the derivative objective: we will explore tangent lines, velocity, and general rates of change and...
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The Derivative
Objective: We will explore tangent lines, velocity, and general rates of
change and explore their relationships.
Tangent Lines
• We are going to look at a secant line (PQ) to a curve and talk about its slope. This is defined as:
0
0 )()(
xx
xfxfmPQ
Definition 2.1.1
• We will now look at what happens as point Q approaches point P. We will look at this as a limit; the limit as x approaches x0. This will be defined as:
0
0tan
)()(lim
0 xx
xfxfm xx
Definition 2.1.1
• We will now look at what happens as point Q approaches point P. We will look at this as a limit, the limit as x approaches x0. This will be defined as:
• and the equation of the tangent line to the curve at the point is:
0
0tan
)()(lim
0 xx
xfxfm xx
)()( 0tan0 xxmxfy
))(,( 00 xfx
Example 1
• Use definition 2.1.1 to find an equation for the tangent line to the parabola at the point P(1, 1).
2xy
Example 1
• Use definition 2.1.1 to find an equation for the tangent line to the parabola at the point P(1, 1).
• P(1, 1) & Q(2, 4)
• P(1, 1) & Q(1.5, 2.25)
• P(1, 1) & Q(1.1, 1.21)
• P(1, 1) & Q(1.01, 1.0201)
2xy
312
14
m
5.215.1
125.2
m
1.211.1
121.1
m
01.210201.1
101.1
m
Example 1
• Use definition 2.1.1 to find an equation for the tangent line to the parabola at the point P(1, 1).
• You can also use a point to the right and a point to the left of the given point to estimate the slope of the tangent line.
• P(2, 4) & Q(0, 0)
2xy
202
04
m
Example 1
• We have the point, we need the slope. We will use our definition of the slope of a tangent line to find this by first substituting a 1 for .
1
)1()(lim
1tan
x
fxfm
x
0x
0
0tan
)()(lim
0 xx
xfxfm xx
Example 1
• Now we use the fact that and find f(1) and substitute where necessary.
1
1lim
1
)1()(lim
2
11tan
x
x
x
fxfm
xx
2)( xxf
Example 1
• Now, we find the limit.
21lim1
1lim
1
)1()(lim
1
2
11tan
x
x
x
x
fxfm
xxx
Example 1
• We now have the point and the slope, so the equation of the tangent line is:
12
)1(21
xy
or
xy
The Difference Quotient
• There is another formula that is commonly used to find the slope of a tangent line. This is called the Difference Quotient and we define h as the difference between and .
• The equation now becomes:
0xxh
h
xfhxfm
h
)()(lim 00
0tan
x 0x
Example 2
• We will use this formula to find the slope of the tangent line in example 1. Again, we start by replacing with a 1.
h
fhfm
h
)1()1(lim
0tan
h
xfhxfm
h
)()(lim 00
0tan
0x
Example 2
• We again use to evaluate
h
h
h
fhfm
hh
1)1(lim
)1()1(lim
2
00tan
2)( xxf )( hxf
Example 2
• Do the algebra to find the limit.
h
hh
h
hm
hh
121lim
1)1(lim
2
0
2
0tan
22lim)2(
lim00
h
h
hhhh
Example 3
• Find the equation of the tangent line to the curve y=2/x at the point (2, 1) on this curve.
Example 3
• Find the equation of the tangent line to the curve y=2/x at the point (2, 1) on this curve.
h
fhfm
h
)2()2(lim
0tan
h
xfhxfm
h
)()(lim 00
0tan
Example 3
• Find the equation of the tangent line to the curve y=2/x at the point (2, 1) on this curve.
hh
h
fhfm
hh
12
2
lim)2()2(
lim00
tan
Example 3
• Find the equation of the tangent line to the curve y=2/x at the point (2, 1) on this curve.
hh
h
fhfm
hh
12
2
lim)2()2(
lim00
tan
2
1
2
1lim
)2(lim
)2(
)2(2lim
000
hhh
h
hh
hhhh
h2 h2
h2
Example 3
• Find the equation of the tangent line to the curve y=2/x at the point (2, 1) on this curve.
22
1
)2(2
11
xy
or
xy
Example 3
• Find the equation of the tangent line to the curve y=2/x at the point (2, 1) on this curve. We will use a different equation this time.
2
1
)2(
)2(
)2(
2
2
12
lim2
xx
x
xx
x
xx
x
0
0tan
)()(lim
0 xx
xfxfm xx
Example 4
• We have been finding the slope of the tangent line at a specific point. We will now find the slope of the tangent line at a general point.
Example 4
• Find the slopes of the tangent lines to the curve at xy .9,4,1 000 xxx
Example 4
• Find the slopes of the tangent lines to the curve at xy .9,4,1 000 xxx
h
xhx
h
xfhxfm
hh
00
0
00
0tan lim
)()(lim
Example 4
• Find the slopes of the tangent lines to the curve at xy .9,4,1 000 xxx
h
xhx
h
xfhxfm
hh
00
0
00
0tan lim
)()(lim
00
0000
0lim
xhx
xhx
h
xhxh
Example 4
• Find the slopes of the tangent lines to the curve at xy .9,4,1 000 xxx
h
xhx
h
xfhxfm
hh
00
0
00
0tan lim
)()(lim
)(lim
00
00
00
0000
0 xhxh
xhx
xhx
xhx
h
xhxh
Example 4
• Find the slopes of the tangent lines to the curve at xy .9,4,1 000 xxx
h
xhx
h
xfhxfm
hh
00
0
00
0tan lim
)()(lim
00000
00
2
1
)()( xxhxh
h
xhxh
xhx
Example 4
• Find the slopes of the tangent lines to the curve atxy .9,4,1 000 xxx
0
0tan
)()(lim
0 xx
xfxfm xx
000
0
0
0 1
))((lim
0 xxxxxx
xx
xx
xxxx
000 2
11lim
0 xxxxx
Example 4
• The slopes of the tangent lines are:
at x = 1 at x = 4 at x = 9
2
1
12
1
4
1
42
1
6
1
92
1
02
1
x
Velocity
• When we talk about the motion of an object, we want its speed and direction. Together, we call this velocity. Movement to the right or up is considered positive velocity and movement to the left or down is considered negative velocity. We will explore these meanings with a position vs. time curve, with the horizontal axis being time (t) and the vertical axis position (s). The movement of the particle will be called Rectilinear Motion.
Position vs. Time Curve
• We will look at two typical position vs. time curves.
Position vs. Time Curve
• We will look at two typical position vs. time curves. • The first is for a car that starts at the origin and
moves only in the positive direction. Movement to the right is considered positive and movement to the left is considered negative. In this case s increases as t increases.
Position vs. Time Curve
• We will look at two typical position vs. time curves. • The second is for a ball that is thrown straight up in
the positive direction and falls straight down in the negative direction.
Displacement and Average Velocity
• The key to describing the velocity of a particle in rectilinear motion is the notion of displacement, or change in position. This differs from distance traveled. Since movement to the right is positive and movement to the left is negative, you could travel 10 units to the right and then 8 units to the left and your displacement would be +2 and your distance traveled would be 18.
Average Velocity vs. Average Speed
• We will define average velocity as:
• And we will define average speed as:
h
tfhtf
time
ntdisplacemevave
)()( 00
time
traveleddistancespeedv
Example 5
• Suppose that is the position function of a particle, where s is in meters and t in seconds. Find the displacements and average velocities of the particle over the time intervals: [0,1] and [1,3].
2231)( tttfs
Example 5
• Suppose that is the position function of a particle, where s is in meters and t in seconds. Find the displacements and average velocities of the particle over the time intervals: [0,1] and [1,3].
• f(0) = 1, f(1) = 2, displacement is 1.• f(1) = 2, f(3) = -8 displacement is -10.
2231)( tttfs
sm /101
12
sm /513
28
Instantaneous Velocity
• Instead of looking at velocity over a time interval, we will now look at velocity at one point and we will call this instantaneous velocity, which describes the behavior of the particle at a specific instant in time.
Example 6
• Consider the particle in Ex. 5, whose position function is . Find the particle’s instantaneous velocity at time t = 2s.
2251)( tttfs
Example 6
• Consider the particle in Ex. 5, whose position function is . Find the particle’s instantaneous velocity at time t = 2s.
• As a first approximation to the particle’s instantaneous velocity at time t = 2, let us recall the average velocity from t = 2 to t = 3 is -5 m/s. To improve this approximation we will compute the average velocity over a succession of smaller and smaller time intervals.
2251)( tttfs
Example 6
• Consider the particle in Ex. 5, whose position function is . Find the particle’s instantaneous velocity at time t = 2s.
• The average velocities in this table appear to be approaching a limit of -3 m/s. Let’s confirm this.
2251)( tttfs
Example 6
• Consider the particle in Ex. 5, whose position function is . Find the particle’s instantaneous velocity at time t = 2s.
• The average velocities in this table appear to be approaching a limit of -3 m/s. Let’s confirm this.
2251)( tttfs
h
hh
h
fhfv
hinst
3])2(2)2(51[)2()2(lim
2
0
3)23(lim23
lim0
2
0
h
h
hhv
hhinst
Example 6
• Consider the particle in Ex. 5, whose position function is . Find the particle’s instantaneous velocity at time t = 2s.
• The average velocities in this table appear to be approaching a limit of -3 m/s. Let’s confirm this.
2251)( tttfs
2
3251lim
)()(lim
2
20
0
0
x
xx
xx
xfxfv
xxxinst
3)12(2
)12)(2(lim
2
)252(lim
2
2
2
x
x
xx
x
xxv
xxinst
Instantaneous Velocity
• We define instantaneous velocity as:
• And instantaneous speed as:
h
tfhtfv
hinst
)()(lim 00
0
|)()(
lim||| 00
0 h
tfhtfvspeed
hinstinst
Slopes and Rates of Change
• Velocity can be viewed as rate of change- the rate of change of position with respect to time. Rates of change occur in other applications as well.
Slopes and Rates of Change
• Velocity can be viewed as rate of change- the rate of change of position with respect to time. Rates of change occur in other applications as well.
• A microbiologist might be interested in the rate of change at which the number of bacteria in a colony changes with time.
Slopes and Rates of Change
• Velocity can be viewed as rate of change- the rate of change of position with respect to time. Rates of change occur in other applications as well.
• A microbiologist might be interested in the rate of change at which the number of bacteria in a colony changes with time.
• An economist might be interested in the rate of change at which production cost changes with the quantity of a product that is manufactured.
Example 7
• Find the rate of change of y with respect to x .
12 xy 15 xy
Example 7
• Find the rate of change of y with respect to x .
• The rate of change for the equation on the left is 2, and the rate of change for the equation on the right is -5.
12 xy 15 xy
Average rate of change
• The average rate of change is the same thing as the slope of the secant line, so we define it as:
or
01
01 )()(
xx
xfxfrave
h
xfhxfrave
)()( 00
Instantaneous rate of change
• The instantaneous rate of change is the same thing as the slope of the tangent line, so we define it as:
or
01
01 )()(lim
01 xx
xfxfr
xxinst
h
xfhxfr
hinst
)()(lim 00
0
Example 9
• Let1) Find the average rate of change of y with respect to x
over the interval [3,5]2) Find the instantaneous rate of change of y with
respect to x when x = - 4.
12 xy
Example 9
• Let1) Find the average rate of change of y with respect to x
over the interval [3,5]
12 xy
82
1026
35
)3()5()()(
01
01
ff
xx
xfxfrave
Example 9
• Let1) Find the instantaneous rate of change of y with
respect to x when x = -4
12 xy
4
)4()(lim
)()(lim
1
1
401
01
101
x
fxf
xx
xfxfr
xxxinst
844
16
4
17)1(lim 1
1
21
1
21
41
xx
x
x
xx
Example 10
• The limiting factor in athletic endurance is cardiac output, that is, the volume of blood that the heart can pump per unit of time during an athletic competition. The figure shows a stress-test graph of cardiac output in liters of blood vs workload for 1 minute of lifting.
Example 10
• Use the secant line shown on the graph below to estimate the average rate of change of cardiac output with respect to workload as the workload increases from 300 to 1200.
Example 10
• Use the secant line shown on the graph below to estimate the average rate of change of cardiac output with respect to workload as the workload increases from 300 to 1200.
mkg
Lrave
0067.3001200
1319
Example 10
• Use the tangent line on the graph below to estimate the instantaneous rate of change of cardiac output with respect to workload at the point where the workload is 300 kg*m.
Example 10
• Use the tangent line on the graph below to estimate the instantaneous rate of change of cardiac output with respect to workload at the point where the workload is 300 kg*m.
mkg
Lrinst
02.0900
725
Homework
• Section 2.1
• Pages 140-141
• 1-10 all
• 11-27 odd
#11a
• Find the average rate of change of y with respect to x over the interval [x0, x1].
1,0,2 102 xxxy
#11a
• Find the average rate of change of y with respect to x over the interval [x0, x1]. Slope of the secant line.
• (0, 0) and (1, 2)
1,0,2 102 xxxy
201
02
#11b
• Find the instantaneous rate of change of y with respect to x at a specified value of x0.
1,0,2 102 xxxy
#11b
• Find the instantaneous rate of change of y with respect to x at a specified value of x0. Tangent line.
1,0,2 102 xxxy
0
0 )()(lim
0 xx
xfxfxx
020
02
0
)0()(lim
2
0
x
x
x
x
fxfx
#11c
• Find the instantaneous rate of change of y with respect to x at an arbitrary value of x0.
1,0,2 102 xxxy
#11c
• Find the instantaneous rate of change of y with respect to x at an arbitrary value of x0.
1,0,2 102 xxxy
0
0 )()(lim
0 xx
xfxfxx
000
00
0
20
2
422))((222
lim0
xxxx
xxxx
xx
xxxx
#11d
• Graph the function and the two lines found in parts a and b.
#12a
• Find the average rate of change of y with respect to x over the interval [x0, x1].
2,1, 103 xxxy
#12a
• Find the average rate of change of y with respect to x over the interval [x0, x1]. Slope of the secant line.
• (1, 1) and (2, 8)
2,1, 103 xxxy
712
18
#12b
• Find the instantaneous rate of change of y with respect to x at the specified value of x0.
2,1, 103 xxxy
#12b
• Find the instantaneous rate of change of y with respect to x at the specified value of x0. Tangent line.
2,1, 103 xxxy
0
0 )()(lim
0 xx
xfxfxx
3)1(
)1)(1(
1
1
1
)1()(lim
23
1
x
xxx
x
x
x
fxfx
#12c
• Find the instantaneous rate of change of y with respect to x at an arbitrary value of x0.
2,1, 103 xxxy
0
0 )()(lim
0 xx
xfxfxx
20
0
200
20
0
30
3
0
0 3)(
))(()()(lim
0
xxx
xxxxxx
xx
xx
xx
xfxfxx