Cal I 9
-
Upload
mani-madiliocious -
Category
Documents
-
view
216 -
download
0
Transcript of Cal I 9
-
8/9/2019 Cal I 9
1/13
MTH- 101 M. Yaseen
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)
-
8/9/2019 Cal I 9
2/13
MTH- 101 M. Yaseen
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.
-
8/9/2019 Cal I 9
3/13
MTH- 101 M. Yaseen
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
-
8/9/2019 Cal I 9
4/13
MTH- 101 M. Yaseen
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:
-
8/9/2019 Cal I 9
5/13
MTH- 101 M. Yaseen
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
!
!
! !
! !
Example (Q 8, Page 91)
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!
-
8/9/2019 Cal I 9
6/13
MTH- 101 M. Yaseen
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
-
8/9/2019 Cal I 9
7/13
MTH- 101 M. Yaseen
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
-
8/9/2019 Cal I 9
8/13
MTH- 101 M. Yaseen
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
!
-
8/9/2019 Cal I 9
9/13
MTH- 101 M. Yaseen
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
-
8/9/2019 Cal I 9
10/13
MTH- 101 M. Yaseen
Example (Q 14, Page 91)
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
!
!
! !
! ! !
Example (Q 14, Page 91)
Find the slope m of the tangent line to the graph of the
function
3 f x x!
-
8/9/2019 Cal I 9
11/13
MTH- 101 M. Yaseen
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
!
! ! !
!
Example (Q 19, Page 91)
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
-
8/9/2019 Cal I 9
12/13
MTH- 101 M. Yaseen
(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
!
!
!
! !
-
8/9/2019 Cal I 9
13/13
MTH- 101 M. Yaseen
Thus the instantaneous speed after 5 seconds is
32 5 160 / secft!
Example (Q 22, Page 91)
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!