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Lecture 09 Suggested Readings: Sec 2.1

Let )(xfy ! be a function. The average rate of change of

y with respect to x over the interval ],[ 21 xx is calculated by

dividing the change in value of y , 2 1 y f x f x( ! , by the

length of the interval 2 1 x x x h( ! ! . That is,

Average rate of change =x

y

(

(

= 2 1 1 12 1

( ) ( ) ( ) ( )f x f x f x h f x

x x h

!

(1)

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Geometric meaning of Average Rate of Change

We can see from (1) that average rate of change of fover

],[ 21 xx is the slope of the line joining the points ))(,( 11 xfxP

and ))(,( 22 xfxQ . But the line joining two points of a curve is

called a secant to the curve. Thus average rate of change of

f from 1x to 2x is the slope of the secant .

Definition (Average Speed)

The average speed of a moving body over any particular time

interval is the amount of distance covered during the interval

divided by the length of the interval.

Consider a function, f x , that represents some quantity that

varies as x varies. For instance, f x may represent the

amount of water in tank after x minutes. Or may be f x is

the distance traveled by a car after x hours.

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Given f x here, what we want to do?

We want to determine that how fast f x is changing at some

point, say 0x x! . This is called the instantaneous rate of

change or just the rate of change.

We know how to find average rate of change

0 0

change in.

chang in

f xAvg rateof change

xf x h f x

h

!

!

Then to estimate the instantaneous rate of change at 0x x!

all we need to do is to choose values of 0x h getting closer

and closer to 0x x! (dont forget to choose them on both sides

of 0x x! ). Thus the instantaneous rate of change is

0 00

limh

f x h f x

hp

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Example (Q 3, Page 91)

Assume that 2 1 21, 3, 3.5y f x x x x x! ! ! ! .

(a) Find the average rate of change y with respect x asx changes from 1x to 2x .

(b) Find the instantaneous rate of change of y withrespect to x at the instant when 0x x!

Solution

(a)

The average rate of change is given by

2 1

2 1

2

3.5 3. .3.5 3

3.5 3.5 1 9 3 13.75

7.50.5 0.5

f x f x f f A R C x x

! !

! ! !

(b)

The instantaneous rate of change at 1 3x ! is:

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1 1

0

2 2

1 1 1 10

1 10

. . lim

1 1lim

lim 2 1 2 1

2 3 1 7

h

h

h

f x h f xI R C

h

x h x h x x

h

x h x

p

p

p

!

!

! !

! !


A particle is moving along a straight line according to the

equation

1 2

2, 1, 1.1

5s t t

t! ! !

Where s is the distance is meters of the particle from its

starting point at the end of t seconds. Find

(a) The average speed st

((

of the particle during the

interval of time from 1t t! to 2t t! .

(b) The instantaneous speed of the particle when 1t t!

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Solution

(a)

1.1 1Avg. Speed

1.1

2 25 1.1 5 1 0.51 0.5

0.100.1 0.1

s ss

t

(! !

(

! ! !

(b)

1 1

0

11

0

1 1

01 1

201 1 1

2

Instantaneou Speed lim

2 2

55lim

2 5 2 5lim

5 5

2 2lim

5 5 5

2 1

85 1

h

h

h

h

f t h f t

h

tt h

h

t t h

h t h t

t h t t

p

p

p

p

!

!

!

! !

! !

Next we are going to take a look at a fairly important

problem of calculus- the tangent line problem

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Before getting into this problem, it would probably be best to

define a tangent line

Definition (Tangent Line)

A tangent line to the function f x at the point x a! is a line

that just touches the graph of the function at the point in

question and is parallel to the graph at that point.

Note that a line and a graph are considered parallel if they

are both moving in the same direction at that point.

In the above graph the line is tangent at the indicated point

because it just touches the graph and is also parallel to the

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graph at that point. Likewise, at the second point shown, the

line does just touch the graph at that point, but is not

parallel to the graph at that point and so its not a tangent

line to the graph at that point.

Consider the following graph of a function y f x! .

The tangent line at 0 0,P x f x and the secant line joining

0 0,P x f x and 0 0,Q x h f x h are also shown in this graph.

The slope of the secant line is:

0 0 0 00 0

f x h f x f x h f xh x h x

!

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If we take Q closer and closer to P by letting 0x h

approach 0x ( . . 0)i e h p , then the secant lines start to look

more and more like the tangent line and so the slopes should

be getting closer and closer. Thus

Slope of the tangent line 0 0

0limh

f x h f xm

hp

! !

Definition

Let f be a function defined at least in some open interval

containing the number 1x , and let 1 1 y f x! . If the limit

1 10

limh

f x h f xm

hp

!

exists, we say that the line in the xy plane containing the

point 1 1,x yand having slope m is the tangent line to the

graph of f at 1 1,x y

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Find the slope m of the tangent line to the graph of the

function

1

f xx

!

at the point1

, 22

Solution

1 1

0

1 1

0

1 1

0 01 1 1 1

2

1

lim

1 1

lim

1lim lim

1 14

14

h

h

h h

f x h f xm

h

x h x

h

x x h

h x x h x x h

x

p

p

p p

!

!

! !

! ! !


Find the slope m of the tangent line to the graph of the

function

3 f x x!

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at the point 7, 2 .

Solution

1 1

0

1 1

0

1 1 1 1

01 1

1 1

01 1

011 1

lim

3 3lim

3 3 3 3lim

3 3

3 3lim

3 3

1 1 1lim

2 3 2 7 33 3

1

4

h

h

h

h

h

f x h f xm

h

x h x

h

x h x x h x

h x h x

x h x

h x h x

x x h x

p

p

p

p

p

!

!

! v

!

! ! !

!


An object falls from rest according to the equation 216s t! ,

where s is the number of feet through which it falls during

the first t seconds after being released. Find

(a) The average speed during the first 5 seconds of fall

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(b) The instantaneous speed at the end of this 5-secondinterval.

Solution

(a)

Since initially the object was at rest, we have to calculate

average speed during the time interval ? A0,5 . So

2

5 0verage peed

5 0

16 5 080 / sec

5

s s

ft

!

! !

The instantaneous speed is given by

0

2 2

0

2 2 2

0

2

0

0

Inst. peed=lim

16 16lim

16 16 32 16lim

16 32lim

lim 16 32 32

h

h

h

h

h

s t h s t

h

t h t

h

t h th t

h

h th

h

h t t

p

p

p

p

p

!

!

!

! !

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Thus the instantaneous speed after 5 seconds is

32 5 160 / secft!


A spherical balloon of radius R meters has volume 34

3V RT!

cubic meters. Find the instantaneous rate of change of V

with respect to R at the moment when 5R ! meters.

Solution

The instantaneous rate of change of V is

0

3 3

0

3 3 2 2 3

0

3 3 2 2 3

0

2 2 2

0

Inst. Rate of Change lim

4 4

3 3lim

4 43 3

3 3lim

4 4 44 43 3 3

lim

4lim 4 4 43

h

h

h

h

h

V R h V R

h

R h R

h

R h R h Rh R

h

R h R h Rh R

h

h R Rh R

T T

T T

T T T T T

T T T T

p

p

p

p

p

!

!

!

!

! !

The instantaneous speed at 5R ! is 2

4 5 100T T!

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