RATES OF CHANGE TANGENTS A tangent can be more precisely defined as follows:
Given a curve y = f(x), and a point P(a, f(a)), then the tangent line to the curve is the line through P with slope
m
=
lim x→a
f(x) – f(a)
1
x–a
(provided this limit exists)
In the earlier section Introduction to Calculus – The Limit Of A Function, we tried to guess the slopes of tangents by picking nearby points to P (a, f(a)) and compute the slope of the secant line PQ, which is given by
mPQ
=
f(x) – f(a) x–a
Then we let Q → P by letting x → a. By doing so, we discovered that the slope of the secant line approached a certain value m. Thus, in terms of limits, we say that ➔ the tangent line is the limiting position of the secant line PQ as Q → P, OR ➔ the slope m of the tangent line PQ is the limiting value of the slope of the secant lines PQ as Q → P. which leads to the mathematical expression
m
=
lim x→a
f(x) – f(a) x–a
Q(x, f(a))
f(x) – f(a) P(a, f(a)) x–a
a
The formula for m can be expressed in another way: Let h = x – a which gives m
=
lim x→a
f(x) – f(a) h
x
Since h = x – a, then x = a + h. Also, notice that h→0 as x→a, since h = x – a. Therefore, (1) can also be written as
m
=
f(a + h) – f(a)
lim
2
h
h→0
Equation (2) is graphically illustrated below:
Q(a+h, f(a+h)) f(a+h) – f(a) P(a, f(a)) h
a
a+h
In summary, for the curve y = f(x), these two equations define the slope of a tangent:
m
=
m
=
lim x→a
lim h→0
f(x) – f(a) x–a f(a + h) – f(a) h
1
2
Let's go over some examples:
Example 1 Find an equation of the tangent line to the curve at the given point: y = 1 – 2x – 3x2
(–2, –7)
Solution The function we're dealing with is y = f(x) = 1 – 2x – 3x2 and the point P(–2, –7). The task is to find an equation of the tangent line to f at P. To do that, the first thing we need to do is compute the slope m of the tangent line. From definition 1, we know that
m
f(x) – f(a)
lim
=
x–a
x→a
where f(x) = 1 – 2x – 3x2 P(a, f(a)) = P(–2, –7) Therefore,
m
=
=
1 – 2x – 3x2 – (–7)
lim
x – (– 2)
x→–2
1 – 2x – 3x2 + 7
lim
=
x+2
x→–2
8 – 2x – 3x2
lim
x+2
x→–2
Next, we factorize the numerator:
=
8 – 2x – 3x2
lim
x+2
x→–2
lim
=
x→–2
(4 – 3x)(2 + x) x+2
Crossing out common factors leaves
lim
(4 – 3x)
x→–2
=
4 – 3(–2)
=
10
The slope of the tangent is 10 Let's use definition 2 to find the slope: Definition 2 states that
m
=
m
=
m
=
m
=
m
=
m
=
lim h→0
f(a + h) – f(a) h
Therefore,
=
lim
1 – 2(–2 + h) – 3(–2 + h) 2 – (–7) h
h→0
lim
1 + 4 – 2h – 3(4 – 4h + h2) + 7 h
h→0
lim
1 + 4 – 2h – 12 + 12h – h2 + 7 h
h→0
lim
–2h + 12h – h2
h→0
h
lim
(10 – h)
h→0
=
lim h→0
10h – h2 h
=
lim
h(10 – h)
h→0
h
10
The slope of the tangent is 10. We have just used definitions 1 and 2 to compute the slope of the tangent to y = f(x) at P(–2, –7). Personally, I prefer using definition 2 for computing slopes; it's generally easier. I believe we have just passed through the hardest aspect of the task. Now for the second and final part, we find an equation of the tangent line to the curve at P(–2, –7). To do this, we use another very important formula: the point-slope formula. In other words, we are to compute the point-slope equation of the tangent. The term “pointslope” should tell you that two vital “ingredients” are required – the slope of the tangent, and the x and y coordinates of the point where the tangent touches the curve. The formula is
y – y1 = m(x – x1)
where m is the slope of the tangent and x1 and y1 are the x and y coordinates respectively of the point P. For the point P(–2, –7), x1 = –2, and y1 = –7. This gives y – (–7) = 10(x – (–2)) y+7
= 10(x + 2)
y+7
= 10x + 20
y+7
= 10x + 20 – 7
y+7
= 10x + 13
Using the point-slope formula, the equation of the tangent line to the curve f(x) = 1 – 2x – 3x2 at the point P(–2, –7) is y + 7 = 10x + 13
Example 2 Find an equation of the tangent line to the curve at the given point: y = 1/ √x
(1, 1)
Solution Step 1 – Find the slope: Using definition 2,
m
f(a + h) – f(a)
lim
=
h
h→0
We have
1 m
√a + h
lim
=
–
h→0
1
1 =
h
–
√1 + h
lim h→0
1
h
1 - √1 + h lim
=
h→0
√1 + h h
We need to simplify the numerator, and we do that by rationalization. Here, we rationalize the numerator:
1 - √1 + h lim
√1 + h
h→0
1 – (1 + h)
1+ √1 + h
×
1+ √1 + h
=
lim
√1 + h + (1 + h)
h→0
h
h
1– 1–h =
=
lim h→0
lim h→0
√1 + h + 1 + h
lim
=
h→0
h
–1 √1 + 0 + 1 + 0
=
–h √1 + h + 1 + h
×
1 h
–1 2
The slope of the tangent is –½ . Using the point-slope formula, the equation of the tangent is y – 1 = –½(x – 1) y – 1 = –½x + ½ y
= –½x + ½ + 1
y
= –½x + 3/2
Therefore, the equation of the tangent line to the curve at (1,1) is y
= –½x + 3/2
Example 3 Find an equation of the tangent line to the curve at the given point: y = 1/ x 2
(–2 , ¼)
Solution Using definition 2,
m
=
f(a + h) – f(a)
lim
h
h→0
Therefore,
m
lim
=
1 (a+ h)2
–
h→0
=
=
=
h
lim
4 – (–2 + h)2 4(–2 + h)2
h→0
h
lim
4 – 4 + 4h – h2 4(–2 + h)2
h→0
4h – h2 2 h→0 4(–2 + h)
×
4– 0 4(–2 + 0)2
=
=
=
h
lim
=
¼
1 h
=
1 2 lim (–2 + h) h→0
–
¼
h
4 – (4 – 4h + h2) 4(–2 + h)2 lim h→0
h
4 – 4 + 4h – h2 4(–2 + h)2
lim h→0
h
h(4 – h) 4h(–2 + h)2 h→0 lim
4– h 4(–2 + h)2 h→0 lim
=
¼
=
The slope is ¼. The equation of the tangent is y – ¼ = ¼(x – (–2)) y – ¼ = ¼(x +2) y – ¼ = ¼x + ½ y
y
= ¼x + ½ + ¼
= ¼x + ¾
Therefore, the equation of the tangent line to the curve at (–2 , ¼) is
y
= ¼x + ¾
Example 4 Find an equation of the tangent line to the curve at the given point: y = x/ (1 – x)
(0 , 0)
Solution Using definition 2,
m
=
lim h→0
f(a + h) – f(a) h
which means
m
=
a+h lim 1 – (a+h) h→0
h
–
0 =
lim
a+h 1–a–h
h→0
h
=
lim
0+h 1–0–h
h→0
h
=
lim
h 1–h
h→0
h
1 1–0
=
lim
=
=
h→0
h 1–h
×
1 h
=
lim h→0
1 1–h
1
The slope is 1. The equation of the tangent is y – 0 = 1(x – 0)
y = x Therefore, the equation of the tangent line to the curve at the origin is
y = x That was easy!!!!
Example 5 (a) Find the slope of the tangent to the curve y = 2/(x+3) at the point where x = a. (b) Find the slopes of the tangent lines at the points whose x-coordinates are (i) –1 (ii) 0
(iii) 1.
Solution We are to find the slope m of the tangent line to the curve y = 2/(x+3) at the point where x = a. In other words, we are required to compute the slope at point (a, f(a)), since x = a. We apply definition 2:
m
=
lim
f(a + h) – f(a) h
h→0
Therefore,
m
2 lim a + h + 3
=
h→0
lim
=
–
2 a+ 3
h
2a + 6 – 2a – 2h – 6 (a+h+3)(a+3)
h→0
h
–2 (a+h+3)(a+3) h→0
=
lim
=
=
=
2(a+3) – 2(a+h+3) (a+h+3)(a+3) lim h→0
h
– 2h h→0 (a+h+3)(a+3) lim
–2 (a+0+3)(a+3)
=
×
1 h
–2 (a+3)(a+3)
–2 (a+3)2
=
This means that, given the curve y = 2/(x+3), the slope m of the tangent at the point (a, f(a)), is given by the formula
m
=
–2 (a+3)2
Now that we have a formula for the slope, we can easily find the slope of tangent lines by simply substituting the x-coordinate of the point. For the point whose x-coordinate is –1,
m
=
–2 (–1+3)2
=
–½
For the point whose x-coordinate is 0,
m
=
–2 (0+3)2
=
– 2/9
For the point whose x-coordinate is 1,
m
=
–2 (1+3)2
=
– 1/8
Example 6 (a) Find the slope of the tangent to the parabola y = 1 + x + x2 at the point where x = a. (b) Find the slopes of the tangent lines at the points whose x-coordinates are (i) –1 (ii) – ½
(iii) 1.
(c) Graph the curve and the three tangents on a common screen.
Solution If y = 1 + x + x2, then the slope of the tangent at point x = a [or in other words, point (a, f(a)))], using definition 2, equals
1 + (a+h) + (a+h)2 – (1+a+a2) h→0 h lim
lim
=
h→0
h + 2ah + h2 h
=
lim h→0
=
1 + a + h + a2 + 2ah + h2 – 1 – a – a2 h
lim h→0
h(1 + 2a + h) h
lim (1 + 2a + h)
=
h→0
1 + 2a
=
Therefore, the slope of the tangent to the parabola y = 1 + x + x2 at the point (a, f(a)) is given by
m = 1 + 2a So, when a = –1 m1
=
1 + 2(–1)
=
when a = – ½ m2 =
1 + 2(– ½)
when a = 1 m3
1 + 2(1)
= =
0 3
m
=3
=
–1
m
=
–1
m=0
Example 7 (a) Find the slope of the tangent to the parabola y = x3 – 4x + 1 at the point where x = a. (b) Find equations of the tangent lines at the points (1, –2) and (2, 1) (c) Graph the curve and both tangents on a common screen.
Solution If f(x) = x3 – 4x + 1, then the slope of the tangent at the point x = a is given by
m
=
=
=
=
= =
f(a + h) – f(a)
lim
h
h→0
(a+h)3 – 4(a+h) + 1 – (a3 – 4a + 1) h→0 h lim
a3 + 3a2h + 3ah2 + h3 – 4a – 4h + 1 – a3 + 4a – 1 h→0 h lim lim h→0
3a2h + 3ah2 + h3 – 4h h
lim (3a2 + 3ah + h2 – 4)
h→0
=
2 2 lim h(3a + 3ah + h – 4) h→0 h
= 3a2 + 3a(0) + (0)2 – 4
3a2 – 4
Hence, the slope of the tangent to the curve at (a, f(a)) is given by
m = 3a2 – 4
The equation of the tangent line at point (1, –2) equals y – (–2) = 3(1)2 – 4(x – 1) y+2
= – 1(x –1)
y+2
= – x +1
y
= –x–1
The equation of the tangent line at point (2, 1) equals y – 1 = 3(2)2 – 4(x – 2) y – 1 = 8(x – 2) y – 1 = 8x – 16
8x –
15
= 8x – 15
y=
y
y=–
x–1
Example 8 (a) Find the slope of the tangent to the parabola y = 1/√5 – 2x at the point where x = a. (b) Find equations of the tangent lines at the points (2, 1) and (–2 , 1/3).
Solution The slope of the tangent to the curve y = 1/√5 – 2x at (a, f(a)) is given by
m
=
lim
f(a + h) – f(a) h
h→0
1 m
=
lim
√5 – 2(a+h)
h→0
–
1 √5 – 2a
=
h
lim h→0
√5 – 2a – √5 – 2(a+h)
1
(√5 – 2(a+h))(√5 – 2a)
h
Then we rationalize the numerator:
lim h→0
√5 – 2a – √5 – 2(a+h)
× √5 – 2a + √5 – 2(a+h)
(√5 – 2(a+h))(√5 – 2a)
√5 – 2a + √5 – 2(a+h)
1 h
lim 5 – 2a + √(5 – 2a)(5 – 2(a+h) – √(5 – 2a)(5 – 2(a+h) – (5 – 2(a+h))
=
h→0
(√5 – 2(a+h))(√5 – 2a)(√5 – 2a + √5 – 2(a+h))h 5 – 2a – (5 – 2a – 2h)
lim
=
h→0
(√5 – 2(a+h))(√5 – 2a)(√5 – 2a + √5 – 2(a+h))h 5 – 2a – 5 + 2a + 2h
lim
=
h→0
(√5 – 2(a+h))(√5 – 2a)(√5 – 2a + √5 – 2(a+h))h 2h
lim
=
h→0
(√5 – 2(a+h))(√5 – 2a)(√5 – 2a + √5 – 2(a+h))h 2
lim
=
h→0
(√5 – 2(a+h))(√5 – 2a)(√5 – 2a + √5 – 2(a+h))
As h→0,
2
lim h→0
=
=
(√5 – 2(a+h))(√5 – 2a)(√5 – 2a + √5 – 2(a+h)) 2 (√(5 – 2a)2)(2√5 – 2a)
=
=
2 2(√(5 – 2a)2 × √5 – 2a)
1 √(5 – 2a)3
Hence, for the point (a, f(a)), the formula that expresses the slope is
m
=
1 √(5 – 2a)3
2 (√5 – 2a)(√5 – 2a)(√5 – 2a + √5 – 2a)
=
2 2(√(5 – 2a)3)
At point (2, 1) Slope = 1√(5 – 2(2))3
= 1
which means y – 1 = 1(x – 2) y–1 = x–2 y = x–1 At point (–2, 1/3) Slope = 1√(5 – 2(1/3))3
= 1/27
which means y – 1/3 = 1/27(x – (–2)) y – 1/3 = x/27 + 2/27 y = x/27 + 2/27 + 1/3 y = x/27 + 11/27 y = 1/27(x + 11)
Example 9 Graph the curve y = sin x in the viewing rectangles [–2, 2] by [–2, 2], [–1, 1] by [–1, 1] and [–0.5, 0.5] by [–0.5, 0.5]. What do you notice about the curve as you zoom in towards the origin?
Solution Let's graph the function y = sin x in the viewing rectangles provided and see what we find:
GRAPH OF y = sin x in [–2, 2] by [–2, 2]
GRAPH OF y = sin x in [–1, 1] by [–1, 1]
Figure 1
Figure 2
So what do you notice from the two graphs (focus om the origin)? In figure 1, there seems to be a slight curve around the origin. But take a look at figure 2. It's like the curve is practically straight. We zoom in more time, and check this out:
GRAPH OF y = sin x in [–0.5, 0.5] by [–0.5, 0.5]
Figure 3 This time it's definitely obvious that the curve has become a straight line completely. Now imagine if we have a tangent at (0. 0). You'd barely be able to differentiate the curve from the tangent. For this very reason, the slope of the tangent to a curve is sometimes referred to as the slope of the curve at the point.
Example 10 (a) Find the slope of the tangent line to the parabola y = x2 + 2x at the point (–3, 3) (i) using Definition 1 (ii) using Equation 2 (b) Find the equation of the tangent line in part (a) (c) Graph the parabola and the tangent line. As a check on your work, zoom in toward the point (–3, 3) until the parabola and the tangent line are indistinguishable.
Solution Using equation 1, the slope of a tangent line is defined as
m
=
lim x→a
f(x) – f(a) x–a
Therefore, the slope m of the tangent to the parabola f(x) = x2 + 2x at the point (–3, 3) equals:
m
=
m
=
m
=
lim
x2 + 2x – ((–3)2 + 2(–3)) x – (–3)
x→–3
lim x→–3
lim x→–3
x2 + 2x – 3 x+3 (x – 1)(x + 3) x+3
=
The slope is – 4. Now let's use Equation 2, which states
m
=
m
=
lim h→0
lim h→0
f(a + h) – f(a) h (–3+h)2 + 2(–3+h) – 3 h
lim (x – 1)
x→–3
=
–3–1
=
–4
m
=
m
=
m
=
lim
9 – 6h + h2 – 6 + 2h – 3 h
h→0
lim
– 6h + 2h + h2
h→0
h
lim (–4 + h)
h→0
=
lim
=
– 4h + h2 h
h→0
–4 + 0
=
=
lim h→0
h(–4 + h) h
–4
So, using definitions 1 and 2, we find that the slope of the tangent line to the curve y = x2 + 2x at the point (–3, 3) is – 4. The equation of the tangent line is y–3
= – 4(x – (– 3))
y–3
= – 4(x + 3)
y–3
= – 4x – 12
y
= – 4x – 12 + 3
y
=
– 4x – 9
The curve and its tangent are graphed below. Observe the curve and the tangent at (-3, 3). Notice the region surrounding this point. You'll see that even without zooming in, the curve and the tangent are almost one and the same. Try zooming in and you'll see:
f(x) = x2 + 2x
y= 9 x– –4 (–3, 3)
Example 11 (a) Find the slope of the tangent line to the parabola y = x3 at the point (–1, 1) (i) using Definition 1 (ii) using Equation 2 (b) Find the equation of the tangent line in part (a) (c) Graph the curve and the tangent line in successively smaller viewing rectangles centered at (-1, 1) until the curve and the line appear to coincide.
Solution From equation 1,
m
lim
=
x→a
f(x) – f(a) x–a
Therefore, the slope m of the tangent to the parabola f(x) = x3 at the point (–1, 1) equals:
m
lim
=
x3 – (– 1)
x→-1 x – (– 1)
x3 + 1
lim
=
x→-1 x + 1
Therefore,
lim
m
=
m
=
m
=
3
=
lim
x3 + 1
=
x→-1 x + 1
lim (x2 – x + 1)
x→-1
=
x→-1
(x + 1)(x2 – x + 1)
lim
x+1
(–1)2 – (–1) + 1
=
1 +1 +1
Now let's use definition 2:
m
m
lim
=
m
=
lim
(–1 + h)3 – (–1) h
h→0
–1 + 3h – 3h2 + h3 + 1 h
h→0
lim
=
h
h→0
=
m
(a + h)3 – (–1)
3h – 3h + h3
h→0
2
=
h
lim (3 – 3h + h2)
h→0
=
lim
h(3 – 3h + h2)
h→0
3 – 3(0) + (0)2
h =
3
Using equations 1 and 2, we find that the slope of the tangent line to the curve y = x3 at the point (–1, 1) is 3, and its equation is y – (– 1)
= 3(x – (– 1))
y+1
= 3(x + 1)
y
=
3x + 2
Let's graph the curve and tangent:
GRAPH OF y = x3 in [–2, 2] by [–5, 5]
Figure 1
GRAPH OF y = x3 in [–1.5, -0.5] by [–5, 5]
Figure 2
GRAPH OF y = x3 in [–1.2, -0.9] by [–5, 5]
GRAPH OF y = x3 in [–1.02, -0.98] by [–5, 5]
Figure 3
Figure 4
I believe the graphs pretty much speak for themselves. An indisputable illustration of the concept that the slope of the tangent to a curve is the slope of the curve at the point. Next, we examine velocities. In the following section you will observe that there is a striking similarity between the tangent problem and the problem of computing velocities. For example, both involve computing slopes by evaluating limits.
VELOCITY In simplest terms, we define velocity as
Velocity
Displacement Time
=
Now let's dig deeper into this definition. Consider this simple illustration:
A
B
Let's assume the ball moves from point A to point B in a straight line. When this happens, the technical term used to describe this motion is displacement (or sometimes directed distance). And it's also worth noting that this displacement occurs over a time period. Suppose we represent this motion with the function s = f(t) where s is the displacement from A to B, at time t. This amounts to saying that displacement is a function of time, and this function s = f(t) is called the position function of the object.
Position at time (a+h)
Position at time a
A
B Time interval h
The position function of the ball at point A is given by s = f(a). At point B, the function describing the ball's position is s = f(a+h). The difference in position (or displacement) over the time interval h equals the Average Velocity of the object:
Displacement Time
=
Average Velocity
f(a+h) – f(a) h
=
Now this is beginning to sound like a tangent problem, wouldn't you say? So, we've established that
Average Velocity
f(a+h) – f(a) h
=
3
When h→0, we're talking Instantaneous Velocity at time a:
Instantaneous Velocity
=
lim
f(a + h) – f(a)
4
h
h→0
Looking at equations 3 and 4, I want you to see the difference between average and instantaneous velocity. Like tangents where we compute slopes of tangents over smaller intervals, the same applies to velocities – we compute average velocities over shorter and shorter time intervals [a, a+h]. Graphically, the instantaneous velocity at time a is expressed as the slope of the tangent line at P: See Illustration. In summary, given the position function s = f(t), the instantaneous velocity of an object v(a) over over time interval h is given by
v(a)
=
lim
f(a + h) – f(a) h
h→0
Let's go over some problems.
Example 12 If a ball is thrown into the air with a velocity of 40ft/s, its height (in feet) after t seconds is given by y = 40t – 16t2. Find the velocity when t = 2.
Solution The motion of the ball in the air is defined by the equation y = 40t – 16t2. We are to find the velocity when t = 2. In this context, the velocity when t = 2 really means instantaneous velocity when t = 2. To do that, we use equation 4:
v(a)
=
lim
f(a + h) – f(a) h
h→0
Since a = 2, this means we are to work out
v(2)
=
v(2)
=
v(2)
= =
lim
f(2 + h) – f(2) h
h→0
lim
40(2 + h) – 16(2+h)2 – (40(2) – 16(2)2)
lim
h – 24h – 16h2
h→0
–24 ft/s
h
=
h
h→0
80 + 40h – 16(4+4h+h2) – 16
h→0
lim
=
lim h→0
=
lim
h
h→0
h(–24 – 16h) h
80 + 40h – 64 – 64h – 16h2 – 16
=
lim (–24 – 16h)
h→0
= –24 – 16(0)
Therefore, after two seconds, the velocity of the ball is 24 ft/s. The minus sign indicates the downward motion of the ball. Remember that it was initially thrown up.
Example 13 The displacement (in meters) of a particle moving in a straight line is given by the equation of motion s = 4t3 + 6t + 2, where t is measured in seconds. Find the velocity of the particle at times t = a, t =1, t = 2, and t = 3.
Solution The motion of the particle is described by the equation f(t) = 4t3 + 6t + 2. The velocity of the particle after a seconds is given by
v(a)
=
v(a)
=
v(a)
=
v(a)
=
v(a)
=
v(a)
=
v(a)
=
lim
f(a + h) – f(a) h
h→0
Therefore,
lim
4(a+h)3 + 6(a+h) + 2 – (4a3 + 6a + 2) h
h→0
lim
4(a3 + 3ah2 + 3a2h + h3) + 6(a+h) + 2 – 4a3 – 6a – 2 h
h→0
lim
4a3 + 12ah2 + 12a2h + 4h3 + 6a+ 6h + 2 – 4a3 – 6a – 2 h
h→0
lim
12ah2 + 12a2h + 4h3 + 6h h
h→0
=
lim
h(12ah + 12a2 + 4h2 + 6)
h→0
h
lim (12ah + 12a2 + 4h2 + 6)
h→0
12a2 + 6
This means that, after a seconds, the velocity of the particle is (12a2 + 6) m/s. And so, After 1 second, v(1) = 12(1)2 + 6 = 18m/s After 2 seconds, v(2) = 12(2)2 + 6 = 54m/s After 3 seconds, v(3) = 12(3)2 + 6 = 114m/s
Example 14 The displacement (in meters) of a particle moving in a straight line is given by s = t2 – 8t + 18, where t is measured in seconds. (a) Find the average velocities over the following time intervals: (i) [3, 4]
(ii) [3.5, 4]
(iii) [4, 5]
(iv) [4, 4.5]
(b) Find the instantaneous velocity when t = 4. (c) Draw the graph of s as a function of t and draw the secant lines whose slopes are the average velocities in part (a) and the tangent line whose slope is the instantaneous velocity in part (b).
Solution The motion of the particle is given by the formula s = f(t) = t2 – 8t + 18. Using this formula, we have the following table:
Time t
Displacement f(t)
3
3
3.5
2.25
4
2
4.5
2.25
5
3
Using this table, we can find the average velocities over specified time intervals using this formula:
Average Velocity
=
f(t2) – f(t1) t2 – t1
Between [3, 4], where t1 = 3, t2 = 4
Average Velocity
2–3 4– 3
=
=
– 1 m/s
Between [3.5, 4], where t1 = 3.5, t2 = 4 Average Velocity
=
2 – 2.25 4 – 3.5
= – 0.5 m/s
Between [4, 5], where t1 = 4, t2 = 5 Average Velocity
3–2 5–4
=
=
1 m/s
=
0.5 m/s
Between [4, 4.5], where t1 = 4, t2 = 4.5 Average Velocity
=
2.25 – 2 4.5 – 4
The instantaneous velocity when t = 4 is given by
v(4)
=
v(4)
=
v(4)
=
v(4)
=
lim
f(4 + h) – f(4) h
h→0
lim
(4 + h)2 – 8(4+h) + 18 – (42 – 8(4) + 18) h
h→0
lim
16 + 8h + h2 – 32 – 8h + 18 – 16 + 32 – 18 h
h→0
lim h
h→0
=
2 lim h h→0 h
= 0
The instantaneous velocity when t = 4 is ZERO. From the graph this is an indication that the particle reached its lowest point after 4 seconds:
LEGEND: Avg. Velocity over [3, 4] Avg. Velocity over [3.5, 4] Avg. Velocity over [4, 4.5] Avg. Velocity over [4, 5] Instantaneous Velocity at t=4
Example 15 If an arrow is shot upwards on the moon with a velocity of 58 m/s, its height (in meters) after t seconds is given by H = 58t – 0.83t2. (a) Find the velocity of the arrow after one second. (b) Find the velocity of the arrow when t = a. (c) When will the arrow hit the moon? (d) With what velocity will the arrow hit the moon?
Solution (i)
The velocity of the arrow after a seconds is given by
v(a)
=
lim
H(a + h) – H(a) h
h→0
which means that, after one second, the velocity of the arrow equals
v(1)
=
v(1)
=
v(1)
=
v(1)
=
lim
H(1 + h) – H(1) h
h→0
h
h→0
58 + 58h – 0.83 – 1.66h – 0.83h2 – 57.17
lim
h
h→0
58 – 0.83 – 57.17 + 58h – 1.66h – 0.83h2 h
h→0
v(1)
=
v(1)
=
v(1)
=
h
h→0
58 + 58h – 0.83(1+2h+h2) – (58 – 0.83)
lim
lim
58(1 + h) – 0.83(1+h)2 – (58(1) - 0.83(1)2)
lim
=
lim
58h – 1.66h – 0.83h
2
lim
=
lim (58 – 1.66 – 0.83h)
(58 – 1.66 – 0.83(0)
=
h→0
h
h→0
h
h→0
h(58 – 1.66 – 0.83h)
=
56.34
56.34 m/s
Therefore, the velocity after one second is 56.34 m/s.
(ii)
Now let's find the velocity when t = a:
v(a)
=
v(a)
=
v(a)
=
v(a)
=
v(a)
=
v(a)
=
lim
58(a + h) – 0.83(a+h)2 – (58a – 0.83a2) h
h→0
lim
58a + 58h – 0.83(a2+2ah+h2) – 58a + 0.83a2 h
h→0
lim
58a + 58h – 0.83a2 – 1.66ah – 0.83h2 – 58a + 0.83a2 h
h→0
lim
58h – 1.66ah – 0.83h2
h→0
h
lim (58 – 1.66a – 0.83h)
h→0
=
=
lim
h(58 – 1.66a – 0.83h)
h→0
h
58 – 1.66a – 0.83(0)
58 – 1.66a
The velocity of the arrow after a seconds is given by the formula v(a) = 58 – 1.66a
(iii)
A brief visualization of the scenario will help us understand the motion of the arrow – at first, it was shot
upwards, then it traveled some distance, and finally headed downwards and eventually hit the surface.
That being said, from the equation of the arrow's motion, we can the say that that the arrow hits the surface when H(t) = 0. This means H(t) = 58t – 0.83t2 = 0 Now we solve for t: 58t – 0.83t2 = 0 58t = 0.83t2 Canceling out t on both sides leaves 58 = 0.83t Therefore,
t
=
58 0.83
= 69.88s
The arrow makes a touchdown on the moon 69.88 seconds after it was shot.
(iv)
Recall that the velocity of the arrow after a seconds is given by v(a) = 58 – 1.66a
Thus, the velocity of the arrow after 69.88 seconds equals v(69.88) = 58 – 1.66(69.88) v(69.88) = 58 – 116.008 = –58.008 ≈ –58 m/s So, the arrow hits the moon 69.88 seconds with a velocity of –58 m/s after being shot. Again, note the minus sign it indicates the downward motion of the arrow.
OTHER RATES OF CHANGE We use the term “Rate of Change” when one quantity changes with respect to another related quantity. This is illustrated in the subsection above, where we expressed velocity as rate of change of displacement with time. Other rates of change include: ➔ POWER – the rate of change of work with respect to time. ➔ DENSITY – the rate of change of mass with respect to volume. ➔ CURRENT – the rate of change of electric charge with respect to time. ➔ COMPRESSIBILITY – the rate of change of volume with respect to pressure. ➔ TEMPERATURE GRADIENT – the rate of change of temperature with respect to position. And so much more. Let's say there are two quantities x and y; where y is dependent on x. We say that y is a function of x and write it as y = f(x). Of course, quantities change from one value to another (obviously!!). So, for a function to be balanced (or in equilibrium), a change in one quantity must be accompanied by a change in the other quantity. In other words, if x changes in value, then y must also change in value. In mathematics, we represent this change in quantity by the Greek symbol Delta (∆). Hence, a change in x from x1 to x2 is expressed as ∆x:
∆x = x2 – x1 On the other hand, a change in y from y1 to y2 is written as ∆y:
∆y = y2 – y1 OR The quotient
∆y = f(x2) – f(x1)
Since y is a function of x
∆y ∆x
=
y2 – y1
f(x2) – f(x1)
=
x2 – x1
5
x2 – x1
is called the Average Rate of Change of y with respect to x over the interval [x1, x2].
Q(x2, f(x2))
∆y P(x1, f(x1))
∆x
x1
x2
Like velocity, where we have average and instantaneous velocity, there's also average rate of change (which has been defined above) and Instantaneous Rate of Change of y with respect to x over the interval [x1, x2] and is defined as
lim
∆y
∆x→0 ∆x
=
lim
x2→x1
f(x2) – f(x1) x2 – x1
6
Like instantaneous velocity, equation 6 above was derived by letting x2 approach x1 (which amounts ∆x approaching zero. Taking the limit of these average rates of change results in instantaneous rate of change. Graphically, average rate of change is interpreted as the slope of a secant line, while instantaneous rate of change is interpreted as the slope of the tangent line to the curve y = f(x).
Example 16 The cost (in dollars) of producing x units of a certain commodity is C(x) = 5000 + 10x + 0.05x2. (a) Find the average rate of change of C with respect to x when the production level is changed (i).
from x = 100 to x = 105
(ii).
From x = 100 to x = 101
(b) Find the instantaneous rate of change of C with respect to x when x = 100. (This is called Marginal Cost).
Solution The question mentions two related quantities: ✔
The cost of production, C (expressed in Dollars)
✔
The quantity of commodities produced, x (expressed in Units)
Both quantities are related by the formula
C(x) = 5000 + 10x + 0.05x2 Recall that we defined average rate of change as
∆y ∆x
f(x2) – f(x1)
=
x2 – x1
In this context, the average rate of change of C with respect to x with reference to the above equation, equals
∆C ∆x
=
C(x2) – C(x1) x2 – x1
So, when the production level is changed from 100 to 105 Units, this means x1 = 100 and x2 = 105. Therefore,
∆C ∆x ∆C ∆x ∆C ∆x
=
= = =
C(105) – C(100) 105 - 100 (5000 + 10(105) + 0.05(105)2) – (5000 + 10(100) + 0.05(100)2) 105 - 100 6601.25 – 6500 5 $ 20.25/Unit
Observe the unit of ∆C/∆x: Dollars per unit. How did we arrive at that unit? Well, it is simply the quotient of the units of the quantities in question. Cost is measured in Dollars, while the production quantity is measured in Units: Therefore, the quotient of these units will simply be Dollars/Unit. So, the average rate of change of C with respect to x when the production level changed from 100 to 105 units is $ 20.25 per Unit. When production levels changes from 100 to 101 units, the rate of change will be
∆C ∆x
=
= =
C(101) – C(100) 101 - 100 6520.05 – 6500 1 $ 20.05/Unit
The average rate of change of C with respect to x when the production level increased by one unit is $ 20.05 per Unit. The instantaneous rate of change of C with respect to x (from equation 6) is
lim
∆C
∆x→0 ∆x
=
lim
x2→x1
C(x2) – C(x1) x2 – x1
=
lim
C(x + h) – C(x)
h→0
h
=
2 2 lim (5000 + 10(x+h) + 0.05(x+h) ) – (5000 + 10x + 0.05x ) h→0 h
=
2 2 2 lim 5000 + 10 x+10h + 0.05(x +2xh+h ) – 5000 – 10x – 0.05x h→0 h
=
2 2 2 lim 5000 + 10x+10h + 0.05x + 0.1xh + 0.05h – 5000 – 10x – 0.05x h→0 h
=
2 lim 10h + 0.1xh + 0.05h h→0 h
=
lim (10 + 0.1x + 0.05h)
h→0 =
(10 + 0.1x) Dollars/Unit
=
lim h(10 + 0.1x + 0.05h) h
h→0 =
10 + 0.1x + 0.05(0)
This means that the instantaneous rate of change of C with respect to x is given by $ (10 + 0.1x)/Unit So, when x = 100, ∆C/∆x = 10 + 0.1(100) = 10 + 10 = $ 20/Unit At a production level of 100 Units, the instantaneous rate of change (or Marginal Cost) is $ 20 per Unit.
Example 17 The population P (in thousands) of the city of San Jose, California, from 1991 to 1997 is given in the table. Year (x)
1991
1993
1995
1997
Population (P)
793
820
839
874
(a) Find the average rate of growth (i).
From 1991 to 1995
(ii).
From 1993 to 1995
(iii).
From 1995 to 1997
In each case, include the units (b) Find the instantaneous rate of change of growth in 1995 by taking the average of two average rates of change. What are its units? (c) Estimate the instantaneous rate of change of growth in 1995 by measuring the slope of a tangent.
Solution The average rate of growth in any given year equals
∆P ∆x
=
P(x2) – P(x1) x2 – x1
Using this formula,
(i)
The average rate of growth from 1991 to 1995 is
∆P ∆x ∆P ∆x (ii)
=
P(1995) – P(1991) 1995 – 1991
=
839000 – 793000 1995 – 1991
=
46000 4
11500 People/Year
The average rate of growth from 1993 to 1995 is
∆P ∆x ∆P ∆x (iii)
=
=
=
P(1995) – P(1993) 1995 – 1993
=
839000 – 820000 1995 – 1993
=
19000 2
9500 People/Year
The average rate of growth from 1995 to 1997 is
∆P ∆x ∆P ∆x
=
=
P(1997) – P(1995) 1997 – 1995 17500 People/Year
=
874000 – 839000 1997 – 1995
=
35000 2
Now let's estimate the instantaneous rate of growth in 1995 by taking the average of any two average rates of change derived in (i), (ii), and (iii) above. Using (i) and (ii):
11500 + 9500 2
=
10500 People/Year
=
14500 People/Year
=
13500 People/Year
Using (i) and (iii):
11500 + 17500 2 Using (ii) and (ii):
9500 + 17500 2
Obviously 1993 and 1997 are relatively closer to 1995 than 1991, so we'll stick to their averages. Based on this, we can say that the average rate of growth in 1995 is somewhere between 13500 and 14500 people/year. Plotting the data on the table produces this:
B
A
C
Observe the tangent at the point where x = 1995. After measuring the sides of triangle ABC, we find that Slope of Tangent
= =
|BC| |AC|
=
852500 – 825000 1996 – 1994
13750 People/Year
Therefore, the rate of growth in San Jose, California, is approximately 13750 People/Year.
Example 18 If a cylindrical tank holds 100,000 gallons of water, which can be drained from the bottom of the tank in an hour, then Torricelli's Law gives the volume V of water remaining in the tank after t minutes as
V(t)
=
100000 1 –
t 60
2
0 ≤ t ≤ 60
Find the rate at which the water is flowing out of the tank (the instantaneous rate of change of V with respect to t) as a function of t. What are its units? For times t = 0, 10, 20, 30, 40, 50, and 60 min, find the flow rate and the amount of water remaining in the tank. Summarize your findings in a sentence or two. At what time is flow rate the greatest? The least?
Solution The question describes the Volume of water remaining in the tank as a function of time:
V(t)
=
2
t
100000 1 –
0 ≤ t ≤ 60
60
To find the rate of water flow (or in other words, the instantaneous rate of change of V with respect to t), we use the formula
∆V ∆t
lim
=
V(t+h) – V(t)
h→0
h
Before we begin evaluating the limit, let's express V in a relatively easier form to ease simplification:
V(t)
=
100000 1 –
2t 60
+
t2 3600
After simplification jumbo, we end up with the simplest possible expression for V:
V(t) = 100000 – (10000t/3) + (250t2/9) Let's move on..
∆V ∆t ∆V ∆t ∆V ∆t ∆V ∆t ∆V ∆t ∆V ∆t ∆V ∆t ∆V ∆t ∆V ∆t ∆V ∆t
=
lim
(100000 – (10000(t+h)/3) + (250(t+h)2/9)) – (100000 – (10000t/3) + (250t2/9))
h→0
h
=
2 2 lim 100000 – (10000(t+h)/3) + (250(t+h) /9) – 100000 + (10000t/3) – (250t /9) h→0 h
=
2 2 lim – (10000(t+h)/3) + (250(t+h) /9) + (10000t/3) – (250t /9) h→0 h
=
2 2 lim – 30000(t+h) + 250(t+h) + 30000t – 250t h→0 9
=
2 2 2 lim – 30000t – 30000h + 250(t + 2ht + h ) + 30000t – 250t h→0 9h
=
2 2 2 lim – 30000t – 30000h + 250t + 500ht + 250h + 30000t – 250t h→0 9h
=
lim
– 30000h + 500ht + 250h2
h→0
9h
=
lim h(– 30000 + 500t + 250h) h→0 9h
=
lim – 30000 + 500t + 250h h→0 9
=
– 30000 + 500t 9
Therefore, the rate of water flow is given by
∆V ∆t
=
– 30000 + 500t 9
Gallons/Minute
×
1 h
Using the equations
V(t)
=
100000 1 –
2
t 60
and
∆V ∆t
=
– 30000 + 500t 9
Gallons/Minute
We have the following table: Time (Min)
Rate of Water Flow Water Remaining (Gallons/Min) (Gallons)
0
– 3333.3
100000
10
– 2777.8
69444.4
20
– 2222.2
44444.4
30
– 1666.7
25000
40
– 1111.1
11111.1
50
– 555.6
2777.8
60
0
0
From the tables, we gather that the rate at which water flows out of the tank decreases steadily until there's nothing left in the tank. The flow rate is greatest in the beginning and lowest at the end.
Example 19 Temperature readings T (in degrees Celsius) were recorded every hour starting midnight on a day in April in Whitefish, Montana. The time x is measured in hours from midnight. The data are given in the following table. x (h)
T (oC)
x (h)
T (oC)
0
6.5
13
16
1
6.1
14
17.3
2
5.6
15
18.2
3
4.9
16
18.8
4
4.2
17
17.6
5
4
18
16
6
4
19
14.1
7
4.8
20
11.5
8
6.1
21
10.2
9
8.3
22
9
10
10
23
7.9
11
12.1
24
7
12
14.3
(a)Use the data in the table to find the average rate of change of Temperature with respect to time (i).
from 8 P.M to 11 P.M
(ii).
from 8 P.M to 10 P.M
(iii).
from 8 P.M to 9 P.M
(b)Estimate the instantaneous rate of change of T with respect to Time at 8 P.M by measuring the slope of a tangent.
Solution In the time column, please note that the values represent the number of hours after midnight. For example, 5 represents 5 hours after midnight meaning 5 A.M, 20 means 20 hours after midnight, which amounts to 8 P.M, and so on. With that in mind, the table could equally mean this: Time
T (oC)
Time
T (oC)
12.00 A.M
6.5
1.00 P.M
16
1.00 A.M
6.1
2.00 P.M
17.3
2.00 A.M
5.6
3.00 P.M
18.2
3.00 A.M
4.9
4.00 P.M
18.8
4.00 A.M
4.2
5.00 P.M
17.6
5.00 A.M
4
6.00 P.M
16
6.00 A.M
4
7.00 P.M
14.1
7.00 A.M
4.8
8.00 P.M
11.5
8.00 A.M
6.1
9.00 P.M
10.2
9.00 A.M
8.3
10.00 P.M
9
10.00 A.M
10
11.00 P.M
7.9
11.00 A.M
12.1
12.00 A.M
7
12.00 P.M
14.3
Now that we've cleared that up. Let's get to the task From the table
(i). From 8 P.M to 11 P.M, the temperature drops from 11.5oC to 7.9oC. So,
∆T ∆x ∆T ∆x
=
Temperature change
=
Change in time 7.9 – 11.5 11 – 8
= –1.2 oC/h
(ii). From 8 P.M to 10 P.M, the temperature drops from 11.5oC to 9.0oC. So,
∆T ∆x
=
9.0 – 11.5 10 – 8
= –1.25 oC/h
(iii). From 8 P.M to 9 P.M, the temperature drops from 11.5oC to 10.2oC. So,
∆T ∆x
=
10.2 – 11.5 9–8
=
–1.3 oC/h
Plotting the data on the table produces the graph below. From it, we find that Slope of Tangent
=
|AB| BC
=
13.5 – 9.5 7–9
=
–2.0 oC/h
A
B
C
Therefore, the instantaneous rate of change of Temperature with respect to Time at 8 P.M is about –2.0 oC/h
Example 20 According to Boyle's law, if the Temperature of a confined gas is held fixed, then the product of the pressure P and the Volume V is a constant. Suppose that, for a certain gas, PV = 800, where P is measured in pounds per square inch and V is measured in cubic inches. (a) Find the average rate of change of P as V increases from 200 in3 to 250 in3. (b) Express V as a function of P and show that the instantaneous rate of change of V with respect to P is inversely proportional to the square of P.
Solution We know that P and V are connected by the formula PV = 800, where 800 is a constant. To start with, let's draw up a table of values using the formula: P
V
16
50
8
100
5.33
150
4
200
3.2
250
2.67
300
2.29
350
2
400
1.78
450
1.6
500
Using the table, the average rate of change of P with respect to V as V increases from 200 in3 to 250 in3 is
∆P ∆V
=
3.2 – 4 250 – 200
=
– 0.016
If we plot the data on the table, we get this:
Now, if PV = 800, then V = 800/P
(V is now a function of P)
Therefore, the instantaneous rate of change of V with respect to P is given by
∆V ∆P
=
(800/P2) – (800/P1)
lim
P2→P1
P2 – P1
Let P2 = P1 + h (where h represents the difference in Pressure) Which means
∆V ∆P
=
lim
h→0
(800/(P1 + h)) – (800/P1) h 800P1 – 800(P1 + h)
∆V ∆P ∆V ∆P ∆V ∆P ∆V ∆P ∆V ∆P
=
=
=
=
=
P1(P1 + h)
lim
h→0
h 800P1 – 800P1 – 800h
lim
h→0
lim
h→0
P1(P1 + h) – 800h P1(P1 + h)
h→0
(P1)2 + P1h
– 800 P
=
2
– 800
1 h
1
×
– 800
lim
×
h – 800
=
(P1)2
1 P
2
Therefore, if PV = 800, we find that
∆V ∆P
= – 800
1 P
2
7
This proves that that the instantaneous rate of change of V with respect to P (∆V/∆P) is inversely proportional to the square of P. From equation 7 above, we can say that
∆V ∆P
∝
1 P2
which leads to
∆V ∆P
=
K
1 P2
(where K = – 800)
The constant K is negative, which means one thing: the rate of change is negative (P decreases as V increases). As you can see from the graph above, we'll always have a negative slope.
This is where we'll conclude this section. Next, we'll be examining Derivatives. We've seen that Rate of change can either be ➔ Average – which is represented by the slope of a secant line , OR ➔ Instantaneous – which is represented by the slope of a tangent line. In other terms, Instantaneous rate of change is found by computing the limit of average rates of change. Each and every single rate o change can be interpreted as the slope of a tangent. Whenever we solve a problem involving tangent lines, it seems we are solving a simple problem in geometry. But if you look at the larger picture, solving a tangent problem means solving a wide range of problems involving rates of change in science and engineering. In fact, the concept of Rate of Change is one of the very fundamentals of Calculus, one you must really understand.
Try the following brush-up exercises.
Concept Check (1). A curve has the equation y = f(x) (a). Write an expression for the slope of the secant line through the points P(3, f(3)) and Q(x, f(x)). (b). Write an expression for the slope of the tangent line at P. (2). Suppose an object moves with position function s = f(t). (a). Write an expression for the average velocity of the object in the time interval from t = a to t = a+h. (b). Write an expression for the instantaneous velocity at time t = a.
Exercises (3).(a) Find the slope of the tangent line to the curve y = 9 – 2x2 at point (2, 1) (b) Find an equation of this tangent line
(4). Find equations of the tangent lines to the curve y = 2/(1 – 3x) at the points with x-coordinates 0 and –1. (5). The displacement (in meters) of an object moving in a straight line is given by s = 1 + 2t + t2/4, where t is ddddmeasured in seconds. (a) Find the average velocity over the following time periods: (i) [1, 3]
(ii) [1, 2]
(iii) [1, 1.5]
(iv) [1, 1.1]
(b) Find the instantaneous velocity when t = 1.
(6). Graph the curve y = (x + 1)/(x – 1) and the tangent lines to this curve at the point (2, 3) and (–1, 0).
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