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2.4
Rates of Change and Tangent Lines
Quick Review
5 ,3 ,2 ,5 .1 BA baBA , ,3 ,1 .2
8D x 3D y
7
4
3
2
In Exercises 1 and 2, find the increments Dx and Dy from point A to point B.
1 ,5 ,3 ,2 .3 BA 3 ,3 ,1 ,3 .4 BA
In Exercises 3 and 4, find the slope of the line determined by the points.
1D ax 3D by
Quick Review
2
3 slope with 3 ,2 through .5
1 ,4 and 6 ,1 through .6
62
3 xy
4
19
4
3 xy
In Exercises 5 – 9, write an equation for the specified line.
24
3 toparallel and 4 ,1 through .7 xy
3
25
3
7 xy
Quick Review
24
3 lar toperpendicu and 4 ,1 through .8 xy
532 toparallel and 3 ,1 through .9 yx3
8
3
4 xy
3
19b
In Exercises 5 – 9, write an equation for the specified line.
3 ,2 through line theof slope the will of alueFor what v .10 b3
7
3
2 xy
?3
5 be ,4 and b
What you’ll learn about Average Rates of Change Tangent to a Curve Slope of a Curve Normal to a Curve Speed Revisited
Essential QuestionHow does the tangent line determine the direction of abody’s motion at every point along its path?
Average Rates of ChangeThe average rate of change of a quantity over a period of time is the amount of change divided by the time it takes.
In general, the average rate of change of a function over an interval is the amount of change divided by the length of the interval.
Also, the average rate of change can be thought of as the slope of a secant line to a curve.
Example Average Rates of Change1. Find the average rate of change of f (x) = 3x
2 – 8 over the interval [1, 3].
13 ff 13
813833 22
13
91
52 2
24 12
Example Instantaneous Rate of Change2. Suppose that the amount of air in a balloon after t hours is given by
V (t) = t 3 – 6t
2 + 35. Estimate the instantaneous rate of change of the volume after 5 hours.
55lim
0
VtVt
55 t
3556535565lim
2323
0
ttt t
t
tttt
159lim
23
0
159lim 2
0
tt
t
15 hourper feet
Tangent to a CurveIn calculus, we often want to define the rate at which the value of a function y = f(x) is changing with respect to x at any particular value x = a to be the slope of the tangent to the curve y = f(x) at x = a.
The problem with this is that we only have one point and our usual definition of slope requires two points.
Tangent to a CurveThe process becomes:1. Start with what can be calculated, namely, the slope of a
secant through P and a point Q nearby on the curve.
2. Find the limiting value of the secant slope (if it exists) as Q approaches P along the curve.
3. Define the slope of the curve at P to be this number and define the tangent to the curve at P to be the line through P with this slope.
Slope of a CurveTo find the tangent to a curve y = f (x) at a point P (a, f (a))calculate the slope of the secant line through P and a point Q (a+h, f (a+h)). Next, investigate the limit of the slope as h→0.
If the limit exists, it is the slope of the curve at P and we define the tangent at P to be the line through P with this slope.
Slope of a Curve at a Point
0
The at the point , is the number
lim
provided the limit exists.
The at is the line through with this slope.
h
y f x P a f a
f a h f am
h
P P
slope of the curve
tangent line to the curve
Example Tangent to a Curve3. Find the slope of the parabola f (x) = 2 x
2 – 8 at the point P (2, 0). Write the equation for the tangent to the parabola at this point.
h
fhfh
22lim
0
h
hh
822822lim
22
0
h
hhh
82lim
2
0
82lim0
hh
8
280 xy
168 xy
Slope of a Curve
0
All of the following mean the same:
1. the slope of ( ) at
2. the slope of the tangent to ( ) at
3. the (instantaneous) rate of change of ( ) with respect to at
4. limh
y f x x a
y f x x a
f x x x a
The expression is the of at .
f a h f a
hf a h f a
f ah
difference quotient
a. Find the slope of the curve at x = a.
h
afhafh
0
lim h
ahah
35
35
lim0
haha
haah
1
33
3535lim
0
haha
hh
1
33
5lim
0
23
5
a 33
5lim
0
ahah
Slope of a Curve
3
5Let .4
xxf
b. Where does the slope equal -5/16?
16
5
3
52
a
163 2 a
3a43 a
7a
Slope of a Curve
3
5Let .4
xxf
4
1 ,
4
5 ,7
4
5 ,1 and
c. What happens to the tangent to the curve at the point (a, 5/(a – 3 )) for the different values of a?
Slope of a Curve
3
5Let .4
xxf
The slope – 5/(a – 3)2 is always negative.
approaches slope the,3 As asteep.ly increasing becomes tangent theand
.3 asagain thissee We a
As a moves away from 3 in either direction, the slope approaches 0 and the tangent becomes increasingly horizontal.
Normal to a CurveThe normal line to a curve at a point is the line perpendicular to the tangent at the point.
The slope of the normal line is the negative reciprocal of the slope of the tangent line.
Example Normal to a Curve5. Write an equation for the normal to the curve y = x
2 – 2x – 3 at x = – 1 .
h
fhfh
11lim
0
h
hhh
31213121lim
22
0
h
hhh
4lim
2
0
4lim
0
h
h4
14
10 xy
4
1
4
1 xy
normalm4
1
,1 0
Speed Revisited2The function 16 is an object's . An object's average speed
along a coordinate axis for a given period of time is the average rate of change
of its position ( ).
It's
y t
y f t
position function
instanta
0
at any time is the of
position with respect to time at time , or lim . h
t
f t h f tt
h
neous speed instantaneous rate of change
Speed Revisited6. Wile E. Coyote drops an anvil from the top of a cliff. Find the
instantaneous rate of speed at 4 seconds.
h
fhfh
44lim
0
h
hh
22
0
416416lim
h
hhh
12816lim
2
0
12816lim
0
h
h
128 secondper feet
Pg. 92, 2.4 #2-40 even