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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

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M 112 short course in calculus

M 112 Short Course in Calculus

Chapter 2 – Rate of Change: The Derivative

Sections 2.1 – Instantaneous Rate of Change

V. J. Motto




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



Example 1 page 89
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
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
Solution (continued)

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



Example 2 page 90
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 continued1
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
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)


Solution1
Solution

What is the domain?

What is the range?


Solution continued2
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
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.


Solution2
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 continued3
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 continued4
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
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.


Solution3
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 continued5
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
An Alternative Solution

Find a model for this data:


Alternative solution continued
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
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
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.

  • Add the characters |x=15

  • When you press ENTER to calculate


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