Math 1101 Questions and answers

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Sample Test Answers for Chapter 5 1

SAMPLE TEST ANSWERS FOR CHAPTER 5

1. (a) The following graph is illustrative.

Answer Key

1. (a) Since the population is believed to be growing exponentially, we calculate the regression line for the natural logarithm of the data: ln N = 0.139t + 2.996. Since r is the slope of this regression line, r = 0.139.

(b) The doubling time t

rd = = =

ln ln .

. 2 2

0 139 4 99

days. (c) The data in the table doubles exactly every 5 days.

2. (a) Answers will vary. The following graph is illustrative.

(b) In practical terms, N(5) is the number of inhabitants (in thousands) 5 years after the state was settled. N(5) = 47.24 thousand inhabitants. (c) i. The carrying capacity for the population in this environment is the limiting value for N. Using a graph or table, we see that the limiting value is 62 thousand inhabitants. ii. The population growing the fastest when the population equals one-half of the

carrying capacity (since it is logistic). Using the graph, N = =

62 2

31 when t = 3.32

years after the state was settled. iii. The graph is concave up from t = 0 until t = 3.32. In practical terms, the population was increasing at an increasing rate during that interval.

3. (a) Answers will vary. The following graph is illustrative.

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(b) In practical terms, N (5) is the number of inhabitants (in thousands) 5 years after

the state was settled. N (5) = 67.2

0.1 + 0.55 = 47.24 thousand inhabitants.

value for N . Using a graph or table, we see that the limiting value is 62 thou-

sand inhabitants.

ii. The population growing the fastest when the population equals one-half of

the carrying capacity (since it is logistic). Using the graph, N = 62 2

= 31,

when t = 3.32 years after the state was settled.

iii. The graph is concave up from t = 0 until t = 3.32. In practical terms, the

population was increasing at an increasing rate during that interval.

2. (a) The following graph is illustrative

(b) According to a graph or table, the carrying capacity for the population is 184 million people. Since the current population is over 280 million, the model does not seem accurate. (c) The U.S. population was growing the fastest when it reaches one-half of 184 million, so when N = 92. This occurred when t = 140, so in the year 1780 + 140 = 1920. (d) If another logistic model has the population growing the fastest at 150 million, then the carrying capacity for the population under this new model is twice 150, so 300 million people.

4. The graph below shows the survivorship curve. Since the graph is more or less a straight line, it is a Type II curve and represents a nearly constant mortality rate.

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(b) According to a graph or table, the carrying capacity for the population is 184 mil-

lion people. Since the current population is over 300 million, the model does not

seem accurate.

2 Sample Test Answers for Chapter 5

lion, so when N = 92. This occured when t = 140, so in the year 1780+140 = 1920.

(d) If another logistic model has the population growing the fastest is 150 million,

then the carrying capacity for the population under this new model is twice 150,

so 300 million people.

3. We calculate differences to identify linear data and ratios to identify exponential data.

Table A The ratios are all 1.16, to two decimal places, so this is exponential data and so

f = P at. The initial data, P , is 6.7, and the growth factor, a, is 1.16, so the formula

is f = 6.7 × 1.16t.

Table B Neither differences nor ratios are constant, so we hope this is power data. Taking

the natural logarithms of both t and g, we find the regression line ln g = 1.160 ln t+

2.041 is a excellent fit, so it is power data for which k = 1.16 and c = e2.041 = 7.7.

The formula is g = 7.7t1.16.

Table C The differences are all 1.73, so this is linear and so h = mt + b. The initial data, b,

is 5.8, and the slope, m, is 1.73, so the formula is h = 1.73t + 5.8.

4. We calculate differences to identify linear data and ratios to identify exponential data.

(a) i. The function f is exponential since all the ratios are 1.1.

ii. Since the ratios are 1.1, a = 1.1. The initial value is 4.55, calculated as 5/1.1, so

f = 4.55 × 1.1t.

(b) i. The function g is power. Neither differences nor ratios are constant, so it is

neither linear nor exponential. It is power since its natural logarithms lie on

the line ln g = 0.138 ln t + 1.609.

ii. A formula for g is ctk where k = 0.138 and c = e1.609 = 5, so g = 5t0.138.

ii. The formula for h is h = 0.5t + 4.5 since the slope is 0.5 and the initial value is

h(1) − m = 5 − 0.5 = 4.5.

5. (a) Since the tax is a linear function of income with slope 0.12, for each additional

dollar of income, the tax you owe increases by 0.12 dollar. If your income increases

by $150, you will owe 150 × 0.12 = 18 dollars more tax.

(b) The decade growth factor is 1.0510 = 1.63, so that is the factor be which the circu-

lation will increase over a decade.

Sample Test Answers for Chapter 5 3

the speed is a power function with power k = 0.3.

6. Since weight W is a power function of its length L, W = cLk for some values of c and

(a) If one lizard is twice as long as a second and the first weighs 3 times as much as

the second, then 3 = 2k. Solving, we find that the power is k = 1.58.

(b) If one lizard is 4 times as long as a second, then their weights are related by a factor

of 4k. Using the value of k from Part (a), the longer lizard is 4k = 41.58 = 8.94 times

as heavy as the shorter.

factor of t where 2 = tk. Using the value of k from Part (a), 2 = t1.58. Solving, we

see that the heavier lizard is t = 1.55 times as long as the lighter.

7. Since terminal velocity T is proportional to the square root of length L, T is a power

function of L with power k = 0.5.

(a) The length of a 4-foot monkey is 4/3 = 1.33 times that of a 3-foot monkey, so the

terminal velocity of a 4-foot monkey is 1.33k = 1.330.5 = 1.15 times that of a 3-foot

monkey.

(b) Since the terminal velocity of a 4-foot monkey is twice that of a certain smaller

mammal with a similar shape, the length of the monkey is t times that of the

smaller mammal where 2 = tk = t0.5, so t = 4. Since the monkey is 4 feet long, the

smaller mammal is 4/4 = 1 foot long.

miles per hour, 98 = c × 40.5, so c = 49 and so the formula is T = 49L0.5. For the

cat, L = 1, so T = 49. This also follows immediately from Part (b).

8. Since the distance D is a power function of the period P , D = cP k.

(a) If the period of one satellite is twice that of another, then its distance is 2k times as

far, so if the distance is 1.59 times, then 2k = 1.59 and so the power is k = 0.67.

(b) If the period of one satellite is 3 times that of a second, then their distances from

the center of the Earth differ by a factor of 3k. Using the value of k from Part (a),

they differ by a factor of 30.67 = 2.09.

then their periods differ by a factor of t where tk = 2. Using the value of k from

Part (a), t0.67 = 2 and so t = 2.81. Their periods differ by a factor of 2.81.

4 Sample Test Answers for Chapter 5

9. (a) To find a formula that models W as a power function of L, we take the natural

logarithms of L and W and find the linear regression line ln W = 3.092 ln L−5.094.

Thus W = cLk where k = 3.09 and c = e−5.094 = 0.0061, so W = 0.0061L3.09.

(b) If one fish is twice as long as another, then their weights differ by a factor of 2k =

23.09 = 8.51.

where 2 = tk = t3.09, so a factor of t = 1.25.

10. (a) The graph is show below. (The horizontal span is 1.45 to 3.38 and the vertical span

is −0.24 to 6.28.) Since the graph of ln P versus ln v is very close to linear, it is

reasonable to model P as a power function of v.

5. Since terminal velocity T is proportional to the square root of length L, T is a power function of L with power k = 0.5. (a) The length of a 4-foot monkey is 4/3 times that of a 3-foot monkey, so the terminal velocity of a 4-foot monkey is (4/3)k = (4/3)0.5 = 1.15 times that of a 3-foot monkey. (b) Since the terminal velocity of a 4-foot monkey is twice that of a certain smaller mammal with a similar shape, the length of the monkey is t times that of the smaller mammal where 2 = tk = t0.5. Thus t = 4. Since the monkey is 4 feet long, the smaller mammal is 4/4 = 1 foot long. (c) For some c, we have T = cL0.5. Since the terminal velocity of a 4-foot monkey is about 98 miles per hour, 98 = c × 40.5, so c = 49. Thus the formula is T = 49L0.5. For the cat, L = 1, so T = 49. This also follows immediately from Part (b).

6. Since the distance D is a power function of the period P, we have D = cPk. (a) If the period of one satellite is twice that of another, then its distance is 2k times as far; so, if the distance is 1.59 times, then 2k = 1.59. Thus the power is k = 0.67. (b) If the period of one satellite is 3 times that of a second, then their distances from the center of the Earth differ by a factor of 3k. Using the value of k from Part (a), they differ by a factor of 30.67 = 2.09. (c) If the distance from the center of the Earth of one satellite is twice that of a second, then their periods differ by a factor of t where tk = 2. Using the value of k from Part (a), we have t0.67 = 2, and so t = 2.81. Their periods differ by a factor of 2.81.

7. (a) To find a formula that models W as a power function of L, we take the natural logarithms of L and W and find the linear regression line ln W = 3.092 ln L – 5.094. Thus W = cLk where k = 3.09 and c = e–5.094 = 0.0061, so W = 0.0061L3.09 (b) If one fish is twice as long as another, then their weights differ by a factor of 2k = 23.09 = 8.51. (c) If one fish is twice as heavy as another, then their lengths differ by a factor of t where 2 = tk = t3.09, so a factor of t = 1.25.

8. (a) The graph is shown below. (The horizontal span is 1.45 to 3.38 and the vertical span is from –0.24 to 6.28.) Since the graph of ln P versus lnυ is very close to linear, it is reasonable to model P as a power function of υ.

(b) Using linear regression on the natural logarithms, we find that ln . ln . .P = −3 024 4 269υ A power model of P as a function of υ is P = cvk where k = 3.024 and c = e–4.269 = 0.014, so the formula is P = 0 014 3 02. ..υ (c) If υ = 35, then P = 0.014 × 353.02 = 644.49 watts.

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(b) Using linear regression on the natural logarithms, we find that ln P = 3.024 ln v −

4.269. A power model of P as a function of v is P = cvk where k = 3.024 and

c = e−4.269 = 0.014, so the formula is P = 0.014v3.02.

(d) To generate 41 watts, we need a wind velocity v for which 41 = P , that is, 41 =

0.014v3.02. Solving, we find that v = 14.06 miles per hour.

(e) If wind speed increases by a factor of 3, then the power generated increases by a

factor of 3k = 33.02 = 27.60.

11. (a) Since S = 4πr2, √ S

4π =

√ 4πr2

4π = √

r2 = r.

(b) Using the formula for V and the expression for r from Part (a),

V = 4 3 πr3 =

4 3 π

(√ S

4π

)3 .

12. (a) Since W is proportional to the cube of L, W = cL3.

Sample Test Answers for Chapter 5 5

(b) Using the formula for W from Part (a) and the given expression for L, function

composition gives

W = cL3 = c(8(1 − e−t))3,

which is a formula expressing weight as a function of time.

and so c = 17/((8(1 − e−3))3) = 0.039.

13. (a) The limiting value is the maximum length, 13 inches.

(b) Since L = 13 − D and L(0) = 0.5, the initial value of D is D(0) = 13 − L(0) =

13 − 0.5 = 12.5.

or decay factor a, and so L = 13 − D = 13 − 12.5at. Now L = 8 when t = 3, so

8 = 13 − 12.5a3. Solving, we find that a = 0.74, so D = 12.5 × 0.74t, which is a

formula for D in terms of t.

(d) To find a formula for L in terms of t, we use our formulas above: L = 13 − D =

13 − 12.5 × 0.74t.

(e) Since W = 0.01L3 and L = 13−D = 13−12.5×0.74t, function composition gives

W = 0.01L3 = 0.01(13 − 12.5 × 0.74t)3.

14. (a) The limiting value of R is the total voting-age population, which is 40,000.

(b) By Part (a), R = 40,000 − D, so R(0) = 40,000 − D(0). Since the initial value of R

is 28,000, we have 28,000 = 40,000 − D(0) and so the initial value of D is 12,000.

growth or decay factor a. When t = 1, R = 33,000 so D = 7,000, and therefore

7,000 = D = 12,000a1. Solving, a = 0.58 and so D = 12,000 × 0.58t.

(d) Since R = 40,000−D and D = 12,000×0.58t, we use function composition to find

a formula for R in terms of t:

R = 40,000 − D = 40,000 − 12,000 × 0.58t.

15. To determine if data are quadratic, we see if the second-order differences are constant.

Table A The first-order differences are −2, 4, 12, and 16 and so the second-order differences

are 6, 8 , and 4, so the data in Table A are not quadratic.

6 Sample Test Answers for Chapter 5

Table B The first-order differences are −2, 4, 10, and 16 and so the second-order differences

are 6, 6 , and 6, so the data in Table A are quadratic, moreover the second-order

difference is 6, the initial first-order difference is −2, and the initial value is 2.

The formula is g = ax2 + bx + c where a = 1 2 × 6 = 3, b = (−2) − 3 = −5, and

c = 2, so the final formula is g = 3x2 − 5x + 2.

16. (a) Using quadratic regression, the equation of the parabola approximately followed

by the cannonball is h = −0.00066d2 + 0.80d + 0.027.

(b) Using the equation from Part (a), 0 = h = −0.00066d2 + 0.80d + 0.027 where

d = 1212.16, so the cannonball will strike the ground 1212.16 feet downrange.

gous number in the formula from Part (a) is 0.80, so the slope s = 0.8.

ii. Comparing the coefficient of x2 and the analogous number in the formula

from Part (a), −16 1 + s2

v20 = −0.00066. Since s = 0.8, −16

1 + 0.82

v20 = −0.00066.

Solving, we find an initial velocity of v0 = 199.39 feet per second.

17. (a) A quadratic model for profit as a function of advertising expenditure is, using

regression, P = −0.035A2 + 46.04A + 221.17.

(b) Graphing the profit function from Part (a), we see that the maximum profit is

P = 15,361.75 dollars and occurs when A = 657.71 is spent on advertising.

18. (a) Graphing the profit function, we see that its maximum occurs when a = 250.00

dollars per month is spent on advertising.

(b) P = 5000 when 5000 = 1 + 50a − 0.1a2. Solving, we find that a = 138.15 or

a = 361.85. Since we’d rather pay less than more, we should spend $138.15 per

month on advertising.

ufacturer spends less on advertising. Hence the manufacturer should decrease the

amount spent on advertising.

19. (a) Using the quadratic formula, the solutions are

−k ± √

k2 − 4k 2

(b) The equation has exactly one solution when k2 − 4k = 0, that is, when k = 0 or

k = 4.

Sample Test Answers for Chapter 5 7

when k < 0 or k > 4.

20. (a) The graph is shown below using a horizontal span of 0 to 5 and a vertical span of

0 to 11.

formula is g = ax2 + bx + c where a = × = 1 2 6 3, b = (–2) – 3 = –5, and c = 2, so the final

formula is g = 3x2 – 5x + 2. 14. (a) Using quadratic regression, we find that the equation of the parabola approximately

followed by the cannonball is h = –0.00066d2 + 0.80d + 0.027. (b) Using the equation from Part (a), we have 0 = h = –0.00066d2 + 0.80d + 0.027. We find the solution d = 1212.16, so the cannonball will strike the ground 1212.16 feet downrange. (c) i. Since the slope of inclination of the cannon, s, is the coefficient of x, the analogous number in the formula from Part (a) is 0.80, so the slope s = 0.8. ii. Comparing the coefficient of x2 and the analogous number in the formula from Part

(a), we have −

+ = −16

1 0 00066

0 2

s υ

. . Since s = 0.8, we have

− +

= −16 1 0 8

0 00066 2

0 2

. . .

υ Solving, we find an initial velocity of υ 0 199 39= . feet per second.

15. (a) A quadratic model for profit as a function of advertising expenditure is, using regression, P = –0.035A2 + 46.04A + 221.17. (b) Graphing the profit function from Part (a), we see that the maximum profit is P = 15,361.75 dollars and occurs when A = 657.71 is spent on advertising.

16. (a) Graphing the profit function, we see that its maximum occurs when a = 250.00 dollars per month is spent on advertising. (b) P = 5000 when 5000 = 1 + 50a – 0.1a2. Solving, we find that a = 138.15 or a = 361.85. Since we'd rather pay less than more, we should spend $138.15 per month on advertising. (c) The graph of profit is decreasing when a = 350, so profit will increase as the manufacturer spends less on advertising. Hence the manufacturer should decrease the amount paid on advertising.

17. (a) Using the quadratic formula, we find that the solutions are

− ± −k k k2 4 2

(b) The equation has exactly one solution when k – 4k = 0, that is, when k = 0 or k = 4. 2 (c) There are two real roots when k2 – 4k > 0. Graphing k2 – 4k, we see that k2 – 4k > 0 when k < 0 or k > 4.

18.

(a) (b) As D increases, it's easier to catch the prey, so more are eaten, so the graph is increasing; on the other hand, the predator will only eat so many prey, so even if the

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(b) As D increases, it’s easier to catch the prey, so more are eaten, so the graph is

increasing; on the other hand, the predator will only eat so many prey, so even

if the prey are very easy to catch, the additional number eaten will be fewer and

fewer.

eaten per day.

(d) The physical significance of the horizontal asymptote is that it indicates that, on

average, the amount consumed by predators will increase to a level of 10.67 prey

per day as the density of prey increases.

21. (a) F = k 1 d2

(b) Near the pole at d = 0, F increases without bound.

closer and closer together, the gravitational attraction increases without bound.

22. (a) Using regression, we find that a quartic model for patronage as a function of time

since 10 a.m. is p = −0.068t4 + 1.644t3 − 12.226t2 + 29.220t + 4.118.

(b) Graphing p versus t and the line p = 10, we see that the graph dips below the

p = 10 line between t = 4.49 and t = 6.97, that is, between about 2:30 p.m. and 5

p.m. (since t = 0 is 10 a.m.).

times are at about 11:45 a.m. and 8:30 p.m.