M 112 Short Course in Calculus

Chapter 2 – Rate of Change: The DerivativeSections 2.1 – Instantaneous Rate of Change

V. J. Motto

Instantaneous Rate of Change

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So we have the function; s(t) = -16t2 + 100 t + 6

Figure 2.1: Average velocities over intervals on either side of t = 1 showing successively smaller intervals

Let’s look over smaller and smaller intervals in the neighborhood of t = 1

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Some basic definitions

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Example 1 (page 89)

The quantity (in mg) of a drug in the blood at time t (in minutes) is given by Q = 25(0.8)t. Estimate the rate of change of the quantity at t = 3 and interpret your answer.

What kind of function is this?What is the domain and range?

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Solution

We estimate the rate of change at t = 3 by computing the average rate of change over intervals near t = 3. We can make our estimate as accurate as we like by choosing our intervals small enough.

Let’s look at the average rate of change over the interval 3 ≤ t ≤ 3.01:

(3.01) (3)

3.01 3.00

Q Q QAverage rate of change

t

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Solution (continued)

3.01 3

(3.01) (3)

3.01 3.00

25(0.8) 25(0.8)

3.01 3.0012.7715 12.80

3.01 3.002.85

Q Q QAverage rate of change

t

A reasonable estimate for the rate of change of the quantity at t = 3 is −2.85 mg/minute.

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Another Basic definition

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Example 2 (page 90)

Estimate f (2) if f(x) = x′ 3.

What does the graph of f look like?What is the domain and range of f?

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Solution (continued)

Since f (2) is the derivative, or rate of change, of f(x) = x′ 3 at 2, we look at the average rate of change over intervals near 2. Using the interval 2 ≤ x ≤ 2.001, we see that

3 3

(2.001) (2)

2.001 2.00

(2.001) (2)

2.001 2.008.0120 8

0.00112.0

f f fAverage rate of change

x

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Figure 2.2: Visualizing the average rate of change of f between a and b

Figure 2.3: Visualizing the instantaneous rate of change of f at a

Visualizing the Derivative of a function

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Example 3 (page 90)

Use a graph of f(x) = x2 to determine whether each of the following quantities is positive, negative, or zero: (a)f (1) ′(b)f (−1) ′(c)f (2) ′(d)f (0)′

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Solution

What is the domain?What is the range?

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Solution (continued)

Figure 2.5 shows tangent line segments to the graph of f(x) = x 2 at the points x = 1, x = −1, x = 2, and x = 0. Since the derivative is the slope of the tangent line at the point, we have:(a) f (1) is positive.′(b) f (−1) is negative.′(c) f (2) is positive (and larger than f (1)).′ ′(d) f (0) = 0 since the graph has a horizontal tangent at x = 0.′

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Example 2 (page 92)

The graph of a function y = f(x) is shown in Figure 2.7. Indicate whether each of the following quantities is positive or negative, and illustrate your answers graphically.

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SolutionSince there are no y-values, I can’t find a model for this function. I must work with the graph.(a) Since f (1) is the slope of the graph at x = 1, we see ′in Figure 2.8 that f (1) is positive.′

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Solution (continued)

(b) The difference quotient is the slope of the secant line between x = 1 and x = 3. We see from Figure 2.9 that this slope is positive.

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Solution (continued)

(c) Since f(4) is the value of the function at x = 4 and f(2) is the value of the function at x = 2, the expression f(4) − f(2) is the change in the function between x = 2 and x = 4. Since f(4) lies below f(2), this change is negative. See Figure 2.10.

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Example 4 (Page 92)

The total acreage of farms in the US1 has decreased since 1980. See Table 2.2.

a)What was the average rate of change in farm land between 1980 and 2000?b)Estimate f (1995) and interpret your answer in ′terms of farm land.

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Solution

(a) Between 1980 and 2000,

million acres per year.

Between 1980 and 2000, the amount of farm land was decreasing at an average rate of 4.7 million per year.

945 1039

2000 198094

2047

Average rate of change

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Solution (continued)

(b) We use the interval from 1995 to 2000 to estimate the instantaneous rate of change at 1995:

million acres per year. In 1995, the amount of farm land was decreasing at a rate of approximately 3.6 million acres per year.

'(1995) 1995

945 963

2000 199518

53.6

f Rate of Change in

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An Alternative Solution

Find a model for this data:

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Alternative Solution (continued)

So we have the function f(x) = 0.05x2 – 5.83x + 1039.31

Average rate of change is still the slope of the secant!But the can actually find the instantaneous rate of change --- the first derivative by using the function f and our calculator: That is f (15) = -4.197. ′

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First Derivative Function TI 83/84

1. Math Button2. Option 8:nDeriv(3. When you press ENTER the function appears on the HOME

Screen. We need to add the parameters.4. nDeriv( y1, x, 15). The first parameter is found using the

VARS button. Find the derivative with respect to the x variable, and the let x = 15. Don’t forget to close the parenthesis.

5. Press the Enter key to calculate

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First Derivative Function TI-891. From the HOME Screen press F3 to get to the calculus functions2. Choose option 1 d( differentiate3. Add the parameters y1(x) by typing all the characters.4. Then add a comma followed by the variable x, and the close the parenthesis.5. Add the characters |x=156. When you press ENTER to calculate

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