Transcript 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
Instantaneous Rate of Change
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So we have the function; s(t) = -16t2 + 100 t + 6
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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
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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:
Average rate of change 
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Q Q(3.01)  Q(3)

t
3.01  3.00
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Solution (continued)
Q Q(3.01)  Q(3)
Average rate of change 

t
3.01  3.00
3.01
3
25(0.8)

25(0.8)
A reasonable estimate

3.01  3.00
for the rate of change
12.7715  12.80

of the quantity at t = 3
3.01  3.00
is −2.85 mg/minute.
 2.85
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Another Basic definition
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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?
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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
Average rate of change 
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f
f (2.001)  f (2)

x
2.001  2.00
(2.001)3  (2)3

2.001  2.00
8.0120  8

0.001
 12.0
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Visualizing the Derivative of a function
Figure 2.2: Visualizing the average
rate of change of f between a and b
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Figure 2.3: Visualizing the instantaneous
rate of change of f at a
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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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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.
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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,
945  1039
2000  1980
94

20
 47
Average rate of change 
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.
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Solution (continued)
(b) We use the interval from 1995 to 2000 to estimate the
instantaneous rate of change at 1995:
f '(1995)  Rate of Change in 1995
945  963

2000  1995
18

5
 3.6
million acres per year. In 1995, the amount of farm land was
decreasing at a rate of approximately 3.6 million acres per year.
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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 Button
2. 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-89
1.
2.
3.
4.
5.
6.
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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