■

15. If v × u = v × w and if v = 0, then u = w. 16. If u = av + bw, then u ⴢ v × w = 0.

804

Chapter 11 / Three-Dimensional Space; Vectors

17–18 Find the area of the parallelogram that has u and v as

adjacent sides. ■ 17. u = i − j + 2k, v = 3 j + k

33. Use the result of Exercise 32 to ﬁnd the volume of the tetrahedron with vertices P (−1, 2, 0), Q(2, 1, −3), R(1, 0, 1), S(3, −2, 3)

18. u = 2i + 3 j, v = −i + 2 j − 2k 19–20 Find the area of the triangle with vertices P , Q, and R. ■

19. P (1, 5, −2), Q(0, 0, 0), R(3, 5, 1) 20. P (2, 0, −3), Q(1, 4, 5), R(7, 2, 9)

34. Let θ be the angle between the vectors u = 2i + 3 j − 6k and v = 2i + 3 j + 6k. (a) Use the dot product to find cos θ . (b) Use the cross product to find sin θ . (c) Confirm that sin2 θ + cos2 θ = 1. F O C U S O N C O N C E P TS

21–24 Find u ⴢ (v × w). ■

23. u = 2, 1, 0 , v = 1, −3, 1 , w = 4, 0, 1

35. Let A, B, C, and D be four distinct points in 3-space. −→ −→ −→ −→ −→ If AB × CD = 0 and AC ⴢ (AB × CD) = 0, explain why the line through A and B must intersect the line through C and D.

25–26 Use a scalar triple product to ﬁnd the volume of the parallelepiped that has u, v, and w as adjacent edges. ■

36. Let A, B, and C be three distinct noncollinear points in 3-space. Describe the set of all points P that satisfy the −→ −→ −→ vector equation AP ⴢ (AB × AC) = 0.

21. u = 2i − 3 j + k, v = 4i + j − 3k, w = j + 5k

22. u = 1, −2, 2 , v = 0, 3, 2 , w = −4, 1, −3 24. u = i, v = i + j, w = i + j + k

25. u = 2, −6, 2 , v = 0, 4, −2 , w = 2, 2, −4

26. u = 3i + j + 2k, v = 4i + 5 j + k, w = i + 2 j + 4k

27. In each part, use a scalar triple product to determine whether the vectors lie in the same plane. (a) u = 1, −2, 1 , v = 3, 0, −2 , w = 5, −4, 0 (b) u = 5i − 2 j + k, v = 4i − j + k, w = i − j (c) u = 4, −8, 1 , v = 2, 1, −2 , w = 3, −4, 12 28. Suppose that u ⴢ (v × w) = 3. Find (a) u ⴢ (w × v) (b) (v × w) ⴢ u (c) w ⴢ (u × v) (d) v ⴢ (u × w) (e) (u × w) ⴢ v (f ) v ⴢ (w × w).

29. Consider the parallelepiped with adjacent edges u = 3i + 2 j + k v = i + j + 2k w = i + 3 j + 3k (a) Find the volume. (b) Find the area of the face determined by u and w. (c) Find the angle between u and the plane containing the face determined by v and w. 30. Show that in 3-space the distance d from a point P to the line L through points A and B can be expressed as −→ − → AP × AB d= − → AB 31. Use the result in Exercise 30 to ﬁnd the distance between the point P and the line through the points A and B. (a) P (−3, 1, 2), A(1, 1, 0), B(−2, 3, −4) (b) P (4, 3), A(2, 1), B(0, 2) 32. It is a theorem of solid geometry that the volume of a tetrahedron is 31 (area of base) · (height). Use this result to prove that the volume of a tetrahedron with adjacent edges given by the vectors u, v, and w is 16 |u ⴢ (v × w)|.

37. What can you say about the angle between nonzero vectors u and v if u ⴢ v = u × v ?

38. Show that if u and v are vectors in 3-space, then u × v 2 = u 2 v 2 − (u ⴢ v)2

[Note: This result is sometimes called Lagrange’s identity.] 39. The accompanying figure shows a force F of 10 lb applied in the positive y-direction to the point Q(1, 1, 1) of a cube whose sides have a length of 1 ft. In each part, find the scalar moment of F about the point P , and describe the direction of rotation, if any, if the cube is free to rotate about P . (a) P is the point (0, 0, 0). (b) P is the point (1, 0, 0). (c) P is the point (1, 0, 1). 40. The accompanying figure shows a force F of 1000 N applied to the corner of a box. (a) Find the scalar moment of F about the point P . (b) Find the direction angles of the vector moment of F about the point P to the nearest degree. z z

Q(1, 1, 1)

10 lb y

1 ft

1000 N

1 ft 2m

1 ft x

Figure Ex-39

Figure Ex-40

1m Q

41. As shown in the accompanying ﬁgure on the next page, a force of 200 N is applied at an angle of 18 ◦ to a point near the end of a monkey wrench. Find the scalar moment of the force about the center of the bolt. [Note: Treat this as a problem in two dimensions.]

11.5 Parametric Equations of Lines

46. (a) Use the result in Exercise 45 to show that u × (v × w) lies in the same plane as v and w, and (u × v) × w lies in the same plane as u and v. (b) Use a geometrical argument to justify the results in part (a).

18° 200 N

200 mm

30 mm

47. In each part, use the result in Exercise 45 to prove the vector identity. (a) (a × b) × (c × d) = (a × b ⴢ d)c − (a × b ⴢ c)d (b) (a × b) × c + (b × c) × a + (c × a) × b = 0

Figure Ex-41

42. Prove parts (b) and (c) of Theorem 11.4.3. 43. Prove parts (d ) and (e) of Theorem 11.4.3.

48. Prove: If a, b, c, and d lie in the same plane when positioned with a common initial point, then

44. Prove part (b) of Theorem 11.4.1 for 3 × 3 determinants. [Note: Just give the proof for the ﬁrst two rows.] Then use (b) to prove (a). C

F O C U S O N C O N C E P TS

45. Expressions of the form u × (v × w)

and

805

(u × v) × w

are called vector triple products. It can be proved with some effort that u × (v × w) = (u ⴢ w)v − (u ⴢ v)w (u × v) × w = (w ⴢ u)v − (w ⴢ v)u These expressions can be summarized with the following mnemonic rule: vector triple product = (outer ⴢ remote)adjacent − (outer ⴢ adjacent)remote See if you can ﬁgure out what the expressions “outer,” “remote,” and “adjacent” mean in this rule, and then use the rule to ﬁnd the two vector triple products of the vectors u = i + 3 j − k, v = i + j + 2k, w = 3i − j + 2k

(a × b) × (c × d) = 0

49. Use a CAS to approximate the minimum area of a triangle if two of its vertices are (2, −1, 0) and (3, 2, 2) and its third vertex is on the curve y = ln x in the xy-plane.

50. If a force F is applied to an object at a point Q, then the line through Q parallel to F is called the line of action of the force. We deﬁned the vector moment of F about a point P −→ to be PQ × F. Show that if Q′ is any point on the line of −→ −→ action of F, then PQ × F = PQ′ × F; that is, it is not essential to use the point of application to compute the vector moment—any point on the line of action will do. [Hint: −→ −→ −−→ Write PQ′ = PQ + QQ′ and use properties of the cross product.]

51. Writing Discuss some of the similarities and differences between the multiplication of real numbers and the cross product of vectors. 52. Writing In your own words, describe what it means to say that the cross-product operation is “coordinate independent,” and state why this fact is signiﬁcant.

✔QUICK CHECK ANSWERS 11.4 1. (a) 7 (b) 0

11.5

2. 8i − 4j − 6k

3. (a) 0, 0, 0 (b) −2, −7, −3 (c) −3, 11, 3 (d) −10, 8, 0

4. (a) −58 (b) 58

PARAMETRIC EQUATIONS OF LINES In this section we will discuss parametric equations of lines in 2-space and 3-space. In 3-space, parametric equations of lines are especially important because they generally provide the most convenient form for representing lines algebraically.

LINES DETERMINED BY A POINT AND A VECTOR A line in 2-space or 3-space can be determined uniquely by specifying a point on the line and a nonzero vector parallel to the line (Figure 11.5.1). For example, consider a line L

Chapter 11 / Three-Dimensional Space; Vectors

806

in 3-space that passes through the point P0 (x0 , y0 , z0 ) and is parallel to the nonzero vector v = a, b, c . Then L consists precisely of those points P (x, y, z) for which the vector −−→ P0 P is parallel to v (Figure 11.5.2). In other words, the point P (x, y, z) is on L if and only −−→ if P0 P is a scalar multiple of v, say −−→ P0 P = tv

L P0(x0, y0) (a, b) v

This equation can be written as z

x − x0 , y − y0 , z − z0 = ta, tb, tc

which implies that P0(x0, y0, z 0)

x − x0 = ta,

(a, b, c) v

y − y0 = tb,

z − z0 = tc

Thus, L can be described by the parametric equations x = x0 + at,

y = y0 + bt,

z = z0 + ct

A similar description applies to lines in 2-space. We summarize these descriptions in the following theorem.

x A unique line L passes through P0 and is parallel to v.

11.5.1

Figure 11.5.1 z

theorem

(a) The line in 2-space that passes through the point P0 (x0 , y0 ) and is parallel to the nonzero vector v = a, b = ai + bj has parametric equations

P(x, y, z)

x = x0 + at,

P0 (x0, y0, z 0) (a, b, c) L v

y = y0 + bt

(1)

(b) The line in 3-space that passes through the point P0 (x0 , y0 , z0 ) and is parallel to the nonzero vector v = a, b, c = ai + bj + ck has parametric equations x = x0 + at,

y = y0 + bt,

z = z0 + ct

(2)

Figure 11.5.2

REMARK

Although it is not stated explicitly, it is understood in Equations (1) and (2) that −⬁ < t < +⬁, which reflects the fact that lines extend indefinitely.

Example 1

Find parametric equations of the line

(a) passing through (4, 2) and parallel to v = −1, 5 ;

(b) passing through (1, 2, −3) and parallel to v = 4i + 5j − 7k; (c) passing through the origin in 3-space and parallel to v = 1, 1, 1 .

Solution (a). From (1) with x0 = 4, y0 = 2, a = −1, and b = 5 we obtain x = 4 − t,

y = 2 + 5t

Solution (b). From (2) we obtain x = 1 + 4t,

y = 2 + 5t,

z = −3 − 7t

Solution (c). From (2) with x0 = 0, y0 = 0, z0 = 0, a = 1, b = 1, and c = 1 we obtain x = t,

y = t,

z=t

11.5 Parametric Equations of Lines

807

Example 2 (a) Find parametric equations of the line L passing through the points P1 (2, 4, −1) and P2 (5, 0, 7). (b) Where does the line intersect the xy-plane? −−→

Solution (a). The vector P1P2 = 3, −4, 8 is parallel to L and the point P1 (2, 4, −1) lies on L, so it follows from (2) that L has parametric equations x = 2 + 3t,

y = 4 − 4t,

z = −1 + 8t

(3)

Had we used P2 as the point on L rather than P1 , we would have obtained the equations x = 5 + 3t,

y = −4t,

z = 7 + 8t

Although these equations look different from those obtained using P1 , the two sets of equations are actually equivalent in that both generate L as t varies from −⬁ to +⬁. To see this, note that if t1 gives a point (x, y, z) = (2 + 3t1 , 4 − 4t1 , −1 + 8t1 ) on L using the ﬁrst set of equations, then t2 = t1 − 1 gives the same point (x, y, z) = (5 + 3t2 , −4t2 , 7 + 8t2 )

= (5 + 3(t1 − 1), −4(t1 − 1), 7 + 8(t1 − 1)) = (2 + 3t1 , 4 − 4t1 , −1 + 8t1 )

on L using the second set of equations. Conversely, if t2 gives a point on L using the second set of equations, then t1 = t2 + 1 gives the same point using the ﬁrst set.

Solution (b). It follows from (3) in part (a) that the line intersects the xy-plane at the point where z = −1 + 8t = 0, that is, when t = 18 . Substituting this value of t in (3) yields the point of intersection (x, y, z) = 19 , 7, 0 . 8 2 Example 3

Let L1 and L2 be the lines L1 : x = 1 + 4t,

L2 : x = 2 + 8t,

y = 5 − 4t,

y = 4 − 3t,

z = −1 + 5t z=5+t

(a) Are the lines parallel? (b) Do the lines intersect?

Solution (a). The line L1 is parallel to the vector 4i − 4j + 5k, and the line L2 is parallel to the vector 8i − 3j + k. These vectors are not parallel since neither is a scalar multiple of the other. Thus, the lines are not parallel. Solution (b). For L1 and L2 to intersect at some point (x0 , y0 , z0 ) these coordinates would have to satisfy the equations of both lines. In other words, there would have to exist values t1 and t2 for the parameters such that x0 = 1 + 4t1 ,

y0 = 5 − 4t1 ,

z0 = −1 + 5t1

and x0 = 2 + 8t2 ,

y0 = 4 − 3t2 ,

z0 = 5 + t2

808

Chapter 11 / Three-Dimensional Space; Vectors

This leads to three conditions on t1 and t2 , 1 + 4t1 = 2 + 8t2 5 − 4t1 = 4 − 3t2

(4)

−1 + 5t1 = 5 + t2 Thus, the lines intersect if there are values of t1 and t2 that satisfy all three equations, and the lines do not intersect if there are no such values. You should be familiar with methods for solving systems of two linear equations in two unknowns; however, this is a system of three linear equations in two unknowns. To determine whether this system has a solution we will solve the ﬁrst two equations for t1 and t2 and then check whether these values satisfy the third equation. We will solve the ﬁrst two equations by the method of elimination. We can eliminate the unknown t1 by adding the equations. This yields the equation

6 = 6 + 5t2

Parallel planes containing skew lines L1 and L2 can be determined by translating each line until it intersects the other.

Figure 11.5.3

from which we obtain t2 = 0. We can now ﬁnd t1 by substituting this value of t2 in either the ﬁrst or second equation. This yields t1 = 41 . However, the values t1 = 41 and t2 = 0 do not satisfy the third equation in (4), so the lines do not intersect. Two lines in 3-space that are not parallel and do not intersect (such as those in Example 3) are called skew lines. As illustrated in Figure 11.5.3, any two skew lines lie in parallel planes. LINE SEGMENTS

Sometimes one is not interested in an entire line, but rather some segment of a line. Parametric equations of a line segment can be obtained by ﬁnding parametric equations for the entire line, and then restricting the parameter appropriately so that only the desired segment is generated.

Example 4 Find parametric equations describing the line segment joining the points P1 (2, 4, −1) and P2 (5, 0, 7).

Solution. From Example 2, the line through the points P1 and P2 has parametric equations x = 2 + 3t, y = 4 − 4t, z = −1 + 8t. With these equations, the point P1 corresponds to t = 0 and P2 to t = 1. Thus, the line segment that joins P1 and P2 is given by x = 2 + 3t,

y = 4 − 4t,

z = −1 + 8t

(0 ≤ t ≤ 1)

VECTOR EQUATIONS OF LINES We will now show how vector notation can be used to express the parametric equations of a line more compactly. Because two vectors are equal if and only if their components are equal, (1) and (2) can be written in vector form as

x, y = x0 + at, y0 + bt x, y, z = x0 + at, y0 + bt, z0 + ct or, equivalently, as x, y = x0 , y0 + t a, b x, y, z = x0 , y0 , z0 + t a, b, c

(5) (6)

11.5 Parametric Equations of Lines

809

For the equation in 2-space we deﬁne the vectors r, r0 , and v as r = x, y ,

r0 = x0 , y0 ,

v = a, b

(7)

and for the equation in 3-space we deﬁne them as r = x, y, z ,

r0 = x0 , y0 , z0 ,

v = a, b, c

(8)

Substituting (7) and (8) in (5) and (6), respectively, yields the equation r = r0 + tv y

tv v x

r = r0 + tv Figure 11.5.4

(9)

in both cases. We call this the vector equation of a line in 2-space or 3-space. In this equation, v is a nonzero vector parallel to the line, and r0 is a vector whose components are the coordinates of a point on the line. We can interpret Equation (9) geometrically by positioning the vectors r0 and v with their initial points at the origin and the vector tv with its initial point at P0 (Figure 11.5.4). The vector tv is a scalar multiple of v and hence is parallel to v and L. Moreover, since the initial point of tv is at the point P0 on L, this vector actually runs along L; hence, the vector r = r0 + tv can be interpreted as the vector from the origin to a point on L. As the parameter t varies from 0 to +⬁, the terminal point of r traces out the portion of L that extends from P0 in the direction of v, and as t varies from 0 to −⬁, the terminal point of r traces out the portion of L that extends from P0 in the direction that is opposite to v. Thus, the entire line is traced as t varies over the interval (−⬁, +⬁), and it is traced in the direction of v as t increases.

Example 5 The equation x, y, z = −1, 0, 2 + t 1, 5, −4 is of form (9) with r0 = −1, 0, 2 and

v = 1, 5, −4

Thus, the equation represents the line in 3-space that passes through the point (−1, 0, 2) and is parallel to the vector 1, 5, −4 . Example 6 Find an equation of the line in 3-space that passes through the points P1 (2, 4, −1) and P2 (5, 0, 7).

Solution. The vector

−−→ P1P2 = 3, −4, 8

is parallel to the line, so it can be used as v in (9). For r0 we can use either the vector from the origin to P1 or the vector from the origin to P2 . Using the former yields r0 = 2, 4, −1 Thus, a vector equation of the line through P1 and P2 is x, y, z = 2, 4, −1 + t 3, −4, 8 If needed, we can express the line parametrically by equating corresponding components on the two sides of this vector equation, in which case we obtain the parametric equations in Example 2 (verify).

810

Chapter 11 / Three-Dimensional Space; Vectors

✔QUICK CHECK EXERCISES 11.5

(See page 812 for answers.)

1. Let L be the line through (2, 5) and parallel to v = 3, −1 . (a) Parametric equations of L are x=

(b) A vector equation of L is x, y =

2. Parametric equations for the line through (5, 3, 7) and parallel to the line x = 3 − t, y = 2, z = 8 + 4t are x=

EXERCISE SET 11.5

4. The line through the points (−3, 8, −4) and (1, 0, 8) intersects the yz-plane at .

5–6 Find parametric equations for the line whose vector equation is given. ■ 5. (a) x, y = 2, −3 + t 1, −4 (b) xi + y j + zk = k + t (i − j + k)

6. (a) xi + y j = (3i − 4 j) + t (2i + j) (b) x, y, z = −1, 0, 2 + t −1, 3, 0

7. (a) xi + y j = (2i − j) + t (4i − j) (b) x, y, z = −1, 2, 4 + t 5, 7, −8

bracket notation and also using i, j, k notation. ■

2. (a) Find parametric equations for the line segments in the unit square in part (a) of the accompanying ﬁgure. (b) Find parametric equations for the line segments in the unit cube shown in part (b) of the accompanying ﬁgure. z

(1, 1)

)

9–10 Express the given parametric equations of a line using

(b)

Figure Ex-1

(

8. (a) x, y = −1, 5 + t 2, 3 (b) xi + y j + zk = (i + j − 2k) + t j

(a)

by inspection. ■ L3

7–8 Find a point P on the line and a vector v parallel to the line

(1, 1, 1)

(1, 1)

CAS

1. (a) Find parametric equations for the lines through the corner of the unit square shown in part (a) of the accompanying ﬁgure. (b) Find parametric equations for the lines through the corner of the unit cube shown in part (b) of the accompanying ﬁgure. L1

Graphing Utility

3. Parametric equations for the line segment joining the points (3, 0, 11) and (2, 6, 7) are

(1, 1, 1)

L2 x

10. (a) x = t, y = −2 + t (b) x = 1 + t, y = −7 + 3t, z = 4 − 5t 11–14 True–False Determine whether the statement is true or false. Explain your answer. In these exercises L0 and L1 are lines in 3-space whose parametric equations are L0 : x = x0 + a0 t, y = y0 + b0 t, z = z0 + c0 t

L1 : x = x1 + a1 t,

y = y1 + b1 t,

z = z1 + c1 t ■

11. By deﬁnition, if L1 and L2 do not intersect, then L1 and L2 are parallel.

(a)

(b)

Figure Ex-2

3–4 Find parametric equations for the line through P1 and P2

and also for the line segment joining those points. ■ 3. (a) P1 (3, −2), P2 (5, 1)

9. (a) x = −3 + t, y = 4 + 5t (b) x = 2 − t, y = −3 + 5t, z = t

(b) P1 (5, −2, 1), P2 (2, 4, 2)

4. (a) P1 (0, 1), P2 (−3, −4) (b) P1 (−1, 3, 5), P2 (−1, 3, 2)

12. If L1 and L2 are parallel, then v0 = a0 , b0 , c0 is a scalar multiple of v1 = a1 , b1 , c1 .

13. If L1 and L2 intersect at a point (x, y, z), then there exists a single value of t such that L0 : x = x0 + a0 t, L1 : x = x1 + a1 t,

are satisﬁed.

y = y0 + b0 t,

y = y1 + b1 t,

z = z0 + c0 t

z = z1 + c1 t

11.5 Parametric Equations of Lines

14. If L0 passes through the origin, then the vectors a0 , b0 , c0 and x0 , y0 , z0 are parallel. 15–22 Find parametric equations of the line that satisﬁes the stated conditions. ■

15. The line through (−5, 2) that is parallel to 2i − 3 j.

16. The line through (0, 3) that is parallel to the line x = −5 + t, y = 1 − 2t.

17. The line that is tangent to the circle x 2 + y 2 = 25 at the point (3, −4). 18. The line that is tangent to the parabola y = x 2 at the point (−2, 4).

19. The line through (−1, 2, 4) that is parallel to 3i − 4 j + k.

811

35–36 Determine whether the points P1 , P2 , and P3 lie on the same line. ■ 35. P1 (6, 9, 7), P2 (9, 2, 0), P3 (0, −5, −3)

36. P1 (1, 0, 1), P2 (3, −4, −3), P3 (4, −6, −5)

37–38 Show that the lines L1 and L2 are the same. ■

37. L1 : x = 3 − t, y = 1 + 2t L2 : x = −1 + 3t, y = 9 − 6t

38. L1 : x = 1 + 3t, y = −2 + t, z = 2t L2 : x = 4 − 6t, y = −1 − 2t, z = 2 − 4t F O C U S O N C O N C E P TS

21. The line through (−2, 0, 5) that is parallel to the line given by x = 1 + 2t, y = 4 − t, z = 6 + 2t.

39. Sketch the vectors r0 = −1, 2 and v = 1, 1 , and then sketch the six vectors r0 ± v, r0 ± 2v, r0 ± 3v. Draw the line L: x = −1 + t, y = 2 + t, and describe the relationship between L and the vectors you sketched. What is the vector equation of L?

23. Where does the line x = 1 + 3t, y = 2 − t intersect (a) the x-axis (b) the y-axis (c) the parabola y = x 2 ?

40. Sketch the vectors r0 = 0, 2, 1 and v = 1, 0, 1 , and then sketch the vectors r0 + v, r0 + 2v, and r0 + 3v. Draw the line L: x = t, y = 2, z = 1 + t, and describe the relationship between L and the vectors you sketched. What is the vector equation of L?

20. The line through (2, −1, 5) that is parallel to −1, 2, 7 .

22. The line through the origin that is parallel to the line given by x = t, y = −1 + t, z = 2.

24. Where does the line x, y = 4t, 3t intersect the circle x 2 + y 2 = 25? 25–26 Find the intersections of the lines with the xy-plane, the xz-plane, and the yz-plane. ■

25. x = −2, y = 4 + 2t, z = −3 + t

26. x = −1 + 2t, y = 3 + t, z = 4 − t

27. Where does the line x = 1 + t, y = 3 − t, z = 2t intersect the cylinder x 2 + y 2 = 16? 28. Where does the line x = 2 − t, y = 3t, z = −1 + 2t intersect the plane 2y + 3z = 6?

29–30 Show that the lines L1 and L2 intersect, and ﬁnd their

point of intersection. ■ 29. L1 : x = 2 + t, y = 2 + 3t, z = 3 + t L2 : x = 2 + t, y = 3 + 4t, z = 4 + 2t

41. Sketch the vectors r0 = −2, 0 and r1 = 1, 3 , and then sketch the vectors 1 r + 23 r1 , 21 r0 + 21 r1 , 23 r0 + 13 r1 3 0 Draw the line segment (1 − t)r0 + tr1 (0 ≤ t ≤ 1). If n is a positive integer, what is the position of the point on this line segment corresponding to t = 1/n, relative to the points (−2, 0) and (1, 3)? 42. Sketch the vectors r0 = 2, 0, 4 and r1 = 0, 4, 0 , and then sketch the vectors 1 r + 43 r1 , 21 r0 + 21 r1 , 43 r0 + 41 r1 4 0 Draw the line segment (1 − t)r0 + tr1 (0 ≤ t ≤ 1). If n is a positive integer, what is the position of the point on this line segment corresponding to t = 1/n, relative to the points (2, 0, 4) and (0, 4, 0)?

30. L1 : x + 1 = 4t, y − 3 = t, z − 1 = 0 L2 : x + 13 = 12t, y − 1 = 6t, z − 2 = 3t

43–44 Describe the line segment represented by the vector equation. ■

31–32 Show that the lines L1 and L2 are skew. ■

44. x, y, z = −2, 1, 4 + t 3, 0, −1 (0 ≤ t ≤ 3)

31. L1 : x = 1 + 7t, y = 3 + t, z = 5 − 3t L2 : x = 4 − t, y = 6, z = 7 + 2t

43. x, y = 1, 0 + t −2, 3 (0 ≤ t ≤ 2)

45. Find the point on the line segment joining P1 (3, 6) and P2 (8, −4) that is 25 of the way from P1 to P2 .

32. L1 : x = 2 + 8t, y = 6 − 8t, z = 10t L2 : x = 3 + 8t, y = 5 − 3t, z = 6 + t

46. Find the point on the line segment joining P1 (1, 4, −3) and P2 (1, 5, −1) that is 23 of the way from P1 to P2 .

33–34 Determine whether the lines L1 and L2 are parallel. ■

47–48 Use the method in Exercise 32 of Section 11.3 to ﬁnd the distance from the point P to the line L, and then check your answer using the method in Exercise 30 of Section 11.4. ■

33. L1 : x = 3 − 2t, y = 4 + t, z = 6 − t L2 : x = 5 − 4t, y = −2 + 2t, z = 7 − 2t 34. L1 : x = 5 + 3t, y = 4 − 2t, z = −2 + 3t L2 : x = −1 + 9t, y = 5 − 6t, z = 3 + 8t

47. P (−2, 1, 1) L: x = 3 − t, y = t, z = 1 + 2t

812

Chapter 11 / Three-Dimensional Space; Vectors

56. Let L1 and L2 be the lines whose parametric equations are

48. P (1, 4, −3) L: x = 2 + t, y = −1 − t, z = 3t

L1 : x = 4t, L2 : x = 1 + t,

49–50 Show that the lines L1 and L2 are parallel, and ﬁnd the

distance between them. ■

50. L1 : x = 2t, y = 3 + 4t, z = 2 − 6t L2 : x = 1 + 3t, y = 6t, z = −9t

51. (a) Find parametric equations for the line through the points (x0 , y0 , z0 ) and (x1 , y1 , z1 ). (b) Find parametric equations for the line through the point (x1 , y1 , z1 ) and parallel to the line x = x0 + at, y = y0 + bt, z = z0 + ct 52. Let L be the line that passes through the point (x0 , y0 , z0 ) and is parallel to the vector v = a, b, c , where a, b, and c are nonzero. Show that a point (x, y, z) lies on the line L if and only if x − x0 y − y0 z − z0 = = a b c These equations, which are called the symmetric equations of L, provide a nonparametric representation of L.

57–58 Find parametric equations of the line that contains the point P and intersects the line L at a right angle, and ﬁnd the distance between P and L. ■

57. P (0, 2, 1) L: x = 2t, y = 1 − t, z = 2 + t

58. P (3, 1, −2) L: x = −2 + 2t, y = 4 + 2t, z = 2 + t

59. Two bugs are walking along lines in 3-space. At time t bug 1 is at the point (x, y, z) on the line x = 4 − t,

z=2+t

Assume that distance is in centimeters and that time is in minutes. (a) Find the distance between the bugs at time t = 0. (b) Use a graphing utility to graph the distance between the bugs as a function of time from t = 0 to t = 5. (c) What does the graph tell you about the distance between the bugs? (d) How close do the bugs get?

54. Consider the lines L1 and L2 whose symmetric equations are y + 23 x−1 z+1 L1 : = = 2 1 2 y−3 z+4 x−4 = = −1 −2 2

55. Let L1 and L2 be the lines whose parametric equations are L1 : x = 1 + 2t, y = 2 − t, z = 4 − 2t L2 : x = 9 + t, y = 5 + 3t, z = −4 − t (a) Show that L1 and L2 intersect at the point (7, −1, −2). (b) Find, to the nearest degree, the acute angle between L1 and L2 at their intersection. (c) Find parametric equations for the line that is perpendicular to L1 and L2 and passes through their point of intersection.

y = 1 + 2t,

and at the same time t bug 2 is at the point (x, y, z) on the line x = t, y = 1 + t, z = 1 + 2t

53. (a) Describe the line whose symmetric equations are x−1 y+3 = =z−5 2 4 (see Exercise 52). (b) Find parametric equations for the line in part (a).

(see Exercise 52). (a) Are L1 and L2 parallel? Perpendicular? (b) Find parametric equations for L1 and L2 . (c) Do L1 and L2 intersect? If so, where?

z = 2 + 2t z = −1 + 4t

(a) Show that L1 and L2 intersect at the point (2, 0, 3). (b) Find, to the nearest degree, the acute angle between L1 and L2 at their intersection. (c) Find parametric equations for the line that is perpendicular to L1 and L2 and passes through their point of intersection.

49. L1 : x = 2 − t, y = 2t, z = 1 + t L2 : x = 1 + 2t, y = 3 − 4t, z = 5 − 2t

L2 :

y = 1 − 2t, y = 1 − t,

60. Suppose that the temperature T at a point (x, y, z) on the line x = t, y = 1 + t, z = 3 − 2t is T = 25x 2 yz. Use a CAS or a calculating utility with a root-ﬁnding capability to approximate the maximum temperature on that portion of the line that extends from the xz-plane to the xy-plane. 61. Writing Give some examples of geometric problems that can be solved using the parametric equations of a line, and describe their solution. For example, how would you ﬁnd the points of intersection of a line and a sphere? 62. Writing Discuss how the vector equation of a line can be used to model the motion of a point that is moving with constant velocity in 3-space.

✔QUICK CHECK ANSWERS 11.5 1. (a) 2 + 3t; 5 − t (b) 2, 5 + t 3, −1

2. 5 − t; 3; 7 + 4t

3. 3 − t; 6t; 11 − 4t; 0 ≤ t ≤ 1

4. (0, 2, 5)

11.6 Planes in 3-Space

11.6

813

PLANES IN 3-SPACE In this section we will use vectors to derive equations of planes in 3-space, and then we will use these equations to solve various geometric problems. PLANES PARALLEL TO THE COORDINATE PLANES The graph of the equation x = a in an xyz-coordinate system consists of all points of the form (a, y, z), where y and z are arbitrary. One such point is (a, 0, 0), and all others are in the plane that passes through this point and is parallel to the yz-plane (Figure 11.6.1). Similarly, the graph of y = b is the plane through (0, b, 0) that is parallel to the xz-plane, and the graph of z = c is the plane through (0, 0, c) that is parallel to the xy-plane. z

y=b (0, 0, c) x=a y

z=c

(0, b, 0)

(a, 0, 0) n x

Figure 11.6.1

PLANES DETERMINED BY A POINT AND A NORMAL VECTOR A plane in 3-space can be determined uniquely by specifying a point in the plane and a vector perpendicular to the plane (Figure 11.6.2). A vector perpendicular to a plane is called a normal to the plane. Suppose that we want to ﬁnd an equation of the plane passing through P0 (x0 , y0 , z0 ) and perpendicular to the vector n = a, b, c . Deﬁne the vectors r0 and r as The colored plane is determined uniquely by the point P and the vector n perpendicular to the plane.

Figure 11.6.2

r0 = x0 , y0 , z0 and

r = x, y, z

It should be evident from Figure 11.6.3 that the plane consists precisely of those points P (x, y, z) for which the vector r − r0 is orthogonal to n; or, expressed as an equation, n ⴢ (r − r0 ) = 0

(1)

If preferred, we can express this vector equation in terms of components as P(x, y, z)

r − r0 P0 (x0 , y0, z 0)

a, b, c ⴢ x − x0 , y − y0 , z − z0 = 0

(2)

a(x − x0 ) + b(y − y0 ) + c(z − z0 ) = 0

(3)

from which we obtain r

r0 O Figure 11.6.3

What does Equation (1) represent if

n = a, b , r0 = x0 , y0 , r = x, y are vectors in an xy -plane in 2-space? Draw a picture.

This is called the point-normal form of the equation of a plane. Formulas (1) and (2) are vector versions of this formula. Example 1 Find an equation of the plane passing through the point (3, −1, 7) and perpendicular to the vector n = 4, 2, −5 .

Solution. From (3), a point-normal form of the equation is 4(x − 3) + 2(y + 1) − 5(z − 7) = 0

(4)

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Chapter 11 / Three-Dimensional Space; Vectors

If preferred, this equation can be written in vector form as 4, 2, −5 ⴢ x − 3, y + 1, z − 7 = 0 Observe that if we multiply out the terms in (3) and simplify, we obtain an equation of the form ax + by + cz + d = 0 (5) For example, Equation (4) in Example 1 can be rewritten as 4x + 2y − 5z + 25 = 0 The following theorem shows that every equation of form (5) represents a plane in 3-space.

11.6.1 theorem If a, b, c, and d are constants, and a, b, and c are not all zero, then the graph of the equation ax + by + cz + d = 0

(6)

is a plane that has the vector n = a, b, c as a normal. proof Since a, b, and c are not all zero, there is at least one point (x0 , y0 , z0 ) whose coordinates satisfy Equation (6). For example, if a = 0, then such a point is (−d /a, 0, 0), and similarly if b = 0 or c = 0 (verify). Thus, let (x0 , y0 , z0 ) be any point whose coordinates satisfy (6); that is, ax0 + by0 + cz0 + d = 0 Subtracting this equation from (6) yields a(x − x0 ) + b(y − y0 ) + c(z − z0 ) = 0 which is the point-normal form of a plane with normal n = a, b, c . ■ Equation (6) is called the general form of the equation of a plane. Example 2

Determine whether the planes 3x − 4y + 5z = 0

and

− 6x + 8y − 10z − 4 = 0

are parallel.

Solution. It is clear geometrically that two planes are parallel if and only if their normals are parallel vectors. A normal to the ﬁrst plane is n1 = 3, −4, 5 and a normal to the second plane is P

n2 = −6, 8, −10 Since n2 is a scalar multiple of n1 , the normals are parallel, and hence so are the planes.

There are infinitely many planes containing P and parallel to v.

Figure 11.6.4

We have seen that a unique plane is determined by a point in the plane and a nonzero vector normal to the plane. In contrast, a unique plane is not determined by a point in the plane and a nonzero vector parallel to the plane (Figure 11.6.4). However, a unique plane

11.6 Planes in 3-Space

815

is determined by a point in the plane and two nonparallel vectors that are parallel to the plane (Figure 11.6.5). A unique plane is also determined by three noncollinear points that lie in the plane (Figure 11.6.6).

P There is a unique plane through P that is parallel to both v and w.

Figure 11.6.5

−−→

Solution. Since the points P1 , P2 , and P3 lie in the plane, the vectors P1P2 = 1, 1, 2 P3 P2

P1 There is a unique plane through three noncollinear points.

Figure 11.6.6

Example 3 Find an equation of the plane through the points P1 (1, 2, −1), P2 (2, 3, 1), and P3 (3, −1, 2). −−→ and P1P3 = 2, −3, 3 are parallel to the plane. Therefore, i j k −−→ −−→ 1 2 = 9i + j − 5k P1P2 × P1P3 = 1 2 −3 3

−−→ −−→ is normal to the plane, since it is orthogonal to both P1P2 and P1P3 . By using this normal and the point P1 (1, 2, −1) in the plane, we obtain the point-normal form 9(x − 1) + (y − 2) − 5(z + 1) = 0 which can be rewritten as

Example 4

9x + y − 5z − 16 = 0

Determine whether the line x = 3 + 8t,

y = 4 + 5t,

z = −3 − t

is parallel to the plane x − 3y + 5z = 12.

Solution. The vector v = 8, 5, −1 is parallel to the line and the vector n = 1, −3, 5 is normal to the plane. For the line and plane to be parallel, the vectors v and n must be orthogonal. But this is not so, since the dot product v ⴢ n = (8)(1) + (5)(−3) + (−1)(5) = −12 is nonzero. Thus, the line and plane are not parallel.

Example 5

Find the intersection of the line and plane in Example 4.

Solution. If we let (x0 , y0 , z0 ) be the point of intersection, then the coordinates of this point satisfy both the equation of the plane and the parametric equations of the line. Thus, x0 − 3y0 + 5z0 = 12

(7)

and for some value of t, say t = t0 , x0 = 3 + 8t0 ,

y0 = 4 + 5t0 ,

z0 = −3 − t0

Substituting (8) in (7) yields (3 + 8t0 ) − 3(4 + 5t0 ) + 5(−3 − t0 ) = 12 Solving for t0 yields t0 = −3 and on substituting this value in (8), we obtain (x0 , y0 , z0 ) = (−21, −11, 0)

(8)

816

Chapter 11 / Three-Dimensional Space; Vectors

u 180° − u

(a)

INTERSECTING PLANES Two distinct intersecting planes determine two positive angles of intersection—an (acute) angle θ that satisﬁes the condition 0 ≤ θ ≤ π/2 and the supplement of that angle (Figure 11.6.7a). If n1 and n2 are normals to the planes, then depending on the directions of n1 and n2 , the angle θ is either the angle between n1 and n2 or the angle between n1 and −n2 (Figure 11.6.7b). In both cases, Theorem 11.3.3 yields the following formula for the acute angle θ between the planes: |n1 ⴢ n2 | cos θ = (9) n1 n2

Example 6 n2

Find the acute angle of intersection between the two planes 2x − 4y + 4z = 6 and 6x + 2y − 3z = 4

Plane 1

Solution. The given equations yield the normals n1 = 2, −4, 4 and n2 = 6, 2, −3 . Thus, Formula (9) yields

Plane 2

cos θ =

(b) Figure 11.6.7

from which we obtain

|−8| 4 |n1 ⴢ n2 | =√ √ = n1 n2 21 36 49 −1

θ = cos

Example 7

4 21

≈ 79 ◦

Find an equation for the line L of intersection of the planes in Example 6.

Solution. First compute v = n1 × n2 = 2, −4, 4 × 6, 2, −3 = 4, 30, 28 . Since

v is orthogonal to n1 , it is parallel to the ﬁrst plane, and since v is orthogonal to n2 , it is parallel to the second plane. That is, v is parallel to L, the intersection of the two planes. To ﬁnd a point on L we observe that L must intersect the xy-plane, z = 0, since v ⴢ 0, 0, 1 = 28 = 0. Substituting z = 0 in the equations of both planes yields 2x − 4y = 6 6x + 2y = 4 P0

with solution x = 1, y = −1. Thus, P (1, −1, 0) is a point on L. A vector equation for L is x, y, z = 1, −1, 0 + t 4, 30, 28 DISTANCE PROBLEMS INVOLVING PLANES

Next we will consider three basic distance problems in 3-space: (a)

(b) Figure 11.6.8

• Find the distance between a point and a plane. • Find the distance between two parallel planes. • Find the distance between two skew lines. The three problems are related. If we can find the distance between a point and a plane, then we can find the distance between parallel planes by computing the distance between one of the planes and an arbitrary point P0 in the other plane (Figure 11.6.8a). Moreover, we can find the distance between two skew lines by computing the distance between parallel planes containing them (Figure 11.6.8b).

11.6 Planes in 3-Space

817

11.6.2 theorem The distance D between a point P0 (x0 , y0 , z0 ) and the plane ax + by + cz + d = 0 is D=

n projn QP0

P0 (x0, y0, z 0) D

Q(x1, y1, z 1)

|ax0 + by0 + cz0 + d| √ a 2 + b2 + c2

(10)

proof Let Q(x1 , y1 , z1 ) be any point in the plane, and position the normal n = a, b, c so that its initial point is at Q. As illustrated in Figure 11.6.9, the distance D is equal to the −−→ length of the orthogonal projection of QP0 on n. Thus, from (12) of Section 11.3, −−→ −−→ −→ QP ⴢ n |− |QP0 ⴢ n| QP0 ⴢ n| −−→ 0 n = n = D = projn QP0 = n 2 n 2 n But

−−→ QP0 = x0 − x1 , y0 − y1 , z0 − z1 −−→ QP0 ⴢ n = a(x0 − x1 ) + b(y0 − y1 ) + c(z0 − z1 ) √ n = a 2 + b2 + c2

Figure 11.6.9

Thus, D=

|a(x0 − x1 ) + b(y0 − y1 ) + c(z0 − z1 )| √ a 2 + b2 + c2

(11)

Since the point Q(x1 , y1 , z1 ) lies in the plane, its coordinates satisfy the equation of the plane; that is, ax1 + by1 + cz1 + d = 0 or d = −ax1 − by1 − cz1 Combining this expression with (11) yields (10). ■ Example 8 There is an analog of Formula (10) in 2-space that can be used to compute the distance between a point and a line (see Exercise 52).

Find the distance D between the point (1, −4, −3) and the plane 2x − 3y + 6z = −1

Solution. Formula (10) requires the plane be rewritten in the form ax + by + cz + d = 0. Thus, we rewrite the equation of the given plane as 2x − 3y + 6z + 1 = 0 from which we obtain a = 2, b = −3, c = 6, and d = 1. Substituting these values and the coordinates of the given point in (10), we obtain D=

3 |−3| |(2)(1) + (−3)(−4) + 6(−3) + 1| = = 2 2 2 7 7 2 + (−3) + 6

Example 9 The planes x + 2y − 2z = 3 and 2x + 4y − 4z = 7 are parallel since their normals, 1, 2, −2 and 2, 4, −4 , are parallel vectors. Find the distance between these planes.

818

Chapter 11 / Three-Dimensional Space; Vectors

Solution. To ﬁnd the distance D between the planes, we can select an arbitrary point in one of the planes and compute its distance to the other plane. By setting y = z = 0 in the equation x + 2y − 2z = 3, we obtain the point P0 (3, 0, 0) in this plane. From (10), the distance from P0 to the plane 2x + 4y − 4z = 7 is D=

1 |(2)(3) + 4(0) + (−4)(0) − 7| = 2 2 2 6 2 + 4 + (−4)

Example 10 It was shown in Example 3 of Section 11.5 that the lines L1 : x = 1 + 4t,

L2 : x = 2 + 8t,

y = 5 − 4t,

y = 4 − 3t,

are skew. Find the distance between them. P2

Q2(2, 4, 5)

D P1

L1 Q1(1, 5, –1)

Figure 11.6.10

z = −1 + 5t z=5+t

Solution. Let P1 and P2 denote parallel planes containing L1 and L2 , respectively (Figure 11.6.10). To find the distance D between L1 and L2 , we will calculate the distance from a point in P1 to the plane P2 . Since L1 lies in plane P1 , we can find a point in P1 by finding a point on the line L1 ; we can do this by substituting any convenient value of t in the parametric equations of L1 . The simplest choice is t = 0, which yields the point Q1 (1, 5, −1). The next step is to find an equation for the plane P2 . For this purpose, observe that the vector u1 = 4, −4, 5 is parallel to line L1 , and therefore also parallel to planes P1 and P2 . Similarly, u2 = 8, −3, 1 is parallel to L2 and hence parallel to P1 and P2 . Therefore, the cross product i j k 5 = 11i + 36j + 20k n = u1 × u2 = 4 −4 8 −3 1

is normal to both P1 and P2 . Using this normal and the point Q2 (2, 4, 5) found by setting t = 0 in the equations of L2 , we obtain an equation for P2 : 11(x − 2) + 36(y − 4) + 20(z − 5) = 0 or 11x + 36y + 20z − 266 = 0 The distance between Q1 (1, 5, −1) and this plane is

95 |(11)(1) + (36)(5) + (20)(−1) − 266| =√ √ 2 2 2 1817 11 + 36 + 20 which is also the distance between L1 and L2 . D=

✔QUICK CHECK EXERCISES 11.6

(See page 821 for answers.)

1. The point-normal form of the equation of the plane through (0, 3, 5) and perpendicular to −4, 1, 7 is . 2. A normal vector for the plane 4x − 2y + 7z − 11 = 0 is .

3. A normal vector for the plane through the points (2, 5, 1), (3, 7, 0), and (2, 5, 2) is .

4. The acute angle of intersection of the planes x + y − 2z = 5 and 3y − 4z = 6 is .

5. The distance between the point (9, 8, 3) and the plane x + y − 2z = 5 is .

11.6 Planes in 3-Space

819

EXERCISE SET 11.6 1. Find equations of the planes P1 , P2 , and P3 that are parallel to the coordinate planes and pass through the corner (3, 4, 5) of the box shown in the accompanying ﬁgure. 2. Find equations of the planes P1 , P2 , and P3 that are parallel to the coordinate planes and pass through the corner (x0 , y0 , z0 ) of the box shown in the accompanying ﬁgure. z

(3, 4, 5) P2

(x0, y0, z 0)

Figure Ex-1

Figure Ex-2

3–6 Find an equation of the plane that passes through the point P and has the vector n as a normal. ■

3. P (2, 6, 1); n = 1, 4, 2

4. P (−1, −1, 2); n = −1, 7, 6 5. P (1, 0, 0); n = 0, 0, 1

6. P (0, 0, 0); n = 2, −3, −4

7–10 Find an equation of the plane indicated in the ﬁgure. ■

10.

1 1 x

15. (a) x = 4 + 2t, y = −t, z = −1 − 4t; 3x + 2y + z − 7 = 0 (b) x = t, y = 2t, z = 3t; x − y + 2z = 5 (c) x = −1 + 2t, y = 4 + t, z = 1 − t; 4x + 2y − 2z = 7 16. (a) x = 3 − t, y = 2 + t, z = 1 − 3t; 2x + 2y − 5 = 0 (b) x = 1 − 2t, y = t, z = −t; 6x − 3y + 3z = 1 (c) x = t, y = 1 − t, z = 2 + t; x+y+z=1 17–18 Determine whether the line and plane intersect; if so,

17. (a) x = t, y = t, z = t; 3x − 2y + z − 5 = 0 (b) x = 2 − t, y = 3 + t, z = t; 2x + y + z = 1 18. (a) x = 3t, y = 5t, z = −t; 2x − y + z + 1 = 0 (b) x = 1 + t, y = −1 + 3t, z = 2 + 4t; x − y + 4z = 7

(b) y = 4x − 2z + 3 x = 41 y + 21 z

ﬁnd the coordinates of the intersection. ■

14. (a) 3x − 2y + z = 4 6x − 4y + 3z = 7 (c) x + 4y + 7z = 3 5x − 3y + z = 0

15–16 Determine whether the line and plane are parallel, perpendicular, or neither. ■

13. (a) 2x − 8y − 6z − 2 = 0 (b) 3x − 2y + z = 1 −x + 4y + 3z − 5 = 0 4x + 5y − 2z = 4 (c) x − y + 3z − 2 = 0 2x + z = 1

19–20 Find the acute angle of intersection of the planes to the nearest degree. ■ 1

1 x

11–12 Find an equation of the plane that passes through the

given points. ■

19. x = 0 and 2x − y + z − 4 = 0 20. x + 2y − 2z = 5 and 6x − 3y + 2z = 8 21–24 True–False Determine whether the statement is true or false. Explain your answer. ■

21. Every plane has exactly two unit normal vectors.

11. (−2, 1, 1), (0, 2, 3), and (1, 0, −1) 12. (3, 2, 1), (2, 1, −1), and (−1, 3, 2)

22. If a plane is parallel to one of the coordinate planes, then its normal vector is parallel to one of the three vectors i, j, or k.

13–14 Determine whether the planes are parallel, perpendicular, or neither. ■

23. If two planes intersect in a line L, then L is parallel to the cross product of the normals to the two planes.

820

Chapter 11 / Three-Dimensional Space; Vectors

24. If a 2 + b2 + c2 = 1, then the distance from P (x0 , y0 , z0 ) to the plane ax + by + cz = 0 is | a, b, c ⴢ x0 , y0 , z0 |. 25–34 Find an equation of the plane that satisﬁes the stated conditions. ■

25. The plane through the origin that is parallel to the plane 4x − 2y + 7z + 12 = 0.

26. The plane that contains the line x = −2 + 3t, y = 4 + 2t, z = 3 − t and is perpendicular to the plane x − 2y + z = 5.

27. The plane through the point (−1, 4, 2) that contains the line of intersection of the planes 4x − y + z − 2 = 0 and 2x + y − 2z − 3 = 0.

28. The plane through (−1, 4, −3) that is perpendicular to the line x − 2 = t, y + 3 = 2t, z = −t.

29. The plane through (1, 2, −1) that is perpendicular to the line of intersection of the planes 2x + y + z = 2 and x + 2y + z = 3.

30. The plane through the points P1 (−2, 1, 4), P2 (1, 0, 3) that is perpendicular to the plane 4x − y + 3z = 2.

31. The plane through (−1, 2, −5) that is perpendicular to the planes 2x − y + z = 1 and x + y − 2z = 3.

32. The plane that contains the point (2, 0, 3) and the line x = −1 + t, y = t, z = −4 + 2t.

33. The plane whose points are equidistant from (2, −1, 1) and (3, 1, 5). 34. The plane that contains the line x = 3t, y = 1 + t, z = 2t and is parallel to the intersection of the planes y + z = −1 and 2x − y + z = 0.

35. Find parametric equations of the line through the point (5, 0, −2) that is parallel to the planes x − 4y + 2z = 0 and 2x + 3y − z +1 = 0.

36. Let L be the line x = 3t + 1, y = −5t, z = t. (a) Show that L lies in the plane 2x + y − z = 2. (b) Show that L is parallel to the plane x + y + 2z = 0. Is the line above, below, or on this plane? 37. Show that the lines x = −2 + t, x = 3 − t,

y = 3 + 2t, y = 4 − 2t,

z=4−t z=t

are parallel and ﬁnd an equation of the plane they determine.

38. Show that the lines L1 : x + 1 = 4t, L2 : x + 13 = 12t,

y − 3 = t, y − 1 = 6t,

z−1=0 z − 2 = 3t

intersect and ﬁnd an equation of the plane they determine.

F O C U S O N C O N C E P TS

39. Do the points (1, 0, −1), (0, 2, 3), (−2, 1, 1), and (4, 2, 3) lie in the same plane? Justify your answer two different ways. 40. Show that if a, b, and c are nonzero, then the plane whose intercepts with the coordinate axes are x = a,

y = b, and z = c is given by the equation x y z + + =1 a b c 41–42 Find parametric equations of the line of intersection of the planes. ■

41. −2x + 3y + 7z + 2 = 0 x + 2y − 3z + 5 = 0

42. 3x − 5y + 2z = 0 z=0

43–44 Find the distance between the point and the plane. ■

43. (1, −2, 3); 2x − 2y + z = 4

44. (0, 1, 5); 3x + 6y − 2z − 5 = 0

45–46 Find the distance between the given parallel planes. ■

45. −2x + y + z = 0 6x − 3y − 3z − 5 = 0

46. x + y + z = 1 x + y + z = −1

47–48 Find the distance between the given skew lines. ■

47. x = 1 + 7t, y = 3 + t, z = 5 − 3t x = 4 − t, y = 6, z = 7 + 2t

48. x = 3 − t, y = 4 + 4t, z = 1 + 2t x = t, y = 3, z = 2t

49. Find an equation of the sphere with center (2, 1, −3) that is tangent to the plane x − 3y + 2z = 4.

50. Locate the point of intersection of the plane 2x + y − z = 0 and the line through (3, 1, 0) that is perpendicular to the plane.

51. Show that the line x = −1 + t, y = 3 + 2t, z = −t and the plane 2x − 2y − 2z + 3 = 0 are parallel, and ﬁnd the distance between them. F O C U S O N C O N C E P TS

52. Formulas (1), (2), (3), (5), and (10), which apply to planes in 3-space, have analogs for lines in 2-space. (a) Draw an analog of Figure 11.6.3 in 2-space to illustrate that the equation of the line that passes through the point P (x0 , y0 ) and is perpendicular to the vector n = a, b can be expressed as n ⴢ (r − r0 ) = 0 where r = x, y and r0 = x0 , y0 . (b) Show that the vector equation in part (a) can be expressed as a(x − x0 ) + b(y − y0 ) = 0 This is called the point-normal form of a line. (c) Using the proof of Theorem 11.6.1 as a guide, show that if a and b are not both zero, then the graph of the equation ax + by + c = 0 is a line that has n = a, b as a normal.

(cont.)

11.7 Quadric Surfaces

(d) Using the proof of Theorem 11.6.2 as a guide, show that the distance D between a point P (x0 , y0 ) and the line ax + by + c = 0 is D=

|ax0 + by0 + c| √ a 2 + b2

is D= √

821

|d1 − d2 |

a 2 + b2 + c2

(b) Use the formula in part (a) to solve Exercise 45.

(e) Use the formula in part (d) to ﬁnd the distance between the point P (−3, 5) and the line y = −2x + 1. 53. (a) Show that the distance D between parallel planes ax + by + cz + d1 = 0 ax + by + cz + d2 = 0

54. Writing Explain why any line in 3-space must lie in some vertical plane. Must any line in 3-space also lie in some horizontal plane? 55. Writing Given two planes, discuss the various possibilities for the set of points they have in common. Then consider the set of points that three planes can have in common.

✔QUICK CHECK ANSWERS 11.6 1. −4x + (y − 3) + 7(z − 5) = 0

11.7

2. 4, −2, 7

3. 2, −1, 0

11 4. cos−1 √ ≈ 26 ◦ 5 6

√

QUADRIC SURFACES In this section we will study an important class of surfaces that are the three-dimensional analogs of the conic sections.

A monkey saddle

Figure 11.7.1

The parenthetical part of Equation (2) is a reminder that the z-coordinate of each point in the trace is z = k . This needs to be stated explicitly because the variable z does not appear in the equation x 2 + y 2 = k .

TRACES OF SURFACES Although the general shape of a curve in 2-space can be obtained by plotting points, this method is not usually helpful for surfaces in 3-space because too many points are required. It is more common to build up the shape of a surface with a network of mesh lines, which are curves obtained by cutting the surface with well-chosen planes. For example, Figure 11.7.1, which was generated by a CAS, shows the graph of z = x 3 − 3xy 2 rendered with a combination of mesh lines and colorization to produce the surface detail. This surface is called a “monkey saddle” because a monkey sitting astride the surface has a place for its two legs and tail. The mesh line that results when a surface is cut by a plane is called the trace of the surface in the plane (Figure 11.7.2). One way to deduce the shape of a surface is by examining its traces in planes parallel to the coordinate planes. For example, consider the surface

z = x2 + y2

(1)

To ﬁnd its trace in the plane z = k, we substitute this value of z into (1), which yields x2 + y2 = k

(z = k)

(2)

If k < 0, this equation has no real solutions, √ so there is no trace. However, if k ≥ 0, then the graph of (2) is a circle of radius k centered at the point (0, 0, k) on the z-axis (Figure 11.7.3a). Thus, for nonnegative values of k the traces parallel to the xy-plane form a family of circles, centered on the z-axis, whose radii start at zero and increase with k. This suggests that the surface has the form shown in Figure 11.7.3b. To obtain more detailed information about the shape of this surface, we can examine the traces of (1) in planes parallel to the yz-plane. Such planes have equations of the form x = k, so we substitute this in (1) to obtain z = k2 + y 2

(x = k)

822

Chapter 11 / Three-Dimensional Space; Vectors z

z z=k

√k

(0, 0, k)

Trac e o f s u r f a ce

(a) Figure 11.7.2

(b)

Figure 11.7.3

which we can rewrite as z − k2 = y 2

(3)

(x = k)

For simplicity, let us start with the case where k = 0 (the trace in the yz-plane), in which case the trace has the equation z = y2

(x = 0)

You should be able to recognize that this is a parabola in the plane x = 0 that has its vertex at the origin, opens in the positive z-direction, and is symmetric about the z-axis (the blue parabola in Figure 11.7.4a). You should also be able to recognize that the −k 2 term in (3) has the effect of translating the parabola z = y 2 in the positive z-direction, so its new vertex in the plane x = k is at the point (k, 0, k 2 ). This is the red parabola in Figure 11.7.4a. Thus, the traces in planes parallel to the yz-plane form a family of parabolas whose vertices move upward as k 2 increases (Figure 11.7.4b). Similarly, the traces in planes parallel to the xz-plane have equations of the form z − k2 = x 2

(y = k)

which again is a family of parabolas whose vertices move upward as k 2 increases (Figure 11.7.4c ). z

(k, 0, k 2 ) Figure 11.7.4

(a)

(b)

(c)

THE QUADRIC SURFACES

In the discussion of Formula (2) in Section 10.5 we noted that a second-degree equation Ax 2 + Bxy + Cy 2 + Dx + Ey + F = 0 represents a conic section (possibly degenerate). The analog of this equation in an xyzcoordinate system is Ax 2 + By 2 + Cz2 + Dxy + Exz + F yz + Gx + Hy + I z + J = 0

(4)

11.7 Quadric Surfaces

823

which is called a second-degree equation in x, y, and z. The graphs of such equations are called quadric surfaces or sometimes quadrics. Six common types of quadric surfaces are shown in Table 11.7.1—ellipsoids, hyperboloids of one sheet, hyperboloids of two sheets, elliptic cones, elliptic paraboloids, and hyperbolic paraboloids. (The constants a, b, and c that appear in the equations in the table are assumed to be positive.) Observe that none of the quadric surfaces in the table have cross-product terms in their equations. This is because of their orientations relative Table 11.7.1 surface

equation

surface

ellipsoid

equation

elliptic cone

z2 =

x2 y2 z2 + 2 + 2 =1 2 a b c

y x

The traces in the coordinate planes are ellipses, as are the traces in those planes that are parallel to the coordinate planes and intersect the surface in more than one point.

The trace in the xy-plane is a point (the origin), and the traces in planes parallel to the xy-plane are ellipses. The traces in the yzand xz-planes are pairs of lines intersecting at the origin. The traces in planes parallel to these are hyperbolas.

y x

hyperboloid of one sheet

x2 y2 + 2 2 a b

elliptic paraboloid 2

y z x + 2 − 2 =1 b c a2

The trace in the xy-plane is an ellipse, as are the traces in planes parallel to the xy-plane. The traces in the yz-plane and xz-plane are hyperbolas, as are the traces in those planes that are parallel to these and do not pass through the x- or y-intercepts. At these intercepts the traces are pairs of intersecting lines.

x2 y2 + 2 2 b a

The trace in the xy-plane is a point (the origin), and the traces in planes parallel to and above the xy-plane are ellipses. The traces in the yz- and xz-planes are parabolas, as are the traces in planes parallel to these.

y x

hyperbolic paraboloid

hyperboloid of two sheets z

x2 z2 y2 − 2 − 2 =1 c2 b a

There is no trace in the xy-plane. In planes parallel to the xy-plane that intersect the surface in more than one point the traces are ellipses. In the yz- and xz-planes, the traces are hyperbolas, as are the traces in those planes that are parallel to these.

x x

x2 y2 − 2 b2 a

The trace in the xy-plane is a pair of lines intersecting at the origin. The traces in planes parallel to the xy-plane are hyperbolas. The hyperbolas above the xy-plane open in the y-direction, and those below in the x-direction. The traces in the yz- and xz-planes are parabolas, as are the traces in planes parallel to these.

824

Chapter 11 / Three-Dimensional Space; Vectors

to the coordinate axes. Later in this section we will discuss other possible orientations that produce equations of the quadric surfaces with no cross-product terms. In the special case where the elliptic cross sections of an elliptic cone or an elliptic paraboloid are circles, the terms circular cone and circular paraboloid are used. TECHNIQUES FOR GRAPHING QUADRIC SURFACES Accurate graphs of quadric surfaces are best left for graphing utilities. However, the techniques that we will now discuss can be used to generate rough sketches of these surfaces that are useful for various purposes. A rough sketch of an ellipsoid

y2 z2 x2 + + =1 a2 b2 c2

(a > 0, b > 0, c > 0)

(5)

can be obtained by ﬁrst plotting the intersections with the coordinate axes, and then sketching the elliptical traces in the coordinate planes. Example 1 illustrates this technique.

Example 1

Sketch the ellipsoid y2 z2 x2 + + =1 4 16 9

(6)

Solution. The x-intercepts can be obtained by setting y = 0 and z = 0 in (6). This Rough sketch

yields x = ±2. Similarly, the y-intercepts are y = ±4, and the z-intercepts are z = ±3. Sketching the elliptical traces in the coordinate planes yields the graph in Figure 11.7.5.

Figure 11.7.5

A rough sketch of a hyperboloid of one sheet y2 z2 x2 + − =1 a2 b2 c2

(a > 0, b > 0, c > 0)

(7)

can be obtained by ﬁrst sketching the elliptical trace in the xy-plane, then the elliptical traces in the planes z = ±c, and then the hyperbolic curves that join the endpoints of the axes of these ellipses. The next example illustrates this technique.

Example 2

Sketch the graph of the hyperboloid of one sheet x2 + y2 −

z2 =1 4

(8)

Solution. The trace in the xy-plane, obtained by setting z = 0 in (8), is x2 + y2 = 1

(z = 0)

which is a circle of radius 1 centered on the z-axis. The traces in the planes z = 2 and z = −2, obtained by setting z = ±2 in (8), are given by Rough sketch

Figure 11.7.6

x2 + y2 = 2

(z = ±2) √ which are circles of radius 2 centered on the z-axis. Joining these circles by the hyperbolic traces in the vertical coordinate planes yields the graph in Figure 11.7.6.

11.7 Quadric Surfaces

825

A rough sketch of the hyperboloid of two sheets x2 y2 z2 − 2 − 2 =1 2 c a b

(a > 0, b > 0, c > 0)

(9)

can be obtained by ﬁrst plotting the intersections with the z-axis, then sketching the elliptical traces in the planes z = ±2c, and then sketching the hyperbolic traces that connect the zaxis intersections and the endpoints of the axes of the ellipses. (It is not essential to use the planes z = ±2c, but these are good choices since they simplify the calculations slightly and have the right spacing for a good sketch.) The next example illustrates this technique. Example 3

Sketch the graph of the hyperboloid of two sheets z2 − x 2 −

y2 =1 4

(10)

Solution. The z-intercepts, obtained by setting x = 0 and y = 0 in (10), are z = ±1. The traces in the planes z = 2 and z = −2, obtained by setting z = ±2 in (10), are given by

Rough sketch

Figure 11.7.7

y2 x2 + =1 (z = ±2) 3 12 Sketching these ellipses and the hyperbolic traces in the vertical coordinate planes yields Figure 11.7.7. A rough sketch of the elliptic cone z2 =

x2 y2 + a2 b2

(a > 0, b > 0)

(11)

can be obtained by ﬁrst sketching the elliptical traces in the planes z = ±1 and then sketching the linear traces that connect the endpoints of the axes of the ellipses. The next example illustrates this technique. Example 4

Sketch the graph of the elliptic cone z2 = x 2 +

y2 4

(12)

Solution. The traces of (12) in the planes z = ±1 are given by

y2 =1 (z = ±1) 4 Sketching these ellipses and the linear traces in the vertical coordinate planes yields the graph in Figure 11.7.8. x2 +

Rough sketch

Figure 11.7.8

A rough sketch of the elliptic paraboloid In the special cases of (11) and (13) where a = b, the traces parallel to the xy -plane are circles. In these cases, we call (11) a circular cone and (13) a circular paraboloid.

x2 y2 + a2 b2

(a > 0, b > 0)

(13)

can be obtained by ﬁrst sketching the elliptical trace in the plane z = 1 and then sketching the parabolic traces in the vertical coordinate planes to connect the origin to the ends of the axes of the ellipse. The next example illustrates this technique.

826

Chapter 11 / Three-Dimensional Space; Vectors

Example 5

Sketch the graph of the elliptic paraboloid z=

y2 x2 + 4 9

(14)

Solution. The trace of (14) in the plane z = 1 is

y2 x2 + =1 (z = 1) 4 9 Sketching this ellipse and the parabolic traces in the vertical coordinate planes yields the graph in Figure 11.7.9.

Rough sketch

Figure 11.7.9

A rough sketch of the hyperbolic paraboloid z=

y2 x2 − 2 2 b a

(a > 0, b > 0)

(15)

can be obtained by ﬁrst sketching the two parabolic traces that pass through the origin (one in the plane x = 0 and the other in the plane y = 0). After the parabolic traces are drawn, sketch the hyperbolic traces in the planes z = ±1 and then ﬁll in any missing edges. The next example illustrates this technique. Example 6

Sketch the graph of the hyperbolic paraboloid z=

x2 y2 − 4 9

(16)

Solution. Setting x = 0 in (16) yields

y2 (x = 0) 4 which is a parabola in the yz-plane with vertex at the origin and opening in the positive z-direction (since z ≥ 0), and setting y = 0 yields z=

x2 (y = 0) 9 which is a parabola in the xz-plane with vertex at the origin and opening in the negative z-direction. The trace in the plane z = 1 is z=−

x2 y2 − =1 (z = 1) 4 9 which is a hyperbola that opens along a line parallel to the y-axis (verify), and the trace in the plane z = −1 is y2 x2 − =1 (z = −1) 9 4 which is a hyperbola that opens along a line parallel to the x-axis. Combining all of the above information leads to the sketch in Figure 11.7.10.

Rough sketch

Figure 11.7.10

REMARK

The hyperbolic paraboloid in Figure 11.7.10 has an interesting behavior at the origin—the trace in the xz-plane has a relative maximum at (0, 0, 0), and the trace in the yz-plane has a relative minimum at (0, 0, 0). Thus, a bug walking on the surface may view the origin as a highest point if traveling along one path, or may view the origin as a lowest point if traveling along a different path. A point with this property is commonly called a saddle point or a minimax point .

11.7 Quadric Surfaces

827

Figure 11.7.11 shows two computer-generated views of the hyperbolic paraboloid in Example 6. The ﬁrst view, which is much like our rough sketch in Figure 11.7.10, has cuts at the top and bottom that are hyperbolic traces parallel to the xy-plane. In the second view the top horizontal cut has been omitted; this helps to emphasize the parabolic traces parallel to the xz-plane.

(h, k)

Figure 11.7.11

(0, 0)

(a)

TRANSLATIONS OF QUADRIC SURFACES

(h, k, l) y

(b) Figure 11.7.12

In Section 10.4 we saw that a conic in an xy-coordinate system can be translated by substituting x − h for x and y − k for y in its equation. To understand why this works, think of the xy-axes as fixed and think of the plane as a transparent sheet of plastic on which all graphs are drawn. When the coordinates of points are modified by substituting (x − h, y − k) for (x, y), the geometric effect is to translate the sheet of plastic (and hence all curves) so that the point on the plastic that was initially at (0, 0) is moved to the point (h, k) (see Figure 11.7.12a). For the analog in three dimensions, think of the xyz-axes as fixed and think of 3-space as a transparent block of plastic in which all surfaces are embedded. When the coordinates of points are modified by substituting (x − h, y − k, z − l) for (x, y, z), the geometric effect is to translate the block of plastic (and hence all surfaces) so that the point in the plastic block that was initially at (0, 0, 0) is moved to the point (h, k, l) (see Figure 11.7.12b). Example 7

Describe the surface z = (x − 1)2 + (y + 2)2 + 3.

Solution. The equation can be rewritten as z − 3 = (x − 1)2 + (y + 2)2 This surface is the paraboloid that results by translating the paraboloid Rough sketch

Figure 11.7.13

z = x2 + y2 in Figure 11.7.3 so that the new “vertex” is at the point (1, −2, 3). A rough sketch of this paraboloid is shown in Figure 11.7.13.

Example 8

Describe the surface 4x 2 + 4y 2 + z2 + 8y − 4z = −4

Solution. Completing the squares yields 4x 2 + 4(y + 1)2 + (z − 2)2 = −4 + 4 + 4 or x 2 + (y + 1)2 +

(z − 2)2 =1 4

828

Chapter 11 / Three-Dimensional Space; Vectors

Thus, the surface is the ellipsoid that results when the ellipsoid z2 =1 4 is translated so that the new “center” is at the point (0, −1, 2). A rough sketch of this ellipsoid is shown in Figure 11.7.14. x2 + y2 +

REFLECTIONS OF SURFACES IN 3-SPACE Rough sketch

Figure 11.7.14

In Figure 11.7.14, the cross section in the yz-plane is shown tangent to both the y - and z-axes. Confirm that this is correct.

Recall that in an xy-coordinate system a point (x, y) is reflected about the x-axis if y is replaced by −y, and it is reflected about the y-axis if x is replaced by −x. In an xyzcoordinate system, a point (x, y, z) is reflected about the xy-plane if z is replaced by −z, it is reflected about the yz-plane if x is replaced by −x, and it is reflected about the xz-plane if y is replaced by −y (Figure 11.7.15). It follows that replacing a variable by its negative in the equation of a surface causes that surface to be reflected about a coordinate plane. z

(−x, y, z)

(y, x, z)

(x, y, z)

(x, −y, z)

(x, y, z)

Plane

(x, y, −z)

Figure 11.7.15

y=x

Figure 11.7.16

Recall also that in an xy-coordinate system a point (x, y) is reflected about the line y = x if x and y are interchanged. However, in an xyz-coordinate system, interchanging x and y reflects the point (x, y, z) about the plane y = x (Figure 11.7.16). Similarly, interchanging x and z reflects the point about the plane x = z, and interchanging y and z reflects it about the plane y = z. Thus, it follows that interchanging two variables in the equation of a surface reflects that surface about a plane that makes a 45 ◦angle with two of the coordinate planes.

Example 9

Figure 11.7.17

Describe the surfaces (a) y 2 = x 2 + z2

(b) z = −(x 2 + y 2 )

Solution (a). The graph of the equation y 2 = x 2 + z2 results from interchanging y and

z in the equation z2 = x 2 + y 2 . Thus, the graph of the equation y 2 = x 2 + z2 can be obtained by reﬂecting the graph of z2 = x 2 + y 2 about the plane y = z. Since the graph of z2 = x 2 + y 2 is a circular cone opening along the z-axis (see Table 11.7.1), it follows that the graph of y 2 = x 2 + z2 is a circular cone opening along the y-axis (Figure 11.7.17).

Solution (b). The graph of the equation z = −(x 2 + y 2 ) can be written as −z = x 2 + y 2 ,

Figure 11.7.18

which can be obtained by replacing z with −z in the equation z = x 2 + y 2 . Since the graph of z = x 2 + y 2 is a circular paraboloid opening in the positive z-direction (see Table 11.7.1), it follows that the graph of z = −(x 2 + y 2 ) is a circular paraboloid opening in the negative z-direction (Figure 11.7.18).

11.7 Quadric Surfaces

829

A TECHNIQUE FOR IDENTIFYING QUADRIC SURFACES The equations of the quadric surfaces in Table 11.7.1 have certain characteristics that make it possible to identify quadric surfaces that are derived from these equations by reflections. These identifying characteristics, which are shown in Table 11.7.2, are based on writing the equation of the quadric surface so that all of the variable terms are on the left side of the equation and there is a 1 or a 0 on the right side. These characteristics do not change when the surface is reflected about a coordinate plane or planes of the form x = y, x = z, or y = z, thereby making it possible to identify the reflected quadric surface from the form of its equation. Table 11.7.2 identifying a quadric surface from the form of its equation

equation

z2 x 2 y2 x 2 y2 x 2 y2 y2 x 2 x 2 y2 y2 z2 z2 x2 + + =1 + − 2 = 1 2 − 2 − 2 = 1 z2 − 2 − 2 = 0 z − 2 − 2 = 0 z − 2 + 2 = 0 2 2 2 2 2 a a b b c b a a b a b c b c a

characteristic

No minus signs

One minus sign

Two minus signs

No linear terms

One linear term; two quadratic terms with the same sign

classification

Ellipsoid

Hyperboloid of one sheet

Hyperboloid of two sheets

Elliptic cone

Elliptic paraboloid

One linear term; two quadratic terms with opposite signs Hyperbolic paraboloid

Example 10 Identify the surfaces (a) 3x 2 − 4y 2 + 12z2 + 12 = 0

(b) 4x 2 − 4y + z2 = 0

Solution (a). The equation can be rewritten as y2 x2 − − z2 = 1 3 4 This equation has a 1 on the right side and two negative terms on the left side, so its graph is a hyperboloid of two sheets.

Solution (b). The equation has one linear term and two quadratic terms with the same sign, so its graph is an elliptic paraboloid.

✔QUICK CHECK EXERCISES 11.7

(See page 832 for answers.)

1. For the surface 4x 2 + y 2 + z2 = 9, classify the indicated trace as an ellipse, hyperbola, or parabola. (a) x = 0 (b) y = 0 (c) z = 1

2. For the surface 4x 2 + z2 − y 2 = 9, classify the indicated trace as an ellipse, hyperbola, or parabola. (a) x = 0 (b) y = 0 (c) z = 1

3. For the surface 4x 2 + y 2 − z = 0, classify the indicated trace as an ellipse, hyperbola, or parabola. (a) x = 0 (b) y = 0 (c) z = 1

4. Classify each surface as an ellipsoid, hyperboloid of one sheet, hyperboloid of two sheets, elliptic cone, elliptic paraboloid, or hyperbolic paraboloid. x2 y2 x2 y2 (a) + −z=0 (b) + + z2 = 1 36 25 36 25 y2 x2 y2 x2 − +z=0 (d) + − z2 = 1 (c) 36 25 36 25 y2 x2 y2 x2 + − z2 = 0 (f ) z2 − − =1 (e) 36 25 36 25

830

Chapter 11 / Three-Dimensional Space; Vectors

EXERCISE SET 11.7 1–2 Identify the quadric surface as an ellipsoid, hyperboloid

of one sheet, hyperboloid of two sheets, elliptic cone, elliptic paraboloid, or hyperbolic paraboloid by matching the equation with one of the forms given in Table 11.7.1. State the values of a, b, and c in each case. ■ y2 y2 x2 − x2 + (b) z = 1. (a) z = 4 9 25 (d) x 2 + y 2 − z2 = 0 (c) x 2 + y 2 − z2 = 16 2 2 (e) 4z = x + 4y (f ) z2 − x 2 − y 2 = 1 2. (a) 6x 2 + 3y 2 + 4z2 = 12 (c) 9x 2 + y 2 − 9z2 = 9 (e) 2z − x 2 − 4y 2 = 0

(b) y 2 − x 2 − z = 0 (d) 4x 2 + y 2 − 4z2 = −4 (f ) 12z2 − 3x 2 = 4y 2

3. Find an equation for and sketch the surface that results when the circular paraboloid z = x 2 + y 2 is reﬂected about the plane (a) z = 0 (b) x = 0 (c) y = 0 (d) y = x (e) x = z (f ) y = z.

4. Find an equation for and sketch the surface that results when the hyperboloid of one sheet x 2 + y 2 − z2 = 1 is reﬂected about the plane (a) z = 0 (b) x = 0 (c) y = 0 (d) y = x (e) x = z (f ) y = z.

F O C U S O N C O N C E P TS

5. The given equations represent quadric surfaces whose orientations are different from those in Table 11.7.1. In each part, identify the quadric surface, and give a verbal description of its orientation (e.g., an elliptic cone opening along the z-axis or a hyperbolic paraboloid straddling the y-axis). z2 y2 x2 x2 y2 z2 (a) 2 − 2 + 2 = 1 (b) 2 − 2 − 2 = 1 c b a a b c 2 2 z2 y z y2 (d) x 2 = 2 + 2 (c) x = 2 + 2 b c b c 2 z2 x x2 z2 (e) y = 2 − 2 (f ) y = − + c a a2 c2 6. For each of the surfaces in Exercise 5, find the equation of the surface that results if the given surface is reflected about the xz-plane and that surface is then reflected about the plane z = 0. 7–8 Find equations of the traces in the coordinate planes and

sketch the traces in an xyz-coordinate system. [Suggestion: If you have trouble sketching a trace directly in three dimensions, start with a sketch in two dimensions by placing the coordinate plane in the plane of the paper, then transfer the sketch to three dimensions.] ■ x2 y2 z2 7. (a) + + =1 (b) z = x 2 + 4y 2 9 25 4 x2 y2 z2 + − =1 (c) 9 16 4

8. (a) y 2 + 9z2 = x y2 (c) z2 = x 2 + 4

(b) 4x 2 − y 2 + 4z2 = 4

9–10 In these exercises, traces of the surfaces in the planes are

conic sections. In each part, ﬁnd an equation of the trace, and state whether it is an ellipse, a parabola, or a hyperbola. ■ 9. (a) (b) (c) (d) (e) (f ) 10. (a) (b) (c) (d) (e) (f )

4x 2 + y 2 + z2 = 4; y = 1 4x 2 + y 2 + z2 = 4; x = 21 9x 2 − y 2 − z2 = 16; x = 2 9x 2 − y 2 − z2 = 16; z = 2 z = 9x 2 + 4y 2 ; y = 2 z = 9x 2 + 4y 2 ; z = 4

9x 2 − y 2 + 4z2 = 9; x = 2 9x 2 − y 2 + 4z2 = 9; y = 4 x 2 + 4y 2 − 9z2 = 0; y = 1 x 2 + 4y 2 − 9z2 = 0; z = 1 z = x 2 − 4y 2 ; x = 1 z = x 2 − 4y 2 ; z = 4

11–14 True–False Determine whether the statement is true or false. Explain your answer. ■

11. A quadric surface is the graph of a fourth-degree polynomial in x, y, and z. 12. Every ellipsoid will intersect the z-axis in exactly two points. 13. Every ellipsoid is a surface of revolution. 14. The hyperbolic paraboloid y2 x2 − 2 2 b a intersects the xy-plane in a pair of intersecting lines. z=

15–26 Identify and sketch the quadric surface. ■

z2 y2 + =1 4 9 x2 y2 z2 17. + − =1 4 9 16 19. 4z2 = x 2 + 4y 2 15. x 2 +

21. 9z2 − 4y 2 − 9x 2 = 36 23. z = y 2 − x 2

25. 4z = x 2 + 2y 2

16. x 2 + 4y 2 + 9z2 = 36 18. x 2 + y 2 − z2 = 9 20. 9x 2 + 4y 2 − 36z2 = 0 x2 z2 22. y 2 − − =1 4 9 24. 16z = y 2 − x 2 26. z − 3x 2 − 3y 2 = 0

27–32 The given equation represents a quadric surface whose orientation is different from that in Table 11.7.1. Identify and sketch the surface. ■

27. x 2 − 3y 2 − 3z2 = 0

29. 2y 2 − x 2 + 2z2 = 8 x2 y2 31. z = − 4 9

28. x − y 2 − 4z2 = 0

30. x 2 − 3y 2 − 3z2 = 9

32. 4x 2 − y 2 + 4z2 = 16

11.7 Quadric Surfaces 33–36 Sketch the surface. ■

x2 + y2 35. z = x 2 + y 2 − 1

33. z =

1 − x2 − y2 36. z = 1 + x 2 + y 2 34. z =

37–40 Identify the surface and make a rough sketch that shows its position and orientation. ■

831

(d) Describe the orientation of the focal axis of the parabola in part (a) relative to the coordinate axes. 47–48 Sketch the region enclosed between the surfaces and describe their curve of intersection. ■

47. The paraboloids z = x 2 + y 2 and z = 4 − x 2 − y 2

37. z = (x + 2)2 + (y − 3)2 − 9

48. The ellipsoid 2x 2 + 2y 2 + z2 = 3 and the paraboloid z = x2 + y2.

39. 9x 2 + y 2 + 4z2 − 18x + 2y + 16z = 10

49–50 Find an equation for the surface generated by revolving the curve about the y-axis. ■

38. 4x 2 − y 2 + 16(z − 2)2 = 100

40. z2 = 4x 2 + y 2 + 8x − 2y + 4z

41–42 Use the ellipsoid 4x 2 + 9y 2 + 18z2 = 72 in these exer-

cises. ■

41. (a) Find√an equation of the elliptical trace in the plane z = 2. (b) Find the lengths of the major and minor axes of the ellipse in part (a). (c) Find the coordinates of the foci of the ellipse in part (a). (d) Describe the orientation of the focal axis of the ellipse in part (a) relative to the coordinate axes. 42. (a) Find an equation of the elliptical trace in the plane x = 3. (b) Find the lengths of the major and minor axes of the ellipse in part (a). (c) Find the coordinates of the foci of the ellipse in part (a). (d) Describe the orientation of the focal axis of the ellipse in part (a) relative to the coordinate axes. 43–46 These exercises refer to the hyperbolic paraboloid z = y2 − x2. ■

43. (a) Find an equation of the hyperbolic trace in the plane z = 4. (b) Find the vertices of the hyperbola in part (a). (c) Find the foci of the hyperbola in part (a). (d) Describe the orientation of the focal axis of the hyperbola in part (a) relative to the coordinate axes.

44. (a) Find an equation of the hyperbolic trace in the plane z = −4. (b) Find the vertices of the hyperbola in part (a). (c) Find the foci of the hyperbola in part (a). (d) Describe the orientation of the focal axis of the hyperbola in part (a) relative to the coordinate axes. 45. (a) Find an equation of the parabolic trace in the plane x = 2. (b) Find the vertex of the parabola in part (a). (c) Find the focus of the parabola in part (a). (d) Describe the orientation of the focal axis of the parabola in part (a) relative to the coordinate axes. 46. (a) Find an equation of the parabolic trace in the plane y = 2. (b) Find the vertex of the parabola in part (a). (c) Find the focus of the parabola in part (a).

49. y = 4x 2 (z = 0)

50. y = 2x (z = 0)

51. Find an equation of the surface consisting of all points P (x, y, z) that are equidistant from the point (0, 0, 1) and the plane z = −1. Identify the surface.

52. Find an equation of the surface consisting of all points P (x, y, z) that are twice as far from the plane z = −1 as from the point (0, 0, 1). Identify the surface. 53. If a sphere

x2 y2 z2 + + =1 a2 a2 a2 of radius a is compressed in the z-direction, then the resulting surface, called an oblate spheroid, has an equation of the form x2 y2 z2 + 2 + 2 =1 2 a a c where c < a. Show that the oblate spheroid has a circular trace of radius a in the xy-plane and an elliptical trace in the xz-plane with major axis of length 2a along the x-axis and minor axis of length 2c along the z-axis.

54. The Earth’s rotation causes a flattening at the poles, so its shape is often modeled as an oblate spheroid rather than a sphere (see Exercise 53 for terminology). One of the models used by global positioning satellites is the World Geodetic System of 1984 (WGS-84), which treats the Earth as an oblate spheroid whose equatorial radius is 6378.1370 km and whose polar radius (the distance from the Earth’s center to the poles) is 6356.5231 km. Use the WGS-84 model to find an equation for the surface of the Earth relative to the coordinate system shown in the accompanying figure. z North Pole

Equator x

Figure Ex-54

55. Use the method of slicing to show that the volume of the ellipsoid x2 y2 z2 + + =1 a2 b2 c2 4 is 3 πabc.

Chapter 11 / Three-Dimensional Space; Vectors

832

56. Writing Discuss some of the connections between conic sections and traces of quadric surfaces.

57. Writing Give a sequence of steps for determining the type of quadric surface that is associated with a quadratic equation in x, y, and z.

✔QUICK CHECK ANSWERS 11.7 1. (a) ellipse (b) ellipse (c) ellipse 2. (a) hyperbola (b) ellipse (c) hyperbola 3. (a) parabola (b) parabola (c) ellipse 4. (a) elliptic paraboloid (b) ellipsoid (c) hyperbolic paraboloid (d) hyperboloid of one sheet (e) elliptic cone (f ) hyperboloid of two sheets

CYLINDRICAL AND SPHERICAL COORDINATES

11.8

In this section we will discuss two new types of coordinate systems in 3-space that are often more useful than rectangular coordinate systems for studying surfaces with symmetries. These new coordinate systems also have important applications in navigation, astronomy, and the study of rotational motion about an axis. CYLINDRICAL AND SPHERICAL COORDINATE SYSTEMS Three coordinates are required to establish the location of a point in 3-space. We have already done this using rectangular coordinates. However, Figure 11.8.1 shows two other possibilities: part (a) of the ﬁgure shows the rectangular coordinates (x, y, z) of a point P , part (b) shows the cylindrical coordinates (r, θ, z) of P , and part (c) shows the spherical coordinates (ρ, θ, φ) of P . In a rectangular coordinate system the coordinates can be any real numbers, but in cylindrical and spherical coordinate systems there are restrictions on the allowable values of the coordinates (as indicated in Figure 11.8.1). z

P( r, u, z)

P( x, y, z) y

z x x

z = z0 y

y = y0 x

Figure 11.8.2

Cylindrical coordinates

Spherical coordinates

(x, y, z)

(r, u, z) (r ≥ 0, 0 ≤ u < 2c)

(r, u, f) (r ≥ 0, 0 ≤ u < 2c, 0 ≤ f ≤ c)

(a)

(b)

(c)

Figure 11.8.1

x = x0

Rectangular coordinates

P(r, u, f)

CONSTANT SURFACES

In rectangular coordinates the surfaces represented by equations of the form x = x0 ,

y = y0 ,

and

z = z0

where x0 , y0 , and z0 are constants, are planes parallel to the yz-plane, xz-plane, and xyplane, respectively (Figure 11.8.2). In cylindrical coordinates the surfaces represented by equations of the form r = r0 , θ = θ0 , and z = z0

11.8 Cylindrical and Spherical Coordinates

833

where r0 , θ0 , and z0 are constants, are shown in Figure 11.8.3:

r = r0

r0 z0

z = z0

• The surface r = r0 is a right circular cylinder of radius r0 centered on the z-axis. • The surface θ = θ0 is a half-plane attached along the z-axis and making an angle

•

u0 u = u0

θ0 with the positive x-axis. The surface z = z0 is a horizontal plane.

In spherical coordinates the surfaces represented by equations of the form

Figure 11.8.3

ρ = ρ0 ,

θ = θ0 ,

and

φ = φ0

where ρ0 , θ0 , and φ0 are constants, are shown in Figure 11.8.4:

f0 f = f0

• The surface ρ = ρ0 consists of all points whose distance ρ from the origin is ρ0 . r = r0

•

• u = u0

Figure 11.8.4

Assuming ρ0 to be nonnegative, this is a sphere of radius ρ0 centered at the origin. As in cylindrical coordinates, the surface θ = θ0 is a half-plane attached along the z-axis, making an angle of θ0 with the positive x-axis. The surface φ = φ0 consists of all points from which a line segment to the origin makes an angle of φ0 with the positive z-axis. If 0 < φ0 < π/2, this will be the nappe of a cone opening up, while if π/2 < φ0 < π, this will be the nappe of a cone opening down. (If φ0 = π/2, then the cone is ﬂat, and the surface is the xy-plane.)

CONVERTING COORDINATES

Just as we needed to convert between rectangular and polar coordinates in 2-space, so we will need to be able to convert between rectangular, cylindrical, and spherical coordinates in 3-space. Table 11.8.1 provides formulas for making these conversions. Table 11.8.1 conversion formulas for coordinate systems

conversion

formulas

Cylindrical to rectangular Rectangular to cylindrical

(r, u, z) → (x, y, z) (x, y, z) → (r, u, z)

x = r cos u, y = r sin u, z = z r = √x 2 + y2, tan u = y/ x, z = z

Spherical to cylindrical Cylindrical to spherical

(r, u, f) → (r, u, z) (r, u, z) → (r, u, f)

r = r sin f, u = u, z = r cos f r = √r 2 + z 2, u = u, tan f = r/ z

Spherical to rectangular Rectangular to spherical

(r, u, f) → (x, y, z) (x, y, z) → (r, u, f)

x = r sin f cos u, y = r sin f sin u, z = r cos f r = √x 2 + y2 + z 2, tan u = y/ x, cos f = z /√x 2 + y2 + z 2

restrictions

r ≥ 0, r ≥ 0 0 ≤ u < 2c 0≤f≤c

The diagrams in Figure 11.8.5 will help you to understand how the formulas in Table 11.8.1 are derived. For example, part (a) of the ﬁgure shows that in converting between rectangular coordinates (x, y, z) and cylindrical coordinates (r, θ, z), we can interpret (r, θ) as polar coordinates of (x, y). Thus, the polar-to-rectangular and rectangular-to-polar

Chapter 11 / Three-Dimensional Space; Vectors

834

(x, y, z) (r, u, z)

r = ρ sin φ,

conversion formulas (1) and (2) of Section 10.2 provide the conversion formulas between rectangular and cylindrical coordinates in the table. Part (b) of Figure 11.8.5 suggests that the spherical coordinates (ρ, θ, φ) of a point P can be converted to cylindrical coordinates (r, θ, z) by the conversion formulas z = ρ cos φ

(1)

Moreover, since the cylindrical coordinates (r, θ, z) of P can be converted to rectangular coordinates (x, y, z) by the conversion formulas

(r, u, 0)

θ = θ,

x = r cos θ,

(a) z

y = r sin θ,

z=z

(2)

we can obtain direct conversion formulas from spherical coordinates to rectangular coordinates by substituting (1) in (2). This yields

P r

(r, u, f) (r, u, z)

y = ρ sin φ sin θ,

z = ρ cos φ

(3)

The other conversion formulas in Table 11.8.1 are left as exercises.

x = ρ sin φ cos θ,

Example 1 (a) Find the rectangular coordinates of the point with cylindrical coordinates (r, θ, z) = (4, π/3, −3)

(b) Find the rectangular coordinates of the point with spherical coordinates

(b)

(ρ, θ, φ) = (4, π/3, π/4)

Comparison of coordinate systems

Figure 11.8.5

Solution (a). Applying the cylindrical-to-rectangular conversion formulas in Table 11.8.1

c/3

√ π = 2 3, z = −3 3 √ Thus, the rectangular coordinates of the point are (x, y, z) = (2, 2 3, −3) (Figure 11.8.6). x = r cos θ = 4 cos

yields π = 2, 3

y = r sin θ = 4 sin

Solution (b). Applying the spherical-to-rectangular conversion formulas in Table 11.8.1 yields

π π √ cos = 2 4 3 π √ π y = ρ sin φ sin θ = 4 sin sin = 6 4 3 √ π z = ρ cos φ = 4 cos = 2 2 4 √ √ √ The rectangular coordinates of the point are (x, y, z) = ( 2, 6, 2 2) (Figure 11.8.7).

cylindrical: (4, c/ 3, −3) rectangular: (2, 2√3, −3)

x = ρ sin φ cos θ = 4 sin

Figure 11.8.6 z

√6

√2

c/4 4

2√2 y

c/3

Since the interval 0 ≤ θ < 2π covers two periods of the tangent function, the conversion formula tan θ = y /x does not completely determine θ . The following example shows how to deal with this ambiguity.

spherical: (4, c/3, c/4) rectangular: (√2, √6, 2√2 ) Figure 11.8.7

Example 2

Find the spherical coordinates of the point that has rectangular coordinates √ (x, y, z) = (4, −4, 4 6)

11.8 Cylindrical and Spherical Coordinates

835

Solution. From the rectangular-to-spherical conversion formulas in Table 11.8.1 we How should θ be chosen if x = 0 ? How should θ be chosen if y = 0 ?

√ √ √ x 2 + y 2 + z2 = 16 + 16 + 96 = 128 = 8 2 y tan θ = = −1 x √ √ z 4 6 3 cos φ = = √ = 2 2 2 2 8 2 x +y +z

obtain

ρ=

4 4

From the restriction 0 ≤ θ < 2π and the computed value of tan θ, the possibilities for θ are θ = 3π/4 and θ = 7π/4. However, the given point has a negative y-coordinate, so we must have θ = 7π/4. Moreover, from the restriction 0 ≤ φ ≤ π and the computed value of cos φ, the only possibility for φ is φ = π/6. Thus, the spherical coordinates of the point √ are (ρ, θ, φ) = (8 2, 7π/4, π/6) (Figure 11.8.8).

8√2 4√6

c/6

EQUATIONS OF SURFACES IN CYLINDRICAL AND SPHERICAL COORDINATES

7c/4 y

Surfaces of revolution about the z-axis of a rectangular coordinate system usually have simpler equations in cylindrical coordinates than in rectangular coordinates, and the equations of surfaces with symmetry about the origin are usually simpler in spherical coordinates than in rectangular coordinates. For example, consider the upper nappe of the circular cone whose equation in rectangular coordinates is z = x2 + y2

rectangular: (4, −4, 4√6) spherical: (8√2, 7c/4, c/6)

(Table 11.8.2). The corresponding equation in cylindrical coordinates can be obtained from the cylindrical-to-rectangular conversion formulas in Table 11.8.1. This yields √ z = (r cos θ)2 + (r sin θ )2 = r 2 = |r| = r

Figure 11.8.8

so the equation of the cone in cylindrical coordinates is z = r. Going a step further, the equation of the cone in spherical coordinates can be obtained from the spherical-tocylindrical conversion formulas from Table 11.8.1. This yields ρ cos φ = ρ sin φ which, if ρ = 0, can be rewritten as

π 4 Geometrically, this tells us that the radial line from the origin to any point on the cone makes an angle of π/4 with the z-axis. tan φ = 1

or φ =

Table 11.8.2

cone

cylinder

sphere

paraboloid

hyperboloid

y x

rectangular

z = √x 2 + y 2

x2 + y2 = 1

x2 + y2 + z2 = 1

z = x2 + y2

x2 + y2 − z2 = 1

cylindrical

z=r

r=1

z2 = 1 − r2

z = r2

z2 = r2 − 1

spherical

f = c/4

r = csc f

r=1

r = cos f csc2 f

r 2 = −sec 2f

836

Chapter 11 / Three-Dimensional Space; Vectors

Example 3 coordinates.

Find equations of the paraboloid z = x 2 + y 2 in cylindrical and spherical

Solution. The rectangular-to-cylindrical conversion formulas in Table 11.8.1 yield Verify the equations given in Table 11.8.2 for the cylinder and hyperboloid in cylindrical and spherical coordinates.

z = r2

(4)

which is the equation in cylindrical coordinates. Now applying the spherical-to-cylindrical conversion formulas to (4) yields ρ cos φ = ρ 2 sin2 φ which we can rewrite as ρ = cos φ csc2 φ Alternatively, we could have obtained this equation directly from the equation in rectangular coordinates by applying the spherical-to-rectangular conversion formulas (verify).

Prime meridian y

z New Orleans

SPHERICAL COORDINATES IN NAVIGATION

East x

West Equator

Figure 11.8.9

Spherical coordinates are related to longitude and latitude coordinates used in navigation. To see why this is so, let us construct a right-hand rectangular coordinate system with its origin at the center of the Earth, its positive z-axis passing through the North Pole, and its positive x-axis passing through the prime meridian (Figure 11.8.9). If we assume the Earth to be a sphere of radius ρ = 4000 miles, then each point on the Earth has spherical coordinates of the form (4000, θ, φ), where φ and θ determine the latitude and longitude of the point. It is common to specify longitudes in degrees east or west of the prime meridian and latitudes in degrees north or south of the equator. However, the next example shows that it is a simple matter to determine φ and θ from such data. Example 4 The city of New Orleans is located at 90 ◦ west longitude and 30 ◦ north latitude. Find its spherical and rectangular coordinates relative to the coordinate axes of Figure 11.8.9. (Assume that distance is in miles.)

Solution. A longitude of 90 ◦ west corresponds to θ = 360 ◦ − 90 ◦ = 270 ◦ or θ = 3π/2

Jon Arnold/Danita Delimont

Modern navigation systems use multiple coordinate representations to calculate position.

radians; and a latitude of 30 ◦ north corresponds to φ = 90 ◦ − 30 ◦ = 60 ◦ or φ = π/3 radians. Thus, the spherical coordinates (ρ, θ, φ) of New Orleans are (4000, 3π/2, π/3). To ﬁnd the rectangular coordinates we apply the spherical-to-rectangular conversion formulas in Table 11.8.1. This yields √ π 3π 3 x = 4000 sin cos = 4000 (0) = 0 mi 3 2 2 √ √ π 3π 3 y = 4000 sin sin = 4000 (−1) = −2000 3 mi 3 2 2 π 1 z = 4000 cos = 4000 = 2000 mi 3 2

✔QUICK CHECK EXERCISES 11.8

(See page 838 for answers.)

1. The conversion formulas from cylindrical coordinates (r, θ, z) to rectangular coordinates (x, y, z) are x=

2. The conversion formulas from spherical coordinates (ρ, θ, φ) to rectangular coordinates (x, y, z) are x=

11.8 Cylindrical and Spherical Coordinates

3. The conversion formulas from spherical coordinates (ρ, θ, φ) to cylindrical coordinates (r, θ, z) are r=

θ=

5. Give an equation of a sphere of radius 5, centered at the origin, in (a) rectangular coordinates (b) cylindrical coordinates (c) spherical coordinates.

4. Let point in 3-space with rectangular coordinates √ P be √the √ ( 2, − 2, 2 3). (a) Cylindrical coordinates for P are (r, θ, z) = . (b) Spherical coordinates for P are (ρ, θ, φ) = .

EXERCISE SET 11.8

Graphing Utility

CAS

1–2 Convert from rectangular to cylindrical coordinates. ■

1. (a) (c) 2. (a) (c)

√ (4 3, 4, −4) (0, 2, 0) √ √ ( 2, − 2, 1) (−4, 4, −7)

(b) (−5, 5,√6) (d) (4, −4 3, 6)

(b) (0, 1, 1) (d) (2, −2, −2)

3–4 Convert from cylindrical to rectangular coordinates. ■

3. (a) (4, π/6, 3) (c) (5, 0, 4) 4. (a) (6, 5π/3, 7) (c) (3, π/2, 5)

(b) (8, 3π/4, −2) (d) (7, π, −9) (b) (1, π/2, 0) (d) (4, π/2, −1)

5–6 Convert from rectangular to spherical coordinates. ■

√ 5. (a) (1, √ 3, −2) (c) (0, 3 3, 3) √ 6. (a) (4, 4, 4 6) (c) (2, 0, 0)

√ (b) (1, −1, √ 2) (d) (−5 3, 5, 0) √ (b) (1, −2) √ − 3,√ (d) ( 3, 1, 2 3)

7–8 Convert from spherical to rectangular coordinates. ■

7. (a) (5, π/6, π/4) (c) (1, π, 0) 8. (a) (1, 2π/3, 3π/4) (c) (8, π/6, π/4)

(b) (7, 0, π/2) (d) (2, 3π/2, π/2) (b) (3, 7π/4, 5π/6) (d) (4, π/2, π/3)

9–10 Convert from cylindrical to spherical coordinates. ■

√ 9. (a) ( 3, π/6, 3) (c) (2, 3π/4, 0) 10. (a) (4, 5π/6, 4) (c) (4, π/2, 3)

(b) (1, π/4, −1) √ (d) (6, 1, −2 3) (b) (2, 0, −2) (d) (6, π, 2)

11–12 Convert from spherical to cylindrical coordinates. ■

11. (a) (5, π/4, 2π/3) (c) (3, 0, 0) 12. (a) (5, √ π/2, 0) (c) ( 2, 3π/4, π) C

837

(b) (1, 7π/6, π) (d) (4, π/6, π/2) (b) (6, 0, 3π/4) (d) (5, 2π/3, 5π/6)

13. Use a CAS or a programmable calculating utility to set up the conversion formulas in Table 11.8.1, and then use the CAS or calculating utility to solve the problems in Exercises 1, 3, 5, 7, 9, and 11.

14. Use a CAS or a programmable calculating utility to set up the conversion formulas in Table 11.8.1, and then use the CAS or calculating utility to solve the problems in Exercises 2, 4, 6, 8, 10, and 12. 15–18 True–False Determine whether the statement is true or

false. Explain your answer. ■ 15. In cylindrical coordinates for a point, r is the distance from the point to the z-axis. 16. In spherical coordinates for a point, ρ is the distance from the point to the origin. 17. The graph of θ = θ0 in cylindrical coordinates is the same as the graph of θ = θ0 in spherical coordinates.

18. The graph of r = f(θ ) in cylindrical coordinates can always be obtained by extrusion of the polar graph of r = f (θ ) in the xy-plane.

19–26 An equation is given in cylindrical coordinates. Express the equation in rectangular coordinates and sketch the graph. ■ 19. r = 3 20. θ = π/4 21. z = r 2

22. z = r cos θ 2

23. r = 4 sin θ

25. r + z = 1

26. r cos 2θ = z

24. r = 2 sec θ

27–34 An equation is given in spherical coordinates. Express the equation in rectangular coordinates and sketch the graph. ■ 27. ρ = 3 28. θ = π/3 29. φ = π/4

30. ρ = 2 sec φ

31. ρ = 4 cos φ

33. ρ sin φ = 2 cos θ

32. ρ sin φ = 1

34. ρ − 2 sin φ cos θ = 0

35–46 An equation of a surface is given in rectangular coordinates. Find an equation of the surface in (a) cylindrical coordinates and (b) spherical coordinates. ■

35. z = 3

37. z = 3x + 3y

36. y = 2 38. z = 3x 2 + 3y 2

39. x 2 + y 2 = 4

40. x 2 + y 2 − 6y = 0

43. 2x + 3y + 4z = 1

44. x 2 + y 2 − z2 = 1

41. x 2 + y 2 + z2 = 9 2

45. x = 16 − z

42. z2 = x 2 − y 2

46. x 2 + y 2 + z2 = 2z

838

Chapter 11 / Three-Dimensional Space; Vectors

F O C U S O N C O N C E P TS

47–50 Describe the region in 3-space that satisﬁes the given

inequalities. ■ 47. r 2 ≤ z ≤ 4 49. 1 ≤ ρ ≤ 3

(a) Find the cylindrical coordinates of the bug after 2 min. (b) Find the rectangular coordinates of the bug after 2 min. (c) Find the spherical coordinates of the bug after 2 min.

48. 0 ≤ r ≤ 2 sin θ , 0 ≤ z ≤ 3 50. 0 ≤ φ ≤ π/6, 0 ≤ ρ ≤ 2

51. St. Petersburg (Leningrad), Russia, is located at 30 ◦ east longitude and 60 ◦ north latitude. Find its spherical and rectangular coordinates relative to the coordinate axes of Figure 11.8.9. Take miles as the unit of distance and assume the Earth to be a sphere of radius 4000 miles. 52. (a) Show that the curve of intersection of the surfaces z = sin θ and r = a (cylindrical coordinates) is an ellipse. (b) Sketch the surface z = sin θ for 0 ≤ θ ≤ π/2.

53. The accompanying ﬁgure shows a right circular cylinder of radius 10 cm spinning at 3 revolutions per minute about the z-axis. At time t = 0 s, a bug at the point (0, 10, 0) begins walking straight up the face of the cylinder at the rate of 0.5 cm/min.

3 rev/min

(0, 10, 0) x

Figure Ex-53

54. Referring to Exercise 53, use a graphing utility to graph the bug’s distance from the origin as a function of time. 55. Writing Discuss some practical applications in which nonrectangular coordinate systems are useful. 56. Writing The terms “zenith” and “azimuth” are used in celestial navigation. How do these terms relate to spherical coordinates?

✔QUICK CHECK ANSWERS 11.8 1. r cos θ; r sin θ ;√z 2. ρ sin φ cos θ ; ρ sin φ sin θ ; ρ cos φ 3. ρ sin φ; θ; ρ cos θ 4. (a) (2, 7π/4, 2 3) (b) (4, 7π/4, π/6) 5. (a) x 2 + y 2 + z2 = 25 (b) r 2 + z2 = 25 (c) ρ = 5

CHAPTER 11 REVIEW EXERCISES 1. (a) What is the difference between a vector and a scalar? Give a physical example of each. (b) How can you determine whether or not two vectors are orthogonal? (c) How can you determine whether or not two vectors are parallel? (d) How can you determine whether or not three vectors with a common initial point lie in the same plane in 3-space? 2. (a) Sketch vectors u and v for which u + v and u − v are orthogonal. (b) How can you use vectors to determine whether four points in 3-space lie in the same plane? (c) If forces F1 = i and F2 = j are applied at a point in 2-space, what force would you apply at that point to cancel the combined effect of F1 and F2 ? (d) Write an equation of the sphere with center (1, −2, 2) that passes through the origin. 3. (a) Draw a picture that shows the direction angles α, β, and γ of a vector.

(b) What are the components of a unit vector in 2-space that makes an angle of 120 ◦ with the vector i (two answers)? (c) How can you use vectors to determine whether a triangle with known vertices P1 , P2 , and P3 has an obtuse angle? (d) True or false: The cross product of orthogonal unit vectors is a unit vector. Explain your reasoning. 4. (a) Make a table that shows all possible cross products of the vectors i, j, and k. (b) Give a geometric interpretation of u × v . (c) Give a geometric interpretation of |u ⴢ (v × w)|. (d) Write an equation of the plane that passes through the origin and is perpendicular to the line x = t, y = 2t, z = −t. 5. In each part, ﬁnd an equation of the sphere with center (−3, 5, −4) and satisfying the given condition. (a) Tangent to the xy-plane (b) Tangent to the xz-plane (c) Tangent to the yz-plane

Chapter 11 Review Exercises z

6. Find the largest and smallest distances between the point P (1, 1, 1) and the sphere 5 in

x 2 + y 2 + z2 − 2y + 6z − 6 = 0

2 in

7. Given the points P (3, 4), Q(1, 1), and R(5, 2), use vector methods to ﬁnd the coordinates of the fourth vertex of the −→ −→ parallelogram whose adjacent sides are PQ and QR. 8. Let u = 3, 5, −1 and v = 2, −2, 3 . Find 1 (a) 2u + 5v (b) v v (c) u (d) u − v .

9. Let a = ci + j and b = 4i + 3 j. Find c so that (a) a and b are orthogonal (b) the angle between a and b is π/4 (c) the angle between a and b is π/6 (d) a and b are parallel. 10. Let r0 = x0 , y0 , z0 and r = x, y, z . Describe the set of all points (x, y, z) for which (a) r ⴢ r0 = 0 (b) (r − r0 ) ⴢ r0 = 0.

11. Show that if u and v are unit vectors and θ is the angle between them, then u − v = 2 sin 21 θ .

12. Find the vector with length 5 and direction angles α = 60 ◦ , β = 120 ◦ , γ = 135 ◦ .

13. Assuming that force is in pounds and distance is in feet, ﬁnd the work done by a constant force F = 3i − 4 j + k acting on a particle that moves on a straight line from P (5, 7, 0) to Q(6, 6, 6).

14. Assuming that force is in newtons and distance is in meters, ﬁnd the work done by the resultant of the constant forces F1 = i − 3 j + k and F2 = i + 2 j + 2k acting on a particle that moves on a straight line from P (−1, −2, 3) to Q(0, 2, 0). 15. (a) Find the area of the triangle with vertices A(1, 0, 1), B(0, 2, 3), and C(2, 1, 0). (b) Use the result in part (a) to ﬁnd the length of the altitude from vertex C to side AB. 16. True or false? Explain your reasoning. (a) If u ⴢ v = 0, then u = 0 or v = 0. (b) If u × v = 0, then u = 0 or v = 0. (c) If u ⴢ v = 0 and u × v = 0, then u = 0 or v = 0. 17. Consider the points

A(1, −1, 2), B(2, −3, 0), C(−1, −2, 0), D(2, 1, −1) (a) Find the volume of the parallelepiped that has the vec− → −→ −→ tors AB, AC, AD as adjacent edges. (b) Find the distance from D to the plane containing A, B, and C. 18. Suppose that a force F with a magnitude of 9 lb is applied to the lever–shaft assembly shown in the accompanying ﬁgure. (a) Express the force F in component form. (b) Find the vector moment of F about the origin.

839

A F

B x

3 in

1 in

Figure Ex-18

19. Let P be the point (4, 1, 2). Find parametric equations for the line through P and parallel to the vector 1, −1, 0 .

20. (a) Find parametric equations for the intersection of the planes 2x + y − z = 3 and x + 2y + z = 3. (b) Find the acute angle between the two planes.

21. Find an equation of the plane that is parallel to the plane x + 5y − z + 8 = 0 and contains the point (1, 1, 4).

22. Find an equation of the plane through the point (4, 3, 0) and parallel to the vectors i + k and 2j − k.

23. What condition must the constants satisfy for the planes a1 x + b1 y + c1 z = d1

and

a2 x + b2 y + c2 z = d2

to be perpendicular? 24. (a) List six common types of quadric surfaces, and describe their traces in planes parallel to the coordinate planes. (b) Give the coordinates of the points that result when the point (x, y, z) is reﬂected about the plane y = x, the plane y = z, and the plane x = z. (c) Describe the intersection of the surfaces r = 5 and z = 1 in cylindrical coordinates. (d) Describe the intersection of the surfaces φ = π/4 and θ = 0 in spherical coordinates. 25. In each part, identify the surface by completing the squares. (a) x 2 + 4y 2 − z2 − 6x + 8y + 4z = 0 (b) x 2 + y 2 + z2 + 6x − 4y + 12z = 0 (c) x 2 + y 2 − z2 − 2x + 4y + 5 = 0 26. In each part, express the equation in cylindrical and spherical coordinates. (a) x 2 + y 2 = z (b) x 2 − y 2 − z2 = 0 27. In each part, express the equation in rectangular coordinates. (a) z = r 2 cos 2θ (b) ρ 2 sin φ cos φ cos θ = 1 28–29 Sketch the solid in 3-space that is described in cylindrical coordinates by the stated inequalities. ■ 28. (a) 1 ≤ r ≤ 2 (b) 2 ≤ z ≤ 3 (c) π/6 ≤ θ ≤ π/3 (d) 1 ≤ r ≤ 2, 2 ≤ z ≤ 3, and π/6 ≤ θ ≤ π/3

29. (a) r 2 + z2 ≤ 4 (b) r ≤ 1 (c) r 2 + z2 ≤ 4 and r > 1

30–31 Sketch the solid in 3-space that is described in spherical coordinates by the stated inequalities. ■

840

Chapter 11 / Three-Dimensional Space; Vectors

30. (a) 0 ≤ ρ ≤ 2 (b) 0 ≤ φ ≤ π/6 (c) 0 ≤ ρ ≤ 2 and 0 ≤ φ ≤ π/6 31. (a) 0 ≤ ρ ≤ 5, 0 ≤ φ ≤ π/2, and 0 ≤ θ ≤ π/2 (b) 0 ≤ φ ≤ π/3 and 0 ≤ ρ ≤ 2 sec φ (c) 0 ≤ ρ ≤ 2 and π/6 ≤ φ ≤ π/3

32. Sketch the surface whose equation in spherical coordinates is ρ = a(1 − cos φ). [Hint: The surface is shaped like a familiar fruit.]

CHAPTER 11 MAKING CONNECTIONS 1. Deﬁne a “rotation operator” R on vectors in the xy-plane by the formula R(xi + yj) = −yi + xj (a) Verify that R rotates vectors 90 ◦ counterclockwise. (b) Prove that R has the following linearity properties: R(cv) = cR(v)

and

R(v + w) = R(v) + R(w)

2. (a) Given a triangle in the xy-plane, assign to each side of the triangle an outward normal vector whose length is the same as that of the corresponding side. Prove that the sum of the resulting three normal vectors is the zero vector. (b) Extend the result of part (a) to a polygon of n sides in the xy-plane. [Hint: Use the results of the preceding exercise.] 3. (a) Given a tetrahedron in 3-space, assign to each face of the tetrahedron an outward normal vector whose length is numerically the same as the area of the corresponding face. Prove that the sum of the resulting four normal vectors is the zero vector. [Hint: Use cross products.] (b) Extend your result from part (a) to a pyramid with a foursided base. [Hint: Divide the base into two triangles and use the result from part (a) on each of the two resulting tetrahedra.] (c) Can you extend the results of parts (a) and (b) to other polyhedra? 4. Given a tetrahedron in 3-space, pick a vertex and label the three faces that meet at that vertex as A, B, and C. Let a, b, and c denote the respective areas of those faces, and let d denote the area of the fourth face of the tetrahedron. Let

α denote the (internal) angle between faces A and B, β the angle between B and C, and γ the angle between A and C. (a) Prove that d 2 = a 2 + b2 + c2 − 2ab cos α − 2bc cos β − 2ac cos γ This result is sometimes referred to as the law of cosines for a tetrahedron. [Hint: Use the result in part (a) of the preceding exercise.] (b) With the result in part (a) as motivation, state and prove a “Theorem of Pythagoras for a Tetrahedron.” 5. Any circle that lies on a sphere can be realized as the intersection of the sphere and a plane. If the plane passes through the center of the sphere, then the circle is referred to as a great circle. Given two points on a sphere, the great circle distance between the two points is the length of the smallest arc of a great circle that contains both points. Assume that

is a sphere of radius ρ centered at the origin in 3-space. If points P and Q lie on and have spherical coordinates (ρ, θ1 , φ1 ) and (ρ, θ2 , φ2 ), respectively, prove that the great circle distance between P and Q is ρ cos−1 (cos φ1 cos φ2 + cos(θ1 − θ2 ) sin φ1 sin φ2 ) 6. A ship at sea is at point A that is 60 ◦ west longitude and 40 ◦ north latitude. The ship travels to point B that is 40 ◦ west longitude and 20 ◦ north latitude. Assuming that the Earth is a sphere with radius 6370 kilometers, ﬁnd the shortest distance the ship can travel in going from A to B, given that the shortest distance between two points on a sphere is along the arc of the great circle joining the points. [Suggestion: Introduce an xyz-coordinate system as in Figure 11.8.9, and use the result of the preceding exercise.]

The design of a roller coaster requires an understanding of the mathematical principles governing the motion of objects that move with varying speed and direction.

12.1

In this chapter we will consider functions whose values are vectors. Such functions provide a unified way of studying parametric curves in 2-space and 3-space and are a basic tool for analyzing the motion of particles along curved paths. We will begin by developing the calculus of vector-valued functions—we will show how to differentiate and integrate such functions, and we will develop some of the basic properties of these operations. We will then apply these calculus tools to define three fundamental vectors that can be used to describe such basic characteristics of curves as curvature and twisting tendencies. Once this is done, we will develop the concepts of velocity and acceleration for such motion, and we will apply these concepts to explain various physical phenomena. Finally, we will use the calculus of vector-valued functions to develop basic principles of gravitational attraction and to derive Kepler’s laws of planetary motion.

INTRODUCTION TO VECTOR-VALUED FUNCTIONS In Section 11.5 we discussed parametric equations of lines in 3-space. In this section we will discuss more general parametric curves in 3-space, and we will show how vector notation can be used to express parametric equations in 2-space and 3-space in a more compact form. This will lead us to consider a new kind of function—namely, functions that associate vectors with real numbers. Such functions have many important applications in physics and engineering. PARAMETRIC CURVES IN 3-SPACE

Recall from Section 10.1 that if f and g are well-behaved functions, then the pair of parametric equations (1) x = f(t), y = g(t) generates a curve in 2-space that is traced in a speciﬁc direction as the parameter t increases. We deﬁned this direction to be the orientation of the curve or the direction of increasing parameter, and we called the curve together with its orientation the graph of the parametric equations or the parametric curve represented by the equations. Analogously, if f, g, and h are three well-behaved functions, then the parametric equations x = f(t),

y = g(t),

z = h(t)

(2)

generate a curve in 3-space that is traced in a speciﬁc direction as t increases. As in 2space, this direction is called the orientation or direction of increasing parameter, and 841

842

Chapter 12 / Vector-Valued Functions z

(0, 3, 2) (t = 1)

the curve together with its orientation is called the graph of the parametric equations or the parametric curve represented by the equations. If no restrictions are stated explicitly or are implied by the equations, then it will be understood that t varies over the interval (−⬁, +⬁).

(1, 0, 0)

Example 1 The parametric equations

x (t = 0)

x = 1 − t, y = 3t, z = 2t

x = 1 − t,

y = 3t,

z = 2t

represent a line in 3-space that passes through the point (1, 0, 0) and is parallel to the vector −1, 3, 2 . Since x decreases as t increases, the line has the orientation shown in Figure 12.1.1.

Figure 12.1.1

(t = c)

Example 2

Describe the parametric curve represented by the equations x = a cos t,

冸 t = i冹冸 t = 6冹

y = a sin t,

z = ct

where a and c are positive constants. y

O (t = 0) x

x = a cos t, y = a sin t, z = ct Figure 12.1.2

Solution. As the parameter t increases, the value of z = ct also increases, so the point

(x, y, z) moves upward. However, as t increases, the point (x, y, z) also moves in a path directly over the circle x = a cos t, y = a sin t in the xy-plane. The combination of these upward and circular motions produces a corkscrew-shaped curve that wraps around a right circular cylinder of radius a centered on the z-axis (Figure 12.1.2). This curve is called a circular helix.

PARAMETRIC CURVES GENERATED WITH TECHNOLOGY

Except in the simplest cases, parametric curves can be difﬁcult to visualize and draw without the help of a graphing utility. For example, the tricuspoid is the graph of the parametric equations x = 2 cos t + cos 2t, y = 2 sin t − sin 2t Ken Eward/Biografx/Photo Researchers, Inc.

The circular helix described in Example 2 occurs in nature. Above is a computer representation of the twin helix DNA molecule (deoxyribonucleic acid). This structure contains all the inherited instructions necessary for the development of a living organism. T E C H N O LO GY M A ST E R Y If you have a CAS, use it to generate the tricuspoid in Figure 12.1.3, and show that this parametric curve is oriented counterclockwise.

Although it would be tedious to plot the tricuspoid by hand, a computer rendering is easy to obtain and reveals the significance of the name of the curve (Figure 12.1.3). However, note that the depiction of the tricuspoid in Figure12.1.3 is incomplete, since the orientation of the curve is not indicated. This is often the case for curves that are generated with a graphing utility. (Some graphing utilities plot parametric curves slowly enough for the orientation to be discerned, or provide a feature for tracing the points along the curve in the direction of increasing parameter.) Parametric curves in 3-space can be difficult to visualize correctly even with the help of a graphing utility. For example, Figure 12.1.4a shows a parametric curve called a torus knot that was produced with a CAS. However, it is unclear from this computer-generated figure whether the points of overlap are intersections or whether one portion of the curve is in front of the other. To resolve the visualization problem, some graphing utilities provide the capability of enclosing the curve within a thin tube, as in Figure 12.1.4b. Such graphs are called tube plots. PARAMETRIC EQUATIONS FOR INTERSECTIONS OF SURFACES

Curves in 3-space often arise as intersections of surfaces. For example, Figure 12.1.5a shows a portion of the intersection of the cylinders z = x 3 and y = x 2 . One method for ﬁnding parametric equations for the curve of intersection is to choose one of the variables as the parameter and use the two equations to express the remaining two variables in terms

12.1 Introduction to Vector-Valued Functions

843

y 3 2

x −2

−1

−1 −2 −3

z = x3

y=x

The tricuspoid

(a)

Figure 12.1.3

(b)

Figure 12.1.4

of that parameter. In particular, if we choose x = t as the parameter and substitute this into the equations z = x 3 and y = x 2 , we obtain the parametric equations x = t,

z = t3

(3)

This curve is called a twisted cubic. The portion of the twisted cubic shown in Figure 12.1.5a corresponds to t ≥ 0; a computer-generated graph of the twisted cubic for positive and negative values of t is shown in Figure 12.1.5b. Some other examples and techniques for ﬁnding intersections of surfaces are discussed in the exercises.

x = t, y = t 2, z = t 3

(a)

VECTOR-VALUED FUNCTIONS The twisted cubic deﬁned by the equations in (3) is the set of points of the form (t, t 2 , t 3 ) for real values of t. If we view each of these points as a terminal point for a vector r whose initial point is at the origin,

y = t 2,

4 8

r = x, y, z = t, t 2 , t 3 = ti + t 2 j + t 3 k z 0 –2 −8

0 x 2

(b) Figure 12.1.5

then we obtain r as a function of the parameter t, that is, r = r(t). Since this function produces a vector, we say that r = r(t) deﬁnes r as a vector-valued function of a real variable, or more simply, a vector-valued function. The vectors that we will consider in this text are either in 2-space or 3-space, so we will say that a vector-valued function is in 2-space or in 3-space according to the kind of vectors that it produces. If r(t) is a vector-valued function in 3-space, then for each allowable value of t the vector r = r(t) can be represented in terms of components as r = r(t) = x(t), y(t), z(t) = x(t)i + y(t)j + z(t)k

(4)

The functions x(t), y(t), and z(t) are called the component functions or the components of r(t). Whereas a vector-valued function in 3space, such as (4), has three components, a vector-valued function in 2space has only two components and hence has the form

r(t) = x(t), y(t) = x(t)i + y(t)j Find the vector-valued function in 2space whose component functions are x(t) = t and y(t) = t 2 .

Example 3 The component functions of r(t) = t, t 2 , t 3 = ti + t 2 j + t 3 k are

x(t) = t,

y(t) = t 2 ,

z(t) = t 3

The domain of a vector-valued function r(t) is the set of allowable values for t. If r(t) is deﬁned in terms of component functions and the domain is not speciﬁed explicitly, then it will be understood that the domain is the intersection of the natural domains of the component functions; this is called the natural domain of r(t).

844

Chapter 12 / Vector-Valued Functions

Example 4

Find the natural domain of √ √ r(t) = ln |t − 1|, et , t = (ln |t − 1|)i + et j + tk

Solution. The natural domains of the component functions x(t) = ln |t − 1|, are

y(t) = et ,

(−⬁, 1) ∪ (1, +⬁),

z(t) =

(−⬁, +⬁),

√ t

[0, +⬁)

respectively. The intersection of these sets is [0, 1) ∪ (1, +⬁) (verify), so the natural domain of r(t) consists of all values of t such that 0≤t <1

or t > 1

GRAPHS OF VECTOR-VALUED FUNCTIONS

If r(t) is a vector-valued function in 2-space or 3-space, then we deﬁne the graph of r(t) to be the parametric curve described by the component functions for r(t). For example, if r(t) = 1 − t, 3t, 2t = (1 − t)i + 3tj + 2tk

(5)

then the graph of r = r(t) is the graph of the parametric equations x = 1 − t,

y = 3t,

z = 2t

Thus, the graph of (5) is the line in Figure 12.1.1. Strictly speaking, we should write (cos t)i and (sin t)j rather than cos ti and sin tj for clarity. However, it is a common practice to omit the parentheses in such cases, since no misinterpretation is possible. Why?

Example 5

Describe the graph of the vector-valued function r(t) = cos t, sin t, t = cos ti + sin tj + tk

Solution. The corresponding parametric equations are x = cos t,

y = sin t,

z=t

Thus, as we saw in Example 2, the graph is a circular helix wrapped around a cylinder of radius 1.

(x(t), y(t), z(t)) r(t) C

Up to now we have considered parametric curves to be paths traced by moving points. However, if a parametric curve is viewed as the graph of a vector-valued function, then we can also imagine the graph to be traced by the tip of a moving vector. For example, if the curve C in 3-space is the graph of r(t) = x(t)i + y(t)j + z(t)k

x As t varies, the tip of the radius vector r(t) traces out the curve C.

Figure 12.1.6

and if we position r(t) so its initial point is at the origin, then its terminal point will fall on the curve C (as shown in Figure 12.1.6). Thus, when r(t) is positioned with its initial point at the origin, its terminal point will trace out the curve C as the parameter t varies, in which case we call r(t) the radius vector or the position vector for C. For simplicity, we will sometimes let the dependence on t be understood and write r rather than r(t) for a radius vector. Example 6

Sketch the graph and a radius vector of

(a) r(t) = cos ti + sin t j,

0 ≤ t ≤ 2π

(b) r(t) = cos ti + sin t j + 2k,

0 ≤ t ≤ 2π

12.1 Introduction to Vector-Valued Functions y

845

Solution (a). The corresponding parametric equations are x = cos t,

(0 ≤ t ≤ 2π)

so the graph is a circle of radius 1, centered at the origin, and oriented counterclockwise. The graph and a radius vector are shown in Figure 12.1.7.

y = sin t

Solution (b). The corresponding parametric equations are x = cos t,

y = sin t,

z=2

(0 ≤ t ≤ 2π)

From the third equation, the tip of the radius vector traces a curve in the plane z = 2, and from the ﬁrst two equations, the curve is a circle of radius 1 centered at the point (0, 0, 2) and traced counterclockwise looking down the z-axis. The graph and a radius vector are shown in Figure 12.1.8.

r = cos t i + sin t j Figure 12.1.7 z

VECTOR FORM OF A LINE SEGMENT

Recall from Formula (9) of Section 11.5 that if r0 is a vector in 2-space or 3-space with its initial point at the origin, then the line that passes through the terminal point of r0 and is parallel to the vector v can be expressed in vector form as

r = r0 + tv

In particular, if r0 and r1 are vectors in 2-space or 3-space with their initial points at the origin, then the line that passes through the terminal points of these vectors can be expressed in vector form as

r = cos t i + sin t j + 2k

r = r0 + t (r1 − r0 )

Figure 12.1.8

r = (1 − t)r0 + tr1

(6–7)

as indicated in Figure 12.1.9. It is common to call either (6) or (7) the two-point vector form of a line and to say, for simplicity, that the line passes through the points r0 and r1 (as opposed to saying that it passes through the terminal points of r0 and r1 ). It is understood in (6) and (7) that t varies from −⬁ to +⬁. However, if we restrict t to vary over the interval 0 ≤ t ≤ 1, then r will vary from r0 to r1 . Thus, the equation

t(r 1 −r 0)

r0 r r1 O

r = (1 − t)r0 + tr1

(0 ≤ t ≤ 1)

(8)

represents the line segment in 2-space or 3-space that is traced from r0 to r1 .

Figure 12.1.9

✔QUICK CHECK EXERCISES 12.1

(See page 847 for answers.)

1. (a) Express the parametric equations √ 1 x = , y = t, z = sin−1 t t as a single vector equation of the form r = x(t)i + y(t)j + z(t)k

2. Describe the graph of r(t) = 1 + 2t, −1 + 3t .

3. Describe the graph of r(t) = sin2 ti + cos2 tj.

4. Find a vector equation for the curve of intersection of the surfaces y = x 2 and z = y in terms of the parameter x = t.

(b) The vector equation in part (a) deﬁnes r = r(t) as a vector-valued function. The domain of r(t) is and r 21 = .

EXERCISE SET 12.1

Graphing Utility

1–4 Find the domain of r(t) and the value of r(t0 ). ■

1. r(t) = cos ti − 3tj; t0 = π

√ 2. r(t) = 3t + 1, t 2 ; t0 = 1 √ 3. r(t) = cos πti − ln t j + t − 2 k; t0 = 3

Chapter 12 / Vector-Valued Functions

846

4. r(t) = 2e−t , sin−1 t, ln(1 − t) ; t0 = 0

23. r(t) = (1 + cos t)i + (3 − sin t) j; 0 ≤ t ≤ 2π

5–6 Express the parametric equations as a single vector equa-

tion of the form r = x(t)i + y(t) j or r = x(t)i + y(t) j + z(t)k ■

5. x = 3 cos t, y = t + sin t

7–8 Find the parametric equations that correspond to the given 2

7. r = 3t i − 2 j √ 8. r = (2t − 1)i − 3 t j + sin 3tk 10. r = 2 sin 3ti − 2 cos 3t j

12. r = 3i + 2 cos t j + 2 sin tk

28. r(t) = 9 cos ti + 4 sin t j + tk

30. r(t) = ti + tj + sin tk; 0 ≤ t ≤ 2π

32. If r(t) = x(t), y(t) is a vector-valued function in 2-space, then the graph of r(t) is a surface in 3-space. 33. If r0 and r1 are vectors in 3-space, then the graph of the vector-valued function r(t) = (1 − t)r0 + tr1

13. r = 2 cos ti − 3 sin t j + k 2

14. r = −3i + (1 − t )j + tk

15. (a) Find the slope of the line in 2-space that is represented by the vector equation r = (1 − 2t)i − (2 − 3t) j. (b) Find the coordinates of the point where the line r = (2 + t)i + (1 − 2t) j + 3tk

intersects the xz-plane.

16. (a) Find the y-intercept of the line in 2-space that is represented by the vector equation r = (3 + 2t)i + 5tj. (b) Find the coordinates of the point where the line r = ti + (1 + 2t) j − 3tk

intersects the plane 3x − y − z = 2.

17–18 Sketch the line segment represented by each vector

equation. ■

(0 ≤ t ≤ 1)

is the straight line segment joining the terminal points of r0 and r1 . 34. The graph of r(t) = 2 cos t, 2 sin t, t is a circular helix. 35–36 Sketch the curve of intersection of the surfaces, and ﬁnd parametric equations for the intersection in terms of parameter x = t. Check your work with a graphing utility by generating the parametric curve over the interval −1 ≤ t ≤ 1. ■

35. z = x 2 + y 2 , x − y = 0 36. y + x = 0, z = 2 − x 2 − y 2

37–38 Sketch the curve of intersection of the surfaces, and ﬁnd a vector equation for the curve in terms of the parameter x = t. ■

37. 9x 2 + y 2 + 9z2 = 81, y = x 2

17. (a) r = (1 − t)i + tj; 0 ≤ t ≤ 1 (b) r = (1 − t)(i + j) + t (i − j); 0 ≤ t ≤ 1

(z > 0)

38. y = x, x + y + z = 1

18. (a) r = (1 − t)(i + j) + tk; 0 ≤ t ≤ 1 (b) r = (1 − t)(i + j + k) + t (i + j); 0 ≤ t ≤ 1 19–20 Write a vector equation for the line segment from P

to Q. ■

39. Show that the graph of

r = t sin ti + t cos t j + t 2 k lies on the paraboloid z = x 2 + y 2 .

40. Show that the graph of

20.

P 4

r = ti +

t >0

F O C U S O N C O N C E P TS

2 x

21–30 Sketch the graph of r(t) and show the direction of in-

creasing t. ■ 21. r(t) = 2i + tj

1+t 1 − t2 j+ k, t t

lies in the plane x − y + z + 1 = 0. 3 y

27. r(t) = 2 cos ti + 2 sin t j + tk

√ t i + (2t + 4) j

31. The natural domain of a vector-valued function is the union of the domains of its component functions.

9–14 Describe the graph of the equation. ■

19.

26. r(t) =

31–34 True–False Determine whether the statement is true or false. Explain your answer. ■

vector equation. ■

9. r = (3 − 2t)i + 5tj

25. r(t) = cosh ti + sinh t j

29. r(t) = ti + t 2 j + 2k

6. x = 2t, y = 2 sin 3t, z = 5 cos 3t

11. r = 2ti − 3 j + (1 + 3t)k

24. r(t) = 2 cos t, 5 sin t ; 0 ≤ t ≤ 2π

22. r(t) = 3t − 4, 6t + 2

41. Show that the graph of r = sin ti + 2 cos t j +

√

3 sin tk

is a circle, and ﬁnd its center and radius. [Hint: Show that the curve lies on both a sphere and a plane.]

12.1 Introduction to Vector-Valued Functions

42. Show that the graph of r = 3 cos ti + 3 sin t j + 3 sin tk is an ellipse, and ﬁnd the lengths of the major and minor axes. [Hint: Show that the graph lies on both a circular cylinder and a plane and use the result in Exercise 44 of Section 10.4.] 43. For the helix r = a cos ti + a sin t j + ctk, ﬁnd the value of c (c > 0) so that the helix will make one complete turn in a distance of 3 units measured along the z-axis. 44. How many revolutions will the circular helix r = a cos ti + a sin t j + 0.2tk make in a distance of 10 units measured along the z-axis? 45. Show that the curve r = t cos ti + t sin t j + tk, t ≥ 0, lies on the cone z = x 2 + y 2 . Describe the curve.

46. Describe the curve r = a cos ti + b sin t j + ctk, where a, b, and c are positive constants such that a = b. 47. In each part, match the vector equation with one of the accompanying graphs, √ and explain your reasoning. (a) r = ti − tj + 2 − t 2 k (b) r = sin πti − tj + tk (c) r = sin ti + cos t j + sin 2tk (d) r = 21 ti + cos 3t j + sin 3tk z

y x

48. Check your conclusions in Exercise 47 by generating the curves with a graphing utility. [Note: Your graphing utility may look at the curve from a different viewpoint. Read the documentation for your graphing utility to determine how to control the viewpoint, and see if you can generate a reasonable facsimile of the graphs shown in the ﬁgure by adjusting the viewpoint and choosing the interval of t-values appropriately.] 49. (a) Find parametric equations for the curve of intersection of the circular cylinder x 2 + y 2 = 9 and the parabolic cylinder z = x 2 in terms of a parameter t for which x = 3 cos t. (b) Use a graphing utility to generate the curve of intersection in part (a). 50. (a) Sketch the graph of r(t) = 2t,

2 1 + t2

(b) Prove that the curve in part (a) is also the graph of the function 8 y= 4 + x2 [The graphs of y = a 3 /(a 2 + x 2 ), where a denotes a constant, were ﬁrst studied by the French mathematician Pierre de Fermat, and later by the Italian mathematicians Guido Grandi and Maria Agnesi. Any such curve is now known as a “witch of Agnesi.” There are a number of theories for the origin of this name. Some suggest there was a mistranslation by either Grandi or Agnesi of some less colorful Latin name into Italian. Others lay the blame on a translation into English of Agnesi’s 1748 treatise, Analytical Institutions.] 51. Writing Consider the curve C of intersection of the cone z = x 2 + y 2 and the plane z = y + 2. Sketch and identify the curve C, and describe a procedure for ﬁnding a vector-valued function r(t) whose graph is C.

II z

847

52. Writing Suppose that r1 (t) and r2 (t) are vector-valued functions in 2-space. Explain why solving the equation r1 (t) = r2 (t) may not produce all of the points where the graphs of these functions intersect.

III

✔QUICK CHECK ANSWERS 12.1 √ √ 1 π 2 i + tj + sin−1 tk (b) 0 < t ≤ 1; 2i + j + k 2. The graph is a line through (1, −1) with direction vector t 2 6 3. The graph is the line segment in the xy-plane from (0, 1) to (1, 0). 4. r = t, t 2 , t 2

1. (a) r =

2i + 3j.

Chapter 12 / Vector-Valued Functions

848

12.2

CALCULUS OF VECTOR-VALUED FUNCTIONS In this section we will deﬁne limits, derivatives, and integrals of vector-valued functions and discuss their properties. LIMITS AND CONTINUITY

Our ﬁrst goal in this section is to develop a notion of what it means for a vector-valued function r(t) in 2-space or 3-space to approach a limiting vector L as t approaches a number a. That is, we want to deﬁne lim r(t) = L (1)

L r(t)

t →a

One way to motivate a reasonable deﬁnition of (1) is to position r(t) and L with their initial points at the origin and interpret this limit to mean that the terminal point of r(t) approaches the terminal point of L as t approaches a or, equivalently, that the vector r(t) approaches the vector L in both length and direction at t approaches a (Figure 12.2.1). Algebraically, this is equivalent to stating that

r(t) approaches L in length and direction if lim r(t) = L.

lim r(t) − L = 0

(2)

t →a

t→a

(Figure 12.2.2). Thus, we make the following deﬁnition.

Figure 12.2.1 y

r(t) − L

12.2.1 definition Let r(t) be a vector-valued function that is deﬁned for all t in some open interval containing the number a, except that r(t) need not be deﬁned at a. We will write lim r(t) = L

t →a

if and only if

lim r(t) − L = 0

r(t)

t →a

|| r(t) − L || is the distance between terminal points for vectors r(t) and L when positioned with the same initial points.

It is clear intuitively that r(t) will approach a limiting vector L as t approaches a if and only if the component functions of r(t) approach the corresponding components of L. This suggests the following theorem, whose formal proof is omitted.

Figure 12.2.2

12.2.2 Note that r(t) − L is a real number for each value of t , so even though this expression involves a vector-valued function, the limit

lim r(t) − L

t →a

is an ordinary limit of a real-valued function.

theorem

(a) If r(t) = x(t), y(t) = x(t)i + y(t)j, then lim r(t) = lim x(t), lim y(t) = lim x(t)i + lim y(t)j t →a

t →a

provided the limits of the component functions exist. Conversely, the limits of the component functions exist provided r(t) approaches a limiting vector as t approaches a.

(b) If r(t) = x(t), y(t), z(t) = x(t)i + y(t)j + z(t)k, then lim r(t) = lim x(t), lim y(t), lim z(t) t →a

t →a

= lim x(t)i + lim y(t)j + lim z(t)k t →a

t →a

provided the limits of the component functions exist. Conversely, the limits of the component functions exist provided r(t) approaches a limiting vector as t approaches a.

12.2 Calculus of Vector-Valued Functions

How would you define the one-sided limits

lim r(t) and

t →0

t → a−

Limits of vector-valued functions have many of the same properties as limits of real-valued functions. For example, assuming that the limits exist, the limit of a sum is the sum of the limits, the limit of a difference is the difference of the limits, and a constant scalar factor can be moved through a limit symbol.

Let r(t) = t 2 i + et j − (2 cos πt)k. Then

lim r(t) = lim t 2 i + lim et j − lim 2 cos πt k = j − 2 k

Example 1

lim r(t)?

t → a+

849

t →0

Alternatively, using the angle bracket notation for vectors, lim r(t) = lim t 2 , et , −2 cos πt = lim t 2 , lim et , lim (−2 cos πt) = 0, 1, −2 t →0

t →0

Motivated by the deﬁnition of continuity for real-valued functions, we deﬁne a vectorvalued function r(t) to be continuous at t = a if lim r(t) = r(a)

t →a

(3)

That is, r(a) is deﬁned, the limit of r(t) as t → a exists, and the two are equal. As in the case for real-valued functions, we say that r(t) is continuous on an interval I if it is continuous at each point of I [with the understanding that at an endpoint in I the two-sided limit in (3) is replaced by the appropriate one-sided limit]. It follows from Theorem 12.2.2 that a vector-valued function is continuous at t = a if and only if its component functions are continuous at t = a.

r(t + h) − r(t)

r(t +

DERIVATIVES The derivative of a vector-valued function is deﬁned by a limit similar to that for the derivative of a real-valued function. r(t) C x

h>0

12.2.3 definition If r(t) is a vector-valued function, we deﬁne the derivative of r with respect to t to be the vector-valued function r′ given by

(a) y

r′ (t) = lim

r(t + h) − r(t)

h→0

(4)

The domain of r′ consists of all values of t in the domain of r(t) for which the limit exists.

r(t) r(t + h)

C x

h<0

(b) y

r′(t)

r(t) C x

r(t + h) − r(t) h

The function r(t) is differentiable at t if the limit in (4) exists. All of the standard notations for derivatives continue to apply. For example, the derivative of r(t) can be expressed as dr d [r(t)], , r′ (t), or r′ dt dt It is important to keep in mind that r′ (t) is a vector, not a number, and hence has a magnitude and a direction for each value of t [except if r′ (t) = 0, in which case r′ (t) has magnitude zero but no speciﬁc direction]. In the next section we will consider the signiﬁcance of the magnitude of r′ (t), but for now our goal is to obtain a geometric interpretation of the direction of r′ (t). For this purpose, consider parts (a) and (b) of Figure 12.2.3. These illustrations show the graph C of r(t) (with its orientation) and the vectors r(t), r(t + h), and r(t + h) − r(t) for positive h and for negative h. In both cases, the vector r(t + h) − r(t) runs along the secant line joining the terminal points of r(t + h) and r(t), but with opposite directions in the two cases. In the case where h is positive the vector r(t + h) − r(t) points in the direction of increasing parameter, and in the case where h is

850

Chapter 12 / Vector-Valued Functions

negative it points in the opposite direction. However, in the case where h is negative the direction gets reversed when we multiply by 1/h, so in both cases the vector 1 r(t + h) − r(t) [r(t + h) − r(t)] = h h points in the direction of increasing parameter and runs along the secant line. As h → 0, the secant line approaches the tangent line at the terminal point of r(t), so we can conclude that the limit r(t + h) − r(t) r′ (t) = lim h→0 h (if it exists and is nonzero) is a vector that is tangent to the curve C at the tip of r(t) and points in the direction of increasing parameter (Figure 12.2.3c). We can summarize all of this as follows. 12.2.4 geometric interpretation of the derivative Suppose that C is the graph of a vector-valued function r(t) in 2-space or 3-space and that r′ (t) exists and is nonzero for a given value of t. If the vector r′ (t) is positioned with its initial point at the terminal point of the radius vector r(t), then r′ (t) is tangent to C and points in the direction of increasing parameter.

Since limits of vector-valued functions can be computed componentwise, it seems reasonable that we should be able to compute derivatives in terms of component functions as well. This is the result of the next theorem. 12.2.5 theorem If r(t) is a vector-valued function, then r is differentiable at t if and only if each of its component functions is differentiable at t, in which case the component functions of r′ (t) are the derivatives of the corresponding component functions of r(t).

proof For simplicity, we give the proof in 2-space; the proof in 3-space is identical, except for the additional component. Assume that r(t) = x(t)i + y(t)j. Then r′ (t) = lim

h→0

r(t + h) − r(t) h

[x(t + h)i + y(t + h)j] − [x(t)i + y(t)j] h

y(t + h) − y(t) x(t + h) − x(t) i + lim j = lim h→0 h→0 h h

= lim

h→0

= x ′ (t)i + y ′ (t)j ■

Example 2

Let r(t) = t 2 i + et j − (2 cos πt)k. Then r′ (t) =

d 2 d d (t )i + (et )j − (2 cos πt)k dt dt dt

= 2ti + et j + (2π sin πt)k DERIVATIVE RULES

Many of the rules for differentiating real-valued functions have analogs in the context of differentiating vector-valued functions. We state some of these in the following theorem.

12.2 Calculus of Vector-Valued Functions

851

12.2.6 theorem (Rules of Differentiation) Let r(t), r1 (t), and r2 (t) be differentiable vector-valued functions that are all in 2-space or all in 3-space, and let f(t) be a differentiable real-valued function, k a scalar, and c a constant vector (that is, a vector whose value does not depend on t). Then the following rules of differentiation hold: d [c] = 0 dt d d [kr(t)] = k [r(t)] (b) dt dt d d d (c) [r1 (t) + r2 (t)] = [r1 (t)] + [r2 (t)] dt dt dt d d d [r1 (t) − r2 (t)] = [r1 (t)] − [r2 (t)] (d ) dt dt dt d d d [f(t)r(t)] = f(t) [r(t)] + [f(t)]r(t) (e) dt dt dt

(a)

The proofs of most of these rules are immediate consequences of Deﬁnition 12.2.3, although the last rule can be seen more easily by application of the product rule for realvalued functions to the component functions. The proof of Theorem 12.2.6 is left as an exercise.

TANGENT LINES TO GRAPHS OF VECTOR-VALUED FUNCTIONS

Motivated by the discussion of the geometric interpretation of the derivative of a vectorvalued function, we make the following deﬁnition.

r′(t0) P

r(t0)

Tangent line

12.2.7 definition Let P be a point on the graph of a vector-valued function r(t), and let r(t0 ) be the radius vector from the origin to P (Figure 12.2.4). If r′ (t0 ) exists and r′ (t0 ) = 0, then we call r′ (t0 ) a tangent vector to the graph of r(t) at r(t0 ), and we call the line through P that is parallel to the tangent vector the tangent line to the graph of r(t) at r(t0 ).

Figure 12.2.4

Let r0 = r(t0 ) and v0 = r ′ (t0 ). It follows from Formula (9) of Section 11.5 that the tangent line to the graph of r(t) at r0 is given by the vector equation (5)

r = r0 + tv0 Example 3

Find parametric equations of the tangent line to the circular helix x = cos t,

y = sin t,

z=t

where t = t0 , and use that result to ﬁnd parametric equations for the tangent line at the point where t = π.

Solution. The vector equation of the helix is r(t) = cos ti + sin t j + tk

852

Chapter 12 / Vector-Valued Functions

so we have r0 = r(t0 ) = cos t0 i + sin t0 j + t0 k −1

v0 = r ′ (t0 ) = (− sin t0 )i + cos t0 j + k

y 0 1

It follows from (5) that the vector equation of the tangent line at t = t0 is r = cos t0 i + sin t0 j + t0 k + t[(− sin t0 )i + cos t0 j + k]

t=c z 3

= (cos t0 − t sin t0 )i + (sin t0 + t cos t0 )j + (t0 + t)k

Thus, the parametric equations of the tangent line at t = t0 are x = cos t0 − t sin t0 ,

t=0

−1 1

0 x

Figure 12.2.5

y = sin t0 + t cos t0 ,

z = t0 + t

In particular, the tangent line at t = π has parametric equations x = −1,

y = −t,

z = π+t

The graph of the helix and this tangent line are shown in Figure 12.2.5.

Example 4

Let

r1 (t) = (tan−1 t)i + (sin t)j + t 2 k

and r2 (t) = (t 2 − t)i + (2t − 2)j + (ln t)k The graphs of r1 (t) and r2 (t) intersect at the origin. Find the degree measure of the acute angle between the tangent lines to the graphs of r1 (t) and r2 (t) at the origin.

Solution. The graph of r1 (t) passes through the origin at t = 0, where its tangent vector is

r1′ (0) =

1 = 1, 1, 0 , cos t, 2t 2 1+t t=0

The graph of r2 (t) passes through the origin at t = 1 (verify), where its tangent vector is 1 ′ r2 (1) = 2t − 1, 2, = 1, 2, 1 t t=1 By Theorem 11.3.3, the angle θ between these two tangent vectors satisﬁes √ 1, 1, 0 ⴢ 1, 2, 1 1+2+0 3 3 cos θ = = √ √ =√ =

1, 1, 0 1, 2, 1

2 2 6 12

It follows that θ = π/6 radians, or 30 ◦ .

DERIVATIVES OF DOT AND CROSS PRODUCTS The following rules, which are derived in the exercises, provide a method for differentiating dot products in 2-space and 3-space and cross products in 3-space. Note that in (6) the order of the factors in each term on the right does not matter, but in (7) it does.

dr2 dr1 d [r1 (t) ⴢ r2 (t)] = r1 (t) ⴢ + ⴢ r2 (t) dt dt dt

(6)

dr2 dr1 d [r1 (t) × r2 (t)] = r1 (t) × + × r2 (t) dt dt dt

(7)

In plane geometry one learns that a tangent line to a circle is perpendicular to the radius at the point of tangency. Consequently, if a point moves along a circle in 2-space that is centered at the origin, then one would expect the radius vector and the tangent vector at any point on the circle to be orthogonal. This is the motivation for the following useful theorem, which is applicable in both 2-space and 3-space.

12.2 Calculus of Vector-Valued Functions

853

12.2.8 theorem If r(t) is a differentiable vector-valued function in 2-space or 3-space and r(t) is constant for all t, then r(t) â´˘ r â&#x20AC;˛ (t) = 0

â&#x20AC;˛

(8)

that is, r(t) and r (t) are orthogonal vectors for all t.

proof It follows from (6) with r1 (t) = r2 (t) = r(t) that

d dr dr [r(t) â´˘ r(t)] = r(t) â´˘ + â´˘ r(t) dt dt dt

or, equivalently,

d dr [ r(t) 2 ] = 2r(t) â´˘ dt dt

(9)

But r(t) 2 is constant, so its derivative is zero. Thus dr =0 dt

2r(t) â´˘ z

from which (8) follows. â&#x2013;

r'(t)

r(t)

Figure 12.2.6

Example 5 Just as a tangent line to a circle in 2-space is perpendicular to the radius at the point of tangency, so a tangent vector to a curve on the surface of a sphere in 3-space that is centered at the origin is orthogonal to the radius vector at the point of tangency (Figure 12.2.6). To see that this is so, suppose that the graph of r(t) lies on the surface of a sphere of positive radius k centered at the origin. For each value of t we have r(t) = k, so by Theorem 12.2.8 r(t) â´˘ r â&#x20AC;˛ (t) = 0 and hence the radius vector r(t) and the tangent vector r â&#x20AC;˛ (t) are orthogonal.

DEFINITE INTEGRALS OF VECTOR-VALUED FUNCTIONS

If r(t) is a vector-valued function that is continuous on the interval a â&#x2030;¤ t â&#x2030;¤ b, then we deďŹ ne the deďŹ nite integral of r(t) over this interval as a limit of Riemann sums, just as in DeďŹ nition 5.5.1, except here the integrand is a vector-valued function. SpeciďŹ cally, we deďŹ ne b n r(t) dt = lim r(tkâ&#x2C6;&#x2014; ) tk (10) max tk â&#x2020;&#x2019; 0

k=1

It follows from (10) that the deďŹ nite integral of r(t) over the interval a â&#x2030;¤ t â&#x2030;¤ b can be expressed as a vector whose components are the deďŹ nite integrals of the component functions of r(t). For example, if r(t) = x(t)i + y(t)j, then b n r(t) dt = lim r(tkâ&#x2C6;&#x2014; ) tk a

max tk â&#x2020;&#x2019; 0

k=1

lim

max tk â&#x2020;&#x2019; 0

lim

max tk â&#x2020;&#x2019; 0

x(tkâ&#x2C6;&#x2014; ) tk

k=1

x(tkâ&#x2C6;&#x2014; ) tk

k=1

x(t) dt i +

y(tkâ&#x2C6;&#x2014; ) tk

k=1

lim

max tk â&#x2020;&#x2019; 0

y(t) dt j

n k=1

y(tkâ&#x2C6;&#x2014; ) tk

854

Chapter 12 / Vector-Valued Functions

Rewrite Formulas (11) and (12) in bracket notation with r(t) = x(t), y(t) and r(t) = x(t), y(t), z(t) respectively.

In general, we have b r(t) dt = a

b a

r(t) dt =

Example 6 1 0

x(t) dt i +

y(t) dt j

y(t) dt j +

z(t) dt k a

Let r(t) = t 2 i + et j â&#x2C6;&#x2019; (2 cos Ď&#x20AC;t)k. Then 1

r(t) dt = t 2 dt i + et dt j â&#x2C6;&#x2019; 0

3 1

t 3

i + et

1 0

jâ&#x2C6;&#x2019;

2 sin Ď&#x20AC;t Ď&#x20AC;

1 0

2-space

(11)

3-space

(12)

2 cos Ď&#x20AC;t dt k

1 i + (e â&#x2C6;&#x2019; 1)j 3

RULES OF INTEGRATION As with differentiation, many of the rules for integrating real-valued functions have analogs for vector-valued functions.

12.2.9 theorem (Rules of Integration) Let r(t), r1 (t), and r2 (t) be vector-valued functions in 2-space or 3-space that are continuous on the interval a â&#x2030;¤ t â&#x2030;¤ b, and let k be a scalar. Then the following rules of integration hold: b b r(t) dt kr(t) dt = k (a) a

(b)

(c)

[r1 (t) + r2 (t)] dt =

[r1 (t) â&#x2C6;&#x2019; r2 (t)] dt =

r1 (t) dt +

r1 (t) dt â&#x2C6;&#x2019;

r2 (t) dt a b

r2 (t) dt a

We omit the proof. ANTIDERIVATIVES OF VECTOR-VALUED FUNCTIONS An antiderivative for a vector-valued function r(t) is a vector-valued function R(t) such that Râ&#x20AC;˛ (t) = r(t) (13)

As in Chapter 5, we express Equation (13) using integral notation as r(t) dt = R(t) + C

(14)

where C represents an arbitrary constant vector. Since differentiation of vector-valued functions can be performed componentwise, it follows that antidifferentiation can be done this way as well. This is illustrated in the next example.

12.2 Calculus of Vector-Valued Functions

855

Example 7

(2ti + 3t 2 j) dt = 2t dt i + 3t 2 dt j = (t 2 + C1 )i + (t 3 + C2 )j

= (t 2 i + t 3 j) + (C1 i + C2 j) = (t 2 i + t 3 j) + C

where C = C1 i + C2 j is an arbitrary vector constant of integration. Most of the familiar integration properties have vector counterparts. For example, vector differentiation and integration are inverse operations in the sense that d r(t) dt = r(t) and râ&#x20AC;˛ (t) dt = r(t) + C (15â&#x20AC;&#x201C;16) dt Moreover, if R(t) is an antiderivative of r(t) on an interval containing t = a and t = b, then we have the following vector form of the Fundamental Theorem of Calculus: Example 8

b a

b r(t) dt = R(t) = R(b) â&#x2C6;&#x2019; R(a)

(17)

Evaluate the deďŹ nite integral

2 0

(2ti + 3t 2 j) dt.

Solution. Integrating the components yields

2 0

(2ti + 3t 2 j) dt = t 2

2 0

i + t3

2 0

j = 4i + 8j

Alternative Solution. The function R(t) = t 2 i + t 3 j is an antiderivative of the integrand since Râ&#x20AC;˛ (t) = 2ti + 3t 2 j. Thus, it follows from (17) that 2 2 2 2 2 3 (2ti + 3t j) dt = R(t) = t i + t j = (4i + 8j) â&#x2C6;&#x2019; (0i + 0j) = 4i + 8j 0

Example 9

Find r(t) given that râ&#x20AC;˛ (t) = 3, 2t and r(1) = 2, 5 .

Solution. Integrating râ&#x20AC;˛ (t) to obtain r(t) yields r(t) =

râ&#x20AC;˛ (t) dt =

3, 2t dt = 3t, t 2 + C

where C is a vector constant of integration. To ďŹ nd the value of C we substitute t = 1 and use the given value of r(1) to obtain r(1) = 3, 1 + C = 2, 5 so that C = â&#x2C6;&#x2019;1, 4 . Thus,

r(t) = 3t, t 2 + â&#x2C6;&#x2019;1, 4 = 3t â&#x2C6;&#x2019; 1, t 2 + 4

856

Chapter 12 / Vector-Valued Functions

â&#x153;&#x201D;QUICK CHECK EXERCISES 12.2

(See page 858 for answers.)

1. (a) lim (t 2 i + 2tj) = t â&#x2020;&#x2019;3

(b) lim cos t, sin t = t â&#x2020;&#x2019; Ď&#x20AC;/4

2. Find râ&#x20AC;˛ (t). (a) r(t) = (4 + 5t)i + (t â&#x2C6;&#x2019; t 2 )j 1 (b) r(t) = , tan t, e2t t 3. Suppose that r1 (0) = 3, 2, 1 , r2 (0) = 1, 2, 3 , r1â&#x20AC;˛ (0) = 0, 0, 0 , and r2â&#x20AC;˛ (0) = â&#x2C6;&#x2019;6, â&#x2C6;&#x2019;4, â&#x2C6;&#x2019;2 . Use this in-

EXERCISE SET 12.2

Graphing Utility

1â&#x20AC;&#x201C;4 Find the limit. â&#x2013;

2. lim+

3. lim (ti â&#x2C6;&#x2019; 3j + t 2 k)

4. lim

1. lim

t â&#x2020;&#x2019; +âŹ

t +1 1 , 3t 2 + 2 t

t â&#x2020;&#x2019;2

t â&#x2020;&#x2019;0

t â&#x2020;&#x2019;1

â&#x2C6;&#x161; sin t ti+ j t

3 ln t , , sin 2t t2 t2 â&#x2C6;&#x2019; 1

5â&#x20AC;&#x201C;6 Determine whether r(t) is continuous at t = 0. Explain

your reasoning. â&#x2013;

(b) r(t) = t 2 i +

5. (a) r(t) = 3 sin ti â&#x2C6;&#x2019; 2tj

1 j + tk t

6. (a) r(t) = e i + j + csc tk â&#x2C6;&#x161; (b) r(t) = 5i â&#x2C6;&#x2019; 3t + 1 j + e2t k 7. Sketch the circle r(t) = cos ti + sin tj, and in each part draw the vector with its correct length. (a) r â&#x20AC;˛ (Ď&#x20AC;/4) (b) r â&#x20AC;˛â&#x20AC;˛ (Ď&#x20AC;) (c) r(2Ď&#x20AC;) â&#x2C6;&#x2019; r(3Ď&#x20AC;/2) 8. Sketch the circle r(t) = cos ti â&#x2C6;&#x2019; sin tj, and in each part draw the vector with its correct length. (a) r â&#x20AC;˛ (Ď&#x20AC;/4) (b) r â&#x20AC;˛â&#x20AC;˛ (Ď&#x20AC;) (c) r(2Ď&#x20AC;) â&#x2C6;&#x2019; r(3Ď&#x20AC;/2) 9. r(t) = 4i â&#x2C6;&#x2019; cos tj 10. r(t) = (tan

t)i + t cos t j â&#x2C6;&#x2019;

â&#x2C6;&#x161;

19â&#x20AC;&#x201C;22 Find parametric equations of the line tangent to the graph of r(t) at the point where t = t0 . â&#x2013;

19. r(t) = t 2 i + (2 â&#x2C6;&#x2019; ln t) j; t0 = 1 20. r(t) = e2t i â&#x2C6;&#x2019; 2 cos 3t j; t0 = 0

21. r(t) = 2 cos Ď&#x20AC;ti + 2 sin Ď&#x20AC;tj + 3tk; t0 = â&#x2C6;&#x2019;t

1 3

22. r(t) = ln ti + e j + t k; t0 = 2

23â&#x20AC;&#x201C;26 Find a vector equation of the line tangent to the graph of r(t) at the point P0 on the curve. â&#x2013; â&#x2C6;&#x161; 23. r(t) = (2t â&#x2C6;&#x2019; 1)i + 3t + 4 j; P0 (â&#x2C6;&#x2019;1, 2)

24. r(t) = 4 cos ti â&#x2C6;&#x2019; 3tj; P0 (2, â&#x2C6;&#x2019;Ď&#x20AC;) 1 25. r(t) = t 2 i â&#x2C6;&#x2019; j + (4 â&#x2C6;&#x2019; t 2 )k; P0 (4, 1, 0) t +1 26. r(t) = sin ti + cosh tj + (tanâ&#x2C6;&#x2019;1 t)k; P0 (0, 1, 0) t â&#x2020;&#x2019;0

11â&#x20AC;&#x201C;14 Find the vector r â&#x20AC;˛ (t0 ); then sketch the graph of r(t) in

2-space and draw the tangent vector r â&#x20AC;˛ (t0 ). â&#x2013; 11. r(t) = t, t 2 ; t0 = 2

17â&#x20AC;&#x201C;18 Use a graphing utility to generate the graph of r(t) and the graph of the tangent line at t0 on the same screen. â&#x2013; 17. r(t) = sin Ď&#x20AC;ti + t 2 j; t0 = 21 18. r(t) = 3 sin ti + 4 cos tj; t0 = Ď&#x20AC;/4

27. Let r(t) = cos ti + sin tj + k. Find (a) lim (r(t) â&#x2C6;&#x2019; r â&#x20AC;˛ (t)) (b) lim (r(t) Ă&#x2014; r â&#x20AC;˛ (t))

9â&#x20AC;&#x201C;10 Find r â&#x20AC;˛ (t). â&#x2013; â&#x2C6;&#x2019;1

formation to evaluate the derivative of each function at t = 0. (a) r(t) = 2r1 (t) â&#x2C6;&#x2019; r2 (t) (b) r(t) = cos tr1 (t) + e2t r2 (t) (c) r(t) = r1 (t) Ă&#x2014; r2 (t) (d) f(t) = r1 (t) â´˘ r2 (t) 1 4. (a) 2t, t 2 , sin Ď&#x20AC;t dt = 0 (b) (ti â&#x2C6;&#x2019; 3t 2 j + et k) dt =

12. r(t) = t 3 i + t 2 j; t0 = 1

13. r(t) = sec ti + tan tj; t0 = 0

14. r(t) = 2 sin ti + 3 cos tj; t0 = Ď&#x20AC;/6 15â&#x20AC;&#x201C;16 Find the vector r â&#x20AC;˛ (t0 ); then sketch the graph of r(t) in

3-space and draw the tangent vector r â&#x20AC;˛ (t0 ). â&#x2013; 15. r(t) = 2 sin ti + j + 2 cos tk; t0 = Ď&#x20AC;/2 16. r(t) = cos ti + sin t j + tk; t0 = Ď&#x20AC;/4

t â&#x2020;&#x2019;0

28. Let r(t) = ti + t 2 j + t 3 k. Find lim r(t) â´˘ (râ&#x20AC;˛ (t) Ă&#x2014; r â&#x20AC;˛â&#x20AC;˛ (t)) t â&#x2020;&#x2019;1

29â&#x20AC;&#x201C;30 Calculate

d d [r1 (t) â´˘ r2 (t)] and [r1 (t) Ă&#x2014; r2 (t)] dt dt ďŹ rst by differentiating the product directly and then by applying Formulas (6) and (7). â&#x2013; 29. r1 (t) = 2ti + 3t 2 j + t 3 k, r2 (t) = t 4 k

30. r1 (t) = cos ti + sin tj + tk, r2 (t) = i + tk

31â&#x20AC;&#x201C;34 Evaluate the indeďŹ nite integral. â&#x2013;

12.2 Calculus of Vector-Valued Functions

31. 33.

(3i + 4tj) dt

32.

tet , ln t dt

34.

1 t 2 i â&#x2C6;&#x2019; 2tj + k t

eâ&#x2C6;&#x2019;t , et , 3t 2 dt

35â&#x20AC;&#x201C;40 Evaluate the deďŹ nite integral. â&#x2013;

35. 37. 38.

Ď&#x20AC;/2

cos 2t, sin 2t dt

(t 2 i + t 3 j) dt

ti + t 2 j dt

â&#x2C6;&#x2019;3 9

39.

36.

i+t

â&#x2C6;&#x2019;1/2

51â&#x20AC;&#x201C;52 Show that the graphs of r1 (t) and r2 (t) intersect at the point P . Find, to the nearest degree, the acute angle between the tangent lines to the graphs of r1 (t) and r2 (t) at the point P . â&#x2013; 51. r1 (t) = t 2 i + tj + 3t 3 k r2 (t) = (t â&#x2C6;&#x2019; 1)i + 41 t 2 j + (5 â&#x2C6;&#x2019; t)k; P (1, 1, 3)

52. r1 (t) = 2eâ&#x2C6;&#x2019;t i + cos tj + (t 2 + 3)k r2 (t) = (1 â&#x2C6;&#x2019; t)i + t 2 j + (t 3 + 4)k; P (2, 1, 3) F O C U S O N C O N C E P TS

(3 â&#x2C6;&#x2019; t)3/2 , (3 + t)3/2 , 1 dt 1/2

857

j) dt

40.

1 0

(e2t i + eâ&#x2C6;&#x2019;t j + tk) dt

41â&#x20AC;&#x201C;44 Trueâ&#x20AC;&#x201C;False Determine whether the statement is true or

false. Explain your answer. â&#x2013; 41. If a vector-valued function r(t) is continuous at t = a, then the limit r(a + h) â&#x2C6;&#x2019; r(a) lim hâ&#x2020;&#x2019;0 h exists. 42. If r(t) is a vector-valued function in 2-space and r(t) is constant, then r(t) and râ&#x20AC;˛ (t) are parallel vectors for all t. 43. If r(t) is a vector-valued function that is continuous on the interval a â&#x2030;¤ t â&#x2030;¤ b, then b r(t) dt a

is a vector.

44. If r(t) is a vector-valued function that is continuous on the interval [a, b], then for a < t < b, t d r(u) du = r(t) dt a

53. Use Formula (7) to derive the differentiation formula d [r(t) Ă&#x2014; r â&#x20AC;˛ (t)] = r(t) Ă&#x2014; r â&#x20AC;˛â&#x20AC;˛ (t) dt 54. Let u = u(t), v = v(t), and w = w(t) be differentiable vector-valued functions. Use Formulas (6) and (7) to show that d [u â´˘ (v Ă&#x2014; w)] dt dw dv du â´˘ [v Ă&#x2014; w] + u â´˘ Ă&#x2014;w +uâ´˘ vĂ&#x2014; = dt dt dt 55. Let u1 , u2 , u3 , v1 , v2 , v3 , w1 , w2 , and w3 be differentiable functions of t. Use Exercise 54 to show that u1 u2 u3 d v1 v2 v3 dt w w w 3 2 1 â&#x20AC;˛ u uâ&#x20AC;˛ uâ&#x20AC;˛ u1 u2 u3 u1 u2 u3 2 3 â&#x20AC;˛ 1 = v1 v2 v3 + v1 v2â&#x20AC;˛ v3â&#x20AC;˛ + v1 v2 v3 w1 w2 w3 w1 w2 w3 w â&#x20AC;˛ w â&#x20AC;˛ w â&#x20AC;˛ 1

56. Prove Theorem 12.2.6 for 2-space.

57. Derive Formulas (6) and (7) for 3-space. 58. Prove Theorem 12.2.9 for 2-space.

45â&#x20AC;&#x201C;48 Solve the vector initial-value problem for y(t) by inte-

grating and using the initial conditions to ďŹ nd the constants of integration. â&#x2013; 45. yâ&#x20AC;˛ (t) = 2ti + 3t 2 j, y(0) = i â&#x2C6;&#x2019; j

46. yâ&#x20AC;˛ (t) = cos ti + sin tj, y(0) = i â&#x2C6;&#x2019; j

47. yâ&#x20AC;˛â&#x20AC;˛ (t) = i + et j, y(0) = 2i, yâ&#x20AC;˛ (0) = j

48. yâ&#x20AC;˛â&#x20AC;˛ (t) = 12t 2 i â&#x2C6;&#x2019; 2tj, y(0) = 2i â&#x2C6;&#x2019; 4j, yâ&#x20AC;˛ (0) = 0 49. (a) Find the points where the curve

r = ti + t 2 j â&#x2C6;&#x2019; 3tk

intersects the plane 2x â&#x2C6;&#x2019; y + z = â&#x2C6;&#x2019;2. (b) For the curve and plane in part (a), ďŹ nd, to the nearest degree, the acute angle that the tangent line to the curve makes with a line normal to the plane at each point of intersection. 50. Find where the tangent line to the curve r = eâ&#x2C6;&#x2019;2t i + cos tj + 3 sin tk at the point (1, 1, 0) intersects the yz-plane.

59. Writing Explain what it means for a vector-valued function r(t) to be differentiable, and discuss geometric interpretations of râ&#x20AC;˛ (t). 60. Writing Let r(t) = t 2 , t 3 + 1 and deďŹ ne Î¸(t) to be the angle between r(t) and râ&#x20AC;˛ (t). The graph of Î¸ = Î¸(t) is shown in the accompanying ďŹ gure. Interpret important features of this graph in terms of information about r(t) and râ&#x20AC;˛ (t). Accompany your discussion with a graph of r(t), highlighting particular instances of the vectors r(t) and râ&#x20AC;˛ (t). u c 9 6 3 â&#x2C6;&#x2019;4 â&#x2C6;&#x2019;3 â&#x2C6;&#x2019;2 â&#x2C6;&#x2019;1

t 1

Figure Ex-60

858

Chapter 12 / Vector-Valued Functions

✔QUICK CHECK ANSWERS 12.2 √

√ 1 2 2 2. (a) r′ (t) = 5i + (1 − 2t)j (b) r′ (t) = − 2 , sec2 t, 2e2t , 1. (a) 9i + 6j (b) 2 2 t t2 1 2 i − t 3 j + et k + C (b) (c) 0 (d) −28 4. (a) 1, , 3 π 2

12.3

3. (a) 6, 4, 2 (b) −4, 0, 4

CHANGE OF PARAMETER; ARC LENGTH We observed in earlier sections that a curve in 2-space or 3-space can be represented parametrically in more than one way. For example, in Section 10.1 we gave two parametric representations of a circle—one in which the circle was traced clockwise and the other in which it was traced counterclockwise. Sometimes it will be desirable to change the parameter for a parametric curve to a different parameter that is better suited for the problem at hand. In this section we will investigate issues associated with changes of parameter, and we will show that arc length plays a special role in parametric representations of curves.

SMOOTH PARAMETRIZATIONS

Graphs of vector-valued functions range from continuous and smooth to discontinuous and wildly erratic. In this text we will not be concerned with graphs of the latter type, so we will need to impose restrictions to eliminate the unwanted behavior. We will say that a curve represented by r(t) is smoothly parametrized by r(t), or that r(t) is a smooth function of t if r ′ (t) is continuous and r ′ (t) = 0 for any allowable value of t. Geometrically, this means that a smoothly parametrized curve can have no abrupt changes in direction as the parameter increases. Mathematically, “smoothness” is a property of the parametrization and not of the curve itself. Exercise 38 gives an example of a curve that is wellbehaved geometrically and has one parametrization that is smooth and another that is not.

Example 1

Determine whether the following vector-valued functions are smooth.

(a) r(t) = a cos ti + a sin tj + ctk (b) r(t) = t 2 i + t 3 j

(a > 0, c > 0)

Solution (a). We have r ′ (t) = −a sin ti + a cos tj + ck

The components are continuous functions, and there is no value of t for which all three of them are zero (verify), so r(t) is a smooth function. The graph of r(t) is the circular helix in Figure 12.1.2.

Solution (b). We have

r(t) = t 2 i + t 3j Figure 12.3.1

r ′ (t) = 2ti + 3t 2 j

Although the components are continuous functions, they are both equal to zero if t = 0, so r(t) is not a smooth function. The graph of r(t), which is shown in Figure 12.3.1, is a semicubical parabola traced in the upward direction (see Example 6 of Section 10.1). Observe that for values of t slightly less than zero the angle between r ′ (t) and i is near π, and for values of t slightly larger than zero the angle is near 0; hence there is a sudden reversal in the direction of the tangent vector as t increases through t = 0 (see Exercise 44).

12.3 Change of Parameter; Arc Length

859

ARC LENGTH FROM THE VECTOR VIEWPOINT

Recall from Theorem 10.1.1 that the arc length L of a parametric curve x = x(t),

y = y(t)

(1)

(a â&#x2030;¤ t â&#x2030;¤ b)

is given by the formula

dx dt

dy dt

(2)

(a â&#x2030;¤ t â&#x2030;¤ b)

(3)

Analogously, the arc length L of a parametric curve x = x(t),

y = y(t),

z = z(t)

in 3-space is given by the formula

dx dt

dy dt

dz dt

(4)

Formulas (2) and (4) have vector forms that we can obtain by letting or r(t) = x(t)i + y(t) j + z(t)k

r(t) = x(t)i + y(t)j 2-space

3-space

It follows that dx dy dr = i+ j or dt dt dt

dr dx dy dz = i+ j+ k dt dt dt dt

2-space

and hence 2 2 dr dy dx = + dt dt dt

3-space

2 2 2 dr dy dz dx = + + dt dt dt dt

2-space

3-space

Substituting these expressions in (2) and (4) leads us to the following theorem.

12.3.1 theorem If C is the graph in 2-space or 3-space of a smooth vector-valued function r(t), then its arc length L from t = a to t = b is L=

Example 2

dr dt dt

(5)

Find the arc length of that portion of the circular helix x = cos t,

y = sin t,

z=t

from t = 0 to t = Ď&#x20AC;.

Solution. Set r(t) = (cos t)i + (sin t)j + tk = cos t, sin t, t . Then râ&#x20AC;˛ (t) = â&#x2C6;&#x2019; sin t, cos t, 1

and râ&#x20AC;˛ (t) =

â&#x2C6;&#x161; (â&#x2C6;&#x2019; sin t)2 + (cos t)2 + 1 = 2

860

Chapter 12 / Vector-Valued Functions

From Theorem 12.3.1 the arc length of the helix is Ď&#x20AC; Ď&#x20AC;â&#x2C6;&#x161; â&#x2C6;&#x161; dr L= 2 dt = 2Ď&#x20AC; dt dt = 0 0 ARC LENGTH AS A PARAMETER

For many purposes the best parameter to use for representing a curve in 2-space or 3-space parametrically is the length of arc measured along the curve from some ďŹ xed reference point. This can be done as follows:

Using Arc Length as a Parameter s=3

Step 1. Select an arbitrary point on the curve C to serve as a reference point. Step 2. Starting from the reference point, choose one direction along the curve to be the positive direction and the other to be the negative direction.

s=2

ire

cti

Step 3. If P is a point on the curve, let s be the â&#x20AC;&#x153;signedâ&#x20AC;? arc length along C from the reference point to P , where s is positive if P is in the positive direction from the reference point and s is negative if P is in the negative direction. Figure 12.3.2 illustrates this idea.

s = â&#x2C6;&#x2019;2

â&#x2C6;&#x2019;d

ire

Reference point

s = â&#x2C6;&#x2019;1

s=1

By this procedure, a unique point P on the curve is determined when a value for s is given. For example, s = 2 determines the point that is 2 units along the curve in the positive direction from the reference point, and s = â&#x2C6;&#x2019; 23 determines the point that is 23 units along the curve in the negative direction from the reference point. Let us now treat s as a variable. As the value of s changes, the corresponding point P moves along C and the coordinates of P become functions of s. Thus, in 2-space the coordinates of P are (x(s), y(s)), and in 3-space they are (x(s), y(s), z(s)). Therefore, in 2-space or 3-space the curve C is given by the parametric equations

s = â&#x2C6;&#x2019;3 C Figure 12.3.2

x = x(s),

y = y(s)

or x = x(s),

y = y(s),

z = z(s)

A parametric representation of a curve with arc length as the parameter is called an arc length parametrization of the curve. Note that a given curve will generally have inďŹ nitely many different arc length parametrizations, since the reference point and orientation can be chosen arbitrarily. Example 3 Find the arc length parametrization of the circle x 2 + y 2 = a 2 with counterclockwise orientation and (a, 0) as the reference point.

Solution. The circle with counterclockwise orientation can be represented by the para-

metric equations P(x, y) s t

(a, 0)

x = a cos t,

y = a sin t

(0 â&#x2030;¤ t â&#x2030;¤ 2Ď&#x20AC;)

(6)

in which t can be interpreted as the angle in radian measure from the positive x-axis to the radius from the origin to the point P (x, y) (Figure 12.3.3). If we take the positive direction for measuring the arc length to be counterclockwise, and we take (a, 0) to be the reference point, then s and t are related by s = at

or t = s /a

Making this change of variable in (6) and noting that s increases from 0 to 2Ď&#x20AC;a as t increases from 0 to 2Ď&#x20AC; yields the following arc length parametrization of the circle: Figure 12.3.3

x = a cos(s /a),

y = a sin(s /a)

(0 â&#x2030;¤ s â&#x2030;¤ 2Ď&#x20AC;a)

12.3 Change of Parameter; Arc Length

861

CHANGE OF PARAMETER

In many situations the solution of a problem can be simpliﬁed by choosing the parameter in a vector-valued function or a parametric curve in the right way. The two most common parameters for curves in 2-space or 3-space are time and arc length. However, there are other useful possibilities as well. For example, in analyzing the motion of a particle in 2-space, it is often desirable to parametrize its trajectory in terms of the angle φ between the tangent vector and the positive x-axis (Figure 12.3.4). Thus, our next objective is to develop methods for changing the parameter in a vector-valued function or parametric curve. This will allow us to move freely between different possible parametrizations. y

r(f)

r(s)

r(t) s

Time as parameter

f as parameter

Arc length as parameter

Figure 12.3.4

A change of parameter in a vector-valued function r(t) is a substitution t = g(τ ) that produces a new vector-valued function r(g(τ )) having the same graph as r(t), but possibly traced differently as the parameter τ increases. Example 4

Find a change of parameter t = g(τ ) for the circle r(t) = cos ti + sin tj

(0 ≤ t ≤ 2π)

such that (a) the circle is traced counterclockwise as τ increases over the interval [0, 1];

(b) the circle is traced clockwise as τ increases over the interval [0, 1].

Solution (a). The given circle is traced counterclockwise as t increases. Thus, if we

choose g to be an increasing function, then it will follow from the relationship t = g(τ ) that t increases when τ increases, thereby ensuring that the circle will be traced counterclockwise as τ increases. We also want to choose g so that t increases from 0 to 2π as τ increases from 0 to 1. A simple choice of g that satisﬁes all of the required criteria is the linear function graphed in Figure 12.3.5a. The equation of this line is

t = g(τ ) = 2πτ

(7)

which is the desired change of parameter. The resulting representation of the circle in terms of the parameter τ is t

t = 2ct

t = 2 c (1 − t)

(a)

(b)

Figure 12.3.5

r(g(τ )) = cos 2πτ i + sin 2πτ j

(0 ≤ τ ≤ 1)

Solution (b). To ensure that the circle is traced clockwise, we will choose g to be a decreasing function such that t decreases from 2π to 0 as τ increases from 0 to 1. A simple choice of g that achieves this is the linear function t = g(τ ) = 2π(1 − τ )

(8)

862

Chapter 12 / Vector-Valued Functions

graphed in Figure 12.3.5b. The resulting representation of the circle in terms of the parameter Ď&#x201E; is r(g(Ď&#x201E; )) = cos(2Ď&#x20AC;(1 â&#x2C6;&#x2019; Ď&#x201E; ))i + sin(2Ď&#x20AC;(1 â&#x2C6;&#x2019; Ď&#x201E; )) j (0 â&#x2030;¤ Ď&#x201E; â&#x2030;¤ 1) which simpliďŹ es to (verify) r(g(Ď&#x201E; )) = cos 2Ď&#x20AC;Ď&#x201E; i â&#x2C6;&#x2019; sin 2Ď&#x20AC;Ď&#x201E; j

(0 â&#x2030;¤ Ď&#x201E; â&#x2030;¤ 1)

When making a change of parameter t = g(Ď&#x201E; ) in a vector-valued function r(t), it will be important to ensure that the new vector-valued function r(g(Ď&#x201E; )) is smooth if r(t) is smooth. To establish conditions under which this happens, we will need the following version of the chain rule for vector-valued functions. The proof is left as an exercise. Strictly speaking, since dr/dt is a vector and dt /dĎ&#x201E; is a scalar, Formula (9) should be written in the form

dr dt dr = dĎ&#x201E; dĎ&#x201E; dt (scalar first). However, reversing the order of the factors makes the formula easier to remember, and we will continue to do so.

12.3.2 theorem (Chain Rule) Let r(t) be a vector-valued function in 2-space or 3space that is differentiable with respect to t. If t = g(Ď&#x201E; ) is a change of parameter in which g is differentiable with respect to Ď&#x201E;, then r(g(Ď&#x201E; )) is differentiable with respect to Ď&#x201E; and dr dt dr = (9) dĎ&#x201E; dt dĎ&#x201E;

A change of parameter t = g(Ď&#x201E; ) in which r(g(Ď&#x201E; )) is smooth if r(t) is smooth is called a smooth change of parameter. It follows from (9) that t = g(Ď&#x201E; ) will be a smooth change of parameter if dt /dĎ&#x201E; is continuous and dt /dĎ&#x201E; = 0 for all values of Ď&#x201E;, since these conditions imply that dr/dĎ&#x201E; is continuous and nonzero if dr/dt is continuous and nonzero. Smooth changes of parameter fall into two categoriesâ&#x20AC;&#x201D;those for which dt /dĎ&#x201E; > 0 for all Ď&#x201E; (called positive changes of parameter) and those for which dt /dĎ&#x201E; < 0 for all Ď&#x201E; (called negative changes of parameter). A positive change of parameter preserves the orientation of a parametric curve, and a negative change of parameter reverses it. Example 5 In Example 4 the change of parameter in Formula (7) is positive since dt /dĎ&#x201E; = 2Ď&#x20AC; > 0, and the change of parameter given by Formula (8) is negative since dt /dĎ&#x201E; = â&#x2C6;&#x2019;2Ď&#x20AC; < 0. The positive change of parameter preserved the orientation of the circle, and the negative change of parameter reversed it.

FINDING ARC LENGTH PARAMETRIZATIONS

Next we will consider the problem of ďŹ nding an arc length parametrization of a vectorvalued function that is expressed initially in terms of some other parameter t. The following theorem will provide a general method for doing this.

s r(t0)

C r(t) x

Figure 12.3.6

12.3.3 theorem Let C be the graph of a smooth vector-valued function r(t) in 2-space or 3-space, and let r(t0 ) be any point on C. Then the following formula deďŹ nes a positive change of parameter from t to s, where s is an arc length parameter having r(t0 ) as its reference point (Figure 12.3.6): t dr du s= (10) du t0

12.3 Change of Parameter; Arc Length

863

proof From (5) with u as the variable of integration instead of t, the integral represents the arc length of that portion of C between r(t0 ) and r(t) if t > t0 and the negative of that arc length if t < t0 . Thus, s is the arc length parameter with r(t0 ) as its reference point and its positive direction in the direction of increasing t. â&#x2013; When needed, Formula (10) can be expressed in component form as

Example 6

dx du

dy du

dz du

2-space

(11)

3-space

(12)

Find the arc length parametrization of the circular helix r = cos ti + sin tj + tk

(13)

that has reference point r(0) = (1, 0, 0) and the same orientation as the given helix.

Solution. Replacing t by u in r for integration purposes and taking t0 = 0 in Formula (10), we obtain

r = cos ui + sin uj + uk

dr = (â&#x2C6;&#x2019; sin u)i + cos uj + k du â&#x2C6;&#x161; dr = (â&#x2C6;&#x2019; sin u)2 + cos2 u + 1 = 2 du t t tâ&#x2C6;&#x161; â&#x2C6;&#x161; dr â&#x2C6;&#x161; du = s= 2 du = 2u = 2t du 0 0 0

â&#x2C6;&#x161; Thus, t = s / 2, so (13) can be reparametrized in terms of s as

s s s r = cos â&#x2C6;&#x161; i + sin â&#x2C6;&#x161; j + â&#x2C6;&#x161; k 2 2 2

We are guaranteed that this reparametrization preserves the orientation of the helix since Formula (10) produces a positive change of parameter.

Example 7 A bug walks along the trunk of a tree following a path modeled by the circular helix in Example 6. The bug starts at the reference point (1, 0, 0) and walks up the helix for a distance of 10 units. What are the bugâ&#x20AC;&#x2122;s ďŹ nal coordinates?

Solution. From Example 6, the arc length parametrization of the helix relative to the reference point (1, 0, 0) is

s s s j+ â&#x2C6;&#x161; k r = cos â&#x2C6;&#x161; i + sin â&#x2C6;&#x161; 2 2 2

864

Chapter 12 / Vector-Valued Functions

or, expressed parametrically,

s x = cos â&#x2C6;&#x161;

s y = sin â&#x2C6;&#x161; , 2

s z= â&#x2C6;&#x161; 2

2 Thus, at s = 10 the coordinates are

10 10 10 cos â&#x2C6;&#x161; , sin â&#x2C6;&#x161; , â&#x2C6;&#x161; â&#x2030;&#x2C6; (0.705, 0.709, 7.07) 2 2 2 Example 8

Recall from Formula (9) of Section 11.5 that the equation (14)

r = r0 + tv

is the vector form of the line that passes through the terminal point of r0 and is parallel to the vector v. Find the arc length parametrization of the line that has reference point r0 and the same orientation as the given line.

Solution. Replacing t by u in (14) for integration purposes and taking t0 = 0 in Formula (10), we obtain

r = r0 + uv

and

dr =v du

Since r0 is constant

It follows from this that

In words, Formula (15) tells us that the line represented by Equation (14) can be reparametrized in terms of arc length with r0 as the reference point by normalizing v and then replacing t by s .

t t t dr du = s=

du =

v u = t v

du 0 0 0

This implies that t = s / v , so (14) can be reparametrized in terms of s as

v r = r0 + s

Example 9

(15)

Find the arc length parametrization of the line x = 2t + 1,

y = 3t â&#x2C6;&#x2019; 2

that has the same orientation as the given line and uses (1, â&#x2C6;&#x2019;2) as the reference point.

Solution. The line passes through the point (1, â&#x2C6;&#x2019;2) and is parallel to v = 2i + 3j. To ďŹ nd the arc length parametrization of the line, we need only rewrite the given equations using v/ v rather than v to determine the direction and replace t by s. Since

2 3 2i + 3j v = â&#x2C6;&#x161; i+ â&#x2C6;&#x161; j = â&#x2C6;&#x161;

13 13 13 it follows that the parametric equations for the line in terms of s are 2 x = â&#x2C6;&#x161; s + 1, 13

3 y = â&#x2C6;&#x161; sâ&#x2C6;&#x2019;2 13

PROPERTIES OF ARC LENGTH PARAMETRIZATIONS

Because arc length parameters for a curve C are intimately related to the geometric characteristics of C, arc length parametrizations have properties that are not enjoyed by other parametrizations. For example, the following theorem shows that if a smooth curve is represented parametrically using an arc length parameter, then the tangent vectors all have length 1.

12.3 Change of Parameter; Arc Length

12.3.4

865

theorem

(a) If C is the graph of a smooth vector-valued function r(t) in 2-space or 3-space, where t is a general parameter, and if s is the arc length parameter for C deďŹ ned by Formula (10), then for every value of t the tangent vector has length dr ds = (16) dt dt (b) If C is the graph of a smooth vector-valued function r(s) in 2-space or 3-space, where s is an arc length parameter, then for every value of s the tangent vector to C has length dr =1 (17) ds

(c) If C is the graph of a smooth vector-valued function r(t) in 2-space or 3-space, and if dr/dt = 1 for every value of t, then for any value of t0 in the domain of r, the parameter s = t â&#x2C6;&#x2019; t0 is an arc length parameter that has its reference point at the point on C where t = t0 .

proof (a) This result follows by applying the Fundamental Theorem of Calculus (Theorem 5.6.3) to Formula (10). proof (b) Let t = s in part (a). proof (c) It follows from Theorem 12.3.3 that the formula t dr du s= du t0

deďŹ nes an arc length parameter for C with reference point r(0). However, dr/du = 1 by hypothesis, so we can rewrite the formula for s as t t s= du = u = t â&#x2C6;&#x2019; t0 â&#x2013; t0

The component forms of Formulas (16) and (17) will be of sufďŹ cient interest in later sections that we provide them here for reference:

Note that Formulas (18) and (19) do not involve t0 , and hence do not depend on where the reference point for s is chosen. This is to be expected since changing the reference point shifts s by a constant (the arc length between the two reference points), and this constant drops out on differentiating.

2 2 dr dx dy ds = = + dt dt dt dt

2 2 2 dr dx ds dy dz = + + dt = dt dt dt dt

2 2 dr dx dy = + =1 ds ds ds

2 2 2 dr dx dy dz = + + =1 ds ds ds ds

2-space

(18)

3-space

(19)

2-space

(20)

3-space

(21)

866

Chapter 12 / Vector-Valued Functions

â&#x153;&#x201D;QUICK CHECK EXERCISES 12.3

(See page 868 for answers.)

1. If r(t) is a smooth vector-valued function, then the integral b dr dt dt a may be interpreted geometrically as the

2. If r(s) is a smooth vector-valued function parametrized by arc length s, then dr = ds

3. If r(t) is a smooth vector-valued function, then the arc length parameter s having r(t0 ) as the reference point may be deďŹ ned by the integral t s= du t0

4. Suppose that r(t) a smooth vector-valued function of t â&#x2C6;&#x161; is â&#x2C6;&#x161; with râ&#x20AC;˛ (1) = 3, â&#x2C6;&#x2019; 3, â&#x2C6;&#x2019;1 , and let r1 (t) be deďŹ ned by the equation r1 (t) = r(2 cos t). Then r1â&#x20AC;˛ (Ď&#x20AC;/3) = .

and the arc length of the graph of r over the interval a â&#x2030;¤ s â&#x2030;¤ b is .

EXERCISE SET 12.3 1â&#x20AC;&#x201C;4 Determine whether r(t) is a smooth function of the parameter t. â&#x2013;

1. r(t) = t 3 i + (3t 2 â&#x2C6;&#x2019; 2t) j + t 2 k

17. If r(t) is a smooth vector-valued function in 2-space, then b

râ&#x20AC;˛ (t) dt a

2. r(t) = cos t 2 i + sin t 2 j + eâ&#x2C6;&#x2019;t k

is a vector.

3. r(t) = teâ&#x2C6;&#x2019;t i + (t 2 â&#x2C6;&#x2019; 2t) j + cos Ď&#x20AC;tk 2

4. r(t) = sin Ď&#x20AC;ti + (2t â&#x2C6;&#x2019; ln t) j + (t â&#x2C6;&#x2019; t)k

5â&#x20AC;&#x201C;8 Find the arc length of the parametric curve. â&#x2013;

5. x = cos3 t, y = sin3 t, z = 2; 0 â&#x2030;¤ t â&#x2030;¤ Ď&#x20AC;/2 6. x = 3 cos t, y = 3 sin t, z = 4t; 0 â&#x2030;¤ t â&#x2030;¤ Ď&#x20AC; â&#x2C6;&#x161; 7. x = et , y = eâ&#x2C6;&#x2019;t , z = 2t; 0 â&#x2030;¤ t â&#x2030;¤ 1

8. x = 21 t, y = 13 (1 â&#x2C6;&#x2019; t)3/2 , z = 13 (1 + t)3/2 ; â&#x2C6;&#x2019;1 â&#x2030;¤ t â&#x2030;¤ 1

18. If the line y = x is parametrized by the vector-valued function r(t), then r(t) is smooth. 19. If r(s) parametrizes the graph of y = |x| in 2-space by arc length, then r(s) is smooth. 20. If a curve C in the plane is parametrized by the smooth vector-valued function r(s), where s is an arc length pa 3 rameter, then

râ&#x20AC;˛ (s) ds = 4 â&#x2C6;&#x2019;1

21. (a) Find the arc length parametrization of the line x = t,

9â&#x20AC;&#x201C;12 Find the arc length of the graph of r(t). â&#x2013;

9. r(t) = t 3 i + tj +

â&#x2C6;&#x161; 1

6t 2 k; 1 â&#x2030;¤ t â&#x2030;¤ 3

10. r(t) = (4 + 3t)i + (2 â&#x2C6;&#x2019; 2t) j + (5 + t)k; 3 â&#x2030;¤ t â&#x2030;¤ 4

11. r(t) = 3 cos ti + 3 sin tj + tk; 0 â&#x2030;¤ t â&#x2030;¤ 2Ď&#x20AC;

12. r(t) = t 2 i + (cos t + t sin t) j + (sin t â&#x2C6;&#x2019; t cos t)k; 0â&#x2030;¤t â&#x2030;¤Ď&#x20AC; 13â&#x20AC;&#x201C;16 Calculate dr/dĎ&#x201E; by the chain rule, and then check your

result by expressing r in terms of Ď&#x201E; and differentiating. â&#x2013; 13. r = ti + t 2 j; t = 4Ď&#x201E; + 1

14. r = 3 cos t, 3 sin t ; t = Ď&#x20AC;Ď&#x201E;

15. r = et i + 4eâ&#x2C6;&#x2019;t j; t = Ď&#x201E; 2

16. r = i + 3t 3/2 j + tk; t = 1/Ď&#x201E;

17â&#x20AC;&#x201C;20 Trueâ&#x20AC;&#x201C;False Determine whether the statement is true or false. Explain your answer. â&#x2013;

y=t

that has the same orientation as the given line and has reference point (0, 0). (b) Find the arc length parametrization of the line x = t,

y = t,

z=t

that has the same orientation as the given line and has reference point (0, 0, 0). 22. Find arc length parametrizations of the lines in Exercise 21 that have the stated reference points but are oriented opposite to the given lines. 23. (a) Find the arc length parametrization of the line x = 1 + t,

y = 3 â&#x2C6;&#x2019; 2t,

z = 4 + 2t

that has the same direction as the given line and has reference point (1, 3, 4). (b) Use the parametric equations obtained in part (a) to ďŹ nd the point on the line that is 25 units from the reference point in the direction of increasing parameter.

12.3 Change of Parameter; Arc Length

24. (a) Find the arc length parametrization of the line x = â&#x2C6;&#x2019;5 + 3t, y = 2t, z = 5 + t that has the same direction as the given line and has reference point (â&#x2C6;&#x2019;5, 0, 5). (b) Use the parametric equations obtained in part (a) to ďŹ nd the point on the line that is 10 units from the reference point in the direction of increasing parameter. 25â&#x20AC;&#x201C;30 Find an arc length parametrization of the curve that has the same orientation as the given curve and for which the reference point corresponds to t = 0. â&#x2013;

25. r(t) = (3 + cos t)i + (2 + sin t) j; 0 â&#x2030;¤ t â&#x2030;¤ 2Ď&#x20AC; 26. r(t) = cos3 ti + sin3 t j; 0 â&#x2030;¤ t â&#x2030;¤ Ď&#x20AC;/2 27. r(t) = 13 t 3 i + 21 t 2 j; t â&#x2030;Ľ 0

28. r(t) = (1 + t)2 i + (1 + t)3 j; 0 â&#x2030;¤ t â&#x2030;¤ 1 29. r(t) = et cos ti + et sin t j; 0 â&#x2030;¤ t â&#x2030;¤ Ď&#x20AC;/2 â&#x2C6;&#x161; 30. r(t) = sin et i + cos et j + 3et k; t â&#x2030;Ľ 0

31. Show that the arc length of the circular â&#x2C6;&#x161; helix x = a cos t, y = a sin t, z = ct for 0 â&#x2030;¤ t â&#x2030;¤ t0 is t0 a 2 + c2 .

32. Use the result in Exercise 31 to show the circular helix r = a cos ti + a sin t j + ctk can be expressed as s

cs r = a cos i + a sin j+ k w w w â&#x2C6;&#x161; where w = a 2 + c2 and s is an arc length parameter with reference point at (a, 0, 0). 33. Find an arc length parametrization of the cycloid x = at â&#x2C6;&#x2019; a sin t (0 â&#x2030;¤ t â&#x2030;¤ 2Ď&#x20AC;) y = a â&#x2C6;&#x2019; a cos t with (0, 0) as the reference point.

34. Show that in cylindrical coordinates a curve given by the parametric equations r = r(t), Î¸ = Î¸ (t), z = z(t) for a â&#x2030;¤ t â&#x2030;¤ b has arc length 2 2 b 2 dr dz dÎ¸ L= + r2 + dt dt dt dt a [Hint: Use the relationships x = r cos Î¸, y = r sin Î¸.]

35. In each part, use the formula in Exercise 34 to ďŹ nd the arc length of the curve. (a) r = e2t , Î¸ = t, z = e2t ; 0 â&#x2030;¤ t â&#x2030;¤ ln 2 (b) r = t 2 , Î¸ = ln t, z = 31 t 3 ; 1 â&#x2030;¤ t â&#x2030;¤ 2

36. Show that in spherical coordinates a curve given by the parametric equations Ď = Ď (t), Î¸ = Î¸ (t), Ď&#x2020; = Ď&#x2020;(t) for a â&#x2030;¤ t â&#x2030;¤ b has arc length 2 2 b 2 dÎ¸ dĎ&#x2020; dĎ L= + Ď 2 sin2 Ď&#x2020; + Ď 2 dt dt dt dt a [Hint: x = Ď sin Ď&#x2020; cos Î¸, y = Ď sin Ď&#x2020; sin Î¸, z = Ď cos Ď&#x2020;.]

37. In each part, use the formula in Exercise 36 to ďŹ nd the arc length of the curve.

867

(a) Ď = eâ&#x2C6;&#x2019;t , Î¸ = 2t, Ď&#x2020; = Ď&#x20AC;/4; 0 â&#x2030;¤ t â&#x2030;¤ 2 (b) Ď = 2t, Î¸ = ln t, Ď&#x2020; = Ď&#x20AC;/6; 1 â&#x2030;¤ t â&#x2030;¤ 5 F O C U S O N C O N C E P TS

38. (a) Sketch the graph of r(t) = ti + t 2 j. Show that r(t) is a smooth vector-valued function but the change of parameter t = Ď&#x201E; 3 produces a vector-valued function that is not smooth, yet has the same graph as r(t). (b) Examine how the two vector-valued functions are traced, and see if you can explain what causes the problem. 39. Find a change of parameter t = g(Ď&#x201E; ) for the semicircle r(t) = cos ti + sin tj (0 â&#x2030;¤ t â&#x2030;¤ Ď&#x20AC;) such that (a) the semicircle is traced counterclockwise as Ď&#x201E; varies over the interval [0, 1] (b) the semicircle is traced clockwise as Ď&#x201E; varies over the interval [0, 1].

40. What change of parameter t = g(Ď&#x201E; ) would you make if you wanted to trace the graph of r(t) (0 â&#x2030;¤ t â&#x2030;¤ 1) in the opposite direction with Ď&#x201E; varying from 0 to 1? 41. As illustrated in the accompanying ďŹ gure, copper cable with a diameter of 21 inch is to be wrapped in a circular helix around a cylinder that has a 12-inch diameter. What length of cable (measured along its centerline) will make one complete turn around the cylinder in a distance of 20 inches (between centerlines) measured parallel to the axis of the cylinder? 12 in

Enlarged cross section

1 2

20 in

Figure Ex-41

42. Let r(t) = cos t, sin t, t 3/2 . Find 2 ds â&#x20AC;˛ (a) r (t)

r â&#x20AC;˛ (t) dt. (c) (b) dt 0 43. Let r(t) = ln ti + 2tj + t 2 k. Find 3 ds (a) r â&#x20AC;˛ (t)

(b)

r â&#x20AC;˛ (t) dt. (c) dt 1 44. Let r(t) = t 2 i + t 3 j (see Figure 12.3.1). Let Î¸(t) be the angle between râ&#x20AC;˛ (t) and i. Show that Î¸ (t) â&#x2020;&#x2019; Ď&#x20AC; as t â&#x2020;&#x2019; 0â&#x2C6;&#x2019;

and

Î¸(t) â&#x2020;&#x2019; 0 as t â&#x2020;&#x2019; 0+

Chapter 12 / Vector-Valued Functions

868

45. Prove: If r(t) is a smoothly parametrized function, then the angles between r ′ (t) and the vectors i, j, and k are continuous functions of t. 46. Prove the vector form of the chain rule for 2-space (Theorem 12.3.2) by expressing r(t) in terms of components.

47. Writing The triangle with vertices (0, 0), (1, 0), and (0, 1) has three “corners.” Discuss whether it is possible to have a smooth vector-valued function whose graph is this triangle. Also discuss whether it is possible to have a differentiable vector-valued function whose graph is this triangle.

✔QUICK CHECK ANSWERS 12.3 1. arc length of the graph of r(t) from t = a to t = b

2. 1; b − a

dr 3. du

4. −3, 3,

√ 3

UNIT TANGENT, NORMAL, AND BINORMAL VECTORS

12.4

In this section we will discuss some of the fundamental geometric properties of vectorvalued functions. Our work here will have important applications to the study of motion along a curved path in 2-space or 3-space and to the study of the geometric properties of curves and surfaces. UNIT TANGENT VECTORS

As a general rule, we will position T(t) with its initial point at the terminal point of r(t), as in Figure 12.4.1. This will ensure that T(t) is actually tangent to the graph of r(t) and not simply parallel to the tangent line.

Recall that if C is the graph of a smooth vector-valued function r(t) in 2-space or 3-space, then the vector r ′ (t) is nonzero, tangent to C, and points in the direction of increasing parameter. Thus, by normalizing r ′ (t) we obtain a unit vector T(t) =

r ′ (t)

(1)

that is tangent to C and points in the direction of increasing parameter. We call T(t) the unit tangent vector to C at t.

T(t)

Example 1 Find the unit tangent vector to the graph of r(t) = t 2 i + t 3 j at the point where t = 2. r(t)

Solution. Since r ′ (t) = 2ti + 3t 2 j

we obtain

Figure 12.4.1

r ′ (2) 3 4i + 12 j 4i + 12 j 1 = √ = √ = √ i+ √ j ′

r (2)

160 4 10 10 10 The graph of r(t) and the vector T(2) are shown in Figure 12.4.2. T(2) =

y 10

(4, 8)

T(2) =

√10

UNIT NORMAL VECTORS r(t) = t 2 i + t 3j

x 8

Figure 12.4.2

Recall from Theorem 12.2.8 that if a vector-valued function r(t) has constant norm, then r(t) and r ′ (t) are orthogonal vectors. In particular, T(t) has constant norm 1, so T(t) and T′ (t) are orthogonal vectors. This implies that T′ (t) is perpendicular to the tangent line to C at t, so we say that T′ (t) is normal to C at t. It follows that if T′ (t) = 0, and if we normalize T′ (t), then we obtain a unit vector N(t) =

T′ (t)

(2)

12.4 Unit Tangent, Normal, and Binormal Vectors

869

that is normal to C and points in the same direction as T′ (t). We call N(t) the principal unit normal vector to C at t, or more simply, the unit normal vector. Observe that the unit normal vector is deﬁned only at points where T′ (t) = 0. Unless stated otherwise, we will assume that this condition is satisﬁed. In particular, this excludes straight lines.

REMARK

In 2-space there are two unit vectors that are orthogonal to T(t), and in 3-space there are infinitely many such vectors (Figure 12.4.3). In both cases the principal unit normal is that particular normal that points in the direction of T′ (t). After the next example we will show that for a nonlinear parametric curve in 2-space the principal unit normal is the one that points “inward” toward the concave side of the curve.

C T(t)

C T(t) y

There are two unit vectors orthogonal to T(t).

There are infinitely many unit vectors orthogonal to T(t).

Figure 12.4.3

Example 2 r(t)

Find T(t) and N(t) for the circular helix x = a cos t,

y = a sin t,

z = ct

where a > 0. t=0

(a, 0, 0)

Solution. The radius vector for the helix is

r(t) = a cos ti + a sin t j + ctk

r(t) = a cos t i + a sin tj + ctk

(Figure 12.4.4). Thus,

Figure 12.4.4

r ′ (t) = (−a sin t)i + a cos t j + ck √

r ′ (t) = (−a sin t)2 + (a cos t)2 + c2 = a 2 + c2

t = 3c

t=0

(a, 0, 0) x

N(t) = −(cos t i + sin tj ) Figure 12.4.5

a sin t r ′ (t) a cos t c = − i+ j+ k ′

r (t)

a 2 + c2 a 2 + c2 a 2 + c2 a cos t a sin t T′ (t) = − i− j a 2 + c2 a 2 + c2

2 a cos t a a2 a sin t ′ − = + − =

T (t) = 2 2 2 2 2 2 2 a + c a +c a +c a + c2 T(t) =

N(t) =

T′ (t) = (− cos t)i − (sin t)j = −(cos ti + sin tj)

T′ (t)

Note that the k component of the principal unit normal N(t) is zero for every value of t, so this vector always lies in a horizontal plane, as illustrated in Figure 12.4.5. We leave it as an exercise to show that this vector actually always points toward the z-axis.

870

Chapter 12 / Vector-Valued Functions

INWARD UNIT NORMAL VECTORS IN 2-SPACE

C n(t)

T(t) f(t)

Our next objective is to show that for a nonlinear parametric curve C in 2-space the unit normal vector always points toward the concave side of C. For this purpose, let φ(t) be the angle from the positive x-axis to T(t), and let n(t) be the unit vector that results when T(t) is rotated counterclockwise through an angle of π/2 (Figure 12.4.6). Since T(t) and n(t) are unit vectors, it follows from Formula (13) of Section 11.2 that these vectors can be expressed as (3) T(t) = cos φ(t)i + sin φ(t)j and

Figure 12.4.6

n(t) = cos[φ(t) + π/2]i + sin[φ(t) + π/2]j = − sin φ(t)i + cos φ(t) j

(4)

Observe that on intervals where φ(t) is increasing the vector n(t) points toward the concave side of C, and on intervals where φ(t) is decreasing it points away from the concave side (Figure 12.4.7). y

C n(t)

T(t)

n(t)

T(t)

C f(t)

f(t)

Figure 12.4.7

f(t) increases as t increases.

f(t) decreases as t increases.

Now let us differentiate T(t) by using Formula (3) and applying the chain rule. This yields dT dφ dφ dT = = [(− sin φ)i + (cos φ) j] dt dφ dt dt and thus from (4) dφ dT = n(t) (5) dt dt But dφ /dt > 0 on intervals where φ(t) is increasing and dφ /dt < 0 on intervals where φ(t) is decreasing. Thus, it follows from (5) that dT/dt has the same direction as n(t) on intervals where φ(t) is increasing and the opposite direction on intervals where φ(t) is decreasing. Therefore, T′ (t) = dT/dt points “inward” toward the concave side of the curve in all cases, and hence so does N(t). For this reason, N(t) is also called the inward unit normal when applied to curves in 2-space. COMPUTING T AND N FOR CURVES PARAMETRIZED BY ARC LENGTH

WARNING Formulas (6) and (7) are only applicable when the curve is parametrized by an arc length parameter s . For other parametrizations Formulas (1) and (2) can be used.

In the case where r(s) is parametrized by arc length, the procedures for computing the unit tangent vector T(s) and the unit normal vector N(s) are simpler than in the general case. For example, we showed in Theorem 12.3.4 that if s is an arc length parameter, then

r ′ (s) = 1. Thus, Formula (1) for the unit tangent vector simpliﬁes to T(s) = r ′ (s)

(6)

and consequently Formula (2) for the unit normal vector simpliﬁes to N(s) =

r ′′ (s)

(7)

12.4 Unit Tangent, Normal, and Binormal Vectors y

871

Example 3 The circle of radius a with counterclockwise orientation and centered at the origin can be represented by the vector-valued function (x, y)

r = a cos ti + a sin tj

s = at t

(0 ≤ t ≤ 2π)

(8)

Parametrize this circle by arc length and ﬁnd T(s) and N(s).

(a, 0)

Solution. In (8) we can interpret t as the angle in radian measure from the positive x-axis to the radius vector (Figure 12.4.8). This angle subtends an arc of length s = at on the circle, so we can reparametrize the circle in terms of s by substituting s /a for t in (8). This yields r(s) = a cos(s /a)i + a sin(s /a) j (0 ≤ s ≤ 2πa)

Figure 12.4.8

To ﬁnd T(s) and N(s) from Formulas (6) and (7), we must compute r ′ (s), r ′′ (s), and

r ′′ (s) . Doing so, we obtain

r ′ (s) = − sin(s /a)i + cos(s /a)j

T(s) s x

N(s)

Thus,

r ′′ (s) = −(1/a) cos(s /a)i − (1/a) sin(s /a) j

r ′′ (s) = (−1/a)2 cos2 (s /a) + (−1/a)2 sin2 (s /a) = 1/a T(s) = r ′ (s) = − sin(s /a)i + cos(s /a)j N(s) = r ′′ (s)/ r ′′ (s) = − cos(s /a)i − sin(s /a) j

so N(s) points toward the center of the circle for all s (Figure 12.4.9). This makes sense geometrically and is also consistent with our earlier observation that in 2-space the unit normal vector is the inward normal.

Figure 12.4.9 Normal plane

r(t)

Rectifying plane

BINORMAL VECTORS IN 3-SPACE

If C is the graph of a vector-valued function r(t) in 3-space, then we deﬁne the binormal vector to C at t to be B(t) = T(t) × N(t) (9)

It follows from properties of the cross product that B(t) is orthogonal to both T(t) and N(t) and is oriented relative to T(t) and N(t) by the right-hand rule. Moreover, T(t) × N(t) is a unit vector since

N T Osculating plane

T(t) × N(t) = T(t)

N(t) sin(π/2) = 1

Figure 12.4.10

N In (10), the vectors B, N, and T are each expressed as the cross product of the other two taken in the counterclockwise direction around the above triangle.

Figure 12.4.11

Thus, {T(t), N(t), B(t)} is a set of three mutually orthogonal unit vectors. Just as the vectors i, j, and k determine a right-handed coordinate system in 3-space, so do the vectors T(t), N(t), and B(t). At each point on a smooth parametric curve C in 3-space, these vectors determine three mutually perpendicular planes that pass through the point— the TB-plane (called the rectifying plane), the TN-plane (called the osculating plane), and the NB-plane (called the normal plane) (Figure 12.4.10). Moreover, one can show that a coordinate system determined by T(t), N(t), and B(t) is right-handed in the sense that each of these vectors is related to the other two by the right-hand rule (Figure 12.4.11): B(t) = T(t) × N(t),

N(t) = B(t) × T(t),

T(t) = N(t) × B(t)

(10)

The coordinate system determined by T(t), N(t), and B(t) is called the TNB-frame or sometimes the Frenet frame in honor of the French mathematician Jean Frédéric Frenet (1816– 1900) who pioneered its application to the study of space curves. Typically, the xyzcoordinate system determined by the unit vectors i, j, and k remains ﬁxed, whereas the TNB-frame changes as its origin moves along the curve C (Figure 12.4.12).

872

Chapter 12 / Vector-Valued Functions B B

r(t) N T

Figure 12.4.12

Formula (9) expresses B(t) in terms of T(t) and N(t). Alternatively, the binormal B(t) can be expressed directly in terms of r(t) as B(t) =

r â&#x20AC;˛ (t) Ă&#x2014; r â&#x20AC;˛â&#x20AC;˛ (t)

(11)

and in the case where the parameter is arc length it can be expressed in terms of r(s) as B(s) =

r â&#x20AC;˛ (s) Ă&#x2014; r â&#x20AC;˛â&#x20AC;˛ (s)

r â&#x20AC;˛â&#x20AC;˛ (s)

(12)

We omit the proof.

â&#x153;&#x201D;QUICK CHECK EXERCISES 12.4

(See page 873 for answers.)

1. If C is the graph of a smooth vector-valued function r(t), then the unit tangent, unit normal, and binormal to C at t are deďŹ ned, respectively, by T(t) =

N(t) =

B(t) =

2. If C is the graph of a smooth vector-valued function r(s) parametrized by arc length, then the deďŹ nitions of the unit tangent and unit normal to C at s simplify, respectively, to and

T(s) =

3. If C is the graph of a smooth vector-valued function r(t), then the unit binormal vector to C at t may be computed directly in terms of râ&#x20AC;˛ (t) and râ&#x20AC;˛â&#x20AC;˛ (t) by the formula B(t) = . When t = s is the arc length parameter, this formula simpliďŹ es to B(s) = .

4. Suppose that C is the graph of a smooth vector-valued function r(s) parametrized by arc length with râ&#x20AC;˛ (0) = 23 , 31 , 23 and râ&#x20AC;˛â&#x20AC;˛ (0) = â&#x2C6;&#x2019;3, 12, â&#x2C6;&#x2019;3 . Then T(0) =

N(s) =

, N(0) =

, B(0) =

EXERCISE SET 12.4 F O C U S O N C O N C E P TS

1. In each part, sketch the unit tangent and normal vectors at the points P , Q, and R, taking into account the orientation of the curve C. (a)

(b)

C P

2. Make a rough sketch that shows the ellipse r(t) = 3 cos ti + 2 sin tj for 0 â&#x2030;¤ t â&#x2030;¤ 2Ď&#x20AC; and the unit tangent and normal vectors at the points t = 0, t = Ď&#x20AC;/4, t = Ď&#x20AC;/2, and t = Ď&#x20AC;.

3. In the marginal note associated with Example 8 of Section 12.3, we observed that a line r = r0 + tv can be parametrized in terms of an arc length parameter s with reference point r0 by normalizing v. Use this result to show that the tangent line to the graph of r(t) at the point t0 can be expressed as r = r(t0 ) + sT(t0 ) where s is an arc length parameter with reference point r(t0 ). 4. Use the result in Exercise 3 to show that the tangent line to the parabola x = t, y = t 2 at the point (1, 1) can be expressed parametrically as s 2s x =1+ â&#x2C6;&#x161; , y =1+ â&#x2C6;&#x161; 5 5

12.5 Curvature 5–12 Find T(t) and N(t) at the given point. ■

5. r(t) = (t 2 − 1)i + tj; t = 1 6. r(t) = 21 t 2 i + 31 t 3 j; t = 1

7. r(t) = 5 cos ti + 5 sin t j; t = π/3

8. r(t) = ln ti + tj; t = e

873

19–20 Find T(t), N(t), and B(t) for the given value of t. Then

ﬁnd equations for the osculating, normal, and rectifying planes at the point that corresponds to that value of t. ■ 19. r(t) = cos ti + sin tj + k; t = π/4 20. r(t) = et i + et cos t j + et sin t k; t = 0

9. r(t) = 4 cos ti + 4 sin tj + tk; t = π/2

21–24 True–False Determine whether the statement is true or false. Explain your answer. ■

11. x = et cos t, y = et sin t, z = et ; t = 0

21. If C is the graph of a smooth vector-valued function r(t) in 2-space, then the unit tangent vector T(t) to C is orthogonal to r(t) and points in the direction of increasing parameter.

10. r(t) = ti + 21 t 2 j + 31 t 3 k; t = 0

12. x = cosh t, y = sinh t, z = t; t = ln 2

for the tangent line to the graph of r(t) at t0 in terms of an arc length parameter s. ■

22. If C is the graph of a smooth vector-valued function r(t) in 2-space, then the angle measured in the counterclockwise direction from the unit tangent vector T(t) to the unit normal vector N(t) is π/2.

13. r(t) = sin ti + cos tj + 21 t 2 k; t0 = 0 √ 14. r(t) = ti + tj + 9 − t 2 k; t0 = 1

23. If the smooth vector-valued function r(s) is parametrized by arc length and r′′ (s) is deﬁned, then r′ (s) and r′′ (s) are orthogonal vectors.

15–18 Use the formula B(t) = T(t) × N(t) to ﬁnd B(t), and then check your answer by using Formula (11) to ﬁnd B(t) directly from r(t). ■

24. The binormal vector B(t) to the graph of a vector-valued function r(t) in 3-space is the dot product of unit tangent and unit normal vectors, T(t) and N(t).

15. r(t) = 3 sin ti + 3 cos t j + 4tk

25. Writing Look up the deﬁnition of “osculating” in a dictionary and discuss why “osculating plane” is an appropriate term for the TN-plane.

13–14 Use the result in Exercise 3 to ﬁnd parametric equations

16. r(t) = e sin ti + e cos t j + 3k

17. r(t) = (sin t − t cos t)i + (cos t + t sin t) j + k 18. r(t) = a cos ti + a sin t j + ctk

(a = 0, c = 0)

26. Writing Discuss some of the advantages of parametrizing a curve by arc length.

✔QUICK CHECK ANSWERS 12.4 r′′ (s) T′ (t) r′ (s) × r′′ (s) r′ (t) r′ (t) × r′′ (t) 3. ; ; T(t) × N(t) 2. r′ (s); ; ′ ′ ′′ ′ ′′

r (t) T (t)

r (s)

r (t) × r (t)

r′′ (s)

2 1 2 1 1 4 1 1 4. , , ; − √ , √ , − √ ; − √ , 0, √ 3 3 3 3 2 3 2 3 2 2 2 1.

12.5

CURVATURE In this section we will consider the problem of obtaining a numerical measure of how sharply a curve in 2-space or 3-space bends. Our results will have applications in geometry and in the study of motion along a curved path. DEFINITION OF CURVATURE

Suppose that C is the graph of a smooth vector-valued function in 2-space or 3-space that is parametrized in terms of arc length. Figure 12.5.1 suggests that for a curve in 2-space the “sharpness” of the bend in C is closely related to dT/ds, which is the rate of change of the unit tangent vector T with respect to s. (Keep in mind that T has constant length, so only its direction changes.) If C is a straight line (no bend), then the direction of T remains constant (Figure 12.5.1a); if C bends slightly, then T undergoes a gradual change of direction (Figure 12.5.1b); and if C bends sharply, then T undergoes a rapid change of direction (Figure 12.5.1c).

874

Chapter 12 / Vector-Valued Functions T

C T

T T

(a)

The situation in 3-space is more complicated because bends in a curve are not limited to a single plane—they can occur in all directions, as illustrated by the complicated tube plot in Figure 12.1.4. To describe the bending characteristics of a curve in 3-space completely, one must take into account dT/ds, dN/ds, and dB/ds. A complete study of this topic would take us too far aﬁeld, so we will limit our discussion to dT/ds, which is the most important of these derivatives in applications.

(b)

C T

12.5.1 definition If C is a smooth curve in 2-space or 3-space that is parametrized by arc length, then the curvature of C, denoted by κ = κ(s) (κ = Greek “kappa”), is deﬁned by dT ′′ κ(s) = (1) ds = r (s)

Observe that κ(s) is a real-valued function of s, since it is the length of dT/ds that measures the curvature. In general, the curvature will vary from point to point along a curve; however, the following example shows that the curvature is constant for circles in 2-space, as you might expect. Example 1 In Example 3 of Section 12.4 we showed that the circle of radius a, centered at the origin, can be parametrized in terms of arc length as s

r(s) = a cos i + a sin j (0 ≤ s ≤ 2πa) a a Thus, s

1 1 r ′′ (s) = − cos i − sin j a a a a and hence from (1) s 2 1 s 2 1 1 − cos + − sin = κ(s) = r ′′ (s) = a a a a a so the circle has constant curvature 1/a.

The next example shows that lines have zero curvature, which is consistent with the fact that they do not bend. Example 2 Recall from Formula (15) of Section 12.3 that a line in 2-space or 3-space can be parametrized in terms of arc length as r = r0 + su where the terminal point of r0 is a point on the line and u is a unit vector parallel to the line. Since u and r0 are constant, their derivatives with respect to s are zero, and hence r ′ (s) =

d dr = [r0 + su] = 0 + u = u ds ds

r ′′ (s) =

d dr ′ = [u] = 0 ds ds

Thus, κ(s) = r ′′ (s) = 0

12.5 Curvature

875

FORMULAS FOR CURVATURE

Formula (1) is only applicable if the curve is parametrized in terms of arc length. The following theorem provides two formulas for curvature in terms of a general parameter t.

12.5.2 theorem If r(t) is a smooth vector-valued function in 2-space or 3-space, then for each value of t at which T′ (t) and r ′′ (t) exist, the curvature κ can be expressed as (a)

κ(t) =

T′ (t)

r ′ (t)

(2)

(b)

κ(t) =

r ′ (t) × r ′′ (t)

r ′ (t) 3

(3)

proof (a) It follows from Formula (1) and Formula (16) of Section 12.3 that dT dT/dt dT/dt T′ (t)

= κ(t) = = = ds ds /dt dr/dt

r ′ (t)

proof (b) It follows from Formula (1) of Section 12.4 that r ′ (t) = r ′ (t) T(t) ′′

′

(4) ′

′

r (t) = r (t) T(t) + r (t) T (t)

(5)

But from Formula (2) of Section 12.4 and part (a) of this theorem we have T′ (t) = T′ (t) N(t) and so

T′ (t) = κ(t) r ′ (t)

T′ (t) = κ(t) r ′ (t) N(t)

Substituting this into (5) yields r ′′ (t) = r ′ (t) ′ T(t) + κ(t) r ′ (t) 2 N(t)

(6)

Thus, from (4) and (6) r ′ (t) × r ′′ (t) = r ′ (t)

r ′ (t) ′ (T(t) × T(t)) + κ(t) r ′ (t) 3 (T(t) × N(t)) But the cross product of a vector with itself is zero, so this equation simpliﬁes to r ′ (t) × r ′′ (t) = κ(t) r ′ (t) 3 (T(t) × N(t)) = κ(t) r ′ (t) 3 B(t) It follows from this equation and the fact that B(t) is a unit vector that

r ′ (t) × r ′′ (t) = κ(t) r ′ (t) 3 Formula (3) now follows. ■ REMARK

Formula (2) is useful if T(t) is known or is easy to obtain; however, Formula (3) will usually be easier to apply, since it involves only r(t) and its derivatives. We also note that cross products were defined only for vectors in 3-space, so to use Formula (3) in 2-space we must first write the 2-space function r(t) = x(t)i + y(t) j as the 3-space function r(t) = x(t)i + y(t) j + 0k with a zero k component.

Example 3

Find κ(t) for the circular helix x = a cos t,

where a > 0.

y = a sin t,

z = ct

876

Chapter 12 / Vector-Valued Functions

Solution. The radius vector for the helix is r(t) = a cos ti + a sin t j + ctk Thus, r ′ (t) = (−a sin t)i + a cos t j + ck

Therefore,

r ′′ (t) = (−a cos t)i + (−a sin t)j i j ′ ′′ r (t) × r (t) = −a sin t a cos t −a cos t −a sin t

r ′ (t) =

and

k 2 c = (ac sin t)i − (ac cos t)j + a k 0

(−a sin t)2 + (a cos t)2 + c2 = a 2 + c2 (ac sin t)2 + (−ac cos t)2 + a 4 √ √ = a 2 c2 + a 4 = a a 2 + c2

r ′ (t) × r ′′ (t) = so

r ′ (t) × r ′′ (t)

a a 2 + c2 a κ(t) = =

3 = 2 ′ 3

r (t)

a + c2 a 2 + c2

Note that κ does not depend on t, which tells us that the helix has constant curvature. y

Example 4 The graph of the vector equation

r = 2 cos ti + 3 sin t j x −2

(0 ≤ t ≤ 2π)

is the ellipse in Figure 12.5.2. Find the curvature of the ellipse at the endpoints of the major and minor axes, and use a graphing utility to generate the graph of κ(t).

Solution. To apply Formula (3), we must treat the ellipse as a curve in the xy-plane of an xyz-coordinate system by adding a zero k component and writing its equation as −3

r = 2 cos ti + 3 sin tj

r = 2 cos ti + 3 sin t j + 0k It is not essential to write the zero k component explicitly as long as you assume it to be there when you calculate a cross product. Thus,

Figure 12.5.2

r ′ (t) = (−2 sin t)i + 3 cos t j

r ′′ (t) = (−2 cos t)i + (−3 sin t)j i j r ′ (t) × r ′′ (t) = −2 sin t 3 cos t −2 cos t −3 sin t

Therefore,

k 0 = [(6 sin2 t) + (6 cos2 t)]k = 6k 0

(−2 sin t)2 + (3 cos t)2 = 4 sin2 t + 9 cos2 t

r ′ (t) × r ′′ (t) = 6

r ′ (t) =

κ(t) =

r ′ (t) × r ′′ (t)

6 =

r ′ (t) 3 [4 sin2 t + 9 cos2 t]3/2

(7)

12.5 Curvature

877

The endpoints of the minor axis are (2, 0) and (−2, 0), which correspond to t = 0 and t = π, respectively. Substituting these values in (7) yields the same curvature at both points, namely, 6 2 6 = κ = κ(0) = κ(π) = 3/2 = 27 9 9 The endpoints of the major axis are (0, 3) and (0, −3), which correspond to t = π/2 and t = 3π/2, respectively; from (7) the curvature at these points is

3 3π 6 =κ κ=κ = 3/2 = 2 2 4 4

Observe that the curvature is greater at the ends of the major axis than at the ends of the minor axis, as you might expect. Figure 12.5.3 shows the graph of κ versus t. This graph illustrates clearly that the curvature is minimum at t = 0 (the right end of the minor axis), increases to a maximum at t = π/2 (the top of the major axis), decreases to a minimum again at t = π (the left end of the minor axis), and continues cyclically in this manner. Figure 12.5.4 provides another way of picturing the curvature. k y

1 0.8 0.6 0.4 y

0.2 t

Radius

4 3

k(t) = x Radius

9 2

k 0.2 0.3 0.4 0.5 0.6 0.7

6 [4 sin2 t + 9 cos2 t]3/ 2

Figure 12.5.3

Figure 12.5.4

RADIUS OF CURVATURE Figure 12.5.5 Osculating circle

Center of curvature

1 r= k P Figure 12.5.6

In the last example we found the curvature at the ends of the minor axis to be 29 and the curvature at the ends of the major axis to be 43 . To obtain a better understanding of the meaning of these numbers, recall from Example 1 that a circle of radius a has a constant curvature of 1/a; thus, the curvature of the ellipse at the ends of the minor axis is the same as that of a circle of radius 29 , and the curvature at the ends of the major axis is the same as that of a circle of radius 43 (Figure 12.5.5). In general, if a curve C in 2-space has nonzero curvature κ at a point P , then the circle of radius ρ = 1/κ sharing a common tangent with C at P , and centered on the concave side of the curve at P , is called the osculating circle or circle of curvature at P (Figure 12.5.6). The osculating circle and the curve C not only touch at P but they have equal curvatures at that point. In this sense, the osculating circle is the circle that best approximates the curve C near P . The radius ρ of the osculating circle at P is called the radius of curvature at P , and the center of the circle is called the center of curvature at P (Figure 12.5.6).

C T f

Figure 12.5.7

AN INTERPRETATION OF CURVATURE IN 2-SPACE A useful geometric interpretation of curvature in 2-space can be obtained by considering the angle φ measured counterclockwise from the direction of the positive x-axis to the unit tangent vector T (Figure 12.5.7). By Formula (13) of Section 11.2, we can express T in terms of φ as T(φ) = cos φi + sin φ j

878

Chapter 12 / Vector-Valued Functions

Thus,

dT = (− sin φ)i + cos φ j dφ dT dT dφ = dφ ds ds

from which we obtain dT dφ dT dφ = = (− sin φ)2 + cos2 φ = dφ κ(s) = ds ds dφ ds ds In summary, we have shown that

dφ κ(s) = ds

f In 2-space, k(s) is the magnitude of the rate of change of f with respect to s.

Figure 12.5.8

T f T f T f f is constant, so the line has zero curvature.

(8)

which tells us that curvature in 2-space can be interpreted as the magnitude of the rate of change of φ with respect to s—the greater the curvature, the more rapidly φ changes with s (Figure 12.5.8). In the case of a straight line, the angle φ is constant (Figure 12.5.9) and consequently κ(s) = |dφ /ds| = 0, which is consistent with the fact that a straight line has zero curvature at every point. FORMULA SUMMARY We conclude this section with a summary of formulas for T, N, and B. These formulas have either been derived in the text or are easily derivable from formulas we have already established. T(s) = r ′ (s) (9)

Figure 12.5.9

N(s) =

1 dT r ′′ (s) r ′′ (s) = ′′ = κ(s) ds

r (s)

κ(s)

(10)

B(s) =

r ′ (s) × r ′′ (s) r ′ (s) × r ′′ (s) = ′′

r (s)

κ(s)

(11)

T(t) =

r ′ (t)

(12)

B(t) =

r ′ (t) × r ′′ (t)

(13)

N(t) = B(t) × T(t)

✔QUICK CHECK EXERCISES 12.5

(14)

(See page 881 for answers.)

1. If C is a smooth curve parametrized by arc length, then the curvature is deﬁned by κ(s) = .

2. Let r(t) be a smooth vector-valued function with curvature κ(t). (a) The curvature may be expressed in terms of T′ (t) and r′ (t) as κ(t) = . (b) The curvature may be expressed directly in terms of r′ (t) and r′′ (t) as κ(t) = .

3. Suppose that C is the graph of a smooth vector-valued function r(s) = x(s), y(s) parametrized by arc length and that the unit tangent T(s) = cos φ(s), sin φ(s) . Then the curvature can be expressed in terms of φ(s) as κ(s) = . 4. Suppose that C is a smooth curve√and that x 2 + y 2 = 4 is the osculating circle to C at P (1, 3). Then the curvature of C at P is .

12.5 Curvature

EXERCISE SET 12.5

Graphing Utility

CAS

F O C U S O N C O N C E P TS

1–2 Use the osculating circle shown in the ﬁgure to estimate

the curvature at the indicated point. ■ y 1. 2.

3/2 3/2 18. r = 1 − 23 s i + 23 s j

0 ≤ s ≤ 23

19–22 True–False Determine whether the statement is true or

false. Explain your answer. ■

y 3

19. A circle of radius 2 has constant curvature 21 .

20. A vertical line in 2-space has undeﬁned curvature.

0.5 x −1

21. If r(s) is parametrized by arc length, then the curvature of the graph of r(s) is the length of r′ (s).

x −3

22. If C is a curve in 2-space, then the osculating circle to C at a point P has radius equal to the curvature of C at P .

−3

3–4 For a plane curve y = f(x) the curvature at (x, f(x))

is a function κ(x). In these exercises the graphs of f(x) and κ(x) are shown. Determine which is which and explain your reasoning. ■ 3. (a)

879

(b)

23. (a) Use Formula (3) to show that in 2-space the curvature of a smooth parametric curve x = x(t),

y = y(t)

|x ′ y ′′ − y ′ x ′′ | (x ′2 + y ′2 )3/2 where primes denote differentiation with respect to t. (b) Use the result in part (a) to show that in 2-space the curvature of the plane curve given by y = f(x) is κ(t) =

|d 2 y /dx 2 | [1 + (dy/dx)2 ]3/2 [Hint: Express y = f(x) parametrically with x = t as the parameter.]

κ(x) =

I x

4. (a)

(b)

II II I

I x

5–12 Use Formula (3) to ﬁnd κ(t). ■

5. r(t) = t 2 i + t 3 j

7. r(t) = e3t i + e−t j

6. r(t) = 4 cos ti + sin t j

8. x = 1 − t 3 , y = t − t 2

9. r(t) = 4 cos ti + 4 sin tj + tk

10. r(t) = ti + 21 t 2 j + 31 t 3 k

24. Use part (b) of Exercise 23 to show that the curvature of y = f(x) can be expressed in terms of the angle of inclination of the tangent line as 2 d y κ(φ) = 2 cos3 φ dx [Hint: tan φ = dy/dx.]

25–28 Use the result in Exercise 23(b) to ﬁnd the curvature at the stated point. ■ 25. y = sin x; x = π/2 26. y = tan x; x = π/4

27. y = e−x ; x = 1

28. y 2 − 4x 2 = 9; (2, 5)

11. x = cosh t, y = sinh t, z = t

29–32 Use the result in Exercise 23(a) to ﬁnd the curvature at the stated point. ■

13–16 Find the curvature and the radius of curvature at the

29. x = t 2 , y = t 3 ; t = 21 31. x = t, y = 1/t; t = 1

12. r(t) = i + tj + t k

stated point. ■ 13. r(t) = 3 cos ti + 4 sin tj + tk; t = π/2

14. r(t) = et i + e−t j + tk; t = 0

15. x = et cos t, y = et sin t, z = et ; t = 0 16. x = sin t, y = cos t, z = 21 t 2 ; t = 0

17–18 Conﬁrm that s is an arc length parameter by showing

that dr/ds = 1, and then apply Formula (1) to ﬁnd κ(s). ■ √ s

17. r = sin 1 + i + cos 1 + j+ 3 1+ k 2 2 2

30. x = e3t , y = e−t ; t = 0

32. x = 2 sin 2t, y = 3 sin t; t = π/2

33. In each part, use the formulas in Exercise 23 to help ﬁnd the radius of curvature at the stated points. Then sketch the graph together with the osculating circles at those points. (a) y = cos x at x = 0 and x = π (b) x = 2 cos t, y = sin t (0 ≤ t ≤ 2π) at t = 0 and t = π/2

34. Use the formula in Exercise 23(a) to ﬁnd κ(t) for the curve x = e−t cos t, y = e−t sin t. Then sketch the graph of κ(t).

880

Chapter 12 / Vector-Valued Functions

35–36 Generate the graph of y = f(x) using a graphing util-

ity, and then make a conjecture about the shape of the graph of y = κ(x). Check your conjecture by generating the graph of y = κ(x). ■ 35. f(x) = xe−x for 0 ≤ x ≤ 5 C

36. f(x) = x 3 − x

for −1 ≤ x ≤ 1

37. (a) If you have a CAS, read the documentation on calculating higher-order derivatives. Then use the CAS and part (b) of Exercise 23 to find κ(x) for f(x) = x 4 − 2x 2 . (b) Use the CAS to generate the graphs of f(x) = x 4 − 2x 2 and κ(x) on the same screen for −2 ≤ x ≤ 2. (c) Find the radius of curvature at each relative extremum. (d) Make a reasonably accurate hand-drawn sketch that shows the graph of f(x) = x 4 − 2x 2 and the osculating circles in their correct proportions at the relative extrema. 38. (a) Use a CAS to graph the parametric curve x = t cos t, y = t sin t for t ≥ 0. (b) Make a conjecture about the behavior of the curvature κ(t) as t → +⬁. (c) Use the CAS and part (a) of Exercise 23 to find κ(t). (d) Check your conjecture by finding the limit of κ(t) as t → +⬁.

39. Use the formula in Exercise 23(a) to show that for a curve in polar coordinates described by r = f(θ ) the curvature is 2 d 2 r dr 2 −r 2 r + 2 dθ dθ κ(θ ) = 2 3/2 dr r2 + dθ

51. (a) Use the result in Exercise 24 to show that for the parabola y = x 2 the curvature κ(φ) at points where the tangent line has an angle of inclination of φ is κ(φ) = |2 cos3 φ| (b) Use the result in part (a) to ﬁnd the radius of curvature of the parabola at the point on the parabola where the tangent line has slope 1. (c) Make a sketch with reasonably accurate proportions that shows the osculating circle at the point on the parabola where the tangent line has slope 1. 52. The evolute of a smooth parametric curve C in 2-space is the curve formed from the centers of curvature of C. The accompanying ﬁgure shows the ellipse x = 3 cos t, y = 2 sin t (0 ≤ t ≤ 2π) and its evolute graphed together. (a) Which points on the evolute correspond to t = 0 and t = π/2? (b) In what direction is the evolute traced as t increases from 0 to 2π? (c) What does the evolute of a circle look like? Explain your reasoning. y 3

x −3

−3

[Hint: Let θ be the parameter and use the relationships x = r cos θ, y = r sin θ.]

40. Use the result in Exercise 39 to show that a circle has constant curvature. 41–44 Use the formula in Exercise 39 to ﬁnd the curvature at

the indicated point. ■ 41. r = 1 + cos θ ; θ = π/2 43. r = sin 3θ ; θ = 0

42. r = e2θ ; θ = 1 44. r = θ; θ = 1

45. Find the radius of curvature of the parabola y 2 = 4px at (0, 0). x

46. At what point(s) does y = e have maximum curvature?

47. At what point(s) does 4x 2 + 9y 2 = 36 have minimum radius of curvature?

48. Find the maximum and minimum values of the radius of curvature for the curve x = cos t, y = sin t, z = cos t.

49. Use the formula in Exercise 39 to show that the curvature of the polar curve r = eaθ is inversely proportional to r.

50. Use the formula in Exercise 39 and √ a CAS to show that the curvature of the lemniscate r = a cos 2θ is directly proportional to r.

Figure Ex-52

F O C U S O N C O N C E P TS

53–57 These exercises are concerned with the problem of

creating a single smooth curve by piecing together two separate smooth curves. If two smooth curves C1 and C2 are joined at a point P to form a curve C, then we will say that C1 and C2 make a smooth transition at P if the curvature of C is continuous at P . ■ 53. Show that the transition at x = 0 from the horizontal line y = 0 for x ≤ 0 to the parabola y = x 2 for x > 0 is not smooth, whereas the transition to y = x 3 for x > 0 is smooth. 54. (a) Sketch the graph of the curve deﬁned piecewise by y = x 2 for x < 0, y = x 4 for x ≥ 0. (b) Show that for the curve in part (a) the transition at x = 0 is not smooth.

55. The accompanying ﬁgure on the next page shows the arc of a circle of radius r with center at (0, r). Find the value of a so that there is a smooth transition from the circle to the parabola y = ax 2 at the point where x = 0.

12.5 Curvature y Arc of circle (0, r)

y = ax

Figure Ex-55

Use the ﬁrst two Frenet–Serret formulas and the fact that r ′ (s) = T if r = r(s) to show that [r ′ (s) × r ′′ (s)] ⴢ r ′′′ (s) r ′ (s) × r ′′ (s) τ= and B = ′′ 2

r (s)

r ′′ (s)

62. (a) Use the chain rule and the ﬁrst two Frenet–Serret formulas in Exercise 61 to show that T′ = κs ′ N

56. Find a, b, and c so that there is a smooth transition at x = 0 from the curve y = ex for x ≤ 0 to the parabola y = ax 2 + bx + c for x > 0. [Hint: The curvature is continuous at those points where y ′′ is continuous.]

dT = κ(s)N(s) ds and use this result to obtain the formulas in (10). 59. (a) Show that dB/ds is perpendicular to B(s). (b) Show that dB/ds is perpendicular to T(s). [Hint: Use the fact that B(s) is perpendicular to both T(s) and N(s), and differentiate B ⴢ T with respect to s.] (c) Use the results in parts (a) and (b) to show that dB/ds is a scalar multiple of N(s). The negative of this scalar is called the torsion of r(s) and is denoted by τ (s). Thus, dB = −τ (s)N(s) ds (d) Show that τ (s) = 0 for all s if the graph of r(s) lies in a plane. [Note: For reasons that we cannot discuss here, the torsion is related to the “twisting” properties of the curve, and τ (s) is regarded as a numerical measure of the tendency for the curve to twist out of the osculating plane.]

+ [3κs ′ s ′′ + κ ′ (s ′ )2 ]N + κτ (s ′ )3 B (d) Use the results in parts (b) and (c) to show that [r ′ (t) × r ′′ (t)] ⴢ r ′′′ (t) τ (t) =

r ′ (t) × r ′′ (t) 2 63–66 Use the formula in Exercise 62(d) to ﬁnd the torsion τ = τ (t). ■

63. The twisted cubic r(t) = 2ti + t 2 j + 13 t 3 k

64. The circular helix r(t) = a cos ti + a sin t j + ctk √ 2tk

65. r(t) = et i + e−t j +

66. r(t) = (t − sin t)i + (1 − cos t) j + tk

67. Writing One property of a twice-differentiable function f(x) is that an inﬂection point on the graph is a point at which the tangent line crosses the graph of f . Consider the analogous issue in 2-space for an osculating circle to a curve C at a point P : What does it mean for the osculating circle to cross (or not to cross) C at P ? Investigate this issue through some examples of your own and write a brief essay, with illustrations, supporting your conclusions. 68. Writing The accompanying ﬁgure is the graph of the radius of curvature versus θ in rectangular coordinates for the cardioid r = 1 + cos θ. In words, explain what the graph tells you about the cardioid. y

60. Let κ be the curvature of C and τ the torsion (deﬁned in Exercise 59). By differentiating N = B × T with respect to s, show that dN/ds = −κT + τ B.

1.5 1

61. The following derivatives, known as the Frenet–Serret formulas, are fundamental in the theory of curves in 3-space: dT/ds = κN dN/ds = −κT + τ B dB/ds = −τ N

0.5 u 6

[Exercise 58] [Exercise 60] [Exercise 59(c)]

2. (a)

′ ′ T (t) r (t) × r′′ (t) (b)

r′ (t)

r′ (t) 3

y = 1/k(u)

✔QUICK CHECK ANSWERS 12.5 dT ′′ 1. ds = r (s)

N′ = −κs ′ T + τ s ′ B

r ′′ (t) = s ′′ T + κ(s ′ )2 N

r ′ (t) = s ′ T and

58–61 Assume that s is an arc length parameter for a smooth

58. Show that

and

where primes denote differentiation with respect to t. (b) Show that Formulas (4) and (6) can be written in the form

57. Assume that f is a function for which f ′′′(x) is deﬁned for all x ≤ 0. Explain why it is always possible to ﬁnd numbers a, b, and c such that there is a smooth transition at x = 0 from the curve y = f(x), x ≤ 0, to the parabola y = ax 2 + bx + c. vector-valued function r(s) in 3-space and that dT/ds and dN/ds exist at each point on the curve. (This implies that dB/ds exists as well, since B = T × N.) ■

881

dφ 3. ds

1 2

Figure Ex-68

882

12.6

Chapter 12 / Vector-Valued Functions

MOTION ALONG A CURVE In earlier sections we considered the motion of a particle along a line. In that situation there are only two directions in which the particle can move—the positive direction or the negative direction. Motion in 2-space or 3-space is more complicated because there are inﬁnitely many directions in which a particle can move. In this section we will show how vectors can be used to analyze motion along curves in 2-space or 3-space.

VELOCITY, ACCELERATION, AND SPEED

Let us assume that the motion of a particle in 2-space or 3-space is described by a smooth vector-valued function r(t) in which the parameter t denotes time; we will call this the position function or trajectory of the particle. As the particle moves along its trajectory, its direction of motion and its speed can vary from instant to instant. Thus, before we can undertake any analysis of such motion, we must have clear answers to the following questions:

• What is the direction of motion of the particle at an instant of time? • What is the speed of the particle at an instant of time? We will deﬁne the direction of motion at time t to be the direction of the unit tangent vector T(t), and we will deﬁne the speed to be ds /dt—the instantaneous rate of change of the arc length traveled by the particle from an arbitrary reference point. Taking this a step further, we will combine the speed and the direction of motion to form the vector r(t)

T(t) v(t) = The length of the velocity vector is the speed of the particle, and the direction of the velocity vector is the direction of motion.

Figure 12.6.1

ds T(t) (1) dt which we call the velocity of the particle at time t. Thus, at each instant of time the velocity vector v(t) points in the direction of motion and has a magnitude that is equal to the speed of the particle (Figure 12.6.1). Recall that for motion along a coordinate line the velocity function is the derivative of the position function. The same is true for motion along a curve, since v(t) =

ds T(t) dt

dr ds ds dr = = T(t) = v(t) dt ds dt dt For motion along a coordinate line, the acceleration function was deﬁned to be the derivative of the velocity function. The deﬁnition is the same for motion along a curve.

12.6.1 definition If r(t) is the position function of a particle moving along a curve in 2-space or 3-space, then the instantaneous velocity, instantaneous acceleration, and instantaneous speed of the particle at time t are deﬁned by velocity = v(t) =

dr dt

acceleration = a(t) = speed = v(t) =

ds dt

(2) dv d 2r = 2 dt dt

(3) (4)

12.6 Motion Along a Curve

883

As shown in Table 12.6.1, the position, velocity, acceleration, and speed can also be expressed in component form. Table 12.6.1 formulas for position, velocity, acceleration, and speed

2-space

3-space

position

r(t) = x(t)i + y(t)j

velocity

v(t) =

dy dx i+ j dt dt

v(t) =

dy dz dx i+ j+ k dt dt dt

acceleration

a(t) =

d 2y d 2x i+ 2 j 2 dt dt

a(t) =

d 2y d 2z d 2x i+ 2 j+ 2k 2 dt dt dt

speed

|| v(t) || =

r(t) = x(t)i + y(t)j + z(t)k

√ 冸 dt 冹 + 冸 dt 冹 dx

|| v(t) || =

√ 冸 dt 冹 + 冸 dt 冹 + 冸 dt 冹 dx

Example 1 A particle moves along a circular path in such a way that its x- and y-coordinates at time t are x = 2 cos t, y = 2 sin t (a) Find the instantaneous velocity and speed of the particle at time t. (b) Sketch the path of the particle, and show the position and velocity vectors at time t = π/4 with the velocity vector drawn so that its initial point is at the tip of the position vector. (c) Show that at each instant the acceleration vector is perpendicular to the velocity vector.

Solution (a). At time t, the position vector is r(t) = 2 cos ti + 2 sin tj so the instantaneous velocity and speed are y

v = −√2 i + √2 j r = √2 i + √2 j c/4

−2

x 2

Figure 12.6.2

dr = −2 sin ti + 2 cos tj dt

v(t) = (−2 sin t)2 + (2 cos t)2 = 2

v(t) =

Solution (b). The graph of the parametric equations is a circle of radius 2 centered at the origin. At time t = π/4 the position and velocity vectors of the particle are √ √ r(π/4) = 2 cos(π/4)i + 2 sin(π/4)j = 2i + 2 j √ √ v(π/4) = −2 sin(π/4)i + 2 cos(π/4)j = − 2 i + 2 j These vectors and the circle are shown in Figure 12.6.2.

Solution (c). At time t, the acceleration vector is dv = −2 cos ti − 2 sin tj dt One way of showing that v(t) and a(t) are perpendicular is to show that their dot product is zero (try it). However, it is easier to observe that a(t) is the negative of r(t), which implies that v(t) and a(t) are perpendicular, since at each point on a circle the radius and tangent line are perpendicular. a(t) =

How could you apply Theorem 12.2.8 to answer part (c) of Example 1?

884

Chapter 12 / Vector-Valued Functions

Since v(t) can be obtained by differentiating r(t), and since a(t) can be obtained by differentiating v(t), it follows that r(t) can be obtained by integrating v(t), and v(t) can be obtained by integrating a(t). However, such integrations do not produce unique functions because constants of integration occur. Typically, initial conditions are required to determine these constants. Example 2 A particle moves through 3-space in such a way that its velocity is v(t) = i + tj + t 2 k Find the coordinates of the particle at time t = 1 given that the particle is at the point (â&#x2C6;&#x2019;1, 2, 4) at time t = 0.

Solution. Integrating the velocity function to obtain the position function yields r(t) =

v(t) dt =

(i + tj + t 2 k) dt = ti +

t2 t3 j+ k+C 2 3

(5)

where C is a vector constant of integration. Since the coordinates of the particle at time t = 0 are (â&#x2C6;&#x2019;1, 2, 4), the position vector at time t = 0 is r(0) = â&#x2C6;&#x2019;i + 2j + 4k

(6)

It follows on substituting t = 0 in (5) and equating the result with (6) that C = â&#x2C6;&#x2019;i + 2 j + 4k Substituting this value of C in (5) and simplifying yields 2

t t r(t) = (t â&#x2C6;&#x2019; 1)i + +2 j+ +4 k 2 3 Thus, at time t = 1 the position vector of the particle is r(1) = 0i +

5 13 j+ k 2 3

. so its coordinates at that instant are 0, 25 , 13 3 DISPLACEMENT AND DISTANCE TRAVELED

If a particle travels along a curve C in 2-space or 3-space, the displacement of the particle over the time interval t1 â&#x2030;¤ t â&#x2030;¤ t2 is commonly denoted by r and is deďŹ ned as (7)

r = r(t2 ) â&#x2C6;&#x2019; r(t1 )

(Figure 12.6.3). The displacement vector, which describes the change in position of the particle during the time interval, can be obtained by integrating the velocity function from t1 to t2 : t2 t2 t2 dr dt = r(t) = r(t2 ) â&#x2C6;&#x2019; r(t1 ) r = Displacement (8) v(t) dt = t1 t1 dt t1 It follows from Theorem 12.3.1 that we can ďŹ nd the distance s traveled by a particle over a time interval t1 â&#x2030;¤ t â&#x2030;¤ t2 by integrating the speed over that interval, since s=

dr dt = dt

v(t) dt

Distance traveled

(9)

12.6 Motion Along a Curve y

885

L r(t1) Î&#x201D;r

r(t2)

Displacement = Î&#x201D;r = r(t2) â&#x2C6;&#x2019; r(t1) x

Figure 12.6.3

Example 3 Suppose that a particle moves along a circular helix in 3-space so that its position vector at time t is r(t) = (4 cos Ď&#x20AC;t)i + (4 sin Ď&#x20AC;t) j + tk

Find the distance traveled and the displacement of the particle during the time interval 1 â&#x2030;¤ t â&#x2030;¤ 5.

Solution. We have dr = (â&#x2C6;&#x2019;4Ď&#x20AC; sin Ď&#x20AC;t)i + (4Ď&#x20AC; cos Ď&#x20AC;t) j + k dt â&#x2C6;&#x161;

v(t) = (â&#x2C6;&#x2019;4Ď&#x20AC; sin Ď&#x20AC;t)2 + (4Ď&#x20AC; cos Ď&#x20AC;t)2 + 1 = 16Ď&#x20AC;2 + 1

v(t) =

Thus, it follows from (9) that the distance traveled by the particle from time t = 1 to t = 5 5 is s= 16Ď&#x20AC;2 + 1 dt = 4 16Ď&#x20AC;2 + 1 1

Moreover, it follows from (8) that the displacement over the time interval is r = r(5) â&#x2C6;&#x2019; r(1) = (4 cos 5Ď&#x20AC;i + 4 sin 5Ď&#x20AC;j + 5k) â&#x2C6;&#x2019; (4 cos Ď&#x20AC;i + 4 sin Ď&#x20AC;j + k) = (â&#x2C6;&#x2019;4i + 5k) â&#x2C6;&#x2019; (â&#x2C6;&#x2019;4i + k) = 4k

which tells us that the change in the position of the particle over the time interval was 4 units straight up. NORMAL AND TANGENTIAL COMPONENTS OF ACCELERATION You know from your experience as an automobile passenger that if a car speeds up rapidly, then your body is thrown back against the backrest of the seat. You also know that if the car rounds a turn in the road, then your body is thrown toward the outside of the curveâ&#x20AC;&#x201D;the greater the curvature in the road, the greater this effect. The explanation of these effects can be understood by resolving the velocity and acceleration components of the motion into vector components that are parallel to the unit tangent and unit normal vectors. The following theorem explains how to do this.

ds 2 k N dt

ĺ&#x2020;¸ ĺ&#x2020;š

12.6.2 theorem If a particle moves along a smooth curve C in 2-space or 3-space, then at each point on the curve velocity and acceleration vectors can be written as a

d 2s T dt 2

N C Figure 12.6.4

v= T

ds T dt

ds T v= dt

d 2s a = 2T+Îş dt

ds dt

(10â&#x20AC;&#x201C;11)

where s is an arc length parameter for the curve, and T, N, and Îş denote the unit tangent vector, unit normal vector, and curvature at the point (Figure 12.6.4).

886

Chapter 12 / Vector-Valued Functions

proof Formula (10) is just a restatement of (1). To obtain (11), we differentiate both sides of (10) with respect to t; this yields

d 2s ds dT d ds T = 2T+ a= dt dt dt dt dt d 2s ds dT ds T+ 2 dt dt ds dt

d 2s ds 2 dT = 2T+ dt dt ds 2 d 2s ds = 2T+ κN dt dt =

Formula (10) of Section 12.5

from which (11) follows. ■ The coefﬁcients of T and N in (11) are commonly denoted by

Formula (14) applies to motion in both 2-space and 3-space. What is interesting is that the 3-space formula does not involve the binormal vector B, so the acceleration vector always lies in the plane of T and N (the osculating plane), even for highly twisting paths of motion (Figure 12.6.5).

aNN

aTT

a = aTT + aNN

aT =

d 2s dt 2

aN = κ

ds dt

(12–13)

in which case Formula (11) is expressed as (14)

a = aT T + aN N

In this formula the scalars aT and aN are called the tangential scalar component of acceleration and the normal scalar component of acceleration, and the vectors aT T and aN N are called the tangential vector component of acceleration and the normal vector component of acceleration. The scalar components of acceleration explain the effect that you experience when a car speeds up rapidly or rounds a turn. The rapid increase in speed produces a large value for d 2 s /dt 2 , which results in a large tangential scalar component of acceleration; and by Newton’s second law this corresponds to a large tangential force on the car in the direction of motion. To understand the effect of rounding a turn, observe that the normal scalar component of acceleration has the curvature κ and the square of the speed ds /dt as factors. Thus, sharp turns or turns taken at high speed both correspond to large normal forces on the car. Although Formulas (12) and (13) provide useful insight into the behavior of particles moving along curved paths, they are not always the best formulas for computations. The following theorem provides some more useful formulas that relate aT , aN , and κ to the velocity v and acceleration a.

Figure 12.6.5

Theorem 12.6.3 applies to motion in 2space and 3-space, but for motion in 2-space you will have to add a zero k component to v to calculate the cross product.

12.6.3 theorem If a particle moves along a smooth curve C in 2-space or 3-space, then at each point on the curve the velocity v and the acceleration a are related to aT , aN , and κ by the formulas

aT =

vⴢa

aN =

v × a

κ=

v × a

v 3

(15–17)

12.6 Motion Along a Curve

aNN

||a||

aT T

proof As illustrated in Figure 12.6.6, let θ be the angle between the vector a and the vector aT T. Thus, aT = a cos θ and aN = a sin θ from which we obtain aT = a cos θ =

aT = ||a|| cos u aN = ||a|| sin u

aN = a sin θ =

Figure 12.6.6

κ=

Recall that for nonlinear smooth curves in 2-space the unit normal vector N is the inward normal (points toward the concave side of the curve). Explain why the same is true for aN N.

887

Example 4 at time t is

a cos θ vⴢa =

v × a

a sin θ =

aN aN 1 v × a

v × a

= = = ■ (ds /dt)2

v 2

v 2 v

v 3

Suppose that a particle moves through 3-space so that its position vector r(t) = ti + t 2 j + t 3 k

(The path is the twisted cubic shown in Figure 12.1.5.) (a) Find the scalar tangential and normal components of acceleration at time t. (b) Find the scalar tangential and normal components of acceleration at time t = 1. (c) Find the vector tangential and normal components of acceleration at time t = 1.

(d) Find the curvature of the path at the point where the particle is located at time t = 1.

Solution (a). We have v(t) = r ′ (t) = i + 2tj + 3t 2 k a(t) = v′ (t) = 2 j + 6tk √

v(t) = 1 + 4t 2 + 9t 4

v(t) ⴢ a(t) = 4t + 18t 3 i j k v(t) × a(t) = 1 2t 3t 2 = 6t 2 i − 6tj + 2k 0 2 6t

Thus, from (15) and (16)

vⴢa 4t + 18t 3 =

1 + 4t 2 + 9t 4

v × a

36t 4 + 36t 2 + 4 9t 4 + 9t 2 + 1 aN = = =2

9t 4 + 4t 2 + 1 1 + 4t 2 + 9t 4

aT =

Solution (b). At time t = 1, the components aT and aN in part (a) are 22 aT = √ ≈ 5.88 14

and aN = 2

19 ≈ 2.33 14

Solution (c). Since T and v have the same direction, T can be obtained by normalizing v, that is, T(t) =

v(t)

888

Chapter 12 / Vector-Valued Functions

At time t = 1 we have T(1) =

i + 2 j + 3k 1 v(1) = = √ (i + 2 j + 3k)

v(1)

i + 2 j + 3k

From this and part (b) we obtain the vector tangential component of acceleration: 11 22 33 22 11 (i + 2 j + 3k) = i+ j+ k aT (1)T(1) = √ T(1) = 7 7 7 7 14 To ﬁnd the normal vector component of acceleration, we rewrite a = aT T + aN N as aN N = a − aT T Thus, at time t = 1 the normal vector component of acceleration is aN (1)N(1) = a(1) − aT (1)T(1)

11 22 33 = (2 j + 6k) − i+ j+ k 7 7 7 =−

11 8 9 i− j+ k 7 7 7

Solution (d). We will apply Formula (17) with t = 1. From part (a)

v(1) = Thus, at time t = 1

Confirm that the value of aN computed in Example 4 agrees with the value that results by applying Formula (18).

√ 14

and v(1) × a(1) = 6i − 6j + 2k

√ 76 1 38

v × a

= √ 3 = ≈ 0.17 κ=

v 3 14 7 ( 14)

In the case where a and aT are known, there is a useful alternative to Formula (16) for aN that does not require the calculation of a cross product. It follows algebraically from Formula (14) (see Exercise 51) or geometrically from Figure 12.6.6 and the Theorem of Pythagoras that aN =

a 2 − aT2

(18)

A MODEL OF PROJECTILE MOTION

Earlier in this text we examined various problems concerned with objects moving vertically in the Earth’s gravitational ﬁeld (see the subsection of Section 5.7 entitled Free-Fall Model and the subsection of Section 8.4 entitled A Model of Free-Fall Motion Retarded by Air Resistance). Now we will consider the motion of a projectile launched along a curved path in the Earth’s gravitational ﬁeld. For this purpose we will need the following vector version of Newton’s Second Law of Motion (6.6.4) F = ma

(19)

and we will need to make three modeling assumptions:

• The mass m of the object is constant. • The only force acting on the object after it is launched is the force of the Earth’s gravity. (Thus, air resistance and the gravitational effect of other planets and celestial objects are ignored.)

• The object remains sufﬁciently close to the Earth that we can assume the force of gravity to be constant.

12.6 Motion Along a Curve y

s0 x Earth

Figure 12.6.7

889

Let us assume that at time t = 0 an object of mass m is launched from a height of s0 above the Earth with an initial velocity vector of v0 . Furthermore, let us introduce an xy-coordinate system as shown in Figure 12.6.7. In this coordinate system the positive y-direction is up, the origin is at the surface of the Earth, and the initial location of the object is (0, s0 ). Our objective is to use basic principles of physics to derive the velocity function v(t) and the position function r(t) from the acceleration function a(t) of the object. Our starting point is the physical observation that the downward force F of the Earthâ&#x20AC;&#x2122;s gravity on an object of mass m is F = â&#x2C6;&#x2019;mgj where g is the acceleration due to gravity. It follows from this fact and Newtonâ&#x20AC;&#x2122;s second law (19) that ma = â&#x2C6;&#x2019;mgj or on canceling m from both sides a = â&#x2C6;&#x2019;gj

(20)

Observe that this acceleration function does not involve t and hence is constant. We can now obtain the velocity function v(t) by integrating this acceleration function and using the initial condition v(0) = v0 to ďŹ nd the constant of integration. Integrating (20) with respect to t and keeping in mind that â&#x2C6;&#x2019;gj is constant yields v(t) = â&#x2C6;&#x2019;gj dt = â&#x2C6;&#x2019;gtj + c1 where c1 is a vector constant of integration. Substituting t = 0 in this equation and using the initial condition v(0) = v0 yields v0 = c1 . Thus, the velocity function of the object is v(t) = â&#x2C6;&#x2019;gtj + v0

(21)

To obtain the position function r(t) of the object, we will integrate the velocity function and use the known initial position of the object to ďŹ nd the constant of integration. For this purpose observe that the object has coordinates (0, s0 ) at time t = 0, so the position vector at that time is r(0) = 0i + s0 j = s0 j (22) This is the initial condition that we will need to ďŹ nd the constant of integration. Integrating (21) with respect to t yields r(t) = (â&#x2C6;&#x2019;gtj + v0 ) dt = â&#x2C6;&#x2019; 21 gt 2 j + tv0 + c2 (23) Observe that the mass m does not appear in Formulas (21) and (24) and hence has no influence on the velocity or the trajectory of the object. This explains the famous observation of Galileo that two objects of different mass that are released from the same height reach the ground at the same time if air resistance is neglected.

where c2 is another vector constant of integration. Substituting t = 0 in (23) and using initial condition (22) yields s0 j = c2

so that (23) can be written as r(t) = â&#x2C6;&#x2019; 21 gt 2 + s0 j + tv0

(24)

This formula expresses the position function of the object in terms of its known initial position and velocity.

PARAMETRIC EQUATIONS OF PROJECTILE MOTION

Formulas (21) and (24) can be used to obtain parametric equations for the position and velocity in terms of the initial speed of the object and the angle that the initial velocity vector makes with the positive x-axis. For this purpose, let v0 = v0 be the initial speed, let Îą be the angle that the initial velocity vector v0 makes with the positive x-axis, let vx and vy be the horizontal and vertical scalar components of v(t) at time t, and let x and y

890

Chapter 12 / Vector-Valued Functions y

be the horizontal and vertical components of r(t) at time t. As illustrated in Figure 12.6.8, the initial velocity vector can be expressed as

v0 = (v0 cos α)i + (v0 sin α) j

(v0 sin a)j

Substituting this expression in (24) and combining like components yields (verify) r(t) = (v0 cos α)ti + s0 + (v0 sin α)t − 21 gt 2 j

a (v0 cos a)i

(25)

(26)

which is equivalent to the parametric equations

x = (v0 cos α)t,

Figure 12.6.8

y = s0 + (v0 sin α)t − 21 gt 2

(27)

Similarly, substituting (25) in (21) and combining like components yields v(t) = (v0 cos α)i + (v0 sin α − gt)j which is equivalent to the parametric equations vx = v0 cos α,

vy = v0 sin α − gt

(28)

The parameter t can be eliminated in (27) by solving the ﬁrst equation for t and substituting in the second equation. We leave it for you to show that this yields

g y = s0 + (tan α)x − (29) x2 2v02 cos2 α which is the equation of a parabola, since the right side is a quadratic polynomial in x. Thus, we have shown that the trajectory of the projectile is a parabolic arc. Example 5 A shell, ﬁred from a cannon, has a muzzle speed (the speed as it leaves the barrel) of 800 ft/s. The barrel makes an angle of 45 ◦ with the horizontal and, for simplicity, the barrel opening is assumed to be at ground level. y

45° Figure 12.6.9

(a) Find parametric equations for the shell’s trajectory relative to the coordinate system in Figure 12.6.9. (b) How high does the shell rise? (c) How far does the shell travel horizontally? (d) What is the speed of the shell at its point of impact with the ground?

Solution (a). From (27) with v0 = 800 ft/s, α = 45 ◦ , s0 = 0 ft (since the shell starts at ground level), and g = 32 ft/s2 , we obtain the parametric equations x = (800 cos 45 ◦ )t,

y = (800 sin 45 ◦ )t − 16t 2

(t ≥ 0)

which simplify to √ x = 400 2t,

√ y = 400 2t − 16t 2

(t ≥ 0)

(30)

Solution (b). The maximum height of the shell is the maximum value of y in (30), which occurs when dy/dt = 0, that is, when √ √ 25 2 400 2 − 32t = 0 or t = 2 Substituting this value of t in (30) yields y = 5000 ft as the maximum height of the shell.

12.6 Motion Along a Curve

891

Solution (c). The shell will hit the ground when y = 0. From (30), this occurs when √ 400 2t − 16t 2 = 0

√ or t (400 2 − 16t) = 0

The solution t = 0 corresponds to the initial position of the shell and the solution √ t = 25 2 to the time of impact. Substituting the latter value in the equation for x in (30) yields x = 20,000 ft as the horizontal distance traveled by the shell.

Solution (d). From (30), the position function of the shell is

√ √ r(t) = 400 2ti + (400 2t − 16t 2 )j

so that the velocity function is

The speed at impact and the muzzle speed of the shell in Example 5 are the same. Is this an expected result? Explain.

√ √ v(t) = r ′ (t) = 400 2i + (400 2 − 32t)j √ From part (c), impact occurs when t = 25 2, so the velocity vector at this point is √ √ √ √ √ √ v(25 2) = 400 2i + [400 2 − 32(25 2)] j = 400 2 i − 400 2 j Thus, the speed at impact is √ √ √

v(25 2) = (400 2)2 + (−400 2)2 = 800 ft/s

✔QUICK CHECK EXERCISES 12.6

(See page 895 for answers.)

1. If r(t) is the position function of a particle, then the velocity, acceleration, and speed of the particle at time t are given, respectively, by v(t) =

a(t) =

ds = dt

2. If r(t) is the position function of a particle, then the displacement of the particle over the time interval t1 ≤ t ≤ t2 , and the distance s traveled by the particle duris ing this time interval is given by the integral .

EXERCISE SET 12.6

Graphing Utility

3. The tangential scalar component of acceleration is given by the formula , and the normal scalar component of acceleration is given by the formula . 4. The projectile motion model r(t) = − 21 gt 2 + s0 j + tv0

describes the motion of an object with constant acceleration a = and velocity function v(t) = . The initial position of the object is and its initial velocity is .

CAS

1–4 In these exercises r(t) is the position vector of a particle

moving in the plane. Find the velocity, acceleration, and speed at an arbitrary time t. Then sketch the path of the particle together with the velocity and acceleration vectors at the indicated time t. ■ 1. r(t) = 3 cos ti + 3 sin tj; t = π/3 2. r(t) = ti + t 2 j; t = 2

3. r(t) = et i + e−t j; t = 0

4. r(t) = (2 + 4t)i + (1 − t) j; t = 1

5–8 Find the velocity, speed, and acceleration at the given time t of a particle moving along the given curve. ■

5. r(t) = ti + 21 t 2 j + 13 t 3 k; t = 1

6. x = 1 + 3t, y = 2 − 4t, z = 7 + t; t = 2

7. x = 2 cos t, y = 2 sin t, z = t; t = π/4

8. r(t) = et sin ti + et cos tj + tk; t = π/2 F O C U S O N C O N C E P TS

9. As illustrated in the accompanying ﬁgure on the next page, suppose that the equations of motion of a particle moving along an elliptic path are x = a cos ωt, y = b sin ωt. (a) Show that the acceleration is directed toward the origin. (cont.)

892

Chapter 12 / Vector-Valued Functions

(b) Show that the magnitude of the acceleration is proportional to the distance from the particle to the origin. y

(b) Use the graph to estimate the maximum and minimum speeds of the particle. (c) Use the graph to estimate the time at which the maximum speed ﬁrst occurs. (d) Find the exact values of the maximum and minimum speeds and the exact time at which the maximum speed ﬁrst occurs.

17–20 Use the given information to ﬁnd the position and velocity vectors of the particle. ■ Figure Ex-9

10. Suppose that a particle vibrates in such a way that its position function is r(t) = 16 sin πti + 4 cos 2πt j, where distance is in millimeters and t is in seconds. (a) Find the velocity and acceleration at time t = 1 s. (b) Show that the particle moves along a parabolic curve. (c) Show that the particle moves back and forth along the curve. 11. What can you say about the trajectory of a particle that moves in 2-space or 3-space with zero acceleration? Justify your answer. 12. Recall from Theorem 12.2.8 that if r(t) is a vectorvalued function in 2-space or 3-space, and if r(t) is constant for all t, then r(t) ⴢ r ′ (t) = 0. (a) Translate this theorem into a statement about the motion of a particle in 2-space or 3-space. (b) Replace r(t) by r ′ (t) in the theorem, and translate the result into a statement about the motion of a particle in 2-space or 3-space.

17. a(t) = − cos ti − sin tj; v(0) = i; r(0) = j

18. a(t) = i + e−t j; v(0) = 2i + j; r(0) = i − j

19. a(t) = sin ti + cos tj + et k; v(0) = k; r(0) = −i + k 20. a(t) = (t + 1)−2 j − e−2t k; v(0) = 3i − j; r(0) = 2k

21. Find, to the nearest degree, the angle between v and a for r = t 3 i + t 2 j when t = 1. 22. Show that the angle between v and a is constant for the position vector r = et cos ti + et sin tj. Find the angle. 23. (a) Suppose that at time t = t0 an electron has a position vector of r = 3.5i − 1.7 j + k, and at a later time t = t1 it has a position vector of r = 4.2i + j − 2.4k. What is the displacement of the electron during the time interval from t0 to t1 ? (b) Suppose that during a certain time interval a proton has a displacement of r = 0.7i + 2.9 j − 1.2k and its ﬁnal position vector is known to be r = 3.6k. What was the initial position vector of the proton?

13. Suppose that the position vector of a particle moving in √ the plane is r = 12 t i + t 3/2 j, t > 0. Find the minimum speed of the particle and its location when it has this speed.

24. Suppose that the position function of a particle moving along a circle in the xy-plane is r = 5 cos 2πti + 5 sin 2πtj. (a) Sketch some typical displacement vectors over the time interval from t = 0 to t = 1. (b) What is the distance traveled by the particle during the time interval?

14. Suppose that the motion of a particle is described by the position vector r = (t − t 2 )i − t 2 j. Find the minimum speed of the particle and its location when it has this speed.

25–28 Find the displacement and the distance traveled over the indicated time interval. ■

15. Suppose that the position function of a particle moving in 2-space is r = sin 3ti − 2 cos 3tj, 0 ≤ t ≤ 2π/3. (a) Use a graphing utility to graph the speed of the particle versus time from t = 0 to t = 2π/3. (b) What are the maximum and minimum speeds of the particle? (c) Use the graph to estimate the time at which the maximum speed ﬁrst occurs. (d) Find the exact time at which the maximum speed ﬁrst occurs. 16. Suppose that the position function of a particle moving in 3-space is r = 3 cos 2ti + sin 2tj + 4tk. (a) Use a graphing utility to graph the speed of the particle versus time from t = 0 to t = π.

25. r = t 2 i + 31 t 3 j; 1 ≤ t ≤ 3 26. r = (1 − 3 sin t)i + 3 cos tj; 0 ≤ t ≤ 3π/2 √ 27. r = et i + e−t j + 2tk; 0 ≤ t ≤ ln 3 28. r = cos 2ti + (1 − cos 2t) j + 3 + 21 cos 2t k; 0 ≤ t ≤ π

29–30 The position vectors r1 and r2 of two particles are given. Show that the particles move along the same path but the speed of the ﬁrst is constant and the speed of the second is not. ■

29. r1 = 2 cos 3ti + 2 sin 3tj r2 = 2 cos(t 2 )i + 2 sin(t 2 ) j

(t ≥ 0)

30. r1 = (3 + 2t)i + tj + (1 − t)k r2 = (5 − 2t 3 )i + (1 − t 3 ) j + t 3 k

12.6 Motion Along a Curve 31–36 The position function of a particle is given. Use Theo-

rem 12.6.3 to ﬁnd (a) the scalar tangential and normal components of acceleration at the stated time t; (b) the vector tangential and normal components of acceleration at the stated time t; (c) the curvature of the path at the point where the particle is located at the stated time t. ■ 31. r = e−t i + et j; t = 0

32. r = cos(t 2 )i + sin(t 2 ) j; t =

√ π/2

33. r = (t 3 − 2t)i + (t 2 − 4) j; t = 1 34. r = et cos ti + et sin tj; t = π/4 35. r = et i + e−2t j + tk; t = 0

36. r = 3 sin ti + 2 cos tj − sin 2tk; t = π/2 37–38 In these exercises v and a are given at a certain instant

of time. Find aT , aN , T, and N at this instant. ■ 37. v = −4 j, a = 2i + 3 j

38. v = 2i + 2 j + k, a = i + 2k

39–40 The speed v of a particle at an arbitrary time t is

given. Find the scalar tangential component of acceleration at the indicated time. ■ 39. v = t 2 + e−3t ; t = 0 40. v = (4t − 1)2 + cos2 πt; t = 41

41. The nuclear accelerator at the Enrico Fermi Laboratory is circular with a radius of 1 km. Find the scalar normal component of acceleration of a proton moving around the accelerator with a constant speed of 2.9 × 105 km/s.

42. Suppose that a particle moves with nonzero acceleration along the curve y = f(x). Use part (b) of Exercise 23 in Section 12.5 to show that the acceleration vector is tangent to the curve at each point where f ′′(x) = 0.

43–44 Use the given information and Exercise 23 of Section 12.5 to ﬁnd the normal scalar component of acceleration as a function of x. ■

43. A particle moves along the parabola y = x 2 with a constant speed of 3 units per second. 44. A particle moves along the curve x = ln y with a constant speed of 2 units per second. 45–46 Use the given information to ﬁnd the normal scalar component of acceleration at time t = 1. ■

893

48. If a particle moves along a smooth curve C in 3-space, then at each point on C the normal scalar component of acceleration for the particle is the product of the curvature of C and speed of the particle at the point. 49. If a particle is moving along a smooth curve C and passes through a point at which the curvature is zero, then the velocity and acceleration vectors have the same direction at that point. 50. The distance traveled by a particle over a time interval is the magnitude of the displacement vector for the particle during that time interval. 51. Derive Formula (18) from Formula (14). 52. An automobile travels at a constant speed around a curve whose radius of curvature is 1000 m. What is the maximum allowable speed if the maximum acceptable value for the normal scalar component of acceleration is 1.5 m/s2 ? 53. If an automobile of mass m rounds a curve, then its inward vector component of acceleration aN N is caused by the frictional force F of the road. Thus, it follows from the vector form of Newton’s second law [Equation (19)] that the frictional force and the normal scalar component of acceleration are related by the equation F = maN N. Thus, 2 ds

F = mκ dt Use this result to ﬁnd the magnitude of the frictional force in newtons exerted by the road on a 500 kg go-cart driven at a speed of 10 km/h around a circular track of radius 15 m. [Note: 1 N = 1 kg·m/s2 .]

54. A shell is ﬁred from ground level with a muzzle speed of 320 ft/s and elevation angle of 60 ◦ . Find (a) parametric equations for the shell’s trajectory (b) the maximum height reached by the shell (c) the horizontal distance traveled by the shell (d) the speed of the shell at impact.

55. A rock is thrown downward from the top of a building, 168 ft high, at an angle of 60 ◦ with the horizontal. How far from the base of the building will the rock land if its initial speed is 80 ft/s? 56. Solve Exercise 55 assuming that the rock is thrown horizontally at a speed of 80 ft/s. 57. A shell is to be ﬁred from ground level at an elevation angle of 30 ◦ . What should the muzzle speed be in order for the maximum height of the shell to be 2500 ft?

45. a(1) = i + 2 j − 2k; aT (1) = 3

58. A shell, ﬁred from ground level at an elevation angle of 45 ◦ , hits the ground 24,500 m away. Calculate the muzzle speed of the shell.

47–50 True–False Determine whether the statement is true or

false. Explain your answer. ■

59. Find two elevation angles that will enable a shell, ﬁred from ground level with a muzzle speed of 800 ft/s, to hit a groundlevel target 10,000 ft away.

47. The velocity and unit tangent vectors for a moving particle are parallel.

60. A ball rolls off a table 4 ft high while moving at a constant speed of 5 ft/s. (cont.)

46. a(1) = 9; aT (1)T(1) = 2i − 2 j + k

894

Chapter 12 / Vector-Valued Functions

(a) How long does it take for the ball to hit the floor after it leaves the table? (b) At what speed does the ball hit the floor? (c) If a ball were dropped from rest at table height just as the rolling ball leaves the table, which ball would hit the ground first? Justify your answer.

(b) The horizontal range R of the shell is the horizontal distance traveled when the shell returns to ground level. Show that R = (v02 sin 2α)/g. For what elevation angle will the range be maximum? What is the maximum range? 66. A shell is ﬁred from ground level with an elevation angle α and a muzzle speed of v0 . Find two angles that can be used to hit a target at ground level that is a distance of three-fourths the maximum range of the shell. Express your answer to the nearest tenth of a degree. [Hint: See Exercise 65(b).]

61. As illustrated in the accompanying figure, a fire hose sprays water with an initial velocity of 40 ft/s at an angle of 60 ◦ with the horizontal. (a) Confirm that the water will clear corner point A. (b) Confirm that the water will hit the roof. (c) How far from corner point A will the water hit the roof?

67. At time t = 0 a baseball that is 5 ft above the ground is hit with a bat. The ball leaves the bat with a speed of 80 ft/s at an angle of 30 ◦ above the horizontal. (a) How long will it take for the baseball to hit the ground? Express your answer to the nearest hundredth of a second. (b) Use the result in part (a) to ﬁnd the horizontal distance traveled by the ball. Express your answer to the nearest tenth of a foot.

A 20 ft

60° 4 ft 15 ft

25 ft

Figure Ex-61

68. Repeat Exercise 67, assuming that the ball leaves the bat with a speed of 70 ft/s at an angle of 60 ◦ above the horizontal.

62. What is the minimum initial velocity that will allow the water in Exercise 61 to hit the roof? 63. As shown in the accompanying ﬁgure, water is sprayed from a hose with an initial velocity of 35 m/s at an angle of 45 ◦ with the horizontal. (a) What is the radius of curvature of the stream at the point where it leaves the hose? (b) What is the maximum height of the stream above the nozzle of the hose? 64. As illustrated in the accompanying ﬁgure, a train is traveling on a curved track. At a point where the train is traveling at a speed of 132 ft/s and the radius of curvature of the track is 3000 ft, the engineer hits the brakes to make the train slow down at a constant rate of 7.5 ft/s2 . (a) Find the magnitude of the acceleration vector at the instant the engineer hits the brakes. (b) Approximate the angle between the acceleration vector and the unit tangent vector T at the instant the engineer hits the brakes.

69. At time t = 0 a skier leaves the end of a ski jump with a speed of v0 ft/s at an angle α with the horizontal (see the accompanying ﬁgure). The skier lands 259 ft down the incline 2.9 s later. (a) Approximate v0 to the nearest ft/s and α to the nearest degree. [Note: Use g = 32 ft/s2 as the acceleration due to gravity.] (b) Use a CAS or a calculating utility with a numerical integration capability to approximate the distance traveled by the skier. y

23° 259

132 ft/s

35 m /s

3000 ft

45°

Figure Ex-63

Figure Ex-64

65. A shell is ﬁred from ground level at an elevation angle of α and a muzzle speed of v0 . (a) Show that the maximum height reached by the shell is maximum height =

(v0 sin α)2 2g

Figure Ex-69

70. At time t = 0 a projectile is ﬁred from a height h above level ground at an elevation angle of α with a speed v. Let R be the horizontal distance to the point where the projectile hits the ground. (a) Show that α and R must satisfy the equation g(sec2 α)R 2 − 2v 2 (tan α)R − 2v 2 h = 0

(cont.)

12.7 Keplerâ&#x20AC;&#x2122;s Laws of Planetary Motion

895

71. Writing Consider the various forces that a passenger in a car would sense while traveling over the crest of a hill or around a curve. Relate these sensations to the tangential and normal vector components of the acceleration vector for the carâ&#x20AC;&#x2122;s motion. Discuss how speeding up or slowing down (e.g., doubling or halving the carâ&#x20AC;&#x2122;s speed) affects these components.

(b) If g, h, and v are constant, then the equation in part (a) deďŹ nes R implicitly as a function of Îą. Let R0 be the maximum value of R and Îą0 the value of Îą when R = R0 . Use implicit differentiation to ďŹ nd dR /dÎą and show that v2 tan Îą0 = gR0 [Hint: Assume that dR /dÎą = 0 when R attains a maximum.] (c) Use the results in parts (a) and (b) to show that v 2 R0 = v + 2gh g v and Îą0 = tanâ&#x2C6;&#x2019;1 v 2 + 2gh

72. Writing The formula r(t) = (v0 cos Îą)ti + (s0 + (v0 sin Îą)t â&#x2C6;&#x2019; 21 gt 2 )j

models a position function for projectile motion [Formula (26)]. Identify the various quantities (v0 , Îą, s0 , and g) in this formula and discuss how the formula is derived, including any assumptions that are made.

â&#x153;&#x201D;QUICK CHECK ANSWERS 12.6 1.

d 2r dr dv ; = 2 ; v(t)

dt dt dt

12.7

2. r(t2 ) â&#x2C6;&#x2019; r(t1 );

t2 t1

v(t) dt

d 2s ; Îş(ds /dt)2 dt 2

4. â&#x2C6;&#x2019;gj; â&#x2C6;&#x2019;gtj + v0 ; s0 j; v0

KEPLERâ&#x20AC;&#x2122;S LAWS OF PLANETARY MOTION One of the great advances in the history of astronomy occurred in the early 1600s when Johannes Keplerâ&#x2C6;&#x2014; deduced from empirical data that all planets in our solar system move in elliptical orbits with the Sun at a focus. Subsequently, Isaac Newton showed mathematically that such planetary motion is the consequence of an inverse-square law of gravitational attraction. In this section we will use the concepts developed in the preceding sections of this chapter to derive three basic laws of planetary motion, known as Keplerâ&#x20AC;&#x2122;s laws. KEPLERâ&#x20AC;&#x2122;S LAWS

In Section 10.6 we stated the following laws of planetary motion that were published by Johannes Kepler in 1609 in his book known as Astronomia Nova.

12.7.1

keplerâ&#x20AC;&#x2122;s laws

â&#x20AC;˘ First law (Law of Orbits). Each planet moves in an elliptical orbit with the Sun â&#x20AC;˘ ÂŠ Science Photo Library/Photo Researchers, Inc.

at a focus. Second law (Law of Areas). Equal areas are swept out in equal times by the line from the Sun to a planet.

â&#x20AC;˘ Third law (Law of Periods). The square of a planetâ&#x20AC;&#x2122;s period (the time it takes the

The planets in our solar system move in accordance with Kepler's three laws.

planet to complete one orbit about the Sun) is proportional to the cube of the semimajor axis of its orbit.

â&#x2C6;&#x2014;

See biography on p. 759.

Chapter 12 / Vector-Valued Functions

896

CENTRAL FORCES

P r

F O Figure 12.7.1

If a particle moves under the influence of a single force that is always directed toward a fixed point O, then the particle is said to be moving in a central force field. The force is called a central force, and the point O is called the center of force. For example, in the simplest model of planetary motion, it is assumed that the only force acting on a planet is the force of the Sun’s gravity, directed toward the center of the Sun. This model, which produces Kepler’s laws, ignores the forces that other celestial objects exert on the planet as well as the minor effect that the planet’s gravity has on the Sun. Central force models are also used to study the motion of comets, asteroids, planetary moons, and artificial satellites. They also have important applications in electromagnetics. Our objective in this section is to develop some basic principles about central force fields and then use those results to derive Kepler’s laws. Suppose that a particle P of mass m moves in a central force field due to a force F that is directed toward a fixed point O, and let r = r(t) be the position vector from O to P (Figure 12.7.1). Let v = v(t) and a = a(t) be the velocity and acceleration functions of the particle, and assume that F and a are related by Newton’s second law (F = ma). Our first objective is to show that the particle P moves in a plane containing the point O. For this purpose observe that a has the same direction as F by Newton’s second law, and this implies that a and r are oppositely directed vectors. Thus, it follows from part (c) of Theorem 11.4.5 that r×a=0 Since the velocity and acceleration of the particle are given by v = dr/dt and a = dv/dt, respectively, we have dv dr d (r × v) = r × + × v = (r × a) + (v × v) = 0 + 0 = 0 dt dt dt Integrating the left and right sides of this equation with respect to t yields

Astronomers call the plane containing the orbit of a planet the ecliptic of the planet.

r×v=b

(1)

(2)

where b is a constant (independent of t). However, b is orthogonal to both r and v, so we can conclude that r = r(t) and v = v(t) lie in a ﬁxed plane containing the point O. NEWTON’S LAW OF UNIVERSAL GRAVITATION

Our next objective is to derive the position function of a particle moving under a central force in a polar coordinate system. For this purpose we will need the following result, known as Newton’s Law of Universal Gravitation.

12.7.2 newton’s law of universal gravitation Every particle of matter in the Universe attracts every other particle of matter in the Universe with a force that is proportional to the product of their masses and inversely proportional to the square of the distance between them. Speciﬁcally, if a particle of mass M and a particle of mass m are at a distance r from each other, then they attract each other with equal and opposite forces, F and −F, of magnitude

r F −F

F =

GMm r2

(3)

where G is a constant called the universal gravitational constant.

M M exerts force F on m, and m exerts force −F on M. Figure 12.7.2

To obtain a formula for the vector force F that mass M exerts on mass m, we will let r be the radius vector from mass M to mass m (Figure 12.7.2). Thus, the distance r between

12.7 Kepler’s Laws of Planetary Motion

897

the masses is r , and the force F can be expressed in terms of r as

r F = F − = F −

r which from (3) can be expressed as

F=−

Observe in Formula (5) that the acceleration a does not involve m. Thus, the mass of a planet has no effect on its acceleration.

GMm r r3

(4)

We start by ﬁnding a formula for the acceleration function. To do this we use Formula (4) and Newton’s second law to obtain GMm ma = − 3 r r from which we obtain a=−

GM r r3

(5)

To obtain a formula for the position function of the mass m, we will need to introduce a coordinate system and make some assumptions about the initial conditions:

• The distance r from m to M is minimum at time t = 0. • The mass m has nonzero position and velocity vectors r0 and v0 at time t = 0. • A polar coordinate system is introduced with its pole at mass M and oriented so θ = 0 • z

b y

v0 x

u r0

u=0

at time t = 0. The vector v0 is perpendicular to the polar axis at time t = 0.

Moreover, to ensure that the polar angle θ increases with t, let us agree to observe this polar coordinate system looking toward the pole from the terminal point of the vector b = r0 × v0 . We will also ﬁnd it useful to superimpose an xyz-coordinate system on the polar coordinate system with the positive z-axis in the direction of b (Figure 12.7.3). For computational purposes, it will be helpful to denote r0 by r0 and v0 by v0 , in which case we can express the vectors r0 and v0 in xyz-coordinates as r0 = r0 i

Figure 12.7.3

and

v0 = v0 j

and the vector b as b = r0 × v0 = r0 i × v0 j = r0 v0 k

(6)

(Figure 12.7.4). It will also be useful to introduce the unit vector u = cos θ i + sin θ j v0 j

r0 i

(7)

which will allow us to express the polar form of the position vector r as r = r cos θi + r sin θ j = r(cos θi + sin θ j) = ru

(8)

and to express the acceleration vector a in terms of u by rewriting (5) as

Figure 12.7.4

GM u (9) r2 We are now ready to derive the position function of the mass m in polar coordinates. For this purpose, recall from (2) that the vector b = r × v is constant, so it follows from (6) that the relationship b = r × v = r0 v0 k (10) a=−

holds for all values of t. Now let us examine b from another point of view. It follows from (8) that dr d du dr v= = (ru) = r + u dt dt dt dt

898

Chapter 12 / Vector-Valued Functions

and hence b = r × v = (ru) × But (7) implies that

du dr du dr du r + u = r 2u × + r u × u = r 2u × dt dt dt dt dt

(11)

du du dθ dθ = = (− sin θi + cos θ j) dt dθ dt dt

so u× Substituting (12) in (11) yields

du dθ = k dt dt

b = r2

(12)

dθ k dt

(13)

Thus, it follows from (7), (9), and (13) that a×b=−

GM (cos θ i + sin θ j) × r2

dθ r2 k dt

du dθ = GM dt dt From this formula and the fact that db/dt = 0 (since b is constant), we obtain = GM(− sin θi + cos θ j)

(14)

d db dv du (v × b) = v × + × b = a × b = GM dt dt dt dt Integrating both sides of this equation with respect to t yields v × b = GMu + C

(15)

where C is a vector constant of integration. This constant can be obtained by evaluating both sides of the equation at t = 0. We leave it as an exercise to show that C = (r0 v02 − GM)i

(16)

v × b = GMu + (r0 v02 − GM)i

(17)

from which it follows that

We can now obtain the position function by computing the scalar triple product r ⴢ (v × b) in two ways. First we use (10) and property (11) of Section 11.4 to obtain r ⴢ (v × b) = (r × v) ⴢ b = b ⴢ b = r02 v02

(18)

and next we use (17) to obtain r ⴢ (v × b) = r ⴢ (GMu) + r ⴢ (r0 v02 − GM)i r

= r ⴢ GM + ru ⴢ (r0 v02 − GM)i r = GMr + r(r0 v02 − GM) cos θ

If we now equate this to (18), we obtain r02 v02 = GMr + r(r0 v02 − GM) cos θ which when solved for r gives

GM +

r02 v02 (r0 v02 − GM) cos θ

or more simply r=

r02 v02 =

GM r0 v02 − 1 cos θ 1+ GM

k 1 + e cos θ

(19)

(20)

12.7 Keplerâ&#x20AC;&#x2122;s Laws of Planetary Motion

where k= Parabola

e=1

Hyperbola

e>1

Ellipse

0<e<1

Circle

e=0

Figure 12.7.5

r02 v02 GM

and e =

r0 v02 â&#x2C6;&#x2019;1 GM

899

(21â&#x20AC;&#x201C;22)

We will leave it as an exercise to show that e â&#x2030;Ľ 0. Accepting this to be so, it follows by comparing (20) to Formula (3) of Section 10.6 that the trajectory is a conic section with eccentricity e, the focus at the pole, and d = k /e. Thus, depending on whether e < 1, e = 1, or e > 1, the trajectory will be, respectively, an ellipse, a parabola, or a hyperbola (Figure 12.7.5). Note from Formula (22) that e depends on r0 and v0 , so the exact form of the trajectory is determined by the mass M and the initial conditions. If the initial conditions are such that e < 1, then the mass m becomes trapped in an elliptical orbit; otherwise the mass m â&#x20AC;&#x153;escapesâ&#x20AC;? and never returns to its initial position. Accordingly, the initial velocity that produces an eccentricity of e = 1 is called the escape speed and is denoted by vesc . Thus, it follows from (22) that 2GM vesc = (23) r0 (verify). KEPLERâ&#x20AC;&#x2122;S FIRST AND SECOND LAWS

It follows from our general discussion of central force ďŹ elds that the planets have elliptical orbits with the Sun at the focus, which is Keplerâ&#x20AC;&#x2122;s ďŹ rst law. To derive Keplerâ&#x20AC;&#x2122;s second law, we begin by equating (10) and (13) to obtain dÎ¸ (24) = r0 v0 dt To prove that the radial line from the center of the Sun to the center of a planet sweeps out equal areas in equal times, let r = f(Î¸) denote the polar equation of the planet, and let A denote the area swept out by the radial line as it varies from any ďŹ xed angle Î¸0 to an angle Î¸. It follows from the area formula in 10.3.4 that A can be expressed as Î¸ 1 A= [f(Ď&#x2020;)]2 dĎ&#x2020; Î¸0 2 r2

where the dummy variable Ď&#x2020; is introduced for the integration to reserve Î¸ for the upper limit. It now follows from Part 2 of the Fundamental Theorem of Calculus and the chain rule that dA dÎ¸ 1 dÎ¸ 1 dÎ¸ dA = = [f(Î¸)]2 = r2 dt dÎ¸ dt 2 dt 2 dt Thus, it follows from (24) that 1 dA = r0 v0 (25) dt 2 which shows that A changes at a constant rate. This implies that equal areas are swept out in equal times. KEPLERâ&#x20AC;&#x2122;S THIRD LAW To derive Keplerâ&#x20AC;&#x2122;s third law, we let a and b be the semimajor and semiminor axes of the elliptical orbit, and we recall that the area of this ellipse is Ď&#x20AC;ab. It follows by integrating (25) that in t units of time the radial line will sweep out an area of A = 21 r0 v0 t. Thus, if T denotes the time required for the planet to make one revolution around the Sun (the period), then the radial line will sweep out the area of the entire ellipse during that time and hence

Ď&#x20AC;ab =

1 r0 v0 T 2

900

Chapter 12 / Vector-Valued Functions

from which we obtain

T2 =

4π2 a 2 b2 r02 v02

(26)

However, it follows from Formula (1) of Section 10.6 and the relationship c2 = a 2 − b2 for an ellipse that c a 2 − b2 e= = a a Thus, b2 = a 2 (1 − e2 ) and hence (26) can be written as T2 =

4π2 a 4 (1 − e2 ) r02 v02

(27)

But comparing Equation (20) to Equation (17) of Section 10.6 shows that k = a(1 − e2 ) Finally, substituting this expression and (21) in (27) yields T2 =

4π2 a 3 4π2 3 4π2 a 3 r02 v02 = a k = GM r02 v02 r02 v02 GM

(28)

Thus, we have proved that T 2 is proportional to a 3 , which is Kepler’s third law. When convenient, Formula (28) can also be expressed as 2π 3/2 T =√ a GM

(29)

ARTIFICIAL SATELLITES

Kepler’s second and third laws and Formula (23) also apply to satellites that orbit a celestial body; we need only interpret M to be the mass of the body exerting the force and m to be the mass of the satellite. Values of GM that are required in many of the formulas in this section have been determined experimentally for various attracting bodies (Table 12.7.1). Table 12.7.1

r Apogee

Figure 12.7.6

u Perigee

attracting body

international system

british engineering system

Earth

GM = 3.99 × 1014 m3/s2 GM = 3.99 × 105 km3/s2

GM = 1.41 × 1016 ft 3/s2 GM = 1.24 × 1012 mi3/ h2

Sun

GM = 1.33 × 1020 m3/s2 GM = 1.33 × 1011 km3/s2

GM = 4.69 × 1021 ft 3/s2 GM = 4.13 × 1017 mi3/ h2

Moon

GM = 4.90 × 1012 m3/s2 GM = 4.90 × 103 km3/s2

GM = 1.73 × 1014 ft 3/s2 GM = 1.53 × 1010 mi3/ h2

Recall that for orbits of planets around the Sun, the point at which the distance between the center of the planet and the center of the Sun is maximum is called the aphelion and the point at which it is minimum the perihelion. For satellites orbiting the Earth, the point at which the maximum distance occurs is called the apogee, and the point at which the minimum distance occurs is called the perigee (Figure 12.7.6). The actual distances between the centers at apogee and perigee are called the apogee distance and the perigee distance. Example 1 A geosynchronous orbit for a satellite is a circular orbit about the equator of the Earth in which the satellite stays ﬁxed over a point on the equator. Use the fact that

12.7 Kepler’s Laws of Planetary Motion

901

the Earth makes one revolution about its axis every 24 hours to ﬁnd the altitude in miles of a communications satellite in geosynchronous orbit. Assume the Earth to be a sphere of radius 4000 mi.

Solution. To remain ﬁxed over a point on the equator, the satellite must have a period of T = 24 h. It follows from (28) or (29) and the Earth value of GM = 1.24 × 1012 mi3 /h2 from Table 12.7.1 that 2 2 12 3 GMT 3 (1.24 × 10 )(24) a= = ≈ 26,250 mi 4π2 4π2 and hence the altitude h of the satellite is h ≈ 26,250 − 4000 = 22,250 mi

✔QUICK CHECK EXERCISES 12.7

(See page 902 for answers.)

1. Let G denote the universal gravitational constant and let M and m denote masses a distance r apart. (a) According to Newton’s Law of Universal Gravitation, M and m attract each other with a force of magnitude . (b) If r is the radius vector from M to m, then the force of attraction that mass M exerts on mass m is . 2. Suppose that a mass m is in an orbit about a mass M and that r0 is the minimum distance from m to M. If G is the universal gravitational constant, then the “escape” speed of m is .

3. For a planet in an elliptical orbit about the Sun, the square of the planet’s period is proportional to what power of the semimajor axis of its orbit? 4. Suppose that a mass m is in an orbit about a mass M and that r0 is the minimum distance from m to M. If v0 is the speed of mass m when it is a distance r0 from M, and if G denotes the universal gravitational constant, then the eccentricity of the orbit is .

EXERCISE SET 12.7 1–14 In exercises that require numerical values, use Table 12.7.1 and the following values, where needed: radius of Earth = 4000 mi = 6440 km radius of Moon = 1080 mi = 1740 km 1 year (Earth year) = 365 days ■ F O C U S O N C O N C E P TS

1. (a) Obtain the value of C given in Formula (16) by setting t = 0 in (15). (b) Use Formulas (7), (17), and (22) to show that v × b = GM[(e + cos θ )i + sin θ j] (c) Show that v × b = v

b . (d) Use the results in parts (b) and (c) to show that the speed of a particle in an elliptical orbit is v0 2 v= e + 2e cos θ + 1 1+e (e) Suppose that a particle is in an elliptical orbit. Use part (d) to conclude that the distance from the particle to the center of force is a minimum if and only if the speed of the particle is a maximum. Similarly,

argue that the distance from the particle to the center of force is a maximum if and only if the speed of the particle is a minimum. 2. Use the result in Exercise 1(d) to show that when a particle in an elliptical orbit with eccentricity e reaches an end of the minor axis, its speed is 1−e v = v0 1+e 3. Use the result in Exercise 1(d) to show that for a particle in an elliptical orbit with eccentricity e, the maximum and minimum speeds are related by 1+e vmax = vmin 1−e 4. Use Formula (22) and the result in Exercise 1(d) to show that the speed v of a particle in a circular orbit of radius r0 is constant and is given by GM v= r0

902

Chapter 12 / Vector-Valued Functions

10. The universal gravitational constant is approximately

5. Suppose that a particle is in an elliptical orbit in a central force ﬁeld in which the center of force is at a focus, and let r = r(t) and v = v(t) be the position and velocity functions of the particle, respectively. Let rmin and rmax denote the minimum and maximum distances from the particle to the center of force, and let vmin and vmax denote the minimum and maximum speeds of the particle. (a) Review the discussion of ellipses in polar coordinates in Section 10.6, and show that if the ellipse has eccentricity e and semimajor axis a, then rmin = a(1 − e) and rmax = a(1 + e). (b) Explain why rmin and rmax occur at points at which r and v are orthogonal. [Hint: First argue that the extreme values of r occur at critical points of the function r 2 = r ⴢ r.] (c) Explain why vmin and vmax occur at points at which r and v are orthogonal. [Hint: First argue that the extreme values of v occur at critical points of the function v 2 = v ⴢ v. Then use Equation (5).] (d) Use Equation (2) and parts (b) and (c) to conclude that rmax vmin = rmin vmax .

G = 6.67 × 10−11 m3 /kg·s2 and the semimajor axis of the Earth’s orbit is approximately a = 149.6 × 106 km Estimate the mass of the Sun in kg. 11. (a) The eccentricity of the Moon’s orbit around the Earth is 0.055, and its semimajor axis is a = 238,900 mi. Find the maximum and minimum distances between the surface of the Earth and the surface of the Moon. (b) Find the period of the Moon’s orbit in days. 12. (a) Vanguard 1 was launched in March 1958 with perigee and apogee altitudes above the Earth of 649 km and 4340 km, respectively. Find the length of the semimajor axis of its orbit. (b) Use the result in part (a) of Exercise 16 in Section 10.6 to ﬁnd the eccentricity of its orbit. (c) Find the period of Vanguard 1 in minutes.

6. Use the results in parts (a) and (d) of Exercise 5 to give a derivation of the equation in Exercise 3.

7. Use the result in Exercise 4 to ﬁnd the speed in km/s of a satellite in a circular orbit that is 200 km above the surface of the Earth. 8. Use the result in Exercise 4 to ﬁnd the speed in mi/h of a communications satellite that is in geosynchronous orbit around the Earth (see Example 1). 9. Find the escape speed in km/s for a space probe in a circular orbit that is 300 km above the surface of the Earth.

13. (a) Suppose that a space probe is in a circular orbit at an altitude of 180 mi above the surface of the Earth. Use the result in Exercise 4 to find its speed. (b) During a very short period of time, a thruster rocket on the space probe is fired to increase the speed of the probe by 600 mi/h in its direction of motion. Find the eccentricity of the resulting elliptical orbit, and use the result in part (a) of Exercise 5 to find the apogee altitude. 14. Show that the quantity e defined by Formula (22) is nonnegative. [Hint: The polar axis was chosen so that r is minimum when θ = 0.]

✔QUICK CHECK ANSWERS 12.7 GMm GMm (b) − 3 r 1. (a) r2 r

2GM r0

3. 3

4. e =

r0 v02 −1 GM

CHAPTER 12 REVIEW EXERCISES 1. In words, what is meant by the graph of a vector-valued function? 2–5 Describe the graph of the equation. ■

2. r = (2 − 3t)i − 4tj

4. r = 3 cos ti + 2 sin tj − k

7. Show that the graph of r(t) = t sin πti + tj + t cos πtk lies on the surface of a cone, and sketch the cone. 8. Find parametric equations for the intersection of the surfaces

3. r = 3 sin 2ti + 3 cos 2tj

5. r = −2i + tj + (t 2 − 1)k

6. Describe the graph of the vector-valued function. (a) r = r0 + t (r1 − r0 ) (b) r = r0 + t (r1 − r0 ) (0 ≤ t ≤ 1) (c) r = r0 + tr ′ (t0 )

y = x2

and

2x 2 + y 2 + 6z2 = 24

and sketch the intersection. 9. In words, give a geometric description of the statement lim r(t) = L. t →a

Chapter 12 Review Exercises

1 â&#x2C6;&#x2019; cos t 10. Evaluate lim eâ&#x2C6;&#x2019;t i + j + t 2k . t â&#x2020;&#x2019;0 t 11. Find parametric equations of the line tangent to the graph of r(t) = (t + cos 2t)i â&#x2C6;&#x2019; (t 2 + t)j + sin tk at the point where t = 0.

12. Suppose that r1 (t) and r2 (t) are smooth vector-valued functions such that r1 (0) = â&#x2C6;&#x2019;1, 1, 2 , r2 (0) = 1, 2, 1 , r1â&#x20AC;˛ (0) = 1, 0, 1 , and r2â&#x20AC;˛ (0) = 4, 0, 2 . Use this information to evaluate the derivative at t = 0 of each function. (a) r(t) = 3r1 (t) + 2r2 (t) (b) r(t) = [ln(t + 1)]r1 (t) (c) r(t) = r1 (t) Ă&#x2014; r2 (t) (d) f(t) = r1 (t) â´˘ r2 (t) 13. Evaluate (cos ti + sin tj) dt. 14. Evaluate

Ď&#x20AC;/3

cos 3t, â&#x2C6;&#x2019; sin 3t dt.

15. Solve the vector initial-value problem yâ&#x20AC;˛ (t) = t 2 i + 2tj,

y(0) = i + j

16. Solve the vector initial-value problem dr = r, r(0) = r0 dt for the unknown vector-valued function r(t). â&#x2C6;&#x161;

i + eâ&#x2C6;&#x2019;

â&#x2C6;&#x161; 2t

j + 2tk

(0 â&#x2030;¤ t â&#x2030;¤

â&#x2C6;&#x161; 2 ln 2)

18. Suppose that r(t) is a smooth vector-valued function of t with râ&#x20AC;˛ (0) = 3i â&#x2C6;&#x2019; j + k and that r1 (t) = r(2 â&#x2C6;&#x2019; et ln 2 ). Find r1â&#x20AC;˛ (1). 19. Find the arc length parametrization of the line through P (â&#x2C6;&#x2019;1, 4, 3) and Q(0, 2, 5) that has reference point P and orients the line in the direction from P to Q. 20. Find an arc length parametrization of the curve r(t) = et cos t, â&#x2C6;&#x2019;et sin t

29. Suppose that r(t) is the position function of a particle moving in 2-space or 3-space. In each part, explain what the given represents physically. quantity t1 dr dr dt (c) r(t)

(a) (b) dt dt t0

30. (a) What does Theorem 12.2.8 tell you about the velocity vector of a particle that moves over a sphere? (b) What does Theorem 12.2.8 tell you about the acceleration vector of a particle that moves with constant speed? (c) Show that the particle with position function r(t) =

1 â&#x2C6;&#x2019; 41 cos2 t cos ti + 1 â&#x2C6;&#x2019; 41 cos2 t sin tj + 21 cos tk

moves over a sphere.

31. As illustrated in the accompanying ďŹ gure, suppose that a particle moves counterclockwise around a circle of radius R centered at the origin at a constant rate of Ď&#x2030; radians per second. This is called uniform circular motion. If we assume that the particle is at the point (R, 0) at time t = 0, then its position function will be r(t) = R cos Ď&#x2030;ti + R sin Ď&#x2030;tj

17. Find the arc length of the graph of r(t) = e

(a) Show that the velocity vector v(t) is always tangent to the circle and that the particle has constant speed v given by v = RĎ&#x2030; (b) Show that the acceleration vector a(t) is always directed toward the center of the circle and has constant magnitude a given by a = RĎ&#x2030;2 (c) Show that the time T required for the particle to make one complete revolution is

(0 â&#x2030;¤ t â&#x2030;¤ Ď&#x20AC;/2)

T =

which has the same orientation and has r(0) as the reference point. 21. Suppose that r(t) is a smooth vector-valued function. State the deďŹ nitions of T(t), N(t), and B(t). 22. Find T(0), N(0), and B(0) for the curve 3 6 r(t) = 2 cos t, 2 cos t + â&#x2C6;&#x161; sin t, cos t â&#x2C6;&#x2019; â&#x2C6;&#x161; sin t 5 5 23. State the deďŹ nition of â&#x20AC;&#x153;curvatureâ&#x20AC;? and explain what it means geometrically. 24. Suppose that r(t) is a smooth curve with râ&#x20AC;˛ (0) = i and râ&#x20AC;˛â&#x20AC;˛ (0) = i + 2j. Find the curvature at t = 0. 25â&#x20AC;&#x201C;28 Find the curvature of the curve at the stated point. â&#x2013;

25. r(t) = 2 cos ti + 3 sin tj â&#x2C6;&#x2019; tk; t = Ď&#x20AC;/2

903

26. r(t) = 2t, e2t , eâ&#x2C6;&#x2019;2t ; t = 0 27. y = cos x; x = Ď&#x20AC;/2 28. y = ln x; x = 1

2Ď&#x20AC; 2Ď&#x20AC;R = Ď&#x2030; v

v(t)

a(t) vt

(R, 0)

Figure Ex-31

32. If a particle of mass m has uniform circular motion (see Exercise 31), then the acceleration vector a(t) is called the centripetal acceleration. According to Newtonâ&#x20AC;&#x2122;s second law, this acceleration must be produced by some force F(t), called the centripetal force, that is related to a(t) by the equation F(t) = ma(t). If this force is not present, then the particle cannot undergo uniform circular motion. (cont.)

904

Chapter 12 / Vector-Valued Functions

(a) Show that the direction of the centripetal force varies with time but that it has constant magnitude F given by

rocket can be expressed in terms b, Î¸, and dÎ¸ /dt as v = b sec2 Î¸

mv 2 R (b) An astronaut with a mass of m = 60 kg orbits the Earth at an altitude of h = 3200 km with a constant speed of v = 6.43 km/s. Find her centripetal acceleration assuming that the radius of the Earth is 6440 km. (c) What centripetal gravitational force in newtons does the Earth exert on the astronaut? F =

33. At time t = 0 a particle at the origin of an xyz-coordinate system has a velocity vector of v0 = i + 2 j â&#x2C6;&#x2019; k. The acceleration function of the particle is a(t) = 2t 2 i + j + cos 2tk. (a) Find the position function of the particle. (b) Find the speed of the particle at time t = 1.

34. Let v = v(t) and a = a(t) be the velocity and acceleration vectors for a particle moving in 2-space or 3-space. Show that the rate of change of its speed can be expressed as d 1 ( v ) = (v â´˘ a) dt

35. Use Formula (23) in Section 12.7 and refer to Table 12.7.1 to ďŹ nd the escape speed (in km/s) for a space probe 600 km above the surface of the Earth.

dÎ¸ dt

Figure Ex-36

37. A player throws a ball with an initial speed of 60 ft/s at an unknown angle Îą with the horizontal from a point that is 4 ft above the ďŹ&#x201A;oor of a gymnasium. Given that the ceiling of the gymnasium is 25 ft high, determine the maximum height h at which the ball can hit a wall that is 60 ft away (see the accompanying ďŹ gure).

25 ft

60 ft/s

36. As illustrated in the accompanying ďŹ gure, the polar coordinates of a rocket are tracked by radar from a point that is b units from the launching pad. Show that the speed v of the

4 ft 60 ft Figure Ex-37

CHAPTER 12 MAKING CONNECTIONS

CAS

2. (a) Use the result in Exercise 1(b) and Exercise 45 of Section 11.4 to show that N(t) can be expressed directly in terms of r(t) as u(t) N(t) =

u(t)

1. (a) Use the formulas N(t) = B(t) Ă&#x2014; T(t) T(t) = râ&#x20AC;˛ (t)/ râ&#x20AC;˛ (t)

and B(t) =

râ&#x20AC;˛ (t) Ă&#x2014; râ&#x20AC;˛â&#x20AC;˛ (t)

where

to show that N(t) can be expressed in terms of r(t) as N(t) =

â&#x20AC;˛

â&#x20AC;˛â&#x20AC;˛

â&#x20AC;˛

r (t) Ă&#x2014; r (t) r (t) Ă&#x2014; â&#x20AC;˛

r â&#x20AC;˛ (t) Ă&#x2014; r â&#x20AC;˛â&#x20AC;˛ (t)

r (t)

(b) Use properties of cross products to show that the formula in part (a) can be expressed as N(t) =

(r â&#x20AC;˛ (t) Ă&#x2014; r â&#x20AC;˛â&#x20AC;˛ (t)) Ă&#x2014; r â&#x20AC;˛ (t)

(c) Use the result in part (b) to ďŹ nd N(t) at the given point. (i) r(t) = (t 2 â&#x2C6;&#x2019; 1)i + tj; t = 1 (ii) r(t) = 4 cos ti + 4 sin tj + tk; t = Ď&#x20AC;/2

u(t) = r â&#x20AC;˛ (t) 2 r â&#x20AC;˛â&#x20AC;˛ (t) â&#x2C6;&#x2019; (r â&#x20AC;˛ (t) â´˘ r â&#x20AC;˛â&#x20AC;˛ (t))r â&#x20AC;˛ (t) (b) Use the result in part (a) to ďŹ nd N(t). (i) r(t) = sin ti + cos tj + tk (ii) r(t) = ti + t 2 j + t 3 k 3. In Making Connections Exercise 1 of Chapter 10 we deďŹ ned the Cornu spiral parametrically as 2

t t Ď&#x20AC;u Ď&#x20AC;u x= du, y = sin du cos 2 2 0 0 This curve, which is graphed in the accompanying ďŹ gure, is used in highway design to create a gradual transition from a straight road (zero curvature) to an exit ramp with positive curvature. (cont.)

Chapter 12 Making Connections

(a) Express the Cornu spiral as a vector-valued function r(t), and then use Theorem 12.3.4 to show that s = t is the arc length parameter with reference point (0, 0). (b) Replace t by s and use Formula (1) of Section 12.5 to show that κ(s) = π|s|. [Note: If s ≥ 0, then the curvature κ(s) = πs increases from 0 at a constant rate with respect to s. This makes the spiral ideal for joining a curved road to a straight road.] (c) What happens to the curvature of the Cornu spiral as s → +⬁? In words, explain why this is consistent with the graph.

905

and B=

sin

c a

i− j+ k cos w w w w

w √ where w = a 2 + c2 and s has reference point (a, 0, 0).

6. Suppose that the position function of a point moving in the xy-plane is r = x(t)i + y(t) j This equation can be expressed in polar coordinates by making the substitution x(t) = r(t) cos θ (t),

y(t) = r(t) sin θ(t)

This yields r = r(t) cos θ (t)i + r(t) sin θ(t) j

which can be expressed as r = r(t)er (t) Figure Ex-3 C

4. In 1975, German engineer Werner Stengel pioneered the use of Cornu spirals (see Exercise 3) in the design of loops for roller coasters with the Revolution at Six Flags Magic Mountain in California. For this design, the top of the loop is a circular arc, joined at either end to Cornu spirals that ease the transitions to horizontal track. The accompanying ﬁgure illustrates this design when the circular arc (in red) is a semicircle, quarter-circle, or a single point, respectively. Suppose that a roller-coaster loop is designed to be 45 feet across at its widest point. For each case in Figure Ex-4, ﬁnd the vertical distance between the level of the horizontal track and the top of the loop. Use the numerical integration capability of your CAS to estimate integrals, as necessary. y

0.8 0.6 0.4 0.2

x 0.2

0.6

0.8 0.6 0.4 0.2

x 0.2

0.6

0.8 0.6 0.4 0.2

x 0.2

0.6

where er (t) = cos θ (t)i + sin θ (t) j. (a) Show that er (t) is a unit vector in the same direction as the radius vector r if r(t) > 0. Also, show that eθ (t) = − sin θ (t)i + cos θ (t) j is the unit vector that results when er (t) is rotated counterclockwise through an angle of π/2. The vector er (t) is called the radial unit vector and the vector eθ (t) is called the transverse unit vector (see the accompanying ﬁgure). (b) Show that the velocity function v = v(t) can be expressed in terms of radial and transverse components as v=

(c) Show that the acceleration function a = a(t) can be expressed in terms of radial and transverse components as 2 2 d 2r dθ d θ dr dθ a= eθ e + r − r + 2 r dt 2 dt dt 2 dt dt y

Figure Ex-4

5. Use the results in Exercise 61 of Section 12.5 and the results in Exercise 32 of Section 12.3 to show that for the circular helix r = a cos ti + a sin t j + ctk with a > 0, the torsion and the binormal vector are c τ= 2 w

dr dθ er + r eθ dt dt

Trajectory

eu er r u

Figure Ex-6

EXPANDING THE CALCULUS HORIZON For a practical application of projectile motion in a whimsical setting, see the module entitled Blammo the Human Cannonball at: www.wiley.com/college/anton

13 PARTIAL DERIVATIVES

Three-dimensional surfaces have high points and low points that are analogous to the peaks and valleys of a mountain range. In this chapter we will use derivatives to locate these points and to study other features of such surfaces.

13.1

In this chapter we will extend many of the basic concepts of calculus to functions of two or more variables, commonly called functions of several variables. We will begin by discussing limits and continuity for functions of two and three variables, then we will define derivatives of such functions, and then we will use these derivatives to study tangent planes, rates of change, slopes of surfaces, and maximization and minimization problems. Although many of the basic ideas that we developed for functions of one variable will carry over in a natural way, functions of several variables are intrinsically more complicated than functions of one variable, so we will need to develop new tools and new ideas to deal with such functions.

FUNCTIONS OF TWO OR MORE VARIABLES In previous sections we studied real-valued functions of a real variable and vector-valued functions of a real variable. In this section we will consider real-valued functions of two or more real variables.

NOTATION AND TERMINOLOGY There are many familiar formulas in which a given variable depends on two or more other variables. For example, the area A of a triangle depends on the base length b and height h by the formula A = 21 bh; the volume V of a rectangular box depends on the length l, the width w, and the height h by the formula V = lwh; and the arithmetic average x¯ of n real numbers, x1 , x2 , . . . , xn , depends on those numbers by the formula

x¯ =

1 (x1 + x2 + · · · + xn ) n

Thus, we say that A is a function of the two variables b and h; V is a function of the three variables l, w, and h; x¯ is a function of the n variables x1 , x2 , . . . , xn . The terminology and notation for functions of two or more variables is similar to that for functions of one variable. For example, the expression

906

z = f(x, y)

13.1 Functions of Two or More Variables

907

means that z is a function of x and y in the sense that a unique value of the dependent variable z is determined by specifying values for the independent variables x and y. Similarly, w = f(x, y, z) expresses w as a function of x, y, and z, and u = f(x1 , x2 , . . . , xn ) expresses u as a function of x1 , x2 , . . . , xn . As with functions of one variable, the independent variables of a function of two or more variables may be restricted to lie in some set D, which we call the domain of f . Sometimes the domain will be determined by physical restrictions on the variables. If the function is deﬁned by a formula and if there are no physical restrictions or other restrictions stated explicitly, then it is understood that the domain consists of all points for which the formula yields a real value for the dependent variable. We call this the natural domain of the function. The following deﬁnitions summarize this discussion.

By extension, one can define the notion of “n-dimensional space” in which a “point” is a sequence of n real numbers (x1 , x2 , . . . , xn ), and a function of n real variables is a rule that assigns a unique real number f(x1 , x2 , . . . , xn ) to each point in some set in this space.

13.1.1 definition A function f of two variables, x and y, is a rule that assigns a unique real number f(x, y) to each point (x, y) in some set D in the xy-plane.

13.1.2 definition A function f of three variables, x, y, and z, is a rule that assigns a unique real number f(x, y, z) to each point (x, y, z) in some set D in threedimensional space.

Example 1 Let f(x, y) = domain of f .

√ y + 1 + ln(x 2 − y). Find f(e, 0) and sketch the natural

Solution. By substitution, √ √ 0 + 1 + ln(e2 − 0) = 1 + ln(e2 ) = 1 + 2 = 3 √ To find the natural domain of f , we note that y + 1 is defined only when y ≥ −1, while ln(x 2 − y) is defined only when 0 < x 2 − y or y < x 2 . Thus, the natural domain of f consists of all points in the xy-plane for which −1 ≤ y < x 2 . To sketch the natural domain, we first sketch the parabola y = x 2 as a “dashed” curve and the line y = −1 as a solid curve. The natural domain of f is then the region lying above or on the line y = −1 and below the parabola y = x 2 (Figure 13.1.1). f(e, 0) =

y = −1 The solid boundary line is included in the domain, while the dashed boundary is not included in the domain.

Figure 13.1.1

Example 2

Let

f(x, y, z) =

1 − x 2 − y 2 − z2

Find f 0, 21 , − 21 and the natural domain of f.

Solution. By substitution,

f 0, 21 , − 21 =

2 2 1 − (0)2 − 21 − − 21 = 21

Because of the square root sign, we must have 0 ≤ 1 − x 2 − y 2 − z2 in order to have a real

908

Chapter 13 / Partial Derivatives

value for f(x, y, z). Rewriting this inequality in the form x 2 + y 2 + z2 ≤ 1 we see that the natural domain of f consists of all points on or within the sphere x 2 + y 2 + z2 = 1 FUNCTIONS DESCRIBED BY TABLES The wind chill index is that temperature (in ◦ F) which would produce the same sensation on exposed skin at a wind speed of 3 mi/h as the temperature and wind speed combination in current weather conditions.

wind speed v (mi/h)

Table 13.1.1 temperature T (°F ) 20

−2

Sometimes it is either desirable or necessary to represent a function of two variables in table form, rather than as an explicit formula. For example, the U.S. National Weather Service uses the formula W = 35.74 + 0.6215T + (0.4275T − 35.75)v 0.16

(1)

to model the wind chill index W (in ◦ F) as a function of the temperature T (in ◦ F) and the wind speed v (in mi/h) for wind speeds greater than 3 mi/h. This formula is sufﬁciently complex that it is difﬁcult to get an intuitive feel for the relationship between the variables. One can get a clearer sense of the relationship by selecting sample values of T and v and constructing a table, such as Table 13.1.1, in which we have rounded the values of W to the nearest integer. For example, if the temperature is 30 ◦ F and the wind speed is 5 mi/h, it feels as if the temperature is 25 ◦ F. If the wind speed increases to 15 mi/h, the temperature then feels as if it has dropped to 19 ◦ F. Note that in this case, an increase in wind speed of 10 mi/h causes a 6 ◦ F decrease in the wind chill index. To estimate wind chill values not displayed in the table, we can use linear interpolation. For example, suppose that the temperature is 30 ◦ F and the wind speed is 7 mi/h. A reasonable estimate for the drop in the 2 wind chill index from its value when the wind speed is 5 mi/h would be 10 · 6 ◦ F = 1.2 ◦ F. (Why?) The resulting estimate in wind chill would then be 25 ◦ − 1.2 ◦ = 23.8 ◦ F. In some cases, tables for functions of two variables arise directly from experimental data, in which case one must either work directly with the table or else use some technique to construct a formula that models the data in the table. Such modeling techniques are developed in statistics and numerical analysis texts. GRAPHS OF FUNCTIONS OF TWO VARIABLES

Recall that for a function f of one variable, the graph of f(x) in the xy-plane was deﬁned to be the graph of the equation y = f(x). Similarly, if f is a function of two variables, we deﬁne the graph of f(x, y) in xyz-space to be the graph of the equation z = f(x, y). In general, such a graph will be a surface in 3-space. Example 3 In each part, describe the graph of the function in an xyz-coordinate system. (a) f(x, y) = 1 − x − 21 y (b) f(x, y) = 1 − x 2 − y 2 (c) f(x, y) = − x 2 + y 2

Solution (a). By deﬁnition, the graph of the given function is the graph of the equation z = 1 − x − 21 y

which is a plane. A triangular portion of the plane can be sketched by plotting the intersections with the coordinate axes and joining them with line segments (Figure 13.1.2a).

Solution (b). By deﬁnition, the graph of the given function is the graph of the equation z=

1 − x2 − y2

(2)

13.1 Functions of Two or More Variables

After squaring both sides, this can be rewritten as

x 2 + y 2 + z2 = 1

z = 1 − x − 12 y

(0, 0, 1)

which represents a sphere of radius 1, centered at the origin. Since (2) imposes the added condition that z ≥ 0, the graph is just the upper hemisphere (Figure 13.1.2b).

(0, 2, 0) (1, 0, 0)

Solution (c). The graph of the given function is the graph of the equation

(a)

After squaring, we obtain

909

√1 − x 2 − y 2

z = − x2 + y2

(3)

z2 = x 2 + y 2

which is the equation of a circular cone (see Table 11.7.1). Since (3) imposes the condition that z ≤ 0, the graph is just the lower nappe of the cone (Figure 13.1.2c).

y −1

1 x

LEVEL CURVES We are all familiar with the topographic (or contour) maps in which a three-dimensional landscape, such as a mountain range, is represented by two-dimensional contour lines or curves of constant elevation. Consider, for example, the model hill and its contour map shown in Figure 13.1.3. The contour map is constructed by passing planes of constant elevation through the hill, projecting the resulting contours onto a ﬂat surface, and labeling the contours with their elevations. In Figure 13.1.3, note how the two gullies appear as indentations in the contour lines and how the curves are close together on the contour map where the hill has a steep slope and become more widely spaced where the slope is gradual.

(b) z

z = −√x 2 + y 2 x

Hundreds of feet

6 5 4 3 2 1

3 45 6

z = f (x, y) A perspective view of a model hill with two gullies

z= k

A contour map of the model hill

Figure 13.1.3 y

Level curve of height k

Figure 13.1.4

f (x, y) = k

Contour maps are also useful for studying functions of two variables. If the surface z = f(x, y) is cut by the horizontal plane z = k, then at all points on the intersection we have f(x, y) = k. The projection of this intersection onto the xy-plane is called the level curve of height k or the level curve with constant k (Figure 13.1.4). A set of level curves for z = f(x, y) is called a contour plot or contour map of f. Example 4 The graph of the function f(x, y) = y 2 − x 2 in xyz-space is the hyperbolic paraboloid (saddle surface) shown in Figure 13.1.5a. The level curves have equations of the form y 2 − x 2 = k. For k > 0 these curves are hyperbolas opening along lines parallel to the y-axis; for k < 0 they are hyperbolas opening along lines parallel to the x-axis; and for k = 0 the level curve consists of the intersecting lines y + x = 0 and y − x = 0 (Figure 13.1.5b).

Chapter 13 / Partial Derivatives y

40 30 20

5.0 4.0 4.0 3.0 2.0

3.0

− 20

−2.0 −3.0

−10

−1.0

−10

−1.0 −2.0 −3.0

− 20

1.0

− 30 − 40

2.0

− 40 − 30

910

−4.0

−5.0

20 30 40

(a)

(b)

Figure 13.1.5

Example 5 Sketch the contour plot of f(x, y) = 4x 2 + y 2 using level curves of height k = 0, 1, 2, 3, 4, 5.

Solution. The graph of the surface z = 4x 2 + y 2 is the paraboloid shown in the left part

of Figure 13.1.6, so we can reasonably expect the contour plot to be a family of ellipses centered at the origin. The level curve of height k has the equation 4x 2 + y 2 = k. If k = 0, then the graph is the single point (0, 0). For k > 0 we can rewrite the equation as x2 y2 + =1 k /4 k

√ √ which represents a family of ellipses with x-intercepts ± k /2 and y-intercepts ± k. The contour plot for the speciﬁed values of k is shown in the right part of Figure 13.1.6. z

z = 4x 2 + y 2

2 1

z=k −2

k=5 k=4 k=3 k=2 k=1 x

−1

−1 y

Figure 13.1.6

−2

In the last two examples we used a formula for f(x, y) to ﬁnd the contour plot of f . Conversely, if we are given a contour plot of some function, then we can use the plot to estimate values of the function. Example 6 Let f(r, L) be the monthly payment on a 5-year car loan as a function of the interest rate r and the loan amount L. Figure 13.1.7 is a contour plot of f(r, L). Use this plot in each part. (a) Estimate the monthly payment on a loan of $3000 at an interest rate of 7%. (b) Estimate the monthly payment on a loan of $5000 at an interest rate of 3%. (c) Estimate the loan amount if the monthly payment is $80 and the interest rate is 3%.

13.1 Functions of Two or More Variables

911

Solution (a). Since the point (7, 3000) appears to lie on the contour labeled 60, we estimate the monthly payment to be $60. Solution (b). Since the point (3, 5000) appears to be midway between the contours labeled 80 and 100, we estimate the monthly payment to be $90. Solution (c). The vertical line x = 3 intersects the contour labeled 80 at a point whose L coordinate appears to be 4500. Hence, we estimate the loan amount to be $4500. 8000

Loan amount L ($)

7000

140

6000

120 5000

100

4000

80 60

3000 1

11 13 15

Interest rate r (%)

Figure 13.1.7

CONTOUR PLOTS USING TECHNOLOGY

Except in the simplest cases, contour plots can be difﬁcult to produce without the help of a graphing utility. Figure 13.1.8 illustrates how graphing technology can be used to display level curves. Figure 13.1.8a shows the graph of f(x, y) = |sin x sin y| plotted over the domain 0 ≤ x ≤ 2π, 0 ≤ y ≤ 2π, and Figure 13.1.8b displays curves of constant elevation on the graph of f . The projections of these curves into the xy-plane are contours of f . Figure 13.1.8c illustrates how the use of color can enhance the display of these contours. z

z > 0.928 < 0.928 < 0.833 < 0.737 < 0.642 < 0.547 < 0.451 < 0.356 < 0.261 < 0.165 < 0.07

5 4

0.928 0.833 0.737 0.642 0.547 0.451 0.356 0.261 0.165 0.07

3 2 y

1 0

f (x, y) = | sin x sin y |

(a)

(b)

(c)

Figure 13.1.8

LEVEL SURFACES The term “level surface” is standard but confusing, since a level surface need not be level in the sense of being horizontal—it is simply a surface on which all values of f are the same.

Observe that the graph of y = f(x) is a curve in 2-space, and the graph of z = f(x, y) is a surface in 3-space, so the number of dimensions required for these graphs is one greater than the number of independent variables. Accordingly, there is no “direct” way to graph a function of three variables since four dimensions are required. However, if k is a constant, then the graph of the equation f(x, y, z) = k will generally be a surface in 3-space (e.g., the graph of x 2 + y 2 + z2 = 1 is a sphere), which we call the level surface with constant k. Some geometric insight into the behavior of the function f can sometimes be obtained by graphing these level surfaces for various values of k.

Chapter 13 / Partial Derivatives

912

Example 7

Describe the level surfaces of (b) f(x, y, z) = z2 − x 2 − y 2

(a) f(x, y, z) = x 2 + y 2 + z2 y

Solution (a). The level surfaces have equations of the form x 2 + y 2 + z2 = k

√ For k > 0 the graph of this equation is a sphere of radius k, centered at the origin; for k = 0 the graph is the single point (0, 0, 0); and for k < 0 there is no level surface (Figure 13.1.9).

x Level surfaces of

Solution (b). The level surfaces have equations of the form

f (x, y, z) = x2 + y2 + z2

z2 − x 2 − y 2 = k

Figure 13.1.9

As discussed in Section 11.7, this equation represents a cone if k = 0, a hyperboloid of two sheets if k > 0, and a hyperboloid of one sheet if k < 0 (Figure 13.1.10).

k<0 k=0 k>0 z

GRAPHING FUNCTIONS OF TWO VARIABLES USING TECHNOLOGY

Level surfaces of

f (x, y, z) = z2 − x2 − y2

Generating surfaces with a graphing utility is more complicated than generating plane curves because there are more factors that must be taken into account. We can only touch on the ideas here, so if you want to use a graphing utility, its documentation will be your main source of information. Graphing utilities can only show a portion of xyz-space in a viewing screen, so the first step in graphing a surface is to determine which portion of xyz-space you want to display. This region is called the viewing box or viewing window. For example, Figure 13.1.11 shows the effect of graphing the paraboloid z = x 2 + y 2 in three different viewing windows. However, within a fixed viewing box, the appearance of the surface is also affected by the viewpoint, that is, the direction from which the surface is viewed, and the distance from the viewer to the surface. For example, Figure 13.1.12 shows the graph of the paraboloid z = x 2 + y 2 from three different viewpoints using the first viewing box in Figure 13.1.11. −2

Figure 13.1.10

−2

y 0

−2

y 0

z 4

z 5

z 10

T E C H N O LO GY M A ST E R Y If you have a graphing utility that can generate surfaces in 3-space, read the documentation and try to duplicate some of the surfaces in Figures 13.1.11 and 13.1.12 and Table 13.1.2.

−2

0 x 2 Figure 13.1.11 Varying the viewing box.

−2 x

−2

−2 x

y 0

x 0

z 4

−2

2 8 z 4

−2

0 y

−2

0 y

0 −2 0 2

Figure 13.1.12 Varying the viewpoint.

13.1 Functions of Two or More Variables

913

Table 13.1.2 shows six surfaces in 3-space along with their associated contour plots. Note that the mesh lines on the surface are traces in vertical planes, whereas the level curves correspond to traces in horizontal planes. Table 13.1.2

surface

contour plot

surface

contour plot

z = 5e x sin y

z = cos y z

4 y 3

y 0 −1

−2 x

−3 −3 −2 −1

0 x

0 −2

z = sin 冸√x2 + y2 冹

z = xye − 2 (x 1

6 4 2 y 0 −2 −4 −6

3 2 1 y 0 −1 −2

−3 −3 −2 −1

0 x

z = xy

z = cos(xy) 3

1 y 0

y 0

y x

−1

+ y2 )

−6 −4 −2 0 2 4 6 x

−1

−2

−3 −3 −2 −1

✔QUICK CHECK EXERCISES 13.1

0 x

3 x

−3 −3 −2 −1

(See page 917 for answers.)

1. The domain of f (x, y) = ln xy is and the domain of g(x, y) = ln x + ln y is . x−y 2. Let f (x, y) = . x+y+1 (a) f (2, 1) = (b) f (1, 2) = (c) f (a, a) = (d) f (y + 1, y) =

3. Let f (x, y) = ex+y . (a) For what values of k will the graph of the level curve f (x, y) = k be nonempty?

(b) Describe the level curves f (x, y) = k for the values of k obtained in part (a). 1 4. Let f (x, y, z) = 2 . x + y 2 + z2 + 1 (a) For what values of k will the graph of the level surface f (x, y, z) = k be nonempty? (b) Describe the level surfaces f (x, y, z) = k for the values of k obtained in part (a).

Chapter 13 / Partial Derivatives

EXERCISE SET 13.1

Graphing Utility

CAS

1–8 These exercises are concerned with functions of two vari-

ables. ■ 1. Let f(x, y) = x 2 y + 1. Find (a) f(2, 1) (b) f(1, 2) (d) f(1, −3) (e) f(3a, a) √ 3 2. Let f(x, y) = x + xy. Find (a) f(t, t 2 ) (b) f(x, x 2 )

3. Let f(x, y) = xy + 3. Find (a) f(x + y, x − y) (b) f(xy, 3x 2 y 3 ). 4. Let g(x) = x sin x. Find (a) g(x /y) (b) g(xy)

5. Find F(g(x), h(y)) if F(x, y) = xexy , g(x) = x 3 , and h(y) = 3y + 1.

6. Find g(u(x, y), v(x, y)) if g(x, y) = y sin(x 2 y), u(x, y) = x 2 y 3 , and v(x, y) = πxy.

7. Let f(x, y) = x + 3x 2 y 2 , x(t) = t 2 , and y(t) = t 3 . Find (a) f(x(t), y(t)) (b) f(x(0), y(0)) (c) f(x(2), y(2)). √ 8. Let g(x, y) = ye−3x , x(t) = ln(t 2 + 1), and y(t) = t. Find g(x(t), y(t)).

9–10 Suppose that the concentration C in mg/L of medica-

tion in a patient’s bloodstream is modeled by the function C(x, t) = 0.2x(e−0.2t − e−t ), where x is the dosage of the medication in mg and t is the number of hours since the beginning of administration of the medication. ■ 9. (a) Estimate the value of C(25, 3) to two decimal places. Include appropriate units and interpret your answer in a physical context. (b) If the dosage is 100 mg, give a formula for the concentration as a function of time t. (c) Give a formula that describes the concentration after 1 hour in terms of the dosage x. 10. (a) Suppose that the medication in the bloodstream reaches an effective level after a half hour. Estimate how much longer the medication remains effective. (b) Suppose the dosage is 100 mg. Estimate the maximum concentration in the bloodstream. 11–14 Refer to Table 13.1.1 to estimate the given quantity. ■

11. The wind chill index when (a) the temperature is 25 ◦ F and the wind speed is 7 mi/h (b) the temperature is 28 ◦ F and the wind speed is 5 mi/h. 12. The wind chill index when (a) the temperature is 35 ◦ F and the wind speed is 14 mi/h (b) the temperature is 32 ◦ F and the wind speed is 15 mi/h. 13. The temperature when (a) the wind chill index is 16 ◦ F and the wind speed is 25 mi/h

(b) the wind chill index is 6 ◦ F and the wind speed is 25 mi/h. 14. The wind speed when (a) the wind chill index is 7 ◦ F and the temperature is 25 ◦ F (b) the wind chill index is 15 ◦ F and the temperature is 30 ◦ F. 15. One method for determining relative humidity is to wet the bulb of a thermometer, whirl it through the air, and then compare the thermometer reading with the actual air temperature. If the relative humidity is less than 100%, the reading on the thermometer will be less than the temperature of the air. This difference in temperature is known as the wet-bulb depression. The accompanying table gives the relative humidity as a function of the air temperature and the wet-bulb depression. Use the table to complete parts (a)–(c). (a) What is the relative humidity if the air temperature is 20 ◦ C and the wet-bulb thermometer reads 16 ◦ C? (b) Estimate the relative humidity if the air temperature is 25 ◦ C and the wet-bulb depression is 3.5 ◦ C. (c) Estimate the relative humidity if the air temperature is 22 ◦ C and the wet-bulb depression is 5 ◦ C. air temperature (°C ) wet-bulb depression (°C)

914

Table Ex-15

16. Use the table in Exercise 15 to complete parts (a)–(c). (a) What is the wet-bulb depression if the air temperature is 30 ◦ C and the relative humidity is 73%? (b) Estimate the relative humidity if the air temperature is 15 ◦ C and the wet-bulb depression is 4.25 ◦ C. (c) Estimate the relative humidity if the air temperature is 26 ◦ C and the wet-bulb depression is 3 ◦ C. 17–20 These exercises involve functions of three variables. ■

17. Let f(x, y, z) = xy 2 z3 + 3. (a) f(2, 1, 2) (c) f(0, 0, 0) (e) f(t, t 2 , −t)

Find (b) f(−3, 2, 1) (d) f(a, a, a) (f ) f(a + b, a − b, b).

18. Let f(x, y, z) = zxy + x. Find (a) f(x + y, x − y, x 2 ) (b) f(xy, y /x, xz).

19. Find F(f(x), g(y), h(z)) if F(x, y, z) = yexyz , f(x) = x 2 , g(y) = y + 1, and h(z) = z2 .

20. Find g(u(x, y, z), v(x, y, z), w(x, y, z)) if g(x, y, z) = z sin xy, u(x, y, z) = x 2 z3 , v(x, y, z) = πxyz, and w(x, y, z) = xy /z.

13.1 Functions of Two or More Variables 21–22 These exercises are concerned with functions of four or

more variables. ■

√ 21. (a) Let f(x,√y, z, t) = x 2 y 3 z + t. Find f( 5, 2, π, 3π). n kxk . (b) Let f(x1 , x2 , . . . , xn ) =

(b) f(x, y) = 2x 2 + 2y 2 (d) f(x, y) = 2y 2 − x

45–46 Refer to Figure 13.1.7 in each part. ■

45. Suppose that $6000 is borrowed at an interest rate of 11%. (a) Estimate the monthly payment on the loan. (b) If the interest rate drops to 9%, estimate how much more can be borrowed without increasing the monthly payment.

k=1

Find f(1, 1, . . . , 1).

22. (a) Let f(u, v, λ, φ) = eu+v cos λ tan φ. Find f(−2, 2, 0, π/4). (b) Let f(x1 , x2 , . . . , xn ) = x12 + x22 + · · · + xn2 . Find f(1, 2, . . . , n). 23–26 Sketch the domain of f. Use solid lines for portions of the boundary included in the domain and dashed lines for portions not included. ■ 23. f(x, y) = ln(1 − x 2 − y 2 ) 24. f(x, y) = x 2 + y 2 − 4

1 25. f(x, y) = x − y2

44. (a) f(x, y) = x 2 − 2xy + y 2 (c) f(x, y) = x 2 − 2x − y 2

915

26. f(x, y) = ln xy

27–28 Describe the domain of f in words. ■ √

− y+2 27. (a) f(x, y) = xe (b) f(x, y, z) = 25 − x 2 − y 2 − z2 (c) f(x, y, z) = exyz √ 4 − x2 (b) f(x, y) = ln(y − 2x) 28. (a) f(x, y) = 2 y +3 xyz (c) f(x, y, z) = x+y+z

29–32 True–False Determine whether the statement is true or

46. Suppose that $3000 is borrowed at an interest rate of 4%. (a) Estimate the monthly payment on the loan. (b) If the interest rate increases to 7%, estimate how much less would need to be borrowed so as not to increase the monthly payment. F O C U S O N C O N C E P TS

47. In each part, match the contour plot with one of the functions f(x, y) = x 2 + y 2 , f(x, y) = x 2 + y 2 , f(x, y) = 1 − x 2 − y 2

by inspection, and explain your reasoning. Larger values of z are indicated by lighter colors in the contour plot, and the concentric contours correspond to equally spaced values of z. (a) 2

(b) 2

y 0

false. Explain your answer. ■ 29. If the domain √ of f(x, y) is the xy-plane, then the domain of f (sin−1 t, t ) is the interval [0, 1].

30. If f(x, y) = y /x, then a contour f(x, y) = m is the straight line y = mx. 31. The natural domain of f(x, y, z) = 1 − x 2 − y 2 is a disk of radius 1 centered at the origin in the xy-plane. 32. Every level surface of f(x, y, z) = x + 2y + 3z is a plane.

−2 −2

48. In each part, match the contour plot with one of the surfaces in the accompanying ﬁgure by inspection, and explain your reasoning. The larger the value of z, the lighter the color in the contour plot. (a)

(b)

33–42 Sketch the graph of f. ■

33. f(x, y) = 3 35. f(x, y) = x 2 + y 2

37. f(x, y) = x 2 − y 2 39. f(x, y) = x 2 + y 2 + 1

41. f(x, y) = y + 1

34. f(x, y) =

9 − x2 − y2 2

36. f(x, y) = x + y

38. f(x, y) = 4 − x 2 − y 2 40. f(x, y) = x 2 + y 2 − 1

42. f(x, y) = x 2

43–44 In each part, select the term that best describes the level curves of the function f. Choose from the terms lines, circles, noncircular ellipses, parabolas, or hyperbolas. ■

43. (a) f(x, y) = 5x 2 − 5y 2 (c) f(x, y) = x 2 + 3y 2

y 0

(b) f(x, y) = y − 4x 2 (d) f(x, y) = 3x 2

−6 −6

(d)

y 0

−2 −2

y 0

−6 −6

Chapter 13 / Partial Derivatives

916

(I)

(II)

(III)

(IV)

(a) Referring to the accompanying weather map, is the wind speed greater in Medicine Hat, Alberta or in Chicago? Explain your reasoning. (b) Estimate the average rate of change in atmospheric pressure (in mb/mi) from Medicine Hat to Chicago, given that the distance between the two cities is approximately 1400 mi.

1012 Medicine Hat 1010 1008

1006

Chicago

1004 1002 1000 x

996

49. In each part, the questions refer to the contour map in the accompanying ﬁgure. (a) Is A or B the higher point? Explain your reasoning. (b) Is the slope steeper at point A or at point B? Explain your reasoning. (c) Starting at A and moving so that y remains constant and x increases, will the elevation begin to increase or decrease? (d) Starting at B and moving so that y remains constant and x increases, will the elevation begin to increase or decrease? (e) Starting at A and moving so that x remains constant and y decreases, will the elevation begin to increase or decrease? (f ) Starting at B and moving so that x remains constant and y decreases, will the elevation begin to increase or decrease? y

B 3

2 1

Pressure in millibars (mb)

Figure Ex-50

51–56 Sketch the level curve z = k for the speciﬁed values

of k. ■

51. 52. 53. 54. 55. 56.

z = x 2 + y 2 ; k = 0, 1, 2, 3, 4 z = y /x; k = −2, −1, 0, 1, 2 z = x 2 + y; k = −2, −1, 0, 1, 2 z = x 2 + 9y 2 ; k = 0, 1, 2, 3, 4 z = x 2 − y 2 ; k = −2, −1, 0, 1, 2 z = y csc x; k = −2, −1, 0, 1, 2

57–60 Sketch the level surface f(x, y, z) = k. ■

57. 58. 59. 60.

f(x, y, z) = 4x 2 + y 2 + 4z2 ; k = 16 f(x, y, z) = x 2 + y 2 − z2 ; k = 0 f(x, y, z) = z − x 2 − y 2 + 4; k = 7 f(x, y, z) = 4x − 2y + z; k = 1

61–64 Describe the level surfaces in words. ■

Elevations in hundreds of feet

A 4

998

1006

Figure Ex-48

2 3

1000

994

998

61. f(x, y, z) = (x − 2)2 + y 2 + z2

62. f(x, y, z) = 3x − y + 2z

Figure Ex-49

50. A curve connecting points of equal atmospheric pressure on a weather map is called an isobar. On a typical weather map the isobars refer to pressure at mean sea level and are given in units of millibars (mb). Mathematically, isobars are level curves for the pressure function p(x, y) deﬁned at the geographic points (x, y) represented on the map. Tightly packed isobars correspond to steep slopes on the graph of the pressure function, and these are usually associated with strong winds—the steeper the slope, the greater the speed of the wind.

64. f(x, y, z) = z − x 2 − y 2

63. f(x, y, z) = x 2 + z2

65. Let f(x, y) = x 2 − 2x 3 + 3xy. Find an equation of the level curve that passes through the point (a) (−1, 1) (b) (0, 0) (c) (2, −1). 66. Let f(x, y) = yex . Find an equation of the level curve that passes through the point (a) (ln 2, 1) (b) (0, 3) (c) (1, −2). 67. Let f(x, y, z) = x 2 + y 2 − z. Find an equation of the level surface that passes through the point (a) (1, −2, 0) (b) (1, 0, 3) (c) (0, 0, 0). 68. Let f(x, y, z) = xyz + 3. Find an equation of the level surface that passes through the point (a) (1, 0, 2) (b) (−2, 4, 1) (c) (0, 0, 0).

13.2 Limits and Continuity

69. If T (x, y) is the temperature at a point (x, y) on a thin metal plate in the xy-plane, then the level curves of T are called isothermal curves. All points on such a curve are at the same temperature. Suppose that a plate occupies the ﬁrst quadrant and T (x, y) = xy. (a) Sketch the isothermal curves on which T = 1, T = 2, and T = 3. (b) An ant, initially at (1, 4), wants to walk on the plate so that the temperature along its path remains constant. What path should the ant take and what is the temperature along that path? 70. If V (x, y) is the voltage or potential at a point (x, y) in the xy-plane, then the level curves of V are called equipotential curves. Along such a curve, the voltage remains constant. Given that 8 V (x, y) = 16 + x 2 + y 2 sketch the equipotential curves at which V = 2.0, V = 1.0, and V = 0.5.

71. Let f(x, y) = x 2 + y 3 . (a) Use a graphing utility to generate the level curve that passes through the point (2, −1). (b) Generate the level curve of height 1. √ 72. Let f(x, y) = 2 xy. (a) Use a graphing utility to generate the level curve that passes through the point (2, 2). (b) Generate the level curve of height 8. C

73. Let f(x, y) = xe−(x +y ) . (a) Use a CAS to generate the graph of f for −2 ≤ x ≤ 2 and −2 ≤ y ≤ 2.

917

(b) Generate a contour plot for the surface, and conﬁrm visually that it is consistent with the surface obtained in part (a). (c) Read the appropriate documentation and explore the effect of generating the graph of f from various viewpoints. C

1 x 74. Let f(x, y) = 10 e sin y. (a) Use a CAS to generate the graph of f for 0 ≤ x ≤ 4 and 0 ≤ y ≤ 2π. (b) Generate a contour plot for the surface, and conﬁrm visually that it is consistent with the surface obtained in part (a). (c) Read the appropriate documentation and explore the effect of generating the graph of f from various viewpoints.

75. In each part, describe in words how the graph of g is related to the graph of f. (a) g(x, y) = f(x − 1, y) (b) g(x, y) = 1 + f(x, y) (c) g(x, y) = −f(x, y + 1) 2

76. (a) Sketch the graph of f(x, y) = e−(x +y ) . (b) Describe in words how the graph of the function 2 2 g(x, y) = e−a(x +y ) is related to the graph of f for positive values of a. 77. Writing Find a few practical examples of functions of two and three variables, and discuss how physical considerations affect their domains. 78. Writing Describe two different ways in which a function f(x, y) can be represented geometrically. Discuss some of the advantages and disadvantages of each representation.

✔QUICK CHECK ANSWERS 13.1 1. points (x, y) in the ﬁrst or third quadrants; points (x, y) in the ﬁrst quadrant 2. (a) 41 (b) − 41 (c) 0 (d) 1/(2y + 2) 3. (a) k > 0 (b) the lines x + y = ln k 4. (a) 0 < k ≤ 1 (b) spheres of radius (1 − k)/k for 0 < k < 1, the single point (0, 0, 0) for k = 1

13.2

LIMITS AND CONTINUITY In this section we will introduce the notions of limit and continuity for functions of two or more variables. We will not go into great detail—our objective is to develop the basic concepts accurately and to obtain results needed in later sections. A more extensive study of these topics is usually given in advanced calculus. LIMITS ALONG CURVES

For a function of one variable there are two one-sided limits at a point x0 , namely, lim f(x)

x → x0 +

and

lim f(x)

x → x0 −

reﬂecting the fact that there are only two directions from which x can approach x0 , the right or the left. For functions of two or three variables the situation is more complicated

Chapter 13 / Partial Derivatives

918 y

because there are infinitely many different curves along which one point can approach another (Figure 13.2.1). Our first objective in this section is to define the limit of f(x, y) as (x, y) approaches a point (x0 , y0 ) along a curve C (and similarly for functions of three variables). If C is a smooth parametric curve in 2-space or 3-space that is represented by the equations

(x 0, y0) (x, y)

x = x(t),

Figure 13.2.1

x = x(t),

y = y(t),

z = z(t)

and if x0 = x(t0 ), y0 = y(t0 ), and z0 = z(t0 ), then the limits lim

(x, y) → (x0 , y0 ) (along C) In words, Formulas (1) and (2) state that a limit of a function f along a parametric curve can be obtained by substituting the parametric equations for the curve into the formula for the function and then computing the limit of the resulting function of one variable at the appropriate point.

y = y(t)

lim

(x, y) → (x0 , y0 ) (along C)

lim

(x, y, z) → (x0 , y0 , z0 ) (along C)

L z = f (x, y) y

lim

(x, y, z) → (x0 , y0 , z0 ) (along C)

f(x, y, z)

are deﬁned by

(x(t), y(t), f (x(t), y(t)))

(x 0 , y0 )

f(x, y) and

f(x, y) = lim f(x(t), y(t)) t → t0

f(x, y, z) = lim f(x(t), y(t), z(t)) t → t0

(1)

(2)

In these formulas the limit of the function of t must be treated as a one-sided limit if (x0 , y0 ) or (x0 , y0 , z0 ) is an endpoint of C. A geometric interpretation of the limit along a curve for a function of two variables is shown in Figure 13.2.2: As the point (x(t), y(t)) moves along the curve C in the xyplane toward (x0 , y0 ), the point (x(t), y(t), f(x(t), y(t))) moves directly above it along the graph of z = f(x, y) with f(x(t), y(t)) approaching the limiting value L. In the ﬁgure we followed a common practice of omitting the zero z-coordinate for points in the xy-plane.

(x(t), y(t)) x

Example 1 lim

(x, y) → (x 0 , y0 ) (along C )

Figure 13.2.2

f (x, y) = L

Figure 13.2.3a shows a computer-generated graph of the function xy f(x, y) = − 2 x + y2

The graph reveals that the surface has a ridge above the line y = −x, which is to be expected since f(x, y) has a constant value of 21 for y = −x, except at (0, 0) where f is undeﬁned (verify). Moreover, the graph suggests that the limit of f(x, y) as (x, y) → (0, 0) along a line through the origin varies with the direction of the line. Find this limit along (a) the x-axis (d) the line y = −x

(b) the y-axis (e) the parabola y = x 2

Solution (a). The x-axis has parametric equations x = t, y = 0, with (0, 0) correspond-

ing to t = 0, so

lim

(x, y) → (0, 0) (along y = 0)

f(x, y) = lim f(t, 0) = lim t →0

t →0

0 − 2 = lim 0 = 0 t →0 t

which is consistent with Figure 13.2.3b.

Solution (b). The y-axis has parametric equations x = 0, y = t, with (0, 0) corresponding to t = 0, so

0 lim f(x, y) = lim f(0, t) = lim − 2 = lim 0 = 0 t →0 t →0 t →0 (x, y) → (0, 0) t (along x = 0)

which is consistent with Figure 13.2.3b.

13.2 Limits and Continuity −1

−1

0 1

−0.5

−1 0

冸z = 冹 1 2

0.5 z

y = −x

0.5

−0.5 y=0 (z = 0)

0 y=x

冸z = − 12 冹

0 −0.5

−1 x=0 (z = 0)

−1 0

冸

(a)

0.5

919

(b)

y = x2 x z=− 1 + x2

冹

(c)

Figure 13.2.3

Solution (c). The line y = x has parametric equations x = t, y = t, with (0, 0) corresponding to t = 0, so lim

(x, y) → (0, 0) (along y = x)

f(x, y) = lim f(t, t) = lim t →0

t →0

−

t2 2t 2

= lim

t →0

1 1 − =− 2 2

which is consistent with Figure 13.2.3b.

Solution (d). The line y = −x has parametric equations x = t, y = −t, with (0, 0) corresponding to t = 0, so lim

(x, y) → (0, 0) (along y = −x)

1 1 t2 = lim = t →0 2 t → 0 2t 2 2

f(x, y) = lim f(t, −t) = lim t →0

which is consistent with Figure 13.2.3b. For uniformity, we have chosen the same parameter t in each part of Example 1. We could have used x or y as the parameter, according to the context. For example, part (b) could be computed using

lim f(0, y)

y→0

and part (e) could be computed using

lim f(x, x 2 )

x→0

Solution (e). The parabola y = x 2 has parametric equations x = t, y = t 2 , with (0, 0) corresponding to t = 0, so lim

(x, y) → (0, 0) (along y = x 2 )

f(x, y) = lim f(t, t 2 ) = lim t →0

t →0

−

t3 2 t + t4

= lim

t →0

−

t 1 + t2

This is consistent with Figure 13.2.3c, which shows the parametric curve t x = t, y = t 2 , z = − 1 + t2 superimposed on the surface. OPEN AND CLOSED SETS Although limits along speciﬁc curves are useful for many purposes, they do not always tell the complete story about the limiting behavior of a function at a point; what is required is a limit concept that accounts for the behavior of the function in an entire vicinity of a point, not just along smooth curves passing through the point. For this purpose, we start by introducing some terminology. Let C be a circle in 2-space that is centered at (x0 , y0 ) and has positive radius δ. The set of points that are enclosed by the circle, but do not lie on the circle, is called the open disk of radius δ centered at (x0 , y0 ), and the set of points that lie on the circle together with those enclosed by the circle is called the closed disk of radius δ centered at (x0 , y0 )

920

Chapter 13 / Partial Derivatives

A closed disk includes all of the points on its bounding circle.

An open disk contains none of the points on its bounding circle.

Figure 13.2.4

An interior point

A boundary point

(Figure 13.2.4). Analogously, if S is a sphere in 3-space that is centered at (x0 , y0 , z0 ) and has positive radius δ, then the set of points that are enclosed by the sphere, but do not lie on the sphere, is called the open ball of radius δ centered at (x0 , y0 , z0 ), and the set of points that lie on the sphere together with those enclosed by the sphere is called the closed ball of radius δ centered at (x0 , y0 , z0 ). Disks and balls are the two-dimensional and three-dimensional analogs of intervals on a line. The notions of “open” and “closed” can be extended to more general sets in 2-space and 3-space. If D is a set of points in 2-space, then a point (x0 , y0 ) is called an interior point of D if there is some open disk centered at (x0 , y0 ) that contains only points of D, and (x0 , y0 ) is called a boundary point of D if every open disk centered at (x0 , y0 ) contains both points in D and points not in D. The same terminology applies to sets in 3-space, but in that case the deﬁnitions use balls rather than disks (Figure 13.2.5). For a set D in either 2-space or 3-space, the set of all interior points is called the interior of D and the set of all boundary points is called the boundary of D. Moreover, just as for disks, we say that D is closed if it contains all of its boundary points and open if it contains none of its boundary points. The set of all points in 2-space and the set of all points in 3-space have no boundary points (why?), so by agreement they are regarded to be both open and closed.

A boundary point

An interior point

Figure 13.2.5

GENERAL LIMITS OF FUNCTIONS OF TWO VARIABLES The statement lim f(x, y) = L (x,y) → (x0 ,y0 )

is intended to convey the idea that the value of f(x, y) can be made as close as we like to the number L by restricting the point (x, y) to be sufﬁciently close to (but different from) the point (x0 , y0 ). This idea has a formal expression in the following deﬁnition and is illustrated in Figure 13.2.6.

13.2.1 definition Let f be a function of two variables, and assume that f is deﬁned at all points of some open disk centered at (x0 , y0 ), except possibly at (x0 , y0 ). We will write lim f(x, y) = L (3) (x,y) → (x0 ,y0 )

When convenient, (3) can also be written as

lim f (x, y) = L

x → x0 y → y0

or as

f(x, y) → L as (x, y) → (x0 , y0 )

if given any number ǫ > 0, we can find a number δ > 0 such that f(x, y) satisfies |f(x, y) − L| < ǫ whenever the distance between (x, y) and (x0 , y0 ) satisfies 0 < (x − x0 )2 + (y − y0 )2 < δ Another illustration of Definition 13.2.1 is shown in the “arrow diagram” of Figure 13.2.7. As in Figure 13.2.6, this figure is intended to convey the idea that the values of f (x, y) can be forced within ǫ units of L on the z-axis by restricting (x, y) to lie within δ units of (x0 , y0 ) in the xy-plane. We used a white dot at (x0 , y0 ) to suggest that the epsilon condition need not hold at this point. We note without proof that the standard properties of limits hold for limits along curves and for general limits of functions of two variables, so that computations involving such limits can be performed in the usual way.

13.2 Limits and Continuity

921

z= L+e L+e f (x, y) L z= L−e

L−e

In Figure 13.2.6, the condition

|f(x, y) − L| < ǫ is satisfied at each point (x, y) within the circular region. However, the fact that this condition is satisfied at the center of the circular region is not relevant to the limit.

This circular region with the center removed consists of all points (x, y) that satisfy 0 < √(x − x0 )2 + (y − y 0 ) 2 < d.

z = f (x, y) y

(x, y) d x

(x0 , y0 )

lim f (x, y) = L (x, y)→(x0, y0)

Figure 13.2.6 y

(x 0 , y 0 ) d

(x, y)

( L−e

f (x, y)

( L+e

Figure 13.2.7

Example 2 lim

(x,y) → (1,4)

[5x 3 y 2 − 9] =

[5x 3 y 2 ] − lim 9 (x,y) → (1,4)

2 lim x lim y − 9

lim

(x,y) → (1,4)

(x,y) → (1,4) 3

(x,y) → (1,4)

= 5(1) (4) − 9 = 71 RELATIONSHIPS BETWEEN GENERAL LIMITS AND LIMITS ALONG SMOOTH CURVES

WARNING In general, one cannot show that

lim

(x,y) → (x0 ,y0 )

f(x, y) = L

by showing that this limit holds along a specific curve, or even some specific family of curves. The problem is there may be some other curve along which the limit does not exist or has a value different from L (see Exercise 34, for example).

Stated informally, if f (x, y) has limit L as (x, y) approaches (x0 , y0 ), then the value of f (x, y) gets closer and closer to L as the distance between (x, y) and (x0 , y0 ) approaches zero. Since this statement imposes no restrictions on the direction in which (x, y) approaches (x0 , y0 ), it is plausible that the function f (x, y) will also have the limit L as (x, y) approaches (x0 , y0 ) along any smooth curve C. This is the implication of the following theorem, which we state without proof. 13.2.2

theorem

(a) If f(x, y) → L as (x, y) → (x0 , y0 ), then f(x, y) → L as (x, y) → (x0 , y0 ) along any smooth curve. (b) If the limit of f(x, y) fails to exist as (x, y) → (x0 , y0 ) along some smooth curve, or if f(x, y) has different limits as (x, y) → (x0 , y0 ) along two different smooth curves, then the limit of f(x, y) does not exist as (x, y) → (x0 , y0 ).

922

Chapter 13 / Partial Derivatives

Example 3 The limit

lim

(x,y) → (0,0)

−

xy x2 + y2

does not exist because in Example 1 we found two different smooth curves along which this limit had different values. Speciﬁcally, lim

(x, y) → (0, 0) (along x = 0)

−

xy =0 x2 + y2

and

lim

(x, y) → (0, 0) (along y = x)

−

1 xy =− x2 + y2 2

CONTINUITY z

y Hole at the origin

Stated informally, a function of one variable is continuous if its graph is an unbroken curve without jumps or holes. To extend this idea to functions of two variables, imagine that the graph of z = f(x, y) is formed from a thin sheet of clay that has been molded into peaks and valleys. We will regard f as being continuous if the clay surface has no tears or holes. The functions graphed in Figure 13.2.8 fail to be continuous because of their behavior at (0, 0). The precise deﬁnition of continuity at a point for functions of two variables is similar to that for functions of one variable—we require the limit of the function and the value of the function to be the same at the point.

13.2.3 definition A function f(x, y) is said to be continuous at (x0 , y0 ) if f(x0 , y0 ) is deﬁned and if lim f(x, y) = f(x0 , y0 ) (x,y) → (x0 ,y0 )

y Infinite at the origin

In addition, if f is continuous at every point in an open set D, then we say that f is continuous on D, and if f is continuous at every point in the xy-plane, then we say that f is continuous everywhere.

The following theorem, which we state without proof, illustrates some of the ways in which continuous functions can be combined to produce new continuous functions. x

y Vertical jump at the origin

Figure 13.2.8

13.2.4

theorem

(a) If g(x) is continuous at x0 and h(y) is continuous at y0 , then f(x, y) = g(x)h(y) is continuous at (x0 , y0 ). (b) If h(x, y) is continuous at (x0 , y0 ) and g(u) is continuous at u = h(x0 , y0 ), then the composition f (x, y) = g(h(x, y)) is continuous at (x0 , y0 ). (c) If f (x, y) is continuous at (x0 , y0 ), and if x(t) and y(t) are continuous at t0 with x(t0 ) = x0 and y(t0 ) = y0 , then the composition f (x(t), y(t)) is continuous at t0 . Example 4 Use Theorem 13.2.4 to show that the functions f(x, y) = 3x 2 y 5 and f(x, y) = sin(3x 2 y 5 ) are continuous everywhere.

Solution. The polynomials g(x) = 3x 2 and h(y) = y 5 are continuous at every real num-

ber, and therefore by part (a) of Theorem 13.2.4, the function f(x, y) = 3x 2 y 5 is continuous at every point (x, y) in the xy-plane. Since 3x 2 y 5 is continuous at every point in the xy-plane and sin u is continuous at every real number u, it follows from part (b) of Theorem 13.2.4 that the composition f(x, y) = sin(3x 2 y 5 ) is continuous everywhere.

13.2 Limits and Continuity

923

Theorem 13.2.4 is one of a whole class of theorems about continuity of functions in two or more variables. The content of these theorems can be summarized informally with three basic principles: Recognizing Continuous Functions

• A composition of continuous functions is continuous. • A sum, difference, or product of continuous functions is continuous. • A quotient of continuous functions is continuous, except where the denominator is zero. By using these principles and Theorem 13.2.4, you should be able to conﬁrm that the following functions are all continuous everywhere: xy xexy + y 2/3 , cosh(xy 3 ) − |xy|, 1 + x2 + y2 Example 5

Evaluate

lim

(x,y) → (−1,2)

xy . x2 + y2

Solution. Since f(x, y) = xy /(x 2 + y 2 ) is continuous at (−1, 2) (why?), it follows from

the deﬁnition of continuity for functions of two variables that lim

(x,y) → (−1,2)

Example 6

xy (−1)(2) 2 = =− 2 2 2 +y (−1) + (2) 5

Since the function f(x, y) =

x3y2 1 − xy

is a quotient of continuous functions, it is continuous except where 1 − xy = 0. Thus, f(x, y) is continuous everywhere except on the hyperbola xy = 1. LIMITS AT DISCONTINUITIES

Sometimes it is easy to recognize when a limit does not exist. For example, it is evident that 1 = +⬁ lim 2 (x,y) → (0,0) x + y 2 which implies that the values of the function approach +⬁ as (x, y) → (0, 0) along any smooth curve (Figure 13.2.9). However, it is not evident whether the limit

lim

z= Figure 13.2.9

1 x2 + y2

(x,y) → (0,0)

(x 2 + y 2 ) ln(x 2 + y 2 )

exists because it is an indeterminate form of type 0 · ⬁. Although L’Hôpital’s rule cannot be applied directly, the following example illustrates a method for ﬁnding this limit by converting to polar coordinates. Example 7

Find

lim

(x,y) → (0,0)

(x 2 + y 2 ) ln(x 2 + y 2 ).

Solution. Let (r, θ) be polar coordinates of the point (x, y) with r ≥ 0. Then we have x = r cos θ,

y = r sin θ,

r2 = x2 + y2

Chapter 13 / Partial Derivatives

924

Moreover, since r ≥ 0 we have r = x 2 + y 2 , so that r → 0+ if and only if (x, y) → (0, 0). Thus, we can rewrite the given limit as lim

(x,y) → (0,0)

(x 2 + y 2 ) ln(x 2 + y 2 ) = lim+ r 2 ln r 2 r →0

2 ln r 1/r 2 2/r = lim+ r → 0 −2/r 3 = lim+ r →0

This converts the limit to an indeterminate form of type ⬁/⬁. L’Hôpital’s rule

= lim+ (−r 2 ) = 0 r →0

REMARK

The graph of f (x, y) = (x 2 + y 2 ) ln(x 2 + y 2 ) in Example 7 is a surface with a hole (sometimes called a puncture) at the origin (Figure 13.2.10). We can remove this discontinuity by defining f (0, 0) to be 0. (See Exercises 39 and 40, which also deal with the notion of a “removable” discontinuity.)

CONTINUITY AT BOUNDARY POINTS x

Recall that in our study of continuity for functions of one variable, we first defined continuity at a point, then continuity on an open interval, and then, by using one-sided limits, we extended the notion of continuity to include the boundary points of the interval. Similarly, for functions of two variables one can extend the notion of continuity of f (x, y) to the boundary of its domain by modifying Definition 13.2.1 appropriately so that (x, y) is restricted to approach (x0 , y0 ) through points lying wholly in the domain of f . We will omit the details.

z = (x2 + y2 ) ln (x2 + y2 ) Figure 13.2.10

Example 8 The graph of the function f (x, y) = 1 − x 2 − y 2 is the upper hemisphere shown in Figure 13.2.11, and the natural domain of f is the closed unit disk

x2 + y2 ≤ 1

y −1

1 x

√1 − x 2 − y 2

The graph of f has no tears or holes, so it passes our “intuitive test” of continuity. In this case the continuity at a point (x0 , y0 ) on the boundary reﬂects the fact that 1 − x 2 − y 2 = 1 − x02 − y02 = 0 lim (x,y) → (x0 ,y0 )

when (x, y) is restricted to points on the closed unit disk x 2 + y 2 ≤ 1. It follows that f is continuous on its domain.

Figure 13.2.11

EXTENSIONS TO THREE VARIABLES All of the results in this section can be extended to functions of three or more variables. For example, the distance between the points (x, y, z) and (x0 , y0 , z0 ) in 3-space is (x − x0 )2 + (y − y0 )2 + (z − z0 )2

so the natural extension of Deﬁnition 13.2.1 to 3-space is as follows:

13.2.5 definition Let f be a function of three variables, and assume that f is deﬁned at all points within a ball centered at (x0 , y0 , z0 ), except possibly at (x0 , y0 , z0 ). We will write f(x, y, z) = L (4) lim (x,y,z) → (x0 ,y0 ,z0 )

if given any number ǫ > 0, we can ﬁnd a number δ > 0 such that f(x, y, z) satisﬁes |f(x, y, z) − L| < ǫ

whenever the distance between (x, y, z) and (x0 , y0 , z0 ) satisﬁes 0 < (x − x0 )2 + (y − y0 )2 + (z − z0 )2 < δ

13.2 Limits and Continuity

925

As with functions of one and two variables, we deﬁne a function f(x, y, z) of three variables to be continuous at a point (x0 , y0 , z0 ) if the limit of the function and the value of the function are the same at this point; that is, lim

(x,y,z) → (x0 ,y0 ,z0 )

f(x, y, z) = f(x0 , y0 , z0 )

Although we will omit the details, the properties of limits and continuity that we discussed for functions of two variables, including the notion of continuity at boundary points, carry over to functions of three variables.

✔QUICK CHECK EXERCISES 13.2 1. Let

(See page 927 for answers.)

(b)

x2 − y2 f (x, y) = 2 x + y2

2. (a)

lim

(x,y) → (3,2)

x cos πy =

EXERCISE SET 13.2 1–6 Use limit laws and continuity properties to evaluate the

limit. ■ 1. 3. 5.

lim

(x,y) → (1,3)

(4xy 2 − x) 3

xy (x,y) → (−1,2) x + y lim

lim

(x,y) → (0,0)

ln(1 + x 2 y 3 )

4x − y (x,y) → (0,0) sin y − 1 lim

lim

(x,y) → (4,−2)

as (x, y) → (0, 0) along the coordinate axes. ■ 3 x+y 7. (a) lim (b) lim 2 2 (x,y) → (0,0) x + 2y (x,y) → (0,0) 2x 2 + y 2 x−y cos xy 8. (a) lim (b) lim (x,y) → (0,0) x 2 + y 2 (x,y) → (0,0) x 2 + y 2 9–12 Evaluate the limit using the substitution z = x 2 + y 2 and

observing that z → 0+ if and only if (x, y) → (0, 0). ■ sin(x 2 + y 2 ) 1 − cos(x 2 + y 2 ) 9. lim 10. lim (x,y) → (0,0) (x,y) → (0,0) x2 + y2 x2 + y2 √2 2 −1/ x +y 2 2 e 11. lim e−1/(x +y ) 12. lim (x,y) → (0,0) (x,y) → (0,0) x2 + y2 13–22 Determine whether the limit exists. If so, ﬁnd its value.

■

x 4 − 16y 4 14. lim (x,y) → (0,0) x 2 + 4y 2 1 − x2 − y2 16. lim (x,y) → (0,0) x2 + y2

(x 2 + y 2 ) sin

4. Determine all values of the constant a such that the function f (x, y) = x 2 − ay 2 + 1 is continuous everywhere. xz2 (x,y,z) → (2,−1,2) x 2 + y 2 + z2 18. lim ln(2x + y − z) 17.

lim

sin(x 2 + y 2 + z2 ) (x,y,z) → (0,0,0) x 2 + y 2 + z2 sin x 2 + y 2 + z2 20. lim (x,y,z) → (0,0,0) x 2 + y 2 + z2 √2 2 2 x +y +z 21. lim e (x,y,z) → (0,0,0) x 2 + y 2 + z2

1 22. lim tan−1 2 (x,y,z) → (0,0,0) x + y 2 + z2

19.

7–8 Show that the limit does not exist by considering the limits

x4 − y4 13. lim (x,y) → (0,0) x 2 + y 2 xy 15. lim (x,y) → (0,0) 3x 2 + 2y 2

lim

(x,y,z) → (2,0,−1)

e2x−y (x,y) → (1,−3) 6. lim x 3 y 3 + 2x

1 = (x,y) → (0,0) x2 + y2 3. A function f (x, y) is continuous at (x0 , y0 ) provided f (x0 , y0 ) exists and provided f (x, y) has limit as (x, y) approaches . (c)

Determine the limit of f (x, y) as (x, y) approaches (0, 0) along the curve C. (a) C: x = 0 (b) C: y = 0 (c) C: y = x (d) C: y = x 2

exy =

lim

(x,y) → (0,1)

lim

23–26 Evaluate the limits by converting to polar coordinates, as in Example 7. ■ 23. lim x 2 + y 2 ln(x 2 + y 2 ) (x,y) → (0,0)

24. 26.

lim

(x,y) → (0,0)

lim

(x,y) → (0,0)

y ln(x 2 + y 2 ) xy 2 x + 2y 2

25.

x2y2 (x,y) → (0,0) x2 + y2 lim

27–28 Evaluate the limits by converting to spherical coordinates (ρ, θ, φ) and by observing that ρ → 0+ if and only if (x, y, z) → (0, 0, 0). ■ xyz 27. lim 2 (x,y,z) → (0,0,0) x + y 2 + z2

926

28.

Chapter 13 / Partial Derivatives

lim

(x,y,z) â&#x2020;&#x2019; (0,0,0)

sin x sin y 2 x + 2y 2 + 3z2

35. (a) Show that the value of xyz x 2 + y 4 + z4

29â&#x20AC;&#x201C;32 Trueâ&#x20AC;&#x201C;False Determine whether the statement is true or

false. Explain your answer. â&#x2013; 29. If D is an open set in 2-space or in 3-space, then every point in D is an interior point of D.

approaches 0 as (x, y, z) â&#x2020;&#x2019; (0, 0, 0) along any line x = at, y = bt, z = ct. (b) Show that the limit

30. If f(x, y) â&#x2020;&#x2019; L as (x, y) approaches (0, 0) along the x-axis, and if f(x, y) â&#x2020;&#x2019; L as (x, y) approaches (0, 0) along the y-axis, then lim(x,y) â&#x2020;&#x2019; (0,0) f(x, y) = L. 31. If f and g are functions of two variables such that f + g and fg are both continuous, then f and g are themselves continuous.

F O C U S O N C O N C E P TS

33. The accompanying ďŹ gure shows a portion of the graph of x2y f(x, y) = 4 x + y2 (a) Based on the graph in the ďŹ gure, does f(x, y) have a limit as (x, y) â&#x2020;&#x2019; (0, 0)? Explain your reasoning. (b) Show that f(x, y) â&#x2020;&#x2019; 0 as (x, y) â&#x2020;&#x2019; (0, 0) along any line y = mx. Does this imply that f(x, y) â&#x2020;&#x2019; 0 as (x, y) â&#x2020;&#x2019; (0, 0)? Explain. (c) Show that f(x, y) â&#x2020;&#x2019; 21 as (x, y) â&#x2020;&#x2019; (0, 0) along the parabola y = x 2 , and conďŹ rm visually that this is consistent with the graph of f(x, y). (d) Based on parts (b) and (c), does f(x, y) have a limit as (x, y) â&#x2020;&#x2019; (0, 0)? Is this consistent with your answer to part (a)? â&#x2C6;&#x2019;0.5

â&#x2C6;&#x2019;0.5 â&#x2C6;&#x2019;1 â&#x2C6;&#x2019;0.5

x2 + 1 . (x,y) â&#x2020;&#x2019; (0,1) x 2 + (y â&#x2C6;&#x2019; 1)2

x2 â&#x2C6;&#x2019; 1 . 37. Find lim tanâ&#x2C6;&#x2019;1 2 (x,y) â&#x2020;&#x2019; (0,1) x + (y â&#x2C6;&#x2019; 1)2 â&#x17D;§ 2 2 â&#x17D;Ş â&#x17D;¨ sin(x + y ) , (x, y) = (0, 0) x2 + y2 38. Let f(x, y) = â&#x17D;Ş â&#x17D;Š 1, (x, y) = (0, 0). Show that f is continuous at (0, 0). lim

tanâ&#x2C6;&#x2019;1

39â&#x20AC;&#x201C;40 A function f (x, y) is said to have a removable discontinuity at (x0 , y0 ) if lim(x,y) â&#x2020;&#x2019; (x0 ,y0 ) f (x, y) exists but f is not continuous at (x0 , y0 ), either because f is not deďŹ ned at (x0 , y0 ) or because f (x0 , y0 ) differs from the value of the limit. Determine whether f (x, y) has a removable discontinuity at (0, 0). â&#x2013;

x2 39. f (x, y) = 2 x + y2 2 x + 7y 2 , 40. f (x) = â&#x2C6;&#x2019;4,

if (x, y) = (0, 0) if (x, y) = (0, 0)

41â&#x20AC;&#x201C;48 Sketch the largest region on which the function f is continuous. â&#x2013; â&#x2C6;&#x161; 41. f(x, y) = y ln(1 + x) 42. f(x, y) = x â&#x2C6;&#x2019; y

0.5 0.5

â&#x2C6;&#x2019;1

xyz x 2 + y 4 + z4

does not exist by letting (x, y, z) â&#x2020;&#x2019; (0, 0, 0) along the curve x = t 2 , y = t, z = t. 36. Find

32. If limx â&#x2020;&#x2019; 0+ f(x) = L = 0, then x2 + y2 lim =0 (x,y) â&#x2020;&#x2019; (0,0) f(x 2 + y 2 )

lim

(x,y,z) â&#x2020;&#x2019; (0,0,0)

0 0.5

Figure Ex-33

34. (a) Show that the value of x3y + y2 approaches 0 as (x, y) â&#x2020;&#x2019; (0, 0) along any straight line y = mx, or along any parabola y = kx 2 . (b) Show that x3y lim 6 (x,y) â&#x2020;&#x2019; (0,0) 2x + y 2 does not exist by letting (x, y) â&#x2020;&#x2019; (0, 0) along the curve y = x 3 . 2x 6

x2y 43. f(x, y) = 25 â&#x2C6;&#x2019; x 2 â&#x2C6;&#x2019; y 2 44. f(x, y) = ln(2x â&#x2C6;&#x2019; y + 1) y 45. f(x, y) = 11x 2 + 3 46. f(x, y) = e1â&#x2C6;&#x2019;xy 47. f(x, y) = sinâ&#x2C6;&#x2019;1 (xy)

48. f(x, y) = tanâ&#x2C6;&#x2019;1 (y â&#x2C6;&#x2019; x)

49â&#x20AC;&#x201C;52 Describe the largest region on which the function f is

continuous. â&#x2013; 49. f(x, y, z) = 3x 2 eyz cos(xyz)

50. f(x, y, z) = ln(4 â&#x2C6;&#x2019; x 2 â&#x2C6;&#x2019; y 2 â&#x2C6;&#x2019; z2 ) y+1 51. f(x, y, z) = 2 x + z2 â&#x2C6;&#x2019; 1 52. f(x, y, z) = sin x 2 + y 2 + 3z2

13.3 Partial Derivatives

53. Writing Describe the procedure you would use to determine whether or not the limit lim

(x,y) → (x0 ,y0 )

927

54. Writing In your own words, state the geometric interpretations of ǫ and δ in the deﬁnition of lim

f(x, y)

(x,y) → (x0 ,y0 )

f(x, y) = L

given in Deﬁnition 13.2.1.

exists.

✔QUICK CHECK ANSWERS 13.2 1. (a) −1 (b) 1 (c) 0 (d) 1

13.3

2. (a) 3 (b) 1 (c) 0

3. f (x0 , y0 ); (x0 , y0 )

4. a ≤ 0

PARTIAL DERIVATIVES In this section we will develop the mathematical tools for studying rates of change that involve two or more independent variables. PARTIAL DERIVATIVES OF FUNCTIONS OF TWO VARIABLES

If z = f(x, y), then one can inquire how the value of z changes if y is held fixed and x is allowed to vary, or if x is held fixed and y is allowed to vary. For example, the ideal gas law in physics states that under appropriate conditions the pressure exerted by a gas is a function of the volume of the gas and its temperature. Thus, a physicist studying gases might be interested in the rate of change of the pressure if the volume is held fixed and the temperature is allowed to vary, or if the temperature is held fixed and the volume is allowed to vary. We now define a derivative that describes such rates of change. Suppose that (x0 , y0 ) is a point in the domain of a function f(x, y). If we fix y = y0 , then f(x, y0 ) is a function of the variable x alone. The value of the derivative d [f(x, y0 )] dx at x0 then gives us a measure of the instantaneous rate of change of f with respect to x at the point (x0 , y0 ). Similarly, the value of the derivative d [f(x0 , y)] dy at y0 gives us a measure of the instantaneous rate of change of f with respect to y at the point (x0 , y0 ). These derivatives are so basic to the study of differential calculus of multivariable functions that they have their own name and notation.

The limits in (1) and (2) show the relationship between partial derivatives and derivatives of functions of one variable. In practice, our usual method for computing partial derivatives is to hold one variable fixed and then differentiate the resulting function using the derivative rules for functions of one variable.

13.3.1 definition If z = f(x, y) and (x0 , y0 ) is a point in the domain of f, then the partial derivative of f with respect to x at (x0 , y0 ) [also called the partial derivative of z with respect to x at (x0 , y0 )] is the derivative at x0 of the function that results when y = y0 is held ﬁxed and x is allowed to vary. This partial derivative is denoted by fx(x0 , y0 ) and is given by f(x0 + x, y0 ) − f(x0 , y0 ) d [f(x, y0 )] = lim fx(x0 , y0 ) = (1) x → 0 dx x x=x0

Similarly, the partial derivative of f with respect to y at (x0 , y0 ) [also called the partial derivative of z with respect to y at (x0 , y0 )] is the derivative at y0 of the function that results when x = x0 is held ﬁxed and y is allowed to vary. This partial derivative is denoted by fy(x0 , y0 ) and is given by f(x0 , y0 + y) − f(x0 , y0 ) d = lim fy(x0 , y0 ) = (2) [f(x0 , y)] y → 0 dy y y=y0

928

Chapter 13 / Partial Derivatives

Example 1

Find fx(1, 3) and fy (1, 3) for the function f(x, y) = 2x 3 y 2 + 2y + 4x.

Solution. Since fx(x, 3) =

d d [f(x, 3)] = [18x 3 + 4x + 6] = 54x 2 + 4 dx dx

we have fx(1, 3) = 54 + 4 = 58. Also, since fy(1, y) =

d d [f(1, y)] = [2y 2 + 2y + 4] = 4y + 2 dy dy

we have fy(1, 3) = 4(3) + 2 = 14. THE PARTIAL DERIVATIVE FUNCTIONS

Formulas (1) and (2) deﬁne the partial derivatives of a function at a speciﬁc point (x0 , y0 ). However, often it will be desirable to omit the subscripts and think of the partial derivatives as functions of the variables x and y. These functions are fx (x, y) = lim

x → 0

f(x + x, y) − f(x, y) x

fy (x, y) = lim

y → 0

f(x, y + y) − f(x, y) y

The following example gives an alternative way of performing the computations in Example 1. Example 2 Find fx (x, y) and fy (x, y) for f(x, y) = 2x 3 y 2 + 2y + 4x, and use those partial derivatives to compute fx (1, 3) and fy (1, 3).

Solution. Keeping y ﬁxed and differentiating with respect to x yields fx(x, y) = T E C H N O LO GY M A ST E R Y Computer algebra systems have specific commands for calculating partial derivatives. If you have a CAS, use it to find the partial derivatives fx (x, y) and fy (x, y) in Example 2.

d [2x 3 y 2 + 2y + 4x] = 6x 2 y 2 + 4 dx

and keeping x ﬁxed and differentiating with respect to y yields fy (x, y) =

d [2x 3 y 2 + 2y + 4x] = 4x 3 y + 2 dy

Thus, fx (1, 3) = 6(12 )(32 ) + 4 = 58

and fy (1, 3) = 4(13 )3 + 2 = 14

which agree with the results in Example 1.

PARTIAL DERIVATIVE NOTATION The symbol ∂ is called a partial derivative sign. It is derived from the Cyrillic alphabet.

If z = f(x, y), then the partial derivatives fx and fy are also denoted by the symbols ∂f , ∂x

∂z ∂x

and

∂f , ∂y

∂z ∂y

Some typical notations for the partial derivatives of z = f(x, y) at a point (x0 , y0 ) are ∂z ∂z ∂f ∂f ∂f (x0 , y0 ), (x0 , y0 ) , , , ∂x x=x0 ,y=y0 ∂x (x0 ,y0 ) ∂x (x0 ,y0 ) ∂x ∂x

13.3 Partial Derivatives

929

Find ∂z/∂x and ∂z/∂y if z = x 4 sin(xy 3 ).

Example 3

Solution. ∂z ∂ 4 ∂ ∂ 4 = [x sin(xy 3 )] = x 4 [sin(xy 3 )] + sin(xy 3 ) · (x ) ∂x ∂x ∂x ∂x = x 4 cos(xy 3 ) · y 3 + sin(xy 3 ) · 4x 3 = x 4 y 3 cos(xy 3 ) + 4x 3 sin(xy 3 )

∂ 4 ∂ ∂ 4 ∂z = [x sin(xy 3 )] = x 4 [sin(xy 3 )] + sin(xy 3 ) · (x ) ∂y ∂y ∂y ∂y = x 4 cos(xy 3 ) · 3xy 2 + sin(xy 3 ) · 0 = 3x 5 y 2 cos(xy 3 )

PARTIAL DERIVATIVES VIEWED AS RATES OF CHANGE AND SLOPES

Recall that if y = f(x), then the value of f ′(x0 ) can be interpreted either as the rate of change of y with respect to x at x0 or as the slope of the tangent line to the graph of f at x0 . Partial derivatives have analogous interpretations. To see that this is so, suppose that C1 is the intersection of the surface z = f(x, y) with the plane y = y0 and that C2 is its intersection with the plane x = x0 (Figure 13.3.1). Thus, fx(x, y0 ) can be interpreted as the rate of change of z with respect to x along the curve C1 , and fy(x0 , y) can be interpreted as the rate of change of z with respect to y along the curve C2 . In particular, fx(x0 , y0 ) is the rate of change of z with respect to x along the curve C1 at the point (x0 , y0 ), and fy(x0 , y0 ) is the rate of change of z with respect to y along the curve C2 at the point (x0 , y0 ). z

z Slope = f y (x 0 , y0 )

Slope = f x (x 0 , y0 )

z = f (x, y) C1

y = y0 (x 0 , y0 )

y0 z = f (x, y)

x = x0 (x 0 , y0 )

Figure 13.3.1

Example 4 In an applied problem, the interpretations of fx (x0 , y0 ) and fy (x0 , y0 ) must be accompanied by the proper units. See Example 4.

Recall that the wind chill temperature index is given by the formula W = 35.74 + 0.6215T + (0.4275T − 35.75)v 0.16

Compute the partial derivative of W with respect to v at the point (T , v) = (25, 10) and interpret this partial derivative as a rate of change.

Solution. Holding T ﬁxed and differentiating with respect to v yields ∂W (T , v) = 0 + 0 + (0.4275T − 35.75)(0.16)v 0.16−1 = (0.4275T − 35.75)(0.16)v −0.84 ∂v Since W is in degrees Fahrenheit and v is in miles per hour, a rate of change of W with respect to v will have units ◦ F/(mi/h) (which may also be written as ◦ F·h/mi). Substituting

930

Chapter 13 / Partial Derivatives

Confirm the conclusion of Example 4 by calculating

W (25, 10 + v) − W (25, 10) v for values of v near 0.

T = 25 and v = 10 gives

◦ F ∂W (25, 10) = (−4.01)10−0.84 ≈ −0.58 ∂v mi/h

as the instantaneous rate of change of W with respect to v at (T , v) = (25, 10). We conclude that if the air temperature is a constant 25 ◦ F and the wind speed changes by a small amount from an initial speed of 10 mi/h, then the ratio of the change in the wind chill index to the change in wind speed should be about −0.58 ◦ F/(mi/h). Geometrically, fx(x0 , y0 ) can be viewed as the slope of the tangent line to the curve C1 at the point (x0 , y0 ), and fy(x0 , y0 ) can be viewed as the slope of the tangent line to the curve C2 at the point (x0 , y0 ) (Figure 13.3.1). We will call fx(x0 , y0 ) the slope of the surface in the x-direction at (x0 , y0 ) and fy(x0 , y0 ) the slope of the surface in the y-direction at (x0 , y0 ).

Example 5

Let f(x, y) = x 2 y + 5y 3 .

(a) Find the slope of the surface z = f(x, y) in the x-direction at the point (1, −2). (b) Find the slope of the surface z = f(x, y) in the y-direction at the point (1, −2).

Solution (a). Differentiating f with respect to x with y held ﬁxed yields fx(x, y) = 2xy Thus, the slope in the x-direction is fx(1, −2) = −4; that is, z is decreasing at the rate of 4 units per unit increase in x.

Solution (b). Differentiating f with respect to y with x held ﬁxed yields fy(x, y) = x 2 + 15y 2 Thus, the slope in the y-direction is fy(1, −2) = 61; that is, z is increasing at the rate of 61 units per unit increase in y. ESTIMATING PARTIAL DERIVATIVES FROM TABULAR DATA

For functions that are presented in tabular form, we can estimate partial derivatives by using adjacent entries within the table. Example 6 Use the values of the wind chill index function W (T , v) displayed in Table 13.3.1 to estimate the partial derivative of W with respect to v at (T , v) = (25, 10). Compare this estimate with the value of the partial derivative obtained in Example 4.

wind speed v (mi/h)

Table 13.3.1 temperature T (°F ) 20

Solution. Since ∂W W (25, 10 + v) − W (25, 10) W (25, 10 + v) − 15 (25, 10) = lim = lim v → 0 v → 0 ∂v v v we can approximate the partial derivative by ∂W W (25, 10 + v) − 15 (25, 10) ≈ ∂v v With v = 5 this approximation is

∂W W (25, 10 + 5) − 15 W (25, 15) − 15 13 − 15 2 ◦F (25, 10) ≈ = = =− ∂v 5 5 5 5 mi/h

13.3 Partial Derivatives

931

and with v = −5 this approximation is

∂W 4 ◦F W (25, 10 − 5) − 15 W (25, 5) − 15 19 − 15 (25, 10) ≈ = = =− ∂v −5 −5 −5 5 mi/h

We will take the average, − 35 = −0.6 ◦ F/(mi/h), of these two approximations as our estimate of (∂W /∂v)(25, 10). This is close to the value ◦ ∂W F (25, 10) = (−4.01)10−0.84 ≈ −0.58 ∂v mi/h

found in Example 4.

IMPLICIT PARTIAL DIFFERENTIATION

冸 23 , 13 , 23 冹

2 2 2 Example slope 2 1 27 Find 2 the of the sphere x + y + z = 1 in the y-direction at the 1 2 points 3 , 3 , 3 and 3 , 3 , − 3 (Figure 13.3.2).

, 13 , 23 lies on the upper hemisphere z = 1 − x 2 − y 2 , and the point 23 , 31 , − 23 lies on the lower hemisphere z = − 1 − x 2 − y 2 . We could ﬁnd the slopes by differentiating each expression for z separately with respect to y and then evaluating the derivatives at x = 23 and y = 13 . However, it is more efﬁcient to differentiate the given equation x 2 + y 2 + z2 = 1

Solution. The point y

冸 23 , 13 , − 23 冹

Figure 13.3.2

Check the results in Example 7 by differentiating the functions

z= and

directly.

1 − x2 − y2

z = − 1 − x2 − y2

implicitly with respect to y, since this will give us both slopes with one differentiation. To perform the implicit differentiation, we view z as a function of x and y and differentiate both sides with respect to y, taking x to be fixed. The computations are as follows: ∂ 2 ∂ [x + y 2 + z2 ] = [1] ∂y ∂y ∂z =0 0 + 2y + 2z ∂y ∂z y =− ∂y z Substituting the y- and z-coordinates of the points 23 , 31 , 23 and 23 , 31 , − 23 in this expres 2 1 2 sion, we find that the slope at the point 3 , 3 , 3 is − 21 and the slope at 23 , 31 , − 23 is 21 . Example 8 Suppose that D = x 2 + y 2 is the length of the diagonal of a rectangle whose sides have lengths x and y that are allowed to vary. Find a formula for the rate of change of D with respect to x if x varies with y held constant, and use this formula to find the rate of change of D with respect to x at the point where x = 3 and y = 4.

Solution. Differentiating both sides of the equation D 2 = x 2 + y 2 with respect to x yields

∂D ∂D = 2x and thus D =x ∂x ∂x Since D = 5 when x = 3 and y = 4, it follows that ∂D 3 ∂D = 3 or = 5 ∂x x=3,y=4 ∂x x=3,y=4 5 2D

Thus, D is increasing at a rate of

3 5

unit per unit increase in x at the point (3, 4).

932

Chapter 13 / Partial Derivatives

PARTIAL DERIVATIVES AND CONTINUITY

In contrast to the case of functions of a single variable, the existence of partial derivatives for a multivariable function does not guarantee the continuity of the function. This fact is shown in the following example. Example 9

Let f(x, y) =

⎧ ⎨−

⎩

xy , (x, y) = (0, 0) + y2 0, (x, y) = (0, 0)

(3)

(a) Show that fx(x, y) and fy(x, y) exist at all points (x, y). (b) Explain why f is not continuous at (0, 0). z

Solution (a). Figure 13.3.3 shows the graph of f . Note that f is similar to the function considered in Example 1 of Section 13.2, except that here we have assigned f a value of 0 at (0, 0). Except at this point, the partial derivatives of f are x2y − y3 (x 2 + y 2 )y − xy(2x) = 2 2 2 2 (x + y ) (x + y 2 )2 (x 2 + y 2 )x − xy(2y) xy 2 − x 3 fy(x, y) = − = 2 2 2 2 (x + y ) (x + y 2 )2 fx(x, y) = −

Figure 13.3.3

(4) (5)

It is not evident from Formula (3) whether f has partial derivatives at (0, 0), and if so, what the values of those derivatives are. To answer that question we will have to use the deﬁnitions of the partial derivatives (Deﬁnition 13.3.1). Applying Formulas (1) and (2) to (3) we obtain 0−0 f( x, 0) − f(0, 0) = lim =0 x → 0 x x → 0 x 0−0 f (0, y) − f(0, 0) = lim =0 fy(0, 0) = lim y → 0 y y → 0 y

fx(0, 0) = lim

This shows that f has partial derivatives at (0, 0) and the values of both partial derivatives are 0 at that point.

Solution (b). We saw in Example 3 of Section 13.2 that lim

(x,y) → (0,0)

−

xy + y2

does not exist. Thus, f is not continuous at (0, 0).

We will study the relationship between the continuity of a function and the properties of its partial derivatives in the next section. PARTIAL DERIVATIVES OF FUNCTIONS WITH MORE THAN TWO VARIABLES

For a function f(x, y, z) of three variables, there are three partial derivatives: fx(x, y, z),

fy(x, y, z),

fz(x, y, z)

The partial derivative fx is calculated by holding y and z constant and differentiating with respect to x. For fy the variables x and z are held constant, and for fz the variables x and y are held constant. If a dependent variable w = f(x, y, z)

13.3 Partial Derivatives

933

is used, then the three partial derivatives of f can be denoted by ∂w , ∂x

∂w , ∂y

and

∂w ∂z

Example 10 If f(x, y, z) = x 3 y 2 z4 + 2xy + z, then fx(x, y, z) = 3x 2 y 2 z4 + 2y fy(x, y, z) = 2x 3 yz4 + 2x fz(x, y, z) = 4x 3 y 2 z3 + 1

fz(−1, 1, 2) = 4(−1)3 (1)2 (2)3 + 1 = −31

Example 11 If f(ρ, θ, φ) = ρ 2 cos φ sin θ, then fρ(ρ, θ, φ) = 2ρ cos φ sin θ

fθ (ρ, θ, φ) = ρ 2 cos φ cos θ

fφ (ρ, θ, φ) = −ρ 2 sin φ sin θ In general, if f(v1 , v2 , . . . , vn ) is a function of n variables, there are n partial derivatives of f , each of which is obtained by holding n − 1 of the variables ﬁxed and differentiating the function f with respect to the remaining variable. If w = f(v1 , v2 , . . . , vn ), then these partial derivatives are denoted by ∂w ∂w ∂w , ,..., ∂v1 ∂v2 ∂vn where ∂w/∂vi is obtained by holding all variables except vi ﬁxed and differentiating with respect to vi . HIGHER-ORDER PARTIAL DERIVATIVES

Suppose that f is a function of two variables x and y. Since the partial derivatives ∂f /∂x and ∂f /∂y are also functions of x and y, these functions may themselves have partial derivatives. This gives rise to four possible second-order partial derivatives of f , which are deﬁned by ∂ 2f ∂ ∂f ∂ ∂f ∂ 2f = = = fxx = fyy ∂x 2 ∂x ∂x ∂y 2 ∂y ∂y Differentiate twice with respect to x.

∂ ∂ 2f = ∂y∂x ∂y

Differentiate twice with respect to y.

∂f ∂x

Differentiate ﬁrst with respect to x and then with respect to y.

= fxy

∂ ∂ 2f = ∂x∂y ∂x

∂f ∂y

= fyx

Differentiate ﬁrst with respect to y and then with respect to x.

The last two cases are called the mixed second-order partial derivatives or the mixed second partials. Also, the derivatives ∂f /∂x and ∂f /∂y are often called the ﬁrst-order partial derivatives when it is necessary to distinguish them from higher-order partial derivatives. Similar conventions apply to the second-order partial derivatives of a function of three variables.

934

Chapter 13 / Partial Derivatives WARNING

Observe that the two notations for the mixed second partials have opposite conventions for the order of differentiation. In the “∂ ” notation the derivatives are taken right to left, and in the “subscript” notation they are taken left to right. The conventions are logical if you insert parentheses:

∂ ∂f ∂ 2f = ∂y∂x ∂y ∂x

Right to left. Differentiate inside the parentheses ﬁrst.

f xy = (fx )y

Left to right. Differentiate inside the parentheses ﬁrst.

Example 12 Find the second-order partial derivatives of f(x, y) = x 2 y 3 + x 4 y.

Solution. We have ∂f = 2xy 3 + 4x 3 y ∂x

and

∂f = 3x 2 y 2 + x 4 ∂y

so that ∂ ∂ 2f = 2 ∂x ∂x

∂ 2f ∂ = 2 ∂y ∂y

∂ 2f ∂ = ∂x∂y ∂x ∂ ∂ 2f = ∂y∂x ∂y

∂f ∂x

∂ (2xy 3 + 4x 3 y) = 2y 3 + 12x 2 y ∂x

∂ ∂f (3x 2 y 2 + x 4 ) = 6x 2 y = ∂y ∂y ∂f ∂ (3x 2 y 2 + x 4 ) = 6xy 2 + 4x 3 = ∂y ∂x ∂f ∂ (2xy 3 + 4x 3 y) = 6xy 2 + 4x 3 = ∂x ∂y

Third-order, fourth-order, and higher-order partial derivatives can be obtained by successive differentiation. Some possibilities are ∂ ∂ 2f ∂ ∂ 3f ∂ 4f ∂ 3f = = = f = fyyyy xxx ∂x 3 ∂x ∂x 2 ∂y 4 ∂y ∂y 3 2 3 ∂ ∂ ∂ 4f ∂f ∂f ∂ 3f = = = fxyy = fxxyy 2 2 2 ∂y ∂x ∂y ∂y∂x ∂y ∂x ∂y ∂y∂x 2 Example 13 Let f(x, y) = y 2 ex + y. Find fxyy .

Solution. fxyy =

∂2 ∂ 3f = ∂y 2 ∂x ∂y 2

∂f ∂x

∂ ∂2 2 x (2yex ) = 2ex (y e ) = ∂y 2 ∂y

EQUALITY OF MIXED PARTIALS

If f is a function of three variables, then the analog of Theorem 13.3.2 holds for each pair of mixed second-order partials if we replace “open disk” by “open ball.” How many second-order partials does f(x, y, z) have?

For a function f(x, y) it might be expected that there would be four distinct second-order partial derivatives: fxx , fxy , fyx , and fyy . However, observe that the mixed second-order partial derivatives in Example 12 are equal. The following theorem (proved in Appendix D) explains why this is so.

13.3.2 theorem Let f be a function of two variables. If fxy and fyx are continuous on some open disk, then fxy = fyx on that disk.

13.3 Partial Derivatives

935

It follows from this theorem that if fxy (x, y) and fyx (x, y) are continuous everywhere, then fxy (x, y) = fyx (x, y) for all values of x and y. Since polynomials are continuous everywhere, this explains why the mixed second-order partials in Example 12 are equal. THE WAVE EQUATION

Consider a string of length L that is stretched taut between x = 0 and x = L on an x-axis, and suppose that the string is set into vibratory motion by “plucking” it at time t = 0 (Figure 13.3.4a). The displacement of a point on the string depends both on its coordinate x and the elapsed time t, and hence is described by a function u(x, t) of two variables. For a ﬁxed value t, the function u(x, t) depends on x alone, and the graph of u versus x describes the shape of the string—think of it as a “snapshot” of the string at time t (Figure 13.3.4b). It follows that at a ﬁxed time t, the partial derivative ∂u/∂x represents the slope of the string at x, and the sign of the second partial derivative ∂ 2 u/∂x 2 tells us whether the string is concave up or concave down at x (Figure 13.3.4c). u

u Slope =

⭸2u < 0 (concave down) ⭸x 2

⭸u ⭸x

(a)

(b)

(c)

Figure 13.3.4

The vibration of a plucked string is governed by the wave equation.

For a fixed value of x, the function u(x, t) depends on t alone, and the graph of u versus t is the position versus time curve of the point on the string with coordinate x. Thus, for a fixed value of x, the partial derivative ∂u/∂t is the velocity of the point with coordinate x, and ∂ 2 u/∂t 2 is the acceleration of that point. It can be proved that under appropriate conditions the function u(x, t) satisfies an equation of the form ∂ 2u ∂ 2u = c2 2 (6) 2 ∂t ∂x where c is a positive constant that depends on the physical characteristics of the string. This equation, which is called the one-dimensional wave equation, involves partial derivatives of the unknown function u(x, t) and hence is classified as a partial differential equation. Techniques for solving partial differential equations are studied in advanced courses and will not be discussed in this text. Example 14 tion (6).

Show that the function u(x, t) = sin(x − ct) is a solution of Equa-

Solution. We have ∂u = cos(x − ct), ∂x ∂u = −c cos(x − ct), ∂t Thus, u(x, t) satisﬁes (6).

∂ 2u = − sin(x − ct) ∂x 2 ∂ 2u = −c2 sin(x − ct) ∂t 2

936

Chapter 13 / Partial Derivatives

✔QUICK CHECK EXERCISES 13.3 1. Let f(x, y) = x sin xy. fy (x, y) = .

(See page 940 for answers.)

Then fx (x, y) =

and

2. The slope of the surface z = xy 2 in the x-direction at the point (2, 3) is , and the slope of this surface in the y-direction at the point (2, 3) is . 3. The volume V of a right circular cone of radius r and height h is given by V = 13 πr 2 h.

EXERCISE SET 13.3

3–10 Evaluate the indicated partial derivatives. ■

4. 5. 6. 7. 8. 9. 10. 11.

12.

13.

4. Find all second-order partial derivatives for the function f(x, y) = x 2 y 3 .

Graphing Utility

1. Let f(x, y) = 3x 3 y 2 . Find (a) fx (x, y) (b) fy (x, y) (d) fx (x, 1) (e) fy (1, y) (g) fx (1, 2) (h) fy (1, 2). 2. Let z = e2x sin y. Find (a) ∂z/∂x (b) ∂z/∂y / (d) ∂z ∂x|(x,0) (e) ∂z/∂y|(0,y) (g) ∂z/∂x|(ln 2,0) (h) ∂z/∂y|(ln 2,0) .

∂z ∂z , ∂x ∂y f(x, y) = 10x 2 y 4 − 6xy 2 + 10x 2 ; fx (x, y), fy (x, y) ∂z ∂z z = (x 2 + 5x − 2y)8 ; , ∂x ∂y 1 f(x, y) = 2 ; fx (x, y), fy (x, y) xy − x 2 y ∂ −7p/q ∂ −7p/q (e ), (e ) ∂p ∂q √ √ ∂ ∂ (xe 15xy ), (xe 15xy ) ∂x ∂y ∂z ∂z z = sin(5x 3 y + 7xy 2 ); , ∂x ∂y f(x, y) = cos(2xy 2 − 3x 2 y 2 ); fx (x, y), fy (x, y) Let f(x, y) = 3x + 2y. (a) Find the slope of the surface z = f(x, y) in the xdirection at the point (4, 2). (b) Find the slope of the surface z = f(x, y) in the ydirection at the point (4, 2). Let f(x, y) = xe−y + 5y. (a) Find the slope of the surface z = f(x, y) in the xdirection at the point (3, 0). (b) Find the slope of the surface z = f(x, y) in the ydirection at the point (3, 0). Let z = sin(y 2 − 4x). (a) Find the rate of change of z with respect to x at the point (2, 1) with y held ﬁxed. (b) Find the rate of change of z with respect to y at the point (2, 1) with x held ﬁxed.

3. z = 9x 2 y − 3x 5 y;

(a) Find a formula for the instantaneous rate of change of V with respect to r if r changes and h remains constant. (b) Find a formula for the instantaneous rate of change of V with respect to h if h changes and r remains constant.

14. Let z = (x + y)−1 . (a) Find the rate of change of z with respect to x at the point (−2, 4) with y held ﬁxed. (b) Find the rate of change of z with respect to y at the point (−2, 4) with x held ﬁxed. F O C U S O N C O N C E P TS

15. Use the information in the accompanying figure to find the values of the first-order partial derivatives of f at the point (1, 2). z

(1, 0, 3)

z = f (x, y) (1, 2, 4) y

(1, 2, 0) (2, 2, 0)

Figure Ex-15

16. The accompanying ﬁgure shows a contour plot for an unspeciﬁed function f(x, y). Make a conjecture about the signs of the partial derivatives fx (x0 , y0 ) and fy (x0 , y0 ), and explain your reasoning. y 8 7 6 5 4 3 2 1

Figure Ex-16

17. Suppose that Nolan throws a baseball to Ryan and that the baseball leaves Nolan’s hand at the same height at which it is caught by Ryan. It we ignore air resistance, the horizontal range r of the baseball is a function of the initial speed v of the ball when it leaves Nolan’s hand and the angle θ above the horizontal at which it is thrown. Use the accompanying table and the method of Example 6 to estimate (a) the partial derivative of r with respect to v when v = 80 ft/s and θ = 40 ◦

13.3 Partial Derivatives

(b) the partial derivative of r with respect to θ when v = 80 ft/s and θ = 40 ◦ .

angle u (degrees)

speed v (ft/s) 80

165

188

212

238

173

197

222

249

176

200

226

253

173

197

222

249

2 3

25. z = 4ex y 27. z = x 3 ln(1 + xy −3/5 ) xy 29. z = 2 x + y2

19. The accompanying ﬁgure shows the graphs of an unspeciﬁed function f(x, y) and its partial derivatives fx (x, y) and fy (x, y). Determine which is which, and explain your reasoning. 0

−2 2

0 −5

−10 2

0 −10 0 −2

I Figure Ex-19

26. z = cos(x 5 y 4 ) 28. z = exy sin 4y 2 x2y3 30. z = √ x+y

31–36 Find fx (x, y) and fy (x, y). ■ Table Ex-17

18. Use the table in Exercise 17 and the method of Example 6 to estimate (a) the partial derivative of r with respect to v when v = 85 ft/s and θ = 45 ◦ (b) the partial derivative of r with respect to θ when v = 85 ft/s and θ = 45 ◦ .

−2

24. There exists a polynomial f(x, y) that satisﬁes the equations fx(x, y) = 3x 2 + y 2 + 2y and fy(x, y) = 2xy + 2y. 25–30 Find ∂z/∂x and ∂z/∂y. ■

937

0 −2

2 0

−2

III

20. What can you say about the signs of ∂z/∂x, ∂ 2 z/∂x 2 , ∂z/∂y, and ∂ 2 z/∂y 2 at the point P in the accompanying ﬁgure? Explain your reasoning. z

z = f (x, y)

31. f(x, y) =

3x 5 y − 7x 3 y

32. f(x, y) =

33. f(x, y) = y −3/2 tan−1 (x /y) √ 34. f(x, y) = x 3 e−y + y 3 sec x

x+y x−y

35. f(x, y) = (y 2 tan x)−4/3 √ 36. f(x, y) = cosh( x) sinh2 (xy 2 )

37–40 Evaluate the indicated partial derivatives. ■

f(x, y) = 9 − x 2 − 7y 3 ; fx (3, 1), fy (3, 1) f(x, y) = x 2 yexy ; ∂f /∂x(1, 1), ∂f /∂y(1, 1) z = x 2 + 4y 2 ; ∂z/∂x(1, 2), ∂z/∂y(1, 2) w = x 2 cos xy; ∂w /∂x 21 , π , ∂w /∂y 21 , π Let f(x, y, z) = x 2 y 4 z3 + xy + z2 + 1. Find (a) fx (x, y, z) (b) fy (x, y, z) (c) fz (x, y, z) (d) fx (1, y, z) (e) fy (1, 2, z) (f ) fz (1, 2, 3). 42. Let w = x 2 y cos z. Find (a) ∂w /∂x(x, y, z) (b) ∂w/∂y(x, y, z) (c) ∂w /∂z(x, y, z) (d) ∂w/∂x(2, y, z) (e) ∂w /∂y(2, 1, z) (f ) ∂w/∂z(2, 1, 0).

37. 38. 39. 40. 41.

43–46 Find fx , fy , and fz . ■

43. f(x, y, z) = z ln(x 2 y cos z) xz −3/2 44. f(x, y, z) = y sec y 1 45. f(x, y, z) = tan−1 xy 2 z3 √ 46. f(x, y, z) = cosh( z ) sinh2 (x 2 yz) 47–50 Find ∂w /∂x, ∂w /∂y, and ∂w /∂z. ■

Figure Ex-20

21–24 True–False Determine whether the statement is true or false. Explain your answer. ■

21. If the line y = 2 is a contour of f(x, y) through (4, 2), then fx (4, 2) = 0. 22. If the plane x = 3 intersects the surface z = f(x, y) in a curve that passes through (3, 4, 16) and satisﬁes z = y 2 , then fy (3, 4) = 8. 23. If the graph of z = f(x, y) is a plane in 3-space, then both fx and fy are constant functions.

x2 − y2 47. w = yez sin xz 48. w = 2 y + z2 50. w = y 3 e2x+3z 49. w = x 2 + y 2 + z2 2 xz 51. Let f(x, y, z) = y e . Find (a) ∂f /∂x|(1,1,1) (b) ∂f /∂y|(1,1,1) (c) ∂f /∂z|(1,1,1) . 52. Let w = x 2 + 4y 2 − z2 . Find (a) ∂w/∂x|(2,1,−1) (b) ∂w/∂y|(2,1,−1) (c) ∂w/∂z|(2,1,−1) . 53. Let f(x, y) = ex cos y. Use a graphing utility to graph the functions fx(0, y) and fy(x, π/2). 54. Let f(x, y) = ex sin y. Use a graphing utility to graph the functions fx (0, y) and fy (x, 0).

938

Chapter 13 / Partial Derivatives

55. A point moves along the intersection of the elliptic paraboloid z = x 2 + 3y 2 and the plane y = 1. At what rate is z changing with respect to x when the point is at (2, 1, 7)? 56. A point moves along the intersection of the elliptic paraboloid z = x 2 + 3y 2 and the plane x = 2. At what rate is z changing with respect to y when the point is at (2, 1, 7)? 57. A point moves along the intersection of the plane y = 3 and the surface z = 29 − x 2 − y 2 . At what rate is z changing with respect to x when the point is at (4, 3, 2)? 58. Find the slope of the tangent line at (−1, 1, 5) to the curve of intersection of the surface z = x 2 + 4y 2 and (a) the plane x = −1 (b) the plane y = 1. 59. The volume V of a right circular cylinder is given by the formula V = πr 2 h, where r is the radius and h is the height. (a) Find a formula for the instantaneous rate of change of V with respect to r if r changes and h remains constant. (b) Find a formula for the instantaneous rate of change of V with respect to h if h changes and r remains constant. (c) Suppose that h has a constant value of 4 in, but r varies. Find the rate of change of V with respect to r at the point where r = 6 in. (d) Suppose that r has a constant value of 8 in, but h varies. Find the instantaneous rate of change of V with respect to h at the point where h = 10 in. 60. The volume V of a right circular cone is given by π 2 2 V = d 4s − d 2 24 where s is the slant height and d is the diameter of the base. (a) Find a formula for the instantaneous rate of change of V with respect to s if d remains constant. (b) Find a formula for the instantaneous rate of change of V with respect to d if s remains constant. (c) Suppose that d has a constant value of 16 cm, but s varies. Find the rate of change of V with respect to s when s = 10 cm. (d) Suppose that s has a constant value of 10 cm, but d varies. Find the rate of change of V with respect to d when d = 16 cm.

61. According to the ideal gas law, the pressure, temperature, and volume of a gas are related by P = kT /V , where k is a constant of proportionality. Suppose that V is measured in cubic inches (in3 ), T is measured in kelvins (K), and that for a certain gas the constant of proportionality is k = 10 in·lb/K. (a) Find the instantaneous rate of change of pressure with respect to temperature if the temperature is 80 K and the volume remains fixed at 50 in3 . (b) Find the instantaneous rate of change of volume with respect to pressure if the volume is 50 in3 and the temperature remains fixed at 80 K. 62. The temperature at a point (x, y) on a metal plate in the xy-plane is T (x, y) = x 3 + 2y 2 + x degrees Celsius. Assume that distance is measured in centimeters and find the

rate at which temperature changes with respect to distance if we start at the point (1, 2) and move (a) to the right and parallel to the x-axis (b) upward and parallel to the y-axis. 63. The length, width, and height of a rectangular box are l = 5, w = 2, and h = 3, respectively. (a) Find the instantaneous rate of change of the volume of the box with respect to the length if w and h are held constant. (b) Find the instantaneous rate of change of the volume of the box with respect to the width if l and h are held constant. (c) Find the instantaneous rate of change of the volume of the box with respect to the height if l and w are held constant. 64. The area A of a triangle is given by A = 21 ab sin θ , where a and b are the lengths of two sides and θ is the angle between these sides. Suppose that a = 5, b = 10, and θ = π/3. (a) Find the rate at which A changes with respect to a if b and θ are held constant. (b) Find the rate at which A changes with respect to θ if a and b are held constant. (c) Find the rate at which b changes with respect to a if A and θ are held constant. 65. The volume of a right circular cone of radius r and height h is V = 31 πr 2 h. Show that if the height remains constant while the radius changes, then the volume satisﬁes ∂V 2V = ∂r r 66. Find parametric equations for the tangent line at (1, 3, 3) to the curve of intersection of the surface z = x 2 y and (a) the plane x = 1 (b) the plane y = 3.

67. (a) By differentiating implicitly, ﬁnd the slope of the hy2 2 perboloid x 2 + √y − z = 1 in the √ x-direction at the points (3, 4, 2 6) and (3, 4, −2 6). (b) Check the results in part (a) by solving for z and differentiating the resulting functions directly.

68. (a) By differentiating implicitly, ﬁnd the slope of the hy2 2 perboloid x 2 + √y − z = 1 in the √ y-direction at the points (3, 4, 2 6 ) and (3, 4, −2 6 ). (b) Check the results in part (a) by solving for z and differentiating the resulting functions directly. 69–72 Calculate ∂z/∂x and ∂z/∂y using implicit differentiation. Leave your answers in terms of x, y, and z. ■

69. (x 2 + y 2 + z2 )3/2 = 1 71. x 2 + z sin xyz = 0

70. ln(2x 2 + y − z3 ) = x

72. exy sinh z − z2 x + 1 = 0

73–76 Find ∂w /∂x, ∂w /∂y, and ∂w /∂z using implicit differentiation. Leave your answers in terms of x, y, z, and w. ■

73. (x 2 + y 2 + z2 + w 2 )3/2 = 4

74. ln(2x 2 + y − z3 + 3w) = z

13.3 Partial Derivatives

75. w 2 + w sin xyz = 1

76. exy sinh w − z2 w + 1 = 0

77–80 Find fx and fy . ■

77. f(x, y) =

79. f(x, y) =

78. f(x, y) =

sin t 3 dt 80. f(x, y) =

et dt

y x2 y3

√ 81. Let z = x cos y. Find (a) ∂ 2 z/∂x 2 (c) ∂ 2 z/∂x∂y

et dt

1 x−y

sin t 3 dt

x+y

(b) ∂ 2 z/∂y 2 (d) ∂ 2 z/∂y∂x.

4 5

82. Let f(x, y) = 4x − 2y + 7x y . Find (a) fxx (b) fyy (c) fxy

(d) fyx .

83. Let f(x, y) = sin(3x 2 + 6y 2 ). Find (a) fxx (b) fyy (c) fxy

(d) fyx .

84. Let f(x, y) = xe . Find (a) fxx (b) fyy

(d) fyx .

85–92 Conﬁrm that the mixed second-order partial derivatives of f are the same. ■

85. f(x, y) = 4x 2 − 8xy 4 + 7y 5 − 3 87. f(x, y) = ex cos y 86. f(x, y) = x 2 + y 2 88. f(x, y) = ex−y

89. f(x, y) = ln(4x − 5y)

90. f(x, y) = ln(x + y ) 91. f(x, y) = (x − y)/(x + y)

92. f(x, y) = (x 2 − y 2 )/(x 2 + y 2 )

93. Express the following derivatives in “∂” notation. (a) fxxx (b) fxyy (c) fyyxx (d) fxyyy 94. Express the derivatives in “subscript” notation. ∂ 4f ∂ 5f ∂ 4f ∂ 3f (c) (d) (b) (a) ∂y 2 ∂x ∂x 4 ∂y 2 ∂x 2 ∂x 2 ∂y 3 95. Given f(x, y) = x 3 y 5 − 2x 2 y + x, find (a) fxxy (b) fyxy (c) fyyy . 96. Given z = (2x − y)5 , find ∂ 3z ∂ 3z (b) (a) ∂y∂x∂y ∂x 2 ∂y 97. Given f(x, y) = y 3 e−5x , find (a) fxyy (0, 1) (b) fxxx (0, 1) 98. Given w = ey cos x, find ∂ 3 w (a) ∂y 2 ∂x (π/4,0)

(c)

∂ 4z . ∂x 2 ∂y 2

∂ 3 w (b) ∂x 2 ∂y (π/4,0)

99. Let f(x, y, z) = x 3 y 5 z7 + xy 2 + y 3 z. Find (a) fxy (b) fyz (c) fxz (d) fzz (e) fzyy (f ) fxxy (g) fzyx (h) fxxyz . 100. Let w = (4x − 3y + 2z)5 . Find ∂ 3w ∂ 2w (b) (a) ∂x∂z ∂x∂y∂z

(c)

∂ 4w . ∂z2 ∂y∂x

939

101. Show that the function satisﬁes Laplace’s equation ∂ 2z ∂ 2z + =0 ∂x 2 ∂y 2 (a) z = x 2 − y 2 + 2xy (b) z = ex sin y + ey cos x (c) z = ln(x 2 + y 2 ) + 2 tan−1 (y /x) 102. Show that the function satisﬁes the heat equation ∂z ∂ 2z = c2 2 (c > 0, constant) ∂t ∂x (a) z = e−t sin(x /c) (b) z = e−t cos(x /c)

103. Show that the function u(x, t) = sin cωt sin ωx satisﬁes the wave equation [Equation (6)] for all real values of ω. 104. In each part, show that u(x, y) and v(x, y) satisfy the Cauchy–Riemann equations ∂u ∂v ∂u ∂v = and =− ∂x ∂y ∂y ∂x (a) u = x 2 − y 2 , v = 2xy (b) u = ex cos y, v = ex sin y 2 2 (c) u = ln(x + y ), v = 2 tan−1 (y /x)

105. Show that if u(x, y) and v(x, y) each have equal mixed second partials, and if u and v satisfy the Cauchy–Riemann equations (Exercise 104), then u, v, and u + v satisfy Laplace’s equation (Exercise 101). 106. When two resistors having resistances R1 ohms and R2 ohms are connected in parallel, their combined resistance R in ohms is R = R1 R2 /(R1 + R2 ). Show that ∂ 2R ∂ 2R 4R 2 = 2 2 (R1 + R2 )4 ∂R1 ∂R2 107–110 Find the indicated partial derivatives. ■

107. f(v, w, x, y) = 4v 2 w 3 x 4 y 5 ; ∂f /∂v, ∂f /∂w, ∂f /∂x, ∂f /∂y 108. w = r cos st + eu sin ur; ∂w /∂r, ∂w /∂s, ∂w /∂t, ∂w/∂u v12 − v22 ; v32 + v42 ∂f /∂v1 , ∂f /∂v2 , ∂f /∂v3 , ∂f /∂v4

109. f(v1 , v2 , v3 , v4 ) =

110. V = xe2x−y + wezw + yw; ∂V /∂x, ∂V /∂y, ∂V /∂z, ∂V /∂w 111. Let u(w, x, y, z) = xeyw sin2 z. Find ∂u ∂u (a) (0, 0, 1, π) (b) (0, 0, 1, π) ∂x ∂y ∂u ∂u (c) (0, 0, 1, π) (d) (0, 0, 1, π) ∂w ∂z 4 ∂ 4u ∂ u . (f ) (e) ∂x∂y∂w∂z ∂w∂z∂y 2 112. Let f(v, w, x, y) = 2v 1/2 w 4 x 1/2 y 2/3 . Find fv (1, −2, 4, 8), fw (1, −2, 4, 8), fx (1, −2, 4, 8), and fy (1, −2, 4, 8).

940

Chapter 13 / Partial Derivatives

113–114 Find ∂w /∂xi for i = 1, 2, . . . , n. ■

113. w = cos(x1 + 2x2 + · · · + nxn ) 1/n n xk 114. w = k=1

115–116 Describe the largest set on which Theorem 13.3.2 can

be used to prove that fxy and fyx are equal on that set. Then conﬁrm by direct computation that fxy = fyx on the given set. 115. (a) f(x, y) = 4x 3 y + 3x 2 y 116. (a) f(x, y) = x 2 + y 2 − 1 (b) f(x, y) = sin(x 2 + y 3 )

(b) f(x, y) = x 3 /y

■

117. Let f(x, y) = 2x 2 − 3xy + y 2 . Find fx (2, −1) and fy (2, −1) by evaluating the limits in Deﬁnition 13.3.1. Then check your work by calculating the derivative in the usual way.

118. Let f(x, y) = (x 2 + y 2 )2/3 . Show that ⎧ 4x ⎨ , (x, y) = (0, 0) fx (x, y) = 3(x 2 + y 2 )1/3 ⎩ 0, (x, y) = (0, 0)

Source: This problem, due to Don Cohen, appeared in Mathematics and Computer Education, Vol. 25, No. 2, 1991, p. 179.

119. Let f(x, y) = (x 3 + y 3 )1/3 . (a) Show that fy (0, 0) = 1. (b) At what points, if any, does fy (x, y) fail to exist? 120. Writing Explain how one might use the graph of the equation z = f(x, y) to determine the signs of fx (x0 , y0 ) and fy (x0 , y0 ) by inspection. 121. Writing Explain how one might use the graphs of some appropriate contours of z = f(x, y) to determine the signs of fx (x0 , y0 ) and fy (x0 , y0 ) by inspection.

✔QUICK CHECK ANSWERS 13.3 1. sin xy + xy cos xy; x 2 cos xy 2. 9; 12 3. (a) 23 πrh (b) 13 πr 2 s 4. fxx (x, y) = 2y 3 , fyy (x, y) = 6x 2 y, fxy (x, y) = fyx (x, y) = 6xy 2

DIFFERENTIABILITY, DIFFERENTIALS, AND LOCAL LINEARITY

13.4

In this section we will extend the notion of differentiability to functions of two or three variables. Our deﬁnition of differentiability will be based on the idea that a function is differentiable at a point provided it can be very closely approximated by a linear function near that point. In the process, we will expand the concept of a “differential” to functions of more than one variable and deﬁne the “local linear approximation” of a function. DIFFERENTIABILITY

Recall that a function f of one variable is called differentiable at x0 if it has a derivative at x0 , that is, if the limit f(x0 + x) − f(x0 ) f ′(x0 ) = lim (1) x → 0 x exists. As a consequence of (1) a differentiable function enjoys a number of other important properties:

• The graph of y = f(x) has a nonvertical tangent line at the point (x0 , f(x0 )); • f may be closely approximated by a linear function near x0 (Section 3.5); • f is continuous at x0 .

(x0 , y0 , f(x0 , y0))

Our primary objective in this section is to extend the notion of differentiability to functions of two or three variables in such a way that the natural analogs of these properties hold. For example, if a function f(x, y) of two variables is differentiable at a point (x0 , y0 ), we want it to be the case that z = f (x, y) y

• the surface z = f(x, y) has a nonvertical tangent plane at the point (x0 , y0 , f(x0 , y0 )) (Figure 13.4.1);

(x0 , y0 ) x

Figure 13.4.1

• the values of f at points near (x0 , y0 ) can be very closely approximated by the values of a linear function;

• f is continuous at (x0 , y0 ).

13.4 Differentiability, Differentials, and Local Linearity

941

One could reasonably conjecture that a function f of two or three variables should be called differentiable at a point if all the first-order partial derivatives of the function exist at that point. Unfortunately, this condition is not strong enough to guarantee that the properties above hold. For instance, we saw in Example 9 of Section 13.3 that the mere existence of both first-order partial derivatives for a function is not sufficient to guarantee the continuity of the function. To determine what else we should include in our definition, it will be helpful to reexamine one of the consequences of differentiability for a single-variable function f (x). Suppose that f (x) is differentiable at x = x0 and let f = f (x0 + x) − f (x0 ) denote the change in f that corresponds to the change x in x from x0 to x0 + x. We saw in Section 3.5 that f ≈ f ′ (x0 ) x provided x is close to 0. In fact, for x close to 0 the error f − f ′ (x0 ) x in this approximation will have magnitude much smaller than that of x because f − f ′ (x0 ) x f (x0 + x) − f (x0 ) ′ lim = lim − f (x0 ) = f ′ (x0 ) − f ′ (x0 ) = 0 x → 0 x → 0 x x Since the magnitude of x is just the distance between the points x0 and x0 + x, we see that when the two points are close together, the magnitude of the error in the approximation will be much smaller than the distance between the two points (Figure 13.4.2). The extension of this idea to functions of two or three variables is the “extra ingredient” needed in our definition of differentiability for multivariable functions. For a function f (x, y), the symbol f , called the increment of f , denotes the change in the value of f (x, y) that results when (x, y) varies from some initial position (x0 , y0 ) to some new position (x0 + x, y0 + y); thus (2)

f = f (x0 + x, y0 + y) − f (x0 , y0 )

(see Figure 13.4.3). [If a dependent variable z = f (x, y) is used, then we will sometimes write z rather than f .] Let us assume that both fx (x0 , y0 ) and fy (x0 , y0 ) exist and (by analogy with the one-variable case) make the approximation Show that if f (x, y) is a linear function, then (3) becomes an equality.

(3)

f ≈ fx (x0 , y0 ) x + fy (x0 , y0 ) y For x and y close to 0, we would like the error

f − fx (x0 , y0 ) x − fy (x0 , y0 ) y in this approximation to be much smaller than the distance ( x)2 + ( y)2 between (x0 , y0 ) and (x0 + x, y0 + y). We can guarantee this by requiring that lim

( x, y) → (0,0)

f − fx (x0 , y0 ) x − fy (x0 , y0 ) y =0 ( x)2 + ( y)2 z

f (x0 + Δ x)

Δf

Δ f − f ′(x0 )Δ x f ′(x0 )Δ x

f (x0 )

f (x 0 + Δx, y0 + Δy) Δy

Δx x

x0 Figure 13.4.2

f (x 0 , y0 )

Δf

x0 + Δ x

Δx

(x 0 , y0 )

Figure 13.4.3

(x 0 + Δx, y0 + Δy)

942

Chapter 13 / Partial Derivatives

Based on these ideas, we can now give our deﬁnition of differentiability for functions of two variables.

13.4.1 definition A function f of two variables is said to be differentiable at (x0 , y0 ) provided fx(x0 , y0 ) and fy(x0 , y0 ) both exist and lim

( x, y) → (0,0)

f − fx (x0 , y0 ) x − fy (x0 , y0 ) y =0 ( x)2 + ( y)2

(4)

As with the one-variable case, veriﬁcation of differentiability using this deﬁnition involves the computation of a limit. Example 1 (0, 0).

Use Deﬁnition 13.4.1 to prove that f (x, y) = x 2 + y 2 is differentiable at

Solution. The increment is f = f (0 + x, 0 + y) − f (0, 0) = ( x)2 + ( y)2 Since fx (x, y) = 2x and fy (x, y) = 2y, we have fx (0, 0) = fy (0, 0) = 0, and (4) becomes ( x)2 + ( y)2 = lim ( x)2 + ( y)2 = 0 2 2 ( x, y) → (0,0) ( x, y) → (0,0) ( x) + ( y) lim

Therefore, f is differentiable at (0, 0).

We now derive an important consequence of limit (4). Deﬁne a function ǫ = ǫ( x, y) =

f − fx (x0 , y0 ) x − fy (x0 , y0 ) y ( x)2 + ( y)2

for ( x, y) = (0, 0)

and deﬁne ǫ(0, 0) to be 0. Equation (4) then implies that lim

( x, y) → (0,0)

ǫ( x, y) = 0

Furthermore, it immediately follows from the deﬁnition of ǫ that f = fx (x0 , y0 ) x + fy (x0 , y0 ) y + ǫ ( x)2 + ( y)2

(5)

In other words, if f is differentiable at (x0 , y0 ), then f may be expressed as shown in (5), where ǫ → 0 as ( x, y) → (0, 0) and where ǫ = 0 if ( x, y) = (0, 0). For functions of three variables we have an analogous deﬁnition of differentiability in terms of the increment f = f (x0 + x, y0 + y, z0 + z) − f (x0 , y0 , z0 ) 13.4.2 definition A function f of three variables is said to be differentiable at (x0 , y0 , z0 ) provided fx (x0 , y0 , z0 ), fy (x0 , y0 , z0 ), and fz (x0 , y0 , z0 ) exist and f − fx (x0 , y0 , z0 ) x − fy (x0 , y0 , z0 ) y − fz (x0 , y0 , z0 ) z =0 ( x, y, z) → (0,0,0) ( x)2 + ( y)2 + ( z)2 (6) lim

13.4 Differentiability, Differentials, and Local Linearity

943

In a manner similar to the two-variable case, we can express the limit (6) in terms of a function ǫ( x, y, z) that vanishes at ( x, y, z) = (0, 0, 0) and is continuous there. The details are left as an exercise for the reader. If a function f of two variables is differentiable at each point of a region R in the xyplane, then we say that f is differentiable on R; and if f is differentiable at every point in the xy-plane, then we say that f is differentiable everywhere. For a function f of three variables we have corresponding conventions. DIFFERENTIABILITY AND CONTINUITY

Recall that we want a function to be continuous at every point at which it is differentiable. The next result shows this to be the case.

13.4.3 theorem that point.

If a function is differentiable at a point, then it is continuous at

proof We will give the proof for f (x, y), a function of two variables, since that will reveal the essential ideas. Assume that f is differentiable at (x0 , y0 ). To prove that f is continuous at (x0 , y0 ) we must show that lim

(x,y) → (x0 ,y0 )

f (x, y) = f (x0 , y0 )

which, on letting x = x0 + x and y = y0 + y, is equivalent to lim

( x, y) → (0,0)

f (x0 + x, y0 + y) = f (x0 , y0 )

By Equation (2) this is equivalent to lim

( x, y) → (0,0)

f = 0

However, from Equation (5) The converse of Theorem 13.4.3 is false. For example, the function

f (x, y) =

x2 + y2

is continuous at (0, 0) but is not differentiable at (0, 0). Why not?

lim

( x, y) → (0,0)

f =

lim

( x, y) → (0,0)

fx (x0 , y0 ) x + fy (x0 , y0 ) y

+ ǫ( x, y) ( x)2 + ( y)2

=0+0+0·0=0 ■

It can be difﬁcult to verify that a function is differentiable at a point directly from the deﬁnition. The next theorem, whose proof is usually studied in more advanced courses, provides simple conditions for a function to be differentiable at a point.

13.4.4 theorem If all ﬁrst-order partial derivatives of f exist and are continuous at a point, then f is differentiable at that point.

For example, consider the function f(x, y, z) = x + yz Since fx(x, y, z) = 1, fy(x, y, z) = z, and fz(x, y, z) = y are deﬁned and continuous everywhere, we conclude from Theorem 13.4.4 that f is differentiable everywhere.

944

Chapter 13 / Partial Derivatives

DIFFERENTIALS As with the one-variable case, the approximations

f ≈ fx (x0 , y0 ) x + fy (x0 , y0 ) y for a function of two variables and the approximation f ≈ fx (x0 , y0 , z0 ) x + fy (x0 , y0 , z0 ) y + fz (x0 , y0 , z0 ) z

(7)

for a function of three variables have a convenient formulation in the language of differentials. If z = f (x, y) is differentiable at a point (x0 , y0 ), we let dz = fx (x0 , y0 ) dx + fy (x0 , y0 ) dy

(8)

denote a new function with dependent variable dz and independent variables dx and dy. We refer to this function (also denoted df ) as the total differential of z at (x0 , y0 ) or as the total differential of f at (x0 , y0 ). Similarly, for a function w = f(x, y, z) of three variables we have the total differential of w at (x0 , y0 , z0 ), dw = fx (x0 , y0 , z0 ) dx + fy (x0 , y0 , z0 ) dy + fz (x0 , y0 , z0 ) dz

(9)

which is also referred to as the total differential of f at (x0 , y0 , z0 ). It is common practice to omit the subscripts and write Equations (8) and (9) as dz = fx (x, y) dx + fy (x, y) dy

(10)

dw = fx (x, y, z) dx + fy(x, y, z) dy + fz (x, y, z) dz

(11)

and

In the two-variable case, the approximation can be written in the form

f ≈ fx (x0 , y0 ) x + fy (x0 , y0 ) y f ≈ df

(12)

for dx = x and dy = y. Equivalently, we can write approximation (12) as z ≈ dz

(13)

In other words, we can estimate the change z in z by the value of the differential dz where dx is the change in x and dy is the change in y. Furthermore, it follows from (4) that if x and y are close to 0, then the magnitude of the error in approximation (13) will be much smaller than the distance ( x)2 + ( y)2 between (x0 , y0 ) and (x0 + x, y0 + y). Example 2 Use (13) to approximate the change in z = xy 2 from its value at (0.5, 1.0) to its value at (0.503, 1.004). Compare the magnitude of the error in this approximation with the distance between the points (0.5, 1.0) and (0.503, 1.004).

Solution. For z = xy 2 we have dz = y 2 dx + 2xy dy. Evaluating this differential at

(x, y) = (0.5, 1.0), dx = x = 0.503 − 0.5 = 0.003, and dy = y = 1.004 − 1.0 = 0.004 yields dz = 1.02 (0.003) + 2(0.5)(1.0)(0.004) = 0.007 Since z = 0.5 at (x, y) = (0.5, 1.0) and z = 0.507032048 at (x, y) = (0.503, 1.004), we have z = 0.507032048 − 0.5 = 0.007032048 and the error in approximating z by dz has magnitude |dz − z| = |0.007 − 0.007032048| = 0.000032048

13.4 Differentiability, Differentials, and Local Linearity

945

Since the distance between (0.5, 1.0) and (0.503, 1.004) = (0.5 + x, 1.0 + y) is √ ( x)2 + ( y)2 = (0.003)2 + (0.004)2 = 0.000025 = 0.005 we have

|dz − z|

( x)2

( y)2

1 0.000032048 = 0.0064096 < 0.005 150

1 Thus, the magnitude of the error in our approximation is less than 150 of the distance between the two points.

With the appropriate changes in notation, the preceding analysis can be extended to functions of three or more variables.

Example 3 The length, width, and height of a rectangular box are measured with an error of at most 5%. Use a total differential to estimate the maximum percentage error that results if these quantities are used to calculate the diagonal of the box.

Solution. The diagonal D of a box with length x, width y, and height z is given by D=

x 2 + y 2 + z2

Let x0 , y0 , z0 , and D0 = x02 + y02 + z02 denote the actual values of the length, width, height, and diagonal of the box. The total differential dD of D at (x0 , y0 , z0 ) is given by x0

dx +

and

dy +

dz x02 + y02 + z02 x02 + y02 + z02 x02 + y02 + z02 If x, y, z, and D = x 2 + y 2 + z2 are the measured and computed values of the length, width, height, and diagonal, respectively, then dD =

x = x − x0 ,

y = y − y0 ,

z = z − z0

x x ≤ 0.05, 0

y y ≤ 0.05, 0

z ≤ 0.05 z 0

We are seeking an estimate for the maximum size of D /D0 . With the aid of Equation (11) we have 1 dD D [x0 x + y0 y + z0 z] = 2 ≈ D0 D0 x0 + y02 + z02

1 2 y 2 z 2 x = 2 + y + z x 0 0 y0 z0 x0 + y02 + z02 0 x0 Since dD 1 D = x 2 + y 2 + z2 0 0 0 0

2 x 2 y 2 z x + y + z 0 0 0 x y0 z0 0 1 2 x 2 y 2 z ≤ 2 x0 + y0 + z0 2 2 x0 y0 z0 x0 + y0 + z0 1 ≤ 2 x02 (0.05) + y02 (0.05) + z02 (0.05) = 0.05 2 2 x0 + y0 + z0

we estimate the maximum percentage error in D to be 5%.

946

Chapter 13 / Partial Derivatives

LOCAL LINEAR APPROXIMATIONS We now show that if a function f is differentiable at a point, then it can be very closely approximated by a linear function near that point. For example, suppose that f (x, y) is differentiable at the point (x0 , y0 ). Then approximation (3) can be written in the form

f (x0 + x, y0 + y) ≈ f (x0 , y0 ) + fx (x0 , y0 ) x + fy (x0 , y0 ) y Show that if f (x, y) is a linear function, then (14) becomes an equality.

If we let x = x0 + x and y = x0 + y, this approximation becomes (14)

f (x, y) ≈ f (x0 , y0 ) + fx (x0 , y0 )(x − x0 ) + fy (x0 , y0 )(y − y0 )

which yields a linear approximation of f (x, y). Since the error in this approximation is equal to the error in approximation (3), we conclude that for (x, y) close to (x0 , y0 ), the error in (14) will be much smaller than the distance between these two points. When f (x, y) is differentiable at (x0 , y0 ) we get Explain why the error in approximation (14) is the same as the error in approximation (3).

(15)

L(x, y) = f (x0 , y0 ) + fx (x0 , y0 )(x − x0 ) + fy (x0 , y0 )(y − y0 ) and refer to L(x, y) as the local linear approximation to f at (x0 , y0 ). Example 4 Let L(x, y) denote the local linear approximation to f(x, y) = at the point (3, 4). Compare the error in approximating f(3.04, 3.98) = (3.04)2 + (3.98)2

x2 + y2

by L(3.04, 3.98) with the distance between the points (3, 4) and (3.04, 3.98).

Solution. We have

with fx(3, 4) = is given by

3 5

fx(x, y) =

x x2

and fy(x, y) =

y x2

+ y2

and fy (3, 4) = 45 . Therefore, the local linear approximation to f at (3, 4) L(x, y) = 5 + 35 (x − 3) + 45 (y − 4)

Consequently, f (3.04, 3.98) ≈ L(3.04, 3.98) = 5 + 35 (0.04) + 45 (−0.02) = 5.008 Since f(3.04, 3.98) =

(3.04)2 + (3.98)2 ≈ 5.00819

the error in the approximation is about 5.00819 − 5.008 = 0.00019. This is less than of the distance (3.04 − 3)2 + (3.98 − 4)2 ≈ 0.045

1 200

between the points (3, 4) and (3.04, 3.98).

For a function f (x, y, z) that is differentiable at (x0 , y0 , z0 ), the local linear approximation is L(x, y, z) = f(x0 , y0 , z0 ) + fx (x0 , y0 , z0 )(x − x0 ) (16) + fy (x0 , y0 , z0 )(y − y0 ) + fz (x0 , y0 , z0 )(z − z0 ) We have formulated our deﬁnitions in this section in such a way that continuity and local linearity are consequences of differentiability. In Section 13.7 we will show that

13.4 Differentiability, Differentials, and Local Linearity

947

if a function f (x, y) is differentiable at a point (x0 , y0 ), then the graph of L(x, y) is a nonvertical tangent plane to the graph of f at the point (x0 , y0 , f (x0 , y0 )).

✔QUICK CHECK EXERCISES 13.4

(See page 949 for answers.)

1. Assume that f (x, y) is differentiable at (x0 , y0 ) and let f denote the change in f from its value at (x0 , y0 ) to its value at (x0 + x, y0 + y). (a) f ≈ (b) The limit that guarantees the error in the approximation in part (a) is very small when both x and y are close to 0 is .

2. Compute the differential of each function. 2 (a) z = xey (b) w = x sin(yz)

3. If f is differentiable at (x0 , y0 ), then the local linear approximation to f at (x0 , y0 ) is L(x) = .

4. Assume that f (1, −2) = 4 and f (x, y) is differentiable at (1, −2) with fx (1, −2) = 2 and fy (1, −2) = −3. Estimate the value of f (0.9, −1.950).

EXERCISE SET 13.4 F O C U S O N C O N C E P TS

1. Suppose that a function f (x, y) is differentiable at the point (3, 4) with fx(3, 4) = 2 and fy(3, 4) = −1. If f (3, 4) = 5, estimate the value of f (3.01, 3.98).

2. Suppose that a function f(x, y) is differentiable at the point (−1, 2) with fx(−1, 2) = 1 and fy(−1, 2) = 3. If f(−1, 2) = 2, estimate the value of f(−0.99, 2.02).

3. Suppose that a function f(x, y, z) is differentiable at the point (1, 2, 3) with fx(1, 2, 3) = 1, fy(1, 2, 3) = 2, and fz(1, 2, 3) = 3. If f(1, 2, 3) = 4, estimate the value of f(1.01, 2.02, 3.03). 4. Suppose that a function f(x, y, z) is differentiable at the point (2, 1, −2), fx(2, 1, −2) = −1, fy(2, 1, −2) = 1, and fz(2, 1, −2) = −2. If f(2, 1, −2) = 0, estimate the value of f(1.98, 0.99, −1.97).

5. Use Definitions 13.4.1 and 13.4.2 to prove that a constant function of two or three variables is differentiable everywhere. 6. Use Definitions 13.4.1 and 13.4.2 to prove that a linear function of two or three variables is differentiable everywhere. 7. Use Definition 13.4.2 to prove that f (x, y, z) = x 2 + y 2 + z2

is differentiable at (0, 0, 0).

8. Use Deﬁnition 13.4.2 to determine all values of r such that f (x, y, z) = (x 2 + y 2 + z2 )r is differentiable at (0, 0, 0). 9–20 Compute the differential dz or dw of the function. ■

9. z = 7x − 2y

10. z = exy

12. z = 5x 2 y 5 − 2x + 4y + 7

11. z = x 3 y 2

13. z = tan−1 xy

15. w = 8x − 3y + 4z

17. w = x 3 y 2 z

14. z = e−3x cos 6y

16. w = exyz

18. w = 4x 2 y 3 z7 − 3xy + z + 5

19. w = tan−1 (xyz)

20. w =

√ √ √ x+ y+ z

21–26 Use a total differential to approximate the change in the values of f from P to Q. Compare your estimate with the actual change in f . ■

21. f(x, y) = x 2 + 2xy − 4x; P (1, 2), Q(1.01, 2.04)

22. f(x, y) = x 1/3 y 1/2 ; P (8, 9), Q(7.78, 9.03) x+y ; P (−1, −2), Q(−1.02, −2.04) 23. f(x, y) = xy √ 24. f(x, y) = ln 1 + xy; P (0, 2), Q(−0.09, 1.98)

25. f(x, y, z) = 2xy 2 z3 ; P (1, −1, 2), Q(0.99, −1.02, 2.02) xyz 26. f (x, y, z) = ; P (−1, −2, 4), x+y+z Q(−1.04, −1.98, 3.97)

27–30 True–False Determine whether the statement is true or false. Explain your answer. ■

27. By deﬁnition, a function f(x, y) is differentiable at (x0 , y0 ) provided both fx (x0 , y0 ) and fy (x0 , y0 ) are deﬁned. 28. For any point (x0 , y0 ) in the domain of a function f(x, y), we have lim f = 0 ( x, y) → (0,0)

where f = f(x0 + x, y0 + y) − f(x0 , y0 )

29. If fx and fy are both continuous at (x0 , y0 ), then so is f . 30. The graph of a local linear approximation to a function f(x, y) is a plane.

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Chapter 13 / Partial Derivatives

31. In the accompanying figure a rectangle with initial length x0 and initial width y0 has been enlarged, resulting in a rectangle with length x0 + x and width y0 + y. What portion of the figure represents the increase in the area of the rectangle? What portion of the figure represents an approximation of the increase in area by a total differential?

y0 + Δy

45. Suppose that a function f(x, y) is differentiable at the point (1, 1) with fx(1, 1) = 2 and f(1, 1) = 3. Let L(x, y) denote the local linear approximation of f at (1, 1). If L(1.1, 0.9) = 3.15, ﬁnd the value of fy (1, 1). 46. Suppose that a function f(x, y) is differentiable at the point (0, −1) with fy(0, −1) = −2 and f(0, −1) = 3. Let L(x, y) denote the local linear approximation of f at (0, −1). If L(0.1, −1.1) = 3.3, ﬁnd the value of fx(0, −1).

47. Suppose that a function f(x, y, z) is differentiable at the point (3, 2, 1) and L(x, y, z) = x − y + 2z − 2 is the local linear approximation to f at (3, 2, 1). Find f(3, 2, 1), fx(3, 2, 1), fy(3, 2, 1), and fz(3, 2, 1).

x0 x 0 + Δx Figure Ex-31

48. Suppose that a function f(x, y, z) is differentiable at the point (0, −1, −2) and L(x, y, z) = x + 2y + 3z + 4 is the local linear approximation to f at (0, −1, −2). Find f(0, −1, −2), fx(0, −1, −2), fy(0, −1, −2), and fz(0, −1, −2).

32. The volume V of a right circular cone of radius r and height h is given by V = 13 πr 2 h. Suppose that the height decreases from 20 in to 19.95 in and the radius increases from 4 in to 4.05 in. Compare the change in volume of the cone with an approximation of this change using a total differential.

49–52 A function f is given along with a local linear approximation L to f at a point P . Use the information given to determine point P . ■

33–40 (a) Find the local linear approximation L to the speciﬁed

50. f(x, y) = x 2 y; L(x, y) = 4y − 4x + 8

function f at the designated point P . (b) Compare the error in approximating f by L at the speciﬁed point Q with the distance between P and Q. ■ 1 ; P (4, 3), Q(3.92, 3.01) 33. f (x, y) = 2 x + y2 34. f (x, y) = x 0.5 y 0.3 ; P (1, 1), Q(1.05, 0.97)

35. f (x, y) = x sin y; P (0, 0), Q(0.003, 0.004) 36. f (x, y) = ln xy; P (1, 2), Q(1.01, 2.02)

37. f (x, y, z) = xyz; P (1, 2, 3), Q(1.001, 2.002, 3.003) x+y 38. f (x, y, z) = ; P (−1, 1, 1), Q(−0.99, 0.99, 1.01) y+z 39. f (x, y, z) = xeyz ; P (1, −1, −1), Q(0.99, −1.01, −0.99) 40. f (x, y, z) = ln(x + yz); P (2, 1, −1), Q(2.02, 0.97, −1.01)

41. In each part, conﬁrm that the stated formula is the local linear approximation at (0, 0). 2x + 1 (a) ex sin y ≈ y ≈ 1 + 2x − y (b) y+1 42. Show that the local linear approximation of the function f (x, y) = x α y β at (1, 1) is x α y β ≈ 1 + α(x − 1) + β(y − 1)

43. In each part, conﬁrm that the stated formula is the local linear approximation at (1, 1, 1). 4x (a) xyz + 2 ≈ x + y + z (b) ≈ 2x − y − z + 2 y +z 44. Based on Exercise 42, what would you conjecture is the local linear approximation to x α y β zγ at (1, 1, 1)? Verify your conjecture by ﬁnding this local linear approximation.

49. f(x, y) = x 2 + y 2 ; L(x, y) = 2y − 2x − 2 51. f(x, y, z) = xy + z2 ; L(x, y, z) = y + 2z − 1

52. f(x, y, z) = xyz; L(x, y, z) = x − y − z − 2

53. The length and width of a rectangle are measured with errors of at most 3% and 5%, respectively. Use differentials to approximate the maximum percentage error in the calculated area. 54. The radius and height of a right circular cone are measured with errors of at most 1% and 4%, respectively. Use differentials to approximate the maximum percentage error in the calculated volume. 55. The length and width of a rectangle are measured with errors of at most r%, where r is small. Use differentials to approximate the maximum percentage error in the calculated length of the diagonal. 56. The legs of a right triangle are measured to be 3 cm and 4 cm, with a maximum error of 0.05 cm in each measurement. Use differentials to approximate the maximum possible error in the calculated value of (a) the hypotenuse and (b) the area of the triangle. 57. The period T of a simple pendulum with small oscillations is calculated from the formula T = 2π L/g, where L is the length of the pendulum and g is the acceleration due to gravity. Suppose that measured values of L and g have errors of at most 0.5% and 0.1%, respectively. Use differentials to approximate the maximum percentage error in the calculated value of T . 58. According to the ideal gas law, the pressure, temperature, and volume of a conﬁned gas are related by P = kT /V , where k is a constant. Use differentials to approximate the

13.5 The Chain Rule

percentage change in pressure if the temperature of a gas is increased 3% and the volume is increased 5%. 59. Suppose that certain measured quantities x and y have errors of at most r% and s%, respectively. For each of the following formulas in x and y, use differentials to approximate the maximum possible error in the calculated result. √ (a) xy (b) x /y (c) x 2 y 3 (d) x 3 y 60. The total resistance R of three resistances R1 , R2 , and R3 , connected in parallel, is given by 1 1 1 1 = + + R R1 R2 R3 Suppose that R1 , R2 , and R3 are measured to be 100 ohms, 200 ohms, and 500 ohms, respectively, with a maximum error of 10% in each. Use differentials to approximate the maximum percentage error in the calculated value of R. 61. The area of a triangle is to be computed from the formula A = 21 ab sin θ , where a and b are the lengths of two sides and θ is the included angle. Suppose that a, b, and θ are measured to be 40 ft, 50 ft, and 30 ◦ , respectively. Use differentials to approximate the maximum error in the calculated value of A if the maximum errors in a, b, and θ are 21 ft, 1 ft, and 2 ◦ , respectively. 4

949

62. The length, width, and height of a rectangular box are measured with errors of at most r% (where r is small). Use differentials to approximate the maximum percentage error in the computed value of the volume. 63. Use Theorem 13.4.4 to prove that f (x, y) = x 2 sin y is differentiable everywhere. 64. Use Theorem 13.4.4 to prove that f (x, y, z) = xy sin z is differentiable everywhere. 65. Suppose that f(x, y) is differentiable at the point (x0 , y0 ) and let z0 = f(x0 , y0 ). Prove that g(x, y, z) = z − f(x, y) is differentiable at (x0 , y0 , z0 ). 66. Suppose that f satisﬁes an equation in the form of (5), where ǫ( x, y) is continuous at ( x, y) = (0, 0) with ǫ(0, 0) = 0. Prove that f is differentiable at (x0 , y0 ).

67. Writing Discuss the similarities and differences between the deﬁnition of “differentiability” for a function of a single variable and the deﬁnition of “differentiability” for a function of two variables.

68. Writing Discuss the use of differentials in the approximation of increments and in the estimation of errors.

✔QUICK CHECK ANSWERS 13.4 f − fx (x0 , y0 ) x − fy (x0 , y0 ) y 2 2 = 0 2. (a) dz = ey dx + 2xyey dy 2 2 ( x) + ( y) (b) dw = sin(yz) dx + xz cos(yz) dy + xy cos(yz) dz 3. f(x0 , y0 ) + fx (x0 , y0 )(x − x0 ) + fy (x0 , y0 )(y − y0 ) 4. 3.65 1. (a) fx (x0 , y0 ) x + fy (x0 , y0 ) y (b)

13.5

lim

( x, y) → (0,0)

THE CHAIN RULE In this section we will derive versions of the chain rule for functions of two or three variables. These new versions will allow us to generate useful relationships among the derivatives and partial derivatives of various functions.

CHAIN RULES FOR DERIVATIVES

If y is a differentiable function of x and x is a differentiable function of t, then the chain rule for functions of one variable states that, under composition, y becomes a differentiable function of t with dy dx dy = dt dx dt We will now derive a version of the chain rule for functions of two variables. Assume that z = f(x, y) is a function of x and y, and suppose that x and y are in turn functions of a single variable t, say x = x(t),

y = y(t)

The composition z = f(x(t), y(t)) then expresses z as a function of the single variable t. Thus, we can ask for the derivative dz/dt and we can inquire about its relationship to the

Chapter 13 / Partial Derivatives

950

derivatives ∂z/∂x, ∂z/∂y, dx /dt, and dy /dt. Letting x, y, and z denote the changes in x, y, and z, respectively, that correspond to a change of t in t, we have z x dx dz = lim , = lim , t → 0 t t → 0 t dt dt It follows from (3) of Section 13.4 that z ≈

and

y dy = lim t → 0 t dt

∂z ∂z x + y ∂x ∂y

(1)

where the partial derivatives are evaluated at (x(t), y(t)). Dividing both sides of (1) by t yields ∂z x ∂z y z ≈ + (2) t ∂x t ∂y t Similarly, we can produce the analog of (2) for functions of three variables as follows: assume that w = f(x, y, z) is a function of x, y, and z, and suppose that x, y, and z are functions of a single variable t. As above we deﬁne w, x, y, and z to be the changes in w, x, y, and z that correspond to a change of t in t. Then (7) in Section 13.4 implies that ∂w ∂w ∂w x + y + z (3) w ≈ ∂x ∂y ∂z and dividing both sides of (3) by t yields ∂w x ∂w y ∂w z w ≈ + + t ∂x t ∂y t ∂z t

(4)

Taking the limit as t → 0 of both sides of (2) and (4) suggests the following results. (A complete proof of the two-variable case can be found in Appendix D.)

13.5.1 theorem (Chain Rules for Derivatives) If x = x(t) and y = y(t) are differentiable at t, and if z = f(x, y) is differentiable at the point (x, y) = (x(t), y(t)), then z = f(x(t), y(t)) is differentiable at t and dz ∂z dx ∂z dy = + dt ∂x dt ∂y dt

(5)

where the ordinary derivatives are evaluated at t and the partial derivatives are evaluated at (x, y). If each of the functions x = x(t), y = y(t), and z = z(t) is differentiable at t, and if w = f(x, y, z) is differentiable at the point (x, y, z) = (x(t), y(t), z(t)), then the function w = f(x(t), y(t), z(t)) is differentiable at t and

z ⭸z ⭸x

∂w dx ∂w dy ∂w dz dw = + + dt ∂x dt ∂y dt ∂z dt

⭸z ⭸y

dx dt

dy dt t

(6)

where the ordinary derivatives are evaluated at t and the partial derivatives are evaluated at (x, y, z).

t dz ⭸z dx ⭸z dy = + dt ⭸x dt ⭸y dt

Figure 13.5.1

Formula (5) can be represented schematically by a “tree diagram” that is constructed as follows (Figure 13.5.1). Starting with z at the top of the tree and moving downward, join each variable by lines (or branches) to those variables on which it depends directly. Thus, z is joined to x and y and these in turn are joined to t. Next, label each branch with a

13.5 The Chain Rule

Create a tree diagram for Formula (6).

951

derivative whose “numerator” contains the variable at the top end of that branch and whose “denominator” contains the variable at the bottom end of that branch. This completes the “tree.” To ﬁnd the formula for dz/dt, follow the two paths through the tree that start with z and end with t. Each such path corresponds to a term in Formula (5).

Example 1

Suppose that z = x 2 y,

x = t 2,

y = t3

Use the chain rule to ﬁnd dz/dt, and check the result by expressing z as a function of t and differentiating directly.

Solution. By the chain rule [Formula (5)], dz ∂z dx ∂z dy = + = (2xy)(2t) + (x 2 )(3t 2 ) dt ∂x dt ∂y dt = (2t 5 )(2t) + (t 4 )(3t 2 ) = 7t 6 Alternatively, we can express z directly as a function of t, z = x 2 y = (t 2 )2 (t 3 ) = t 7 and then differentiate to obtain dz/dt = 7t 6 . However, this procedure may not always be convenient.

Example 2

Suppose that w = x 2 + y 2 + z2 ,

x = cos θ,

y = sin θ,

z = tan θ

Use the chain rule to ﬁnd dw/dθ when θ = π/4.

Solution. From Formula (6) with θ in the place of t, we obtain dw ∂w dx ∂w dy ∂w dz = + + dθ ∂x dθ ∂y dθ ∂z dθ =

1 2 1 (x + y 2 + z2 )−1/2 (2x)(− sin θ) + (x 2 + y 2 + z2 )−1/2 (2y)(cos θ ) 2 2 1 + (x 2 + y 2 + z2 )−1/2 (2z)(sec2 θ) 2

When θ = π/4, we have π 1 1 π π = √ , y = sin = √ , z = tan = 1 4 4 4 2 2 √ √ Substituting x = 1/ 2, y = 1/ 2, z = 1, θ = π/4 in the formula for dw/dθ yields √ √ dw 1 1 1 1 1 1 1 1 = 2) − 2) ( + ( + (2)(2) √ √ √ √ √ dθ θ =π/4 2 2 2 2 2 2 2 2 √ = 2 x = cos

Confirm the result of Example 2 by expressing w directly as a function of θ .

952

Chapter 13 / Partial Derivatives REMARK

There are many variations in derivative notations, each of which gives the chain rule a different look. If z = f(x, y), where x and y are functions of t , then some possibilities are

dz dx dy = fx + fy dt dt dt

∂f dx ∂f dy df = + dt ∂x dt ∂y dt df = fx x ′ (t) + fy y ′ (t) dt

CHAIN RULES FOR PARTIAL DERIVATIVES

In Formula (5) the variables x and y are each functions of a single variable t. We now consider the case where x and y are each functions of two variables. Let z = f(x, y) and suppose that x and y are functions of u and v, say x = x(u, v),

y = y(u, v)

The composition z = f(x(u, v), y(u, v)) expresses z as a function of the two variables u and v. Thus, we can ask for the partial derivatives ∂z/∂u and ∂z/∂v; and we can inquire about the relationship between these derivatives and the derivatives ∂z/∂x, ∂z/∂y, ∂x /∂u, ∂x /∂v, ∂y /∂u, and ∂y /∂v. Similarly, if w = f(x, y, z) and x, y, and z are each functions of u and v, then the composition w = f(x(u, v), y(u, v), z(u, v)) expresses w as a function of u and v. Thus we can also ask for the derivatives ∂w/∂u and ∂w/∂v; and we can investigate the relationship between these derivatives, the partial derivatives ∂w/∂x, ∂w/∂y, and ∂w/∂z, and the partial derivatives of x, y, and z with respect to u and v.

13.5.2 theorem (Chain Rules for Partial Derivatives) If x = x(u, v) and y = y(u, v) have ﬁrst-order partial derivatives at the point (u, v), and if z = f(x, y) is differentiable at the point (x, y) = (x(u, v), y(u, v)), then z = f(x(u, v), y(u, v)) has ﬁrstorder partial derivatives at the point (u, v) given by ∂z ∂z ∂x ∂z ∂y = + ∂u ∂x ∂u ∂y ∂u

and

∂z ∂z ∂x ∂z ∂y = + ∂v ∂x ∂v ∂y ∂v

(7–8)

If each function x = x(u, v), y = y(u, v), and z = z(u, v) has ﬁrst-order partial derivatives at the point (u, v), and if the function w = f(x, y, z) is differentiable at the point (x, y, z) = (x(u, v), y(u, v), z(u, v)), then w = f(x(u, v), y(u, v), z(u, v)) has ﬁrstorder partial derivatives at the point (u, v) given by ∂w ∂x ∂w ∂y ∂w ∂z ∂w = + + ∂u ∂x ∂u ∂y ∂u ∂z ∂u

and

∂w ∂x ∂w ∂y ∂w ∂z ∂w = + + ∂v ∂x ∂v ∂y ∂v ∂z ∂v (9–10)

proof We will prove Formula (7); the other formulas are derived similarly. If v is held ﬁxed, then x = x(u, v) and y = y(u, v) become functions of u alone. Thus, we are back to the case of Theorem 13.5.1. If we apply that theorem with u in place of t, and if we use ∂ rather than d to indicate that the variable v is ﬁxed, we obtain ∂z ∂x ∂z ∂y ∂z ■ = + ∂u ∂x ∂u ∂y ∂u

13.5 The Chain Rule

Figures 13.5.2 and 13.5.3 show tree diagrams for the formulas in Theorem 13.5.2. As illustrated in Figure 13.5.2, the formula for ∂z/∂u can be obtained by tracing all paths through the tree that start with z and end with u, and the formula for ∂z/∂v can be obtained by tracing all paths through the tree that start with z and end with v. Figure 13.5.3 displays analogous results for ∂w/∂u and ∂w/∂v.

z ⭸z ⭸x

⭸z ⭸y

⭸x ⭸u

⭸x ⭸v

⭸y ⭸u

y ⭸y ⭸v

953

Example 3

Given that z = exy ,

x = 2u + v,

y = u/v

ﬁnd ∂z/∂u and ∂z/∂v using the chain rule.

⭸z ⭸z ⭸x ⭸z ⭸y = + ⭸u ⭸x ⭸u ⭸y ⭸u

Solution. ∂z ∂x ∂z ∂y 1 ∂z x xy e = + = (yexy )(2) + (xexy ) = 2y + ∂u ∂x ∂u ∂y ∂u v v

2u 2u + v (2u+v)(u/v) 4u = = + + 1 e(2u+v)(u/v) e v v v u ∂z ∂x ∂z ∂y ∂z = + = (yexy )(1) + (xexy ) − 2 ∂v ∂x ∂v ∂y ∂v v u u u = y − x 2 exy = − (2u + v) 2 e(2u+v)(u/v) v v v

z ⭸z ⭸x

⭸z ⭸y

⭸x ⭸u

⭸x ⭸v

⭸y ⭸u

⭸y ⭸v

⭸z ⭸z ⭸x ⭸z ⭸y = + ⭸v ⭸x ⭸v ⭸y ⭸v

=−

2u2 (2u+v)(u/v) e v2

Figure 13.5.2

Example 4

⭸w ⭸x

⭸w ⭸y

x ⭸x ⭸v

⭸y ⭸u u

⭸y ⭸v v

w = exyz ,

⭸w ⭸z

⭸x ⭸u

⭸w ⭸w ⭸x ⭸w ⭸y ⭸w ⭸z = + + ⭸u ⭸x ⭸u ⭸y ⭸u ⭸z ⭸u ⭸w ⭸w ⭸x ⭸w ⭸y ⭸w ⭸z = + + ⭸x ⭸v ⭸y ⭸v ⭸z ⭸v ⭸v

x = 3u + v,

y = 3u − v,

z = u2 v

Use appropriate forms of the chain rule to ﬁnd ∂w/∂u and ∂w/∂v. z

⭸z ⭸u

Suppose that

⭸z ⭸v v

Solution. From the tree diagram and corresponding formulas in Figure 13.5.3 we obtain ∂w = yzexyz (3) + xzexyz (3) + xyexyz (2uv) = exyz (3yz + 3xz + 2xyuv) ∂u and ∂w = yzexyz (1) + xzexyz (−1) + xyexyz (u2 ) = exyz (yz − xz + xyu2 ) ∂v If desired, we can express ∂w/∂u and ∂w/∂v in terms of u and v alone by replacing x, y, and z by their expressions in terms of u and v.

Figure 13.5.3

OTHER VERSIONS OF THE CHAIN RULE Although we will not prove it, the chain rule extends to functions w = f (v1 , v2 , . . . , vn ) of n variables. For example, if each vi is a function of t, i = 1, 2, . . . , n, the relevant formula is dw ∂w dv1 ∂w dv2 ∂w dvn (11) = + + ··· + dt ∂v1 dt ∂v2 dt ∂vn dt

Note that (11) is a natural extension of Formulas (5) and (6) in Theorem 13.5.1. There are inﬁnitely many variations of the chain rule, depending on the number of variables and the choice of independent and dependent variables. A good working procedure is to use tree diagrams to derive new versions of the chain rule as needed.

954

Chapter 13 / Partial Derivatives

Example 5

Suppose that w = x 2 + y 2 − z2 and x = ρ sin φ cos θ,

y = ρ sin φ sin θ,

z = ρ cos φ

Use appropriate forms of the chain rule to ﬁnd ∂w/∂ρ and ∂w/∂θ .

Solution. From the tree diagram and corresponding formulas in Figure 13.5.4 we obtain

w ⭸w ⭸x

⭸w ⭸y

y ⭸x ⭸u

⭸x ⭸r

⭸x ⭸f f u

⭸y ⭸r r

∂w = 2x sin φ cos θ + 2y sin φ sin θ − 2z cos φ ∂ρ = 2ρ sin2 φ cos2 θ + 2ρ sin2 φ sin2 θ − 2ρ cos2 φ

⭸w ⭸z z

⭸y ⭸u

⭸y ⭸f f u

⭸z ⭸r

⭸z ⭸f

= 2ρ sin2 φ(cos2 θ + sin2 θ ) − 2ρ cos2 φ

= 2ρ(sin2 φ − cos2 φ) = −2ρ cos 2φ

∂w = (2x)(−ρ sin φ sin θ ) + (2y)ρ sin φ cos θ ∂θ = −2ρ 2 sin2 φ sin θ cos θ + 2ρ 2 sin2 φ sin θ cos θ

⭸w ⭸w ⭸x ⭸w ⭸y ⭸w ⭸z = + + ⭸r ⭸x ⭸r ⭸y ⭸r ⭸z ⭸r ⭸w ⭸y ⭸w ⭸w ⭸x = + ⭸x ⭸u ⭸y ⭸u ⭸u

=0 This result is explained by the fact that w does not vary with θ. You can see this directly by expressing the variables x, y, and z in terms of ρ, φ, and θ in the formula for w. (Verify that w = −ρ 2 cos 2φ.)

Figure 13.5.4

Example 6

Suppose that w = xy + yz,

y = sin x,

z = ex

Use an appropriate form of the chain rule to ﬁnd dw/dx. ⭸w ⭸x

⭸w ⭸y

⭸w ⭸z

Solution. From the tree diagram and corresponding formulas in Figure 13.5.5 we obtain

dy dx

dz dx x

⭸w ⭸w dy ⭸w dz dw = + + ⭸x ⭸y dx ⭸z dx dx Figure 13.5.5

WARNING

dw = y + (x + z) cos x + yex dx = sin x + (x + ex ) cos x + ex sin x This result can also be obtained by ﬁrst expressing w explicitly in terms of x as w = x sin x + ex sin x and then differentiating with respect to x; however, such direct substitution is not always possible. The symbol ∂z, unlike the differential dz, has no meaning of its own. For example, if we were to “cancel” partial symbols in the chain-rule formula

∂z ∂z ∂x ∂z ∂y = + ∂u ∂x ∂u ∂y ∂u we would obtain

∂z ∂z ∂z + = ∂u ∂u ∂u which is false in cases where ∂z/∂u = 0.

One of the principal uses of the chain rule for functions of a single variable was to compute formulas for the derivatives of compositions of functions. Theorems 13.5.1 and 13.5.2 are important not so much for the computation of formulas but because they allow us

13.5 The Chain Rule

955

to express relationships among various derivatives. As an illustration, we revisit the topic of implicit differentiation. IMPLICIT DIFFERENTIATION

Consider the special case where z = f (x, y) is a function of x and y and y is a differentiable function of x. Equation (5) then becomes dz ∂f dx ∂f dy ∂f ∂f dy = + = + dx ∂x dx ∂y dx ∂x ∂y dx

(12)

This result can be used to ﬁnd derivatives of functions that are deﬁned implicitly. For example, suppose that the equation (13)

f (x, y) = c

deﬁnes y implicitly as a differentiable function of x and we are interested in ﬁnding dy/dx. Differentiating both sides of (13) with respect to x and applying (12) yields ∂f ∂f dy + =0 ∂x ∂y dx Thus, if ∂f /∂y = 0, we obtain

dy ∂f /∂x =− dx ∂f /∂y

In summary, we have the following result.

Show that the function y = x is defined implicitly by the equation

13.5.3 theorem If the equation f(x, y) = c deﬁnes y implicitly as a differentiable function of x, and if ∂f /∂y = 0, then

x 2 − 2xy + y 2 = 0

∂f /∂x dy =− dx ∂f /∂y

but that Theorem 13.5.3 is not applicable for finding dy/dx .

Example 7

Given that

(14)

x3 + y2x − 3 = 0

ﬁnd dy/dx using (14), and check the result using implicit differentiation.

Solution. By (14) with f(x, y) = x 3 + y 2 x − 3,

∂f /∂x 3x 2 + y 2 dy =− =− dx ∂f /∂y 2yx

Alternatively, differentiating implicitly yields dy −0=0 3x 2 + y 2 + x 2y dx

3x 2 + y 2 dy =− dx 2yx

which agrees with the result obtained by (14). The chain rule also applies to implicit partial differentiation. Consider the case where w = f(x, y, z) is a function of x, y, and z and z is a differentiable function of x and y. It follows from Theorem 13.5.2 that ∂w ∂f ∂f ∂z (15) = + ∂x ∂x ∂z ∂x

956

Chapter 13 / Partial Derivatives

If the equation (16)

f(x, y, z) = c

deﬁnes z implicitly as a differentiable function of x and y, then taking the partial derivative of each side of (16) with respect to x and applying (15) gives ∂f ∂f ∂z + =0 ∂x ∂z ∂x If ∂f /∂z = 0, then

∂f /∂x ∂z =− ∂x ∂f /∂z

A similar result holds for ∂z/∂y.

13.5.4 theorem If the equation f(x, y, z) = c deﬁnes z implicitly as a differentiable function of x and y, and if ∂f /∂z = 0, then ∂f /∂x ∂z =− ∂x ∂f /∂z

and

∂f /∂y ∂z =− ∂y ∂f /∂z

8 Consider the sphere x 2 + y 2 + z2 = 1. Find ∂z/∂x and ∂z/∂y at the point 2 Example 1 2 , , . 3 3 3

Note the similarity between the expression for ∂z/∂y found in Example 8 and that found in Example 7 of Section 13.3.

Solution. By Theorem 13.5.4 with f(x, y, z) = x 2 + y 2 + z2 ,

∂f /∂x 2x x ∂z ∂f /∂y 2y y ∂z =− =− =− and =− =− =− / / ∂x ∂f ∂z 2z z ∂y ∂f ∂z 2z z 2 1 2 At the point 3 , 3 , 3 , evaluating these derivatives gives ∂z/∂x = −1 and ∂z/∂y = − 21 .

✔QUICK CHECK EXERCISES 13.5

(See page 959 for answers.)

1. Suppose that z = xy 2 and x and y are differentiable functions of t with x = 1, y = −1, dx /dt = −2, and dy /dt = 3 when t = −1. Then dz/dt = when t = −1. 2. Suppose that C is the graph of the equation f(x, y) = 1 and that this equation deﬁnes y implicitly as a differentiable function of x. If the point (2, 1) belongs to C with fx (2, 1) = 3 and fy (2, 1) = −1, then the tangent line to C at the point (2, 1) has slope .

3. A rectangle is growing in such a way that when its length is 5 ft and its width is 2 ft, the length is increasing at a rate

of 3 ft/s and its width is increasing at a rate of 4 ft/s. At this instant the area of the rectangle is growing at a rate of . 4. Suppose that z = x /y, where x and y are differentiable functions of u and v such that x = 3, y = 1, ∂x /∂u = 4, ∂x /∂v = −2, ∂y /∂u = 1, and ∂y /∂v = −1 when u = 2 and v = 1. When u = 2 and v = 1, ∂z/∂u = and ∂z/∂v = .

EXERCISE SET 13.5 1–6 Use an appropriate form of the chain rule to ﬁnd dz/dt. ■ 2 3

1. z = 3x y ; x = t , y = t √ 2. z = ln(2x 2 + y); x = t, y = t 2/3

3. z = 3 cos x − sin xy; x = 1/t, y = 3t 4. z = 1 + x − 2xy 4 ; x = ln t, y = t 5. z = e1−xy ; x = t 1/3 , y = t 3

13.5 The Chain Rule

6. z = cosh2 xy; x = t /2, y = et 7–10 Use an appropriate form of the chain rule to ﬁnd dw /dt. ■ 2 3 4

7. w = 5x y z ; x = t , y = t , z = t

8. w = ln(3x 2 − 2y + 4z3 ); x = t 1/2 , y = t 2/3 , z = t −2 9. w = 5 cos xy − sin xz; x = 1/t, y = t, z = t 3 10. w = 1 + x − 2yz4 x; x = ln t, y = t, z = 4t F O C U S O N C O N C E P TS

11. Suppose that w = x 3 y 2 z4 ; x = t 2 ,

y = t + 2,

z = 2t 4

Find the rate of change of w with respect to t at t = 1 by using the chain rule, and then check your work by expressing w as a function of t and differentiating. 12. Suppose that w = x sin yz2 ; x = cos t, y = t 2 ,

z = et

Find the rate of change of w with respect to t at t = 0 by using the chain rule, and then check your work by expressing w as a function of t and differentiating. 13. Suppose that z = f(x, y) is differentiable at the point (4, 8) with fx (4, 8) = 3 and fy (4, 8) = −1. If x = t 2 and y = t 3 , ﬁnd dz/dt when t = 2.

14. Suppose that w = f(x, y, z) is differentiable at the point (1, 0, 2) with fx (1, 0, 2) = 1, fy (1, 0, 2) = 2, and fz (1, 0, 2) = 3. If x = t, y = sin(πt), and z = t 2 + 1, ﬁnd dw/dt when t = 1.

15. Explain how the product rule for functions of a single variable may be viewed as a consequence of the chain rule applied to a particular function of two variables. 16. A student attempts to differentiate the function x x using the power rule, mistakenly getting x · x x−1 . A second student attempts to differentiate x x by treating it as an exponential function, mistakenly getting (ln x)x x . Use the chain rule to explain why the correct derivative is the sum of these two incorrect results.

17–22 Use appropriate forms of the chain rule to ﬁnd ∂z/∂u

and ∂z/∂v. ■ 17. z = 8x 2 y − 2x + 3y; x = uv, y = u − v 18. z = x 2 − y tan x; x = u/v, y = u2 v 2 19. z = x /y; x = 2 cos u, y = 3 sin v

20. z = 3x − 2y; x = u + v ln u, y = u2 − v ln v √ 2 21. z = ex y ; x = uv, y = 1/v

957

23. Let T = x 2 y − xy 3 + 2; x = r cos θ , y = r sin θ . Find ∂T /∂r and ∂T /∂θ. 2 24. Let R = e2s−t ; s = 3φ, t = φ 1/2 . Find dR /dφ. 25. Let t = u/v; u = x 2 − y 2 , v = 4xy 3 . Find ∂t /∂x and ∂t /∂y. 26. Let w = rs /(r 2 + s 2 ); r = uv, s = u − 2v. Find ∂w /∂u and ∂w /∂v. 27. Let z = ln(x 2 + 1), where x = r cos θ . Find ∂z/∂r and ∂z/∂θ. 28. Let u = rs 2 ln t, r = x 2 , s = 4y + 1, t = xy 3 . Find ∂u/∂x and ∂u/∂y. 29. Let w = 4x 2 + 4y 2 + z2 , x = ρ sin φ cos θ, y = ρ sin φ sin θ, z = ρ cos φ. Find ∂w/∂ρ, ∂w /∂φ, and ∂w/∂θ. √ 30. Let w = 3xy 2 z3 , y = 3x 2 + 2, z = x − 1. Find dw/dx. dw 31. Use a chain rule to find the value of if ds s=1/4 √ w = r 2 − r tan θ; r = s, θ = πs. 32. Use a chain rule to find the values of ∂f ∂f and ∂u u=1,v=−2 ∂v u=1,v=−2 √ if f(x, y) = x 2 y 2 − x + 2y; x = u, y = uv 3 . 33. Use a chain rule to find the values of ∂z ∂z and ∂r r=2,θ =π/6 ∂θ r=2,θ=π/6

if z = xyex /y ; x = r cos θ, y = r sin θ . dz if z = x 2 y; x = t 2 , y = t + 7. 34. Use a chain rule to ﬁnd dt t=3 35. Let a and b denote two sides of a triangle and let θ denote the included angle. Suppose that a, b, and θ vary with time in such a way that the area of the triangle remains constant. At a certain instant a = 5 cm, b = 4 cm, and θ = π/6 radians, and at that instant both a and b are increasing at a rate of 3 cm/s. Estimate the rate at which θ is changing at that instant. 36. The voltage, V (in volts), across a circuit is given by Ohm’s law: V = I R, where I is the current (in amperes) ﬂowing through the circuit and R is the resistance (in ohms). If two circuits with resistances R1 and R2 are connected in parallel, then their combined resistance, R, is given by 1 1 1 = + R R1 R2 Suppose that the current is 3 amperes and is increasing at 10−2 ampere/s, R1 is 2 ohms and is increasing at 0.4 ohm/s, and R2 is 5 ohms and is decreasing at 0.7 ohm/s. Estimate the rate at which the voltage is changing.

22. z = cos x sin y; x = u − v, y = u2 + v 2

37–40 True–False Determine whether the statement is true or

23–30 Use appropriate forms of the chain rule to ﬁnd the derivatives. ■

37. The symbols ∂z and ∂x are deﬁned in such a way that the partial derivative ∂z/∂x can be interpreted as a ratio.

false. Explain your answer. ■

958

Chapter 13 / Partial Derivatives

38. If z is a differentiable function of x1 , x2 , and x3 , and if xi is a differentiable function of t for i = 1, 2, 3, then z is a differentiable function of t and 3 dz ∂z dxi = dt ∂xi dt i=1

39. If z is a differentiable function of x and y, and if x and y are twice differentiable functions of t, then z is a twice differentiable function of t and d 2z ∂z d 2 x ∂z d 2 y = + 2 2 dt ∂x dt ∂y dt 2 40. If f(x, y) is a differentiable function of x and y, and if the line y = x is a contour of f , then fy (t, t) = −fx (t, t) for all real numbers t. 41–44 Use Theorem 13.5.3 to ﬁnd dy/dx and check your result

using implicit differentiation. ■ 41. x 2 y 3 + cos y = 0

43. exy + yey = 1

42. x 3 − 3xy 2 + y 3 = 5 √ 44. x − xy + 3y = 4

45–48 Find ∂z/∂x and ∂z/∂y by implicit differentiation, and

conﬁrm that the results obtained agree with those predicted by the formulas in Theorem 13.5.4. ■ 45. x 2 − 3yz2 + xyz − 2 = 0 x

47. ye − 5 sin 3z = 3z

46. ln(1 + z) + xy 2 + z = 1

48. exy cos yz − eyz sin xz + 2 = 0

49. (a) Suppose that z = f(u) and u = g(x, y). Draw a tree diagram, and use it to construct chain rules that express ∂z/∂x and ∂z/∂y in terms of dz/du, ∂u/∂x, and ∂u/∂y. (b) Show that ∂ 2z dz ∂ 2 u d 2 z ∂u 2 = + 2 ∂x 2 du ∂x 2 du ∂x 2 2 2 ∂ z dz ∂ u d z ∂u 2 = + 2 ∂y 2 du ∂y 2 du ∂y 2 2 2 dz ∂ u d z ∂u ∂u ∂ z = + 2 ∂y∂x du ∂y∂x du ∂x ∂y 50. (a) Let z = f(x 2 − y 2 ). Use the result in Exercise 49(a) to ∂z ∂z show that y +x =0 ∂x ∂y (b) Let z = f(xy). Use the result in Exercise 49(a) to show ∂z ∂z that x −y =0 ∂x ∂y (c) Conﬁrm the result in part (a) in the case where z = sin(x 2 − y 2 ). (d) Conﬁrm the result in part (b) in the case where z = exy .

51. Let f be a differentiable function of one variable, and let z = f(x + 2y). Show that ∂z ∂z 2 − =0 ∂x ∂y 52. Let f be a differentiable function of one variable, and let z = f(x 2 + y 2 ). Show that ∂z ∂z −x =0 y ∂x ∂y

53. Let f be a differentiable function of one variable, and let w = f(u), where u = x + 2y + 3z. Show that ∂w ∂w ∂w dw + + =6 ∂x ∂y ∂z du

54. Let f be a differentiable function of one variable, and let w = f(ρ), where ρ = (x 2 + y 2 + z2 )1/2 . Show that ∂w 2 ∂w 2 ∂w 2 dw 2 + + = ∂x ∂y ∂z dρ 55. Let z = f(x − y, y − x). Show that ∂z/∂x + ∂z/∂y = 0.

56. Let f be a differentiable function of three variables and suppose that w = f(x − y, y − z, z − x). Show that ∂w ∂w ∂w + + =0 ∂x ∂y ∂z

57. Suppose that the equation z = f(x, y) is expressed in the polar form z = g(r, θ ) by making the substitution x = r cos θ and y = r sin θ . (a) View r and θ as functions of x and y and use implicit differentiation to show that ∂r ∂θ sin θ = cos θ and =− ∂x ∂x r (b) View r and θ as functions of x and y and use implicit differentiation to show that ∂r ∂θ cos θ = sin θ and = ∂y ∂y r (c) Use the results in parts (a) and (b) to show that ∂z ∂z 1 ∂z = cos θ − sin θ ∂x ∂r r ∂θ ∂z 1 ∂z ∂z = sin θ + cos θ ∂y ∂r r ∂θ (d) Use the result in part (c) to show that 2 2 2 ∂z ∂z ∂z 1 ∂z 2 + = + 2 ∂x ∂y ∂r r ∂θ (e) Use the result in part (c) to show that if z = f(x, y) satisﬁes Laplace’s equation ∂ 2z ∂ 2z + =0 ∂x 2 ∂y 2 then z = g(r, θ ) satisﬁes the equation

∂ 2z 1 ∂ 2z 1 ∂z + + =0 ∂r 2 r 2 ∂θ 2 r ∂r and conversely. The latter equation is called the polar form of Laplace’s equation. 58. Show that the function z = tan−1

2xy x2 − y2

satisﬁes Laplace’s equation; then make the substitution x = r cos θ, y = r sin θ, and show that the resulting function of r and θ satisﬁes the polar form of Laplace’s equation given in part (e) of Exercise 57.

13.5 The Chain Rule

59. (a) Show that if u(x, y) and v(x, y) satisfy the Cauchy– Riemann equations (Exercise 104, Section 13.3), and if x = r cos θ and y = r sin θ, then ∂u 1 ∂v ∂v 1 ∂u = and =− ∂r r ∂θ ∂r r ∂θ This is called the polar form of the Cauchy–Riemann equations. (b) Show that the functions 2

u = ln(x + y ),

−1

v = 2 tan (y /x)

satisfy the Cauchy–Riemann equations; then make the substitution x = r cos θ , y = r sin θ , and show that the resulting functions of r and θ satisfy the polar form of the Cauchy–Riemann equations. 60. Recall from Formula (6) of Section 13.3 that under appropriate conditions a plucked string satisfies the wave equation 2 ∂ 2u 2∂ u = c ∂t 2 ∂x 2 where c is a positive constant. (a) Show that a function of the form u(x, t) = f(x + ct) satisfies the wave equation. (b) Show that a function of the form u(x, t) = g(x − ct) satisfies the wave equation. (c) Show that a function of the form

u(x, t) = f(x + ct) + g(x − ct) satisfies the wave equation. (d) It can be proved that every solution of the wave equation is expressible in the form stated in part (c). Confirm that u(x, t) = sin t sin x satisfies the wave equation in which c = 1, and then use appropriate trigonometric identities to express this function in the form f(x + t) + g(x − t).

61. Let f be a differentiable function of three variables, and let w = f(x, y, z), x = ρ sin φ cos θ , y = ρ sin φ sin θ , and z = ρ cos φ. Express ∂w/∂ρ, ∂w /∂φ, and ∂w /∂θ in terms of ∂w/∂x, ∂w /∂y, and ∂w/∂z.

959

64. Suppose that w is a differentiable function of x1 , x2 , and x3 , and x1 = a1 y1 + b1 y2 x2 = a2 y1 + b2 y2 x3 = a3 y1 + b3 y2

where the a’s and b’s are constants. Express ∂w/∂y1 and ∂w/∂y2 in terms of ∂w /∂x1 , ∂w/∂x2 , and ∂w/∂x3 . 65. (a) Let w be a differentiable function of x1 , x2 , x3 , and x4 , and let each xi be a differentiable function of t. Find a chain-rule formula for dw/dt. (b) Let w be a differentiable function of x1 , x2 , x3 , and x4 , and let each xi be a differentiable function of v1 , v2 , and v3 . Find chain-rule formulas for ∂w/∂v1 , ∂w/∂v2 , and ∂w/∂v3 . 66. Let w = (x12 + x22 + · · · + xn2 )k , where n ≥ 2. For what values of k does ∂ 2w ∂ 2w ∂ 2w + + · · · + =0 ∂xn2 ∂x12 ∂x22 hold? 67. Derive the identity g(x) d f(t) dt = f(g(x))g ′ (x) − f(h(x))h′ (x) dx h(x) by letting u = g(x) and v = h(x) and then differentiating u the function F(u, v) = f(t) dt v

with respect to x.

68. Prove: If f , fx , and fy are continuous on a circular region containing A(x0 , y0 ) and B(x1 , y1 ), then there is a point (x ∗ , y ∗ ) on the line segment joining A and B such that f(x1 , y1 ) − f(x0 , y0 ) = fx (x ∗ , y ∗ )(x1 − x0 ) + fy (x ∗ , y ∗ )(y1 − y0 ) This result is the two-dimensional version of the MeanValue Theorem. [Hint: Express the line segment joining A and B in parametric form and use the Mean-Value Theorem for functions of one variable.]

62. Let w = f(x, y, z) be differentiable, where z = g(x, y). Taking x and y as the independent variables, express each of the following in terms of ∂f /∂x, ∂f /∂y, ∂f /∂z, ∂z/∂x, and ∂z/∂y. (a) ∂w/∂x (b) ∂w/∂y

69. Prove: If fx (x, y) = 0 and fy (x, y) = 0 throughout a circular region, then f(x, y) is constant on that region. [Hint: Use the result of Exercise 68.]

63. Let w = ln(er + es + et + eu ). Show that

71. Writing Compare the use of the formula ∂f /∂x dy =− dx ∂f /∂y with the process of implicit differentiation.

wrstu = −6e

r+s+t+u−4w

[Hint: Take advantage of the relationship ew = er + es + et + eu .]

✔QUICK CHECK ANSWERS 13.5 1. −8

2. 3

3. 26 ft2 /s

4. 1; 1

70. Writing Use differentials to give an informal justiﬁcation for the chain rules for derivatives.

Chapter 13 / Partial Derivatives

960

DIRECTIONAL DERIVATIVES AND GRADIENTS

13.6

The partial derivatives fx(x, y) and fy(x, y) represent the rates of change of f(x, y) in directions parallel to the x- and y-axes. In this section we will investigate rates of change of f(x, y) in other directions.

DIRECTIONAL DERIVATIVES

In this section we extend the concept of a partial derivative to the more general notion of a directional derivative. We have seen that the partial derivatives of a function give the instantaneous rates of change of that function in directions parallel to the coordinate axes. Directional derivatives allow us to compute the rates of change of a function with respect to distance in any direction. Suppose that we wish to compute the instantaneous rate of change of a function f (x, y) with respect to distance from a point (x0 , y0 ) in some direction. Since there are inﬁnitely many different directions from (x0 , y0 ) in which we could move, we need a convenient method for describing a speciﬁc direction starting at (x0 , y0 ). One way to do this is to use a unit vector u = u1 i + u2 j

z y

(x 0 , y0 ) u

that has its initial point at (x0 , y0 ) and points in the desired direction (Figure 13.6.1). This vector determines a line l in the xy-plane that can be expressed parametrically as

Figure 13.6.1

x = x0 + su1 ,

y = y0 + su2

(1)

Since u is a unit vector, s is the arc length parameter that has its reference point at (x0 , y0 ) and has positive values in the direction of u. For s = 0, the point (x, y) is at the reference point (x0 , y0 ), and as s increases, the point (x, y) moves along l in the direction of u. On the line l the variable z = f (x0 + su1 , y0 + su2 ) is a function of the parameter s. The value of the derivative dz/ds at s = 0 then gives an instantaneous rate of change of f(x, y) with respect to distance from (x0 , y0 ) in the direction of u.

13.6.1 definition If f(x, y) is a function of x and y, and if u = u1 i + u2 j is a unit vector, then the directional derivative of f in the direction of u at (x0 , y0 ) is denoted by Du f(x0 , y0 ) and is deﬁned by Slope in u direction = rate of change of z with respect to s z

Du f(x0 , y0 ) =

z = f (x, y)

provided this derivative exists.

d f(x0 + su1 , y0 + su2 ) s=0 ds

(2)

s x

(x 0 , y0 ) u

Figure 13.6.2

l (x, y)

Geometrically, Du f(x0 , y0 ) can be interpreted as the slope of the surface z = f (x, y) in the direction of u at the point (x0 , y0 , f(x0 , y0 )) (Figure 13.6.2). Usually the value of Du f(x0 , y0 ) will depend on both the point (x0 , y0 ) and the direction u. Thus, at a ﬁxed point the slope of the surface may vary with the direction (Figure 13.6.3). Analytically, the directional derivative represents the instantaneous rate of change of f (x, y) with respect to distance in the direction of u at the point (x0 , y0 ). Example 1

Let f(x, y) = xy. Find and interpret Du f(1, 2) for the unit vector √ 3 1 i+ j u= 2 2

13.6 Directional Derivatives and Gradients z

961

Solution. It follows from Equation (2) that

z = f (x, y)

â&#x2C6;&#x161; d s 3s ,2 + f 1+ Du f(1, 2) = ds 2 2 y

(x 0 , y0 ) u

s=0

Since f

x The slope of the surface varies with the direction of u.

â&#x2C6;&#x161;

s 3s ,2 + 1+ 2 2

â&#x2C6;&#x161;

3s = 1+ 2

â&#x2C6;&#x161; s 1 â&#x2C6;&#x161; 3 2 = s + + 3 s+2 2 4 2

â&#x2C6;&#x161; 1 â&#x2C6;&#x161; 3 2 d s + + 3 s+2 Du f(1, 2) = ds 4 2 s=0 â&#x2C6;&#x161; 1 â&#x2C6;&#x161; 3 1 â&#x2C6;&#x161; = + 3 = s+ + 3 2 2 2

we have

Figure 13.6.3

s=0

1 2

â&#x2C6;&#x161;

Since + 3 â&#x2030;&#x2C6; 2.23, we conclude that if we move a small distance from the point (1, 2) in the direction of u, the function f(x, y) = xy will increase by about 2.23 times the distance moved. The deďŹ nition of a directional derivative for a function f(x, y, z) of three variables is similar to DeďŹ nition 13.6.1.

13.6.2 definition If u = u1 i + u2 j + u3 k is a unit vector, and if f(x, y, z) is a function of x, y, and z, then the directional derivative of f in the direction of u at (x0 , y0 , z0 ) is denoted by Du f(x0 , y0 , z0 ) and is deďŹ ned by Du f(x0 , y0 , z0 ) = provided this derivative exists.

d f(x0 + su1 , y0 + su2 , z0 + su3 ) s=0 ds

(3)

Although Equation (3) does not have a convenient geometric interpretation, we can still interpret directional derivatives for functions of three variables in terms of instantaneous rates of change in a speciďŹ ed direction. For a function that is differentiable at a point, directional derivatives exist in every direction from the point and can be computed directly in terms of the ďŹ rst-order partial derivatives of the function.

13.6.3

theorem

(a) If f(x, y) is differentiable at (x0 , y0 ), and if u = u1 i + u2 j is a unit vector, then the directional derivative Du f (x0 , y0 ) exists and is given by Du f(x0 , y0 ) = fx(x0 , y0 )u1 + fy(x0 , y0 )u2

(4)

(b) If f(x, y, z) is differentiable at (x0 , y0 , z0 ), and if u = u1 i + u2 j + u3 k is a unit vector, then the directional derivative Du f (x0 , y0 , z0 ) exists and is given by Du f(x0 , y0 , z0 ) = fx(x0 , y0 , z0 )u1 + fy(x0 , y0 , z0 )u2 + fz(x0 , y0 , z0 )u3

(5)

962

Chapter 13 / Partial Derivatives

proof We will give the proof of part (a); the proof of part (b) is similar and will be omitted. The function z = f (x0 + su1 , y0 + su2 ) is the composition of the function z = f (x, y) with the functions x = x(s) = x0 + su1

and

y = y(s) = y0 + su2

As such, the chain rule in Formula (5) of Section 13.5 immediately gives d f(x0 + su1 , y0 + su2 ) s=0 ds dz = = fx(x0 , y0 )u1 + fy(x0 , y0 )u2 ■ ds s=0

Du f(x0 , y0 ) =

We can use Theorem 13.6.3 to conﬁrm the result of Example 1. For f(x, y) = xy we have fx(1, 2) = 2 and fy(1, 2) = 1 (verify). With √ 1 3 i+ j u= 2 2 Equation (4) becomes √ 3 1 1 √ Du f (1, 2) = 2 + = 3+ 2 2 2 which agrees with our solution in Example 1. Recall from Formula (13) of Section 11.2 that a unit vector u in the xy-plane can be expressed as (6) u = cos φ i + sin φ j where φ is the angle from the positive x-axis to u. Thus, Formula (4) can also be expressed as Du f(x0 , y0 ) = fx(x0 , y0 ) cos φ + fy(x0 , y0 ) sin φ (7) Example 2 Find the directional derivative of f(x, y) = exy at (−2, 0) in the direction of the unit vector that makes an angle of π/3 with the positive x-axis.

Solution. The partial derivatives of f are fx(x, y) = yexy , fy(x, y) = xexy fx(−2, 0) = 0,

fy(−2, 0) = −2

The unit vector u that makes an angle of π/3 with the positive x-axis is √ 1 3 u = cos(π/3)i + sin(π/3) j = i + j 2 2 Thus, from (7)

Note that in Example 3 we used a unit vector to specify the direction of the directional derivative. This is required in order to apply either Formula (4) or Formula (5).

Du f (−2, 0) = fx (−2, 0) cos(π/3) + fy (−2, 0) sin(π/3) √ √ = 0(1/2) + (−2)( 3/2) = − 3

Example 3 Find the directional derivative of f (x, y, z) = x 2 y − yz3 + z at the point (1, −2, 0) in the direction of the vector a = 2i + j − 2k.

13.6 Directional Derivatives and Gradients

963

Solution. The partial derivatives of f are fx(x, y, z) = 2xy,

fy(x, y, z) = x 2 − z3 , fz(x, y, z) = −3yz2 + 1

fx(1, −2, 0) = −4, fy(1, −2, 0) = 1,

fz(1, −2, 0) = 1

Since a is not a unit vector, we normalize it, getting u= Formula (5) then yields

a 1 1 2 2 = √ (2i + j − 2k) = i + j − k a 3 3 3 9

1 2 2 + − = −3 Du f (1, −2, 0) = (−4) 3 3 3 THE GRADIENT

Formula (4) can be expressed in the form of a dot product as Du f (x0 , y0 ) = (fx (x0 , y0 )i + fy (x0 , y0 )j) ⴢ (u1 i + u2 j) = (fx (x0 , y0 )i + fy (x0 , y0 )j) ⴢ u

Similarly, Formula (5) can be expressed as Du f (x0 , y0 , z0 ) = (fx (x0 , y0 , z0 )i + fy (x0 , y0 , z0 )j + fz (x0 , y0 , z0 )k) ⴢ u In both cases the directional derivative is obtained by dotting the direction vector u with a new vector constructed from the ﬁrst-order partial derivatives of f .

13.6.4 (a) Remember that ∇f is not a product of ∇ and f . Think of ∇ as an “operator” that acts on a function f to produce the gradient ∇f .

definition

If f is a function of x and y, then the gradient of f is deﬁned by ∇f(x, y) = fx(x, y)i + fy(x, y)j

(8)

(b) If f is a function of x, y, and z, then the gradient of f is deﬁned by ∇f(x, y, z) = fx(x, y, z)i + fy(x, y, z)j + fz(x, y, z)k

(9)

The symbol ∇ (read “del”) is an inverted delta. (It is sometimes called a “nabla” because of its similarity in form to an ancient Hebrew ten-stringed harp of that name.) Formulas (4) and (5) can now be written as

Slope = ∇f . u z

z = f (x, y) y

∇f x

(x, y)

Figure 13.6.4

Du f (x0 , y0 ) = ∇f(x0 , y0 ) ⴢ u

(10)

Du f (x0 , y0 , z0 ) = ∇f(x0 , y0 , z0 ) ⴢ u

(11)

and

respectively. For example, using Formula (11) our solution to Example 3 would take the form Du f (1, −2, 0) = ∇f(1, −2, 0) ⴢ u = (−4i + j + k) ⴢ 23 i + 13 j − 23 k = (−4) 23 + 13 − 23 = −3

Formula (10) can be interpreted to mean that the slope of the surface z = f (x, y) at the point (x0 , y0 ) in the direction of u is the dot product of the gradient with u (Figure 13.6.4).

964

Chapter 13 / Partial Derivatives

PROPERTIES OF THE GRADIENT The gradient is not merely a notational device to simplify the formula for the directional derivative; we will see that the length and direction of the gradient ∇f provide important information about the function f and the surface z = f(x, y). For example, suppose that ∇f(x, y) = 0, and let us use Formula (4) of Section 11.3 to rewrite (10) as

Du f(x, y) = ∇f(x, y) ⴢ u = ∇f(x, y) u cos θ = ∇f(x, y) cos θ

(12)

where θ is the angle between ∇f(x, y) and u. Equation (12) tells us that the maximum value of Du f at the point (x, y) is ∇f(x, y) , and this maximum occurs when θ = 0, that is, when u is in the direction of ∇f(x, y). Geometrically, this means:

At (x, y), the surface z = f(x, y) has its maximum slope in the direction of the gradient, and the maximum slope is ∇f(x, y) . That is, the function f(x, y) increases most rapidly in the direction of its gradient (Figure 13.6.5). Similarly, (12) tells us that the minimum value of Du f at the point (x, y) is − ∇f(x, y) , and this minimum occurs when θ = π, that is, when u is oppositely directed to ∇f(x, y). Geometrically, this means:

z Increasing most rapidly

Decreasing most rapidly y

∇f (x, y) −∇f x

Figure 13.6.5

At (x, y), the surface z = f(x, y) has its minimum slope in the direction that is opposite to the gradient, and the minimum slope is − ∇f(x, y) . That is, the function f(x, y) decreases most rapidly in the direction opposite to its gradient (Figure 13.6.5). Finally, in the case where ∇f(x, y) = 0, it follows from (12) that Du f(x, y) = 0 in all directions at the point (x, y). This typically occurs where the surface z = f(x, y) has a “relative maximum,” a “relative minimum,” or a saddle point. A similar analysis applies to functions of three variables. As a consequence, we have the following result.

13.6.5 theorem Let f be a function of either two variables or three variables, and let P denote the point P (x0 , y0 ) or P (x0 , y0 , z0 ), respectively. Assume that f is differentiable at P . (a) If ∇f = 0 at P , then all directional derivatives of f at P are zero. (b) If ∇f = 0 at P , then among all possible directional derivatives of f at P , the derivative in the direction of ∇f at P has the largest value. The value of this largest directional derivative is ∇f at P . (c) If ∇f = 0 at P , then among all possible directional derivatives of f at P , the derivative in the direction opposite to that of ∇f at P has the smallest value. The value of this smallest directional derivative is − ∇f at P .

Example 4 Let f(x, y) = x 2 ey . Find the maximum value of a directional derivative at (−2, 0), and ﬁnd the unit vector in the direction in which the maximum value occurs.

13.6 Directional Derivatives and Gradients

965

Solution. Since ∇f(x, y) = fx(x, y)i + fy(x, y) j = 2xey i + x 2 ey j the gradient of f at (−2, 0) is ∇f(−2, 0) = −4i + 4 j What would be the minimum value of a directional derivative of

f(x, y) = x 2 ey at (−2, 0) ?

By Theorem 13.6.5, the maximum value of the directional derivative is √ √ ∇f(−2, 0) = (−4)2 + 42 = 32 = 4 2

This maximum occurs in the direction of ∇f(−2, 0). The unit vector in this direction is 1 1 1 ∇f(−2, 0) = √ (−4i + 4 j) = − √ i + √ j u= ∇f(−2, 0) 4 2 2 2 GRADIENTS ARE NORMAL TO LEVEL CURVES We have seen that the gradient points in the direction in which a function increases most rapidly. For a function f(x, y) of two variables, we will now consider how this direction of maximum rate of increase can be determined from a contour map of the function. Suppose that (x0 , y0 ) is a point on a level curve f(x, y) = c of f , and assume that this curve can be smoothly parametrized as x = x(s), y = y(s) (13)

where s is an arc length parameter. Recall from Formula (6) of Section 12.4 that the unit tangent vector to (13) is dy dx i+ j T = T(s) = ds ds

Since T gives a direction along which f is nearly constant, we would expect the instantaneous rate of change of f with respect to distance in the direction of T to be 0. That is, we would expect that DT f(x, y) = ∇f(x, y) ⴢ T(s) = 0 To show this to be the case, we differentiate both sides of the equation f(x, y) = c with respect to s. Assuming that f is differentiable at (x, y), we can use the chain rule to obtain ∂f dy ∂f dx + =0 ∂x ds ∂y ds which we can rewrite as ∂f dx dy ∂f i+ j ⴢ i+ j =0 ∂x ∂y ds ds or, alternatively, as

∇f(x, y) ⴢ T = 0

Therefore, if ∇f(x, y) = 0, then ∇f(x, y) should be normal to the level curve f (x, y) = c at any point (x, y) on the curve. It is proved in advanced courses that if f(x, y) has continuous ﬁrst-order partial derivatives, and if ∇f(x0 , y0 ) = 0, then near (x0 , y0 ) the graph of f(x, y) = c is indeed a smooth curve through (x0 , y0 ). Furthermore, we also know from Theorem 13.4.4 that f will be differentiable at (x0 , y0 ). We therefore have the following result. Show that the level curves for

f(x, y) = x 2 + y 2 are circles and verify Theorem 13.6.6 at (x0 , y0 ) = (3, 4).

13.6.6 theorem Assume that f(x, y) has continuous ﬁrst-order partial derivatives in an open disk centered at (x0 , y0 ) and that ∇f(x0 , y0 ) = 0. Then ∇f(x0 , y0 ) is normal to the level curve of f through (x0 , y0 ).

When we examine a contour map, we instinctively regard the distance between adjacent contours to be measured in a normal direction. If the contours correspond to equally spaced

966

Chapter 13 / Partial Derivatives

values of f , then the closer together the contours appear to be, the more rapidly the values of f will be changing in that normal direction. It follows from Theorems 13.6.5 and 13.6.6 that this rate of change of f is given by ∇f(x, y) . Thus, the closer together the contours appear to be, the greater the length of the gradient of f . Example 5 A contour plot of a function f is given in Figure 13.6.6a. Sketch the directions of the gradient of f at the points P , Q, and R. At which of these three points does the gradient vector have maximum length? Minimum length?

2.5

2 1.5

10 14 18 22

1.5

0.5 0 −2

0.5

Q −1.5

−1

−0.5

0.5

0 −2

Q −1.5

−1

−0.5

0.5

Vectors not to scale

(a)

(b)

Figure 13.6.6

Solution. It follows from Theorems 13.6.5 and 13.6.6 that the directions of the gradient vectors will be as given in Figure 13.6.6b. Based on the density of the contour lines, we would guess that the gradient of f has maximum length at R and minimum length at P , with the length at Q somewhere in between. REMARK

f (x, y) = c u

∇f (x0, y0)

(x0, y0) −∇f (x0, y0)

If (x0 , y0 ) is a point on the level curve f(x, y) = c, then the slope of the surface z = f(x, y) at that point in the direction of u is

Du f(x0 , y0 ) = ∇f(x0 , y0 ) ⴢ u If u is tangent to the level curve at (x0 , y0 ), then f(x, y) is neither increasing nor decreasing in that direction, so Du f(x0 , y0 ) = 0. Thus, ∇f(x0 , y0 ), −∇f(x0 , y0 ), and the tangent vector u mark the directions of maximum slope, minimum slope, and zero slope at a point (x0 , y0 ) on a level curve (Figure 13.6.7). Good skiers use these facts intuitively to control their speed by zigzagging down ski slopes—they ski across the slope with their skis tangential to a level curve to stop their downhill motion, and they point their skis down the slope and normal to the level curve to obtain the most rapid descent.

Figure 13.6.7

AN APPLICATION OF GRADIENTS There are numerous applications in which the motion of an object must be controlled so that it moves toward a heat source. For example, in medical applications the operation of certain diagnostic equipment is designed to locate heat sources generated by tumors or infections, and in military applications the trajectories of heat-seeking missiles are controlled to seek and destroy enemy aircraft. The following example illustrates how gradients are used to solve such problems. UPI Photo/Michael Ammons/Air Force/Landov LLC

Heat-seeking missiles such as "Stinger" and "Sidewinder" use infrared sensors to measure gradients.

Example 6 A heat-seeking particle is located at the point (2, 3) on a ﬂat metal plate whose temperature at a point (x, y) is T (x, y) = 10 − 8x 2 − 2y 2

13.6 Directional Derivatives and Gradients

967

Find an equation for the trajectory of the particle if it moves continuously in the direction of maximum temperature increase.

Solution. Assume that the trajectory is represented parametrically by the equations x = x(t),

y = y(t)

where the particle is at the point (2, 3) at time t = 0. Because the particle moves in the direction of maximum temperature increase, its direction of motion at time t is in the direction of the gradient of T (x, y), and hence its velocity vector v(t) at time t points in the direction of the gradient. Thus, there is a scalar k that depends on t such that v(t) = k∇T (x, y) from which we obtain

dx dy i+ j = k(−16xi − 4y j) dt dt

Equating components yields dx = −16kx, dt and dividing to eliminate k yields

4 3

dy = −4ky dt

−4ky y dy = = dx −16kx 4x Thus, we can obtain the trajectory by solving the initial-value problem

2 1

dy y − = 0, y(2) = 3 dx 4x The differential equation is a separable ﬁrst-order equation and hence can be solved by the method of separation of variables discussed in Section 8.2. We leave it for you to show that the solution of the initial-value problem is

0 −1 −2 −3 −4 −3 −2 −1 Figure 13.6.8

3 1/4 y= √ x 4 2 The graph of the trajectory and a contour plot of the temperature function are shown in Figure 13.6.8.

✔QUICK CHECK EXERCISES 13.6

(See page 971 for answers.)

1. The gradient of f(x, y, z) = xy 2 z3 at the point (1, 1, 1) is . 2. Suppose that the differentiable function f(x, y) has the property that √ s 3 s f 2+ = 3ses ,1 + 2 2 The directional derivative of f in the direction of √ 3 1 i+ j u= 2 2 at (2, 1) is .

3. If the gradient of f(x, y) at the origin is 6i + 8j, then the directional derivative of f in the direction of a = 3i + 4j at the origin is . The slope of the tangent line to the level curve of f through the origin at (0, 0) is . 4. If the gradient of f(x, y, z) at (1, 2, 3) is 2i − 2j + k, then the maximum value for a directional derivative of f at (1, 2, 3) is and the minimum value for a directional derivative at this point is .

968

Chapter 13 / Partial Derivatives

EXERCISE SET 13.6

Graphing Utility

CAS

y x+y Find a unit vector u for which Du f(2, 3) = 0.

1–8 Find Du f at P . ■

26. Let

1 1 1. f(x, y) = (1 + xy)3/2 ; P (3, 1); u = √ i + √ j 2 2 2. f(x, y) = sin(5x − 3y); P (3, 5); u = 53 i − 45 j

27. Find the directional derivative of

3. f(x, y) = ln(1 + x 2 + y); P (0, 0); 1 3 u = −√ i − √ j 10 10 cx + dy ; P (3, 4); u = 45 i + 53 j 4. f(x, y) = x−y 5. f(x, y, z) = 4x 5 y 2 z3 ; P (2, −1, 1); u = 13 i + 23 j − 23 k 6. f(x, y, z) = ye

+ z ; P (0, 2, 3); u =

2 i 7

3 7

− j+

f(x, y) =

6 k 7

7. f(x, y, z) = ln(x + 2y + 3z ); P (−1, 2, 4); 3 4 u = − 13 i − 13 j − 12 k 13 8. f(x, y, z) = sin xyz; P 21 , 13 , π ; 1 1 1 u = √ i− √ j+ √ k 3 3 3

9–18 Find the directional derivative of f at P in the direction of a. ■

9. f(x, y) = 4x 3 y 2 ; P (2, 1); a = 4i − 3 j

10. f(x, y) = 9x 3 − 2y 3 ; P (1, 0); a = i − j 2

11. f(x, y) = y ln x; P (1, 4); a = −3i + 3 j 12. f(x, y) = ex cos y; P (0, π/4); a = 5i − 2 j

13. f(x, y) = tan−1 (y /x); P (−2, 2); a = −i − j

14. f(x, y) = xey − yex ; P (0, 0); a = 5i − 2 j

15. f(x, y, z) = xy + z2 ; P (−3, 0, 4); a = i + j + k √ 16. f(x, y, z) = y − x 2 + z2 ; P (−3, 1, 4); a = 2i − 2 j − k z−x 17. f(x, y, z) = ; P (1, 0, −3); a = −6i + 3 j − 2k z+y 18. f(x, y, z) = ex+y+3z ; P (−2, 2, −1); a = 20i − 4 j + 5k

y x+z at P (2, 1, −1) in the direction from P to Q(−1, 2, 0). f(x, y, z) =

28. Find the directional derivative of the function

f(x, y, z) = x 3 y 2 z5 − 2xz + yz + 3x

at P (−1, −2, 1) in the direction of the negative z-axis. F O C U S O N C O N C E P TS

29. Suppose that Du f(1, 2) = −5 and Dv f(1, 2) = 10, where u = 53 i − 45 j and v = 45 i + 53 j. Find (a) fx (1, 2) (b) fy (1, 2) (c) the directional derivative of f at (1, 2) in the direction of the origin. 30. Given that fx (−5, 1) = −3 and fy (−5, 1) = 2, find the directional derivative of f at P (−5, 1) in the direction of the vector from P to Q(−4, 3). 31. The accompanying figure shows some level curves of an unspecified function f(x, y). Which of the three vectors shown in the figure is most likely to be ∇f ? Explain. 32. The accompanying figure shows some level curves of an unspecified function f(x, y). Of the gradients at P and Q, which probably has the greater length? Explain. y

20 II

20 30

III

P x

19–22 Find the directional derivative of f at P in the direc-

tion of a vector making the counterclockwise angle θ with the positive x-axis. ■ √ 19. f(x, y) = xy; P (1, 4); θ = π/3 x−y 20. f(x, y) = ; P (−1, −2); θ = π/2 x+y 21. f(x, y) = tan(2x + y); P (π/6, π/3); θ = 7π/4 22. f(x, y) = sinh x cosh y; P (0, 0); θ = π 23. Find the directional derivative of x f(x, y) = x+y at P (1, 0) in the direction of Q(−1, −1).

24. Find the directional derivative of f(x, y) = e−x sec y at P (0, π/4) in the direction of the origin. √ 25. Find the directional derivative of f(x, y) = xyey at P (1, 1) in the direction of the negative y-axis.

Figure Ex-31

Figure Ex-32

33–40 Find ∇z or ∇w. ■

33. z = sin(7y 2 − 7xy) 6x + 7y 35. z = 6x − 7y 37. w = −x 9 − y 3 + z12 39. w = ln x 2 + y 2 + z2

34. z = 7 sin(6x /y) 6xe3y 36. z = x + 8y 38. w = xe8y sin 6z

40. w = e−5x sec x 2 yz

41–46 Find the gradient of f at the indicated point. ■

41. f(x, y) = 5x 2 + y 4 ; (4, 2)

√ 42. f(x, y) = 5 sin x 2 + cos 3y; ( π/2, 0)

43. f(x, y) = (x 2 + xy)3 ; (−1, −1) 44. f(x, y) = (x 2 + y 2 )−1/2 ; (3, 4)

13.6 Directional Derivatives and Gradients

45. f(x, y, z) = y ln(x + y + z); (−3, 4, 0) 46. f(x, y, z) = y 2 z tan3 x; (π/4, −3, 1) 47–50 Sketch the level curve of f(x, y) that passes through P

and draw the gradient vector at P . ■ 47. f(x, y) = 4x − 2y + 3; P (1, 2) 48. f(x, y) = y /x 2 ; P (−2, 2) 49. f(x, y) = x 2 + 4y 2 ; P (−2, 0)

50. f(x, y) = x 2 − y 2 ; P (2, −1)

51. Find a unit vector u that is normal at P (1, −2) to the level curve of f(x, y) = 4x 2 y through P .

52. Find a unit vector u that is normal at P (2, −3) to the level curve of f(x, y) = 3x 2 y − xy through P . 53–60 Find a unit vector in the direction in which f increases most rapidly at P , and ﬁnd the rate of change of f at P in that direction. ■

53. f(x, y) = 4x 3 y 2 ; P (−1, 1)

54. f(x, y) = 3x − ln y; P (2, 4) 55. f(x, y) = x 2 + y 2 ; P (4, −3) x ; P (0, 2) 56. f(x, y) = x+y

57. f(x, y, z) = x 3 z2 + y 3 z + z − 1; P (1, 1, −1) √ 58. f(x, y, z) = x − 3y + 4z; P (0, −3, 0) x z 59. f(x, y, z) = + 2 ; P (1, 2, −2) z y x −1 ; P (4, 2, 2) 60. f(x, y, z) = tan y+z 61–66 Find a unit vector in the direction in which f decreases

most rapidly at P , and ﬁnd the rate of change of f at P in that direction. ■

969

69. If u is a ﬁxed unit vector and Du f(x, y) = 0 for all points (x, y), then f is a constant function. 70. If the displacement vector from (x0 , y0 ) to (x1 , y1 ) is a positive multiple of ∇f(x0 , y0 ), then f(x0 , y0 ) ≤ f(x1 , y1 ). F O C U S O N C O N C E P TS

71. Given that ∇f(4, −5) = 2i − j, ﬁnd the directional derivative of the function f at the point (4, −5) in the direction of a = 5i + 2 j.

72. Given that ∇f(x0 , y0 ) = i − 2 j and Du f(x0 , y0 ) = −2, find u (two answers). 73. The accompanying figure shows some level curves of an unspecified function f(x, y). (a) Use the available information to approximate the length of the vector ∇f(1, 2), and sketch the approximation. Explain how you approximated the length and determined the direction of the vector. (b) Sketch an approximation of the vector −∇f(4, 4). y

5 4

3 1

1 2 1

x 2

Figure Ex-73

74. The accompanying ﬁgure shows a topographic map of a hill and a point P at the bottom of the hill. Suppose that you want to climb from the point P toward the top of the hill in such a way that you are always ascending in the direction of steepest slope. Sketch the projection of your path on the contour map. This is called the path of steepest ascent. Explain how you determined the path.

61. f(x, y) = 20 − x 2 − y 2 ; P (−1, −3) 62. f(x, y) = exy ; P (2, 3)

63. f(x, y) = cos(3x − y); P (π/6, π/4) x−y 64. f(x, y) = ; P (3, 1) x+y x+z ; P (5, 7, 6) z−y 66. f(x, y, z) = 4exy cos z; P (0, 1, π/4)

65. f(x, y, z) =

67–70 True–False Determine whether the statement is true or false. Explain your answer. In each exercise, assume that f denotes a differentiable function of two variables whose domain is the xy-plane. ■

67. If v = 2u, then the directional derivative of f in the direction of v at a point (x0 , y0 ) is twice the directional derivative of f in the direction of u at the point (x0 , y0 ). 68. If y = x 2 is a contour of f , then fx (0, 0) = 0.

200

300 400

500

100 0 ft P

Figure Ex-74

75. Let z = 3x 2 − y 2 . Find all points at which ∇z = 6.

76. Given that z = 3x + y 2 , ﬁnd ∇ ∇z at the point (5, 2).

77. A particle moves along a path C given by the equations x = t and y = −t 2 . If z = x 2 + y 2 , ﬁnd dz/ds along C at the instant when the particle is at the point (2, −4).

78. The temperature (in degrees Celsius) at a point (x, y) on a metal plate in the xy-plane is xy T (x, y) = (cont.) 1 + x2 + y2

970

Chapter 13 / Partial Derivatives

(a) Find the rate of change of temperature at (1, 1) in the direction of a = 2i − j. (b) An ant at (1, 1) wants to walk in the direction in which the temperature drops most rapidly. Find a unit vector in that direction. 79. If the electric potential at a point (x, y) in the xy-plane is V (x, y), then the electric intensity vector at the point (x, y) is E = −∇V (x, y). Suppose that V (x, y) = e−2x cos 2y. (a) Find the electric intensity vector at (π/4, 0). (b) Show that at each point in the plane, the electric potential decreases most rapidly in the direction of the vector E.

87–88 A heat-seeking particle is located at the point P on a ﬂat metal plate whose temperature at a point (x, y) is T (x, y). Find parametric equations for the trajectory of the particle if it moves continuously in the direction of maximum temperature increase.

80. On a certain mountain, the elevation z above a point (x, y) in an xy-plane at sea level is z = 2000 − 0.02x 2 − 0.04y 2 , where x, y, and z are in meters. The positive x-axis points east, and the positive y-axis north. A climber is at the point (−20, 5, 1991). (a) If the climber uses a compass reading to walk due west, will she begin to ascend or descend? (b) If the climber uses a compass reading to walk northeast, will she ascend or descend? At what rate? (c) In what compass direction should the climber begin walking to travel a level path (two answers)?

89. Use a graphing utility to generate the trajectory of the particle together with some representative level curves of the temperature function in Exercise 87.

81. Given that the directional derivative of f(x, y, z) at the point (3, −2, 1) in the direction of a = 2i − j − 2k is −5 and that ∇f(3, −2, 1) = 5, ﬁnd ∇f(3, −2, 1).

82. The temperature (in degrees Celsius) at a point (x, y, z) in a metal solid is xyz T (x, y, z) = 1 + x 2 + y 2 + z2 (a) Find the rate of change of temperature with respect to distance at (1, 1, 1) in the direction of the origin. (b) Find the direction in which the temperature rises most rapidly at the point (1, 1, 1). (Express your answer as a unit vector.) (c) Find the rate at which the temperature rises moving from (1, 1, 1) in the direction obtained in part (b). 83. Let r = x 2 + y 2 . r (a) Show that ∇r = , where r = xi + yj. r f ′(r) (b) Show that ∇f(r) = f ′(r)∇r = r. r 84. Use the formula in part (b) of Exercise 83 to ﬁnd (a) ∇f(r) if f(r) = re−3r (b) f(r) if ∇f(r) = 3r 2 r and f(2) = 1. 85. Let ur be a unit vector whose counterclockwise angle from the positive x-axis is θ, and let uθ be a unit vector 90 ◦ counterclockwise from ur . Show that if z = f(x, y), x = r cos θ , and y = r sin θ , then ∂z 1 ∂z ∇z = ur + uθ ∂r r ∂θ [Hint: Use part (c) of Exercise 57, Section 13.5.]

86. Prove: If f and g are differentiable, then (a) ∇(f + g) = ∇f + ∇g (b) ∇(cf ) = c∇f (c constant)

■ 2

87. T (x, y) = 5 − 4x − y ; P (1, 4)

88. T (x, y) = 100 − x 2 − 2y 2 ; P (5, 3)

90. Use a graphing utility to generate the trajectory of the particle together with some representative level curves of the temperature function in Exercise 88. C

91. (a) Use a CAS to graph f(x, y) = (x 2 + 3y 2 )e−(x +y ) . (b) At how many points do you think it is true that Du f(x, y) = 0 for all unit vectors u? (c) Use a CAS to ﬁnd ∇f . (d) Use a CAS to solve the equation ∇f(x, y) = 0 for x and y. (e) Use the result in part (d) together with Theorem 13.6.5 to check your conjecture in part (b). 92. Prove: If x = x(t) and y = y(t) are differentiable at t, and if z = f(x, y) is differentiable at the point (x(t), y(t)), then dz = ∇z ⴢ r′ (t) dt where r(t) = x(t)i + y(t) j.

93. Prove: If f , fx , and fy are continuous on a circular region, and if ∇f(x, y) = 0 throughout the region, then f(x, y) is constant on the region. [Hint: See Exercise 69, Section 13.5.] 94. Prove: If the function f is differentiable at the point (x, y) and if Du f(x, y) = 0 in two nonparallel directions, then Du f(x, y) = 0 in all directions.

95. Given that the functions u = u(x, y, z), v = v(x, y, z), w = w(x, y, z), and f(u, v, w) are all differentiable, show that ∂f ∂f ∂f ∇f(u, v, w) = ∇u + ∇v + ∇w ∂u ∂v ∂w 96. Writing Let f denote a differentiable function of two variables. Write a short paragraph that discusses the connections between directional derivatives of f and slopes of tangent lines to the graph of f .

97. Writing Let f denote a differentiable function of two variables. Although we have defined what it means to say that f is differentiable, we have not defined the “derivative” of f . Write a short paragraph that discusses the merits of defining the derivative of f to be the gradient ∇f .

13.7 Tangent Planes and Normal Vectors

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✔QUICK CHECK ANSWERS 13.6 1. 1, 2, 3

2. 3

3. 10; − 43

4. 3; −3

TANGENT PLANES AND NORMAL VECTORS

13.7

In this section we will discuss tangent planes to surfaces in three-dimensional space. We will be concerned with three main questions: What is a tangent plane? When do tangent planes exist? How do we ﬁnd equations of tangent planes?

T Tangent line

Figure 13.7.1

Tangent lines

F(x, y, z) = c

TANGENT PLANES AND NORMAL VECTORS TO LEVEL SURFACES F (x, y, z) = c We begin by considering the problem of finding tangent planes to level surfaces of a function F (x, y, z). These surfaces are represented by equations of the form F (x, y, z) = c. We will assume that F has continuous first-order partial derivatives, since this has an important geometric consequence. Fix c, and suppose that P0 (x0 , y0 , z0 ) satisfies the equation F (x, y, z) = c. In advanced courses it is proved that if F has continuous first-order partial derivatives, and if ∇F (x0 , y0 , z0 ) = 0, then near P0 the graph of F (x, y, z) = c is indeed a “surface” rather than some possibly exotic-looking set of points in 3-space. We will base our concept of a tangent plane to a level surface S: F (x, y, z) = c on the more elementary notion of a tangent line to a curve C in 3-space (Figure 13.7.1). Intuitively, we would expect a tangent plane to S at a point P0 to be composed of the tangent lines at P0 of all curves on S that pass through P0 (Figure 13.7.2). Suppose C is a curve on S through P0 that is parametrized by x = x(t), y = y(t), z = z(t) with x0 = x(t0 ), y0 = y(t0 ), and z0 = z(t0 ). The tangent line l to C through P0 is then parallel to the vector

r′ = x ′ (t0 )i + y ′ (t0 ) j + z′ (t0 )k

S Tangent plane y

where we assume that r′ = 0 (Deﬁnition 12.2.7). Since C is on the surface F (x, y, z) = c, we have (1) c = F (x(t), y(t), z(t)) Computing the derivative at t0 of both sides of (1), we have by the chain rule that

(x 0 , y0 )

0 = Fx (x0 , y0 , z0 )x ′ (t0 ) + Fy (x0 , y0 , z0 )y ′ (t0 ) + Fz (x0 , y0 , z0 )z′ (t0 )

x All tangent lines at P0 lie in the tangent plane.

We can write this equation in vector form as 0 = (Fx (x0 , y0 , z0 )i + Fy (x0 , y0 , z0 )j + Fz (x0 , y0 , z0 )k) ⴢ (x ′ (t0 )i + y ′ (t0 )j + z′ (t0 )k)

Figure 13.7.2

∇F(x 0 , y0, z 0 )

S (x 0 , y0, z 0 ) F(x, y, z) = c

Figure 13.7.3

0 = ∇F (x0 , y0 , z0 ) ⴢ r′

(2)

It follows that if ∇F (x0 , y0 , z0 ) = 0, then ∇F (x0 , y0 , z0 ) is normal to line l. Therefore, the tangent line l to C at P0 is contained in the plane through P0 with normal vector ∇F (x0 , y0 , z0 ). Since C was arbitrary, we conclude that the same is true for any curve on S through P0 (Figure 13.7.3). Thus, it makes sense to deﬁne the tangent plane to S at P0 to be the plane through P0 whose normal vector is n = ∇F (x0 , y0 , z0 ) = Fx (x0 , y0 , z0 ), Fy (x0 , y0 , z0 ), Fz (x0 , y0 , z0 ) Using the point-normal form [see Formula (3) in Section 11.6], we have the following deﬁnition.

Chapter 13 / Partial Derivatives

972

13.7.1 definition Assume that F (x, y, z) has continuous ﬁrst-order partial derivatives and that P0 (x0 , y0 , z0 ) is a point on the level surface S: F (x, y, z) = c. If ∇F (x0 , y0 , z0 ) = 0, then n = ∇F (x0 , y0 , z0 ) is a normal vector to S at P0 and the tangent plane to S at P0 is the plane with equation Definition 13.7.1 can be viewed as an extension of Theorem 13.6.6 from curves to surfaces.

Fx (x0 , y0 , z0 )(x − x0 ) + Fy (x0 , y0 , z0 )(y − y0 ) + Fz (x0 , y0 , z0 )(z − z0 ) = 0

(3)

The line through the point P0 parallel to the normal vector n is perpendicular to the tangent plane (3). We will call this the normal line, or sometimes more simply the normal to the surface F (x, y, z) = c at P0 . It follows that this line can be expressed parametrically as x = x0 + Fx (x0 , y0 , z0 )t, Example 1

y = y0 + Fy (x0 , y0 , z0 )t,

z = z0 + Fz (x0 , y0 , z0 )t

(4)

Consider the ellipsoid x 2 + 4y 2 + z2 = 18.

(a) Find an equation of the tangent plane to the ellipsoid at the point (1, 2, 1). (b) Find parametric equations of the line that is normal to the ellipsoid at the point (1, 2, 1). (c) Find the acute angle that the tangent plane at the point (1, 2, 1) makes with the xy-plane.

Solution (a). We apply Deﬁnition 13.7.1 with F (x, y, z) = x 2 + 4y 2 + z2 and (x0 , y0 , z0 ) = (1, 2, 1). Since

∇F (x, y, z) = Fx (x, y, z), Fy (x, y, z), Fz (x, y, z) = 2x, 8y, 2z we have

n = ∇F (1, 2, 1) = 2, 16, 2

Hence, from (3) the equation of the tangent plane is 2(x − 1) + 16(y − 2) + 2(z − 1) = 0

or x + 8y + z = 18

Solution (b). Since n = 2, 16, 2 at the point (1, 2, 1), it follows from (4) that parametric equations for the normal line to the ellipsoid at the point (1, 2, 1) are

x 0

−4

x = 1 + 2t,

(1, 2, 1) z 0

y Figure 13.7.4

z = 1 + 2t

−4 −2

y = 2 + 16t,

Solution (c). To ﬁnd the acute angle θ between the tangent plane and the xy-plane, we will apply Formula (9) of Section 11.6 with n1 = n = 2, 16, 2 and n2 = 0, 0, 1 . This yields 2 | 2, 16, 2 ⴢ 0, 0, 1 | 1 = √ cos θ = =√ 2, 16, 2 0, 0, 1 (2 66 )(1) 66 Thus, 1 θ = cos−1 √ ≈ 83 ◦ 66 (Figure 13.7.4).

TANGENT PLANES TO SURFACES OF THE FORM z = f (x, y)

To ﬁnd a tangent plane to a surface of the form z = f(x, y), we can use Equation (3) with the function F (x, y, z) = z − f(x, y).

13.7 Tangent Planes and Normal Vectors

973

Example 2 Find an equation for the tangent plane and parametric equations for the normal line to the surface z = x 2 y at the point (2, 1, 4).

Solution. Let F (x, y, z) = z − x 2 y. Then F (x, y, z) = 0 on the surface, so we can ﬁnd the ﬁnd the gradient of F at the point (2, 1, 4):

∇F (x, y, z) = −2xyi − x 2 j + k ∇F (2, 1, 4) = −4i − 4 j + k

From (3) the tangent plane has equation −4(x − 2) − 4(y − 1) + 1(z − 4) = 0

− 4x − 4y + z = −8

and the normal line has equations x = 2 − 4t,

y = 1 − 4t,

z=4+t

Suppose that f(x, y) is differentiable at a point (x0 , y0 ) and that z0 = f(x0 , y0 ). It can be shown that the procedure of Example 2 can be used to ﬁnd the tangent plane to the surface z = f(x, y) at the point (x0 , y0 , z0 ). This yields an alternative equation for a tangent plane to the graph of a differentiable function.

13.7.2 theorem If f(x, y) is differentiable at the point (x0 , y0 ), then the tangent plane to the surface z = f(x, y) at the point P0 (x0 , y0 , f (x0 , y0 )) [or (x0 , y0 )] is the plane z = f(x0 , y0 ) + fx(x0 , y0 )(x − x0 ) + fy(x0 , y0 )(y − y0 ) (5)

proof Consider the function F (x, y, z) = z − f(x, y). Since F (x, y, z) = 0 on the surface, we will apply (3) to this function. The partial derivatives of F are Fx (x, y, z) = −fx (x, y),

Fy (x, y, z) = −fy (x, y),

Fz (x, y, z) = 1

Since the point at which we evaluate these derivatives lies on the surface, it will have the form (x0 , y0 , f(x0 , y0 )). Thus, (3) gives 0 = Fx (x0 , y0 , z0 )(x − x0 ) + Fy (x0 , y0 , z0 )(y − y0 ) + Fz (x0 , y0 , z0 )(z − f(x0 , y0 )) = −fx (x0 , y0 )(x − x0 ) − fy (x0 , y0 )(y − y0 ) + 1(z − f(x0 , y0 )) which is equivalent to (5). ■ Recall from Section 13.4 that if a function f(x, y) is differentiable at a point (x0 , y0 ), then the local linear approximation L(x, y) to f at (x0 , y0 ) has the equation L(x, y) = f(x0 , y0 ) + fx (x0 , y0 )(x − x0 ) + fy (x0 , y0 )(y − y0 ) Notice that the equation z = L(x, y) is identical to that of the tangent plane to f(x, y) at the point (x0 , y0 ). Thus, the graph of the local linear approximation to f(x, y) at the point (x0 , y0 ) is the tangent plane to the surface z = f(x, y) at the point (x0 , y0 ). TANGENT PLANES AND TOTAL DIFFERENTIALS

Recall that for a function z = f(x, y) of two variables, the approximation by differentials is z = f = f(x, y) − f(x0 , y0 ) ≈ dz = fx(x0 , y0 )(x − x0 ) + fy(x0 , y0 )(y − y0 )

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Chapter 13 / Partial Derivatives

Note that the tangent plane in Figure 13.7.5 is analogous to the tangent line in Figure 13.4.2.

The tangent plane provides a geometric interpretation of this approximation. We see in Figure 13.7.5 that z is the change in z along the surface z = f(x, y) from the point P0 (x0 , y0 , f(x0 , y0 )) to the point P (x, y, f(x, y)), and dz is the change in z along the tangent plane from P0 to Q(x, y, L(x, y)). The small vertical displacement at (x, y) between the surface and the plane represents the error in the local linear approximation to f at (x0 , y0 ). We have seen that near (x0 , y0 ) this error term has magnitude much smaller than the distance between (x, y) and (x0 , y0 ). z

Δz

z = f (x, y) dz P0 (x 0, y 0, f (x 0, y 0 ))

Tangent plane at P0

z = L(x, y)

(x 0, y 0)

(x, y) dy = Δy = y − y 0 dx = Δ x = x − x 0

Figure 13.7.5

USING GRADIENTS TO FIND TANGENT LINES TO INTERSECTIONS OF SURFACES

In general, the intersection of two surfaces F(x, y, z) = 0 and G(x, y, z) = 0 will be a curve in 3-space. If (x0 , y0 , z0 ) is a point on this curve, then ∇F(x0 , y0 , z0 ) will be normal to the surface F(x, y, z) = 0 at (x0 , y0 , z0 ) and ∇G(x0 , y0 , z0 ) will be normal to the surface G(x, y, z) = 0 at (x0 , y0 , z0 ). Thus, if the curve of intersection can be smoothly parametrized, then its unit tangent vector T at (x0 , y0 , z0 ) will be orthogonal to both ∇F(x0 , y0 , z0 ) and ∇G(x0 , y0 , z0 ) (Figure 13.7.6). Consequently, if

F(x, y, z) = 0 ∇G

T ∇F

∇F(x0 , y0 , z0 ) × ∇G(x0 , y0 , z0 ) = 0

(x 0 , y0, z 0 ) G(x, y, z) = 0

then this cross product will be parallel to T and hence will be tangent to the curve of intersection. This tangent vector can be used to determine the direction of the tangent line to the curve of intersection at the point (x0 , y0 , z0 ).

Figure 13.7.6

Example 3 Find parametric equations of the tangent line to the curve of intersection of the paraboloid z = x 2 + y 2 and the ellipsoid 3x 2 + 2y 2 + z2 = 9 at the point (1, 1, 2) (Figure 13.7.7).

Solution. We begin by rewriting the equations of the surfaces as x2 + y2 − z = 0

and 3x 2 + 2y 2 + z2 − 9 = 0

and we take (1, 1, 2)

and G(x, y, z) = 3x 2 + 2y 2 + z2 − 9

We will need the gradients of these functions at the point (1, 1, 2). The computations are

x y

Figure 13.7.7

F(x, y, z) = x 2 + y 2 − z

∇F(x, y, z) = 2xi + 2y j − k, ∇G(x, y, z) = 6xi + 4y j + 2zk ∇F(1, 1, 2) = 2i + 2 j − k, ∇G(1, 1, 2) = 6i + 4 j + 4k Thus, a tangent vector at (1, 1, 2) to the curve of intersection is i j k 2 −1 = 12i − 14 j − 4k ∇F(1, 1, 2) × ∇G(1, 1, 2) = 2 6 4 4

13.7 Tangent Planes and Normal Vectors

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Since any scalar multiple of this vector will do just as well, we can multiply by 21 to reduce the size of the coefﬁcients and use the vector of 6i − 7 j − 2k to determine the direction of the tangent line. This vector and the point (1, 1, 2) yield the parametric equations x = 1 + 6t,

✔QUICK CHECK EXERCISES 13.7

2. Suppose that f (x, y) is differentiable at the point (3, 1) with f (3, 1) = 4, fx (3, 1) = 2, and fy (3, 1) = −3. An equation for the tangent plane to the graph of f at the point (3, 1, 4) is , and parametric equations for the normal line to the graph of f through the point (3, 1, 4) are x=

EXERCISE SET 13.7

y= C

z = 2 − 2t

(See page 977 for answers.)

1. Suppose that f (1, 0, −1) = 2, and f (x, y, z) is differentiable at (1, 0, −1) with ∇f (1, 0, −1) = 2, 1, 1 . An equation for the tangent plane to the level surface f (x, y, z) = 2 at the point (1, 0, −1) is , and parametric equations for the normal line to the level surface through the point (1, 0, −1) are x=

y = 1 − 7t,

√ 3. An equation for the tangent plane to the graph of z = x 2 y at the point (2, 4, 8) is , and parametric equations √ for the normal line to the graph of z = x 2 y through the point (2, 4, 8) are x= 2

4. The sphere x + y + z = 9 and the plane x + y + z = 5 intersect in a circle that passes through the point (2, 1, 2). Parametric equations for the tangent line to this circle at (2, 1, 2) are x=

CAS

1. Consider the ellipsoid x 2 + y 2 + 4z2 = 12. (a) Find an equation of the tangent plane to the ellipsoid at the point (2, 2, 1). (b) Find parametric equations of the line that is normal to the ellipsoid at the point (2, 2, 1). (c) Find the acute angle that the tangent plane at the point (2, 2, 1) makes with the xy-plane. 2. Consider the surface xz − yz3 + yz2 = 2. (a) Find an equation of the tangent plane to the surface at the point (2, −1, 1). (b) Find parametric equations of the line that is normal to the surface at the point (2, −1, 1). (c) Find the acute angle that the tangent plane at the point (2, −1, 1) makes with the xy-plane. 3–12 Find an equation for the tangent plane and parametric

equations for the normal line to the surface at the point P . ■ 3. x 2 + y 2 + z2 = 25; P (−3, 0, 4) 4. x 2 y − 4z2 = −7; P (−3, 1, −2) 2

5. x − xyz = 56; P (−4, 5, 2)

6. z = x 2 + y 2 ; P (2, −3, 13) 3 2

7. z = 4x y + 2y; P (1, −2, 12) 8. z = 21 x 7 y −2 ; P (2, 4, 4)

9. z = xe−y ; P (1, 0, 1) 10. z = ln x 2 + y 2 ; P (−1, 0, 0) 11. z = e3y sin 3x; P (π/6, 0, 1)

12. z = x 1/2 + y 1/2 ; P (4, 9, 5)

F O C U S O N C O N C E P TS

13. Find all points on the surface at which the tangent plane is horizontal. (a) z = x 3 y 2 (b) z = x 2 − xy + y 2 − 2x + 4y

14. Find a point on the surface z = 3x 2 − y 2 at which the tangent plane is parallel to the plane 6x + 4y − z = 5.

15. Find a point on the surface z = 8 − 3x 2 − 2y 2 at which the tangent plane is perpendicular to the line x = 2 − 3t, y = 7 + 8t, z = 5 − t. 16. Show that the surfaces z = x 2 + y 2 and

1 (x 2 10

+ y2) +

5 2

intersect at (3, 4, 5) and have a common tangent plane at that point.

17. (a) Find all points of intersection of the line x = −1 + t,

y = 2 + t,

z = 2t + 7

and the surface z = x2 + y2 (b) At each point of intersection, ﬁnd the cosine of the acute angle between the given line and the line normal to the surface. 18. Show that if f is differentiable and z = xf(x /y), then all tangent planes to the graph of this equation pass through the origin.

976

Chapter 13 / Partial Derivatives

19–22 True–False Determine whether the statement is true or

−4 −2

false. Explain your answer. ■ 19. If the tangent plane to the level surface of F (x, y, z) at the point P0 (x0 , y0 , z0 ) is also tangent to a level surface of G(x, y, z) at P0 , then ∇F (x0 , y0 , z0 ) = ∇G(x0 , y0 , z0 ).

20. If the tangent plane to the graph of z = f (x, y) at the point (1, 1, 2) has equation x − y + 2z = 4, then fx (1, 1) = 1 and fy (1, 1) = −1. 21. If the tangent plane to the graph of z = f (x, y) at the point (1, 2, 1) has equation 2x + y − z = 3, then the local linear approximation to f at (1, 2) is given by the function L(x, y) = 1 + 2(x − 1) + (y − 2).

22. The normal line to the surface z = f(x, y) at the point P0 (x0 , y0 , f(x0 , y0 )) has a direction vector given by fx (x0 , y0 )i + fy (x0 , y0 )j − k. 23–24 Find two unit vectors that are normal to the given surface at the point P . ■ z+x = z2 ; P (3, 5, 1) 23. y−1

24. sin xz − 4 cos yz = 4; P (π, π, 1)

25. Show that every line that is normal to the sphere x 2 + y 2 + z2 = 1 passes through the origin.

26. Find all points on the ellipsoid 2x 2 + 3y 2 + 4z2 = 9 at which the plane tangent to the ellipsoid is parallel to the plane x − 2y + 3z = 5.

27. Find all points on the surface x 2 + y 2 − z2 = 1 at which the normal line is parallel to the line through P (1, −2, 1) and Q(4, 0, −1).

28. Show that the ellipsoid 2x 2 + 3y 2 + z2 = 9 and the sphere x 2 + y 2 + z2 − 6x − 8y − 8z + 24 = 0 have a common tangent plane at the point (1, 1, 2).

29. Find parametric equations for the tangent line to the curve of intersection of the paraboloid z = x 2 + y 2 and the ellipsoid x 2 + 4y 2 + z2 = 9 at the point (1, −1, 2).

30. Find parametric equations for the tangent line to the curve of intersection of the cone z = x 2 + y 2 and the plane x + 2y + 2z = 20 at the point (4, 3, 5).

31. Find parametric equations for the tangent line to the curve of intersection of the cylinders x 2 + z2 = 25 and y 2 + z2 = 25 at the point (3, −3, 4).

32. The accompanying ﬁgure shows the intersection of the surfaces z = 8 − x 2 − y 2 and 4x + 2y − z = 0. (a) Find parametric equations for the tangent line to the curve of intersection at the point (0, 2, 4). (b) Use a CAS to generate a reasonable facsimile of the ﬁgure. You need not generate the colors, but try to obtain a similar viewpoint.

0 2 4

20 0 −20 −2 4

−4

Figure Ex-32

33. Show that the equation of the plane that is tangent to the ellipsoid x2 y2 z2 + 2 + 2 =1 2 a b c at (x0 , y0 , z0 ) can be written in the form x0 x y0 y z0 z + 2 + 2 =1 a2 b c 34. Show that the equation of the plane that is tangent to the paraboloid x2 y2 z= 2 + 2 a b at (x0 , y0 , z0 ) can be written in the form 2x0 x 2y0 y z + z0 = 2 + 2 a b 35. Prove: If the surfaces z = f(x, y) and z = g(x, y) intersect at P (x0 , y0 , z0 ), and if f and g are differentiable at (x0 , y0 ), then the normal lines at P are perpendicular if and only if fx (x0 , y0 )gx (x0 , y0 ) + fy (x0 , y0 )gy (x0 , y0 ) = −1 36. Use the result in Exercise 35 to show that the normal lines to the cones z = x 2 + y 2 and z = − x 2 + y 2 are perpendicular to the normal lines to the sphere x 2 + y 2 + z2 = a 2 at every point of intersection (see Figure Ex-38). 37. Two surfaces f(x, y, z) = 0 and g(x, y, z) = 0 are said to be orthogonal at a point P of intersection if ∇f and ∇g are nonzero at P and the normal lines to the surfaces are perpendicular at P . Show that if ∇f(x0 , y0 , z0 ) = 0 and ∇g(x0 , y0 , z0 ) = 0, then the surfaces f(x, y, z) = 0 and g(x, y, z) = 0 are orthogonal at the point (x0 , y0 , z0 ) if and only if f x gx + f y gy + f z gz = 0 at this point. [Note: This is a more general version of the result in Exercise 35.] 38. Use the result of Exercise 37 to show that the sphere x 2 + y 2 + z2 = a 2 and the cone z2 = x 2 + y 2 are orthogonal at every point of intersection (see the accompanying ﬁgure).

Figure Ex-38