Transcript Slide 1

Homework
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Homework Assignment #13
Read Section 3.5
Page 158, Exercises: 1 – 45 (EOO)
Rogawski Calculus
Copyright © 2008 W. H. Freeman and Company
Homework, Page 158
1. Find the ROC of the area of a square with respect to the length
of its side s when s = 3 and s = 5.
A  s2 
dA
dA
dA
 2s 
 2  3  6 
 2  5   10
ds
ds s 3
ds s 5
A  3  6, A  5   10
Rogawski Calculus
Copyright © 2008 W. H. Freeman and Company
Homework, Page 158
5. Calculate the ROC dV/dr, where V is the volume of a cylinder
whose height is equal to its radius.
V   r 2h  h  r  V   r 3
dV
 3 r 2
dr
Rogawski Calculus
Copyright © 2008 W. H. Freeman and Company
Homework, Page 158
9. Refer to Figure 10.
(a) Estimate the average
velocity over [0.5, 1].
vavg 
50  25
 50km / hr
1  0.5
(b) Is the average velocity
greater over [1, 2] or [2, 3]?
75  50
150  75
vavg 
 25km / hr , vavg 
 75km / hr
2 1
32
Average velocity is greater over [2, 3].
(c) At what time is velocity maximum?
Velocity appears to be maximum at t  1.25 hr.
Rogawski Calculus
Copyright © 2008 W. H. Freeman and Company
Homework, Page 158
13. A stone is tossed vertically upward with an initial velocity of
25 ft/s from the top of a 30-ft high building.
a. What is the height of the stone after 0.25 s?
1 2
1
2
s  t   h0  v0t  gt  s  0.25   30  25  0.25    32  0.25 
2
2
s  0.25   35.25 ft
b. Find the velocity of the stone after 1 s.
v  t   v0  gt  v 1  25   32 1  v 1  7 ft/s
c. When does the stone hit the ground?
0  30  25t  16t 2  t  2.358 sec
Rogawski Calculus
Copyright © 2008 W. H. Freeman and Company
Homework, Page 158
17. The Earth exerts a gravitational force of F (r) = (2.99 x 1016)/r 2
N on an object of mass 75 kg where r is the distance from the center
of the Earth. Assuming the radius of the Earth is 6.77 x 106 m,
calculate the ROC of the force with respect to distance at the Earth’s
surface.
2.99  1016
16
2
16
3

F r  

2.99

10
r

F
r


2
2.99

10
r






r2
F   6.77  106   2  2.99  1016  6.77  106   1.927  104 N / m
3
F   6.77  106   1.927  104 N / m
Rogawski Calculus
Copyright © 2008 W. H. Freeman and Company
Homework, Page 158
21. By Faraday’s law, if a conducting wire of length l meters moves
at a velocity of v m/s perpendicular to a magnetic field of strength B
(in teslas), a voltage of size V = –Blv is induced in the wire. Assume
that B = 2 and l = 0.5.
(a) Find the rate of change dV/dv.
V   Blv    2  0.5 v  v 
dV
 1
dv
(b) Find the ROC of V with respect to t, if v = 4t + 9
dV
V   Blv    2  0.5 4t  9   4t  9 
 4
dv
Rogawski Calculus
Copyright © 2008 W. H. Freeman and Company
Homework, Page 158
25. Ethan finds that with h hours of tutoring, he is able to answer
correctly S(h) percent of the problems on a math exam. What is the
meaning of the derivative S′(h)? Which would you expect to be
larger, S′(3) or S′(30)? Explain
The derivative S′(h) would be the rate of change in the percentage
correct after h hours of tutoring. I would expect S′(3) to be greater
than S′(30) since the percentage increase would first be rapid and
then slow down as the percent correct approaches 100%.
Rogawski Calculus
Copyright © 2008 W. H. Freeman and Company
Homework, Page 158
29. According to a formula used for determining drug dosage, a
person’s body surface area (BSA) (in m2) is given by the formula
BSA  hw / 60 , h is height in cm and w is weight in kg. Calculate
the ROC of BSA with respect to weight for a person of height h =
180. What is the ROC for w = 70 and w = 80? Does BSA increase
more rapidly with respect to weight for higher or lower weights?
180w 6 5w
5w
5 w
d
5 1
BSA 




BSA 
60
60
10
10
dw
10 2 w
d
5
d
5
BSA 

BSA  70  
 0.0134
dw
dw
20 w
20 70
d
5
BSA  80  
 0.0125  As shown, BSA increases more
dw
20 80
rapidly for lower body weight.
Rogawski Calculus
Copyright © 2008 W. H. Freeman and Company
Homework, Page 158
33. A ball is tossed up vertically from ground level and returns to
Earth 4 s later. What was the initial velocity of the ball and how
high did it go?
1 2
2
2
s  t   h0  v0t  gt  s  t   v0t  16t  s  4   0  v0  4   16  4 
2
4v0  256  v0  64 ft/s  s  2   64  2   16  2   64
2
smax  64 ft
Rogawski Calculus
Copyright © 2008 W. H. Freeman and Company
Homework, Page 158
37. Show that for an object rising and falling according to Galileo’s
formula, the average velocity over any time interval [t1, t2] is equal
to the average of the instantaneous velocities at t1 and t2.
1
1
s  t1   h0  v0t1  gt12  s  t2   h0  v0t2  gt2 2
2
2
1 2 
1 2

h  v t  gt  h  v t  gt
s  t2   s  t1   0 0 2 2 2   0 0 1 2 1 
vavg 

t2  t1
t2  t1


1
1
v0  t2  t1   g t2 2  t12
v0  t2  t1   g  t2  t1  t2  t1 
2
2


t2  t1
t2  t1
1
 v0  g  t2  t1 
2
Rogawski Calculus
Copyright © 2008 W. H. Freeman and Company
Homework, Page 158
37. Continued.
v  t1   v0  gt1  v  t2   v0  gt2
vavg 
v  t2   v  t1 
v0  gt2    v0  gt1  2v0  g  t2  t1 



2
1
 v0  g  t2  t1 
2
2
2
Rogawski Calculus
Copyright © 2008 W. H. Freeman and Company
Homework, Page 158
41. Let F (s) = 1.1s + 0.03s2 be the stopping distance. Calculate
F (65) and estimate the increase in stopping distance if speed is
increased from 65 to 66 mph. Compare your estimate to the actual.
F  s   1.1s  0.03s  F  65   1.1 65   0.03  65   198.25 ft
2
2
F   s   1.1  0.06s  F   65   1.1  0.06  65   5 ft
F  66   1.1 66   0.03  66   203.28 ft  198.25 ft  5.03 ft
2
Rogawski Calculus
Copyright © 2008 W. H. Freeman and Company
Homework, Page 158
45. The demand for a commodity generally decreases as the price
increases. Suppose that demand for oil is D(p) =900/p barrels, where
p is the price per barrel in dollars. Find the demand when p = $40.
Estimate the decrease in demand if p rises to $41 and the increase if p
falls to $39.
900
900
1
D  p 
 900 p  D  40  
 22.5
p
40
D  p   900 p 2  D  40   900  40   0.5625
2
The demand per person at $40 per barrel is 22.5 barrels/year. If the
price increases to $41, the demand will fall by 0.5625 barrels/year
and if the price decreases to $39, the demand will increase by 0.5625
barrels per year.
Rogawski Calculus
Copyright © 2008 W. H. Freeman and Company
Jon Rogawski
Calculus, ET
First Edition
Chapter 3: Differentiation
Section 3.5: Higher Derivatives
Rogawski Calculus
Copyright © 2008 W. H. Freeman and Company
As shown in the graph and table below, the population of Sweden
increased at a decreasing rate from 1993 to 1997. P′ (t) is positive,
as the slope is positive, but P″(t) is negative as the rate of increase
is decreasing.
Rogawski Calculus
Copyright © 2008 W. H. Freeman and Company
Higher derivatives are derivatives of derivatives. The second
derivative is the derivative of the first derivative, the third
derivative is the derivative of the second derivative, and so forth.
Symbolically, we write the derivatives as f ′, f ″, f ′″, f (4), …,f (n), …
dy d 2 y d 3 y d 4 y
In Leibniz notation, we write: , 2 , 3 , 4 ,   
dx dx dx dx
Rogawski Calculus
Copyright © 2008 W. H. Freeman and Company
Table 2 tabulates the six derivatives of f (x) = x5. In general, an
nth order function with all positive, whole number exponents, will
have n +1 derivatives, the last of which will be zero.
Rogawski Calculus
Copyright © 2008 W. H. Freeman and Company
Example, Page 165
Calculate the second and third derivatives.
4. y  4t 3  9t 2  7
Rogawski Calculus
Copyright © 2008 W. H. Freeman and Company
Example, Page 165
Calculate the second and third derivatives.
12. y 
1
1 x
Rogawski Calculus
Copyright © 2008 W. H. Freeman and Company
Example, Page 165
Calculate the second and third derivatives.
14. y  x4e x
Rogawski Calculus
Copyright © 2008 W. H. Freeman and Company
Figure 2 shows the graphs of height versus time and velocity versus
time for a ball tossed vertically into the air with an initial velocity of
40 ft/s. Notice that the velocity plots as a negative linear function.
This indicates a constant negative acceleration due to gravity. For
most calculations, this acceleration may be estimated at 32 ft/s or
9.8 m/s. In general, given position as a function of time, s (t), velocity
is the first derivative of position and acceleration is the second
derivative or v (t) = s′ (t) and a (t) = v′ (t) = s″ (t).
Rogawski Calculus
Copyright © 2008 W. H. Freeman and Company
The larger the magnitude of the second derivative, the more quickly
the slope of the tangent changes. If the second derivative is zero,
then the function is linear and the slope is unchanging. The sign of
the second derivative tells us if the slope is increasing (positive) or
decreasing (negative). In general, if the graph of the function lies
above the graph of the tangent, the second derivative is positive, and
if the graph of the function lies below the graph of the tangent, the
second derivative is negative.
Rogawski Calculus
Copyright © 2008 W. H. Freeman and Company
Example, Page 165
Find a general formula for f (n) (x).
34. y  x2
Rogawski Calculus
Copyright © 2008 W. H. Freeman and Company
Figure 4, on the left, shows the graph of a function in blue and graphs
of two tangent lines in red. On the right are the graphs of the first
and second derivatives, in red and green, respectively.
Rogawski Calculus
Copyright © 2008 W. H. Freeman and Company
Example, Page 165
In their 1997 study, researchers related the traffic speed S on a twolane road to traffic density Q (number of cars per mile of road) by
1
the formula: S  2,882Q  0.052Q  31.73 for 60 < Q < 400.
dS
 0.
48.  a  Explain intuitively why we should expect that
dQ
d 2S
dS
d 2S
 0 and
0
 b  Show that 2  0. Then use the fact that
2
dQ
dQ
dQ
to justify: A one - unit increment increase in traffic density slows down
traffic more when Q is small than when Q is big.
Rogawski Calculus
Copyright © 2008 W. H. Freeman and Company
Homework
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
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Homework Assignment #14
Read Section 3.6
Page 165, Exercises: 1 – 49 (EOO)
Rogawski Calculus
Copyright © 2008 W. H. Freeman and Company