CHAPTER 3
The Derivative SECTION 3.1 3.1.1 Let f (x) = (a) (b) (c) (d)
1 . x2
Find the average rate of change of y with respect to x over the interval [2, 3]. Find the instantaneous rate of change of y with respect to x at the point x = 2. Find the instantaneous rate of change of y with respect to x at a general point x0 . Sketch the graph of y = f (x) together with the secant and tangent lines whose slopes are given by the results in parts (a) and (b).
3.1.2 Let f (x) = x2 + 1. (a) (b) (c) (d)
Find the average rate of change of y with respect to x over the interval [−2, −1]. Find the instantaneous rate of change of y with respect to x at the point x = −2. Find the instantaneous rate of change of y with respect to x at a general point x0 . Sketch the graph of y = f (x) together with the secant and tangent lines whose slopes are given by the results in parts (a) and (b).
3.1.3 Let f (x) = (a) (b) (c) (d)
Find the average rate of change of y with respect to x over the interval [3, 5]. Find the instantaneous rate of change of y with respect to x at the point x = 3. Find the instantaneous rate of change of y with respect to x at a general point x. Sketch the graph of y = f (x) together with the secant and tangent lines whose slopes are given by the results in parts (a) and (b).
3.1.4 Let f (x) = (a) (b) (c) (d)
1 . x−2
1 . x+1
Find the average rate of change of y with respect to x over the given interval [1, 3]. Find the instantaneous rate of change of y with respect to x at the point x = 1. Find the instantaneous rate of change of y with respect to x at the general point x0 . Sketch the graph of y = f (x) together with the secant and tangent lines whose slopes are given by the results in parts (a) and (b).
3.1.5 Let f (x) =
2 . 3−x
(a) Find the slope of the tangent to the graph of f at a general point x0 using the method of Section 3.1. (b) Use the result in part (a) to find the slope of the tangent at x0 = 1. 3.1.6 Let f (x) =
3 . x−1
(a) Find the slope of the tangent to the graph of f at a general point x0 using the method of Section 3.1. (b) Use the result in part (a) to find the slope of the tangent at x0 = 4.
1
2
Questions, Section 3.1
3.1.7 Let f (x) =
1 . x2
(a) Find the slope of the tangent to the graph of f at a general point x0 using the method of section 3.1. (b) Use the result in part (a) to find the slope of the tangent at x0 = −2. 3.1.8 Let f (x) = 3x2 . (a) Find the slope of the tangent to the graph of f at a general point x0 using the method of section 3.1. (b) Use the result in part (a) to find the slope of the tangent at x = 3. 3.1.9 A rock is dropped from a height of 144 feet and falls toward the earth in a straight line. In t seconds, the rock drops a distance of s = 16t2 feet. (a) What is the average velocity of the rock while it is falling? (b) Use the method of 3.1 to find the instantaneous velocity of the rock when it hits the ground. 3.1.10 A rock is dropped from a height of 64 feet and falls toward the earth in a straight line. In t seconds, the rock drops a distance of s = 16t2 feet. (a) What is the average velocity of the rock while it is falling? (b) Use the method of Section 3.1 to find the instantaneous velocity of the rock when it hits the ground. 3.1.11 A particle moves in a straight line from its initial position so that after t seconds, its distance is given by s = t2 + t feet from its initial position. (a) Find the average velocity of the particle over the interval [1, 3] seconds. (b) Use the method of Section 3.1 to find the instantaneous velocity of the particle at t = 1 second. 3.1.12 A particle moves in a straight line from its initial position so that after t seconds, its distance t is given by s = feet from its initial position. t+2 (a) Find the average velocity of the particle over the interval [2, 3] seconds. (b) Use the method of Section 3.1 to find the instantaneous velocity of the particle at t = 2 seconds. 3.1.13 Let f (x) = x2 . Use the method of Section 3.1 to show that the slope of the tangent to the graph of f at x = x0 is 2x0 . 3.1.14 Let f (x) = ax2 + b, where a and b are constant. Use the method of Section 3.1 to show that the slope of the tangent to the graph of f at x = x0 is 2ax0 . 3.1.15 Let f (x) = ax3 + b, where a and b are constants. Use the method of Section 3.1 to show that the slope of the tangent to the graph of f at x = x0 is 3ax20 . 3.1.16 A particle moves in a straight line from its initial position so that after t seconds, its distance is given by s = 16t2 feet. Use the method of Section 3.1 to show that the instantaneous velocity of the particle at t = t0 seconds is 32t0 .
Questions, Section 3.1
3
3.1.17 A particle moves in a straight line from its initial position so that after t seconds, its distance is given by s = 4 − 16t2 feet. Use the method of Section 3.1 to show that the instantaneous velocity of the particle at t = t0 seconds is v = −32t0 . 3.1.18 The figure shows the position versus time curves of four different particles moving on a straight line. For each particle, determine if its instantaneous velocity is increasing or decreasing with time. (a)
(b)
(c)
(d)
3.1.19 The figure shows the position versus time curve for a certain particle moving along a straight line. Estimate each of the following from the graph. (a) (b) (c) (d)
The The The The
average velocity over the interval 0 ≤ t ≤ 4.6 value of t at which the instantaneous velocity is zero values of t at which the instantaneous velocity is maximum; minimum instantaneous velocity when t = 5 seconds
3.1.20 Find the slope of the graph of f at the x-value specified by the given x0 . f (x) = x4 − 2
x0 = 6
4
Questions, Section 3.1
3.1.21 Find the slope of the graph of f at the x-value specified by the given x0 . √ f (x) = 7 x x0 = 9 3.1.22 Find the instantaneous rate of change of f (x) =
4 at x0 = 1. x2
3.1.23 Find the instantaneous rate of change of f (x) = 3x2 − 2 at x0 = 2. 3.1.24 Find the instantaneous rate of change of f (x) = x2 − 4x + 7 at x0 = 3.
SOLUTIONS SECTION 3.1
3.1.1 (a) msec
1 1 − 2 f (x) − f (2) 5 (3) (2)2 = = =− 3−2 1 36
Thus, on the average, y decreases 5 units per 36 units increase in x over the interval [2, 3]. 1 1 1 1 − 2 − x21 2 x21 4 = lim (b) mtan = lim x1 →2 x1 = 2 x1 →2 x1 = 2 = lim
4 − x21 (x1 − 2)(x1 + 2) (x1 + 2) 1 = lim − = lim − =− x1 →2 − 2) x1 →2 4x21 (x1 − 2) 4x1 − 2 4
x1 →2 4x2 1 (x1
Thus, y is decreasing at the point x = 2 at a rate of 1 unit per 4 units increase in x. 1 1 − 2 (x1 ) (x0 )2 (c) mtan = lim x1 →x0 x1 − x0 = lim
x1 →x0
x20 − x21 −(x1 + x0 ) 2 = lim =− 3 2 2 2 2 →x x x1 x0 (x1 − x0 ) x1 x0 x0 1 0
Thus, the instantaneous rate of change of y with respect to x at x = x0 is −
2 . x30
(d)
3.1.2 (a) msec =
f (−1) − f (−2) [(−1)2 + (−1)] − [(−2)2 + 1] = = −3 −1 − (−2) 1
Thus, on the average, y decreases 3 units per unit increase in x over the interval [−2, −1]. f (x1 ) − f (−2) [x21 + 1] − [(−2)2 + 1] = lim x1 →−2 x1 →−2 x1 − (−2) x1 + 2
(b) mtan = lim
x21 − 4 = lim x1 − 2 = −4 x1 →−2 x1 + 2 x1 →−2
= lim
Thus, y is decreasing at the point x = −2 at a rate of 4 units per unit increase in x.
5
6
Solutions, Section 3.1
(c) mtan = lim
x1 →x0
[x21 + 1] − [x20 + 1] x2 − x20 = lim 1 x1 →x0 x1 − x0 x1 − x0
= lim (x1 + x0 ) = 2x0 x1 →x0
Thus, the instantaneous rate of change of y with respect to x at x = x0 is 2x0 . (d)
3.1.3 (a) msec
1 1 1 − −1 1 f (5) − f (3) 3 − 2 5 − 2 3 = =− = = 5−3 2 3 5−3
Thus, on the average, y decreases 1 unit per 3 units increase in x over the interval [3, 5]. 1 1 1 − −1 1 − (x1 − 2) x −2 3−2 x −2 = lim 1 = lim (b) mtan = lim 1 x1 →3 x1 →3 x1 →3 (x1 − 2)(x1 − 3) x1 − 3 x1 − 3 = lim
x1 →3
−1 = −1 x1 − 2
Thus, y is decreasing at the point x = 3 at a rate of 1 unit per unit increase in x. 1 1 − (x0 − 2)(x1 − 2) lim x1 − 2 x0 − 2 (c) mtan = lim = x1 →x0 x1 − x0 x1 → x0 (x1 − 2)(x0 − 2)(x1 − x0 ) = lim
x1 →x0
−1 −1 = (x1 − 2)(x0 − 2) (x0 − 2)2
Thus, the instantaneous rate of change of y with respect to x at x = x0 is − (d)
1 . (x0 − 2)2
Solutions, Section 3.1
3.1.4 (a) msec
7
1 1 − f (2) − f (1) 2 + 1 1 + 1 = −1 = = 2−1 1 6
Thus, on the average, y decreases one unit per six units increase in x over the interval [1, 2]. 1 1 − f (x1 ) − f (1) 2 − (x1 + 1) (x1 + 1) 1 + 1 = lim (b) mtan = lim = lim x1 →1 x1 →1 x1 →1 2(x1 + 1)(x1 − 1) x1 − 1 x1 − 1 = lim − x1 →1
1 −1 (x1 − 1) = lim − =− 2(x1 + 1)(x1 − 1) x1 →1 2(x1 + 1) 4
Thus, y is decreasing at the point x = 1 at a rate of 1 unit per 4 units increase in x. 1 1 − (x0 + 1) − (x1 + 1) x1 + 1 x0 + 1 (c) mtan = lim = lim x1 →x0 x1 →x0 (x1 + 1)(x0 + 1)(x1 − x0 ) x1 − x0 = lim
x1 →x0
1 −1 =− (x1 + 1)(x0 + 1) (x0 + 1)2
Thus, the instantaneous rate of change of y with respect to x at x = x0 is (d)
2 2 − 2(3 − x0 ) − 2(3 − x1 ) 3 − x1 3 − x0 3.1.5 (a) mtan = lim = lim x1 →x0 x1 →x0 (3 − x0 )(3 − x1 )(x1 − x0 ) x1 − x0 = lim
x1 →x0
2 2 = (3 − x0 )(3 − x1 ) (3 − x0 )2
(b) mtan , when x0 = 1 is
2 =1 (3 − 1)
3 3 − 3(x0 − 1) − 3(x1 − 1) x1 − 1 x0 − 1 3.1.6 (a) mtan = lim = lim x1 →x0 x1 →x0 (x1 − 1)(x1 − x0 )(x0 − 1) x1 − x0 = lim
x1 →x0
−3 −3 = (x1 − 1)(x0 − 1) (x0 − 1)2
(b) mtan , when x0 = 4, is
−3 1 =− 2 (4 − 1) 3
−1 . (x0 + 1)2
8
Solutions, Section 3.1
1 1 − 2 x2 − x21 x21 x0 3.1.7 (a) mtan = lim = lim 2 20 x1 −x0 x1 − x0 x1 −x0 x x (x1 − x0 ) 1 0 −(x1 + x0 ) 2 =− 3 x0 x21 x20
= lim
x1 −x0
(b) mtan when x0 = −2, is −
3.1.8 (a) mtan = lim
x1 →x0
2 1 =− (−2)3 4
3(x1 )2 − 3(x0 )2 = lim 3(x1 + x0 ) = 3x20 x1 →x0 x1 − x0
(b) mtan , when x0 = 3, is 3(3)2 = 27 3.1.9 (a) The rock will hit the ground when 16t2 = 144, t = 3 seconds, so the average velocity is 16(3)2 − 16(0)2 = 48 feet per second 3−0 f (t1 ) − f (3) 16t21 − 16(3)2 = lim t1 →3 t1 →3 t1 − 3 t1 − 3
(b) The instantaneous velocity = lim
= lim 16(t1 + 3) = 16(6) = 96 feet per second t1 →3
3.1.10 (a) The rock will hit the ground when 16t2 = 64, t = 2 seconds so the average velocity is 16(2)2 − 16(0)2 = 32 feet per second. 2−0 f (t1 ) − f (2) 16t21 − 16(2)2 = lim t1 →2 t1 →2 t1 − 2 t1 − 2
(b) The instantaneous velocity = lim
= lim 16(t1 + 2) = 16(4) = 64 feet per second t1 →2
3.1.11 (a) average velocity =
[(3)2 + (3)] − [(1)2 + (1)] f (3) − f (1) = = 5 feet per second 3−1 2
(b) The instantaneous velocity at t = 1 second is f (t1 ) − f (1) (t2 + t1 ) − (12 + 1) = lim 1 t1 →1 t1 →1 t1 − 1 t1 − 1 lim
t21 + t1 − 2 = lim (t1 + 2) = 3 feet per second t1 →1 t1 →1 t1 − 1
= lim
3 2 − f (3) − f (2) 3 + 2 2 + 2 = 1 feet per second = 3.1.12 (a) average velocity = 3−2 1 10 (b) The instantaneous velocity at t = 2 seconds is t1 1 − f (t1 ) − f (2) 2t1 − (t1 + 2) t1 + 2 2 lim = lim = lim t1 →2 t1 →2 t1 →2 2(t1 + 2)(t1 − 2) t1 − 2 t1 − 2 = lim
t1 →2
1 1 = feet per second 2(t1 + 2) 8
Solutions, Section 3.1
9
3.1.13 mtan = lim
x21 − x20 = lim (x1 + x0 ) = 2x0 x1 →x0 x1 − x0
3.1.14 mtan = lim
(ax21 + b) − (ax20 + b) a(x21 − x20 ) = lim = lim a(x1 + x0 ) = 2ax0 x1 →x0 x1 − x0 x1 →x0 x1 − x0
3.1.15 mtan = lim
(a(x1 )3 + b) − (a(x0 )3 + b) a[(x1 )3 − (x0 )3 ] = lim x1 →x0 x1 − x0 x1 − x0
x1 →x0
x1 →x0
x1 →x0
= lim a(x21 + x1 x0 + x20 ) = 3ax20 x1 →x0
3.1.16 Instantaneous velocity at t = t0 seconds is lim
t1 →t0
16t21 − 16t20 = lim 16(t1 + t0 ) = 32t0 feet per second t1 →t0 t1 − t0
3.1.17 Instantaneous velocity at t = t0 seconds is lim
t1 →t0
4 − 16t21 − (4 − 16t20 ) −16(t21 − t20 ) = lim t1 →t0 t1 − t0 t1 − t0 = lim −16(t1 + t0 ) = 32t0 feet per second t1 →t0
3.1.18 (a) decreasing
(b) increasing
(c) increasing
(d) decreasing
f (4.6) − f (0) 10 − 10 = =0 4.6 − 0 4.6 (b) The instantaneous velocity is zero when the slope of the tangent line to the curve is zero. The values of t at which the tangent line is horizontal are t ≈ 2.5, 6.
3.1.19 (a)
(c) The velocity is a maximum when t ≈ 2.25. The velocity is a minimum when t ≈ 6. (d) When t = 5, the instantaneous velocity ≈ − 1. 3.1.20 864
3.1.21 1.1667
3.1.22 −8
3.1.23 12
3.1.24 2
10
Questions, Section 3.2
SECTION 3.2 3.2.1 Use the definition of the derivative to calculate f (x) if f (x) = 3x2 − x and find the equation of the tangent to the graph of f at x = 1. 3.2.2 Use the definition of the derivative to calculate f (x) if f (x) = 2x2 − x + 1. 3.2.3 Use the definition of the derivative to calculate f (x) if f (x) = 2x3 + 1 and find the equation of the tangent line and the normal line to the graph of f at x = 1. 3.2.4 Use the definition of the derivative to calculate f (x) if f (x) = x3 − 3x and find the equation of the tangent line and the normal line to the graph of f at x = 2. 3.2.5 Use the definition of the derivative to calculate f (x) if f (x) = the tangent line and the normal line to the graph of f at x = 2. 3.2.6 Let y = 3.2.7 Let y =
√
3x + 1. Use the definition of the derivative to find
√
2x and find the equation of
dy . dx
dy 1 . Use the definition of the derivative to find . x+2 dx
3.2.8 Use the definition of the derivative to calculate f (x) if f (x) =
2 . 3−x
3.2.9 Given that f (0) = 4 and f (0) = −1, find an equation for the tangent line to the graph of y = f (x) at the point where x = 0. 3.2.10 Given that f (2) = −1 and f (2) = 5, find an equation for the tangent line to the graph of y = f (x) at the point where x = 2. 1 3.2.11 Use the definition of the derivative to calculate f (x) if f (x) = √ and find the equation of 2x the tangent line and the normal line to the graph of f at x = 2. 4 3.2.12 The volume of a sphere is given by πr3 where r is the radius of the sphere. Use the method 3 of Section 3.2 to find the instantaneous rate of change of V with respect to r when r = 4. 3.2.13 The surface area of a sphere is given by S = 4πr2 where r is the radius of the sphere. Use the method of Section 3.2 to find the instantaneous rate of change of S with respect to r when r = 4. π 3.2.14 The volume of a sphere is given by V = D3 where D is the diameter of the sphere. Use the 6 method of Section 3.2 to find the instantaneous rate of change of V with respect to D when D = 2. x2 − 5 x ≤ 1 3.2.15 Show that f (x) = is continuous but not differentiable at x = 1. Sketch the x−5 x>1 graph of f .
Questions, Section 3.2
11
3.2.16 Sketch the graph of the derivative of the function whose graph is shown.
3.2.17 It has been observed that some large colonies of bacteria tend to grow at a rate proportional to the number of bacteria present. The graph shows bacteria count P (in thousands) versus time t (in seconds). dP (a) Estimate P and when t = 2 sec. dt dP (b) This model for bacterial growth can be expressed as = kP where k is the constant of dt proportionality. Use the results in part (a) to estimate the value of k.
3.2.18 Use a graphing utility to show that y =
√ 6
x2 does not have a derivative at x = 0.
3.2.19 Use a graphing utility to show that y = x − 2 does not have a derivative everywhere. 3.2.20 Let f (x) = 4x . Estimate f (1) using a graphing utility. 3.2.21 Let f (x) = cos x. Estimate f (1) using a graphing utility. 3.2.22 The cost of manufacturing an object is given by C = f (x) dollars. Estimate the cost of making the 401st item if f (400) = 12. 3.2.23 The cost of manufacturing an object is given by C = f (x) dollars. Estimate the cost of making the 801st item if f (800) = 72. 3.2.24 The cost of manufacturing an object is given by C = f (x) dollars. Estimate the cost of making the 57th item if f (56) = 24.
SOLUTIONS SECTION 3.2 [3(x + h)2 − (x + h)] − (3x2 − x) h→0 h
3.2.1 f (x) = lim
6xh + 3h2 − h = lim (6x + 3h − 1) = 6x − 1 h→0 h→0 h
= lim
so the slope of the tangent at (1, 2) is f (1) = 6(1) − 1 = 5, thus, the equation of the tangent to f at (1, 2) is y − 2 = 5(x − 1) or y = 5x − 3. [2(x + h)2 − (x + h) + 1] − (2x2 − x + 1) h→0 h
3.2.2 f (x) = lim
4xh + 2h2 − h = lim (4x + 2h − 1) = 4x − 1 h→0 h→0 h
= lim
[2(x + h)3 + 1] − (2x3 + 1) h→0 h
3.2.3 f (x) = lim
6x2 h + 6xh2 + 2h3 = lim (6x2 + 6xh + 2h2 ) = 6x2 h→0 h→0 h
= lim
so the slope of the tangent at (1, 3) is f (1) = 6, thus, the equation of the tangent to the graph 1 of f at (1, 3) is y − 3 = 6(x − 1) or y = 6x − 3; the slope of the normal at (1, 3) is − so the 6 19 1 1 equation of the normal to the graph of f at (1, 3) is y − 3 = − (x − 1) of y = − x + . 6 6 6 [(x + h)3 − 3(x + h)] − (x3 − 3x) h→0 h
3.2.4 f (x) = lim
3x2 h + 3xh2 + h3 − 3h = lim (3x2 + 3xh + h2 − 3) h→0 h→0 h 2 = 3x − 3
= lim
so the slope of the tangent at (2, 2) is f (2) = 3(2)2 − 3 = 9, thus, the equation of the tangent to the graph of f at (2, 2) is y − 2 = 9(x − 2) or y = 9x − 16; the slope of the normal at 1 1 (2, 2) is − so the equation of the normal to the graph of f at (2, 2) is y − 2 = − (x − 2) or 9 9 1 20 y =− x+ . 9 9 √ √ √ 2(x + h) − 2x 2(x + h) − 2x 2(x + h) + 2x √ 3.2.5 f (x) = lim = lim h→0 h→0 h h 2(x + h) + 2x 2(x + h) − 2x 2h = lim √ √ h→0 h→0 h 2(x + h) + 2x h 2(x + h) + 2x 2 1 √ =√ = lim h→0 2x 2(x + h) + 2x = lim
so the slope of the tangent at (2, 2) is f (2) =
1 2(2)
=
1 , thus, the equation of the tangent 2
1 1 to the graph of f at (2, 2) is y − 2 = (x − 2) or y = x + 1; the slope of the normal at (2, 2) 2 2 1 = −2 so the equation of the normal to the graph of f at (2, 2) is y − 2 = −2(x − 2) is − (1/2) or y = −2x + 6.
12
Solutions, Section 3.2
3.2.6
13
√ 3(x + h) + 1 − 3x + 1 h √ √ 3(x + h) + 1 − 3x + 1 3(x + h) + 1 + 3x + 1 = lim √ h→0 h 3(x + h) + 1 + 3x + 1
dy = lim dx h→0
[3(x + h) + 1] − (3x + 1) √ h 3(x + h) + 1 + 3x + 1 3h = lim √ h→0 3(x + h) + 1 + 3x + 1 h 3 3 = √ = lim √ h→0 2 3x + 1 3(x + h) + 1 + 3x + 1 = lim
h→0
3.2.7
1 1 − dy (x + h) + 2 x + 2 = lim dx h→0 h = lim −
h h[(x + h) + 2][x + 2]
= lim −
1 1 =− [(x + h) + 2][x + 2] (x + 1)2
h→0
h→0
2 2 − 2h 3 − (x + h) 3 − x = lim 3.2.8 f (x) = lim h→0 h→0 h(3 − x)(3 − x − h) h = lim
h→0
2 2 = (3 − x)(3 − x − h) (3 − x)2
3.2.9 The slope of the tangent line at (0, 4) is −1, thus the equation of the tangent to the graph of f at (0, 4) is y − 4 = −1(x − 0) or y = −x + 4. 3.2.10 The slope of the tangent line at (2, −1) is 5, thus the equation of the tangent to the graph of f at (2, −1) is y − (−1) = 5(x − 2) or y = 5x − 11.
3.2.11 f (x) = lim
h→0
1 2(x + h) h
1 −√ 2x
√ = lim
h→0
√ 2x − 2(x + h) 2x + 2(x + h) √ √ 2x + 2(x + h) h 2x 2(x + h)
−2h √ = lim √ h→0 2x + 2(x + h) h 2x 2(x + h) −2 1 √ =− = lim √ h→0 (2x)3/2 2x 2(x + h) 2x + 2(x + h) 1 1 = − , thus, the equation of the 8 (4)3/2 1 1 3 1 tangent to the graph of f at (2, 1/2) is y − = − (x − 2) or y = − x + ; the slope of the 2 8 8 4 1 normal at (2, 1/2) is − = 8 so the equation of the normal to the graph of f at (2, 1/2) is 1 − 8 31 1 y − = 8(x − 2) or y = 8x − . 2 2 so the slope of the tangent at (2, 1/2) is f (2) = −
14
Solutions, Section 3.2
4 4π 4 π(r + h)3 − πr3 (3r2 h + 3rh2 + h3 ) 3 3 = lim 3 3.2.12 f (r) = lim h→0 h→0 h h 4π(3r2 + 3rh + h2 ) = 4πr2 h→0 3
= lim
so the instantaneous rate of change of V with respect to r is f (4) = 4π(4)2 = 64π. 4π(r + h)2 − 4πr2 4π(2rh + h2 ) = lim = lim 4π(2r + h) = 8πr h→0 h→0 h→0 h h
3.2.13 f (r) = lim
so the instantaneous rate of change of S with respect to r at r = 4 is f (4) = 8π(4) = 32π. π π π (3D2 h + 3Dh2 + h3 ) (D + h)3 − D3 6 6 6 = lim 3.2.14 f (D) = lim h→0 h→0 h h
π π(3D2 + 3Dh + h2 ) = D2 h→0 6 2
= lim
thus, the instantaneous rate at which V changes with respect to D when D = 2 is f (2) = π 2 (2) = 2π. 2 3.2.15
lim f (x) = lim+ f (x) = f (1) = −4 so f is continuous at x = 1
x→1−
x→1
f (1 + h) − f (1) [(1 + h)2 − 5] + 4 2h + h2 = lim = lim = lim− (2 + h) = 2 h h h h→0− h→0− h→0− h→0 f (1 + h) − f (1) [(1 + h) − 5] + 4 h lim = lim =1 lim h h h→0+ h→0+ h h→0+ lim
so f (1) does not exist.
3.2.16
Solutions, Section 3.2
15
3.2.17 P is approximately 7.4 thousand dP is approximately 7.4 dt dP = kP , 7.4 = k7.4 so k = 1. dt 3.2.18
As x → 0− the tangent line is negative, while as x → 0+ the tangent line is positive. Since the tangent line’s slope does not approach 0 as x approaches 0, the derivative at x = 0 does not exist. 3.2.19
There is no tangent line at x = 2. Hence, there is no derivative at x = 2. 3.2.20 5.545
3.2.21 −0.84
3.2.22 $12
3.2.23 $72
3.2.24 $24
16
Questions, Section 3.3
SECTION 3.3 3.3.1 Find
dy 3x3 + 5x2 + if y = dx x
3.3.3 Find f (2) if f (x) =
√
.
3.3.2 Find
dy x2 + 3x if y = . dx 7 − 2x
−8 1 5 + x . x2 5
3.3.4 Find
dy if y = −2(x2 − 5x)(3 + x7 ). dx
x
3.3.5 Find f (s) is f (s) = (3s2 + 4)(s2 − 9s). 3.3.6 Find f (x) if f (x) =
2x + 1 . x2 + 3x
3.3.7 If f (3) = 2, f (3) = −1, g(3) = 3, g (3) = 0, find F (3) (a) F (x) = 2f (x) − g(x) (b) F (x) =
1 f (x)g(x) 2
(c) F (x) =
1 f (x) . 3 g(x)
3.3.8 Find
d2 y 1 5 if y = − − 2 . dt2 t t
3.3.9 Find f (u) is f (u) =
u2 − 5 . 3u2 − 1
dy if y = (x2 − 2)(x3 + 5x). dx 1 3 dv 2 3.3.11 Find if v = π ah − h , a is a constant. dh 3 3.3.10 Find
3.3.12 Find f (x) if f (x) = (x2 + 1)(x3 − 2x2 + x). 3.3.13 Find equations for the tangents and normals to the graph of y = 4 − 3x − x2 at those points where the curve intersects the x-axis. 3.3.14 Find equations for the tangents and normals to the graph of y = 6 − x − x2 at the points where the curve intersects the x-axis. 3.3.15 Find the points on the graph of y = 2x3 − 3x2 − 12x + 20 at which the tangent is parallel to the x-axis. 3.3.16 Show that the parabola y = −x2 and the line x − 4y − 18 = 0 intersect at right angles at one of their points of intersection. 3.3.17 Find the equation of the tangents and normals to the graph of y =
x+1 at x = 2. x−1
3.3.18 Find the equation of the tangent and normal to the graph of y = 10 − 3x − x2 at the point where the curve intersects the x-axis. 3.3.19 Show that the parabola y = x2 and the line x + 2y − 3 = 0 intersect at right angles at one of their points of intersection.
Questions, Section 3.3
17
dy if y = π 7 . dx
3.3.20 Find
dy if y = 12x10 . dx
3.3.21 Find
3.3.22 Find
√ dy if y = 7 x − 2x. dx
3.3.23 y = x−6 + 3x2 . Find
3.3.24 y = x6 − 2x + 4. Find
d2 y . dx2
3.3.25 y = x8 . Find y (1).
d2 y . dx2
SOLUTIONS SECTION 3.3 3.3.1 y = 3x2 + 5x + x−1/2 so
3.3.2
dy 1 = 6x + 5 − x−3/2 dx 2
d d (7 − 2x) [x2 + 3x] − (x2 + 3x) [y − 2x] dy dx dx = dx (7 − 2x)2 =
(7 − 2x)(2x + 3) − (x2 + 3x)(−2) 21 + 14x − 2x2 = (7 − 2x)2 (7 − 2x)2
1 3.3.3 f (x) = −8x−2 + x5 so f (x) = 16x−3 + x4 and f (x) = −48x−4 + 4x3 , then 5 f (2) = −48(2)−4 + 4(2)3 = 29 3.3.4
dy d d = −2 (x2 − 5x) [3 + x7 ] + (3 + x7 ) [x2 − 5x] dx dx dx = −2[(x2 − 5x)(7x6 ) + (3 + x7 )(2x − 5)] = −18x8 + 80x7 − 12x + 30 d 2 d [s − 9s] + (s2 − 9s) [3s2 + 4] ds ds = (3s2 + 4)(2s − 9) + (s2 − 9s)(6s) = 12s3 − 27s2 − 46s − 36
3.3.5 f (s) = (3s2 + 4)
(x2 + 3x)
3.3.6 f (x) = =
d d [2x + 1] − (2x + 1) [x2 + 3x] dx dx (x2 + 3x)2
(x2 + 3x)(2) − (2x + 1)(2x + 3) −2x2 − 2x − 3 = (x2 + 3x)2 (x2 + 3x)2
3.3.7 (a) F (x) = 2f (x) − g (x); F (3) = 2f (3) − g (3) = 2(−1) − 0 = −2 (b) F (x) =
1 [f (x)g (x) = g(x)f (x)]; 2
1 1 3 [f (3)g (3) + g(3)f (3)] = [(2)(0) + (3)(−1)] = − 2 2 2
1 g(x)f (x) − f (x)g (x) (c) F (x) = 3 g(x) 2
1 (3)(−1) − (2)(0) 1 1 g(3)f (3) − f (3)g (3) (d) F (x) = = =− 2 2 3 g(3) 3 (3) 9 F (3) =
3.3.8 y = −t−1 − 5t−2 so
3.3.9 f (u) = =
(3u2 − 1)
dy d2 y = t−2 + 10t−3 and 2 = −2t−3 − 30t−4 dt dt
d 2 d [u − 5] − (u2 − 5) [3u2 − 1] du du (3u2 − 1)2
(3u2 − 1)(2u) − (u2 − 5)(6u) 28u = 2 2 (3u − 1) (3u2 − 1)2
18
Solutions, Section 3.3
3.3.10
3.3.11
19
d d dy = (x2 − 2) [x3 + 5x] + (x3 + 5x) [x2 − 2] dx dx dx = (x2 − 2)(3x2 + 5) + (x3 + 5x)(2x) = 5x4 + 9x2 − 10 dv = π(2ah − h2 ) dh d 3 d [x − 2x2 + x] + (x3 − 2x2 + x) [x2 + 1] dx dx = (x2 + 1)(3x2 − 4x + 1) + (x3 − 2x2 + x)(2x)
3.3.12 f (x) = (x2 + 1)
= 5x4 − 8x3 + 6x2 − 4x + 1 3.3.13 The curve intersects the x-axis when 4 − 3x − x2 = 0; x = −4 or x = 1. f (x) = −3 − 2x so the slope of the tangent at x = −4 is f (−4) = −3 − 2(−4) = 5 and at x = 1 is f (1) = −3 − 2(1) = −5. The equation of the tangent to y at (−4, 0) is y − 0 = 5(x + 4) or y = 5x + 20 and at (1, 0) 1 is y − 0 = −5(x − 1) or y = −5x + 5; the slope of the normal at x = −4 is − and at x = 1 is 5 1 1 1 4 1 = so the equation of the normal at (−4, 0) is y − 0 = − (x + 4) or y = − x − and − −5 5 5 5 5 1 1 1 at (1, 0) is y − 0 = (x − 1) or y = x − . 5 5 5 3.3.14 The curve intersects the x-axis when 6 − x − x2 = 0; x = −3 or x = 2. f (x) = −1 − 2x so the slope of the tangent at x = −3 is f (−3) = −1 − 2(−3) = 5 and at x = 2 is f (2) = −1 − 2(2) = −5. The equation of the tangent to y at (−3, 0) is y − 0 = 5(x + 3) or y = 5x + 15 and at (2, 0) 1 is y − 0 = −5(x − 2) or y = −5x + 10; the slope of the normal at x = −3 is − and at x = 2 is 5 1 1 1 1 3 − = so the equation of the normal at (−3, 0) is y − 0 = − (x + 3) or y = − x − and −5 5 5 5 5 1 1 2 at (2, 0) is y − 0 = (x − 2) or y = x − . 5 5 5 3.3.15 The tangent is parallel to the x-axis when f (x) = 6x2 − 6x − 12 = 0, thus, 6(x − 2)(x + 1) = 0, x = 2 or x = −1 so the points on the graph of y are (−1, 27) and (2, 0). 3.3.16 Substitute y = −x2 into x − 4y − 18 = 0 to get 4x2 + x − 18 = 0. Solve for x to get x = 2 1 9 9 or x = − . Place x − 4y − 18 = 0 into the slope intercept form to get y = x − , thus the 4 4 2 dy slope of the line is m1 = 1/4. Differentiate y = −x2 to get = −2x so that the slope of a dx line drawn tangent to y = −x2 at x = x0 is m2 = −2x0 . When x = 2, m1 = 1/4 and m2 = −4, so the graphs intersect at right angles since m1 m2 = (1/4)(−4) = −1.
x+1 d x+1 is m1 = 3.3.17 The slope of the tangent to y = ; x−1 dx x − 1 x=2
(x − 1) − (x + 1) d x+1 −2 d x+1 = = so m = = −2. When x = 2, y = 3 1 dx x − 1 (x − 1)2 (x − 1)2 dx x − 1 x=2 so the equation of the tangent line is y − 3 = −2(x − 2) or y = −2x + 7. The slope of the normal 1 1 1 1 x+1 is m2 = − = , so the equation of the normal is y − 3 = (x − 2) or =− to y = x−1 m1 −2 2 2 1 y = x + 2. 2
20
Solutions, Section 3.3
3.3.18 The curve intersects the x axis when 10 − 3x − x2 = 0; x = −5 or x = 2. f (x) = −3 − 2x so the slope of the tangent at x = −5 is f (−5) = 7 and at x = 2, is f (2) = −7. The equation of the tangent at x = −5 is y = 7x + 35 and at x = 2 is y = −7x + 14. The slope of the normal 1 −1 −1 1 = − and the slope of the normal at x = 2 is at x = −5 is = . The equation f (2) 7 f (−5) 7 5 1 2 1 of the normal at x = −5 is y = − x − and at x = 2 is y = x − . 7 7 7 7 3.3.19 Substitute y = x2 into x + 2y − 3 = 0 to get 2x2 + x − 3 = 0. Solve for x to get x = 1 or 1 3 3 x = − . Place x + 2y − 3 = 0 into the slope intercept form to get y = − x + , thus, the 2 2 2 1 dy slope of the line is m1 = − . Differentiate y = x2 to get = 2x so that the slope of a line dx 2 1 2 drawn tangent to y = x at x = x0 is m2 = 2x0 . When x0 = 1, m1 = − and m2 = 2, so the 2 1 (2) = −1. graphs intersect at right angles since m1 m2 = − 2 3.3.20 120x9 3.3.22
√ 7 x −2 2x
3.3.24
dy d2 y = 6x5 − 2, = 30x4 dx dx2
3.3.21 0 3.3.23
d2 y dy = −6x−7 + 6x, = 42x−8 + 6 dx2 dx
3.3.25 y = 8x7 , y = 56x6 , y = 336x5 , y (1) = 336(1)5 = 336
Questions, Section 3.4
21
SECTION 3.4 3.4.1 Find f (x) if f (x) = x tan x. 3.4.3 Find
3.4.2 Find f (x) if f (x) = x sin x.
sin x dy if y = 2 . dx x
3.4.5 Find f (x) if f (x) =
3.4.4 Find
cot x . 1 + csc x
3.4.6 Find f (x) if f (x) = (5x2 + 7) cos x.
csc x 3.4.7 Differentiate y = √ . x 3.4.9 Find
dy if y = (x3 + 7x) tan x. dx
3.4.11 Find f (θ) if f (θ) = 3.4.12 Find
dy if y = sec x tan x. dx
3.4.8 Find
dy cos x if y = . dx 1 − sin x
3.4.10 Find y (x) if y = 12 sin x + 5 cos x +
x4 . 4
1 . 1 − 2 cos θ
dy if y = 2x sin x − 2 cos x + x2 cos x. dx
3.4.13 Find f (θ) if f (θ) =
1 + sin θ . 1 − sin θ
3.4.14 Find
3.4.15 Show by use of a trigonometric identity that
1 + tan t dy . if y = 1 − tan t dt
d [tan x − x] = tan2 x. dx
3.4.16 Show by use of a trigonometric identity that d [x + cot x] = − cot2 x dx 3.4.17 A 12 foot long ladder leans against a wall at an angle θ with the horizontal as shown in the figure. The top of the ladder is x feet above the ground. If the bottom of the ladder is pushed toward the wall, find the rate at which x changes with θ when θ = 60◦ . Express the answer in units of feet/degree.
3.4.18 An airplane is flying on a horizontal path at a height of 4500 ft, as shown in the figure. At what rate is the distance s between the airplane and the fixed point P changing with θ when θ = 30◦ . Express the answer in units of feet/degree.
22
Questions, Section 3.4
3.4.19 f (x) = 5 cos x + 3 sin x. Find f (x). 3.4.20 f (x) = x2 sin x + cos x. Find f (x). 3.4.21 f (x) = x tan x. Find f (x). 3.4.22 f (x) = 3.4.23 Find
sec x . Find f (x). x3
dy if y = cot x. dx
SOLUTIONS SECTION 3.4 3.4.1 f (x) = x(sec2 x) + tan x(1) = x sec2 x + tan x 3.4.2 f (x) = x(cos x) + sin x(1) = x cos x + sin x; f (x) = x(− sin x) + cos x(1) + cos x = 2 cos x − x sin x 3.4.3
x2 (cos x) − sin x(2x) dy x2 cos x − 2x sin x x cos x − 2 sin x = = = x4 x4 x3 dx
3.4.4
dy = sec x(sec2 x) + tan x(sec x tan x) = sec3 x + sec x tan2 x dx (1 + csc x)(− csc2 x) − cot x(− csc x cot x) (1 + csc x)2
3.4.5 f (x) = =
− csc2 x − csc3 x + csc x cot2 x csc x(− csc x − csc2 x + cot2 x) = 2 (1 + csc x)2 (1 + csc x)
=
csc x(− csc x − 1) − csc x = 1 + csc x (1 + csc x)2
3.4.6 f (x) = (5x2 + 7)(− sin x) + cos x(10x) = −(5x2 + 7) sin x + 10x cos x 1/2
3.4.7
dy = dx
x
(− csc x cot x) − csc x
1 −1/2 x 2
x
1 −x1/2 csc x cot x − x−1/2 csc x −2x csc x cot x − csc x 2 = = x 2x3/2 3.4.8
dy − sin x + sin2 x + cos2 x (1 − sin x)(− sin x) − cos x(− cos x) = = dx (1 − sin x)2 (1 − sin x)2 =
3.4.9
− sin x + 1 1 = 2 (1 − sin x) 1 − sin x
dy = (x3 + 7x)(sec2 x) + tan x(3x2 + 7) dx = (x3 + 7x) sec2 x + (3x2 + 7) tan x
3.4.10 y = 12 cos x − 5 sin x + x3 , y = −12 sin x − 5 cos x + 3x2 3.4.11 f (θ) = − 3.4.12
2 sin θ (reciprocal rule) (1 − 2 cos θ)2
dy = 2x(cos x) + 2 sin x(1) − 2(− sin x) + x2 (− sin x) + cos x(2x) dx = 4x cos x + 4 sin x − x2 sin x
3.4.13 f (θ) =
(1 − sin θ)(cos θ) − (1 + sin θ)(− cos θ) 2 cos θ = 2 (1 − sin θ) (1 − sin θ)2
23
24
Solutions, Section 3.4
3.4.14
dy 2 sec2 t (1 − tan t)(sec2 t) − (1 + tan t)(− sec2 t) = = 2 dt (1 − tan t) (1 − tan t)2
3.4.15
d [tan x − x] = sec2 x − 1 = tan2 x dx
3.4.16
d [x + cot x] = 1 − csc2 x = − cot2 x dx x 12 x = 12 sin θ
3.4.17
sin θ =
dx = 12 cos θ dθ
◦
θ = 60
dx = 12 cos 60◦ = 12 12 = 6 ft/degree dθ θ=60◦ s 4500 s = 4500 csc θ
3.4.18
csc θ =
ds = −4500 csc θ cot θ dθ
θ = 30◦
√ ds = −45000 csc 30◦ cot 30◦ = −4500(2)( 3) dθ θ=30◦ √ = −9000 3 ft/degree
3.4.19 −5 sin x + 3 cos x 3.4.20 x2 cos x + 2x sin x − sin x 3.4.21 tan x + x sec2 x 3.4.22
x3 sec x tan x − 3x2 sec x x sec x tan x − 3 sec x = x6 x4
3.4.23 − csc2 x
Questions, Section 3.5
25
SECTION 3.5 3.5.1 Find f (x) where f (x) = x2 (sin 2x)3 . 3.5.2 Find f (x) where f (x) =
3 . (x2 − 2x + 2)3
3.5.3 Find f (x) where f (x) = sin(tan 2x). 3.5.4 Find f (θ) where f (θ) = (θ + sin 2θ)2 . 3.5.5 Find f (θ) where f (θ) = sin2 (2θ2 − θ)3 . 3.5.6 Find f
π where f (x) = cos3 2x. 12
3.5.8 Find f (x) where f (x) = csc3 4x.
3.5.7 Find f
π 8
where f (x) = sin2 2x.
3.5.9 Find f (x) where f (x) = sec2 (3x − x2 ).
3.5.10 Find f (x) where f (x) = (x2 − 3)3 (x2 + 1)2 . 3.5.11 Find
dy where y = (x + 4)4 (3x + 2)3 . dx
3.5.12 Find
dy where y dx
x+1 x−1
3.5.13 Find y (π) where y =
2 .
−1 1 + sin x . x
3.5.14 Find f (t) where f (t) =
1 1 + 2 t t
4 .
3.5.15 Find equations for the tangent and normal lines to the graph of f (x) = sin(4 − x2 ) at x = 2. 3.5.16 Find equations for the tangent and normal lines to the graph of f (x) = x cos 4x at x = π/4. 3.5.17 Find f (x) where f (x) = (x4 + 3x)52 . 3.5.18 Find f (x) where f (x) =
√
x5 + 2x + 3.
3.5.19 y = x4 cos(6x). Find dy/dx. 3.5.20 Find f (x) if f (x) =
√
x4 + 3x2 + 2.
3.5.21 Find f (x) if f (x) = cos(x7 ). 3.5.22 Given that f (1) = 2, f (1) = 4 and g(x) = (f (x))−3 , find g (1). 3.5.23 Find (f ◦ g) (0) if f (0) = 4, g(0) = 0 and g (0) = 2. 3.5.24 Find f (x) if f (x) =
1 . (x − 2)6
SOLUTIONS SECTION 3.5 3.5.1 f (x) = x2
d d [(sin 2x)3 ] + (sin 2x)3 [x2 ] dx dx
= x2 (3)(sin 2x)2
d [sin 2x] + (sin 2x)3 (2x) dx
d [2x] + 2x(sin 2x)3 dx = 6x2 cos 2x(sin 2x)2 + 2x(sin 2x)3 = 3x2 (sin 2x)2 cos 2x
= 2x(sin 2x)2 (3x cos 2x + sin 2x) 3.5.2 f (x) = 3(−3)(x2 − 2x + 2)−4
d 2 [x − 2x + 2] dx
= −9(x2 − 2x + 2)−4 (2x − 2) = 3.5.3 f (x) = cos(tan 2x)
18(1 − x) − 2x + 2)4
(x2
d [tan 2x] dx
d [2x] dx = sec2 2x cos(tan 2x)(2) = cos(tan 2x)(sec2 2x)
= 2 sec2 2x cos(tan 2x) d [θ + sin 2θ] dθ d = 2(θ + sin 2θ) 1 + cos 2θ [2θ] dθ = 2(θ + sin 2θ)(1 + 2 cos 2θ)
3.5.4 f (θ) = 2(θ + sin 2θ)
3.5.5 f (θ) = 2 sin(2θ2 − θ)3
d [sin(2θ2 − θ)3 ] dθ
= 2 sin(2θ2 − θ)3 cos(2θ2 − θ)3
d [(2θ2 − θ)3 ] dθ
d [2θ2 − θ] dθ = 6(2θ2 − θ)2 sin(2θ2 − θ)3 cos(2θ2 − θ)3 (4θ − 1) = 2 sin(2θ2 − θ)3 cos(2θ2 − θ)3 (3)(2θ2 − θ)2
= 6(4θ − 1)(2θ2 − θ)2 sin(2θ2 − θ)3 cos(2θ2 − θ)3 or 3(4θ − 1)(2θ2 − θ)2 sin 2(2θ2 − θ)3 3.5.6 f (x) = 3 cos2 2x
d [cos 2x] dx
d [2x] dx = −3 cos2 2x sin 2x(2) π 9 =− = −6 cos2 2x sin 2x, so f 12 4 = 3 cos2 2x(− sin 2x)
26
Solutions, Section 3.5
27
3.5.7 f (x) = 2 sin 2x
d [sin 2x] dx
= 2 sin 2x cos 2x
d [2x] dx
= 4 sin 2x cos 2x, so f 3.5.8 f (x) = 3 csc2 4x
π 8
=2
d [csc 4x] dx
= 3 csc2 4x(− csc 4x cot 4x)
d [4x] dx
= −3 csc3 4x cot 4x(4) = −12 csc3 4x cot 4x 3.5.9 f (x) = 2 sec(3x − x2 )
d [sec(3x − x2 )] dx
d [3x − x2 ] dx = 2 sec2 (3x − x2 ) tan(3x − x2 )(3 − 2x)
= 2 sec2 (3x − x2 ) tan(3x − x2 )
= 2(3 − 2x) sec2 (3x − x2 ) tan(3x − x2 ) 3.5.10 f (x) = (x2 − 3)3
d d [(x2 + 1)2 ] + (x2 + 1)2 [(x2 − 3)3 ] dx dx
d d 2 [x + 1] + (x2 + 1)2 (3)(x2 − 3)2 [x2 − 3] dx dx = 2(x2 + 1)(x2 − 3)3 (2x) + 3(x2 + 1)2 (x2 − 3)2 (2x) = (x2 − 3)3 (2)(x2 + 1)
= 4x(x2 + 1)(x2 − 3)3 + 6x(x2 + 1)2 (x2 − 3)2 = 2x(x2 + 1)(x2 − 3)2 (5x2 − 3) 3.5.11
d d dy = (x + 4)4 (3)(3x + 2)2 [3x + 2] + (3x + 2)3 (4)(x + 4)3 [x + 4] dx dx dx = 3(x + 4)4 (3x + 2)(3) + 4(3x + 2)3 (x + 4)3 (1) = (x + 4)3 (3x + 2)2 (21x + 44)
3.5.12
dy =2 dx =
x+1 x−1
d x+1 x+1 (x − 1)(1) − (x + 1)(1) =2 dx x − 1 x−1 (x − 1)2
−4(x + 1) 4(x + 1) = 3 (x − 1) (1 − x)3
3.5.13 y (x) = −(x−1 + sin x)−2
d −1 [x + sin x] dx
= −(x−1 + sin x)−2 (−x−2 + cos x) −2 1 1 − 2 + cos π = π 2 + 1 so y (π) + − + sin π π π
28
Solutions, Section 3.5
3.5.14 f (t) = 4(t−1 + t−2 )3 = 4(t
−1
+t
d −1 [t + t−2 ] dt
−2 3
) (−t
−2
− 2t
−3
) = −4
1 1 + 2 t t
3
1 2 + 3 2 t t
d [4 − x2 ] = cos(4 − x2 )(−2x) = −2x cos(4 − x2 ) dx so the slope of the tangent to the graph of f at x = 2 is f (2) = −4 cos 0 = −4, thus, the equation of the tangent to f at (2, 0) is y − 0 = −4(x − 2) or y = −4x + 8; the slope of the 1 1 1 = so the equation of the normal to f at (2, 0) is y −0 = (x−2) normal to f at x = 2 is − 4 4 −4 1 1 or y = x − . 4 2
3.5.15 f (x) = cos(4 − x2 )
d d d [cos 4x] + cos 4x [x] = x(− sin 4x) [4x] + cos 4x(1) = −4x sin 4x + cos 4x dx dx dx π π 4π 4π 4π so the slope of the tangent to the graph of f at x = is f =− sin + cos = −1, 4 4 4 4π π π 4 π is y − − = − x− or y = −x; the ,− thus the equation of the tangent to f at 4 4 4 4 π π 1 slope of the normal to f at x = π/4 is − = 1 so the equation of the normal to f at ,− −1 4 4 π π π is y − − = x− or y = x − . 4 4 2
3.5.16 f (x) = x
3.5.17 f (x) = 52(x4 + 3x)51
d 4 (x + 3x) = 52(x4 + 3x)51 (4x3 + 3) dx
1 5x4 + 2 d 5 3.5.18 f (x) = √ (x + 2x + 3) = √ 2 x5 + 2x + 3 dx 2 x5 + 2x + 3 dy = x4 (−6 sin(6x)) + 4x3 (cos(6x)) = −6x4 sin(6x) + 4x3 cos(6x) dx √ x4 + 3x2 + 2(2x3 + 3x) 1 4 2 −1/2 3 3.5.20 f (x) = (x + 3x + 2) (4x + 6x) = 2 x4 + 3x2 + 2 3.5.19
3.5.21 f (x) = −7x6 sin(x7 ) 3.5.22 g(x) = (f (x))−3 d f (x) = −3(f (x))−4 f (x) so dx 3 g (1) = −3(f (1))−4 f (1) = −3(2)−4 (4) = − 4
g (x) = −3(f (x))−4
3.5.23 (f ◦ g) (x) f (g(x))g (x) so (f ◦ g) (0) = f (g(0))g (0) = f (0)g (0) = 8 3.5.24 f (x) = (x − 2)−6 f (x) = −6(x − 2)−7 =
−6 (x − 2)7
Questions, Section 3.6
SECTION 3.6 3.6.1 Find dy/dx by differentiating implicitly. x4 + 2xy + 4x = 7 3.6.2 Find dy/dx by differentiating implicitly. x3 − y 2 = 3x 3.6.3 Find dy/dx by differentiating implicitly. 3xy = x2 3.6.4 Find dy/dx by differentiating implicitly. x sin y = x2 3.6.5 Find dy/dx by differentiating implicitly. sin(xy 2 ) = x 3.6.6 Find dy/dx by differentiating implicitly. xy = y 2 3.6.7 Find dy/dx by differentiating implicitly. xy 2 = x2 3.6.8 Find dy/dx by differentiating implicitly. tan(xy) = x 3.6.9 Find dy/dx by differentiating implicitly. x2 − y 2 = 169 3.6.10 Find dy/dx by differentiating implicitly. sin y = cos x
29
SOLUTIONS SECTION 3.6 3.6.1 4x3 + 2xy + 2y + 4 = 0
3.6.2 3x2 − 2yy = 3
2xy = −(4x3 + 2y + 4) y = −
−2yy = 3 − 3x2
2x3 + y + 2 4x3 + 2y + 4 =− x 2x
y =
−3 + 3x2 2x
3.6.3 3xy + 3y = 2x
3.6.4 x cos y y + sin y = 2x
3xy = 2x − 3y
x cos y y = 2x − sin y
y =
2x − 3y 3x
y =
3.6.5 cos(xy 2 )(2xyy + y 2 ) = 1
3.6.6 xy + y = 2xy xy − 2xy = −y
1 2xyy + y = cos(xy 2 )
2xyy = y =
2
(x − 2y)y = −y
1 − y 2 cos(xy 2 ) cos(xy 2 )
y =
3.6.8 sec2 (xy)(xy + y) = 1
2xyy = 2x − y 2
xy + y =
2x − y 2 2xy
xy = y =
3.6.9 2x − 2yy = 0
1 sec2 (xy)
1 −y sec2 (xy)
1 y − 2 x sec (xy) x
3.6.10 cos y y = − sin x
−2yy = −2x y =
−y x − 2y
1 − y 2 cos(xy 2 ) 2xy cos(xy 2 )
3.6.7 2xyy + y 2 = 2x
y =
2x − sin y x cos y
y = −
x y
30
sin x cos y
Questions, Section 3.7
31
SECTION 3.7 3.7.1 A shark, looking for dinner, is swimming parallel to a straight beach and 90 feet offshore. The shark is swimming at the constant speed of 30 feet per second. At time t = 0, the shark is directly opposite a lifeguard station. How fast is the shark moving away from the lifeguard station when the distance between them is 150 feet? 3.7.2 A ladder 13 feet long is leaning against a wall. If the base of the ladder is moving away from the wall at the rate of 1/2 foot per second, at what rate will the top of the ladder be moving when the base of the ladder is 5 feet from the wall? 3.7.3 A spherical balloon is inflated so that its volume is increasing at the rate of 3 cubic feet per minute. How fast is the radius of the balloon increasing at the instant the radius is 1/2 foot?
4 V = πr3 3 3.7.4 Sand is falling into a conical pile so that the radius of the base of the pile is always equal to one half its altitude. If the sand is falling at the rate of 10 cubic feet per minute, how fast is the altitude of the pile increasing when the pile is 5 feet deep?
1 V = πr2 h 3 3.7.5 A metal cone contracts as it cools. Assume the height of the cone is 16 cm and the radius at the base of the cone is 4 cm. If the height of the cone is decreasing at 4.0 × 10−5 cm per second, at what rate is the volume of the cone decreasing when its height is 15 cm?
1 2 V = πr h 3 3.7.6 A spherical balloon is inflated so that its volume is increasing at the rate of 20 cubic feet per minute. How fast is the surface area of the balloon increasing when the radius is 4 feet? [Use 4 V = πr3 and S = 4πr2 .] 3 3.7.7 Two ships leave port at noon. One ship sails north at 6 miles per hour and the other sails east at 8 miles per hour. At what rate are the two ships separating 2 hours later? 3.7.8 A conical funnel is 14 inches in diameter and 12 inches deep. A liquid is flowing out at the rate of 40 cubic inches per second. How fast is the depth of the liquid falling when the level is 6 inches deep?
1 V = πr2 h 3 3.7.9 A baseball diamond is a square 90 feet on each side. A player is running from home to first base at the rate of 25 feet per second. At what rate is his distance from second base changing when he has run half way to first base? 3.7.10 A ship, proceeding southward on a straight course at the rate of 12 miles/hr is, at noon, 40 miles due north of a second ship, which is sailing west at 15 miles/hr. (a) How fast are the ships approaching each other 1 hour later? (b) Are the ships approaching each other or are they receding from each other at 2 o’clock and at what rate?
32
Questions, Section 3.7
3.7.11 An angler has a fish at the end of his line, which is being reeled in at the rate of 2 feet per second from a bridge 30 feet above the water. At what speed is the fish moving through the water towards the bridge when the amount of line out is 50 feet? (Assume the fish is at the surface of the water and that there is no sag in the line.) 3.7.12 A kite is 150 feet high and is moving horizontally away from a boy at the rate of 20 feet per second. How fast is the string being paid out when the kite is 250 feet from him? 3.7.13 A kite is flying horizontally at a constant height of 250 feet above the girl flying the kite. At a certain instant, the angle which the string makes with the girl is 30◦ and decreasing. If the string is paying out at 16 feet per second, how fast is the angle decreasing? Express your answer in degrees per second. 3.7.14 Consider a rectangle where the sides are changing but the area is always 100 square inches. If one side changes at the rate of 3 inches per second, when it is 20 inches long, how fast is the other side changing? 3.7.15 The sides of an equilateral triangle are increasing at the rate of 5 centimeters per hour. At what rate is the area increasing when the side is 10 centimeters? 3.7.16 A circular cylinder has a radius r and a height h feet. If the height and radius both increase at the constant rate of 10 feet per minute, at what rate is the lateral surface area increasing? (S = 2πrh) 3.7.17 A straw is used to drink soda from the bottom of a cylindrical shaped cup. The diameter of the cup is 3 inches. The liquid is being consumed at the rate of 3 cubic inches per second. How fast is the level of the soda dropping? [V = πr2 h] 3.7.18 The edge of a cube of side x is contracting. At a certain instant, the rate of change of the surface area is equal to 6 times the rate of change of its edge. Find the length of the edge. 3.7.19 An aircraft is climbing at a 30◦ angle to the horizontal. Find the aircraft’s speed if it is gaining altitude at the rate of 200 miles per hour. 4 3 πr , where r is the radius. Given that the radius 3 increases at 0.001 m/s when r = 1 m, find the rate at which the volume increases.
3.7.20 The volume of a sphere is given by V =
3.7.21 A ship is sailing 6 east of being 500 directly offshore and 500 miles from shore at 4 mi/hr east parallel to the coast. How fast is the ship moving away from the observer? 3.7.22 A ship is sailing 60 east of being directly offshore and 40 miles from shore at 4 mi/hr east parallel to the coast. How fast is the ship moving away from the observer?
SOLUTIONS SECTION 3.7 3.7.1 Let x and r be the distances shown on the diagram. We want to find
dr given that dt r=150
dx dr x dx dr dx = 2x , or = . When r = 150, = 30. From r2 = (90)2 + x2 , we get 2r dt dt dt r dt dt dr 120 x2 = (150)2 −(90)2 = 22500−8100 = 14400 and x = 120, thus = (30) = 24 ft/sec. dt r=150 150
dx dy 1 given that 3.7.2 Let x and y be as shown on the diagram. We want to find = . From dt x=5 dt 2 dx dy x dx dy + 2x = 0, or, =− . When x = 5, y 2 = (13)2 − (5)2 = dt dt dt y dt 5 dy 5 1 =− 169 − 25 = 144, so y = 12 and = − ft/sec., i.e., the top of the ladder dt x=5 12 2 24 y 2 + x2 = (13)2 , we get 2y
is moving down at the rate of
5 ft/sec. 24
dr dr dV dV 4 4 given that = 3. So, from V = πr3 , we get = π(3)r2 dt 3 dt 3 dt dt r=1/2 1 dV dr 1 3 dr = and = or 2 (3) = ft/min. 2 dt 4πr dt dt r=1/2 π 1 4π 2
3.7.3 We want to find
33
34
Solutions, Section 3.7
2 1 2 1 h dV h π 3 dh = 10. Since V = πr h and r = , then V = π h , given that h= 3.7.4 Find dt h=5 dt 3 2 3 2 12 π dh 4 dV dh 4 dh 8 dV = (3)h2 , and = , thus = ft/min. (10) = 2 2 dt 12 dt dt πh dt dt π(5) 5π
1 dh 1 dv given that 3.7.5 Find = −4.0 × 10−5 cm/sec. Since V = πr2 h and r = h, then dt 3 4 dt h=15cm 2 1 h dh πh3 dV π π π 2 dh dV V = π h= = , = (3)h2 = h (15)2 (−4.0 × 10−5 ) = 4 48 dt 48 dt 16 dt dt n=15 16 3 −1.8×10−3 cm3 /sec, i.e., the volume of the cone is decreasing at the rate of 1.8×10−3 cm3 /sec.
dS 3.7.6 Find dt
given that r=4
4 4 dr dV dV = 20. From V = πr3 and S = 4πr2 , we get = π(3)r2 dt 3 dt 3 dt
dr dr dr 1 dV 1 dS dS 2 dV dS = 4π(2)r . Then, solving for , = = or = so, and 2 dt dt dt dt 4πr dt 8πr dt dt r dt dS 2 = (20) = 10, thus, the surface area is increasing at the rate of 10 sq. ft/min. dt 4 r=4
3.7.7 Let A and B be the two ships and x, y, and r be their distances as shown on the diagram. dy dr dx hrs, given that We want to find = 6 and = 8. From r2 = x2 + y 2 , we get dt t=2 dt dt
dx dr dx dy dr 1 dy x . When t = 2 hours, B will have sailed 16 miles 2r = 2x + 2y or = +y dt dt dt dt r dt dt east and A will have sailed 12 miles north of port, thus, r2 = (16)2 + (12)2 = 400, r = 20. dr 1 = [16(8) + 12(6)] = 10, thus, the ships are separating at the rate of 10 miles per hour. dt t=2 20
Solutions, Section 3.7
35
dh given that 3.7.8 Let r and h be the dimensions shown on the diagram. We want to find dt h=6 2 dV r 7 7h π 7h h = = −40. By similar triangles (see figure), = or r = , thus, V = dt h 12 12 3 12 49π 3 dV 144 160 49π dh 144 dV dh 2 dh h . = (−40) = − = (3)h or = , thus, . The dt 432 dt dt 49πh2 dt dt 49π(6)2 49π 432 h=6
160 inches per second. depth of the liquid is falling at the rate of − 49π
3.7.9 Let x, y, and r be the distances as shown in the diagram and let the baseball diamond be posi dx dx dr = −25. Thus x2 + y 2 = r2 , 2x = given that tioned as shown. We want to find dt dt dt x=45
√ dr x dx dr [y is constant] so = when x = 45, r = (45)2 + (90)2 = 2025 + 8100 = 2r dt dt r dt √ √ √ 45 dr = √ (−25) = −5 5 feet per second. 10125 = 45 5 feet, so, dt 45 5
36
Solutions, Section 3.7
3.7.10 (a) Let A be one ship and B be the other. Let x, y, and r be the dimensions shown in dr the figure at a certain instant of time. We want to find , since x and y are both dt t=1
functions of time, let x = 15t and y = 40 − 12t, then, r = x + y 2 = (15t)2 + (40 − 12t)2 ; 2
2
1 dr dr = [369t − 480]. When t = 1, r2 = [15(1)]2 + = 2(15t)(15) + 2(40 − 12t)(−12) or dt r dt √ dr 111 1 [40 − 12(1)]2 = 1009, r = 1009 and [369(1) − 480] = − √ ≈ −3.49 = √ dt 1009 1009 111 miles/hr. The ships are approaching each other at the rate of √ ≈ 3.49 miles/hr 1 1009 hour later.
2r
(b) At 2 o’clock, t = 2 hours, so, r2 = [15(2)]2 + [40 − 12(2)]2 = 1156, r = 34.
dr = dt t=2
258 129 129 1 [369(2) − 480] = or . At 2 o’clock, the ships are receding at the rate of 34 34 17 17 miles/hour. 3.7.11 Let x and r be as shown on the diagram. We want to find
dx dr = −2. From given that dt r=50 dt
dr dx r dr dx = 2r and = . When r = 50, x2 = (50)2 −(30)2 = 2500−900 = dt dt dt x dt 50 dx 100 5 = 1600, x = 40 and (−2) = − = − . The fish is moving towards the bridge at dt r=50 40 40 2 x2 +(30)2 = r2 , 2x
the rate of
5 ft/sec when the amount of line out is 50 feet. 2
Solutions, Section 3.7
37
dx dr given that 3.7.12 Let x and r be as shown on the diagram. We want to find = 20. From dt dt r=250 dr dx dr x dx = 2x or = . When r = 250, x2 = (250)2 − (150)2 = dt dt dt r dt 200 dr (20) = 16. The string is being paid out = 62500 − 22500 = 40000; x = 200 and 250 dt r=250 r2 = (150)2 + x2 , we get 2r
at the rate of 16 ft/sec when the kite is 250 away from the boy.
250 dθ dr = 16. From sin θ = , given that dt θ=30◦ dt r 250 dr dθ dθ sec θ dr 250 250 = − 2 cos θ , = −250 2 . From sin θ = , r = . When θ = 30◦ , dt r dt dt r dt r sin θ dθ 250 sec 30◦ = 500. So r= (16) = 0.0185 deg/sec., i.e., the angle of the string is = −250 dt (500)2 sin 30 decreasing at the rate of 0.0185 deg/sec.
3.7.13 Let θ and r be as shown in the diagram. Find
3.7.14 Let xy = 100. We want to find, say
dy dx dy dx = 3; x +y = 0 and given that dt dt dt dt x=20
y dx dy dy 5 3 =− . From 20y = 100, y = 5, and = − (3) = − inch/sec, thus, the rate of x dt dt 20 4 dt 3 change of the second side is opposite to that of the first side and is at the rate of inch/sec. 4
38
Solutions, Section 3.7
√ √ 3 3 2 1 dA x = x . We want to find 3.7.15 Let x = side of the triangle. The area is A = (x) 2 2 4 dt x=10 √ √ √ dA 3 3 dx dx dA = 5, given that = = (2)x . Thus, (10)(5) = 25 3 or the area is dt 4 dt dt x=10 2 dt √ increasing at the rate of 25 3 cm2 /hr when the side is 10 cm.
3.7.16 We want to find
dS dh dr dS dh dr or given that = = 10, from S = 2πrh, = 2π r +h dt dt dt dt dt dt
dS = 2π[r(10) + h(10)] = 20π(r + h) ft2 min. dt dV dh = −3 in3 /sec. From V = πr2 h = given that 3.7.17 With d and h as shown in the figure find dt dt 2 π dV 4 d π dh dh 4 dV 4 h = d2 h (−3) = π = d2 and = = or −0.42 in/sec., i.e., 2 4 dt 4 dt dt πd2 dt π(3)2 3π 4 in/sec. or 0.42 in/sec. the level of the soda is dropping at the rate of 3π
dx dx dS dx dS dS = 6 . From S = 6x2 , we get = 6(2)x , but =6 , 3.7.18 We want to find x given that dt dt dt dt dt dt dx dx dx so 6 = 12x , t (2x − 1) = 0; x = 1/2. dt dt dt
Solutions, Section 3.7
39
dh ds given that = 200 mph. From the 3.7.19 With s and h as shown in the figure, we want to find dt dt h ds dh figure, s = = 2h so =2 = 2(200) = 400 mph. sin 30◦ dt dt
3.7.20 V = V =
4 2 πr r 3 4 π(1)2 (0.001) = 0.004 m/s 3
3.7.21 x2 + y 2 = h2 2xx = 2hh xx (6)(4) h = = = 0.048 mi/hr h 500 √ 3.7.22 x2 + y 2 = h2 h = 402 + 602 = 72.1 2xx = 2hh xx (60)(4) h = = = 3.33 mi/hr h 72.1
40
Questions, Section 3.3
SECTION 3.8 3.8.1 Let y = x3 − 1. (a) Find ∆y if ∆x = 1 and the initial value of x is x = 1. (b) Find dy if dx = 1 and the initial value of x is x = 1. (c) Make a sketch of y = x3 − 1 and show ∆y and dy in the picture. 1 2 x + 1. 2 (a) Find ∆y if ∆x = 1 and the initial value of x is x = 1.
3.8.2 Let y =
(b) Find dy if dx = 1 and the initial value of x is x = 1. 1 (c) Make a sketch of y = x2 + 1 and show ∆y and dy in the picture. 2 √ 3.8.3 Use a differential to approximate 4 14. 3.8.4 Use a differential to approximate 3.8.5 Use a differential to approximate 3.8.6 Use a differential to approximate
√ 3 √ 5 √ 3
9. 29. 10.
3.8.7 Use a differential to approximate (1.98)4 . 3.8.8 Use a differential to approximate cos 58◦ . 3.8.9 Use a differential to approximate sin 31◦ . 3.8.10 Use a differential to approximate tan 43◦ . 3.8.11 The surface area of a sphere is given by S = 4πr2 where r is the radius of the sphere. The radius is measured to be 3 cm with an error of ±0.1 cm. (a) Use differentials to estimate the error in the calculated surface area. (b) Estimate the percentage error in the radius and surface area. 3.8.12 The surface area S of a cube is to be computed from a measured value of its side x. Estimate the maximum permissible percentage error in the side measurement if the percentage error in the surface area must be kept to within ±4%. 3.8.13 A circular hole 6 inches in diameter and 10 feet deep is to be drilled out of a glacier. The diameter of the hole is exact but the depth of the hole is measured with an error of ±1%. π Estimate the percentage error in the volume of ice removed. (V = d2 h is the volume of a 4 cylinder of diameter d and height h.) 3.8.14 The pressure P , the volume V , and the temperature T of an enclosed gas are related by the Ideal Gas Law, P V = kT where k is a constant. With the temperature held constant, the volume of the gas is calculated from a measured value of its pressure. Estimate the maximum permissible error in the pressure measurement if the percentage error in the volume must be kept to within ±2%.
Questions, Section 3.3
41
k , where r is the distance from the r2 magnetic source and k is a constant. r is measured to be 3 cm with a possible error of ±6%.
3.8.15 The magnetic force F acting on a particle is given by F =
(a) Use differentials to estimate the error in the calculated value of F . (b) Estimate the percentage error in F and r. 3.8.16 When a cubical block of metal is heated, each edge increases by 0.1% per degree increase in temperature. Use differentials to estimate the percentage increase in the surface area and volume of the block per degree increase in temperature. 3.8.17 When a spherical ball of metal is heated, the radius of the sphere increases by 0.1% per degree increase in temperature. Use differentials to estimate the percentage increase in the surface area and volume of the ball per degree increase in temperature. 4 3 2 S = 4πr and V = πr 3 3.8.18 The area of a circle is to be computed from a measured value of its diameter. Estimate the maximum permissible percentage error in the measurement if the percentage error in the area must be kept within 0.5%. 3.8.19 Let y = x4 and dx = 0.01 at x = 2. Find dy. 3.8.20 Let y = 3x2 and dx = 0.02 at x = 4. Find dy. 3.8.21 Let y = 5x4 and dx = 0.001 at x = 3. Find dy. 3.8.22 Find the formula for dy if y = x9 . 3.8.23 Find the formula for dy if y =
√
x + 3.
3.8.24 Find the formula for dy if y = tan x. 3.8.25 Find the formula for ∆y if y = x2 − 3. 3.8.26 Find the formula for ∆y if y = tan x.
SOLUTIONS SECTION 3.8 3.8.1 (a) ∆y = (x + ∆x)3 − 1 − (x3 − 1) = (1 + 1)3 − 1 − (13 − 1) = 7 (b) dy = 3x2 dx = 3(1)2 (1) = 3 (c)
3.8.2 (a) ∆y =
1 1 1 3 1 (x + ∆x)2 + 1 − x2 + 1 = (1 + 1)2 + 1 − (1)2 + 1 = 2 2 2 2 2
(b) dy = x dx = (1)(1) = 1 (c)
3.8.3 Let f (x) =
√ 4
x, x0 = 16, ∆x = −2, then f (x) =
1 −3/4 x and x0 + ∆x = 14 so 4
f (14) ≈ f (16) + f (16)(−2) ≈ 3.8.4 Let f (x) =
√ 3
√ 4
1 1 31 (−2) = 2 − = 16 16 4(16)3/4
16 +
x, x0 = 8, ∆x = 1, then, f (x) =
1 −2/3 x and x0 + ∆x = 9, so 3
f (9) ≈ f (8) + f (8)(1) ≈ 3.8.5 Let f (x) =
√ 5
√ 3
8+
1 1 25 (1) = 2 + = 12 12 3(8)2/3
x, x0 = 32, ∆x = −3, then f (x) =
1 −4/5 and x0 + ∆x = 29, so x 5
f (29) ≈ f (32) + f (32)(−3) ≈
√ 5
32 +
157 1 3 = (−3) = 2 − 4/5 80 80 5(32)
42
Solutions, Section 3.8
3.8.6 Let f (x) =
43
√ 3
x, x0 = 8, ∆x = 2, then f (x) =
1 −2/3 x and x0 + ∆x = 10, so 3
f (10) ≈ f (8) + f (8)(2) ≈
√ 3
8+
13 1 1 (2) = 2 + = 6 6 3(8)2/3
3.8.7 Let f (x) = x4 , so x0 = 2, ∆x = −0.02, then f (x) = 4x3 and x0 + ∆x = 1.98, so f (1.98) ≈ f (2) + f (2)(−0.02) ≈ (2)4 + 4(2)3 (−0.02) = 16 − 0.64 = 15.36 π −π radians, ∆x = −2◦ = radians, then, f (x) = − sin x and 3.8.8 Let f (x) = cos x, x0 = 60◦ = 3 90 29π radians, so x0 + ∆x = 90 π π −π π 29π π −π + f = cos − sin =f f 3 3 90 3 3 90 90 √ 1 3 π = + ≈ 0.5302 2 2 90 π π 3.8.9 Let f (x) = sin x, x0 = 30◦ = radians, ∆x = 1◦ = , then f (x) = cos x and x0 + ∆x = 6 180 31π , so 180 π π π π π π 31π ≈f + f ≈ sin + cos f 180 6 6 180 6 6 180 √ 1 3 π ≈ + ≈ 0.515 2 180 2 π −π 3.8.10 Let f (x) = tan x, x0 = 45◦ = radians, ∆x = −2◦ = radians, then f (x) = sec2 x and 4 90 43π , so x0 + ∆x = 180 π −π π −π 43π π 2 π ≈f +f ≈ tan + sec f 180 4 4 90 4 4 90 −π ≈ 1 + (2) ≈ 0.930 90 3.8.11 (a) ds = 8πr dr = 8π(3)(±0.1) = ±2.4π
dr ±0.1 = = ±0.033 so the percentage error is r 3 8πr dr dr ds = = 2(±0.033) = =2 ≈ ±3.3%; the relative error in the surface area is ≈ 2 s 4πr r ±0.066 so the percentage error in the surface area is ≈ ±6.6%.
(b) The relative error in the radius is ≈
dS dS dx 12x dx 3.8.12 The relative error in S is ≈ =2 where S = 6x2 and dS = 12x dx, thus = so 2 S S 6x x dS 1 1 dx and the percentage error is ≈ (±4%) = ±2%. = x 2 S 2
44
Solutions, Section 3.8
π 1 dV where V = d2 h, but d is exactly 6 inches or foot so V 4 2 π 2 dh dh π dV ±0.01 π 1 π = h and dV = dh, so = 16 = = ±0.001, thus the V = h= π 4 2 16 16 V h 10 h 16 percentage error is V ≈ ±0.1%.
3.8.13 The relative error in V is ≈
kT − 2 dP kT dV dP kT 3.8.14 If V = = P ≈ relative error =− , then dV = − 2 dP (T held constant) so kT V P P P P dP dV = −(±2%) = ±2%. in P , thus =− P V 2k −2k(±0.06) ±0.04k = ±0.0044k dr = = r3 (3)3 9 dr 0.06 (b) The relative error in r ≈ =± = ±0.02 so the percentage error in r is ±2%; the r 3 −2k dr 3 dF dr = −2(±0.02) = ±0.04 so the percentage = r relative error in F ≈ = −2 k r F r2 error in F is ±4%.
3.8.15 (a) dF = −
3.8.16 The surface area of the block is S = 6x2 and the relative error in the measurement of the surface 12xdx 2dx dS = = area is approximately = 2(±0.001) = ±0.002 so the percentage error is x S 6x2 S ≈ ±0.2%; the volume of the block is V = x3 and the relative error in the measurement of the dx dV 3x2 dx =3 volume is ≈ = = 3(±0.001) = ±0.003 so the percentage error in V ≈ ±0.3%. V x3 x dr 4π(2r dr) dS = 2 = = S 4πr2 r dr 2(±0.001) = ±0.002, so the percentage error is ±0.2% where is the relative increase r in radius of the sphere; the relative increase in the volume of the sphere is approximately 4 π(3r2 dr) dr dV 3 = =3 = 3(±0.001) = ±0.003, so the percentage error is ±0.3%. 4 3 r V πr 3
3.8.17 The relative increase in the surface area of the sphere is ≈
dA 1 (πD/2)dD dD dD 1 dA πD2 where D is the diameter of the circle; = =2 ; = but 4 A πD2 /4 D D 2 A dD dA 0.005 dD ≈ ±0.005 so ≈ ± , ≈ ±0.0025; maximum permissible percentage error in D 2 D A
3.8.18 A =
D ≈ ±0.25%. 3.8.19 dy = 4x3 dx 3.8.20 dy = 6xdx 3.8.21 dy = 20x3 dx
dy = 4(2)3 (0.01) = 0.32 dy = 6(4)(0.02) = 0.48 dy = 20(3)3 (0.001) = 0.54
Solutions, Section 3.8
45
3.8.22 dy = 9x8 dx
√ x 1 3.8.23 dy = √ dx = dx 2x 2 x
3.8.24 dy = sec2 x dx
3.8.25 ∆y = (x + ∆x)2 − x2
3.8.26 ∆y = tan(x + ∆x) − tan x
46
Supplementary Exercises
SUPPLEMENTARY EXERCISES, CHAPTER 3 In Exercises 1–4, use Definition 3.2.1 to find f (x). 2. f (x) = (x − a)2
1. f (x) = kx (k constant) 3. f (x) =
√
9 − 4x
4. f (x) =
d 3 [|x| ] . 5. Use Definition 3.2.1 to find dx x=0 6. Suppose f (x) =
(a constant)
x x+1
x2 − 1
x≤1 k(x − 1) x > 1
For what values of k is f (a) continuous
(b) differentiable?
7. Suppose f (3) = −1 and f (3) = 5. Find an equation for the tangent line to the graph of f at x = 3. 8. Let f (x) = x2 . Show that for any distinct values of a and b, the slope of the tangent line to 1 y = f (x) at x = (a + b) is equal to the slope of the secant line through the points (a, a2 ) and 2 (b, b2 ). 9. Given the following table of values at x = 1 and x = −2, find the indicated derivatives in parts (a)–(l). x 1 −2
(a) (c) (e) (g) (i) (k)
f (x) 1 −2
d 2 2 [f (x) − 3g(xx )] dx x=1
d f (x) dx g(x) x=−2 d [f g(x)] dx x=1 d [g(f (x))] dx x=−2 d [f (g(4 − 6x))] dx x=1 d [ f (x)] dx x=1
10. Use a graphing utility to show y =
f (x) 3 −5
g(x) −2 1
(b) (d) (f ) (h) (j) (l)
g (x) −1 7 d [f (x)g(x)] dx x=1
d g(x) dx f (x) x=−2 d [f (g(x))] dx x=−2 d [g(g(x))] dx x=−2 d 3 [g (x)] dx x=1 1 d f − x dx 2 x=−2
|x2 − 9| is not differentiable everywhere.
Chapter 3
47
In Exercises 11–16, find f (x) and determine those values of x for which f (x) = 0. 11. f (x) = (2x + 7)6 (x − 2)5 13. f (x) = 15. f (x) =
√
3x + 1 (x − 1)2
3(5x − 1)1/3 3x − 5
(x − 3)4 x2 + 2x 3 3x + 1 14. f (x) = x2 12. f (x) =
16. f (x) =
√ √ x 3 x2 + x + 1
17. Suppose that f (x) = 1/x for all x = 0. (a) Use the chain rule to show that for any nonzero constant a, d(f (ax))/dx = d(f (x))/dx. (b) If y = f (sin x) and v = f (1/x), find dy/dx and dv/dx.
In Exercises 18–27, find the indicated derivatives. 18. 20. 22.
d dx
√
2
2 − 5x x2
19.
dz if z = 4 sin2 r cos2 r dr r=π/6 −2 x du if u = dx x−1
24. d(sec2 x − tan2 x)/dx 26. F (x) if F (x) =
(1/x) + 2x 1 (1/x2 ) + 1 2
dy 3x2 + 7 if y = 2 x −1 dx
√ 21. g (2) if g(x) = 1/ 2x 23. 25.
√ dw if w = 5 v 3 − 4 v dv dy if y = tan t and t = cos(2x) dx x=π/4
27. Φ (x) if Φ(x) =
x2 − 4x √ 5 x
28. Find all values of x for which the tangent to y = x − (1/x) is parallel to the line 2x − y = 5. 29. Find all values of x for which the tangent to y = 2x3 − x2 is perpendicular to the line x + 4y = 10. 30. Find all values of x for which the tangent to y = (x + 2)2 passes through the origin. 31. Find all values of x for which the tangent to y = x − sin 2x is horizontal. 32. Find all values of x for which the tangent to y = 3x − tan x is parallel to the line y − x = 2.
In Exercises 33–35, find ∆x, ∆y, and dy. 33. y = 1/(x − 1); x decreases from 2 to 1.5. 34. y = tan x; x increases from −π/4 to 0. 35. y =
√
25 − x2 ; x increases from 0 to 3.
48
Supplementary Exercises
36. Use a differential to approximate √ (a) 3 −8.25
(b) cot 46◦ .
37. Let V and S denote the volume and surface area of a cube. Find the rate of change of V with respect to S. 38. The amount of water in a tank t minutes after it has started to drain is given by W = 100(t − 15)2 gal. (a) At what rate is the water running out at the end of 5 min? (b) What is the average rate at which the water flows out during the first 5 min? 39. Verify that the function y = cos x − 3 sin x satisfies y + y + y + y = 0.
SOLUTIONS SUPPLEMENTARY EXERCISES, CHAPTER 3 1. f (x) = lim
h−0
k(x + h) − kx = lim k = k h→0 h
(x + h − a)2 − (x − a)2 = lim [2(x − a) + h] = 2(x − a) h→0 h→0 h √ 9 − 4(x + h) − 9 − 4x [9 − 4(x + h)] − [9 − 4x] 3. f (x) = lim = lim h→0 h→0 h( 9 − 4(x + h) + h 9 − 4x)
2. f (x) = lim
= lim h→0
−4 9 − 4(x + h) +
√
9 − 4x
= −√
2 9 − 4x
x x+h − (x + h)(x + 1) − x(x + h + 1) 4. f (x) = lim x + h + 1 x + 1 = lim h→0 h→0 h h(x + 1)(x + h + 1) 1 1 = h→0 (x + 1)(x + h + 1) (x + 1)2
= lim
5.
|0 + h|3 − |0|3 |h|3 d (|x|3 ) = lim = lim = lim h|h| = 0 h→0 h→0 h h→0 dx h x=0
6. (a) f is continuous everywhere for all k, except perhaps at x = 1; lim f (x) = lim (x2 − 1) = 0, x→1−
x→1−
lim f (x) = lim k(x−1) = 0, and f (1) = 0 thus lim f (x) = f (1) for all k, so f is continuous +
x→1+
x→1
x→1
for all k (b) f is differentiable everywhere for all k, except perhaps at x = 1. Using the theorem that precedes Exercise 71, Section 3.3, lim− f (x) = lim− 2x = 2 and lim+ f (x) = lim+ k = k; x→1
x→1
x→1
x→1
these limits are equal if k = 2, so f is differentiable if k = 2. 7. y − (−1) = 5(x − 3), y = 5x − 16 8. f (x) = 2x so mtan = f
a+b 2
= a + b, but msec =
9. (a) 2f (x)f (x) − 3g (x2 )(2x)|x=1 = 12 (c)
b2 − a2 = b + a if a = b so mtan = msec b−a
(b) f (x)g (x) + f (x)g(x)|x=1 = −7
g(x)f (x) − f (x)g (x) =9 g 2 (x) x=−2
(d)
f (x)g (x) − g(x)f (x) 9 =− 2 f (x) 4 x=−2
(e) f (g(x))g (x)|x=1 = f (g(1))g (1) = f (−2)(−1) = 5 (f )
f (g(x))g (x)|x=−2 = f (g(−2))g (−2) = f (1)(7) = 21
(g) g (f (x))f (x)|x=−2 = g (f (−2))f (−2) = g (−2)(−5) = −35 (h) g (g(x))g (x)|x=−2 = g (g(−2))g (−2) = g (1)(7) = −7 (i)
f (g(4 − 6x)g (4 − 6x)(−6)|x=1 = f (g(−2))g (−2)(−6) = f (1)(7)(−6) = −126
49
50
Supplementary Exercises
3g 2 (x)g (x)|x=1 = 3(−2)2 (−1) = −12 1 1 3 [f (x)]−1/2 f (x) (k) = (1)−1/2 (3) = 2 2 2 x=1
(j)
(l)
f (−x/2)(−1/2)|x=−2 = −
3 2
10.
The graph changes direction abruptly at x = −3 and x = 3. The function does not have a derivative at x = −3 or at x = 3. 11. f (x) = (2x + 7)6 5(x − 2)4 + (x − 2)5 6(2x + 7)5 (2) = (2x + 7)5 (x − 2)4 [5(2x + 7) + 12(x − 2)] = (2x + 7)5 (x − 2)4 (22x + 11) = 11(2x + 7)5 (x − 2)4 (2x + 1) so f (x) = 0 if x = −7/2, 2, −1/2. 12. f (x) =
(x2 + 2x)4(x − 3)3 − (x − 3)4 (2x + 2) (x2 + 2x)2
=
(x − 3)3 [4(x2 + 2x) − (x − 3)(2x + 2)] (x2 + 2x)2
=
2(x − 3)3 (x2 + 6x + 3) (x − 3)3 (2x2 + 12x + 6) = (x2 + 2x)2 (x2 + 2x)2
so f (x) = 0 if x − 3 = 0 or if x2√+ 6x + 3 = 0; the solution of x − 3 = 0 is x = 3, and the solution of x2 + 6x + 3 = 0 is x = −3 ± 6. 1 13. f (x) = (3x + 1)1/2 2(x − 1) + (x − 1)2 (3x + 1)−1/2 (3) 2 =
1 (x − 1)(15x + 1) √ (3x + 1)−1/2 (x − 1)[4(3x + 1) + 3(x − 1)] = 2 2 3x + 1
so f (x) = 0 if x = −1/15, 1.
3x + 2 14. f (x) = 3 x2
15. f (x) = 3 · =
2
x2 (3) − (3x + 1)(2x) 3(3x + 2)(3x + 1)2 = − so f (x) = 0 if x = −2/3, −1/3. x4 x7
(3x − 5)(1/3)(5x − 1)−2/3 (5) − (5x − 1)1/3 (3) (3x − 5)2
−2(15x + 8) (5x − 1)−2/3 [5(3x − 5) − 9(5x − 1)] = (3x − 5)2 (3x − 5)2 (5x − 1)2/3
so f (x) = 0 if x = −8/15.
Chapter 3
51
16. f (x) = x1/2 (1/3)(x2 + x + 1)−2/3 (2x + 1) + (x2 + x + 1)1/3 (1/2)x−1/2 =
1 −1/2 2 x (x + x + 1)−2/3 [2x(2x + 1) + 3(x2 + x + 1)] 6
=
7x2 + 5x + 3 6x1/2 (x2 + x + 1)2/3
but 7x2 + 5x + 3 = 0 has no real solutions so there are no values of x for which f (x) = 0. 1 d d d 1 (a) = = [f (x)] [f (ax)] = f (ax) (ax) = ax x dx dx dx dy d 1 (b) = f (sin x) (sin x) = (cos x) = cot x; dx dx sin x
17. (a) by the chain rule;
1 dy d (−1/x2 ) = −1/x = f (1/x) (1/x) = (1/x) dx dx
√
2 2x−2 − x−1 5
√ 2 = −2 2x−3 + x−2 5
18.
d dx
19.
(x2 − 1)(6x) − (3x2 + 7)(2x) 20x dy = =− 2 dx (x2 − 1)2 (x − 1)2
dz 20. z = (2 sin r cos r)2 = sin2 2r so = 2(sin 2r)(cos 2r)(2) = 2 sin 4r dt √ dz = 2 sin(2π/3) = 3 and dr r=π/6 1 21. g(x) = (2x)−1/2 so g (x) = − (2x)−3/2 (2) = −1/(2x)3/2 and g (2) = −1/43/2 = −1/8 2 22. u =
x−1 x
2
= (1 − x−1 )2 so
du = 2(1 − x−1 )(x−2 ) = 2(x − 1)/x3 dx
1/5 −4/5 dw 1 1 3 v − v 1/4 3v 2 − v −3/4 23. w = v 3 − v 1/4 so = dv 5 4 24. 25.
d d (sec2 x − tan2 x) = (1) = 0 dx dx dy dy dt = = sec2 t(−2 sin 2x), if x = π/4 then t = cos(π/2) = 0 dx dt dx dy = sec2 (0)(−2 sin(π/2)) = −2 so dx x=π/4
2x + 4x3 2x(1 + 2x2 ) = = 2x so F (x) = 2 1 + 2x2 1 + 2x2 √ 1 3 1/2 −1/2 3/2 1/2 x − 2x = (1/10)(3x − 4)/ x 27. Φ(x) = (x − 4x )/5, x = 0 so Φ (x) = 5 2 26. F (x) =
52
Supplementary Exercises
28. y = 1 + x−2 , and the slope of 2x − y = 5 is 2 so we want 1 + x−2 = 2 which gives x2 = 1, x = ±1. 29. y = 6x2 − 2x, and the slope of x + 4y = 10 is −1/4 so we want 6x2 − 2x = 4 which results in x = −2/3, 1. 30. y = 2(x + 2) so at (x0 , f (x0 )) the tangent line is y − f (x0 ) = 2(x0 + 2)(x − x0 ), or y − (x0 + 2)2 = 2(x0 + 2)(x − x0 ). But if the line passes through the origin then x = 0, y = 0 must satisfy the latter equation thus −(x0 + 2)2 = −2x0 (x0 + 2) which leads to (x0 + 2)(x0 − 2) = 0 so x0 = −2, 2. 31. y = 1−2 cos 2x; the tangent is horizontal where 1−2 cos 2x = 0 so cos 2x = 1/2, 2x = ±π/3+2kπ, x = ±π/6 + kπ where k = 0, ±1, ±2 . . . . 2 2 2 32. y = 3 − sec √ x, and the slope of y − x = 2 is 1 so we want 3 − sec x = 1 which gives sec x = 2, sec x = ± 2, x = π/4 + kπ/2 where k = 0, ±1, ±2, . . . .
33. ∆x = 1.5 − 2 = −0.5, ∆y = y|x=1.5 − y|x=2 = 2 − 1 = 1, dy 1 dy = dx = − (−0.5) = 0.5 dx (2 − 1)2 x=2
34. ∆x = 0 − (−π/4) = π/4, ∆y = y|x=0 − y|x=−π/4 = 0 − (−1) = 1, dy = sec2 (−π/4)(π/4) = π/2 35. ∆x = 3 − 0 = 3, ∆y = y|x=3 − y|x=0 = dy 0 dy = dx = − √ (3) = 0 dx 25 − 02
√
16 −
√
25 = −1,
x=0
√ 36. (a) Consider y = f (x) = 3 x with x = −8 and dx = −0.25 = −1/4, then f (−8.25) ≈ f (−8) + dy, √ √ 1 3 −8.25 ≈ 3 −8 + (−8)−2/3 (−1/4) = −2 − 1/48 = −97/48. 3 (b) Consider y = f (x) = cot x (x in radians) with x = 45◦ = π/4 radians and dx = 1◦ = π/180 radians, then f (π/4 + 180/π) ≈ f (π/4) + dy, cot 46◦ ≈ cot 45◦ + (− csc2 45◦ )(π/180) = 1 − π/90. 37. V = x3 and S = 6x2 where s is the length of an edge thus x = (S/6)1/2 so V = (S/6)3/2 and dV /dS = (3/2)(S/6)1/2 (1/6) = S/6/4. 38. (a) dW/dt|t=5 = 200(t − 15)|t=5 = −2000 so water is running out at the rate of 2000 gal/min. (b) average rate of change of W = (W |t=5 − W |t=0 )/5 = (10,000 − 22,500)/5 = −2500 so water flows out at an average rate of 2500 gal/min during the first 5 minutes. 39. y = cos x − 3 sin x, y = − sin x − 3 cos x, y = − cos x + 3 sin x, y = sin x + 3 cos x so y + y + y + y = (−3 − 1 + 3 + 1) sin x + (1 − 3 − 1 + 3) cos x = 0