Transcript x - Piazza

Chapter 6
Applications of
the Derivative
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6- 1
Section 6.1
Absolute
Extrema
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Figure 1
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Figure 2
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Figure 3
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Figure 4
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Your Turn 1
2/3
5/3
Find the absolute extrema of the function f ( x)  3x  3x
on the interval [0,8].
Solution: First look for critical numbers in the interval (0,8).
2 1/3
5 2/3
f ( x)  3  x  3  x
3
3
 2 x 1/3  5 x 2/3
 x 1/3 (2  5x)
2  5x
 1/3
x
Set f ( x)  0 and solve for x.
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Factor.
Continued
Your Turn 1 continued
Notice that f ( x)  0
at x = 2/5 and 2/5 is in the
interval (0,8) . The derivative
is undefined at x = 0 but the
function is defined there, so 0
is also a critical number.
Evaluate the function at the
critical numbers and the
endpoints.
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x - Value Extrema Candidates Value
of the Function
0
0
2/5
0.977 Absolute Maximum
8
-84
Absolute Minimum
Your Turn 2
Find the locations and values of the absolute extrema, if they
exist, for the function f ( x)   x 4  4 x3  8x 2  20.
Solution: In this example, the extreme value theorem does not
apply since the domain is an open interval (−∞,∞), which has
no endpoints. Begin as before by finding any critical numbers.
f ( x)  4 x3  12 x 2  16 x  0
4 x( x 2  3x  4)  0
4 x( x  4)( x  1)  0
x = 0 or x = − 4 or x = 1.
Evaluate the function at the critical numbers.
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Continued
Your Turn 2 continued
For an open interval, rather
than evaluating the function at
the endpoints, we evaluate
the limit of the function when
the endpoints are approached.
Because the negative x4
term dominates the other
terms as x becomes small,
it has no absolute minimum.
The absolute maximum 148,
occurs at x = − 4.
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x - Value Extrema Candidates Value
of the Function
-4
148
0
20
1
23
Absolute Maximum
Figure 5
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Figure 6
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Section 6.2
Applications of
Extrema
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Your Turn 1
Find two nonnegative numbers x and y for which x + 3y = 30,
such that x2y is maximized.
Solution: Step 1, reading and understanding the problem, is up
to you. Step 2 does not apply in this example; there is nothing
to draw. We proceed to Step 3, in which we decide
what is to be maximized and assign a variable to that quantity.
Here x2y, is to be maximized, so let
M = x2y.
According to Step 3, we must express M in terms of just one
variable, which can be done using x + 3y = 30 the equation by
solving for either x or y. Solving for y gives
Continued
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Your Turn 1 Continued
x + 3y = 30
30  x
x
y
 10  .
3
3
Substitute for y in the expression for M to get
3
x
x
M  x 2 (10  )  10 x 2  .
3
3
We are now ready for Step 4, when we find the domain of the
function. Because of the nonnegativity requirement, x must be
at least 0. Since y must also be at least 0, we require
Continued
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Your Turn 1 Continued
x
10   0
3
x
  10
3
x
 10
3
x  30.
Thus x is confined to the interval [0,30].
Moving on to Step 5, we find the critical points for M by
finding dM
dx
dM
 0 for x.
and then solving
dx
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Your Turn 1 Continued
dM
3 2
 20 x  x  20 x  x 2  0
dx
3
x(20  x)  0
Factor out the x.
x = 0 or x = 20
Finally, at Step 6, we find M for the critical numbers and
endpoint of the domain. The other endpoint has already been
included as a critical number. We see in the table that the
maximum value of the function occurs when x = 20,
y = 10/3 and the maximum value of x2y is 1333.3
Continued
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Your Turn 1 continued
x – value
Extrema Candidates
M
0
0
20
1333.3 Absolute Maximum
30
0
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Your Turn 2 - Figure 7
A math professor participating in the sport of
orienteering must get to a specific tree in the woods
as fast as possible. He can get there by traveling east
along the trail for 300 m and then north through the
woods for 800 m. He can run 160 m per minute along
the trail but only 40 m per minute through the woods.
Running directly through the woods toward the tree
minimizes the distance, but he will be going slowly
the whole time. He could instead run 300 m along
the trail before entering the woods, maximizing the
total distance but minimizing the time in the woods.
Perhaps the fastest route is a combination, as shown
in Figure 7. Find the path that will get him to the tree
in the minimum time.
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Your Turn 2 Continued
Solution : We are trying to minimize the total amount of time,
which is the sum of the time on the trail and the time through
the woods. We must express this time as a function of x. Since
Time = Distance/ Speed, the total time is
300  x
8002  x 2
T ( x) 

.
160
40
Notice in this equation that 0 ≤ x ≤ 300.
1
1 1
T ( x)  
   (8002  x 2 ) 1/2 (2 x)  0
160 40  2 
Continued
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Your Turn 2 Continued
1
x

160 40 8002  x 2
x
T(x)
0
21.88
16 x  4 8002  x 2
207
21.24
300
21.36
4 x  8002  x 2
16 x 2  8002  x 2
15 x 2  8002
2
800
x2 
15
800
x
 207
15
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Absolute Minimum
Your Turn 2 Continued
We see from the table that the time is minimized when
x = 207 m. That is, when the professor goes 93 m along the
trail and then heads into the woods.
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Figure 8
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Figure 9
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Figure 10
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Figure 11
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Figure 12
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Figure 13
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Section 6.3
Further Business Applications:
Economic
Lot Size; Economic Order
Quantity;
Elasticity of Demand
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Figure 14
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Figure 14
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Figure 15
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Section 6.4
Implicit
Differentiation
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Your Turn 1
Find
dy
if x 2  y 2  xy.
dx
Solution: Differentiate with respect to x on both sides of the
equation. x 2  y 2  xy
d 2
d
2
( x  y )  ( xy )
dx
dx
Now differentiate each term on the left side of the equation.
Think of xy as the product (x)(y) and use the product rule and
the chain rule.
2x  2 y
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dy
dy
 x  y 1
dx
dx
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Continued
Your Turn 1 Continued
dy
Now solve this result for
.
dx
dy
dy
2 y  x  y  2x
dx
dx
dy
(2 y  x)  y  2 x
dx
dy y  2 x

dx 2 y  x
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Figure 16
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Figure 17
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Your Turn 3
4
4
2
2
The graph of y  x  y  x  0
is called the devil’s curve. Find the equation of the tangent line
at the point (1, 1).
Solution: We can calculate by implicit differentiation.
dy
dy
3
4y
 4x  2 y  2x  0
dx
dx
dy
dy
4 y3  2 y
 4 x3  2 x
dx
dx
dy
(4 y 3  2 y )  4 x3  2 x
dx
3
Chain Rule
Continued
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Your Turn 3 Continued
dy 4 x3  2 x 2(2 x3  x) (2 x3  x)
 3


3
dx 4 y  2 y 2(2 y  y ) (2 y 3  y )
To find the slope of the tangent line at the point (1, 1), let x = 1
and y = 1 The slope is
2(1)3  1
m
 1.
3
2(1)  1
The equation of the tangent line is then found by using the
point-slope form of the equation of a line.
y  y1  m( x  x1 )
y  1  1( x  1)
y 1  x 1
yx
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Section 6.5
Related Rates
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Your Turn 1
Suppose x and y are both functions of t and x3  2 xy  y 2  1.
dx
dy
If x  1, y  2, and
 6, then find
.
dt
dt
Solution: We start by taking the derivative of the relationship,
using the product and chain rules. Keep in mind that both x and
y are functions of t. The result is
dx
dy
dx
dy
 2x  2 y  2 y
 0.
dt
dt
dt
dt
dx
Now substitute x  1, y  2, and
=  6 to get
dt
3x 2
Continued
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Your Turn 1 Continued
dy
dy
 2(2)(6)  2(2)  0
dt
dt
dy
dy
18  2  24  4  0
dt
dt
3(1) 2  6  2(1)
dy
6  2  0
dt
dy
2  6
dt
dy
 3.
dt
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Figure 18
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Figure 19
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Your Turn 2
A 25-ft ladder is placed against a
building. The base of the ladder is
slipping away from the building at a
rate of 3 ft per minute. Find the rate
at which the top of the ladder is
sliding down the building when the
bottom of the ladder is 7 ft
from the base of the building.
Continued
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Your Turn 2 Continued
Solution: Starting with Step 1, let y be the height of the top of
the ladder above the ground, and let x be the distance of the
base of the ladder from the base of the building. We
are trying to find dy/dt when x = 7. To perform Step 2, use the
Pythagorean theorem to write x 2  y 2  252.
Both x and y are functions of time t (in minutes) after the
moment that the ladder starts slipping. According to Step 3,
take the derivative of both sides of the equation with respect to
time, getting d ( x 2  y 2 )  d (252 )
dt
dt
dx
dy
Continued
2x  2 y
 0.
dt
dt
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Your Turn 2 Continued
To complete Step 4, we need to find the values of x, y, and dx/dt.
Once we find these, we can substitute them into Equation to find
dy/dt.
Since the base is sliding at the rate of 3 ft per minute, dx/dt = 3.
Also, the base of the ladder is 7 ft from the base of the building,
so x =7 . Use this to find y.
x 2  y 2  252
7 2  y 2  625
49  y 2  625
y 2  576
y  24
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Continued
Your Turn 2 Continued
In summary, x = 7, y = 24, and dx/dt = 3.
dx
dy
 2y
0
dt
dt
dy
2(7)(3)  2(24)  0
dt
dy
42  48  0
dt
dy
48  42
dt
dy
42
7
    ft/min
dt
48
8
2x
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Figure 20
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Figure 21
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Section 6.6
Differentials:
Linear
Approximation
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Figure 22
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Your Turn 1
Find dy if y  300 x 2/3 , x  8, and dx  0.05.
2
Solution: dy  300     x 5/3dx
 3
dy  200 x 5/3dx
 200(8)5/3 (0.05)
 1
 200  5  (0.05)
2 
5

16
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Your Turn 2
Approximate 99.
Solution: We will use the linear approximation
f ( x  x)  f ( x)  f ( x)dx
for a small value of ∆x to form this estimate.
We first choose a number x that is close to 99 for which we
know its square root.
Since 100  10, we let f ( x)  x , x  100, and x  1.
f ( x  x)  x 
1
2 x
dx
f (99)  f (100  1)  100 
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1
1
 9.95
 (1)  10 
20
2 100
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Your Turn 3
In a precision manufacturing process, ball bearings must be
made with a radius of 1.25 mm, with a maximum error in the
radius of ± 0.025 mm. Estimate the maximum error in the
volume of the ball bearing.
4 3
Solution: The formula for the volume of a sphere is V   r .
3
If an error of ∆r is made in measuring the radius of the sphere,
the maximum error in the volume is dV  4 r 2 dr.
Replacing r with 1.25 and dr = ∆r with ± 0.025 gives
dV  4 (1.25)2 (0.025)  0.5 mm3 .
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