Solution Guide for Chapter 5:
A Survey of Other
Common Functions
5.1 POWER FUNCTIONS
E-1. More skid marks: Parts (a) and (b) of Exercise 5 do not use homogeneity, so we need to
consider only Part (c). We want to slow to a speed that will cut the emergency stopping
distance in half. Since we are originally traveling at 60 miles per hour, the stopping
1
1
distance is L =
602 , since the speed is 60. The new stopping distance will be L, so
30h
2
if S is the new reduced speed, then
1
1
1
1 2
2
S = L=
60 .
30h
2
2 30h
r
1 2
1 2
2
60 ,
Multiplying both sides by 30h and cancelling yields S = 60 . Thus S =
2
2
which equals 42.43 miles per hour. This is between 42 and 43 miles per hour.
E-2. Kepler’s third law again: We have that P is a power function of D, with power 1.5, so
P = cD1.5 , where P is the period (time needed to complete one orbit) and D is the mean
distance from the sun. The constant c does not depend on which planet is used. We
measure P in years and D in millions of miles.
(a) For Earth P = 1 year, so 1 = cd1.5 , where d is the distance of Earth from the sun
1
in million of miles; dividing gives c = 1.5 . If D is the distance for Neptune,
d
we know that D = 30d, and we’d like to find the period P . On the other hand,
1
P = cD1.5 = 1.5 (30d)1.5 = 301.5 , since the d1.5 terms cancel. As 301.5 is 164.32, the
d
period of Neptune is about 164 years.
398
Solution Guide for Chapter 5
(b) We know that P = cD1.5 with the same c for any planet. For Earth, P = 1 year and
1
D = 93 million miles, so 1 = c × 931.5 , and so c =
. For Mercury we have
931.5
88
P = cD1.5 but we know both P and c. Now 88 days is
year, so for Mercury,
365
88
P =
. Putting this together with the value of c calculated above, we find
365
88
1
= 1.5 D1.5 .
365
93
88
365
and cancelling yields D =
931.5 . Raising each side
1/1.5
88
1.5
to the 1/1.5 power, we get D =
93
, which equals 36.03, or about
365
36, million miles.
1.5
1.5
Multiplying by 93
E-3. Calculating a limit: To calculate the limx→∞
2 × 3x + x1000
, we use the hint to rewrite
3x
the fraction as a sum of two fractions:
2 × 3x
x1000
2 × 3x + x1000
=
+
.
3x
3x
3x
In the first fraction, the 3x terms cancel, leaving only the 2. For the second fraction, we
know from the discussion in the text that
xp
=0
x→∞ ax
lim
when a is greater than 1. In our case, p = 1000 and a = 3, so that limit is zero. Thus
2 × 3x + x1000
limx→∞
= 2.
3x
E-4. Algebraic justification of homogeneity: We want to show that if x is multiplied by t,
then y = Axk is multiplied by tk . Multiplying x by t is the same as replacing x with
xt, so y = A(xt)k which equals y = Axk tk . This value for y is the same as the original
y = Axk except that it is now multiplied by tk .
E-5. Uniqueness of homogeneity: Suppose that a function f has the property that f (tx) =
tk f (x) for all positive x and t. This means that in particular this is true when x = 1.
When x = 1, the property is f (t) = tk f (1) for all t. Writing c = f (1), we have f (t) = ctk ,
so f is a power function with power k.
S-1. Graph of a power function: Graphing a power function with negative k, such as x−2 ,
shows that the graph is decreasing. See Key Idea 5.1.
S-2. Graph of a power function: Graphing a power function with positive k, such as x2 ,
shows that the graph is increasing. See Key Idea 5.1.
SECTION 5.1
399
Power Functions
S-3. Homogeneity: In this case f is a power function with power k = 1.47. By the homogeneity property, if x is increased by a factor of t, then f is increased by a factor of t1.47 .
In this case, x is tripled, so t = 3. Therefore f is increased by a factor of 31.47 , that is, by
a factor of 5.03.
S-4. Homogeneity: In this case f is a power function with power k = 2.53. By the homogeneity property, if x is increased by a factor of t, then f is increased by a factor of t2.53 .
To find the t tripling the value of f , we need to solve the equation t2.53 = 3. Solving by
the crossing-graphs method or algebraically, we see that t = 1.54, so x must be increased
by a factor of 1.54.
S-5. Homogeneity: In this case f is a power function with power k = 3.11. By the homogeneity property, if x is increased by a factor of t, then f is increased by a factor of t3.11 .
5.5
Since x increases from 3.6 to 5.5, x is increased by a factor of
and so f is increased by
3.6
3.11
5.5
a factor of
= 3.74, that is, f (5.5) is 3.74 times as large as f (3.6).
3.6
S-6. Homogeneity: In this case f is a power function with power k = 3.11. By the homogeneity property, if z is increased by a factor of t, then f is increased by a factor of t3.11 .
To find the t increasing the value of f by a factor of 9, we need to solve the equation
t3.11 = 9. Solving by the crossing-graphs method or algebraically, we see that t = 2.03,
so z must be increased by a factor of 2.03. Thus y is 2.03 times z.
S-7. Homogeneity: In this case f is a power function with power k. By the homogeneity
property, if x is increased by a factor of t, then f is increased by a factor of tk . Since x
6.6
increases from 1.76 to 6.6, 1.76 is increased by a factor of t =
. Thus f is increased
1.76
k
6.6
by a factor of tk =
. Since f (6.6) is 6.2 times as large as f (1.76),
1.76
k
6.2 = t =
6.6
1.76
k
.
Solving for k yields k = 1.38.
S-8. Constant term: Since f (x) = cx4.2 and f (4) = 8, 8 = c × 44.2 . Thus c =
8
, which
44.2
equals 0.02, so c = 0.02.
S-9. Constant term: Since f (x) = cx−1.32 and f (5) = 11, 11 = c × 5−1.32 . Thus c =
which equals 92.05, so c = 92.05.
11
,
5−1.32
400
Solution Guide for Chapter 5
S-10. Exponential versus power functions: Despite the values of these functions which would
indicate that x1023 grows much faster than 1.0002x , we already know that increasing exponential functions grow faster than power functions, so 1.0002x is eventually larger
than x1023 .
S-11. Power: From 1 to 2, 1 is increased by a factor of t =
2
= 2. Thus f is increased by a
1
factor of tk = 2k . Since f (2) is 10 times the size of f (1),
10 = tk = 2k .
Solving for k yields k = 3.32.
S-12. Power: From 2 to 1, 2 is multiplied by t =
1
= 0.5. Thus f is multiplied by tk = 0.5k .
2
Since f (1) is 10 times the size of f (2),
10 = tk = 0.5k .
Solving for k yields k = −3.32.
1. The role of the constant term in power functions with positive power: We use a
horizontal span of 0 to 5 as suggested, and the table of values below leads us to choose
a vertical span of 0 to 100. In the right-hand figure, x2 is the bottom graph, 2x2 is the
next one up, and 4x2 is the top graph.
Larger values of c make the function increase faster.
SECTION 5.1
Power Functions
401
2. The role of the constant term in power functions with a negative power: We use a
horizontal span of 0 to 5 as suggested, and the table of values below leads us to choose
a vertical span of 0 to 5. In the right-hand figure, x−2 is the graph on bottom, 2x−2 is
the next graph up, and 4x−2 is the top graph.
The graph shows that larger values of c make the function decrease more slowly.
3. Speed and stride length: We use the homogeneity property of power functions to solve
this. If one animal has a stride length 3 times that of another, then it will run faster by
a factor of
3Function power = 31.7 = 6.47,
or about 6.5 times as fast.
4. Weight and length: Since we are dealing with a power function, we know that when
the length is doubled, the weight will be increased by a factor of 2k , where k is the
power. On the other hand, we are told that in this case a fish twice as long weighs 8
times as much. This means that
2k = 8.
We note that 23 = 8, and so the power is k = 3. (Alternatively, one might solve the
equation 2k = 8 using the crossing graphs method.)
5. Length of skid marks versus speed:
(a) We are on dry concrete, so the value of the friction coefficient is h = 0.85. Thus the
function we are dealing with is
L=
1
S2
S2 =
.
30 × 0.85
25.5
If the speed is S = 55 miles per hour, then the skid marks will have length
L=
552
= 118.63 feet.
25.5
402
Solution Guide for Chapter 5
(b) Again we are on dry concrete pavement, so we use the same function as in Part
(a). We need to know what speed will result in skid marks which are 230 feet long.
That is, we need to solve the equation
S2
= 230.
25.5
We will solve this using the crossing graphs method. We use a horizontal span of
0 to 80 miles per hour, and the table of values for L below leads us to choose a
vertical span of 0 to 300. We have graphed L and the target length of 230 (thick
line). We see that the intersection occurs at a speed of S = 76.58 miles per hour.
The driver would appear to be in danger of receiving a citation.
(c) Since we now do not know the value of h, we must use the homogeneity property
to solve this problem. We want to change the distance by a factor of 0.5. Since
the power for the function is 2, changing the speed by a factor of f changes the
distance L by a factor of f 2 . We want to cut the speed in half, so we want to choose
f so that
f 2 = 0.5.
This may be solved by the crossing graphs method or simply by noting that the
solution is
f=
√
0.5 = 0.7071.
Hence we need to change our speed from 60 miles per hour to
60 × 0.7071 = 42.43 miles per hour.
The new speed should be between 42 and 43 miles per hour.
6. Binary stars:
(a) The mass of the Alpha Centauri pair is given by M = s3 a−3 p−2 where s = 17.6,
a = 0.76, and p = 80.1, so M = 17.63 0.76−3 80.1−2 . This equals 1.94, so the mass of
the Alpha Centauri pair is 1.94 solar masses.
SECTION 5.1
Power Functions
403
(b) If the parallax and period remain the same, then the mass is a power function of s
with power 3. If the separation angle s is doubled, then the mass is changed by a
factor of 23 = 8.
(c) If the separation and period remain the same, then the mass is a power function
of a with power −3. If the parallax angle is doubled, then the mass is changed by
1
a factor of 2−3 = .
8
(d) If the parallax angle and separation remain the same, then the mass is a power
function of p with power −2. If the period is doubled, then the mass is changed
1
by a factor of 2−2 = .
4
7. Life expectancy of stars:
(a) Since E is a power function with negative exponent, as M increases, E decreases,
so a more massive star has a shorter life expectancy than a less massive star.
(b) Since Spica has a mass of 7.3 solar masses, M = 7.3, and therefore E = M −2.5 =
7.3−2.5 , which equals 0.0069, or about 0.007, of a solar lifetime. Since Earth’s life
expectancy is about 10 billion years, Spica’s life expectancy is about 10 × 0.0069 =
0.069 billion years or about 70 million years.
(c) In functional notation, the life expectancy of a main-sequence star with mass M =
0.5 is E(0.5). Its value is E = 0.5−2.5 or about 5.66 solar lifetimes. This is about
56.6 billion years.
(d) Because Vega has a life expectancy of 6.36 billion years and 1 solar lifetime is 10
billion years, Vega has a life expectancy of 0.636 of a solar lifetime. Since E =
0.636, we have the equation M −2.5 = 0.636, which we can solve for M . This yields
M = 1.20, so the mass of Vega is 1.20 solar masses.
(e) If one main-sequence star is twice as massive as another, then M is changed by a
factor of 2, and therefore E is changed by a factor of 2−2.5 = 0.18. Thus the life
expectancy of the larger star is 0.18 times that of the smaller.
8. Height of tsunami waves:
(a) We measure H, D, d, and h in feet. If a height of a tsunami wave is 3 feet at an ocean
depth of 15,000 feet, then H = 3 and D = 15,000. In water 25 feet deep, d = 25,
D
15,000
and so the new height is given by h = HR0.25 where H = 3 and R =
=
.
d
25
0.25
15,000
Thus the new height is h = 3
= 14.85 feet.
25
(b) If R is doubled, then the new height of a tsunami wave is 20.25 = 1.19 times the
height before R is doubled.
404
Solution Guide for Chapter 5
9. Tsunami waves and breakwaters: The wave height h beyond the breakwater is a power
function of the width ratio R with power 0.5.
(a) We measure H, W , w, and h in feet. Since the wave has height 8 feet in a channel
of width 5000 before the breakwater, H = 8 and W = 5000. Since the breakwater
w
3000
has width 3000 feet, w = 3000, and so R =
=
= 0.6. The height of the
W
5000
wave beyond the breakwater is h = HR0.5 = 8 × 0.60.5 = 6.20 feet.
1
(b) If the channel width is cut in half, then R is multiplied by . Thus the new height
2
0.5
1
is
= 0.71 times the original height.
2
10. Bores: Bore velocity V is a power function with power 0.5, so if the height h changes
by a factor of t, then V changes by a factor of t0.5 .
1
(a) If the height is reduced by half, then t = , so the height changes by a factor of
2
0.5
1
= 0.71.
2
1
(b) If the velocity is reduced by half, then = t0.5 . Solving for t yields t = 0.25, so the
2
height is reduced to one-quarter of the initial height.
(c) If one bore has height three times that of another, then t = 3, so the velocity of the
first is 30.5 = 1.73 times that of the other.
11. Terminal velocity: Terminal velocity T is a power function with power 0.5, so if the
length L changes by a factor of t, then T changes by a factor of t0.5 .
(a) The man is 36 times as long as the mouse, so t = 36. The terminal velocity of the
man is therefore 360.5 = 6 times that of the mouse.
(b) If the terminal velocity of the man is 120 miles per hour, then, from Part (a), that
120
of the mouse is
= 20 miles per hour.
6
7
(c) Since a squirrel is 7 inches long, it is only
times the length of the 6-foot man, so
72
0.5
7
the terminal velocities differ by a factor of
. Thus the terminal velocity of
72
0.5
7
a squirrel is 120 ×
= 37.42 miles per hour.
72
12. Dropping rocks on other planets:
(a) The distance D is a power function of time t with power 2. If t goes from 2 to
6 seconds, it increases by a factor of 3. Then, by the homogeneity property, D
increases by a factor of 32 = 9. Thus the distance fallen increases by a factor of 9
from 2 seconds to 6 seconds into the drop.
SECTION 5.1
Power Functions
405
(b) Since c = 6.4 for Mars, the formula for D is D = 6.4t2 . We want to find t so that
D = 70, so we want to solve the equation 6.4t2 = 70. We solve this using the
crossing graphs method. The table of values below leads us to choose a horizontal
span of 0 to 4 and a vertical span of 0 to 105. Below we have graphed D and the
target distance of 70 (thick line). We see that the intersection occurs at a time of
t = 3.31 seconds, and this is how long it takes for the rock to strike the ground.
(c) The formula for D is D = ct2 , and we know that if D = 70 then t = 2.2. Thus
70
70 = c × 2.22 . Dividing both sides by 2.22 , we find that c =
= 14.46. Thus
2.22
c = 14.46 for Venus.
13. Newton’s law of gravitation:
(a) If the distance d is changed by a factor of 0.5, then the gravitational force will be
changed by a factor of
0.5Function power = 0.5−2 = 4.
Hence if the distance between the asteroids is halved, the force of gravity will be
4 times as strong.
If the distance is reduced to one quarter of its original value, it is changed by a
factor of 0.25. That gives a change in force by a factor of
0.25Function power = 0.25−2 = 16.
Thus if the distance is reduced to one quarter of its original value, the force of
gravity will be 16 times stronger.
(b) We want to know the value of c that will make
2,000,000 = c × 300−2 .
This is a linear equation in c which we can solve by dividing each side by 300−2 :
c=
2,000,000
= 1.8 × 1011 .
300−2
406
Solution Guide for Chapter 5
Hence we know that, for these asteroids, the gravitational force is given by
F = 1.8 × 1011 × d−2 .
When d = 800, we get that the force is
F = 1.8 × 1011 × 800−2 = 281,250 newtons.
This could also be solved used the homogeneity property of power functions. The
first step is to note that increasing the distance from 300 to 800 kilometers is an
increase by a factor of 800/300 = 2.67.
(c) We use a horizontal span of 0 to 1000 as suggested. The table of values below for
F = 1.8 × 1011 × d−2 leads us to choose a vertical span of 0 to 20,000,000 = 2 × 107 .
In the graph below, the horizontal axis is distance between the centers, and the
vertical axis is force of gravity.
The graph shows that when the asteroids are relatively close, gravitational force is
very strong. (It is in fact so strong that when bodies of planetary size get too close,
the force of gravity will tear one or both of them to pieces.) As they move farther
apart, the gravitational force decreases toward zero.
14. Geostationary orbits:
1
(a) We want the period to be
th that of the moon. If the period is changed by a
28
1
factor of , then the distance is changed by a factor of
28
1
28
Function power
=
1
28
2/3
.
Thus the distance from the center of the Earth to the satellite should be
2/3
1
239,000 ×
= 25, 919.46 miles,
28
or about 25,919 miles.
SECTION 5.1
Power Functions
407
(b) Since the radius of the Earth is 3963 miles, the satellite should be about 25,919 −
3963 = 21,956 miles above the surface of the Earth.
15. Giant ants and spiders:
(a) Weight is a power function of length, with power 3. Thus, if the length of the ant
is increased by a factor of 500, then its weight is increased by a factor of 5003 =
125,000,000.
(b) Cross-sectional area of limbs is a power function of length, with power 2. Thus, if
the length of the ant is increased by a factor of 500, then the cross-sectional area of
its legs is increased by a factor of 5002 = 250,000.
(c) Using the hint and the results of Parts (a) and (b), we see that the pressure on a leg
125,000,000
increases by a factor of
= 500. The poor ant will be flattened.
250,000
16. Kepler’s third law: We have that P is a power function of D, with power 1.5.
(a) If D is increased by a factor of 30, then P is increased by a factor of 301.5 = 164.32.
Thus, since the period for Earth is 1 year, the time required for Neptune to complete an orbit is about 164 years.
(b) We first find the factor t by which we must multiply the distance of the Earth to
get the distance of Mercury. To do this, note from the hint that the corresponding
88
factor for the period is
. Then the homogeneity property for power functions
365
88
tells us that t1.5 =
. We solve this equation using the crossing graphs method.
365
88
We enter the function t1.5 and the constant function
. We expect the factor t to
365
be less than 1, and the table of values below leads us to choose a horizontal span
of 0 to 0.5 and a vertical span of 0 to 0.4. From the graph below we see that the
intersection occurs at t = 0.3874. (This equation can also be solved using the laws
1/1.5
88
of exponents to get t =
= 0.3874.) Now that we have t, we multiply it
365
by the distance of the Earth to get the distance of Mercury as 93 × 0.3874 = 36.03,
or about 36, million miles.
408
Solution Guide for Chapter 5
17. Time to failure: If the building collapses in 30 seconds at one intensity, put t1 = 30 and
denote that intensity as I1 . Tripling that intensity means putting I2 = 3I1 , so
t1
=
t2
Since t1 = 30, we have
I2
I1
2
=
3I1
I1
2
= 9.
30
= 9. Solving for t2 yields 3.33 seconds for the building to
t2
collapse.
5.2 MODELING DATA WITH POWER FUNCTIONS
E-1. Finding power functions: We wish to find power functions passing through certain
points, so the functions will be of the form y = cxk and we are to determine c and k.
(a) For the power function to pass through both (2, 7) and (6, 56), two equations must
hold:
7
= c × 2k
56
= c × 6k .
Dividing the second equation by the first gives
56
c × 6k
=
=
7
c × 2k
k
6
2
or
8 = 3k .
We can solve this using logarithms:
ln 8
=
ln 3k
ln 8 = k ln 3
ln 8
= k
ln 3
1.89 = k.
We put this value back into the first equation to solve for c:
7 = c × 21.89
7
= c
21.89
1.89 = c.
The power function is therefore y = 1.89x1.89 . It is purely coincidental that the c
and k values are the same.
SECTION 5.2
Modeling Data with Power Functions
409
(b) For the power function to pass through both (a, 12) and (10a, 1200), two equations
must hold:
12
1200
= c × ak
= c × (10a)k .
Dividing the second equation by the first gives
c × (10a)k
1200
=
=
12
c × ak
10a
a
k
or
100 = 10k .
We could solve using logarithms, but since we know that 102 = 100, clearly k = 2.
To do this using common logarithms, for example, we compute as follows:
log 100 =
log 10k
log 100 = k log 10
log 100
= k
log 10
2 = k.
We put this value back into the first equation to solve for c. (Note that c must depend on a, since different values of a will require use of different power functions.)
We get:
12 = c × a2
12
= c.
a2
12
The power function is therefore y =
x2 .
a2
E-2. Finding more power functions: We wish to find power functions passing through certain points, so the functions will be of the form y = cxk and we are to determine c and
k.
(a) For the power function to pass through both (3, 7) and (4, 9), two equations must
hold:
7
= c × 3k
9
= c × 4k .
Dividing the second equation by the first gives
k
9
c × 4k
4
=
=
.
k
7
c×3
3
410
Solution Guide for Chapter 5
We can solve this using logarithms:
k
9
4
= ln
7
3
9
4
ln
= k ln
7
3
ln(9/7)
= k
ln(4/3)
0.87 = k.
ln
We put this value back into the first equation to solve for c:
= c × 30.87
7
7
= c
30.87
2.69 = c.
The power function is therefore y = 2.69x0.87 . As usual, you can check your answer
using a table of values.
(b) For the power function to pass through both (2, 5) and (6, 1), two equations must
hold:
5
= c × 2k
1
= c × 6k .
Dividing the second equation by the first gives
1
c × 6k
=
=
5
c × 2k
k
6
2
or
0.2 = 3k .
We can solve this using logarithms:
ln 0.2
= ln 3k
ln 0.2 = k ln 3
ln 0.2
= k
ln 3
−1.46 = k.
We put this value back into the first equation to solve for c:
5
5
2−1.46
13.76
= c × 2−1.46
= c
= c.
SECTION 5.2
Modeling Data with Power Functions
411
The power function is therefore y = 13.76x−1.46 . As usual, you can check your
answer using a table of values.
(c) For the power function to pass through both (3, 2) and (5, 3), two equations must
hold:
2
= c × 3k
3
= c × 5k .
Dividing the second equation by the first gives
3
c × 5k
=
=
2
c × 3k
k
5
.
3
We can solve this using logarithms:
k
3
5
= ln
2
3
5
3
= k ln
ln
2
3
ln(3/2)
= k
ln(5/3)
0.79 = k.
ln
We put this value back into the first equation to solve for c:
2
= c × 30.79
2
= c
30.79
0.84 = c.
The power function is therefore y = 0.84x0.79 . As usual, you can check your answer
using a table of values.
E-3. Linear to power functions: We wish to show that if ln y = ln c + k ln x, then y = cxk . To
do this, we exponentiate both sides of the equation and use the properties of exponents:
ln y
=
ln c + k ln x
eln y
= eln c+k ln x
y
= eln c × ek ln x
y
= c(eln x )k
y
= cxk .
412
Solution Guide for Chapter 5
E-4. Finding a power function through general points: Even though the points involved in
this exercise are not given by explicit numbers, we can proceed in the same way we did
earlier with numbers. For the power function to pass through both (x1 , y1 ) and (x2 , y2 ),
two equations must hold:
y1
= c × (x1 )k
y2
= c × (x2 )k .
Dividing the second equation by the first gives
y2
c × (x2 )k
=
=
y1
c × (x1 )k
x2
x1
k
.
We can solve this for k using logarithms:
y2
y1
y2
ln
y1
ln(y2 /y1 )
ln(x2 /x1 )
x2
x1
x2
= k ln
x1
ln
=
k
ln
= k.
We put this value back into the first equation to solve for c:
ln(y2 /y1 )
y1
y1
ln(y2 /y1 )
ln(x2 /x1 )
= c × x1ln(x2 /x1 )
= c.
x1
The power function is therefore


y1
y=
ln(y2 /y1 )
ln(x2 /x1 )
ln(y2 /y1 )
 x ln(x2 /x1 ) .
x1
Amazing, isn’t it! You can check your formula by using it to find the answers to one of
the previous exercises.
S-1. Logarithmic conversion of both x and y: If we start from power data and logarithmically convert both rows, then the resulting data is linear.
S-2. Formula conversion: Since ln f = 3 ln x + 2, ln f is a linear function of ln x. Therefore f
is a power function of x, so f = cxk . The exponent k is the same as the slope of the ln f
function, so k = 3. The number c is calculated as c = e2 = 7.39, so the formula for f is
f = 7.39x3 .
S-3. Getting the power: The slope of the regression line for the logarithm of the data is the
same as the power, so that power is 3.
SECTION 5.2
Modeling Data with Power Functions
413
S-4. Getting c: The value of c is eb , where b is the initial value of the regression line for the
logarithm of the data. Here b = 4 and so c = e4 = 54.60.
S-5. Modeling power data: To find the power model, we convert the data into linear data
using the logarithm. To do this, we put the original data in the 3rd and 4th columns,
the logarithms in the 1st and 2nd columns (see the figure on the left below), and then
calculate the linear relation using linear regression. This is shown below on the right.
The regression calculation shows that ln f = 1.30 ln x + 1.28. Thus the power model
uses k = 1.30 and c = e1.28 = 3.60, and so f = 3.60x1.30 . This fits the original data,
as shown in the figure below, where we have graphed the original data from the 3rd
and 4th columns together with the power model. This can also be seen from a table of
values.
414
Solution Guide for Chapter 5
S-6. Modeling almost power data: To find the power model, we convert the data into linear
data using the logarithm. To do this, we put the original data in the 3rd and 4th columns,
the logarithms in the 1st and 2nd columns (see the figure on the left below), and then
calculate the linear relation using linear regression. This is shown below on the right.
The regression calculation shows that ln f = −2.74 ln x + 2.16. Thus the power model
uses k = −2.74 and c = e2.16 = 8.67, and so f = 8.67x−2.74 . This more or less fits the
original data, as shown in the figure below, where we have graphed the original data
from the 3rd and 4th columns together with the power model.
S-7. Modeling almost power data: To find the power model, we convert the data into linear
data using the logarithm. To do this, we put the original data in the 3rd and 4th columns,
the logarithms in the 1st and 2nd columns (see the figure on the left below), and then
calculate the linear relation using linear regression. This is shown below on the right.
SECTION 5.2
Modeling Data with Power Functions
415
The regression calculation shows that ln f = 4.54 ln x + 0.802. Thus the power model
uses k = 4.54 and c = e0.802 = 2.23, and so f = 2.23x4.54 . This closely fits the original
data, as shown in the figure below, where we have graphed the original data from the
3rd and 4th columns together with the power model.
S-8. Modeling almost power data: To find the power model, we convert the data into linear
data using the logarithm. To do this, we put the original data in the 3rd and 4th columns,
the logarithms in the 1st and 2nd columns (see the figure on the left below), and then
calculate the linear relation using linear regression. This is shown below on the right.
The regression calculation shows that ln f = 1.73 ln x + 1.62. Thus the power model uses
k = 1.73 and c = e1.62 = 5.05, and so f = 5.05x1.73 . This more or less fits the original
data, as shown in the figure below, where we have graphed the original data from the
3rd and 4th columns together with the power model.
416
Solution Guide for Chapter 5
S-9. Modeling almost power data: To find the power model, we convert the data into linear
data using the logarithm. To do this, we put the original data in the 3rd and 4th columns,
the logarithms in the 1st and 2nd columns (see the figure on the left below), and then
calculate the linear relation using linear regression. This is shown below on the right.
The regression calculation shows that ln f = −0.90 ln x + 0.719. Thus the power model
uses k = −0.90 and c = e0.719 = 2.05, and so f = 2.05x−0.90 . This closely fits the original
data, as shown in the figure below, where we have graphed the original data from the
3rd and 4th columns together with the power model.
S-10. The common logarithm: To find the power model using the common logarithm, we
convert the data into linear data using the common logarithm. To do this, we put the
original data in the 3rd and 4th columns, the common logarithms in the 1st and 2nd
columns (see the figure on the left below), and then calculate the linear relation using
linear regression. This is shown below on the right.
SECTION 5.2
Modeling Data with Power Functions
417
The regression calculation shows that log f = −2.74 log x + 0.94. Thus the power model
uses k = −2.74 and c = 100.94 = 8.71, and so f = 8.71x−2.74 . This is almost the same
as the calculation using the natural logarithm and differs only due to rounding of the
0.94 term. This model more or less fits the original data as well as the model found in
Exercise S-6, as shown in the figure below, where we have graphed the original data
from the 3rd and 4th columns together with the power model.
S-11. Modeling almost power data: To find the power model, we convert the data into linear
data using the logarithm. To do this, we put the original data in the 3rd and 4th columns,
the logarithms in the 1st and 2nd columns (see the figure on the left below), and then
calculate the linear relation using linear regression. This is shown below on the right.
The regression calculation shows that ln f = −1.59 ln x + 0.83. Thus the power model
uses k = −1.59 and c = e0.83 = 2.29, and so f = 2.29x−1.59 .
418
Solution Guide for Chapter 5
S-12. Modeling almost power data: To find the power model, we convert the data into linear
data using the logarithm. To do this, we put the original data in the 3rd and 4th columns,
the logarithms in the 1st and 2nd columns (see the figure on the left below), and then
calculate the linear relation using linear regression. This is shown below on the right.
The regression calculation shows that ln f = 2.00 ln x − 0.69. Thus the power model uses
k = 2 and c = e−0.69 = 0.50, and so f = 0.5x2 .
1. Hydroplaning:
(a) From the figure below, the equation of the regression line is ln V = 0.50 ln p + 2.34.
(b) From Part (a), the power is 0.5, and the coefficient c is e2.34 = 10.4. Thus the
formula is V = 10.4p0.5 .
(c) We want to compare V (35) with V (60). From the formula in Part (b) we get
V (35) = 10.4 × 350.5 = 61.5, while V (60) = 10.4 × 600.5 = 80.6. Thus the critical speed for hydroplaning is 61.5 mph for the car and 80.6 mph for the bus. Since
both are traveling at 65 mph, the car is in danger of hydroplaning, while the bus
is not. Thus, if both apply their brakes, the car might hydroplane and hit the bus
from the rear.
SECTION 5.2
Modeling Data with Power Functions
419
2. Urban travel times:
(a) To find the power model, we convert the data into linear data using the logarithm.
To do this, we put the original data in the 3rd and 4th columns, the logarithms in
the 1st and 2nd columns (see the figure on the left below), and then calculate the
linear relation using linear regression. This is shown below on the right.
The regression calculation shows that ln T = 0.18 ln N + 1.22. Thus the power
model uses k = 0.18 and c = e1.22 = 3.39, and so T = 3.39N 0.18 is the power
model.
(b) Pittsburgh has a population of N = 1804 thousand and a travel time of T = 12.6
minutes. On the other hand, the power model predicts for N = 1804 a travel time
of 3.39 × 18040.18 = 13.07 minutes. Thus the actual driving time in Pittsburgh is
slightly less than would be expected.
(c) To reduce your driving time by 25% is the same as multiplying it by a factor of 0.75.
Using the homogeneity property with power 0.18, we know that if N changes by a
factor of t, then T must change by a factor of t0.18 . We want T to change by a factor
of 0.75, so we want to solve 0.75 = t0.18 . Solving, for example by computing that
t = (0.75)1/0.18 = 0.20, we see that N needs to change by a factor of 0.20. Thus to
reduce driving time by 25%, the population of the smaller city should be 20% that
of the original city, that is, the population of the new city should be 80% smaller.
420
Solution Guide for Chapter 5
3. Mass-luminosity relation:
(a) To find the power model, we convert the data into linear data using the logarithm.
To do this, we put the original data in the 3rd and 4th columns, the logarithms in
the 1st and 2nd columns (see the figure on the left below), and then calculate the
linear relation using linear regression. This is shown below on the right.
The regression calculation shows that ln L = 3.50 ln M + 0.02. Thus the power
model uses k = 3.5 and c = e0.02 = 1 to one decimal place, and so L = M 3.5 is the
power model.
(b) Since Kruger 60 has a mass of 0.11 solar mass, its relative luminosity is L(0.11) in
functional notation. The value of L(0.11) is 0.113.5 = 4.41 × 10−4 or 0.000441, or
about 0.0004.
(c) Since Wolf 359 has a relative luminosity of about 0.0001, for that star L = 0.0001
and so M 3.5 = 0.0001. Solving for M , for example by M = 0.00011/3.5 , shows that
M = 0.072, or about 0.07, of a solar mass.
(d) If one star is 3 times as massive an another, then M changes by a factor of 3 and so,
by the homogeneity property with power 3.5, L changes by a factor of 33.5 = 46.77.
Thus the more massive star has a relative luminosity 46.77, or about 47, times that
of the less massive star.
SECTION 5.2
Modeling Data with Power Functions
421
4. Growth rate versus weight:
(a) The plot of ln r against ln W is shown below. The plot is reasonably close to linear, except for the third point, which represents the Roe deer entry. Hence it is
reasonable to model r as a power function of W .
(b) As the figure below at the left indicates, the regression line for the data plotted in
Part (a) is ln r = −0.37 ln W + 0.79. So the power is k = −0.37, and c = e0.79 = 2.2.
Thus the formula is r = 2.2W −0.37 . The graph is shown below at the right. The
horizontal span is −1 to 10, and the vertical span is −1 to 8.
422
Solution Guide for Chapter 5
5. Speed in flight versus length:
(a) Table 5.3 indicates that larger animals fly faster, since as L increases so does F .
(b) As the figure below at the left indicates, the regression line for ln F against ln L is
ln F = 0.34 ln L + 3.026. This gives k = 0.34 and c = e3.026 = 20.61, so the desired
formula is F = 20.61L0.34 .
(c) The graph of F = 20.61L0.34 is shown below at the right. The horizontal span is 0
to 300, and the vertical span is 0 to 150.
(d) The graph is concave down and increasing. This means that the flying speed F
increases as the length L increases, but at a decreasing rate.
(e) If L is increased by a factor of 10, then F is increased by a factor of 100.34 = 2.19.
Thus the longer bird should fly 2.19 times as fast.
6. Speed swimming versus length:
(a) As the figure below at the left indicates, the formula for the regression line of ln S
against ln L is ln S = 1.0 ln L − 0.60.
(b) Since the slope of the regression line is 1.0 then for the power k we get k = 1.0.
Since the vertical intercept of the regression line is −0.60 then c = e−0.60 = 0.55.
Thus the power function is S = 0.55L1 = 0.55L. This is a linear function.
(c) The two graphs are shown below at the right using the same window as in the
previous exercise.
SECTION 5.2
Modeling Data with Power Functions
423
(d) Using the graphs to evaluate the functions at L = 12 inches, we find that an animal
that is one foot long can fly at 48.2 feet per second but can swim at only 6.6 feet
per second. So an animal one foot long can fly much faster than it can swim.
(e) If an animal is 20 feet long, then L = 20 × 12 = 240 inches. Using the graphs to
evaluate the functions at L = 240, we find that F = 133.4 and S = 132 (both in
feet per second). So, F and S are about equal for an animal 20 feet long, and hence
flying is not a significant improvement over swimming for such an animal.
(f) The formula we found in Part (b) predicts that an 85 foot (or 85 × 12 = 1020 inch)
whale will swim at a speed of 0.55 × 1020 = 561 feet per second, which is far
greater than the actual speed. No, the trend does not continue indefinitely.
7. Metabolism:
(a) As the figure below indicates, the formula for the regression line of ln B against
ln W is ln B = 0.75 ln W + 3.69. Since the slope of the regression line is 0.75, then
k = 0.75. Since the vertical intercept of the regression line is 3.69, then c = e3.69 =
40.04. So the power function is B = 40.04W 0.75 .
(b)
i. By Part (a), the basic energy needed for survival (B in our notation) is proportional not to the weight W but to W 0.75 . Thus to see how an animal is eating
relative to its energy requirements, we should divide its intake by W 0.75 , not
by W .
ii. The metabolic weight of a 2.76-pound animal is 2.760.75 = 2.14, and the metabolic
weight of a 126.8-pound animal is 126.80.75 = 37.79.
iii. We use the metabolic weights found in Part ii above. For the rabbit the ratio
0.18
2.8
is
= 0.08, and for the sheep it is
= 0.07. So the daily consumption
2.14
37.79
levels of the two animals are about the same on this basis.
424
Solution Guide for Chapter 5
8. Metabolism and surface area:
(a) As the figure below indicates, the formula for the regression line of ln B against
ln W is ln B = 0.73 ln W + 3.30. This gives k = 0.73 and c = e3.30 = 27.11, so
B = 27.11W 0.73 for these marsupials.
(b) The power is nearly the same as in the formula from the previous exercise, but the
coefficient c is lower for the marsupials. The marsupials have a lower metabolic
rate for a given weight than warm-blooded animals in general.
(c) A plot of ln A against ln W is in the figure below.
(d) As the figure below at the left indicates, the regression line is ln A = 0.64 ln W +
4.66. The graph is below at the right.
(e) We evaluate ln A when ln W = ln 0.26 = −1.35 using the plot in Part (d), as shown
in the figure below. The result is ln A = 3.80. Thus the formula predicts a surface
area of A = e3.80 , or about 45 square inches. However, the actual area is A = 95
SECTION 5.2
Modeling Data with Power Functions
425
square inches. Thus, the sugar glider has a much greater surface area than would
be predicted given its weight. This is because the membranes used for gliding add
significantly to the surface area.
(f) From the formula for the regression line of ln A, we have k = 0.64 and c = e4.66 =
105.64, so A = 105.64W 0.64 .
2
(g) Note that = 0.67 to two decimal places, and this is close to 0.64.
3
(h) As the figure below indicates, the regression line of ln B against ln A is ln B =
1.15 ln A − 2.03. Thus the slope is 1.15, so B is proportional to A1.15 . Over a range
of areas, this is close to A = A1 . So the basal metabolic rate is close to being
proportional to surface area.
9. Proportions of trees:
(a) The plot of ln h against ln d is in the left-hand figure below.
(b) As shown in the right-hand figure below, the regression line is ln h = 0.66 ln d +
3.58. The graph of this line is shown in the left-hand figure below.
426
Solution Guide for Chapter 5
(c) The plains cottonwood is taller for its diameter since the point in the plot corresponding to the cottonwood lies above the regression line, while that for the
willow lies below the line.
(d) From the regression line, k = 0.66 and c = e3.58 = 35.87. Thus h = 35.87d0.66 .
(e)
i. The powers of d for the trees and the critical height function are about the
2
same, since = 0.67 to two decimal places. However, the coefficient is 35.87
3
for the trees, while the critical height function has a coefficient of 140, which
is about four times as large.
ii. No tree from the table is taller than its critical buckling height. This can be
determined by looking at a table of values for 140d2/3 .
10. Weight versus length:
(a) The figure below indicates that the regression line for ln W against ln L is given by
ln W = 3.0 ln L − 4.88 if we round the slope to one decimal place.
From the regression line, k = 3.0 and c = e−4.88 = 0.0076, so W = 0.0076L3 .
(b) Here W (50) is the weight, in grams, of a North Sea plaice of length L = 50
centimeters. Using the formula from Part (a), we see that the value of W (50) is
0.0076 × 503 = 950 grams. (Looking at the table shows that this is an underestimate.)
(c) We use the homogeneity property of power functions. The power here is 3, so if
one plaice were 2 times as long as another then it should be 23 = 8 times as heavy.
SECTION 5.2
Modeling Data with Power Functions
427
11. Self-thinning:
(a) As time went by, the density decreased, which means that there were fewer plants
in the plot.
(b) As the figure below indicates, the regression line for ln w against ln p is given by
ln w = −1.48 ln p + 8.47. From the regression line we get that k = −1.48 and
c = e8.47 = 4769.52, so w = 4769.52p−1.48 .
(c) We use the homogeneity property of power functions. The power here is −1.48,
1
so if the density p changes by a factor of then the weight w changes by a factor
2
−1.48
1
of
= 2.79.
2
(d) The total yield is y = w × p = 4769.52p−1.48 × p = 4769.52p−0.48 . The power is
negative, so y decreases as p increases. On the other hand, as time increases, p
decreases, as noted in Part (a). As p decreases, y increases; so, as time increases,
the yield also increases.
12. Species-area relation:
(a) As the figure at the left below indicates, the regression line is ln S = 0.31 ln A+1.08,
so k = 0.31 and c = e1.08 = 2.94. The formula is S = 2.94A0.31 .
(b) Based on the given table of data, we use a horizontal span of 0 to 45,000 and a
vertical span of 0 to 100. The graph is shown below at the right. It is increasing
and concave down. This means that, as the area increases, the number of species
increases, but at a decreasing rate.
428
Solution Guide for Chapter 5
(c) If A increases by a factor of 10, then S increases by a factor of 100.31 = 2.04, or
about 2. Thus if one island is 10 times larger than another, then it will have about
2 times as many species.
(d) Suppose the power is k. By the homogeneity property, when the area A increases
by a factor of 10, the number of species S increases by a factor of 10k . Thus if one
island is 10 times larger than another, then it will have 10k times as many species.
13. Cost of transport:
(a) In general, as the weight increases, the cost of transport decreases. The only exception is the white rat.
(b) The plot of ln C against ln W is in the left-hand figure below.
(c) As shown in the right-hand figure below, the regression line is ln C = −0.40 ln W +
2.15.
The graph of this line is shown below.
SECTION 5.2
Modeling Data with Power Functions
429
(d) We evaluate ln C when ln W = ln (20,790) = 9.94 using the plot in Part (c), as
shown in the figure below. The result is ln C = −1.83. Thus the formula predicts
that the cost of transport will be C = e−1.83 = 0.16. However, the actual cost of
transport is C = 0.43. Thus, the penguin has a higher cost of transport than would
be predicted given its weight. This does confirm the stereotype of penguins as
awkward waddlers.
(e) From the regression line, k = −0.40 and c = e2.15 = 8.58, so the formula is C =
8.58W −0.40 .
14. Parallax angle:
(a) Placing the original data in the 3rd and 4th columns and ln p and ln d in the 1st and
2nd columns, respectively, we graph the logarithm of the data in the following
figure on the left.
Since these data line up almost perfectly, it is reasonable to model the original data
with a power function.
(b) Calculating the regression line for ln d against ln p, as shown in the figure above on
the right, yields ln d = −1.00 ln p + 1.18. Thus k = −1 and c = e1.18 = 3.25. Thus
3.25
the power function model is d = 3.25p−1 , which can also be written as d =
.
p
(c) If one star has a parallax angle twice that of another, then, by the homogeneity
property with power −1, the one star’s distance is 2−1 = 0.5, that is, one-half, the
distance of the other star.
430
Solution Guide for Chapter 5
(d) Since Mergez has a parallax angle of 0.052, so p = 0.052, the distance of Mergez
from the sun is expressed as d(0.052) in functional notation, and it has a value of
d = 3.25 × 0.052−1 = 62.5 light-years.
(e) If Sabik is 69 light-years from the sun, then d = 69. Thus 69 = 3.25p−1 , so p−1 =
6.9
= 21.23. Thus p = 0.047 second of arc.
3.25
15. Exponential growth rate and generation time:
(a) The plot of ln r against ln T is shown below. The plot is reasonably close to linear.
Hence it is reasonable to model r as a power function of T .
(b) As the figure below indicates, the regression line for the data plotted in Part (a) is
ln r = −1.0 ln T − 0.17 if we round the slope to one decimal place.
(c) From Part (b) the power is k = −1.0, and c = e−0.17 = 0.8, rounding to one
decimal place. Thus the formula is r = 0.8T −1 . The graph is shown below using a
horizontal span from 0 to 10 and a vertical span from 0 to 1.
SECTION 5.2
Modeling Data with Power Functions
431
(d) From Example 5.6, the generation time in years of an organism is modeled by
1.1L0.8 , if L is the length in feet. In this exercise we measure generation time T in
days, so we multiply the model by 365:
T = 365 × 1.1L0.8 = 401.5L0.8 .
This relationship holds over a wide range of organisms. From Part (c) we know
1
that, for our group of lower organisms, r = 0.8T −1 . Since T −1 = , this can be
T
0.8
written as r =
. We substitute into this formula for r the model for T in terms
T
of L to get
0.8
0.8
r=
=
.
T
401.5L0.8
0.8
is about 0.002, we get
Since
401.5
r=
0.002
.
L0.8
This is the desired formula for r in terms of L.
16. Using the common logarithm:
(a) We proceed as in Exercise 10, except that we take the common logarithm, not the
natural logarithm, of the data for length and weight. The figure below indicates
that the regression line for log W against log L is given by log W = 3.0 log L − 2.12
if we round the slope to one decimal place.
From the regression line, k = 3.0. To find the coefficient c we compute c =
10−2.12 = 0.0076. (We use 10x here since the regression line involves the common logarithm.) Thus W = 0.0076L3 . This is the same formula as in Exercise
10.
(b) The rest of the exercise is solved as shown in the solution to Exercise 10 above.
17. Laboratory experiment: Answers will vary. Please see the website for more information.
432
Solution Guide for Chapter 5
18. Laboratory experiment: Answers will vary. Please see the website for more information.
5.3 COMBINING AND DECOMPOSING FUNCTIONS
E-1. Further investigation of the case c = −0.75: In the plot we used a vertical span of −1 to
1. It appears that the two branches do come together.
E-2. Other values of c: Choosing c = −2.1 gives a sequence that increases rapidly. Other
choices of c with c < −2 confirm that the sequence increases without bound then.
E-3. Graphical analysis of another sequence: Choosing c = 0.5 gives a sequence that decreases to 0, as the graph on the left below illustrates. (We included the first 5 iterates
and used a vertical span of 0 to 0.1.) For the choice c = 3.1 the sequence seems to split
into two sequences, as the graph on the right below illustrates. (We included the first 50
iterates and used a vertical span of 0 to 1.)
SECTION 5.3
Combining and Decomposing Functions
433
For the choice c = 3.6 the sequence appears to be chaotic, as the graph below illustrates.
(We included the first 500 iterates and used a vertical span of 0 to 1.)
S-1. Formulas for composed functions: In each case we have a formula for w as a function
of s and a formula for s as a function of t. To find a formula for w as a function of t, we
simply replace s by its expression in terms of t.
(a) Since s = t − 3 and w = s2 + 1, w can also be written as w = s2 + 1 = (t − 3)2 + 1.
This expresses w as a function of t.
s
t2 + 2
s
, w can also be written as w =
= 2
,
s+1
s+1
(t + 2) + 1
which expresses w as a function of t. This expression can be simplified to w =
t2 + 2
.
t2 + 3
√
√
(c) Since s = et − 1 and w = 2s + 3, w can also be written as w = 2s + 3 =
p
2(et − 1) + 3, which expresses w as a function of t. This expression can be sim√
plified to w = 2et + 1 since 2(et − 1) + 3 = 2et − 2 + 3 = 2et + 1.
(b) Since s = t2 + 2 and w =
S-2. Formulas for composed functions:
2
(a) Here f (x) = 3x + 1 and g(x) = , so f (g(x)) = f
x
2
g(3x + 1) =
.
3x + 1
2
2
= 3 + 1 and g(f (x)) =
x
x
(b) Here f (x) = x2 + x and g(x) = x − 1, so f (g(x)) = f (x − 1) = (x − 1)2 + (x − 1)
and g(f (x)) = g(x2 + x) = (x2 + x) − 1.
(c) Here f (x) =
1
1
and g(x) = , so
x
x
1
1
f (g(x)) = f
= 1.
x
x
This equals x, so f (g(x)) = x. Since g = f , g(f (x)) = x also.
S-3. Limiting values: To find the limiting value of 7 + a × 0.6t , we note that a × 0.6t is a
function which represents exponential decay, so the limiting value of a × 0.6t is 0. Thus
the limiting value of 7 + a × 0.6t is 7.
434
Solution Guide for Chapter 5
S-4. Multiplying functions: We know that f is the product of weight and height, so f =
1
weight × height. Since weight is given by t + and height is given by t2 , f can be
t
1
2
expressed as f = t +
× t .
t
S-5. Adding functions: Since the function f is the sum of two temperatures, its formula may
be written as the sum of the formulas for the two temperatures. Thus f = (t2 + 3) +
t
.
2
t +1
S-6. Decomposing functions: To write the expression (x + 1)2 + 1 in terms of compositions
of f and g where f (x) = x2 and g(x) = x + 1, we look for these formulas inside the
expression:
(x + 1)2 + 1
=
(g(x))2 + 1
= f (g(x)) + 1.
This can also be written as g(f (g(x)).
S-7. Composing a function with itself: Since f (x) = x2 + 1, f (f (x)) = f (x2 + 1) = (x2 +
1)2 + 1. Also f (f (f (x))) = f (f (x2 + 1)), and using the previous form for f (f (x)) gives
f (f (f (x))) = ((x2 + 1)2 + 1)2 + 1.
S-8. Decomposing functions: The cost is C = 30 + 17n where n is the number of books
bought. The initial fee is the part of the formula which doesn’t depend on n, so it is the
initial value of C, which is C(0) = 30 dollars. The cost of each book is $17 since each
additional book adds $17 to C.
S-9. Decomposing functions: If the population is given by 128 × 1.07t , then the initial population is N (0) = 128. The population grows by a factor of 1.07 each year, which represents a growth of 7% each year.
S-10. Combining functions: Since f (x) = x2 − 1 and g(x) = 1 − x, f (g(x)) + g(f (x)) = f (1 −
x)+g(x2 −1) = ((1−x)2 −1)+(1−(x2 −1)). This can be simplified as (1−x)2 −(x2 −1) =
1 − 2x + x2 − x2 + 1 = 2 − 2x.
S-11. Combining functions: Because r(t) = 1 + 2t and A(r) = πr2 , we have A(r(t)) = π(1 +
2t)2 . Thus the formula is A = π(1 + 2t)2 .
S-12. Decomposing functions: The net profit is I = 0.25n − 5, and this is a linear function
with slope 0.25. This means that the net profit increases by $0.25 for each additional
glass of lemonade that the child sells.
SECTION 5.3
Combining and Decomposing Functions
435
1. A skydiver:
(a) We want to find the initial value of D. Here D = T − v and T = 176, so the initial
value of D is 176 − Initial value of v. Now v starts at 0, so the initial value of v is
0; thus the initial value of D is 176 − 0 = 176 feet per second.
(b) Let t be time in seconds since the jump. Now D is an exponential function of time,
so it can be written as D = P at . Here P is the initial value of D, so from Part
(a), P = 176. Thus D = 176at . On the other hand, v(2) = 54.75, r
so D(2) = 176 −
121.25
54.75 = 121.25, and thus 176a2 = 121.25. Solving for a yields a =
= 0.83,
176
so an exponential formula for D is D = 176 × 0.83t .
(c) To find a formula for v, note that we have D = T − v. Thus D + v = T and so
v = T − D. On the other hand, T = 176 and D = 176 × 0.83t , so a formula for v is
v = 176 − 176 × 0.83t .
(d) The velocity 4 seconds into the fall is expressed as v(4) in functional notation. Its
value is v(4) = 176 − 176 × 0.834 = 92.47 feet per second.
2. Present value: We want to find a formula for the present value P . Since the present
value P is the product of the future value F and the discount rate D, P = F × D. To
get the formula in terms of the interest rate and the life of the investment, we use the
1
formula D =
and substitute it for D to get
(1 + r)t
P =F ×
1
.
(1 + r)t
3. Immigration: We have a population N given by a formula
N=
v t
(a − 1),
r
where v is a positive number, r is a number which is negative, and a is a positive
number less than 1. Now at represents exponential decay (since a is less than 1), so at
v
v
has a limiting value of 0. Thus the limiting value of the population N is (0 − 1) = − .
r
r
4. Correcting respiration rate for temperature:
(a) If the temperature T is greater than 20 degrees Celsius, then T − 20 is positive. In
this case C = 1.07T −20 is 1.07 raised to a positive power, and so C is greater than
1. Since the correction factor is multiplied with the energy loss O, the effect of the
correction factor will be to increase O.
436
Solution Guide for Chapter 5
(b) If the temperature T is equal to 20 degrees Celsius, then C = 1.07T −20 = 1.0720−20 =
1.070 = 1, so the correction factor will have no effect.
(c) If the temperature T is less than 20 degrees Celsius, then T − 20 is negative. In
this case C = 1.07T −20 is 1.07 raised to a negative power, and so C is less than 1.
Since the correction factor is multiplied with the energy loss O, the effect of the
correction factor will be to decrease O.
5. Biomass of haddock:
(a) Now D is the difference between the maximum length, which is 21 inches, and the
length L, so D(t) = 21 − L(t) for all t. The initial value of L is 4 inches, so L(0) = 4,
and therefore the initial value of D is D(0) = 21 − L(0) = 21 − 4 = 17 inches.
(b) To find a formula for L, we start with D = 21 − L. This says that D + L = 21 and
therefore L = 21 − D. To obtain a formula for L in terms of t we need a formula
for D in terms of t.
We know from Part (a) that D is an exponential function with initial value 17,
so D = 17at for some a. Since a 6-year-old haddock is about 15.8 inches long,
when t = 6, then D = 21 − L = 21 − 15.8 = 5.2. Using the formula for D, we
5.2
= a6 ,
have 5.2 = 17a6 . Solving for a gives a = 0.82. (For example, since
17
1/6
5.2
a=
= 0.82). Thus D = 17 × 0.82t . Since L = 21 − D, then using the
17
formula for D above, we find that L = 21 − 17 × 0.82t .
(c) Weight W is written as function of L by the formula W = 0.000293L3 , and L is
written as a function of t by the formula in Part (b) above, so W can be written as
a function of t by function composition:
W
=
0.000293L3
W
=
0.000293(21 − 17 × 0.82t )3 .
(d) The total biomass B is the product of the number of fish N and the weight per fish
W . Each of N and W can be expressed as functions of the age t, so B can be also:
B
= N ×W
B
=
1000e−0.2t × 0.000293(21 − 17 × 0.82t )3 .
(e) The maximum value of the biomass B can be determined by graphing B as a function t using the formula from Part (d). We are to consider ages up to 10 years, so
the horizontal span should be from 0 to 10. Looking at a table of values (see the
SECTION 5.3
Combining and Decomposing Functions
437
figure on the left below), we see that a vertical span of 0 to 400 should nicely display the graph. The figure on the right below shows the graph with the maximum
calculated.
The maximum occurs at t = 5.89 with B = 350.31. So the biomass is at a maximum
when the cohort is 5.89 years old, or about 5.9 years old.
6. Traffic flow:
(a) Here R = k/km , where k is the mean traffic density, while km is the density at
which the flow rate is a maximum.
i. If R < 1, then k < km , so the traffic on the highway is flowing at a mean traffic
density that is less than the density at which the flow rate is a maximum (so
the traffic is relatively light, and the road can carry more vehicles).
ii. If R > 1, then k > km , so the traffic on the highway is flowing at a mean traffic
density that is greater than the density at the flow rate is a maximum (so the
traffic is heavy, restricting the flow of vehicles).
(b) We have a formula that expresses mean speed u as a function of relative density
R, namely
2
u = uf e−0.5R ,
and we have a formula that expresses relative density R as a function of mean
traffic density k, namely
R=
k
.
km
Function composition allows us to find a formula that directly expresses mean
speed u as a function of mean traffic density k:
2
u
= uf e−0.5R
u
= uf e−0.5(k/km ) .
2
(c) If km = 122 and uf = 75 then the formula from Part (b) becomes
2
u = 75e−0.5(k/122) .
438
Solution Guide for Chapter 5
To graph this for k from 0 to 250 vehicles per mile, we use a table of values (see
the left figure below) to choose a vertical span of 0 to 100. The right figure below
shows the graph of u against k.
The graph is concave down at first, then concave up, with a point of inflection
at about k = 122, which represents the point of steepest descent. Thus the mean
density at which the mean speed is dropping the fastest is the mean traffic density.
(d) Traffic is considered seriously congested when u drops to 35. We use the crossinggraphs method to solve the equation u = 35, that is
2
75e−0.5(k/122) = 35,
in the graph below, for which we used a horizontal span of 0 to 250 and a vertical
span of 0 to 75.
The graph shows that u drops to 35 when k = 150.62, that is, when the mean traffic
density reaches 150.62 vehicles per mile.
7. Waiting at a stop sign: In this exercise T = 5, so the formula for the average delay D,
in seconds, at a stop sign to enter a highway is given by
D=
e5q − 1 − 5q
,
q
where q is the flow rate, in cars per second, of traffic on the highway.
SECTION 5.3
Combining and Decomposing Functions
(a) If the flow rate is 500 cars per hour, then q =
439
500
= 0.14 car per second, and
60 × 60
so the delay is
D=
e5×0.14 − 1 − 5 × 0.14
= 2.24 seconds.
0.14
(b) The service rate s, in cars per second, is a function of D given by the formula
s = D−1 . To represent s as a function of the flow rate q, we substitute the formula
for D in terms of q into the formula for s in terms of D:
s = D−1
5q
−1
e − 1 − 5q
s =
q
q
.
s =
e5q − 1 − 5q
(c) If the service rate is 5 cars per minute, then s =
5
= 0.083 car per second. To
60
solve the equation
s =
0.083
=
q
e5q − 1 − 5q
q
e5q − 1 − 5q
we can use the crossing-graphs method as shown below with a horizontal span of
q from 0 to 1 cars per second and a vertical span of s from 0 to 0.15 car per second.
This shows that when s = 0.083, q = 0.42 car per second, or 25.2 cars per minute.
8. Average traffic spacing:
(a) Here P = 0.87t and this represents exponential decay; in particular the limiting
value of P is zero. This means that the probability of finding very large headways
is small.
(b) The headway h is h = f /v where f is the spacing and v is the speed. We want
the probability Q that the spacing is at least f feet. As noted in the problem, Q
is the same as the probability that the headway is at least f /v seconds. On the
other hand, P (t) = 0.87t is the probability that the headway is at least t seconds,
so Q = P (f /v) = 0.87f /v .
440
Solution Guide for Chapter 5
(c) If the average speed is 88 feet per second, then v = 88. If the spacing between two
vehicles is at least 40 feet then f = 40, and so the probability Q of this occurring is
Q = 0.8740/88 = 0.94 or 94%, so a spacing of 40 feet or more is highly likely.
9. Probability of extinction:
(a) We are given that the probability of extinction by time t is
P (t) =
d(ert − 1)
bert − d
and that d = 0.24, b = 0.72, and so r = b − d = 0.48. Substituting these values, we
have
P (t) =
0.24(e0.48t − 1)
.
0.72e0.48t − 0.24
Scanning a table of values, for example the table below, we see quickly that the
limiting value of P is 0.33, so there is a 0.33 or 33% probability that the population
will eventually become extinct.
(b) We have a formula for Q expressing Q as a function of P , namely Q = P k , and
we already have a formula expressing P in terms of t, b, d, and r from Part (a).
To find a formula for Q in terms of these latter variables and k , we substitute the
expression for P from Part (a) into the formula for Q in terms of P :
Q = Pk
rt
k
d(e − 1)
Q =
.
bert − d
(c) We are told that if b > d, then Q can be expressed as
Q=
d(1 − at )
b − dat
k
,
where a < 1. Since a < 1, the limiting value of at is zero, so the limiting value of Q
k
d
is
. This is the probability that a population starting with k individuals will
b
eventually become extinct.
SECTION 5.3
Combining and Decomposing Functions
441
d
1
(d) We use the formula from Part (c). If b is twice as large as d, then = = 0.5, and
b
2
k
d
k
= 0.5 .
so the probability of eventual extinction is
b
(e) As a function of k, the limiting value of 0.5k is zero since 0.5 < 1. In practical terms,
this means that for a large initial population, the probability that the population
will eventually become extinct is very small.
10. Decay of litter:
L
and both L and k are constant, so R is
k
a constant. Thus the initial value of D is the difference between R and the initial
(a) We are given that D = R − A. Now R =
value of A, that is, D(0) = R − A(0). On the other hand, A is the amount of litter
present at time t and we are assuming that at time t = 0 the forest floor is clear of
litter, so A(0) = 0. Thus the initial value of D is R − A(0) = R − 0 = R.
(b) Here D is an exponential function with initial value R, by Part (a), and now we
are told that the yearly decay factor is a = e−k , so D = R(e−k )t , which can also be
written as D = Re−kt .
(c) By Parts (a) and (b), we have that D = R − A and D = Re−kt , so R − A = Re−kt .
Solving for A, we add A to each side of the equation, yielding R = Re−kt + A.
Subtracting Re−kt from each side of the equation gives R − Re−kt = A.
11. Applications of the litter formula:
(a) When t =
3
, then, replacing t by 3/k in the formula for A, we find that
k
A = R − Re−kt = R − Re−k×3/k = R − Re−3 = R − 0.05R = 0.95R,
so the amount of litter present A is 95% of R, the limiting value.
3
(b) By Part (a), we know that 95% of the limiting value is reached when t = . In this
k
3
3
case k = 0.003 per year, so 95% of the limiting value is reached in t = =
=
k
0.003
1000 years.
3
(c) To reach 95% of the limiting value for forest litter in the Congo takes t = = 0.75
4
3
of a year; in the Sierra Nevada Mountains it takes t =
= 47.62 years.
0.063
442
Solution Guide for Chapter 5
12. High school graduates: For this problem, let t be the number of years since 1986, so,
carefully noting that the years listed in the table are not always consecutive (!), we see
that the table may be written:
t
0
1
3
4
5
6
8
10
12
N
2.79
2.65
2.45
2.36
2.28
2.40
2.52
2.66
2.80
Here N is the number of high school graduates in the United States for the given year,
in millions.
(a) A plot of N as a function of t is given in the figure below. Clearly the graph does
not lie close to a single line, so a linear model is not appropriate.
(b) The calculation of the regression line parameters for the part of the data between
1986 and 1991 (so t from 0 to 5) is shown in the figure below on the left. Over this
period, the linear regression model is N = −0.101t + 2.769.
(c) The calculation of the regression line parameters for the part of the data between
1992 and 1998 (so t from 6 to 12) is shown in the figure above on the right. Over
this period, the linear regression model is N = 0.067t + 1.992.
SECTION 5.3
Combining and Decomposing Functions
443
(d) For t between 0 and 5, the regression model for N is given by N = −0.101t+2.769,
whereas for t between 6 and 12, the regression model for N is given by N =
0.067t + 1.992. We write N as a single piecewise-defined function using
N=
−0.101t + 2.769 for 0 ≤ t ≤ 5
0.067t + 1.992 for 6 ≤ t ≤ 12
(e) The figure below on the left is the graph of N using the formula from Part (d) and
the same window as for the graph in Part (a). On the right we have added the data
points so as to see what a nice fit the formula from Part (d) gives.
(f) For 1993 we use t = 7. Now N (7) = 0.067 × 7 + 1.992 = 2.461 million. The actual
value of 2.34 million is a little less than would be expected from the formula.
13. Gray wolves in Michigan:
(a) Since the gray wolves recolonized in the Upper Peninsula of Michigan it is reasonable to assume that this area is both suitable habitat and has far fewer wolves than
it had historically. Under these conditions, we would expect exponential growth.
(b) We use regression to find an exponential model since the data is not exactly exponential. We find the model W = 10.94 × 1.39t , where t is the number of years since
1990 and W is the number of wolves.
(c) The figure below is a graph of the data, with W as a function of t, together with
the exponential model from Part (b).
444
Solution Guide for Chapter 5
The data looks as if it follows two different exponential functions, neither of which
is the exponential model we graphed. It would be better to use a piecewisedefined function.
(d) We find an exponential model for 1990 through 1996 by omitting later data. Using
regression we get the exponential model W = 7.95 × 1.59t , where t is time in years
since 1990. This formula holds for t from 0 through 6. We use the remaining data,
from 1997 through 2000, to calculate the exponential model for W as a function of
T , where T is time in years since 1997. Using regression we obtain W = 112.22 ×
1.24T . This formula holds for T from 0 through 3, which is the same as t from 7
through 10. In fact, T = t − 7, and so the second exponential model can also be
written as W = 112.22 × 1.24t−7 .
(e) A single formula for W as a piecewise-defined function, using the formulas from
Part (d), is
W =
7.95 × 1.59t
112.22 × 1.24t−7
for 0 ≤ t ≤ 6
for 7 ≤ t ≤ 10
Graphing these two functions as a single piecewise-defined function shows an
excellent fit with the data, as in the figure below.
14. Mother-to-child transmission of HIV:
(a) The age at which the additional risk of HIV transmission from breastfeeding B
is the same as the additional risk from artificial feeding A is when B = A, so
2 = 5 × 0.785t . Solving by any method, for example using crossing graphs, yields
t = 3.79 months.
(b) Breastfeeding carries the smaller additional risk for ages at or below 3.79, since for
those ages the graph of additional risk of death from artificial feeding is above the
graph of additional risk of death from breastfeeding.
(c) The optimal plan for feeding a child, in the absence of any additional information,
would be to breastfeed until age 3.79 months (about 3 months plus 3 weeks), and
then to use artificial feeding.
SECTION 5.3
Combining and Decomposing Functions
445
(d) A piecewise-defined function R of t giving the additional risk of death under the
optimal plan from Part (c) would equal 2 for ages t from 0 to 3.79, and after that
equal 5 × 0.785t , so
2 for 0 ≤ t ≤ 3.79
5 × 0.785t for 3.79 ≤ t
R=
15. Quarterly pine pulpwood prices:
(a) The graph should start in the first quarter at $18.50, decrease steadily (so in a
straight line) to $14 at the second quarter, and then rise steadily to $18 by the
Price, dollars
fourth quarter:
20
18
16
14
12
10
8
6
4
2
0
1
2
3
4
Quarter
(b) Since the price P is decreasing steadily, it has a constant rate of change and so is a
linear function of the quarter.
(c) From the first to second quarter, P decreases by $4.50 over one quarter from a first
quarter value of $18. Thus the slope of P is −4.50. Now P = 18.50 when t = 1,
so the value when t = 0 would be 18.50 − (−4.5) = 23, and thus the formula is
P = −4.50t + 23. This formula is only valid for t from 1 to 2.
(d) Since the price P is increasing steadily, it has a constant rate of change and so is a
linear function of t.
(e) From the second to fourth quarter, P increases by $4 over two quarters from $14.
4
Thus the slope of P is = 2. Now P = 14 when t = 2, so the value when t = 0
2
would be 14 − 2 × 2 = 10, and so the formula is P = 2t + 10. This formula is only
valid for t from 2 to 4.
(f) Using the formulas from Parts (c) and (e),we can write a single formula for P as a
piecewise-defined function of t as follows:
P =
−4.5t + 23 for 1 ≤ t ≤ 2
2t + 10 for 2 ≤ t ≤ 4
Solution Guide for Chapter 5
16. Quarterly sawtimber prices:
(a) The graph should show a price of $54 at the beginning of the first quarter, a decrease to $49 in the second quarter, returning to $54 at the third quarter, then a
steady price of $54 for the whole fourth quarter. One possible graph is shown
below:
Price, dollars
446
56
54
52
50
48
46
44
42
40
1
2
3
4
Quarter
(b) If 5t2 is the correct shape of the graph for the first three quarters, we want a translation to move the minimum from t = 0 to t = 2. Thus the translation should be
t − 2, and so we should look at 5(t − 2)2 . This graph is fine except that the values
are too low. For example, the value at t = 2 is 0, not 49. Adding 49 gives the
correct values, so the function is P = 5(t − 2)2 + 49 dollars for t from 1 to 3.
(c) Now P is a steady $54 during the third to fourth quarters, so P = 54 dollars for t
from 3 to 4.
(d) Using the formulas from Parts (b) and (c), we can write a single formula for P (in
dollars) as a piecewise-defined function of t:
P =
5(t − 2)2 + 49 for 1 ≤ t ≤ 3
54 for 3 ≤ t ≤ 4
SECTION 5.4
Quadratic Functions and Parabolas
447
5.4 QUADRATIC FUNCTIONS AND PARABOLAS
E-1. Solving equations by completing the square:
(a) To solve x2 + 6x = 8 by completing the square, we add
b2
to each side, where
4
62
= 9 to each side: x2 + 6x + 9 = 8 + 9. The left side is
4
√
√
(x + 3)2 , so we now have (x + 3)2 = 17. Thus x + 3 = ± 17, and so x = −3 ± 17
b = 6; thus we need to add
are the solutions.
(b) To solve x2 − 2x − 5 = 0 by completing the square, we add 5 to each side and then
(−2)2
b2
add
where b = −2, that is, we add
= 1. Thus we add 5 + 1 = 6 to each
4
4
side, which gives x2 − 2x + 1 = 6. The left side is (x − 1)2 , so (x − 1)2 = 6. Thus
√
√
x − 1 = ± 6, and so x = 1 ± 6 are the solutions.
(c) To solve 2x2 = x + 4 by completing the square, we subtract x from each side of the
equation, in order to collect the x-terms, and then we divide both sides by 2, since
that is the coefficient of x2 :
2x2
= x+4
2x2 − x = 4
1
x2 − x = 2.
2
b2
1
where b = − , that
4
2
1
1
1
1
(−1/2)2
2
=
. This yields x − x +
= 2 + . Now the left side
is, we add
16
2
16
16
r
2 4
2
1
33
1
33
1
33
is x −
and the right is
, so x −
=
. Thus x − = ±
, and
4 r
16
4
16
4
16
√
1
33
1
33
so x = ±
are the solutions. Thus can be simplified to x = ±
since
4
16
4
4
√
16 = 4.
Now we proceed, as in Parts (a) and (b), to add to each side
(d) To solve this equation, we need to first simplify it by expanding the terms and
collecting the x2 and x terms together:
(x + 1)2 + (x + 2)2
=
(x + 4)2
(x2 + 2x + 1) + (x2 + 4x + 4)
= x2 + 8x + 16
2x2 + 6x + 5
= x2 + 8x + 16
x2 − 2x = 11.
As in Part (b), we add 1 to each side to complete the square: x2 − 2x + 1 = 12. Since
√
x2 − 2x + 1 = (x − 1)2 , we have (x − 1)2 = 12. Thus x − 1 = ± 12, and therefore
√
x = 1 ± 12 are the solutions.
448
Solution Guide for Chapter 5
E-2. Special form for quadratics: To show that every quadratic function can be written as
a(x + b)2 + c, we will start with an arbitrary quadratic function Ax2 + Bx + C and try to
complete the square. In the end we will be able to calculate a, b and c in terms of the A,
B, and C values.
To begin, we factor out the coefficient A of x2 as follows:
2
Ax + Bx + C
B
C
Ax2 + Bx + C = A
= A x2 + x +
.
A
A
A
Now we concentrate on the term x2 +
B
C
B
x + . To complete the square of x2 + x,
A
A
A
B2
, as noted in the text. In order to leave values unchanged, we must
4A2
both add and subtract this term(!):
2
B
C
B
B2
B2
C
B
C
B2
−
+
=
x
+
+ −
.
= x2 + x +
x2 + x +
2
2
A
A
A
4A
4A
A
2A
A 4A
we need to add
Finally, we complete the calculations, beginning from the first equation above:
C
B
2
2
Ax + Bx + C = A x + x +
A
A
!
2
B
C
B2
= A
x+
+ −
2A
A 4A
2
B
C
B2
= A x+
+A
−
2A
A 4A
2
B2
B
+C −
.
= A x+
2A
4
This is in the form a(x + b)2 + c with a = A, b =
B2
B
, and c = C −
.
2A
4
E-3. Complex solutions: The quadratic formula says that the solutions to ax2 + bx + c = 0
are
x=
−b ±
√
b2 − 4ac
.
2a
(a) To solve x2 + 2x + 3 = 0 we use a = 1, b = 2, and c = 3, so the solutions are
√
−2 ± 22 − 4 × 1 × 3
x =
2×1
√
√
√
−2 ± −8
−2 ± 2 2i
=
= −1 ± 2i
=
2
2
p
√ √
√
√
since −8 = 22 × 2 × (−1) = 2 2 −1 = ±2 2i.
(b) To solve x2 = x − 8, we subtract x − 8 in order to rewrite the equation in the form
ax2 + bx + c = 0. Subtracting yields x2 − x + 8 = 0, so we use a = 1, b = −1, and
c = 8. Thus the solutions are
x =
=
p
(−1)2 − 4 × 1 × 8
2×1
√
√
1 ± −31
1 ± 31i
=
2
2
−(−1) ±
SECTION 5.4
since
√
Quadratic Functions and Parabolas
449
√
−31 = ± 31i.
(c) To solve 2x2 + 3x + 4 = 0 we use a = 2, b = 3, and c = 4, so the solutions are
√
−3 ± 32 − 4 × 2 × 4
x =
2×2
√
√
−3 ± 23i
−3 ± −23
=
=
4
4
√
√
since −23 = ± 23i.
1+i
in the form a + bi, we use the suggestion
1−i
that we multiply top and bottom of this fraction by 1 + i:
E-4. Rearranging complex numbers: To write
1+i
1−i
=
=
1+i 1+i
(1 + i)(1 + i)
×
=
1−i 1+i
(1 − i)(1 + i)
1 + 2i + i2
1 + 2i − 1
2i
=
=
= i.
2
1−i
1 − (−1)
2
E-5. High powers of i: To calculate i233 , we start with low powers and look for a pattern.
Now i2 = −1, i3 = i2 i = −i, and i4 = (i2 )2 = (−1)2 = 1, so i5 = i4 i = i and so on,
that is, the powers repeat in cycles of 4 as i, −1, −1, 1, ... . Now 233 = 58 × 4 + 1,
so i233 = i58×4+1 = i58×4 i1 = (i4 )58 i = i, or observe that 58 × 4 + 1 takes you to the
beginning of the cycle, so that power of i is i.
a + bi
in the form
c + di
p + qi, we use the same idea as in Exercise E-4 and multiply both top and bottom of the
E-6. Standard form for complex numbers: To write a complex number
fraction by c − di:
a + bi
c + di
=
=
a + bi c − di
(a + bi)(c − di)
×
=
c + di c − di
(c + di)(c − di)
ac + bci − adi − bdi2
ac + bd + (bc − ad)i
ac + bd bc − ad
=
= 2
+ 2
i.
2
2
2
2
2
c + cdi − cdi − d i
c +d
c + d2
c + d2
ac + bd
bc − ad
and q = 2
.
2
2
c +d
c + d2
1
1
E-7. The square roots of i: To show that ± √ + √ i are the square roots of i, it suffices
2
2
to show that the square of that expression equals i:
This is now in the form p + qi where p =
2
1
1
± √ +√ i
2
2
2
1
1
1
+2× √ × √ i + √ i
2
2
2
1
1
1 2
1
1
+ 2 i + i = + i + (−1) = i.
2
2
2
2
2
=
=
1
√
2
2
450
Solution Guide for Chapter 5
E-8. Complex conjugates: We know that for a complex number z = a + bi, the complex
conjugate is z = a − bi.
(a) If z = a + bi and w = c + di, then z + w = a + bi + c + di = (a + c) + (b + d)i. Thus
z + w = (a + c) + (b + d)i = (a + c) − (b + d)i.
On the other hand,
z + w = (a − bi) + (c − di) = (a + c) − (b + d)i,
so z + w = z + w.
(b) For z = a + bi and w = c + di,
zw = (a + bi)(c + di) = ac + bci + adi + bdi2 = (ac − bd) + (bc + ad)i,
and so zw = (ac − bd) − (bc + ad)i. On the other hand,
z w = (a − bi)(c − di) = ac − bci − adi + bdi2 = (ac − bd) − (bc + ad)i,
so zw = z w.
(c) If z is a zero of P , then az 3 + bz 2 + cz + d = 0. Now real numbers such as a, b, c,
d, or 0 are unchanged by complex conjugation, so 0 = 0 = az 3 + bz 2 + cz + d. By
Parts (a) and (b), 0 = az 3 + bz 2 + cz + d = az 3 + bz 2 + cz + d, and so z is also a root
of P .
S-1. The rate of change: The rate of change of a quadratic function is a linear function.
S-2. Testing for quadratic data: To see if the data in the table are quadratic, we calculate the
second-order differences:
x
y
First-order difference
Second-order difference
1
0
3
4
2
3
7
4
3
10
11
4
4
21
15
5
36
Since the second-order differences are equal, the data are quadratic.
SECTION 5.4
451
Quadratic Functions and Parabolas
S-3. Testing for quadratic data: To see if the data in the table are quadratic, we calculate the
second-order differences:
x
y
First-order difference
Second-order difference
1
1
4
0
2
5
4
11
3
9
15
−2
4
24
13
5
37
Since the second-order differences are not equal, the data are not quadratic.
S-4. Quadratic formula: The quadratic formula gives the solutions of ax2 + bx + c = 0 as
√
−b ± b2 − 4ac
.
x=
2a
To solve 3x2 − 5x + 1 = 0, we use a = 3, b = −5, and c = 1, so
p
√
−(−5) ± (−5)2 − 4 × 3 × 1
5 ± 13
=
x=
2×3
6
are the solutions. This expression gives the values x = 1.43 and x = 0.23.
S-5. Quadratic formula: The quadratic formula gives the solutions of ax2 + bx + c = 0 as
√
−b ± b2 − 4ac
.
x=
2a
To solve −2x2 + 2x + 5 = 0, we use a = −2, b = 2, and c = 5, so
√
√
−2 ± 44
−2 ± 22 − 4 × −2 × 5
=
x=
−4
2 × (−2)
are the solutions. This expression can be simplified to x =
1±
√
2
11
1
= ±
2
√
11
and
2
gives the values x = −1.16 and x = 2.16.
S-6. Quadratic formula: The quadratic formula gives the solutions of ax2 + bx + c = 0 as
√
−b ± b2 − 4ac
x=
.
2a
To solve 5x2 − 8 = 0, we use a = 5, b = 0, and c = −8, so
√
√
−0 ± 02 − 4 × 5 × −8
± 160
x=
=
10
2×5
are the solutions. This expression gives the values x = ±1.26.
452
Solution Guide for Chapter 5
S-7. Single-graph method: To solve the quadratic equation 3x2 − 5x + 1 = 0 using the singlegraph method, we graph the function with a suitable window. Using a window with a
horizontal span of −2 to 2 and a vertical span of −5 to 5, we can find the roots. One root
is shown in the figure on the left below, while the other is shown on the right:
These graphs show that the solutions are x = 1.43 and x = 0.23.
S-8. Quadratic regression: To find a quadratic model for this data set using regression, we
enter the data in the first two columns, as shown in the figure below on the left, and
use the calculator to find directly the quadratic regression coefficients, as shown in the
figure on the right below.
This shows that the quadratic model is f (x) = 0.85x2 + 0.90x + 0.21.
SECTION 5.4
Quadratic Functions and Parabolas
453
S-9. Quadratic regression: To find a quadratic model for this data set using regression, we
enter the data in the first two columns, as shown in the figure below on the left, and
use the calculator to find directly the quadratic regression coefficients, as shown in the
figure on the right below.
This shows that the quadratic model is f (x) = −0.35x2 + 3.36x − 2.52.
S-10. Quadratic regression: To find a quadratic model for this data set using regression, we
enter the data in the first two columns, as shown in the figure below on the left, and
use the calculator to directly find the quadratic regression coefficients, as shown in the
figure on the right below.
This shows that the quadratic model is f (x) = −0.51x2 + 3.78x + 2.90.
454
Solution Guide for Chapter 5
S-11. Quadratic regression: To find a quadratic model for this data set using regression, we
enter the data in the first two columns, as shown in the figure below on the left, and
use the calculator to find directly the quadratic regression coefficients, as shown in the
figure on the right below.
This shows that the quadratic model is f (x) = −x2 + 2x − 5.
S-12. Quadratic regression: To find a quadratic model for this data set using regression, we
enter the data in the first two columns, as shown in the figure below on the left, and
use the calculator to find directly the quadratic regression coefficients, as shown in the
figure on the right below.
This shows that the quadratic model is f (x) = 1.02x2 + 0.02x − 0.03.
SECTION 5.4
Quadratic Functions and Parabolas
455
1. Sales growth:
(a) To make a graph of G versus s we use a horizontal span of 0 to 4 since the model
is valid up to a sales level of 4 thousand dollars. Using a table of values, we
determine a vertical span of 0 to 1.3, and show the graph below.
(b) When the sales level is $2260, s = 2.26 since s is the sales level in thousands. The
rate of growth in represented in functional notation by G(2.26). Using the formula
to evaluate this, we find that the growth is G(2.26) = 1.2 × 2.26 − 0.3(2.26)2 = 1.18
thousand dollars per year. This can also be found from a table of values or the
graph.
(c) The rate of growth G is maximized at the maximum point of the graph. This is
shown on the graph below:
The maximum growth rate is achieved when s = 2.00, that is, at a sales level of
$2000.
456
Solution Guide for Chapter 5
2. Marine fishery:
(a) To make a graph of G versus n we use a horizontal span of 0 to 1.7 since the model
is valid up to a population size of 1.7 million tons. Using a table of values, we
determine a vertical span of −0.10 to 0.15, and show the graph below.
(b) The value G(1.62) = 0.3 × 1.62 − 0.2(1.62)2 = −0.039 (it could also be found using
a table or the graph). This means that at a population n of 1.62 million tons of fish,
the growth rate G is −0.039 million tons of fish per year, that is, the population is
declining by about 39,000 tons of fish per year.
(c) The rate of growth G is maximized at the maximum point of the graph. This is
shown on the graph below:
The maximum growth rate is achieved when n = 0.75, that is, at a population
level of about 0.75 million tons, or about 750,000 tons, of fish.
SECTION 5.4
Quadratic Functions and Parabolas
457
3. Cox’s formula:
(a) To graph the quadratic function 4v 2 + 5v − 2 using a horizontal span from 0 to 10, a
table of values suggests a vertical span from 0 to 400. The table of values is shown
below on the left while the graph is shown on the right.
(b) If the velocity v is 0.25 foot per second, then the value of 4v 2 + 5v − 2 is −0.5. This
would mean that the right side of Cox’s formula is negative, whereas the left side
Hd
and H, d, and L are all positive numbers. Thus it is not
of Cox’s formula is
L
possible for v to be 0.25 foot per second.
(c) If d = 4, L = 1000, and H = 50, then Cox’s formula implies that
50 × 4
4v 2 + 5v − 2
=
.
1000
1200
Evaluating the left-hand side and multiplying by 1200, we see that 4v 2 + 5v − 2 =
240. Solving for v shows that v = 7.18 feet per second.
(d) To find the largest pipe diameter d for which v does not exceed 10 feet per second,
we calculate d when v = 10, L = 1000, and H = 50:
50d
4 × 102 + 5 × 10 − 2
=
.
1000
1200
Evaluating, we see that d = 7.47 inches is the largest such pipe diameter.
458
Solution Guide for Chapter 5
4. Poiseuille’s law for fluid velocities:
(a) The tube’s radius in R, whereas r is the distance from the centerline of the tube.
Thus for a point along the walls of the tube, r = R. The velocity v is 0 when r = R,
since then R2 − r2 = 0.
(b) The velocity v is greatest when r = 0, that is, in the center of the tube, so that is
where the fluid flows most rapidly.
(c) Choosing, for example, k = 2 and R = 7, Poiseuille’s law becomes v = 2(49 −
r2 ). A graph is shown below using a horizontal span from 0 to R = 7 and a
corresponding vertical span from 0 to 100.
(d) The graph shows the greatest flow at the center of the tube (when r = 0), dropping
off gradually, and then faster, until the flow is 0 when r = 7 = R at the walls of
the tube.
(e) Any point in the tube is no farther in distance than R from the centerline.
5. Surveying vertical curves:
(a) Since the rate of change of grade is constant, a quadratic model is appropriate.
(b) In this case, g1 = 1.35, g2 = −1.75, L = 5, and E = 1040.63, so the formula for the
elevations of the vertical curve is
y
−1.75 − 1.35 2
1.35 × 5
x + 1.35x + 1040.63 −
2×5
2
= −0.31x2 + 1.35x + 1037.255.
=
(c) Using the formula from Part (b), the stations correspond to x = 0 through x = 5.
The six stations are given in the table below:
Station number
Elevation
0
1037.26
Here the elevations are in feet.
1
1038.30
2
1038.72
3
1038.52
4
1037.70
5
1036.26
SECTION 5.4
Quadratic Functions and Parabolas
459
(d) The highest point of the road on the vertical curve is at the maximum point on
the graph of y. From the graph (see the figure below – we used a window with
a horizontal span from 0 to 5 and a vertical span from 1035 to 1040) we find the
maximum at station x = 2.1774, that is 217.74 feet along the vertical curve with an
elevation of E = 1038.72 feet.
6. Quadratic data: We compute the first-order and second-order differences:
x
P
First-order difference
Second-order difference
0
6
−1
4
1
5
3
4
2
8
7
4
3
15
11
4
26
Since the second-order differences are constant, the data can be modeled by a quadratic
function.
7. Data that is not quadratic: We compute the first-order and second-order differences:
x
P
First-order difference
Second-order difference
0
5
3
6
1
8
9
12
2
17
21
18
3
38
39
4
77
Since the second-order differences are not constant, the data cannot be modeled by a
quadratic function.
8. A quadratic model: We compute the first-order and second-order differences:
x
Q
First-order difference
Second-order difference
0
5
1
6
1
6
7
6
2
13
13
6
3
26
19
4
45
Since the second-order differences are constant, the data can be modeled by a quadratic
function. To find the formula for the quadratic function, we use the standard form
460
Solution Guide for Chapter 5
Q = ax2 + bx + c and the relations
1
× second-order difference
2
= initial first-order difference − a
a =
b
c = initial value.
Thus a =
1
2
× 6 = 3, b = 1 − a = 1 − 3 = −2, and c = 5, so Q = 3x2 − 2x + 5 is the
formula.
9. Linear and quadratic data: We compute the first-order and second-order differences in
Table A:
x
f
First-order difference
Second-order difference
0
10
7
2
1
17
9
2
2
26
11
2
3
37
13
4
50
Since the second-order differences are constant, the data can be modeled by a quadratic
function. To find the formula for the quadratic function, we use the standard form
f = ax2 + bx + c and the relations
1
× second-order difference
2
= initial first-order difference − a
a =
b
c = initial value.
Thus a =
1
2
× 2 = 1, b = 7 − a = 7 − 1 = 6, and c = 10, so f = x2 + 6x + 10 is the
formula.
Successive differences in Table B are always 7, so Table B represents a linear function
with slope 7 (since x increases by single units). The initial value is 10, so g = 7x + 10.
10. A leaking can:
(a) We compute the first-order and second-order differences:
D
H
First-order difference
Second-order difference
0
1
0.25
0.5
1
1.25
0.75
0.5
2
2
1.25
0.5
3
3.25
1.75
4
5
Since the second-order differences are constant, the data can be modeled by a
quadratic function.
SECTION 5.4
Quadratic Functions and Parabolas
461
(b) To find the formula for the quadratic function, we use the standard form H =
aD2 + bD + c and the relations
1
× second-order difference
2
= initial first-order difference − a
a =
b
c = initial value.
Thus a = 21 ×0.5 = 0.25, b = 0.25−a = 0.25−0.25 = 0, and c = 1, so H = 0.25D2 +1
is the formula.
(c) We are given that H = 4, and we want to find D. Thus we want to solve the
equation H(D) = 4 for D. We do this using the crossing graphs method. (We
could use quadratic formula instead.) From the original table of data we see that
the solution lies between D = 3 and D = 4. In the figure below we have graphed
H and the constant 4 using a horizontal span of 0 to 4 and a vertical span of 0 to
5. We see that the intersection occurs at D = 3.46. Thus the stream will strike the
ground 3.46 inches from the base of the can when the depth is 4 inches.
(d) When D = 5 the depth is H(5) = 0.25 × 52 + 1 = 7.25 inches.
11. Traffic accidents:
(a) Using quadratic regression as shown in the figure below, we see that the quadratic
model is R = 7.79s2 − 514.36s + 8733.57.
462
Solution Guide for Chapter 5
(b) Using the model from Part (a), we find that R(50) = 7.79 × 502 − 514.36 × 50 +
8733.57 = 2490.57, which means that commercial vehicles driving at night on urban streets at 50 miles per hour have traffic accidents at a rate of about 2491 per
100,000,000 vehicle-miles.
(c) The speed at which traffic accidents are minimized corresponds to the minimum
point on the graph. This is calculated in the figure below:
This shows that the minimum is when s = 33.01 or about 33 miles per hour.
12. Vehicles parked:
(a) Using t as hours since midnight (using a 24-hour clock), we see the data displayed
in the first two columns of the figure on the left below. The quadratic regression
coefficients are shown on the right.
If V is the number of vehicles parked, in thousands, then the quadratic regression
model is V = −0.163t2 + 3.950t − 16.228.
(b) In functional notation, the number of vehicles parked at 2 p.m. is V (14). The value
of V (14) is 7.12 using the model from Part (a), so the model estimates that about
7120 vehicles will be parked at 2 p.m.
SECTION 5.4
Quadratic Functions and Parabolas
463
(c) The number of vehicles parked V is greatest at the maximum point of the graph
shown below:
Here the maximum is calculated as t = 12.12, so the greatest number of vehicles
are parked at about 7 minutes past noon.
13. Women employed outside the home:
(a) Entering the data, and using t as the number of years since 1942, we see the data
shown in the figure below on the left and the calculation of the quadratic regression coefficients shown on the right.
Using N as the number of women employed outside the home, in millions, the
quadratic model is N = −0.735t2 + 3.107t + 16.154.
(b) In functional notation, the number of women working outside the home in 1947 is
N (5). The quadratic model from Part (a) gives that value as N (5) = 13.31 million.
(c) The table below summarizes the quadratic prediction versus the actual numbers
for 1947 and 1948:
Year
1947
1948
t
5
6
Quadratic prediction
13.31
8.34
Actual value
16.90
17.58
Here the figures are in millions. Clearly the predictions of the quadratic model are
far from accurate, so a quadratic model is probably not appropriate for this time
period. Another way to see this is to observe that the original data table shows a
464
Solution Guide for Chapter 5
maximum in 1944, and the additional information in Part (c) shows that after 1946
the number started to increase. A quadratic model can’t have both a maximum
and a minimum.
14. Resistance in copper wire:
(a) The second-order differences are all equal to 0.0012.
(b) Using quadratic regression, we calculate a quadratic model as
C(t)/C(0) = 0.000006t2 + 0.00387t + 1.
Using this formula in a table, we can see that the table values are exactly the same
as the original data. The figures below show the quadratic regression coefficients
(on the left) and the table generated using the formula (on the right).
(c) When electric resistance is double that at 0 degrees, C(t)/C(0) = 2. Solving 2 =
0.000006t2 + 0.00387t + 1 yields t = 197.76, so the electric resistance doubles at
about 198 degrees.
(d) We are considering an electric appliance to be used at 72 degrees with a resistance
to be within 10% of that predicted at 72 degrees
i. At 72 degrees, we predict the resistance ratio to be C(72)/C(0) = 1.3097 using
the formula from Part (b).
ii. To find the resistance ratios representing ±10%, note that 10% of 1.3097 is
about 0.1310, so the resistance ratios should be within 1.3097 ± 0.1310, that is,
the ratios should be between 1.4407 and 1.1787.
iii. To find the temperature range for the appliance to ensure that it operates
within the 10% tolerance, we need to find the values of t such that C(t)/C(0)
is 1.1787 and 1.4407. Solving for each of these, the temperature range is from
t = 43.27 to t = 98.76 degrees Fahrenheit. This range is quite reasonable for
use inside a heated and air-conditioned home.
SECTION 5.4
Quadratic Functions and Parabolas
465
15. A falling rock:
(a) We want to find t so that 16t2 +3t = 400. To solve this using the quadratic formula,
we first subtract 400 from each side to get 16t2 + 3t − 400 = 0. This is of the form
at2 + bt + c = 0 with a = 16, b = 3, and c = −400. Plugging these values into the
quadratic formula, we get:
√
b2 − 4ac
t=
2a
√
−b − b2 − 4ac
t=
2a
−b +
=
=
p
32 − 4 × 16 × (−400)
= 4.91
2 × 16
p
−3 − 32 − 4 × 16 × (−400)
= −5.09.
2 × 16
−3 +
The second of these times has no physical meaning, but from the first we see that
it takes the rock 4.91 seconds to fall 400 feet.
(b) We use the crossing graphs method to find t so that D(t) = 400. In the figure
below we have graphed D = 16t2 + 3t and the constant 400 using a horizontal
span of 0 to 6 and a vertical span of 0 to 600. We see that the intersection occurs at
t = 4.91. Thus, it takes the rock 4.91 seconds to fall 400 feet.
16. The water jug revisited: We want to find t so that 0.265t2 − 4.055t + 15.5 = 0. This is
of the form at2 + bt + c = 0 with a = 0.265, b = −4.055, and c = 15.5. Plugging these
values into the quadratic formula, we get:
p
√
4.055 + (−4.055)2 − 4 × 0.265 × 15.5
−b + b2 − 4ac
t=
=
= 7.87
2a
2 × 0.265
p
√
4.055 − (−4.055)2 − 4 × 0.265 × 15.5
−b − b2 − 4ac
t=
=
= 7.44.
2a
2 × 0.265
The earlier time, 7.44 minutes, represents the point at which the jug is drained. Thus,
the later time has no physical meaning.
466
Solution Guide for Chapter 5
17. Profit:
(a) Total cost C is simply the fixed cost of $2000 plus the variable cost of $30 per
widget times the number of widgets. Because N is the number of widgets, C =
30N + 2000.
(b) The table of price p as a function of number of widgets N shows a constant decrease in price of 0.50 per 50 additional widgets sold, so p should be a linear func−0.50
= −0.01 dollar per widget, so p = −0.01N + b for
tion of N . The slope is
50
some initial value b. Now when N = 200, then p = 41, according to the table, so
41 = −0.01 × 200 + b, and thus b = 43. Thus the formula is p = −0.01N + 43.
(c) The total revenue R is Price × Quantity sold, so R = p × N = (−0.01N + 43)N
using the formula for p from Part (b).
(d) The profit P is Revenue − Total cost, so P = R − C = (−0.01N + 43)N − (30N +
2000). This is a quadratic function since, if multiplied out, there would be a term
−0.01N 2 .
(e) The manufacturer breaks even when P = 0. Solving the equation (−0.01N +
43)N − (30N + 2000) = 0, we find two solutions: N is 178.30 or 1121.70 widgets.
18. More on profit:
(a) We calculate second-order differences in the table below.
Number N
Profit P
First-order differences
Second-order differences
200
900
475
−50
250
1375
425
−50
300
1800
375
−50
350
2175
325
400
2500
Since the first-order differences are not constant (and the data are evenly spaced),
P should not be modeled by a linear function. Since the second-order differences
are constant, a quadratic model is appropriate. Calculating using quadratic regression, we find P = −0.01N 2 + 14N − 1500.
(b) The fixed costs are the costs when nothing is produced. Now P (0) = −1500, and
P (0) = R(0)−C(0). But R(0) is the revenue when nothing is produced (so nothing
sold), so R(0) = 0. Thus C(0) = 1500 and so the fixed costs are $1500.
SECTION 5.5
Higher-degree Polynomials and Rational Functions
467
(c) Graphing the profit function P = −.01N 2 + 14N − 1500 using a horizontal span
of N from 0 to 1000 widgets and a corresponding vertical span of P from −1500 to
3500, we see from the graph below that the maximum profit of P = $3400 occurs
when N = 700 widgets are manufactured.
5.5 HIGHER-DEGREE POLYNOMIALS AND RATIONAL FUNCTIONS
E-1. Finding the degree of a polynomial: Since the polynomial has no complex zeros and
no repeated zeros, the degree is exactly the number of real zeros, that is, the number of
times the graph crosses the horizontal axis. Thus the degree of this polynomial is 5.
E-2. Choosing a model: If the data set shows three maxima and two minima, then the
smallest-degree polynomial that could possibly fit the data exactly would be degree
6.
E-3. Finding the polynomial: The polynomial has zeros 2, 3, and 4, so by the factor theorem,
the polynomial is divisible by (x − 2)(x − 3)(x − 4), which also has degree 3 and leading
coefficient 1. Thus the polynomial must be precisely (x − 2)(x − 3)(x − 4). This can be
written out as x3 − 9x2 + 26x − 24.
E-4. Calculating limits: To calculate the limiting values of these rational functions, we use
the fact that we need consider only the leading terms.
(a)
5x4 + 4x3 + 7
5x4
5
5
= lim
= lim
= = 2.5.
4
2
x→∞ 2x − x + 4
x→∞ 2x4
x→∞ 2
2
lim
(b)
3x4 + 4x2 − 9
3x4
3
= lim
= lim
= 0.
5
3
x→∞ 2x + 4x − 8
x→∞ 2x5
x→∞ 2x
lim
(c)
ax5 + bx2 + c
ax5
a
a
= lim
= lim
= .
5
2
5
x→∞ dx − ex + f
x→∞ dx
x→∞ d
d
lim
468
Solution Guide for Chapter 5
E-5. Getting a polynomial from points: In the three cases below we are looking for a quadratic
ax2 + bx + c to pass through the three given points. Substituting the values for x into the
quadratic gives three linear equations in terms of a, b and c which we solve by elimination (see pages 244-245 of the text for details about solving by elimination in general).
(a) For y = ax2 + bx + c to pass through (1, 5), (−1, 3), and (2, 9), the following equations need to hold:
a+b+c =
5
a−b+c =
3
4a + 2b + c =
9
Subtracting the first equation from the second, and subtracting 4 times the first
equation from the third, yield:
a+b+c =
−2b
5
= −2
−2b − 3c = −11
The second equation shows that b = 1. Substituting into the third equation, we find
−2−3c = −11 and so c = 3. Substituting into the first equation, we get a+1+3 = 5
and so a = 1. Thus the desired quadratic is x2 + x + 3.
(b) For y = ax2 +bx+c to pass through (1, 5), (2, 4), and (3, 19), the following equations
need to hold:
a+b+c =
5
4a + 2b + c =
4
9a + 3b + c =
19
Subtracting 4 times the first equation from the second, and subtracting 9 times the
first equation from the third, yield:
a+b+c =
5
−2b − 3c = −16
−6b − 8c = −26
Subtracting 3 times the second equation from the third yields:
a+b+c =
5
−2b − 3c = −16
c =
22
SECTION 5.5
Higher-degree Polynomials and Rational Functions
469
Thus c = 22. Substituting into the second equation, we find −2b − 3 × 22 = −16
and so b = −25. Substituting into the first equation, we get a − 25 + 22 = 5 and so
a = 8. Thus the desired quadratic is 8x2 − 25x + 22.
(c) For y = ax2 + bx + c to pass through (1, −2), (0, −5), and (−1, −2), the following
equations need to hold:
a + b + c = −2
c = −5
a − b + c = −2
Thus c = −5. Substituting into the first and third equations yields:
a+b
= 3
a−b
= 3
Adding the equations, we find 2a = 6, so a = 3. Substituting in either equation
shows that b = 0. Thus the desired quadratic is 3x2 − 5.
E-6. Lagrangian polynomials:
(a) Since the Lagrangian polynomials Pi are similarly constructed, we consider P1 and
compute P1 (x2 ):
P1 (x2 ) =
(x2 − x2 )(x2 − x3 )
= 0.
(x1 − x2 )(x1 − x3 )
In like fashion P1 (x3 ) is also zero due to the (x − x3 ) term in the numerator. In
general Pi (xj ) = 0 whenever i 6= j.
(b) We compute P1 (x1 ):
P1 (x1 ) =
(x1 − x2 )(x1 − x3 )
= 1.
(x1 − x2 )(x1 − x3 )
Similarly Pi (xi ) = 1 for all i.
(c) Let P = y1 P1 + y2 P2 + y3 P3 and consider P (x1 ):
P (x1 ) = y1 P1 (x1 ) + y2 P2 (x1 ) + y3 P3 (x1 ) = y1 × 1 + y2 × 0 + y3 × 0 = y1 .
In similar fashion P (x2 ) = y2 and P (x3 ) = y3 .
(d) Using x1 = 1, x2 = 2, and x3 = 3, we have
P1 (x)
=
P2 (x)
=
P3 (x)
=
(x − 2)(x − 3)
(x − 2)(x − 3)
1
5
= x2 − x + 3
=
(1 − 2)(1 − 3)
2
2
2
(x − 1)(x − 3)
(x − 1)(x − 3)
=
= −x2 + 4x − 3
(2 − 1)(2 − 3)
−1
(x − 1)(x − 2)
(x − 1)(x − 2)
1
3
=
= x2 − x + 1.
(3 − 1)(3 − 2)
2
2
2
470
Solution Guide for Chapter 5
Using y1 = 2, y2 = 5, and y3 = 6, then using Part (c), we find that the quadratic
passing through the points is
P
= y1 P1 + y2 P2 + y3 P3
1 2 3
1 2 5
= 2
x − x + 3 + 5 −x2 + 4x − 3 + 6
x − x+1
2
2
2
2
= x2 − 5x + 6 − 5x2 + 20x − 15 + 3x2 − 9x + 6
= −x2 + 6x − 3.
It is easy to check that P (1) = 2, P (2) = 5, and P (3) = 6.
S-1. Recognizing polynomials: A polynomial is a function which can be written as a sum of
power functions where each power is a non-negative whole number.
(a) x8 − 17x + 1 is a polynomial of degree 8.
(b)
√
x + 8 is not a polynomial since
√
x = x0.5 has a fractional power.
(c) 9.7x − 53.1x4 is a polynomial of degree 4.
(d) x3.2 − x2.3 is not a polynomial since the powers are not integers.
S-2. Testing for polynomial data: To test evenly-spaced data to determine if they represent
a polynomial of degree n, we need to show that the nth-order differences are constant.
S-3. Rational functions: A rational function is a function which can be written as a ratio of
polynomial functions.
S-4. Degree from maxima and minima: The smallest-degree polynomial that can have both
a maximum and a minimum is a cubic, that is, the degree is 3.
S-5. Cubic regression: To use cubic regression, we enter the data in the first two columns
(see the figure on the left below) and then use the calculator to determine the cubic
regression coefficients (see the figure on the right below).
The cubic regression model is y = 0.07x3 − 1.11x2 + 4.92x − 2.95.
SECTION 5.5
Higher-degree Polynomials and Rational Functions
471
S-6. Cubic regression: To use cubic regression, we enter the data in the first two columns
(see the figure on the left below) and then use the calculator to determine the cubic
regression coefficients (see the figure on the right below).
The cubic regression model is y = 0.03x3 − 0.46x2 + 3.06x − 1.14.
S-7. Quartic regression: Entering the data as in Exercise S-5, we use the calculator to determine the quartic regression coefficients (see the figure below)
The quartic regression model is y = −0.01x4 + 0.20x3 − 2.00x2 + 7.26x − 4.58.
S-8. Quartic regression: Entering the data as in Exercise S-6, we use the calculator to determine the quartic regression coefficients (see the figure below)
The quartic regression model is y = −0.01x4 + 0.22x3 − 1.86x2 + 6.72x − 3.70.
472
Solution Guide for Chapter 5
x
are the values of x for which the denominator
x2 − 3x + 2
x2 − 3x + 2 is zero but the numerator x is not zero. In this case, x2 − 3x + 2 = 0 when
S-9. Finding poles: The poles of
x = 2 and x = 1, so these are the poles.
2x2 + 1
are the limiting values
x2 − 1
of this expression. Using a table of values, for example, or calculating the limits more
S-10. Horizontal asymptotes: The horizontal asymptotes of
formally, it can be determined that the limit is 2, so the horizontal asymptote is the line
y = 2.
3x − 1
are the values of x for which the denominator
x2 + 2x
x2 + 2x is zero but the numerator 3x − 1 is not zero. In this case, x2 + 2x = 0 when x = 0
S-11. Finding poles: The poles of
and x = −2, so these are the poles.
6x4 − 5
are the limiting values
2x4 + x
of this expression. Using a table of values, for example, or calculating the limits more
S-12. Horizontal asymptotes: The horizontal asymptotes of
formally, it can be determined that the limit is 3, so the horizontal asymptote is the line
y = 3.
1. Production:
(a) To include values of n up to 1.5 thousand units, we use a horizontal span from 0
to 1.5. A corresponding vertical span from 0 to 2.2 yields the following graph
(b) In functional notation, the amount produced if the input is 1.45 thousand units is
T (1.45). Using the formula, graph, or a table of values, we see that T (1.45) = 1.66
thousand units of product.
SECTION 5.5
Higher-degree Polynomials and Rational Functions
473
(c) Tracing on the graph, we estimate the point of inflection at n = 0.5 and T = 1.
This is shown in the figure below on the left.
(d) The maximum point on the point is when n = 1.15 and T = 2.08, as shown in
the figure above on the right. Thus the maximum amount produced is T = 2.08
thousand units.
2. Speed of sound in the North Atlantic:
(a) Since the highest powers of T and D are 3, V is a cubic polynomial of either T or
D alone.
(b) Letting D = 1000, the formula for V becomes
V
=
1447.733 + 4.7713T − 0.05435T 2 + 0.0002374T 3 + 0.0163 × 1000
+
1.675 × 10−7 × 10002 − 7.319 × 10−13 × 10003 T.
This can be simplified to V = 1464.2005 + 4.7705861T − 0.05435T 2 + 0.0002374T 3 .
(c) Here is the graph of V as a function of T using a horizontal span of T from 0 to 30
degrees Celsius and a vertical span of V from 1400 to 1600 meters per second:
(d) The graph is concave down and increasing, so as T increases, so does V , but less
and less so.
474
Solution Guide for Chapter 5
3. Traffic accidents:
(a) Calculating the cubic regression coefficients, we find a cubic model for the data:
C = 0.000422s3 − 0.0294s2 + 0.562s − 1.65. Here we rounded to two decimal places
after the first non-zero digit, as specified in the problem.
(b) Using the cubic model from Part (a), we can calculate C(42) as 1.36. This means
that if commercial vehicles travel at 42 miles per hour at night on urban streets,
the cost due to traffic accidents will be 1.36 cents per vehicle-mile.
(c) Graphing the cubic model, we find the minimum from the graph. Using a horizontal span of s from 20 to 50 miles per hour and a corresponding vertical span of
C from −2 to 4, the graph is:
This shows the minimum to be when s = 33 miles per hour.
4. Poiseuille’s law for rate of fluid flow:
(a) If R increases by 10%, that means that R is multiplied by a factor of 1.10. Since F
is a power function with power 4, F changes by a factor of 1.104 = 1.4641, which
is an increase 46.41%.
3
1
-inch pipe has a radius 1.5 times that of a -inch pipe, so the flow rate is
4
2
1.54 = 5.06 times as much in the larger pipe.
4
1
1
(c) If an artery has its normal radius, then the blood flow will be
= 0.0625
2
2
times as much, so the flow will be only 6.25% of the usual blood flow.
(b) A
SECTION 5.5
Higher-degree Polynomials and Rational Functions
475
5. Population genetics:
1
1
λ + can be solved by any of a number of tech2
4
niques. For example, using crossing graphs with a horizontal span of λ from −1
(a) The quadratic equation λ2 =
to 1 and a vertical span from −1 to 1, we find the two solutions (see the figures
below).
Here we see that the solutions are λ = −0.31 and λ = 0.81. Since λ1 is the larger
solution, λ1 = 0.81.
(b) According to the problem, the proportion of the population which will be heterozygous (so not homozygous) after 5 generations is given by λ51 = 0.815 =
0.3487 or 34.87%. Thus the homozygous proportion will be 100% − 34.87% =
65.13%, or about 65.1%.
(c) As in Part (b), the proportion of the population which will be heterozygous (so
20
= 0.0148 or 1.48%.
not homozygous) after 20 generations is given by λ20
1 = 0.81
Thus the homozygous proportion will be 100% − 1.48% = 98.52%, or about 98.5%.
6. Population genetics — first cousins:
(a) Solving this equation, for example using crossing graphs, we find only a single
solution, λ = 0.92, so that solution must be λ1 .
(b) In general we expect three solutions to a cubic equation, although we know that
there are at most three solutions.
(c) According to the problem, the proportion of the population which will be heterozygous (so not homozygous) after 5 generations is given by λ51 = 0.925 =
0.6591 or 65.91%. Thus the homozygous proportion will be 100% − 65.91% =
34.09%, or about 34.1%.
(d) As in Part (c), the proportion of the population which will be heterozygous (so not
20
homozygous) after 20 generations is given by λ20
= 0.1887 or 18.87%.
1 = 0.92
Thus the homozygous proportion will be 100%−18.87% = 81.13%, or about 81.1%.
476
Solution Guide for Chapter 5
7. Population genetics — second cousins:
1 2
1
1
λ + λ+
can be solved by any of a number of
8
32
64
techniques. For example, using crossing graphs with a horizontal span of λ from
(a) The quartic equation λ4 =
−0.75 to 0.75 and a vertical span from −0 − .10 to 0.20, we find two solutions (see
the figures below).
Here we see that the solutions are λ = −0.39 and λ = 0.50. Since λ1 is the larger
solution, λ1 = 0.50.
(b) In general we expect four solutions to a quartic equation, although we know that
there are at most four solutions.
(c) According to the problem, the deviation of the population from its equilibrium
structure after 5 generations is given by λ51 = 0.55 = 0.0313 or 3.13%. The deviation is about 3.1%.
(d) As in Part (c), the deviation of the population from its equilibrium structure after
20
20 generations is given by λ20
= 9.54 × 10−7 = 0.000000954 or 0.0000954%.
1 = 0.5
The deviation is less than 0.0001%.
8. Forming a pen:
(a) We use a horizontal span of 0 to 15 and a vertical span of 0 to 100. The graph is
shown below at the left.
(b) From the graph we see that, near the pole, F gets larger and larger as w gets
smaller and smaller. This says that for small widths the amount of fence required
SECTION 5.5
Higher-degree Polynomials and Rational Functions
477
gets larger and larger as the width gets smaller and smaller. This makes sense
because a very narrow pen would have to be very long to enclose 80 square feet
of area.
(c) From the graph above at the right we see that the minimum occurs at w = 6.32
feet, so the length of the sides perpendicular to the building is about 6.32 feet.
80
The length of the side parallel to the building is
, or about 12.65 feet. Thus the
w
rectangle is about 6.32 feet by 12.65 feet.
9. Inventory:
(a) We use a horizontal span of 0 to 10 and a vertical span of 0 to 8500. The graph is
shown below.
(b) From the graph we see that, near the pole, E gets larger and larger as Q gets
smaller and smaller. This says that the yearly inventory expense gets larger and
larger as the order size gets smaller and smaller. This makes sense: if the order
sizes are very small then the dealer will have to place many orders, and so the
total cost will be very high.
10. Power for flying:
(a) Using the values of a and b for the budgerigar, we find that the required power is
P =
u3
600
+
.
7800
u
The graph is shown below on the left, using a horizontal span of u from 25 to 45,
and a corresponding vertical span of P from 20 to 30.
478
Solution Guide for Chapter 5
(b) Using the graph, a table of values, or the formula from Part (a), we calculate P (39)
to be 22.99. In practical terms, this means that if a budgerigar flies at a speed of
39 kilometers per hour, then the power required for level flight is about 23 cubic
centimeters of oxygen per gram per hour.
(c) From the graph on the right above, we see that the minimum required power is
found at 35.34, or about 35, kilometers per hour.
11. Traffic signals:
(a) The graph of
n=1+
90
v
+
30
v
is shown below on the left, using a horizontal span of v from 30 to 80 and a vertical
span of n from 4.4 to 5.1.
(b) In functional notation, the length of the yellow light when the approach speed is
45 feet per second is n(45). The value of n(45) is obtained using the formula given,
a table of values, or the graph; we find that n(45) = 4.5 seconds.
(c) At v tends towards 0, n increases sharply. In practical terms, as the approach speed
becomes very slow, the time needed to cross the intersection increases greatly.
Therefore the yellow light needs to be much longer.
(d) Finding the minimum point on the graph, as shown in the figure above on the
right, we see that the minimum occurs when v = 51.96 feet per second and n =
4.46 seconds. The minimum length of time for a yellow light is 4.46 seconds.
SECTION 5.5
Higher-degree Polynomials and Rational Functions
479
12. Gliding falcon:
7.5
u3
+
is shown below on the left, using a horizontal span
4000
u
of u from 0 to 15 and a vertical span of s from 0 to 3.
(a) The graph of s =
(b) In functional notation, when the airspeed is 5 meters per second the sinking speed
is s(5). The value of s(5) is obtained using the formula given, a table of values, or
the graph; we find that s(5) = 1.53 meters per second.
(c) As u tends towards 0, s increases sharply. In practical terms, as the airspeed becomes very slow, the sinking speed becomes much greater, that is, the falcon glides
down very quickly.
(d) From the graph on the right above, we see that the minimum sinking speed is
found when the airspeed is 10 meters per second.
13. Gliding pigeon:
u3
25
+
is shown below on the left using a horizontal span of
2500
u
u from 0 to 15 and a vertical span of s from 0 to 9.
(a) The graph of s =
(b) Using the formula, a table of values, or a graph, we find that s(10) = 2.9 meters
per second. In practical terms, a pigeon gliding with an airspeed of 10 meters per
second will sink at a rate of 2.9 meters per second.
(c) The graph above on the right shows the performance diagrams for the falcon and
the pigeon (the darker graph is that of the falcon). Since the falcon’s graph lies
480
Solution Guide for Chapter 5
below that of the pigeon, for the same airspeed, the falcon sinks much more slowly,
so the falcon is the more efficient glider.
14. Average transit vehicle speed:
(a) Using the formula given for S with D = 3, T = 0.05, and a = d = 12,600, we find
that the formula is
3C
S=
0.05C + 3 +
C2
1
2×12,600
+
1
2×12,600
=
3C
0.05C + 3 + C 2
1
12,600
.
(b) If cruising speed is an estimate of transit speed, then we expect S to be very close
to C. The graph on the left below shows the two functions S = C, which is a line,
and the function given in Part (a) for S. The window used is a horizontal span of
C from 0 to 60 miles per hour and a vertical span of S from 0 to 60 miles per hour.
The cruising speed is a very poor estimate of transit speed, and the estimate becomes less accurate with increasing cruising speed.
(c) To find the cruising speed C for which S = 30, we need to solve the equation
3C
30 =
. The solution is found using the crossing-graphs
0.05C + 3 + C 2 (1/12,600)
method – see the figure above on the right, which uses a horizontal span from 0
to 80 and a vertical span from 0 to 60. The graph shows the solution C = 67.16, or
about 67 miles per hour.
15. Queues:
a2
has poles at s = 0 and s − a, that is s = a. Thus
s(s − a)
if the arrival rate a is very close to the gap rate s, then the queue length L becomes
(a) The rational function L =
very long.
a
a
and that the gap length s is 3, so w =
. To
s(s − a)
3(3 − a)
a
determine a such that w = 2, we need to solve 2 =
. Using any method,
3(3 − a)
for example crossing graphs, we find that a = 2.57 cars per minute.
(b) We are given that w =
SECTION 5.5
Higher-degree Polynomials and Rational Functions
481
16. Travel time in London: In central London v0 = 28 and a = 0.008, so the formula for
travel time is
t=
60d
.
28 − 0.008q
(a) With d = 2 miles and a traffic flow of q = 1500 vehicles per hour, the travel time is
60 × 2
t=
= 7.5 minutes.
28 − 0.008 × 1500
(b) This model is valid only when aq < v0 , so we may assume that the largest traffic
v0
28
flow that central London can handle is when aq = v0 , so q =
=
= 3500
a
0.008
vehicles per hour.
(c) The rational function t has a pole when 28 − 0.008q = 0, that is, when q = 3500, so
as the flow approaches 3500 vehicles per hour, travel time becomes very large.
17. Change in London travel time: Using the values v0 = 28, a = 0.008, and d = 2 for
central London, the formula for the additional travel time if one additional vehicle is
added to flow is
t0 =
60 × 0.008 × 2
.
(28 − 0.008q)2
(a) If q = 1000 then t0 = 0.0024, so the addition of one vehicle increases travel time
(for all vehicles) by 0.0024 minute, or about 0.14 second.
(b) If q = 3400 then t0 = 1.5, so the addition of one vehicle increases travel time (for
all vehicles) by 1.5 minutes.
(c) When traffic is light, additional cars have little effect on travel time, but when traffic flow is closer to 3500 vehicles per hour, additional traffic flow greatly increases
travel time.
18. Catch equation:
(a) Assuming that a lake is stocked with N0 = 1000 fish and the natural mortality rate
is M = 0.1, then the catch equation is
C = 1000
F
.
0.1 + F
(b) If F = 1 means 100% of the fish are caught in a year, then F = 2 would mean that
200% are caught in a year, which can only be reasonably interpreted as meaning
that all the fish are caught within one-half a year.
(c) The horizontal asymptote of C is the limiting value. Using a table or graph, or
formally calculating a limit, we see that the limiting value is 1000, so the horizontal
asymptote is C = 1000.
482
Solution Guide for Chapter 5
(d) As F gets larger and larger, the time required to catch all of the fish takes less and
less time. The meaning of the horizontal asymptote is that if you catch all the fish
as they are stocked, you will catch all 1000 fish.
(e) The vertical asymptote for C is F = −0.1, since F = −0.1 makes the denominator
zero and so is a pole for C. Now F = −0.1 indicates a negative rate of fishing and
so makes no sense.
19. Hill’s law:
(a) Using the given values for a fast twitch vertebrate muscle of F` = 300, a = 81, and
b = 6.75 in Hill’s law, we obtain
S=
6.75(300 − F )
.
F + 81
(b) Graphing S using a horizontal span of F from 0 to 300 and a vertical of S from 0
to 25, we obtain the figure below.
(c) As the force F increases to 300, the muscle’s contraction speed S decreases towards
0.
(d) Here S = 0 when F = 300, so the muscle does not contract at all when the maximum force is applied to the muscle.
(e) The horizontal asymptote of S is the limiting value. Using a table or graph, or
formally calculating a limit, we see that the limiting value of S is −6.75. Thus S
has a horizontal asymptote at S = −6.75. This makes no sense in terms of the
muscle: as F gets large, it becomes greater than the maximum force; moreover, a
negative value of S would not represent a muscle contraction.
(f) Here S has a vertical asymptote at F = −81. A negative force is not the situation
for which the equation applies.
Chapter 5 Review Exercises
483
20. The Michaelis-Menton relation: We are asked to graph the Michaelis-Menton relation
Vs
v=
for two different values of V and of Km . Using a horizontal span of s from
s + Km
50s
0 to 200 and a vertical span of v from 0 to 200, we have graphed v =
, so V = 50
s + 20
150s
and Km = 20, on the left below, while we gave graphed v =
, so V = 150 and
s+5
Km = 5, on the right below
(a) Calculating the horizontal asymptotes for the graphs, we get that the first one is
v = 50 and the second one is v = 150, so in each case the horizontal asymptote is
v =V.
(b) In each case, the value of s which makes v(s) =
V
is Km . This can be verified by
2
calculating:
v(Km ) =
V × Km
V × Km
V
=
= .
Km + K m
2Km
2
(c) If you graph v as a function of s, you can find the horizontal asymptote, and that
value of v will be V . Having found V , you can find where the graph crosses the
V
line v = . The value of s at that point is Km .
2
Chapter 5 Review Exercises
1. Homogeneity: In this case f is a power function with power k = 3.2. By the homogeneity property, if x is increased by a factor of t, then f is increased by a factor of t3.2 .
In this case, x is doubled, so t = 2. Therefore f is increased by a factor of 23.2 , that is,
by a factor of 9.19.
2. Power: From 1 to 10, 1 is increased by a factor of t = 10. Thus f is increased by a factor
of tk = 10k . Since f (10) is twice the size of f (1),
2 = tk = 10k .
Solving for k yields k = 0.30.
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Solution Guide for Chapter 5
3. Flow rate: The radius R is a power function with power 0.25, so if the flow rate F
changes by a factor of t, then R changes by a factor of t0.25 .
(a) If the flow rate is doubled, then t = 2, so the radius changes by a factor of 20.25 =
1.19.
(b) If the radius is tripled, then 3 = t0.25 . Solving for t yields t = 81, so the flow rate is
multiplied by 81.
4. Falling object:
(a) For Earth the formula is T = 0.25s0.5 . We put in s = 16 and find T = 0.25 × 160.5 =
1. Thus it takes 1 second for an object to fall 16 feet on Earth.
(b) We know that T = 5 when s = 20, so 5 = c × 200.5 . Thus c =
5
= 1.12.
200.5
(c) The time T is a power function with power 0.5. By the homogeneity property, if s
is increased by a factor of t, then T is increased by a factor of t0.5 . In this case, s is
multiplied by 4 in going from 10 feet to 40 feet, so t = 4. Therefore T is increased
by a factor of 40.5 , that is, by a factor of 2. Hence it takes an object 2×1 = 2 seconds
to fall 40 feet. This can also be found by first solving for c using the method of Part
(b) and then plugging s = 40 into the resulting formula.
5. Power formula: The slope of ln f as a linear function of ln x is the same as the power,
so the value of k is 5.
6. Modeling almost power data: To find the power model, we convert the data into linear data using the logarithm. To do this, we put the original data in the 3rd and 4th
columns, the logarithms in the 1st and 2nd columns, and then calculate the linear relation using linear regression. The regression calculation shows that ln f = −1.20 ln x +
1.252. Thus the power model uses k = −1.2 and c = e1.252 = 3.50, and so f = 3.5x−1.2 .
7. Gas cost:
(a) To find the power model, we convert the data into linear data using the logarithm.
To do this, we put the original data in the 3rd and 4th columns, the logarithms
in the 1st and 2nd columns, and then calculate the linear relation using linear
regression. The regression calculation shows that ln D = −1.00 ln g + 3.092. Thus
the power model uses k = −1 and c = e3.092 = 22.02, and so D = 22.02g −1 .
Chapter 5 Review Exercises
485
(b) The distance D is a power function with power −1. Now an increase of 50% in
the price of gas corresponds to multiplying g by 1.5 or 3/2. By the homogeneity
property, if g is multiplied by 1.5, then D is multiplied by 1.5−1 = (3/2)−1 = 2/3.
Thus the distance is multiplied by 2/3, or reduced by 33.33%.
(c) If a gallon of gas costs $1 then the distance can you drive on $1 worth of gas is the
distance you can drive on 1 gallon of gas, and that is your gas mileage. Thus we
want to find the value of D when g = 1. Now if we put g = 1 into the formula
from Part (a) we find D = 22.02 × 1−1 = 22.02. Thus your gas mileage is 22.02, or
about 22, miles per gallon. Another way to do this is to note that the gas mileage is
the product of the cost g with the distance D (check the units!), and by the formula
D = 22.02g −1 from Part (a) we have g × D = g × 22.02g −1 = 22.02.
8. Falling rocks:
(a) To find the power model, we convert the data into linear data using the logarithm.
To do this, we put the original data in the 3rd and 4th columns, the logarithms
in the 1st and 2nd columns, and then calculate the linear relation using linear
regression. The regression calculation shows that ln T = 0.50 ln s − 1.616. Thus the
power model uses k = 0.5 and c = e−1.616 = 0.20, and so T = 0.2s0.5 .
(b) We put s = 70 into the formula from Part (a) and find T = 0.2 × 700.5 = 1.67. Thus
it takes 1.67 seconds for a rock to fall 70 feet.
(c) We want to find the value of s for which T = 2, so (by the formula from Part (a))
we want to solve the equation 0.2s0.5 = 2 for s. Using a table of values or a graph
(or solving by hand), we find s = 100. Thus a rock falls 100 feet in 2 seconds.
(d) The time T is a power function with power 0.5. By the homogeneity property, if s
is multiplied by 2, then T is multiplied by 20.5 = 1.41. Thus the time is multiplied
by 1.41.
9. Composing functions: Because x(t) = t − 1 and y(x) = 3x2 + 5x, we have y(x(t)) =
3(t − 1)2 + 5(t − 1). Thus the formula is y = 3(t − 1)2 + 5(t − 1).
10. Decomposing functions: The balance is B = 120 × 1.02t , and this is an exponential
function. The initial value is 120, so the balance at the start of 2005 was $120. The
growth factor is 1.02, so the yearly percentage growth rate is 2%, and this is the annual
interest rate.
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Solution Guide for Chapter 5
11. Population growth:
(a) We show the graph below with a horizontal span of 0 to 200 and a vertical span of
0 to 0.015.
(b) Because the population level N is between 100 and 200, we use the formula
N
R = 0.02 1 −
.
200
Putting in N = 150 gives
150
R = 0.02 1 −
= 0.005.
200
Thus the per capita growth rate is 0.005 per year.
12. Volume:
(a) We put 1.5 + 0.1t in place of V in the formula for R and find
R=
3(1.5 + 0.1t)
4π
1/3
.
(b) We put t = 2 into the formula from Part (a) and find
R=
3(1.5 + 0.1 × 2)
4π
1/3
= 0.74.
Thus the radius is 0.74 inch.
13. Quadratic formula: The quadratic formula gives the solutions of ax2 + bx + c = 0 as
x=
−b ±
√
b2 − 4ac
.
2a
To solve 2x2 − x − 1 = 0, we use a = 2, b = −1, and c = −1. The solutions are
p
√
−(−1) ± (−1)2 − 4 × 2 × (−1)
1±3
1± 9
x=
=
.
=
4
4
2×2
This expression gives the values x = 1 and x = −1/2.
Chapter 5 Review Exercises
487
14. Quadratic regression: To find a quadratic model for this data set using regression,
we enter the data in the first two columns and use the calculator to find directly the
quadratic regression coefficients. This gives the quadratic model f (x) = −3x2 + 5x − 2.
15. Data that are not quadratic: To show that the data in the table are not exactly quadratic,
we calculate the second-order differences:
x
f (x)
First-order difference
Second-order difference
0
0
0
−6
1
0
−6
−12
2
−6
−18
−18
3
−24
−36
4
−60
Since the second-order differences are not equal, the data are not quadratic.
16. Chemical reaction:
(a) Quadratic regression gives the model R = 0.005x2 − 0.75x + 25.
(b) Using the model from Part (a), we find that R(24) = 0.005 × 242 − 0.75 × 24 + 25 =
9.88 moles per cubic meter per second. This means that the reaction rate at a
concentration of 24 moles per cubic meter is 9.88 moles per cubic meter per second.
(c) We want to find the value of x for which R = 6. We can do this using the crossinggraphs method or the quadratic formula. The result is s = 32.28 moles per cubic
meter. Thus the reaction rate is 6 moles per cubic meter per second at a concentration of 32.28 moles per cubic meter.
17. Cubic regression: To use cubic regression, we enter the data in the first two columns
and then use the calculator to determine the cubic regression coefficients. This gives
the cubic model y = 0.1x3 − 0.2x2 + 3.
2x − 5
are the values of x for which the denominator
x2 + 4x + 3
2
x + 4x + 3 is zero but the numerator 2x − 5 is not zero. In this case, x2 + 4x + 3 = 0
18. Finding poles: The poles of
when x = −1 and x = −3, so these are the poles.
19. Expanding balloon:
(a) Calculating the cubic regression coefficients, we find a cubic model for the data:
V = 0.03t3 + 0.50t2 + 2.51t + 4.19.
(b) The volume after 5 seconds is expressed in functional notation as V (5). Plugging
t = 5 into the model from Part (a) gives the value 32.99 cubic inches.
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Solution Guide for Chapter 5
20. Travel time:
(a) The graph is shown below. We used a horizontal span of 0 to 70 and a vertical
span of 0 to 50.
(b) The value of T (25) can be found by putting s = 25 into the formula given or by
using a table of values (or the graph). The result is T (25) = 4 hours. This means
that 4 hours is the time required to drive 100 miles if the average speed is 25 miles
per hour.
(c) At s tends towards 0, T increases sharply. In practical terms, as the average speed
becomes very slow, the time needed to drive 100 miles increases greatly.