This presentation is the property of its rightful owner.
1 / 27

# M 112 Short Course in Calculus PowerPoint PPT Presentation

M 112 Short Course in Calculus. Chapter 2 – Rate of Change: The Derivative Sections 2.1 – Instantaneous Rate of Change V. J. Motto. Instantaneous Rate of Change. So we have the function; s(t) = -16t 2 + 100 t + 6. Let’s look over smaller and smaller intervals in the

M 112 Short Course in Calculus

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

## M 112 Short Course in Calculus

Chapter 2 – Rate of Change: The Derivative

Sections 2.1 – Instantaneous Rate of Change

V. J. Motto

Instantaneous Rate of Change

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

Let’s look over smaller and smaller intervals in the

neighborhood of t = 1

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

showing successively smaller intervals

Some basic definitions

### 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?

### 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:

### Solution (continued)

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

Another Basic definition

### Example 2 (page 90)

Estimate f′ (2) if f(x) = x3.

What does the graph of f look like?

What is the domain and range of f?

### Solution (continued)

Since f′(2) is the derivative, or rate of change, of f(x) = x3 at 2, we look at the average rate of change over intervals near 2. Using the interval 2 ≤ x ≤ 2.001, we see that

Visualizing the Derivative of a function

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

### Example 3 (page 90)

Use a graph of f(x) = x2to determine whether each of the following quantities is positive, negative, or zero:

• f′(1)

• f′(−1)

• f′(2)

• f′(0)

### Solution

What is the domain?

What is the range?

### 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.

### 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.

### Solution

Since 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.

### 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.

### 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.

### Example 4 (Page 92)

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

• What was the average rate of change in farm land between 1980 and 2000?

• Estimate f′(1995) and interpret your answer in terms of farm land.

### Solution

• 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.

### 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.

### An Alternative Solution

Find a model for this data:

### 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.

### First Derivative Function TI 83/84

• Math Button

• Option 8:nDeriv(

• When you press ENTER the function appears on the HOME Screen. We need to add the parameters.

• 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.

• Press the Enter key to calculate

### First Derivative Function TI-89

• From the HOME Screen press F3 to get to the calculus functions

• Choose option 1 d( differentiate

• Add the parameters y1(x) by typing all the characters.

• Then add a comma followed by the variable x, and the close the parenthesis.