Chapter 9
Ingredients of Multivariable Change:
Models, Graphs, Rates
Section 9.1 Multivariable Functions and Contour Graphs
1. a. P(1.2, s) is the profit in dollars from the sale of a yard of fabric as a function of s, the
selling price per yard, when the production cost is $1.20 per yard.
b. P(c, 4.5) is the profit in dollars from the sale of a yard of fabric as a function of c, the
production cost per yard, when the selling price is $4.50 per yard.
c. When the production cost is $1.20 per yard and the selling price is $4.50 per yard, the
profit is $3.00 for each yard sold.
d.
2. a. T(36,000, d) is the amount of income tax in dollars owed by a household as a function of d,
the number of dependents, when the adjusted income is $36,000.
b. T(i, 4) is the amount of income tax in dollars owed by a household as a function of i, the
adjusted income in dollars, when the number of dependents is 4.
c. When the adjusted income is $36,000 and the number of dependents is 4, the amount of
income tax owed by household is $10,000.
d.
3. a. P(100,000, m) is the probability of the senator voting in favor of a the bill as a function of
the amount m, in millions of dollars, invested by the tobacco industry lobbying against the
bill, when the senator receives 100,000 letters supporting the bill.
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512
Chapter 9: Ingredients of Multivariable Change: Models, Graphs, Rates
Calculus Concepts
b. P(l, 53) is the probability of the senator voting in favor of the bill as a function of l, the
number of letters supporting the bill received by the senator, when the tobacco industry
spends $53 million lobbying against the bill.
c.
4. a. N(25, s) is the number of skiers on a Saturday at a ski resort in Utah as function of s, the
number of inches of fresh snow that fell since the previous Saturday, when the price of an
all-day lift ticket is $25.
b. N(p, 6) is the number of skiers on a Saturday at a ski resort in Utah as function of p, the
price of an all day lift ticket, when six inches of fresh snow fell since the previous
Saturday.
c.
7.
5.
8.
6.
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9.
10. a.
b. Points on the contour curve correspond to possible combinations of the month and the
pressure that produce the mean monthly temperature at noon Greenwich Mean Time.
c. One possible answer: Because these data were collected in the Southern Hemisphere,
where December and January are summer months.
d. In general, temperatures are higher at 850 millibars than at 30 millibars. Higher air
pressure corresponds to elevations closer to the earth’s surface.
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Chapter 9: Ingredients of Multivariable Change: Models, Graphs, Rates
Calculus Concepts
11. a. There will be 11.97 hours of daylight.
b. There will be 12.31 hours of daylight.
c. Answers will vary. For example, if your college is at the 45th parallel north, it will receive
9.19 hours of daylight each day during January.
d.
12. a. There are 22 women with 2 daughters and 2 sons.
b. There are 2 women with 5 daughters. There are 5 women with 4 daughters and 1 son.
There are 21 women with 3 daughters and 2 sons. There 27 with 2 daughters and 3 sons.
There are 10 women with 1 daughter and 4 sons. There are 3 women with 5 sons. Thus
there are 68 women with 5 children.
c.
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Calculus Concepts
Section 9.1: Multivariable Functions and Contour Graphs
513
14
13.
15.
16.
120
100
80
60
40
20
0
0
2
4
6
8
17.
18.
19.a. P(c, s) = P(c, 100, 10, s) = 0.175c + 0.027s2 − 0.730s + 108.958 for a supermarket with s
thousand square feet of sales space and a customer base with a per capita income of $c
thousand
b.
10
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Chapter 9: Ingredients of Multivariable Change: Models, Graphs, Rates
20. a.
b.
Calculus Concepts
V(12, 32) = 0.002198(12)1.739925(32)1.133187 ≈ 8.42 cubic feet
c. If volume remains constant, the
height decreases as the dbh
increases.
21. a. For example, the BMI for a person who is 5’8” and 150 pounds is 22.8 points.
b.
c. Replace the output with a constant K and solve for one variable in terms of the
other. We choose to solve for w.
K=
w=
0.4536w
0.00064516h 2
K (0.00064516)h 2
pounds, where h is the height in inches and K is the BMI
0.4536
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Calculus Concepts
Section 9.1: Multivariable Functions and Contour Graphs
513
d.
This graph is the upside-down mirror
reflection of the graph in part b. It is
also more accurately drawn.
22. a.
b. Replace the output with a constant K and solve for one of the variables. We choose to
solve for p.
K=
15.44 g1.905
p1.247
Kp1.247 = 15.44 g1.905
p1.247 =
15.44 g1.905
K
1
 15.44 g1.905  1.247
p=
billion dollars


K


when the average air fare is p dollars and there are K thousand passengers
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Chapter 9: Ingredients of Multivariable Change: Models, Graphs, Rates
Calculus Concepts
c.
This graph is the upside-down mirror
reflection of the graph in part a. It is
also more accurately drawn.
23. a. Solve for one of the variables in 10.65 + 113
. w + 104
. s − 5.83ws = K (we show both).
s (1.04 − 5.83w) = K − 10.65 − 1.13w
s=
K − 10.65 − 1.13w
1.04 − 5.83w
w(1.13 − 5.83s) = K − 10.65 − 1.04s
or
w=
K − 10.65 − 1.04 s
1.13 − 5.83s
b.
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Calculus Concepts
Section 9.1: Multivariable Functions and Contour Graphs
515
24. b
a. D =
11.56P
, where K represents the carrying capacity of a wheat crop in people per hectare.
K
25. a. The center of the 184 thousand-contour curve corresponds to approximately 10 thousand
tons of lower fat cheese and 55 thousand tons of regular cheese since is approximately (10,
55). This point corresponds to a maximum because the contour curves decrease in every
direction away from the point.
b. The contour graph shows that the maximum revenue is greater than 184 million guilders
but less than 214 million guilders (or else we would see the 214-contour curve). From the
three-dimensional graph, it appears that the maximum revenue is near 200 million guilders.
One possible approximation is 190 million guilders.
26. a. Estimates will vary. The input values are r ≈ 35 milliliters per gram and t ≈ 10 minutes, and
the output value is approximately 38%. Thus, approximately 38% of pigment can be
removed by washing for 10 minutes in 35 milliliters of water per gram of sunflower heads
when the water temperature is 75°C.
b.
375 milliliters
= 25 milliliters per gram
15 grams
At (25, 10), the output is between 37 and 38, so we estimate that approximately 37.5% of
the pigment is removed.
c. The line t = 10 intersects the contour graph in two places, r ≈ 21 and r ≈ 42, so either 21 or
42 milliliters per gram is needed. Thus either (21)(9) = 189 or (42)(9) = 378 milliliters of
water is needed.
27. a. Estimates will vary. The input values are P ≈ 10 hours and H ≈ 70%, and the output value
is approximately 12.5 days. When C. grandis is exposed to 70% relative humidity and 10
hours of light, it takes approximately 12.5 days to develop.
b. Estimates will vary. The input values are P ≈ 10 hours and H ≈ 60%, and the output value
is approximately 11.5 days. When C. grandis is exposed to 60% relative humidity and 10
hours of light, it takes approximately 11.5 days to develop.
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Calculus Concepts
28. a. Estimates will vary. The input values are approximately 14 days and 20% moisture, and
the adhesiveness output value is approximately 5.6.
b. Estimates will vary. The input values are approximately 12 days and 17% moisture, and
the adhesiveness output value is approximately 7.
29. a. f(x, y) decreases when y increases because the contour curves have smaller numbers when
y increases from 2 and x is 1.5; therefore, f(x, y) increases when y decreases.
b. Because the contour curves are more closely spaced to the left of (2.5, 2.5) than they are
directly below that point, the function decreases more quickly as x decreases than it does as
y decreases.
c. The change is greater when (2, 2) shifts to (1, 2.5), causing the contour values to change
from approximately 21 to approximately 14, than it is when (1, 0) shifts to (4, 1), causing
virtually no change in contour values.
30. a. This question is difficult to answer from the contour curve. An examination of the threedimensional graph shows that the graph slopes upward from (10, 10) as s decreases. As w
increases from the point (10, 10) the graph slopes downward but then begins sloping
upward.
b. At (15, 2.5), K(s, w) will decrease more rapidly when w increases because the contour
curves are more closely spaced above that point than to the left of that point.
c. The change appears to be greater from (5, 5) to (6, 3) because the contour curves appear to
be more closely spaced in that direction.
31. a. The three dimensional graph has two peaks the approximately the same height one
around (0.7, 0, 0.25) and one around (−0.7, 0. 0.25) separated by a valley.
b. The point (0.4, 0.4) lies between the 0 and −0.05 contours. The point (0. 0.3) lies near the
−0.1 contour. Thus the descent is greater from (0.7, 0.1) to (0, 0.3).
c. Moving up from (0, 0.1), we encounter increasingly negative contours. Moving right from
(0, 0.1) we encounter increasingly positive contours. Thus the function output increases as
x increases from (0, 0.1).
d. The point (−0.2, −0.3) lies near the −0.05 contour. Thus we are looking for a point lying on
the −0.05 + 0.15 = 0.1 contour. There are infinitely many such points. Two possibilities are
(−0.7, 0.38) and (0.94, 0).
32.
a. If x remains constant and y increases, then we need to consider moving upward on a
vertical line. On any vertical line, we encounter increasingly positive contours as we move
up. Thus for constant x, the function output increases as y increases.
b. When x decreases by 12 to −1.5, y must increase by approximately 3 to 8. In other words,
the point (−1.5, 8) lies on approximately the same contour as the point (−1, 5). (Answers
will vary because of the difficulty of estimating the position of the 2.5 contour curve.)
c. When y increases from −2 to −0.5, x must increase by approximately 1.5 in order to remain
on the same contour. In other words, the point (−2, −2) lies on approximately the same
contour as the point (−0.5, −0.5).
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Calculus Concepts
Section 9.1: Multivariable Functions and Contour Graphs
517
33. a.
b.
34. a.
The temperature dropped 36ºF.
Possible optimal placements are shown on the graph below.
b. We estimate the average available wind power in mid-Texas to be 125 watts per square
meter and that in western Nebraska to be 275 watts per square meter. Thus, the difference
is approximately 150 watts per square meter.
35. To locate a relative maximum on a contour graph look for contour curves that form a simple,
closed curve (or if completed appear to form a closed curve). Estimate the point in the center of
this closed curve. Moving away from the “center” check that contour levels are decreasing. If
the levels are decreasing by a constant amount d, the maximum value at the point can estimated
between the largest indicated contour value k and k+d.
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Chapter 9: Ingredients of Multivariable Change: Models, Graphs, Rates
Calculus Concepts
On a table of data, locate a number whose eight surrounding numbers are all less than it. It may
help to spot this number if you first quickly sketch a couple of contour curves. Once you have
located a number that appears to be a maximum value, you describe its location by following
its column and row to the edges of the table to determine the input values that produce that
maximum.
Keep in mind that for a contour graph as well as a table, there may exist more than one relative
maximum.
36.
Given a starting point p on a three dimensional surface, the path of steepest descent is the path
that results in the greatest decrease in output over the shortest distance traveled on the surface.
(Using geography as an example, the course water chooses to run down a mountain is the path
of steepest descent.)
To sketch a path of steepest descent on a contour graph…
1) locate the starting point,
2) find the closest point on the next lower contour curve,
3) cross this contour curve perpendicularly, and repeat steps 2 and 3 until there are no
lower contours.
Your sketch may produce a straight line, but more often it will produce a curve and sometimes
a squiggly line. Keep in mind that the path of steepest descent will meet all intersecting
contour levels at right angles.
Section 9.2 Cross-sectional Models and Rates of Change
1. a.
b.
c.
air temperature
row
A( p, 95) = 0.0238 p 2 − 2.3455 p + 151.31 °F when the dew point is p°F.
2. a. dew point
b. column
c. A plot of apparent temperature against air temperature for 70ºF dew point reveals a linear
trend. A model for apparent temperature is A(70, t ) = 1.244t − 14.716 degrees Fahrenheit
when the air temperature is tºF.
8 of the sky is covered by clouds 60% of the time.
3. a. At least 10
b. The time of day is given as 9 A.M. so the variable to be held constant is hour of the day.
The input variable for the cross-sectional model is fraction of sky covered.
c. column
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Calculus Concepts
Section 9.2: Cross-sectional Models and Rates of Change
519
d. A plot of the frequency of cloud cover over Minneapolis in January against the fraction of
sky covered at 9:00 a.m. indicates that a linear model would be a good fit.
C (9, f ) = −1.651 f 3 + 2.686 f 2 −
1.597 f + 1.019
where f is the fraction (expressed as a decimal) of the sky covered by clouds.
4. a. Half is the same as
5
, so looking in across the row for input ≥ 5 /10 , we find the entry
10
corresponding to 3:00 P.M. The sky is at least half covered by clouds at 3:00 P.M. 64% of
the time.
b. Because time of day is being used as input, fraction of sky covered must be held constant.
c. row
d. Using hour of the day (aligned as 0,3,6,9,12,14,18,21) as input and the top row, labeled
overcast, (aligned as 45, 48, 47, 45, 41, 39, 41, 38) as output, a cubic model for the
frequency of an overcast sky in Minneapolis in January is
f (t ) = 0.005t 3 − 0.171t 2 + 1.056t + 45.485 percent t hours after midnight.
5. a. no; In order to model monthly payments for a 52-month loan, the table must include a
column for a 52-month loan given different interest rates. The table shown does not
include such a column.
b. yes; row; In order to model monthly payments for a loan at 9%, the table must include a
row for 9% interest given different lifetimes. The table shown includes such a row.
c. A possible cross-sectional model for monthly payments for a $1000 loan at 9% is
p (m,9) = 0.017 m 2 − 2.091m + 85.977 dollars when m months is the length of the loan
(m between 24 and 60). Payments on a 52 month loan will be
p (52,9) = 0.017(52)2 − 2.091(52) + 85.97 ≈ 22.65 dollars each month.
6. a. yes; column; In order to model monthly payments for a 42-month loan, the table must have
a column for 42-months. This table has such a column.
b. no; In order to model monthly payments for a loan at 10.5% the table must have a row for
10.5%. This table does not have such a row.
c. Using interest rate as input and the column for 42-months as output, we observe a plot of
the data to have a linear trend. A model for the monthly payment of a 42-month $1,000
loan is p (42, r ) = 0.468r + 23.646 dollars when r% is the interest rate. Using this model,
we calculate monthly payments for a loan at 10.5% to be
p (42,10.5) = 0.468(10.5) + 23.646 ≈ 28.56 dollars.
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7. a. Using price per pound above $1.50 as input and the data in the row labeled 4 as output, we
choose a quadratic as a possible model. A model for the per capita consumption of peaches
by people living in households with $40,000 yearly income is
C ( p, 4) = 0.893 p 2 − 1.304 p + 7.811 pounds per person per year where p and 1.50
dollars per pound is the price of peaches.
b. When peaches sell for $1.55 per pound, p = 1.55 − 1.50 = 0.05 . The per capita
consumption at this price by someone in a household with $40,000 yearly income is
C (0.05, 4) = 0.893(0.05)2 − 1.304(0.05) + 7.811 ≈ 7.7 pounds per person per year.
8. a. First, $1.80 − $1.50 = $0.30 , so we must consider the column labeled 0.30. using $ yearly
income (aligned as in table) as input and the entries from the 0.30 column as output, we
note that a plot of the data has the characteristics of a natural logarithmic model. A model
for per capita consumption of peaches at $1.80 per pound is
C (0.30, h) = 4.696 + 2.025ln h pounds per person per year where h is the household
yearly income in tens of thousands of dollars.
b. A yearly household income of $35,000 is represented by h = 3.5 . The consumption of
peaches by someone in a household with a $35,000 income when peaches are sold for
$1.80 is C (0.30, 3.5) = 4.696 + 2.025ln(3.5) ≈ 7.2 pounds per person per year.
9. a. Since there is a column for $1.00 coke products but not a row for 90 cent Pepsi products,
we must hold the cost of Coke products constant.
b. We use the column for $1.00 Coke products.
c. Using cost of Pepsi products (aligned to cents, i.e. 50, 75, 100, 125, 150) as input and the
data from the $1.00 Coke products column as output, we obtain the following linear model
for daily sales of Coke products: S (100, p ) = 1.96 p + 25 cans when Pepsi products are
sold for p cents per can. Using this model to estimate the number of Coke products sold at
$1.00 when Pepsi products sell for $0.90, we calculate S (100,90) = 1.96(90) + 25 ≈ 201
cans.
10. a. Since the table does not have a column for $1.40 Coke products, but does have a row for
$0.50 Pepsi products, we hold the cost of Pepsi products constant at $0.50.
b. The cross-section is represented by the $0.50 Pepsi product row.
c. use the cost of Coke products in cents (i.e., 50, 75, 100, 125, 150) as input and the data
from the row labeled $0.50. A plot of the data indicates a quadratic model is appropriate.
Daily sales of Coke products when the cost of Pepsi products is 50 cents can be modeled as
S (c,50) = −0.005c 2 + 0.067c + 165.8 cans when Coke products are sold for c cents per
can. Using this model, we estimate daily sales of coke products to be
S (140, 50) = −0.005(140) 2 + 0.067(140) + 165.8 ≈ 78 cans when Coke products cost
$1.40 per can.
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Calculus Concepts
Section 9.2: Cross-sectional Models and Rates of Change
521
11. a. 3.88 million people
b. See figure on text page A-71.
c. We find a cross-sectional model
− 0.0506a + 5.036
when 15 ≤ a ≤ 30

million people of age a
P (2005, a) = 
2
− 0.0064a + 0.548a − 7.240 when 30 ≤ a ≤ 50
in 2005.
∂
P (2005, a )
∂a
= −0.0065 ⋅ 2(35) + 0.548 ≈ 0.1 million people per year of age a.
a = 35
d. We find a cross-sectional model
(
)
P ( y, 50) = − 1.778 ⋅ 10 −4 y 3 + 0.004 y 2 + 0.106 y + 2.405 million people of age 50 y years
after 1990.
∂
P ( y, 50)
∂y
y = 30
people per year
(
)(
)
= − 1.778 ⋅ 10 − 4 3 ⋅ (30) 2 + 0.004(2 ⋅ (30) ) + 0.106 ≈ −0.2 million
e. The answer to part c shows that moving down off the 400-contour curve at (2005, 35) will
result in a an increase of output. The answer to part d shows that moving right off the 400contour curve at (2020, 50) will result in a decrease of output.
12. a. 4.48 million people
b. Using year (aligned to years after 1990) as input at the row for 20 years of age as output,
we obtain the following cubic model for U.S. population (of a given age):
P ( y, 20) = −0.0003 y 3 + 0.0128 y 2 − 0.123 y + 3.988 million people of age 20 y years
∂
P ( y, 20) = −0.0082 y 2 + 0.026 y − 0.123 million people per year
∂y
evaluated at y = 20 (i.e. 2010) yields approximately 0.06 million people per year. The
after 1990.
population of 20-year-olds will be increasing by approximately 60 thousand people per
year in 2010.
c. No; The people who are 20-years-old in 2010 will be 21-years-old in 2011. This change
would show up as a diagonal move on the table. The rate of change in part b estimates a
horizontal move on the table and estimates how many more 19-year-olds (in 2010) will be
turning 20 than 20-year-olds who will be turning 21.
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Calculus Concepts
13. a. -0.35 kg/day (loss of 0.35 kilograms per day)
b. W(68,t) represents the cross-sectional model of the first column in the table.
W (68, t ) = −0.00009t 3 + 0.003t 2 − 0.013t + 0.572 kg/day where tºC is the temperature.
A pig weighing 68 kg will (on average) gain/loose W(68,t) kilograms during a day when
the temperature is tºC.
c. The graph of W(68,t) is concave up, increasing until approximately t = 10 . After t = 10 ,
the graph is concave down. It reaches a maximum just after t = 21.1 . A pig weighing
68 kilograms will gain the most weight when the temperature is near 21.1ºC. For
temperatures in excess of 21.1ºC, the weight gain diminishes more and more rapidly until
weight is actually lost at temperatures in excess of approximately 37.5ºC.
14. a. A plot of the average daily weight gain/loss against the air temperature for a 68-kg pig
indicates that a quadratic model would be a good fit.
W(68, t) = −0.0024t2 + 0.0880t + 0.1377 kg/day when the temperature is t °C.
b.
dW (68, t )
= −0.0048t + 0.0880 kg per day per °C
dt
dW (68,26.7)
= −0.0048(26.7) + 0.0880 ≈ −0.04 kg per day per °C
dt
When the temperature is 26.7°C, the rate at which a 68-kilogram pig is gaining weight is
decreasing by approximately 0.04 kilogram per day per °C. In other words, if the
temperature is increased to 27.7°C, then the amount of weight the pig will gain each day
decreases by approximately 0.04 kilogram.
When t = 26.7,
c. A plot of the average daily weight gain/losses in the column corresponding to an air
temperature of 26.7°C indicates that a linear model would be a good fit.
W(w, 26.7) = −0.0031w + 1.0347 kg/day when the mean live weight is w kg.
d.
15. a.
b.
dW ( w,26.7)
= −0.0031 kg per day per kg
dw
dW (68,26.7)
When w = 68,
= −0.0031 kg per day per kg
dw
When the temperature is 26.7°C, the amount of weight a 68-kilogram pig gains each day is
changing by approximately -0.003 kilograms per kilogram. In other words, if the pig gains
1 kilogram, then the rate at which it is gaining weight will decrease by approximately
0.003 kilograms per day.
Approximately 0.52° F per ° F
First, determine a cross-sectional model–
A ( p, 85) = 0.0194 p 2 − 2.0052 p + 135.832° F when the dew point is p°F . Second,
determine the derivative of this model
d ( p,85)
= 0.0388 p − 2.0052 Third, evaluate the
dp
derivative for p = 65° F
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Calculus Concepts
Section 9.2: Cross-sectional Models and Rates of Change
523
16. a. Approximately 1.15ºF per ºF.
First determine a cross-sectional model-- A(65, t ) = 1.147t − 9.114 ºF when the air
temperature is tºF. Second, determine the derivative of this model
∂ (65, t )
= 1.147
∂t
degrees per degree. The rate of change of A(65, t ) is constant for all t.
b. While the units are the same, they are describing different types of temperatures.
17. a. A(14,000, r) = 14,000r2 + 28,000r + 14,000 dollars
where 100%r is the annual percentage yield (i.e., r is in decimals).
b.
dA(14,000, r )
= 28,000r + 28,000 dollars per 100 percentage points
dr
dA(14,000, 0.127)
= 28,000(0.127) + 28,000 = $31,556 / 100 percentage points
dr
c. A(14,000, r) = 1.4r2 + 280r + 1400 dollars, where r% is the annual percentage yield
The magnitudes of the coefficients are reduced in this model, the r 2 coefficient by a factor
of 10,000, the r coefficient by a factor of 100, and the constant by a factor of 10.
d.
dA(14,000, r )
= 2.8r + 280 dollars per percentage point
dr
dA(14,000, 12.7)
= 2.8(12.7) + 280 = $315.56 per percentage point
dr
The derivative tells us approximately how much the output will change when the input
increases by one unit. If the input is a percentage expressed as a decimal, then an increase
in one unit corresponds to 100 percentage points. For example, if r = 0.127 and is
increased by 1 to r = 1.127 the corresponding percentages are 12.7% and 112.7%.
This answer is equivalent to the one found in part b.
18. a. A quadratic model is A(10, r) = 102,803.26r2 + 6348.82r + 1064.99 dollars,
where 100%r is the annual percentage yield (i.e., r is in decimals).
b.
dA(10, r )
= 205,606.52r + 6348.82 dollars per 100 percentage points
dr
dA(10, 0.07)
= 205,606.52(0.07) + 6348.82 ≈ $20,741 per 100 percentage points
dr
A $1000 investment after 10 years in an account paying 7% is increasing at rate of
approximately $20,741 per 100 percentage points increase in the interest rate. The partial
rate of change appears to be 100 times what we expect because it is with respect to 100
percentage points not 1 percentage point.
c. A quadratic model is A(10, r) = 10.2803r2 + 63.4882r + 1064.99 dollars, where r% is the
annual percentage yield.
d.
dA(10, r )
= 20.5606r + 63.4882 dollars per percentage point
dr
dA(10, 7)
= 20.5606(7) + 63.4882 ≈ $207.41 per percentage point
dr
524
Chapter 9: Ingredients of Multivariable Change: Models, Graphs, Rates
Calculus Concepts
A $1000 investment after 10 years in an account paying 7% is increasing at rate of
approximately $207.41 per percentage point increase in the interest rate. This is the same
as the rate found in part b.
19.
Derivatives of cross-sectional models of a three-dimensional function can be used to pinpoint optimal points on those cross-sections. These optimal points in turn may e used to
estimate the point at which a critical point may appear on the three-dimensional function.
20.
When using cross-sections to analyze a three-dimensional function, we restrict ourselves to
the points shown on the cross-section. We may be missing the true critical points by not
looking at the cross-sections that contain them.
Section 9.3 Partial Rates of Change
1.
∂W
pounds per inch
∂h
2.
∂P
dollars per karat
∂k
3.
∂T
°F per degree of latitude
∂t g = 23
4.
∂T
dollars per dollar made
∂i d = 4
5.
∂R
dollars per cow for c = 100
∂c b = 2
6.
∂G
grade point average per point on SAT for s = 1048
∂s h = 3.5
7. a.
∂P
is the rate of change of the probability that the senator will vote for the bill
∂m l = 100,000
with respect to the amount spent by the tobacco industry on lobbying when the senator
receives 100,000 letters in opposition to the bill. We expect this rate of change to be
negative because if the number of letters is constant but lobbying funding against the bill
increases, the probability that the senator votes for the bill is likely to decline.
b.
∂P
is the rate of change of the probability that the senator votes for the bill with
∂l m = 53
respect to the number of letters received when $53 million is spent on lobbying efforts. We
expect this rate of chance to be positive because if the number of letters increases (while
Copyright © Houghton Mifflin Company. All rights reserved.
Calculus Concepts
Section 9.3: Partial Rates of Change
525
lobbying funding remains constant) the probability of the senator voting in favor of the bill
is likely to increase.
8. a.
b.
9. a.
∂N
is the rate of change of the number of skiers on a Saturday at a ski resort in Utah
∂s p = 25
with respect to the number of inches of fresh snow received since the previous Saturday
when the price of an all-day lift ticket is $25. We expect this rate of change to be positive
because if the number the number of inches of fresh snow received increases, the number
of skiers is likely to increase.
∂N
is the rate of change of the number of skiers on a Saturday at a ski resort in Utah
∂p s = 6
with respect to the price of an all-day lift ticket when the number of inches of fresh snow
received since the previous Saturday is 6 inches. We expect this rate of change to be
negative because if the price of an all-day lift ticket increases, the number of skiers is
likely to decrease.
∂f
= 3(2) x + 5 y (1) + 0
∂x
= 6 x + 5x
∂f
= 0 + 5x(1) + 2(3y 2 )
∂
y
b.
= 5x + 6y 2
c.
∂f
∂x
= 6x + 5(7)
y =7
= 6x + 35
∂g
= 3k 2 m5 − 2(1)m
10. a. ∂k
= 3k 2 m5 − 2m
b.
c.
∂g
= k 3 (5m 4 ) − 2k(1)
∂m
= 5k 3m 4 − 2k
∂g
= 5(2)3 m 4 − 2(2)
∂m k = 2
= 40m 4 − 4
∂f
= 5(3x 2 ) + 3(2x)y 3 + 9(1)y + 14(1) + 0
11. a. ∂x
= 15x 2 + 6xy 3 +9y + 14
526
Chapter 9: Ingredients of Multivariable Change: Models, Graphs, Rates
Calculus Concepts
∂f
= 0 + 3x 2 (3y 2 ) + 9x(1) + 0 + 0
b. ∂y
= 9x 2 y 2 + 9x
∂f
= 15x 2 + 6x(2) 3 +9(2) + 14
c. ∂x y =2
= 15x 2 + 48x + 32
∂K
12. a. ∂a
= 5(1)b3 + 0
= 5b3
∂K
= 5a(3b 2 ) + 7(1.4 b ) ln(1.4)
b. ∂b
= 15ab 2 + 7(1.45 ) ln(1.4)
∂K
= 15(6)b 2 + 7(1.45 ) ln(1.4)
c. ∂b a =6
= 90b2 + 7(1.45 ) ln(1.4)
13. a.
s
 1
M t = s  + 0 + 0 =
 t
t
b.
M s = ln t + 3.75
c.
M s t = 3 = ln 3 + 3.75
14. a.
g x = 2(3.2 yz 2 ) x + y (2.9) x y −1 + 0 = 6.4 xyz 2 + (2.9 y ) x y −1
b.
g y = 3.2 x 2 z 2 + ln y (2.9) x y + 0 = 3.2 x 2 z 2 + (2.9ln y ) x y
c.
g z = 2(3.2 x 2 y ) z + 0 + 1 = 6.4 x 2 yz + 1
15. a.
∂h 1
1
= − 2( st − tr ) 1 (t − 0) = − 2t ( st − tr )
∂s t
t
b.
∂h
1
−s 1
= −1( s) t − 2 + − 2( st − tr ) 1 ( s − r ) =
+ − 2( st − tr )( s − r )
∂t
r
t2 r
c.
∂h
−t
= 0 + −1(t )r − 2 − 2( st − tr ) 1 (0 − t ) =
+ 2t ( st − tr )
∂r
r2
d.
∂h
−2
=
+ 2(2)[1(2) − 2( −1)] = −2 + 4(4) = 14
∂r ( s, t , r ) = (1, 2, −1) ( −1) 2
16. a.
 1 
∂k
x
= x
+ ln( x + y )
 + (1) ln( x + y ) =
∂x
x+y
 x + y
Copyright © Houghton Mifflin Company. All rights reserved.
Calculus Concepts
Section 9.3: Partial Rates of Change
b.
 1 
∂k
x
= x
=
∂y
 x + y x + y
c.
∂k
3
=
∂y x = 3 3 + y
d.
∂k
3
3
=
=
∂y ( x , y ) = (3,2) 3 + 2 5
17. a.
b.
f x = 2 y + 8(2 x) y 3 + 0 + 0
= 2 y + 16 xy 3
f y = 2 x + 8 x 2 (3 y 2 ) + 5(ln e 2 )e 2 y + 0
= 2 x + 24 x 2 y 2 + 10e 2 y
note: ln e 2 = 2
f xx = 0 + 16 y 3
= 16 y 3
f xy = 2 + 16 x(3 y 2 )
c.
= 2 + 48 xy 2
f yx = 2 + 24(2 x) y 2 + 0
= 2 + 48 xy 2
f yy = 0 + 24 x 2 92 y ) + 10(ln e2 )e2 y
= 48 x 2 y + 20e 2 y
 16 y 3
d. 
2
 2 + 48 xy
18.
a.
2 + 48 xy 2 

48 x 2 y + 20e 2 y 
∂G
= ln r + 12r (7t 6 ) + 0 − r
∂t
= ln r + 84rt 6 − r
∂G
1
= t   + 12t 7 − 4(ln 8)8r + 0
∂r
r
b.
t
= + 12t 7 − 4(ln 8)8r − t
r
527
528
Chapter 9: Ingredients of Multivariable Change: Models, Graphs, Rates
Grr = −
t
+ 0 − 4(ln 8)(ln 8)8r + 0
2
r
t
− 4(ln 8)2 8r
2
r
1
Grt = + 84t 6 + 0 − 1
r
1
c.
= + 84t 6 − 1
r
Gtt = 0 + 84r (6t 5 ) + 0
=
note:
Calculus Concepts
t
= tr −1
r
d t 
t
−2
so
  = −1tr = − 2
dr  r 
r
= 504rt 5
Gtr =
1
+ 84t 6 − 1
r
 t
2 r
 − r 2 − 4(ln 8) 8
d. 
 1 + 84t 6 − 1
 r
x
y
−1
 − 2y
x
x3
19. 

−1 1
y 2 + 2
x
 y
1 

y
x2 

2x 

y 3 
2
x
x
20.
y
x  4e 2 x − 3 y
21. 

y − 6e 2 x −3 y

+
y
 324(3x − y + 4)


-54(3x-y+4)
x
1

+ 84t 6 − 1
r

5
504rt 

−108(3x − y + 4) 


18(3x − y + 4) 
y
− 6e 2 x −3 y 


2 x−3y 
9e

Copyright © Houghton Mifflin Company. All rights reserved.
Calculus Concepts
Section 9.3: Partial Rates of Change
x
x
22.
 − y 2 x −2


 2yx −1
y
529
y
2yx −1 


2 ln(x) 
23.a. A(14,000, r ) = 14,000(1 + r ) 2 dollars. This function gives the value of an investment after 2
years when the APY is 100r%.
b.
∂A
dA(14,000, r )
=
= 28,000(1 + r ) dollars per 100 percentage points
∂r P= 14,000
dr
∂A
= $31,556 per 100 percentage points
∂r ( P , r ) = (14,000, 0.127)
The value of an investment of $14,000 after 2 years in an account with A.P.R. of 12.7%, is
increasing by $315.56 per additional percentage point.
The slope of the line tangent to a graph of
A(14,000, r) at r = 0.127 is $31,566 per
100 percentage points.
c.
(Note: It is difficult to distinguish the
graph of A from the tangent line in this
figure because the graph of A appears
nearly linear in this close-up view. This is
an illustration of the principle of local
linearity discussed in Chapter 1.)
A(10, r ) = 1000e10r dollars is the investment worth of $1000 after 10 years when the
interest rate is 100r%
∂A
dA(10, r )
b.
=
= 10,000e10r dollars per 100 percentage points
∂r t = 10
dr
24. a.
∂A
= 10,000e10( 0.07) ≈ $20,137.5 per 100 percentage points
∂r ( t , r ) = (10, 0.07)
A $1000 investment after 10 years in an account paying 7% is increasing at rate of
approximately $20,138 per 100 percentage points increase in the interest rate.
530
Chapter 9: Ingredients of Multivariable Change: Models, Graphs, Rates
Calculus Concepts
c. The answer to part b is the slope of a line
tangent to the graph of the cross-sectional
function A(10, r) at r = 0.07.
25. a. A cross-sectional model for the amount of UVA radiation with respect to the latitude for
the month of March is R(March, l) = −0.0073l2 − 0.0346l + 49.7492 watts/m2/month where
l is the number of degrees of latitude north of the equator (negative values of l correspond
to locations south of the equator).
An alternative model a sine model: R(March, l) = 35.9104 sin(−0.0232l + 1.5088) +
15.5463 watts/m2/month where l is the number of degrees of latitude north of the equator.
A cross-sectional model for the amount of UVA radiation with respect to the month at 50°
south of the equator is R(m, −50) = 27.1052sin(0.4846m + 1.7070) + 32.9106
watts/m2/month where m is the number of months since the beginning of the year.
An alternative model is the piecewise continuous model
 0.389m3 − 3.321m 2 − 3.425m + 61.857
when 0 ≤ m < 7

watts per square meter per month

R (m, −50 ) = 
3
2
 −0.565m + 16.008m − 137.689m + 381.183 when 7 ≤ m ≤ 12

watts per square meter per month

b. Quadratic model:
∂R
= −0.0146l − 0.0342 watts/m2/month/degree of latitude
∂l m = March
where l is the number of degrees of latitude north of the equator.
∂R
= −0.0146(−50) − 0.0342 ≈ 0.70 watt/m2/month/degree of latitude
∂l ( m,l ) = (March,−50)
Sine model:
∂R
= 35.2303( −0.0236) cos( −0.0236l + 15069
.
) watts/m2/month/degree of
∂l m = March
latitude where l is the number of degrees of latitude north of the equator.
∂R
= 35.2303( −0.0236) cos( −0.0236( −50) + 15069
.
)
∂l ( m,l ) = (March,−50)
≈ 0.75 watt/m2/month/degree of latitude
Copyright © Houghton Mifflin Company. All rights reserved.
Calculus Concepts
Section 9.3: Partial Rates of Change
531
c. Sine model:
∂R
= 27.1052(0.4846)cos(0.4846m + 1.7070) watts/m2/month/month where m is the
∂m l = −50
number of degrees of latitude north of the equator.
∂R
= 27.1052(0.4846) cos(0.4846(3) + 1.7070)
∂m ( m,l ) = (3,−50)
≈ −13.13 watts/m2/month/month
Piecewise continuous model:
∂R
∂m l
= −50
 3(0.389)m 2 − 2(3.321)m − 3.425
when 0 ≤ m < 7

2
 watts per square meter per month
=
2
3(−0.565)m + 2(16.008)m − 137.689 when 7 < m ≤ 12

2
 watts per square meter per month
∂R
= 3(0.389)32 − 2(3.321)3 − 3.425 ≈ −12.85 watts/m2/month/month
∂m ( m,l ) = (3,−50)
d.
26. a. A possible cross-sectional model for sales in January is
S(1, y) = 17.2492y3 −352.4066y2 + 2569.7029y + 6101.0808 cars sold in the month of
January y years after 1960.
A possible cross-sectional model for sales in June is
S(6, y) = 6.4458y4 −132.8399y3 + 764.6789y2 +41.6033y + 13,635.3054 cars sold in the
month of June y years after 1960.
b. For January:
∂S
∂y
=
m = January
January 1965:
dS (1, y )
= 51.7476 y 2 − 704.8132 y + 2569.7029 cars per year in the
dy
month of January y years after 1960
∂S
≈ 339 cars per year
∂y ( m, y ) = (January,5)
532
Chapter 9: Ingredients of Multivariable Change: Models, Graphs, Rates
Calculus Concepts
For June:
∂S
∂y
=
m = June
June 1965:
dS (6, y )
= 25.7832 y 3 − 398.5197 y 2 + 1529.3578 y + 41.6033 cars per
dy
year in the month of June y years after 1960
∂S
≈ 948 cars per year
∂y ( m, y ) = (June, 5)
∂S
≈ 1367 cars per year. In
∂y ( m, y ) = (January, 2)
January 1962, car sales are increasing at a rate of approximately 1367 cars per year.
c. Using the first equation found in part b,
27. a.
A logarithmic model is C(0.2, i) = 4.805 + 2.002 ln i peaches per person, at a price
of $1.70 per pound when the yearly income is i tens of thousands of dollars.
∂C
= 0.67 peaches per person per ten thousand dollars
∂i (3)
b. An exponential model is C(p,3) = 7.177(0.896 p ) peaches per person, at a yearly income
of $30,000 and a price per pound of (p +1.5) $.
∂C
= −0.77 peaches per person per $ per pound
∂p (0.2)
c. Using the provided model to answer part a results in
∂C
≈ 0.8 pound/person/year per
∂i thousand dollars of income
∂C
Using the provided model to answer part b results in ∂p ≈ −0.82 pound/person/year per
dollar per pound
d. Parts (a) and (b) were easier to use since they were single variable functions. I would
expect that part (c) was more accurate since we were given the appropriate model.
28. a. Part a answer: The needed cross-sectional model is
S (c,1.25) = −50.286c 2 + 7.771c + 312.6 cans sold when the price of Coke products is $c
∂S
dS (c, 125
. )
=
= −100.571c + 7.771 cans sold per dollar of Coke price
∂c p = 1.25
dc
∂S
= −100.571(0.75) + 7.771 ≈ −67.7 cans sold per dollar of Coke price
∂c ( c, p) = ( 0.75, 1.25)
Copyright © Houghton Mifflin Company. All rights reserved.
Calculus Concepts
b.
Section 9.3: Partial Rates of Change
533
∂S
= 196.42 − 1.04c cans sold per dollar of Pepsi price
∂p
∂S
= 196.42 − 104
. (0.75) = 195.64 cans sold per dollar of Pepsi price
∂p ( c, p) = ( 0.75,1.25)
c. Using the provided equation ,
∂S
= −67
∂c (c, p) = (0.75, 1.25)
∂S
= 195.64
∂p (c,p)=(0.75,1.25)
d. Parts (a) and (b) were easier to use since they were single variable functions. I would
expect that part (c) was more accurate since we were given the appropriate model.
e.
∂S
= −100.4(1.30) + 9.6 − 1.04(120
. ) ≈ −122 cans sold/dollar of Coke price
∂c ( c, p) = (1.30,1.20)
When the price of Pepsi products is $1.20 and the price of Coke products is $1.30, if the
price of Coke products increases by $1 while the price of Pepsi products remains constant,
then the daily sale of Coke products will decline by approximately 122 cans.
∂S
= 196.42 − 104
. (1.30) = 195 cans sold per dollar of Pepsi price
∂p ( c, p) = (1.30,1.20)
When the price of Pepsi products is $1.20 and the price of Coke products is $1.30, if the
price of Pepsi products increases by $1 while the price of Coke products remains constant,
then the daily sale of Coke products will increase by approximately 195 cans.
29. a.
∂H
= (10.45 + 10 v − v)(−1) = −10.45 − 10 v + v kilogram-calories per square meter
∂t
per hour per degree Celsius
∂H
 5

= ( 12 v − 1)(33 − t ) = (33 − t ) 
− 1 kilogram-calories per square meter per
∂v
 v

hour per meter per second
−1
2
b.
∂H
should be positive because an increase in wind speed (when temperature is constant)
∂v
should increase heat loss.
c.
∂H
 5

= (33 − 12) 
− 1 ≈ 2.48 kilogram-calories per square meter per
∂v ( v , t ) = ( 20, 12)
 20 
hour per meter per second
534
Chapter 9: Ingredients of Multivariable Change: Models, Graphs, Rates
Calculus Concepts
d.
∂H
should be negative because an increase in temperature (when wind speed is constant)
∂t
should decrease heat loss.
e.
∂H
∂t
30. a.
( v , t ) = (20, 12)
= −10.45 − 10 20 + 20 ≈ −35.17 kilogram-calories per square meter
per hour per degree Celsius
∂K 9.25
=
animals/hectare per kilogram/hectare/year
∂P
A
b.
∂K
9.25
=
≈ 0.000145 animal/hectare per kilogram/hectare/year
∂P ( P , A) = (15,000, 64,000) 64,000
c.
∂K − 9.25 P
=
animals/hectare per megajoule/animal
∂A
A2
d.
∂K
−9.25(15,000)
=
64,0002
∂ A ( P , A) = (15,000, 64,000)
≈ −0.000034 animal/hectare per megajoule/animal
e.
P
 0
18.5 P
− 9.25
K PP = 0, K AA =
, K AP = K PA =
; The matrix is P 
3
2
A
A
A  − 9.225
 A
A
− 9.25 
A2 
18.5 P 
A3 
K PP is zero, which means that as P increases, the rate of change of K with respect to P is
constant. K AA is positive for positive values of P and A, which means that as A increases,
the rate of change of K with respect to A is increasing. (Note that we consider P and A to
be only positive.) The mixed second partials are always negative, which means that as P
increases, the rate of change of K with respect to A is decreasing or that as A increases, the
rate of change of K with respect to P is decreasing.
31. a. We expect food intake to increase as either milk production or size increases. Therefore,
we expect both partial derivatives to be positive.
b.
∂I
= −1.244 + 0.1794 s + 0.21491m kilograms per day per unit of size index
∂s
This equation is the rate of change of the amount of organic matter eaten with respect to
the size of the cow (when the amount of milk produced is constant).
∂I
= −0.20988 + 0.071894m + 0.214915s kilograms per day per kilogram of milk
∂m
This equation is the rate of change of the amount of organic matter eaten with respect to
the amount of milk produced (when the size of the cow is constant).
c.
∂I
= −0.20988 + 0.071894(6) + 0.214915(2) ≈ 0.65 kilogram per day per
∂m ( s , m) = (2, 6)
kilogram of milk per day
Copyright © Houghton Mifflin Company. All rights reserved.
Calculus Concepts
d.
e.
∂I
∂s
( s , m ) = (2, 6)
∂2 I
∂s 2
= 0.1794;
Section 9.3: Partial Rates of Change
535
= −1.244 + 0.1794(2) + 0.21491(6) ≈ 0.40 kilogram per day per
unit of size index
∂2 I
= 0.071894;
∂m 2
s
m
s  01794
.
0.214915
m 0.214915 0.071894
∂2 I
∂2 I
=
= 0.214915
∂m∂s ∂s∂m
Because all second partials are positive, we know that the rates of change in the s and m
directions increase as both s and m increase. This indicates that the surface is concave up
in the s and m directions.
32. a.
b.
∂A
= 4.26 − 0.1g units per percentage point of glucose and maltose
∂g
∂A
= 5.69 − 0.28m − 0.07 h units per percentage point of moisture
∂m
∂A
= 0.67 − 0.06 s units per percentage point of seed
∂s
∂A
= 2.48 − 0.1h − 0.07 m units per day
∂h
∂A
= 4.26 − 0.1g units of adhesiveness measure per percentage point of glucose/maltose
∂g
c. You need to know values of m and h.
d.
Agg = −01
. , Agm = 0 , Ags = 0 , Agh = 0
Amg = 0 , Amm = −0.28 , Ams = 0 , Amh = −0.07
Asg = 0 , Asm = 0 , Ass = −0.06 , Ash = 0
Ahg = 0 , Ahm = −0.07 , Ahs = 0 , Ahh = −0.1
The second partials matrix is
g
m
s
h
0
0
0 
 −0.1

−0.28
0
−0.07 
m  0
s  0
0
−0.06
0 


h  0
−0.07
0
−0.1 
g
536
Chapter 9: Ingredients of Multivariable Change: Models, Graphs, Rates
33. a.
Calculus Concepts
∂A
∂A
= 1000re rt dollars per year and
= 1000te rt dollars per 100 percentage points
∂t
∂r
t
r
b. The second partials matrix is t  1000r 2e rt
1000e rt (rt + 1)  . Note that we use the


2 rt
r 1000ert (rt + 1)

1000
t
e

Product Rule to find the mixed partials. Evaluating this matrix at t = 30 and r = 0.047, we
9871.25 
 9.05
have 
.
9871.25 3,686,359.86 
When a $1000 investment has been earning 4.7% compounded continuously for 30 years,
(1) the rate at which the amount is growing with respect to time is increasing with respect
to time by $9.05 per year per year.
(2) the rate at which the amount is growing with respect to time is increasing with respect
to the interest rate by $9871.25 per year per 100 percentage points.
(3) the rate at which the amount is growing with respect to the interest rate is increasing
with respect to time by $9871.25 per 100 percentage points per year.
(4) the rate at which the amount is growing with respect to the interest rate is increasing
with respect to the rate by $3,686,359.86 per 100 percentage points per 100 percentage
points.
34. a.
b.
A P = (1 + r ) 2 dollars per dollar of principal
Ar = 2 P(1 + r ) dollars per 100 percentage points
A PP = 0 dollars per (dollar of principal)2 ; Arr = 2 P dollars per (100 percentage points)2
A Pr = ArP = 2(1 + r ) dollars per dollar invested per 100 percentage points
For P = 10,000 and r = 0.09:
A PP = 0 ; The rate at which the amount is changing with respect to the amount invested is
constant.
Arr = 20,000 dollars per (100 percentage points); When the investment is $10,000 at 9%
APY, the rate at which the investment is growing as the interest rate grows is increasing by
$20,000 per 100 percentage points. In other words, if the interest rate is increased to 10%,
the rate of change of the amount with respect to the interest rate will increase by
approximately $20,000 per 100 percentage points.
A Pr = ArP = 2(1 + r ) =$2.18 per dollar invested per 100 percentage points.
When the investment is $10,000 at 9% APY, the rate of change of the amount with respect
to the amount invested will increased by approximately $2.18 per dollar invested if the
interest rate increases to 10%; or the rate of change of the amount with respect to the
interest rate will increase by approximately $2.18 per 100 percentage points if the amount
invested increases to $10,001.
P
r
2.18  .
The second partials matrix for P = 10,000 and r = 0.09 is P  0

r 2.18 20,000
Copyright © Houghton Mifflin Company. All rights reserved.
Calculus Concepts
35. a.
b.
c.
Section 9.3: Partial Rates of Change
∂A
= (1 + r ) t ln(1 + r ) million dollars per year
∂t
∂A
= t (1 + r ) t −1 million dollars per 100 percentage points
∂r
When t = 5 and r = 0.15,
∂A
≈ 0.28 million dollars per year
∂t
d.
36. a.
b.
c.
d.
−n
∂A 12  
r 
=
1 −  1 +   dollars of loan amount per dollar of monthly payment
∂m r   12  


When r = 0.12, n = 180, and m = 500,
∂A
≈ $83.32 per dollar of the monthly payment.
∂m
(
)
r
12m ln 1 + 12
∂A 12m 
r − n ln 1 + r ( −1)  =
r
=
0
−
1
+
1 + 12

12
12
∂n
r 
r
when the period is n months.
∂A
When r = 0.11, n = 36, and m = 250,
≈ $179.18 per month
∂n
(
)
(
)
(
)
−n
dollars per month
537
538
Chapter 9: Ingredients of Multivariable Change: Models, Graphs, Rates
Calculus Concepts
37. At a relative maximum, the partial derivatives must be equal to zero. At the instant of the
high point, the output is neither increasing nor decreasing regardless of the input direction.
38. It does not matter whether you hold the variable constant before or after the derivative is
taken. According to the constant multiplier rule, the constant portion will just be rewritten
when the derivative is taken. Therefore, it doesn’t matter it the constant is substituted before
or after.
Section 9.4 Compensating for Change
1.
∂f
∂f
= 30 xy 3 , = 45 x 2 y 2
∂x
∂y
(
2.
)
dc − f k −c(ln 39)(39k )
=
=
= −c(ln 39)
dk
fc
(39 k )
2 2
dx − f y − 45 x y
−3x
=
=
=
3
dy
fx
2y
30 xy
3.
∂g 59.3
∂g
+49n ,
=
= 49m
m
∂m
∂n
4.
∂A
∂A
= 83.2te rt ,
= 83.2 re rt
∂r
∂t
dr − At −83.2re rt − r
=
=
=
dt
Ar
t
83.2te rt
dm − g n
−49m
=
=
59.3
dn
gm
+49n
m
5.
∂S
∂S
= (39k ) ,
= c(ln 39)(39 k )
∂c
∂k
∂g
∂g
= 1.05y
= x(ln1.05)1.05y
∂x
∂y
− gy
dx
=
= − x ln1.05
dy
gx
When y = 5 and g(x, y) = 100, x =
Alternatively,
100
dx
 100 
≈ 78.35 , so
= −
ln1.05 ≈ −3.82 .
5
5 
1.05
dy
 1.05 
dy
1
≈
≈ −0.26 .
dx −3.82
dx
≈ −3.82
dy
dy
≈ −0.26
dx
Copyright © Houghton Mifflin Company. All rights reserved.
Calculus Concepts
6.
Section 9.4: Compensating for Change
539
∂f
∂f
s
 2
= ln(2t ) ,
= s  − 1.34e −1.34t = − 1.34e −1.34t
 2t 
∂s
∂t
t
−1.34t
s
− s + 1.34e −1.34t
ds − ft − t − 1.34e
=
=
= t
ln(2t )
ln(2t )
dt
fs
−1.34(3.6675)
−1
ds 3.6675 + 1.34e
When s = 1 and f(s, t) = 2, t ≈ 3.6675, so
≈
≈ −0.132 .
dt
ln(2 ⋅ 3.6675)
dt
≈ −7.582 .
Alternatively,
ds
)
(
dt
≈ −7.582
ds
ds
≈ −0.132
dt
7.
∂f
= 5.6ab3 − 3.6a
∂a
∂f
= 8.4a 2 b 2 + 12
∂b
da − f b −(8.4a 2b2 + 12)
=
=
db
fa
5.6ab3 − 3.6a
When b = 0.9 and f(a, b) = 15, a ≈ 4.173, so
da
≈ −64.82 .
db
Alternatively,
db
≈ −0.015
da
da
≈ −64.82
db
540
8.
Chapter 9: Ingredients of Multivariable Change: Models, Graphs, Rates
Calculus Concepts
∂f
∂f
= −20n(3.67 − m)e−0.2m − 2n(3.67 − m)2 e −0.2m ,
= 10(3.67 − m)2 e−0.2 m
∂m
∂n
dm − f n
−10(3.67 − m)2 e−0.2m
5(3.67 − m)
5(3.67 − m)
=
=
=
=
−
0.2
m
2
−
0.2
m
dn
f m −20n(3.67 − m)e
10n + n(3.67 − m) n(13.67 − m)
− 2n(3.67 − m) e
When n = 20 and f(m, n) = 80, m ≈ 2.83061 or m ≈ 4.6799, so
dm
dm
≈ 0.019 or
≈ −0.028 .
dn
dn
dm
≈ −0.028
dn
or
dm
≈ 0.019
dn
dn
dn
≈ 51.654 or
≈ −35.608
dm
dm
9. f(2, 1) = 21
∂f
∂f
= 6m + 2n , = 2m + 10n
∂m
∂n
dm − f m − ( 2m + 10n )
=
=
dn
fn
6m + 2n
dm −14
dn
=
= −1 and ∆n ≈
∆m = (−1)(0.2) = −0.2.
dn 14
dm
The value of n should decrease be approximately 0.2 in order to compensate for an increase of
0.2 in m.
When m = 2, n = 1, and ∆m = 0.2,
10. f(4.2, 3.7) ≈ 459,696.4256
∂f
= (96h 2 + 30h − 10)(43k + 15)
∂h
∂f
= (32h 3 + 15h 2 − 10h + 47)(43)
∂k
When h = 4.2, k = 3.7, and ∆k = 0.6,
dh − f k
=
≈ −0.36041 and
dk
fh
dh
( ∆k ) ≈ ( −0.36041)(0.6) ≈ −0.2162.
dk
The value of h should decrease be approximately 0.2162 in order to compensate for an increase
of 0.6 in k.
∆h ≈
Copyright © Houghton Mifflin Company. All rights reserved.
Calculus Concepts
Section 9.4: Compensating for Change
541
11. f(3.5, 1148) ≈ 3.7217
∂f
∂f
= 0.00091s[0.103(ln 2.505)(2.505h )],
= 0.00091[0.103(2.5h ) + 1]
∂h
∂s
ds − f h
When h = 3.5, s = 1148, and ∆h = −0.5,
=
≈ −758.2811 and
dh
fs
ds
∆h ≈ ( −758.2811)( −0.5) ≈ 379.14.
dh
The input s should increase by approximately 379.14 in order to compensate for a decrease of
0.05 in h.
∆s ≈
12. W(10, 60) ≈ 581,004
∂W
∂W
= 2.8r 2 (ln1.08)1.08h +59r(.6h-3.3),
= 5.6r (1.08h ) + 59(.3h 2 − 3.3h + 72)
∂h
∂r
dh −Wr
When h = 60, r = 10, and ∆r = −1.3,
=
≈ −3.058 and
dr Wh
dh
( ∆r ) ≈ ( −3.058)( −1.3) ≈ 3.975.
dr
The value of h should increase by approximately 3.975 in order to compensate for a decrease
of 1.3 in r.
∆h ≈
13. a. A(6, 250) ≈ 7.16
The average cost is approximately $7.16.
b.
c.
∂A
= (−0.02c 2 + 0.35c + 0.99)(ln 0.99897)(0.99897 n )
∂n
When c = 6 and n = 250,
∂A
= (−0.02c 2 + 0.35c + 0.99)(ln 0.99897)(0.99897 n ) ≈ −0.001888
∂n
the average cost is changing at a rate of approximately −$0.002 per shirt.
∂A
= (−0.04c + 0.35)(0.99897 n ) + 0.46
∂c
shirts per color
dn − Ac
−[(−0.04c + 0.35)(0.99897 n ) + 0.46]
=
=
2
n
dc
An
(−0.02c + 0.35c + 0.99)(ln 0.99897)(0.99897 )
dn
We expect
to be positive because if the number of colors increases, the order size
dc
would also need to increase to keep average cost constant.
dn
≈ 450 shirts per color. For each additional color, the order
dc
size would need to increase by approximately 450 shirts. Similarly, if the number of colors
decreases by 1, the order size could decrease by approximately 450 shirts and the average
cost would remain constant.
d. When c = 4 and n = 500,
14. a. A(90, 0.85) ≈ 107.26; The apparent temperature is approximately 107°F.
542
Chapter 9: Ingredients of Multivariable Change: Models, Graphs, Rates
Calculus Concepts
dh − At −(0.885 + 1.20h)
∂A
∂A
per F
= 0.885 + 1.20h,
= −78.7 + 1.20t ;
=
=
dt
Ah
∂t
∂h
−78.7 + 1.20t
b.
We expect dh to be negative because the humidity should decrease when the temperature
dt
increases in order to keep the apparent temperature constant.
c.
dh
≈ −0.06502 and
dt
dh
∆h ≈
(∆t) ≈ (−0.06502)(2) ≈ −0.13.
dt
The relative humidity should decrease by approximately 13 percentage points.
When t = 90, h = 0.85, and ∆t = 2,
dh
≈ −0.06502 and
dt
dh
∆h ≈
(∆t) ≈ (−0.06502)(−3.5) ≈ 0.23.
dt
The relative humidity should increase by approximately 23 percentage points.
d. When t = 90, h = 0.85, and ∆t = −3.5,
15. a, b.
c.
∂p
∂p
= −9.6544 + 0.14736t ,
= 1.9836 − 0.05916r
∂t
∂r
dt − pr −(1.9836 − 0.05916r )
=
=
°C per milliliter
dr
pt
−9.6544 + 0.14736t
d. We can solve for r in p(86.5, r) = 53 to get r ≈ 23.125 or r ≈ 43.934.
dt
When t = 86.5 and r = 23.125,
≈ −0.199 °C per milliliter
dr
dt
When t = 86.5 and r = 43.934,
≈ 0.199 °C per milliliter
dr
It is also possible to calculate the slope formula as
dr − pt −(−9.6544 + 0.14736t )
=
=
milliliter per °C and the two specific slopes as
dt
pr
1.9836 − 0.05916r
follows:
dr
≈ −5 milliliter per °C
When t = 86.5 and r = 23.125,
dt
dt
≈ 5 milliliter per °C
When t = 86.5 and r = 43.934,
dr
To illustrate the tangent lines whose slopes these values represent, we must draw the
parabola sideways with one tangent line on top and one on the bottom.
Copyright © Houghton Mifflin Company. All rights reserved.
Calculus Concepts
Section 9.4: Compensating for Change
543
16. a. V(24, 40) = 58.88 cubic feet.
b.
c.
∂V
2d
∂V
363.676d 2 h −2
=
,
=
∂d 0.691 + 363.676h −1 ∂h (0.691 + 363.676h −1 )2
2d


−

−1
−
1
dh −Vd
h  = −0.691 − 363.676h
=
=  0.691 + 363.676
dd Vh
363.676d 2 h −2
181.838dh −2
(0.691 + 363.676h −1 ) 2
dh
When d = 24 and h = 40,
≈ −3.59 feet per dbh
dd
d. If dbh is 25.5 inches, then ∆d = 1.5 inches.
dh
∆h ≈
(∆d ) ≈ (−3.59)(1.5) ≈ −5.4 ft, so the height of the tree is approximately 40 – 5.4 =
dd
34.6 ft.
If dbh is 23 inches, then ∆d = −1 inches.
dh
∆h ≈
(∆d ) ≈ (−3.59)(−1) ≈ 3.6 ft, so the height of the tree is approximately 40 + 3.6 =
dd
43.6 ft.
e. The answer to part c is the slope of the line drawn in part .
17. a. B(67, 129) ≈ 20.044 points
Solving B(h, w) = 20.044 for w, we get w =
b.
(20.044)(0.00064516)h 2
pounds
0.45
dw 2(20.044)(0.00064516)h
=
dh
0.45
dw 2(20.044)(0.00064516)(67)
When h = 67,
=
≈ 3.85 pounds per inch
dh
0.45
c. The answer to part b agrees with the answer given in Example 1.
544
18.
Chapter 9: Ingredients of Multivariable Change: Models, Graphs, Rates
Calculus Concepts
∂A
∂A
= 0.6416(0.425)w −0.575h 0.725,
= 0.6416(0.725)w 0.425 h −0.275
∂w
∂h
∂A
−
When w = 130, h = 71, and ∆h = 2, ∆w = ∂h ∆h ≈ −6.25 pounds.
∂A
∂w
The person must lose approximately 6.25 pounds in order for the skin surface area to remain
constant if the person grows 2 inches.
19. a. m(10,000, 5) ≈ $193.31
b.
∂m
0.005
∂m
0.005 A
=
,
=
(ln 0.9419)0.9419t
2
t
t
∂A 1 − 0.9419
∂t (1 − 0.9419 )
− mt
∆t ≈ −$2212.65
When A = 10,000, t = 5, and ∆t = −1, ∆ A =
mA
The amount you could borrow is approximately $10,000 − $2212.65 = $7787.35.
20. a,b.
7000
c. Using the cross-sectional models f (3.5, t ) = −836.2t + 22,538.156 students and
f (a,18.5) = 1122.873a + 3074.555 students, where t is the tuition in thousands of dollars
and a is the average financial aid per student in thousands of dollars, we estimate that the
tuition needs to change by
 −1122.873

− fa
thousand tuition dollars per  (−$0.5 thousand in aid)
∆a ≈ 

−
836.2
ft
thousand dollars of aid


≈ −$0.7 thousand = − $700
21. a. From the table, we estimate that Coke would need to lower its prices by more than $0.50
per can.
Copyright © Houghton Mifflin Company. All rights reserved.
Calculus Concepts
Section 9.4: Compensating for Change
545
b. Using the equations S (1.00, P) = 196 P + 25 cans of Coke products and
S (c,1.25) = −50.286C 2 + 7.771C + 312.6 cans of Coke products where $P is the price of
Pepsi products and $C is the price of Coke products, we estimate the change in Coke prices
−S
 −196 
as ∆C ≈ P ∆P = 
 (−0.25) ≈ −$0.53 . Coke would need to lower its price from
SC
 −92.8 
$1.00 a can to approximately $0.47 a can.
c. Rather than lower its prices so drastically, Coke should probably consider such alternatives
as more advertising on campus.
24. a.
To remain on the 7.4
pounds per person per
year contour, the price
will need to decrease by
approximately 40 cents
per pound.
b. Using the cross-sectional models f (0.5, i ) = 4.605 + 2.002ln i pounds per person per year
and f ( p, 4) = 0.893 p 2 − 1.304 p + 7.811 pounds per person per year, where i is the yearly
income in tens of thousands of dollars and p is the price per pound in dollars over $1.50,
we estimate that the price of peaches needs to change by
− fi
 −0.5005

∆i ≈ 
dollars per ten thousand dollars of income  (−$0.5 thousand) ≈ −$0.61
fp
 −0.4107

The price will need to decrease by approximately 61 cents per pound. (Note that models
may vary and that no continuous model is a good fit for the cross section with i = 4.)
c.
2
and C p = −(ln 2.718)(2.718− p ) . At i = 4 and p = 0.49,
i
−Ci
−0.5
(∆p) ≈
(−0.5) ≈ −$0.41 .
Cp
−0.6126
Ci =
The price will need to decrease by approximately 41 cents per pound.
d. One possible answer: The method in part b is the most time consuming and the least
accurate because of the poor fit of one of the models. The method in part a is the fastest,
and it produces an excellent approximation. The method in part c produces an
approximation as accurate as the model given. Because the partials in part c are easily
obtained, it is significantly less time consuming than the method in part b.
23.
fx
represents the opposite of the slopes of contour curves along the contour graph of f(x, y)
fy
where y is graphed along the vertical axis and x is graphed along the horizontal axis.
546
Chapter 9: Ingredients of Multivariable Change: Models, Graphs, Rates
Calculus Concepts
Chapter 9 Concept Review
1. a.
b. C(3, 71) = 2.9 cm
When apples are stored for 3 months and blanched at 71°C, the applesauce flows 2.9 cm
down a vertical surface in 30 seconds.
c. Find a model for the consistometer value as function of storage time, using storage time as
input and the values in the 35°C column as output. A piecewise continuous model is
 −0.05s 2 + 0.35s + 3 cm when 0 ≤ s < 2
C ( s,35) = 
2
 −0.05s + 0.15s + 3.4 cm when 2 ≤ s ≤ 4
2 wks
4.3 wks/month
≈ 0.465 month
Assuming that 1 month is 4.3 weeks, 2 weeks corresponds to s ≈
C(0.465, 35) ≈ 3.2 centimeters
d.
∂C(4,t)
is the rate of change of the consistometer value with respect to the blanching
∂t
temperature when the storage time is 4 months.
e. Find a model for the consistometer value as function of the blanching temperature, using
temperature as the input and the values in the row corresponding to a 4-month storage time
as the output. A quadratic model is C (4, t ) = (6.9444 ⋅ 10−4 )t 2 − 0.0869t + 5.4124
centimeters for 35 ≤ t ≤ 83 where t is the blanching temperature in degrees Celsius.
∂C(4,t)
= 2(6.9444 ⋅10 −4 )t − 0.0869 cm per °C
∂t
∂C(4, 45)
= 2(6.9444⋅10 −4 )(45) − 0.0869 ≈ −0.0244 cm per °C
∂t
When the storage time is a constant 4 months and the blanching temperature is 45°C, the
consistometer value is decreasing by approximately 0.024 cm per °C. That is, if the
blanching temperature is increased to 46°C, the consistometer value should decrease by
approximately 0.024 cm.
Copyright © Houghton Mifflin Company. All rights reserved.
Calculus Concepts
Chapter 9: Concept Review
547
f.
2. a. E(24, 60) ≈ 466 eggs
A female insect kept at 24°C and 60% relative humidity will lay approximately 466 eggs in
30 days.
b.
c.
∂E
= 23.1412 − 0.1874h − 0.4023t
∂h
∂E
When t = 27°C and h = 77%,
= 23.1412 − 0.1874(77) − 0.4023(27) ≈ 2.2 eggs per
∂h
percentage point of humidity. When the temperature is held constant at 27°C and the
relative humidity is 77%, the number of eggs is decreasing by approximately 2.2 eggs per
percentage point of relative humidity. That is, if the relative humidity were increased to
78%, the number of eggs would decrease by approximately 2.
∂E
= 299.7038 −10.4420t − 0.4023h
∂t
dh − Et −(299.7038 − 10.4420t − 0.4023h)
=
=
percentage points per °C
dt
Eh
23.1412 − 0.1874h − 0.4023t
d. When h = 63% and t = 25°C,
dh
≈ −10.4 percentage points per °C.
dt
dh
∆t ≈ (−10.4)(−0.5) = 5.2 percentage points
dt
The humidity should increase by approximately 5.2%.
∆h ≈
e. On the contour curve corresponding to the egg production for t =25°C and h = 63% (the
490.1601 egg contour curve), the slope of the tangent line at that point is −10.4 percentage
points per °C. The h value on the tangent line when t = 24.5°C is h ≈ 63 + 5.2 = 68.2%.
This is an approximation to the value of h that corresponds to t = 24.5°C on the 490.1601
egg contour curve.
3. a. Inputs are t ≈ 26 °C and h ≈ 56%. The output is approximately 485 eggs.
b. An increase in temperature by 3°C will result in a greater change in the number of eggs
laid.
c. The number of eggs laid will decrease more rapidly when the temperature decreases than
when the humidity decreases.
d. By sketching a tangent line on Figure 9.42 and calculating its slope, you should obtain an
estimate of approximately 13 percentage points per ºC.
548
Chapter 9: Ingredients of Multivariable Change: Models, Graphs, Rates
e.
Calculus Concepts
dh − Et −24.1532
=
=
≈ −12.9 percentage points per C
dt
Eh
1.8672
When the temperature is 24ºC and the humidity is 62%, the change in humidity needed to
compensate for a small change in temperature (so that the number of eggs remains
constant) can be estimated as ∆h ≈ (−12.9)∆t percentage points.
∂E
= 23.1412 − 0.1874h − 0.4023t
∂h
∂E
= 299.7038 −10.4420t − 0.4023h
∂t
Ett = −10.4420 , Eth = −0.4023 , Ehh = −0.1874 , and Eht = −0.4023
t
4. From Activity 2 we have
h
The second partials matrix for any values of h and t is t  −10.442 −0.4023 .
h  −0.4023 −0.1874 
Copyright © Houghton Mifflin Company. All rights reserved.