Test bank for calculus for the life sciences 1st edition by schreiber

Not enough information to decide; not a function if the closing price was ever the same on July 1 for different years, function if the closing price was never the same on July 1 for diff

Problem Set 1.1 - Real Numbers and

Functions

1 a Function; D : {3, 4, 5, 6}, R : {4, 7, 9}

b Not a function, just a set of numbers

2 a Not a function; two values for 7

b Function; D : R, R : R

3 a Not a function; multiple values for all

x values

b Function; D : x 6= 0, R : {−1, 1}

4 a Function; D : years when IBM stock

has a closing price on July 1, R : closing

prices of IBM stock on July 1

b Not enough information to decide; not

a function if the closing price was ever the

same on July 1 for different years, function if

the closing price was never the same on July

1 for different years In case it is a function,

D : closing prices for Apple stock on July

1, R : years when Apple stock has a closing

16 D : x > 1/2, f (1) = 1, 1/2 and 0 arenot in D

35 The side length is given by P/4, so the

area is A(P ) = (P/4)2 = P2/16

36 The radius is given by C/(2π), so the

area is A(C) = π(C/(2π))2 = C2/(4π)

37 The x mg decreases to 0.33x during the

first four hours; then 2 · 325 = 650 mg is

added After another four hours, the total

650+0.33x will decrease to 0.33(650+0.33x)

38 The x mg decreases to 0.32x during the

first 24 hours; then 30 mg is added After

another 24 hours, the total 30 + 0.32x will

decrease to 0.32(30 + 0.32x)

39 We assume that the maximum height is

2 (in), and after the cut the height is 0.5 (in)

c The domain is [0, R] = [0, 1.2 · 10−2]

d

42 a D : [0, a]

d The cost for the first 50% is given by

150 · 50/(200 − 50) = 50 million dollars Thecost for 100% is 150 · 100/(200 − 100) =

150 million dollars, so the second half costs150−50 = 100 million dollars; twice as much

as the first half

45 a D : 0 ≤ n ≤ 12.5

b

c First, D(3) = 2 · 3A/25 = 100, so A =2500/6 Thus D(5) = 2 · 5 · (2500/6)/25 =500/3 (mg)

46 a D : 0 ≤ n ≤ 16

c First, D(6) = (6/18)A = 120, so A =

360 Thus D(8) = (8/20)360 = 144 (mg)

47 a D : 0 ≤ w ≤ 150, if we consider the

meaning of the function (that a child can’t

get a larger amount of drug than an adult)

b

c D(70) = (70/150)A = 90, thus A =1350/7 (mg)

48 a

b

49 2 = p2/q2, thus p2 = 2q2 Thus p2 iseven, and then p is even So p = 2k; but then2q2 = p2 = (2k)2 = 4k2, so q2 = 2k2 Thus

q2 is even and then q is even We arrived tothe contradiction, because we assumed thatp/q cannot be reduced

Problem Set 1.2 - Data Fitting withLinear and Periodic Functions

1 y = 5x

4 − 2

2 y = x

3 +23

18 For a vertical line, x = constant Thusthe equation is x = h This is not a function.

19 The slope is 4/2 = 2, the y-intercept is

b Her life expectancy is E = 70 (A = 0)

c We have to solve 0 = −5A/6 + 70; we

d Using the line from part c, the estimate

is 0.28 ft/sec at 12 feet, and −0.65 ft/sec at

20 feet The second result is clearly out ofthe scope of the model

46 a The slope is (8.9 − 1.7)/(7.3 − 1.3) =7.2/6 = 1.2 = 6/5 The equation is y =6x/5 + 7/50

b We have to solve the equation 12 =6x/5 + 7/50, the solution is x ≈ 9.88 cm

47 a

b The amplitude is about (77.5−35)/2 =21.25 (degrees), the period is 12 (months)

c a = 21.25, b = 2π/12 = π/6

c Not a power function.

3 Not a power function

4 Not a power function

12 We know that y = a 6x and x = b t,thus 6 · 104/2 · 102 = 300 = y2/y1 =

a 6x2/(a 6x1) = x2/x1 = b t2/(b t1) = t2/t1.This means t increases 300-fold

13 This means y = a 10x3 = b x3, so y ∝ x3

as well Then x ∝ y1/3

14 The proportions state that x =

a 100y and y = b 45z, thus 12/95 =

x2/x1 = a 100y2/(a 100y1) = y2/y1 =

b 45z2/(b 45z1) = z2/z1 Thus z decreases

by the factor 12/95

15 Using the general transitive property,

24 All of them are power functions.

25 From S = 4πr2, we get r = pS/(4π)

Thus V = 4πr3/3 = 4π(pS/(4π) )3/3 =

S3/2/(6√

π) If S is quadrupled, r is bled

dou-26 The volume is V = r2π · 5r = 5πr3; the

oz = 15000 lb (i.e 100 times his weight)

36 The assumption is D = aS0.91 We canfind a from the equation 300 = a(10)0.91;

A = 0.00299d2.99

38 The assumption is C = aB1.6 We can

find a from the equation 500 = a(1000)1.6;

we obtain a = 500/(1000)1.6 ≈ 0.0079 Thus

C = 0.0079B1.6

39 The assumption is L = aW0.95for

pump-kins, L = aW2.2 for snake gourds We

can find the a values from the equations

10 = a(10)0.95(pumpkins) and 10 = a(10)2.2

(snake gourds) ; we obtain a = 10/(10)0.95≈

1.12 (pumpkins) and a = 10/(10)2.2 ≈ 0.063

(snake gourds) Thus L = 1.12W0.95

(pump-kins) and L = 0.063W2.2 (snake gourds)

This also means W = 0.888L1.05 and W =

3.51L0.45

40 a The volume is grown by a factor of

33 = 27, so we need a 27 · 30, 000 = 810, 000watt furnace

b The surface area is grown by a factor

of 32 = 9, so we need a 9 · 30, 000 = 270, 000watt furnace

41 We have 1724 = F2/F1 = Lb2/Lb1 =(L2/L1)b = 12b, which means that b ≈ 3.Weight is proportional to volume

42 In this case, b = 2, because surface area

is proportional to size square This wouldmean he needs 122 = 144 times as muchfood

43 a We obtain that A ∝ M2/3, which plies M/A ∝ M1/3

im-b

Problem Set 1.4 - Exponential Growth

8 It seems ex > 2x for all x.

21 The future value is 1000·e0.01≈ 1010.05.

22 The future value is 1000 · e0.1 ≈ 1105.17

23 The future value is 1000·(1+0.16/12)4 ≈1054.41

24 The future value is 1000·(1+0.08/12)6 ≈1040.67

25 2 = ba0 = b, 5 = 2a1, so f (x) = 2(5/2)x

26 32 = ba−2, 8 = ba2, so (by multiplyingthe equations) 256 = b2, and then b = 16(−16 is not good, see the original equations).Also, a2 = 8/16 = 1/2, so a = 1/√

2 Weobtain f (x) = 16(1/√

and then f (x) = (√

2)x

29 t ≈ 168 seconds.

30 t ≈ 3.9, which means around 1854

31 a Compounded once a year, we

ob-tain 100 · (1 + 0.2) = 120, twice a year:

100 · (1 + 0.2/2)2 = 121, four times a year:

100 · (1 + 0.2/4)4 = 121.55

b Compounded n times a year, we obtain

100 · (1 + 0.2/n)n

c For n = 100, 122.116, for n = 1000,122.138, for n = 10000, 122.14 The values

approach 100e1/5 ≈ 122.14

32 a Compounded once a year, we

ob-tain 1000 · (1 + 0.05) = 1050, twice a year:

33 a Once a day: N = (1 + 10)365, twice

a day: N = (1 + 10/2)730, four times a day:

N = (1 + 10/4)1460

b n times a day: (1 + 10/n)365n

c The values approach e3650

34 a Once a day: N = 20(1 + 5)365, twice

a day: N = 20(1 + 5/2)730, four times a day:

N = (1 + 5/4)1460

b n times a day: 20(1 + 5/n)365n

c The values approach 20e1825

35 a At t = 0, Country #3 has the largestpopulation size (20 million)

1 − (0.99)20≈ 0.18, so about 18% is lost

c Comparing H1(10) ≈ 18.1, H2(10) ≈13.9 and H3(10) ≈ 12.3, we see that Beer

#1 has the highest froth after 10 seconds

37 a After 14 days, the amount of T4 inthe body is 100(1/2)14/7 = 25 mcg

b After t days, the amount of T4 in thebody is 100(1/2)t/7 mcg

c t ≈ 23.3 (days)

38 a After 30 hours, the amount of T3 in

the body is 100(1/2)30/10= 12.5 mcg

b After t hours, the amount of T3 in thebody is 100(1/2)t/10 mcg

c t ≈ 33.2 (hours)

39 The remaining amount is 500·(1/2)t/5730

40 The size of the bacteria population is

given by 20(2)t/9.3, where t is measured in

hours Thus after three days, the size is

d The population size was 104.96 million

43 a It is 1 − (1 − 0.1)10 ≈ 0.65, so about65%

b It is 1 − (1 − 0.01)100 ≈ 0.63, so about63%

c It is 1 − (1 − 1/N )N For large values of

N , these seem to approach 1 − 1/e ≈ 0.632

44 a The size is 0.1(2)t/2.9 (cm3)

b t ≈ 6.7 (days)

45 a The size is 0.5(1/2)t/5.7 (cm3)

b t ≈ 18.9 (days)

Problem Set 1.5 - Function Building

so the period is 2π2 The amplitude is 1.

d The period is p = 2π again, the tude is 2

(x + 1)2, domain is x 6= −1, x 6= 2

31 After simplification, we obtain f + g =

2x − 3

x2− x + 2,domain is x 6= −1, x 6= 2

do-34 The data suggests that the period of

T is approximately 12 hours, which gives

B = 2π/12 = π/6 The amplitude is

A = (5.8 − 2.1)/2 = 1.85 The vertical shift

is D = (5.8 + 2.1)/2 = 3.95 The high tide

occurs at around 6:00AM, so C = −6

35 The data suggests that the period of T

is approximately 12 hours, which gives B =

2π/12 = π/6 The amplitude can be

1 + ax.

c

37 a y = bx/(1 + ax) and t = 1/x, z = 1/ygives us 1/z = b(1/t)/(1 + a(1/t)) Multipli-cation of both numerator and denominator

by t on the right side gives 1/z = b/(t + a);taking reciprocals of both sides gives z =t/b + a/b

b Technology finds that z = 0.503 +0.499t; this means b = 1/0.499 ≈ 2, anda/b = 0.503, so a ≈ 1 The approximation

is y = 2x/(1 + x)

38 The air pollution is given by L(p(t)) =0.07p(1 + 0.02t3)2+ 3, so when t = 4, weobtain L = 0.2 ppm

39 a When t = 2, h(2) = 4, and we obtain

70x4− 40 · 3004

110x4

42 We have to solve the equation 12 = H =

12.17 + 1.5 sin(2πn/365 − 1.5); this is the

same as sin(2πn/365 − 1.5) = −0.1133 We

get n = 81 and n = 276, i.e March 22 and

October 3

43 We have 13 hours of daylight when n =

121 and n = 236, i.e on May 1 and August

24

44 a The model we obtain is the

follow-ing combination of the previous 2 models:

V (t) = 0.5(0.99(1/2)t/5.7+ 0.01(2)t/2.9)

b The tumor size is decreasing for about

11 days, but after that the proliferating cells

”win out”

45 Let x = 0 correspond to the year 1844

The data suggests that the period of y is proximately (1997 − 1844)/14 ≈ 11 years,which gives c = 2π/11 The amplitude isabout b = (150 − 0)/2 = 75 The verticalshift is a = (150 + 0)/2 = 75 The first max-imum occurs at around x = 11/2, so d = π

ap-46 Let x = 0 correspond to the year 1823(the first minimum value) The data sug-gests that the period of y is approximately

10 years, which gives c = 2π/10 The plitude is about b = (40000 − 0)/2 = 20000.The vertical shift is a = (40000 + 0)/2 =

am-20000 The first maximum occurs at around

10 Set y = e2x+1; then ln y = 2x + 1 and

13 Set y = ex2 Then ln y = x2, and

36 a We have to solve the equation 0.02 =0.055(220C − 11000)/(320C) The result is

C = 106.1 (cocoons per thousandth acre)

b Set F = 0.055(220C − 11000)/(320C)and solve for C; the resulting function is

C = 605/(12.1 − 320F ) (cocoons per sandth acre)

thou-37 The doubling time can be found by ing 700 = 350(1.12)T This is the same

solv-as 2 = 1.12T; take the natural logarithm

of both sides and divide to obtain T =

ln 2/ ln 1.12 ≈ 6.12 years

38 The doubling time can be found by ing 2 = (1.026)T Take the natural log-arithm of both sides and divide to obtain

solv-T = ln 2/ ln 1.026 ≈ 27 years

39 We solve the equation 0.5 = 0.1(2)t/2.9.Division by 0.1 gives 5 = 2t/2.9 Take thebase two logarithm of both sides, then mul-tiply by 2.9 to obtain t = 2.9 log25 ≈ 6.73days

40 We have to solve the equation 0.1 =0.5(1/2)t/5.7 Division by 0.5 gives 1/5 =(1/2)t/5.7 = 2−t/5.7 Take the base two loga-rithm of both sides, then multiply by −5.7

to obtain t = −5.7 log2(1/5) ≈ 13.24 days

41 The first equation gives ln 28 = ln c +

m ln 0.4, the second ln 100 = ln c + m ln 0.6.Subtract the first equation from the sec-ond: ln 100 − ln 28 = m(ln 0.6 − ln 0.4), so

m ≈ 3.14; then c = 28/(0.4)3.14≈ 497.4 Weobtain W = 497L3.14

42 a We solve the equation Q0/2 =

Q0(0.85)t Taking natural logarithms of

both sides (after dividing them by Q0) gives

rt, and then r = 0.0012

47 The figure shows the plot of (t, ln x).The best fitting line is −9.13 + 0.005t Be-cause x = cert, we obtain that ln x = ln c +

rt, and then r = 0.005

48 The figure shows the plot of (t, ln x).

The best fitting line is −28.92 + 0.015t

Be-cause x = cert, we obtain that ln x = ln c +

5 a1 = 1, a2 = 4, a3 = 2, a4 = 8, a5 = 5

21 The equilibria can be found by solving

x = 2x(1 − x); subtracting x from both sides

and factoring gives 0 = x(1−2x) This

prod-uct is zero when x = 0 or x = 1/2

22 The equilibria can be found by solving

x = x(2 − x); subtracting x from both sides

and factoring gives 0 = x(1 − x) This

prod-uct is zero when x = 0 or x = 1

23 The equilibria can be found by solving

x = 3x/(1 + x); subtracting x from bothsides and factoring gives 0 = x(3/(1+x)−1).This product is zero when x = 0 or x = 2

24 The equilibria can be found by solving

x = 3x/(1 + x); subtracting x from bothsides and factoring gives 0 = x(3/(1+x)−1).This product is zero when x = 0 or x = 2

25 The equilibria can be found by solving

x = 1 + x/2; we get x = 2

26 The equilibria can be found by solving

x = 1/(1 + x); multiplication and

rearrange-ment gives x2+x−1 = 0 Thus the equilibria

are x = (−1 +√

5)/2 or x = (−1 −√

5)/2

27 Equilibria are 0 and 5

28 Equilibria are 0 and ≈ 8

29 Equilibria are 0, ≈ 1, and ≈ −1

30 Equilibria are 0, ≈ 3, and ≈ −3

31 Let the original amount be A At theend of the first hour, the amount is A/2 Atthe end of the second hour, the amount isA/2/2 = A/4 At the end of the third hour,the amount is A/8 At the end of 4 hours,

we get A/16 This is 6.25% At the end of

n hours, the drug present is A/2n

32 a a1 = 500, an = 0.2an−1+ 500

b The values are 500, 0.2·500+500 = 600,0.2 · 600 + 500 = 620, 0.2 · 620 + 500 = 624,

0.2 · 624 + 500 = 624.8

c The equilibrium is given by x = 0.2x +

500, which gives x = 500/0.8 = 625 mg

33 a First, a1 = 0 Then a2 = (1 − c)A +

(1 − c)a1 = (1 − c)(A + a1) Continuing,

a3 = (1 − c)A + (1 − c)a2 = (1 − c)(A + a2)

We obtain that an = (1 − c)(A + an−1)

b The equilibrium is the solution of x =(1 − c)(A + x), which is (1 − c)A/c

c The equilibrium value is bigger than Awhen (1 − c)/c > 1; i.e when 1 − c > c,

which is c < 1/2

34 a First, a1 = A Then a2 = (1−c)a1+A

Continuing, a3 = (1 − c)a2+ A We obtain

b The model gives a very good fit

c This means the difference equation is

xn+1 = axn− 150 We choose a1 = 1249 anditerate We obtain the values 1219, 1186,

equa-b = 20.0, if a1 = 2, c = 1.0

37 a a1 = 1, a2 = 1 + 1/1 = 2, a3 =

1 + 1/2 = 3/2, a4 = 1 + 1/(3/2) = 5/3,

a5 = 1 + 1/(5/3) = 8/5

b The equilibria are the solutions of

x = 1 + 1/x; multiplication by x gives the

quadratic x2 − x − 1 = 0 with solutions

38 a See figure in book; generally, the next

generation will consist of the previous eration, plus the addition coming from onecycle before Thus an+1= an+ an−1

x1/(1 + nx1)

40 a With the assumption given, xn+1 =number of a alleles/total number of alleles

= (xn(1 − xn)N/2)/(xn(1 − xn)N +(1−xn)2N ) = (xn/2)/(xn+(1−xn)) = xn/2

b x1 = 1/2, x2 = 1/4, x3 = 1/8, x4 =1/16, x5 = 1/32, x6 = 1/64, x7 = 1/128,

x8 = 1/256, x9 = 1/512, x10= 1/1024

c Equilibrium is given by x = x/2, i.e

x = 0

d a disappears faster in this case

41 a With the assumption given, xn+1 =number of a alleles/total number of alleles

= (xn(1 − xn)N/3)/(2xn(1 − xn)N/3+(1 − xn)2N ) = (xn/3)/(2xn/3 + (1 − xn)) =

xn/(3 − xn)

b x1 = 1/2, x2 = 1/5, x3 = 1/14, x4 =1/41, x5 = 1/122, x6 = 1/365, x7 = 1/1094,

x8 = 1/3281, x9 = 1/9842, x10 = 1/29525

c Equilibria are given by x = x/(3 − x),i.e x = 0 and x = 2.

d a disappears faster in this case

42 a With the assumption given, xn+1 =

number of a alleles/total number of alleles

d a disappears faster in this case

43 As expected, a disappears more rapidly

when it kills a bigger proportion of Aa types

Review Questions

1 a We need that log x + 1 ≥ 0, which

means log x ≥ −1, so x ≥ 1/10 is the

do-main The range is [0, ∞)

b For the inverse, we have to solve y =

The period is t = 24 hours, so C = 2π/24

Finally, the maximum is at t = 17, so D =

17 The function approximating the

so the sum-of-squares is 0.09 + 0.16 + 0.04 =0.29 The second line is a better fit

8 a Interest compounded once a yeargives 1000(1 + 0.1) = 1100; twice a year: