Chapter 2_3.pptx

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Chapter 2.3 1

Chapter 2.3

Derivative Notation and Numerical Estimates

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02p152b

Chapter 2.3 2

To recap from last section:

• Average rate of change of a

function measures how a function changes, on average, over an

interval. It is the slope of a secant.

• _{Instantaneous rate of change}_(or

just, rate of change) of a function measures how a function changes at a point. It is the slope of the

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02p152b

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Chapter 2.3 4

If f(x) is the output for an input variable, x,

then the instantaneous rate of change of f(x) at an input value of x = a, is written as

This value is the slope of the tangent line to f(x) at x = a.

And it is defined as the derivative of f(x) at x = a.

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Chapter 2.3 5

is one way to write the derivative of f(x) with respect to the input variable x at x = a.

Another way

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Chapter 2.3 6

is the slope of the tangent line to f(x) at x = a.

And it is defined as the derivative of f(x) at x = a.

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Chapter 2.3 7

f( x)

The slope of this tangent line is the derivative of f(x) at .

It is the instantaneous rate of change of f(x) at .

It is written as:

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Chapter 2.3 8 2

f( x)

The slope of this tangent line is the derivative of f(x) at .

It is the instantaneous rate of change of f(x) at .

It is written as:

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Chapter 2.3 9

Similar notation is used for other names for the input and output.

For example, when t is the input, and s(t) is the output, then the derivative of s(t) at t = a, is written

Like any slope,

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Chapter 2.3 10

�

′

(

6 )

=

��

|

_�₌₆

=

4.51 $

�

��

s(t) billion dollars

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Chapter 2.3 11

Problem 1:

The profit, P(d), in units of thousands of dollars, is a function of the price, d, in units of dollars, at which an item

sells. We write this as P(d). Describe the following and indicate their units: A.

B.

C. D.

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Chapter 2.3 12

Problem 1 Solved:

A. selling the item for $55

produces a profit of $32,500

B. when the item is selling at $45 profit is

increasing at profit is increasing as price

increases

C. when the item is selling at $51 profit is neither increasing nor

decreasing, it has reached a maximum value D. when the item is selling at $55

profit is decreasing at profit

is decreasing as price increases

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Chapter 2.3 13

Problem 2:

In the following graph, c(t) is the number of customers served in a fast-food restaurant as a function of time, aligned to 7AM.

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Chapter 2.3 14

A. Approximate the average rate of change between 7AM and 11AM. B. Approximate the average rate of

change between noon and 1 PM. C. Approximate . Interpret.

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Chapter 2.3 15

D. Approximate . Interpret.

E. How many inflection points can you approximate?

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Chapter 2.3 16

F. Over what input intervals is the rate at which customers are

entering the shop increasing?

G. Over what intervals is the rate at which customers are entering the shop decreasing?

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Chapter 2.3 19

( 0, 16)

( 4,63)

( 5, 80) ₍_{6, 78}₎

A. avg rate of change between 7AM and 11AM: = 11.8

B. avg rate of change between Noon and 1PM: = -2.0

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Chapter 2.3 20

( 1, 0)

( 4,63)

( 0, 57) ( 8, 40)

C. = 21.0

Problem 2 Solved (cont):

D. = -2.1

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Chapter 2.3 21

��  4��

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Chapter 2.3 22 F. increasing on these intervals: ,

G. decreasing on these intervals: ,

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Chapter 2.3 23 H. = 0 at, approximately, , , and

I. on these intervals: ,

J. on these intervals: ,

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Chapter 2.3 24

The percent rate of change of a

function is the rate of change of a function divided by the value of the function at the point of tangency, converted to a percentage:

% �� h��= �� h�� 

�� h��

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Chapter 2.3 25

For a function, f(t), the percent rate of

change is the ratio of the derivative at a

point to the functional value at the point, that is:

% �� h��= �

′

(�)

� (� )

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Chapter 2.3 26

What are the units for the percent rate

of change of a function?

% �� h��= �

′ (�) � (� ) ��= � ′ (� )�� � (� ) �� � ′(� )��= ��  ��

�

(

�

)

��

=

��

% �� h�� =%

�� 

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Chapter 2.3 27

Example:

A worker receives a $10,000 / year raise.

What is the percentage rate of change if:

A. her salary was = $50,000 B. her salary was = $150,000

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Chapter 2.3 28

Part A:

If her salary increased by $10,000 per year and her salary was $50,000,

then

% �� h��= �

′

(�)

� (� )

% �� h��= $10,000 / ��

$50 ,000

% �� h��=20 %

��

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Chapter 2.3 29

Part B:

If her salary increased by $10,000 per year and her salary was $150,000, then

% �� h��= �

′

(�)

� (� )

% �� h��= $10,000 / ��

$ 150 ,000

% �� h��=6.6 %´

��

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Chapter 2.3 30

Problem 3:

The following graph represents a bank account balance as a function of time.

F(t) (dollars)

t (years)

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Chapter 2.3 31

A. Graphically estimate how quickly the balance is growing 10 years after the initial deposit. That is,

determine the rate of change of the balance at 10 years. Do so by

estimating the slope of the tangent line at 10 yrs.

B. Approximate the percentage rate of change of the balance at 10

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Chapter 2.3 32

Problem 3 Solved:

( 12, 2400)

( 4,1700)

��=(2400−1700) $ (12−4) �� =

$ 87.5

��

( 10, 2240)

tangent line

F(t) (dollars)

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Chapter 2.3 33

�� h��=�′ (10)= � �

��

|

_�₌₁₀

�� h��= (2400− 1700) $

(12− 4) �� =

$ 87.5

��

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Chapter 2.3 34

% �� h��= �� h�� 

�� h��

% �� h��= $87.5/ ��

$ 2240 =.0391

% �� h��=3.91%/ ��

% �� h��=�

′

(10 )

� (10)

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Chapter 2.3 35

Earlier, we determined the slope of a tangent line, that is, derivative of a

function, at a point. During that process we saw that

• as a sequence of points, P₁, P₂, and P₃, more closely approached the

point at which we required evaluating the derivative,

• the slope of the secants connecting the points more nearly approximated the slope of the tangent at the point.

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Chapter 2.3 36

point at which we require slope of tangent line = derivative

P₁ P₂

P₃

S₁ S₂ S₃

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Chapter 2.3 37

S₁ S₂ S₃

T

point at which we require slope of tangent line = derivative

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Chapter 2.3 38

Similarly, when we know the functional relationship between an input and an output, we can numerically estimate

the derivative at a point by evaluating

the slope of secant lines for points progressively closer to the point at which we would like to know the derivative.

This is similar to the way we

numerically estimated end behavior of a function in Chapter 1.2.

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Chapter 2.3 39

Example:

A corporation invests $32B in a market which generates a future value, F(t), where t is in years, given by:

A. How rapidly is the investment growing in the middle of year 4?

B. What is the percent rate of change in this investment at that time?

�

(

�

)

₌32

(

1.12�

)

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02p161b

Chapter 2.3 40

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Chapter 2.3 41

We wish to calculate the slope of

secant lines connecting the ordered pair (3.5, F(3.5)) to points

progressively closer to (3.5, F(3.5)). • We will do this for points

approaching (3.5, F(3.5)) from both sides (t  3.5 from left, and t  3.5 from right).

• And we will use the Table capabilities of the TI-84.

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Chapter 2.3 42

t slope of secant

t  3.5 from left

3.49 3.499 3.4999

t  3.5 from right

3.51 3.501 3.5001

t slope of secant

3.49 3.499 3.4999

3.51 3.501 3.5001

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Chapter 2.3 43

First, enter the investment growth function in Y1:

Any ordered pair on the graph of the function will therefore be in the form:

The ordered pair for an input of 3.5 is:

So a general slope formula can be entered in Y2 as:

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Chapter 2.3 44

Now select TABLE (2nd GRAPH)

Entering values into the X column will automatically generate results in the

Y1 and Y2 columns.

While the table display may only show 3 or 4 decimal digits, resting the cursor on a value in the table will show the

value with more decimal digits at the bottom of the display window.

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Chapter 2.3 45

t Y1 = F(t) Y2 = slope of secant

3.49 47.525

3.499 47.573

3.4999 47.578

3.49 47.525

3.499 47.573

3.4999 47.578

Approaching t = 3.5 years from the left, the derivative of the function at t = 3.5 years is $5.392 B/yr

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Chapter 2.3 46

3.51 47.633

3.501 47.584

3.5001 47.579

3.51 47.633

3.501 47.584

3.5001 47.579

Approaching t = 3.5 years from the right, the derivative of the function at t = 3.5 years is $5.392 B/yr

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Chapter 2.3 47

The slope of the tangent line to the function at t = 3.5 years, that is, the

derivative of the function at t = 3.5 years is

�

′

(

3.5 )

=

$

5.392 �

/

��

The investment growth function, , is therefore growing at a rate of $5.392

B/year in the middle of the 4th year of the

investment.

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Chapter 2.3 48

Recall, the percent rate of change is

the ratio of the derivative at a point to the functional value at the point. For

function :

% �� h��= �

′

(� )

� (�)

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Chapter 2.3 49

Confirm that:

% �� h��= � ′

(3.5)

� (3.5) =

$ 5.392 �/ ��

$ 47.579 � =11.3 %/ ��

We have already shown:

Therefore:

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Chapter 2.3 50

Problem 4:

A. How rapidly is the investment

growing in the middle of year 11? That is, find

B. What is the percent rate of change in this investment at that time? That is, find

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Chapter 2.3 51

10.49 105.06 11.913

10.499 105.17 11.920

10.4999 105.18 11.920

10.51 105.30 11.927

10.501 105.19 11.921

10.5001 105.18 11.920

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Chapter 2.3 52

� (� )=32(1.12� )

�

₍

10.5 ₎

₌

$

105.18 �

�′

(10.5)=$ 11.92 �/ ��

% �� h��= �

′

(10.5 )

� (10.5) =

$11.92 � / ��

$ 105.18 � =11.3 %/ ��

Note the % rate of change, 11.3%, is the same at 3.5 years and 11.5 years. Later we will see why this is always true for exponential growth.



�′

(3.5)=$5.392 �/ ��

Compare to:

Chapter 2_3.pptx

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