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

Test Bank for Calculus for The Life Sciences 1st Edition by Schreiber Full file at https://TestbankDirect.eu/ Problem Set 1.1 - Real Numbers and Functions b Function; D : R, R : R a Not a function; two values for −2 < x < a Function; D : {3, 4, 5, 6}, R : {4, 7, 9} b Function; D : x = ±2, R : y = b Not a function, just a set of numbers a Not a function; two values for b Function; D : R, R : R a Not a function; multiple values for all x values b Function; D : x = 0, R : {−1, 1} a Function; D : years when IBM stock has a closing price on July 1, R : closing prices of IBM stock on July 10 a Not a function; two values for −1 < x < b Not a function; two values for −3 < x < 11 a Not a function; two values for x > −3 b Function; D : R, R : y ≥ −80 12 a Function; D : −2 ≤ x ≤ 3, R : −8 ≤ y ≤ b Function; D : R, R : R b Not enough information to decide; not a function if the closing price was ever the same on July for different years, function if the closing price was never the same on July 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 price on July a Not a function; two values for certain x values 13 D : R, f (0) = 3, f (1) = 4, f (−2) = −5 14 D : R, f (1) = 6, f (0) = −2, f (−2) = 15 D : x = −3, f (2) = 0, f (0) = −2, −3 is not in D 16 D : x > 1/2, f (1) = 1, 1/2 and are not in D 17 D : R, f (3) = 4, f (1) = 2, f (0) = b Function; D : R, R : y ≥ −4 18 D : R, f (−6) = 3, f (5) = 6, f (16) = a Function; D : R, R : R 19 F (x) = x2 b Function; D : R, R : {5} 20 S(x) = 2x + a Function; D : R, R : y ≥ b Function; D : R, R : R a Not a function; two values for x ≥ Full file at https://TestbankDirect.eu/ 21 M (x) = 3x − 7, M (5) = 8, M (0) = −7, M (−3) = −16 22 T (x) = x2 + Test Bank for Calculus for The Life Sciences 1st Edition by Schreiber Full file at https://TestbankDirect.eu/ 23 D : [−6, 6], R : [−6, 5], increasing: [0, 6], decreasing: [−6, 0] 24 D : [−6, 6], R : [−5, 5], increasing: [−6, 0], decreasing: [0, 6] 25 D : x = 2, R : {5}, constant: on D 26 D : [−4, 7], R : [−4, 5], increasing: [3, 5], decreasing: [1, 3] and [5, 7], constant: [−4, 1] 27 D : [−5, 3)∪(3, ∞), R : [−3, 6)∪(6, ∞), increasing: [0, 3) and (3, ∞), decreasing: [−2, 0], constant: [−5, −2] 32 f (x) = 5x + , D : x = −1 x+1 28 D : x = 2, R : y = 1, increasing: (−∞, 2) and (2, ∞) 29 f (x) = 3x − 5, D : R 33 f (x) = (5 + x)2 − x2 = 25 + 10x 30 f (x) = x2 − 5x, D : R 31 f (x) = √ − x, D : (−∞, 5] Full file at https://TestbankDirect.eu/ 34 f (x) = x2 + (10 + x)2 = 2x2 + 20x + 100 Test Bank for Calculus for The Life Sciences 1st Edition by Schreiber Full file at https://TestbankDirect.eu/ 40 35 The side length is given by P/4, so the area is A(P ) = (P/4)2 = P /16 36 The radius is given by C/(2π), so the area is A(C) = π(C/(2π))2 = C /(4π) 37 The x mg decreases to 0.33x during the first four hours; then · 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) 41 a At the center, r = 0, so S(0) = CR2 = 1.76·105 ·(1.2·10−2 )2 ≈ 25.3(cm/sec) b Midway between the artery’s wall and the central axis, r = R/2 = 0.6 · 10−2 Thus S(R/2) = C(R2 − r2 ) = C(R2 − (R/2)2 ) = 1.76 · 105 · ((1.2 · 10−2 )2 − (0.6 · 10−2 )2 ) ≈ 19(cm/sec) c The domain is [0, R] = [0, 1.2 · 10−2 ] d 39 We assume that the maximum height is (in), and after the cut the height is 0.5 (in) 42 a D : [0, a] Full file at https://TestbankDirect.eu/ Test Bank for Calculus for The Life Sciences 1st Edition by Schreiber Full file at https://TestbankDirect.eu/ b 43 a D : (−∞, 0) ∪ (0, ∞) b D : {1, 2, 3, 4, } c d The cost for the first 50% is given by 150 · 50/(200 − 50) = 50 million dollars The cost for 100% is 150 · 100/(200 − 100) = 150 million dollars, so the second half costs 150−50 = 100 million dollars; twice as much as the first half c 45 a D : ≤ n ≤ 12.5 b d The time will decrease and approach minutes It will never be less than or equal to 44 a D : x = 200 b For ≤ x ≤ 100 Full file at https://TestbankDirect.eu/ c First, D(3) = · 3A/25 = 100, so A = 2500/6 Thus D(5) = · · (2500/6)/25 = 500/3 (mg) 46 a D : ≤ n ≤ 16 Test Bank for Calculus for The Life Sciences 1st Edition by Schreiber Full file at https://TestbankDirect.eu/ b b c First, D(6) = (6/18)A = 120, so A = 360 Thus D(8) = (8/20)360 = 144 (mg) 47 a D : ≤ 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 49 = p2 /q , thus p2 = 2q Thus p2 is even, and then p is even So p = 2k; but then 2q = p2 = (2k)2 = 4k , so q = 2k Thus q is even and then q is even We arrived to the contradiction, because we assumed that p/q cannot be reduced Problem Set 1.2 - Data Fitting with Linear and Periodic Functions y = 5x −2 y = x + 3 c D(70) = (70/150)A = 90, thus A = 1350/7 (mg) 48 a Full file at https://TestbankDirect.eu/ Test Bank for Calculus for The Life Sciences 1st Edition by Schreiber Full file at https://TestbankDirect.eu/ y = y = 2x +2 y = x + 2x − 40 y = −3x + Full file at https://TestbankDirect.eu/ Test Bank for Calculus for The Life Sciences 1st Edition by Schreiber Full file at https://TestbankDirect.eu/ 14 Slope is 0, so y = 15 Slope is 0, so y = 16 Slope is 0, so y = 17 The equation can also be written as y = m(x − h) + k = mx + k − mh The slope is m, and the point (h, k) satisfies the equation: k = mh + k − mh 18 For a vertical line, x = constant Thus the equation is x = h This is not a function 19 The slope is 4/2 = 2, the y-intercept is −4 The equation is y = 2x − 20 The slope is −2/4 = −1/2, the yintercept is The equation is y = −x/2+2 21 y = cos x 10 22 y = −3 cos x 23 y = −2 cos πx 24 y = cos πx 25 D (the slope is (8 − 2)/10) 26 E (the slope is (7 − 2)/10) 27 A (the slope is (6 − 2)/10) 11 y − = 3(x − 1), after simplification y = 3x 12 y − (−2) = (2/5)(x − 5), after simplification y = 2x/5 − 28 B (the slope is (−2 − 2)/10) 29 C (the slope is (−3 − 2)/10) 30 F (the slope is (−5 − 2)/10) 13 Slope is (1 − 2)/(0 − (−1)) = −1, so y − = · (x − 0), i.e y = −x + Full file at https://TestbankDirect.eu/ 31 Amplitude is 1/2, period is 2π Test Bank for Calculus for The Life Sciences 1st Edition by Schreiber Full file at https://TestbankDirect.eu/ 35 No amplitude, period is π/2 32 Amplitude is 2, period is 2π 36 No amplitude, period is π/(1/2) = 2π 33 Amplitude is 2, period is 2π/2π = 37 The slope is (20 − 50)/(60 − 24) = −30/36 = −5/6, and the equation is given by E = −5A/6 + 70 a We have to solve the equation 30 = −5A/6 + 70 We obtain A = 48 b Her life expectancy is E = 70 (A = 0) 34 Amplitude is 3, period is 2π/3π = 2/3 Full file at https://TestbankDirect.eu/ c We have to solve = −5A/6 + 70; we Test Bank for Calculus for The Life Sciences 1st Edition by Schreiber Full file at https://TestbankDirect.eu/ obtain 84 38 a The slope is (110 − 97)/(500 − 100) = 13/400, and the equation is given by N = 13x/400 + (97 − 13/4) = 13x/400 + 375/4 b We obtain N = 103.5 for x = 300; we get x = 2500/13 ≈ 192.3 39 B; the estimate is 0.231·150−3 = 31.65 b 40 C; the estimate is 0.29 · + = 2.16 41 a The data is close to linear 43 a The data does not seem to be linear b b 42 a The data is close to linear Full file at https://TestbankDirect.eu/ Test Bank for Calculus for The Life Sciences 1st Edition by Schreiber Full file at https://TestbankDirect.eu/ 10 44 a t = A/10 (hours) b Here A = 1000 · 0.03 = 30 (mL), and we know t = A/10, thus t = hours 45 a 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 of the 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 The first two data points give us the slope (1.11 − 1.55)/(2.0 − 0.7) = −0.44/1.3, equation is y − 1.55 = −(0.44/1.3)(x − 0.7) The first and last data points give us the slope (0.22 − 1.55)/(11.2 − 0.7) = −1.33/10.5, equation is y − 1.55 = −(1.33/10.5)(x−0.7) The second line seems to be a better fit 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 y = −0.1165x + 1.681 Full file at https://TestbankDirect.eu/ c a = 21.25, b = 2π/12 = π/6 Test Bank for Calculus for The Life Sciences 1st Edition by Schreiber Full file at https://TestbankDirect.eu/ 24 19 23 a f (t + p) = f (t), so f (t − + p) + = f (t − 1) + The period is p = T , the amplitude is a = A b The period is p = T again with the same method, the amplitude is 4A c Using the periodicity of f and some algebra, we get that −2f (3t) = −2f (3t+T ) = −2f (3(t + T /3)), so the period is T /3; the amplitude is a = 2A 20 a (f + g)(1) = f (1) + g(1) = − = b (f − g)(2) = f (2) − g(2) = − = c (f g)(2) = f (2)g(2) = · = d (f /g)(0) = f (0)/g(0) = 1/3 e (f ◦ g)(2) = f (g(2)) = f (1) = 21 a (f + g)(−1) = f (−1) + g(−1) = −3 + = −3 d The period is p = T again with the same method, the amplitude is a = 2A 24 One possibility is f (x) = x4 and g(x) = 2x2 − Another is f (x) = (x − 1)4 and g(x) = 2x2 25 One possibility is f (x) = − sin x √ x and g(x) = 26 One possibility is f (x) = ex and g(x) = −x2 b f is not defined at c (f g)(9) = f (9)g(9) = 17 · 70 = 1190 d (f /g)(99) = f (99)/g(99) = 197/9700 e (f ◦ g)(0) = f (g(0)) = f (−2) = −5 22 a f (t + p) = f (t), so f (t − + p) + = f (t − 1) + The period is p = 2π, the amplitude is 27 One possibility is f (x) = ex and g(x) = − x2 28 One possibility is f (x) = x + and g(x) = |x + 1|2 29 One possibility is f (x) = x3 + and g(x) = x2 − x−2 + x2 − x − 2, domain x+1 (x − 2)(x2 − x − 2) is x = −1 f g = = x+1 (x − 2)(x − 2)(x + 1) = (x − 2)2 , domain x+1 x−2 = is x = −1 f /g = (x + 1)(x2 − x − 2) , domain is x = −1, x = (x + 1)2 30 f + g = b The period is p = 2π again with the same method, the amplitude is c f (t/π) = f (t/π + 2π) = f ((t + 2π )/π), so the period is 2π The amplitude is d The period is p = 2π again, the amplitude is Full file at https://TestbankDirect.eu/ √ x+5 Test Bank for Calculus for The Life Sciences 1st Edition by Schreiber Full file at https://TestbankDirect.eu/ 25 31 After simplification, we obtain f + g = 2x2 − x − (2x − 3)(x + 1) +x2 −x−2 = + x+1 (x + 1) x2 − x − = x2 + x − 5, domain is x = (2x2 − x − 3)(x2 − x − 2) −1 f g = = x+1 (2x − 3)(x2 − x − 2), domain is x = −1 2x − 2x2 − x − = , f /g = (x + 1)(x2 − x − 2) x2 − x + domain is x = −1, x = √ √ 32 f + g = − √ x + 4√− x2 , domain is −2 ≤ x ≤ 1, f g = 1√ −x 4− √x , domain is −2 ≤ x ≤ 1, f /g = − x/ − x2 , domain is −2 < x ≤ √ 33 f + g = −√ x2 + sin(πx), domain is − x2 sin(πx), domain −2 ≤ x ≤ 2, f g = 4√ is −2 ≤ x ≤ 2, f /g = − x2 / sin(πx), domain is −2 < x < 2, x = −1, x = 0, x = 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 approximated by A = (5.8 − (−0.4))/2 = 3.1 The vertical shift is D = (5.8 + (−0.4))/2 = 2.7 The high tide occurs at around 11:00AM, so C = −11 36 a The assumption is − t = atx, which implies t = 1/(1 + ax) b f (x) = b(1 − t) = b(1 − bax + ax ) = + ax Full file at https://TestbankDirect.eu/ c 37 a y = bx/(1 + ax) and t = 1/x, z = 1/y gives us 1/z = b(1/t)/(1 + a(1/t)) Multiplication 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, and a/b = 0.503, so a ≈ The approximation is y = 2x/(1 + x) 38 The air pollution is given by L(p(t)) = 0.07 (1 + 0.02t3 )2 + 3, so when t = 4, we obtain L = 0.2 ppm 39 a When t = 2, h(2) = 4, and we obtain V = π43 /12 = 16π/3 b V ◦ h = V (h(t)) = π(2t)3 /12 = 2πt3 /3 c We need that ≤ 2t ≤ 6, so ≤ t ≤ 40 a When t = 2, r(2) = 6, and we obtain S = 4π62 = 144π b S ◦ r = S(r(t)) = 4π(3t)2 = 36πt2 c We need that < 3t < 8, which gives < t < 8/3 41 The value of r is 4/365; thus the composite per-capita growth rate function Test Bank for Calculus for The Life Sciences 1st Edition by Schreiber Full file at https://TestbankDirect.eu/ 26 is G(x) = (g ◦ f )(x) = g(f (x)) = 40 40(3004 + x4 ) (1 − )= (1 − )= 365 f (x) 365 110x4 70x4 − 40 · 3004 365 110x4 The data suggests that the period of y is approximately (1997 − 1844)/14 ≈ 11 years, which gives c = 2π/11 The amplitude is about b = (150 − 0)/2 = 75 The vertical shift is a = (150 + 0)/2 = 75 The first maximum occurs at around x = 11/2, so d = π 46 Let x = correspond to the year 1823 (the first minimum value) The data suggests that the period of y is approximately 10 years, which gives c = 2π/10 The amplitude is about b = (40000 − 0)/2 = 20000 The vertical shift is a = (40000 + 0)/2 = 20000 The first maximum occurs at around x = 5, so d = π 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 Problem Set 1.6 - Inverse Functions and Logarithms Not one-to-one, f (2) = f (4) 43 We have 13 hours of daylight when n = 121 and n = 236, i.e on May and August 24 44 a The model we obtain is the following combination of the previous 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” One-to-one; f −1 (0) = 2, f −1 (14) = −3, f −1 (22) = 44, and f −1 (6) = One-to-one; f −1 (0) = 11.9, f −1 (1) = 17, f −1 (4) = −2, f −1 (2) = 4, and f −1 (6) = Not one-to-one, f (3) = f (4) One-to-one Not one-to-one Seems to be not one-to-one One-to-one 45 Let x = correspond to the year 1844 Full file at https://TestbankDirect.eu/ Set y = x/(1 + x); then y + yx = x, so y = x − yx = x(1 − y) and x = y/(1 − y) Change the role of x and y: we obtain f −1 (x) = x/(1−x) The domain is R, x = 1, the range is R, y = −1 Test Bank for Calculus for The Life Sciences 1st Edition by Schreiber Full file at https://TestbankDirect.eu/ 27 10 Set y = e2x+1 ; then ln y = 2x + and x = (ln y −1)/2 Thus f −1 (x) = (ln x−1)/2 The domain is x > 0, the range is R 11 Set y = (x+1)3 −2; then y +2 = (x+1)3 √ y + = x + Thus f −1 (x) = and √ x + − The domain is R, the range is R x 12 √ Set y = e Then ln y = x , and ln y = x because √ x ≥ by assumption Thus f −1 (x) = ln x The domain is x ≥ 1, the range is y ≥ x2 13 √ Set y = e Then ln y = x , and − ln y = x because x√ ≤ by assumption Thus f −1 (x) = − ln x The domain is x ≥ 1, the range is y ≤ √ 14 Set y = ln x Then y = ln x and 2 ey = x Thus f −1 (x) = ex The domain is x ≥ 0, the range is y ≥ 15 a log 10 = 1, because 101 = 10 b log 0.001 = −3, because 10−3 = 0.001 16 a ln e2 = b 33 log3 x = 3log3 x = x3 −2 c 5−2 log5 x = 5log5 x 3 d 23 log1/2 x = 2log1/2 x = (1/2)− log1/2 x = −3 (1/2)log1/2 x = x−3 e 3− log1/3 x = (1/3)log1/3 x = x 21 a log2 8x = log2 23x = 3x b log3 81x = log3 34x = 4x c log4 64x = log4 43x = 3x d log1/2 32x = log1/2 (1/2)−5x = −5x e log3 9−x = log3 3−2x = −2x 22 a e4 ln x = eln x = x4 b e3 ln(x +1) c e−2 ln(x = eln(x −1) +1)3 = eln(x = (x2 + 1)3 −1)−2 = (x2 − 1)−2 d e−3 ln(1/x) = eln x = x3 e e− ln(1/(x b ln e−4 = −4 = x−2 +1)) = eln(x +1) = x2 + x 23 a 5x = eln = ex ln 17 a log5 125 = 3, because 53 = 125 b log8 64 = 2, because 82 = 64 x b (1/2)x = eln(1/2) = ex ln(1/2) = e−x ln c 51/x = eln x2 e xe 18 a x = 105 d 4x = eln b x = e18 e 3x = eln 19 a x = e3 1/x b 3x+2 = eln 20 a 28 log2 x = 2log2 x = x8 Full file at https://TestbankDirect.eu/ ln e ln = ex = ex 24 a 31−x = eln b x = 104.5 = e(ln 5)/x 1−x x+2 c 21/x+e = eln = e(1−x) ln = e(x+2) ln 1/x+e = e(1/x+e) ln Test Bank for Calculus for The Life Sciences 1st Edition by Schreiber Full file at https://TestbankDirect.eu/ 28 d 4x = eln x2 = ex e 3−3x−2 = eln −3x−2 ln = e(3x+2) ln(1/3) 25 a log(x + 1) = ln(x + 1)/ ln 10 b log(ex + e) = ln(ex + e)/ ln 10 = (1 + ln(x + 1))/ ln 10 2 c log2 (x − 2) = ln(x − 2)/ ln d log7 (2x − 3) = ln(2x − 3)/ ln 26 log 100 + log + 1/2 = 5/2 √ 10 = log 102 + log 101/2 = 27 ln e + ln + ln e542 = + + 542 = 543 28 log8 + log8 16 + log8 82.3 = log8 82/3 + log8 84/3 + log8 82.3 = 2/3 + 4/3 + 2.3 = 4.3 29 10log 0.5 = 0.5 b We solve U = 1.2078C/(1 + 0.0506C) for C; the resulting function is C = U/(1.2078 − 0.0506U ) (mg/l) 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 thousandth acre) 37 The doubling time can be found by solving 700 = 350(1.12)T This is the same as = 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 solving = (1.026)T Take the natural logarithm of both sides and divide to obtain T = ln 2/ ln 1.026 ≈ 27 years 30 ln elog 1000 = log 1000 = log 103 = 31 32 33 34 35 a We have to solve the equation = 1.2078C/(1 + 0.0506C) The result is C = 0.864155 (mg/l) Full file at https://TestbankDirect.eu/ 39 We solve the equation 0.5 = 0.1(2)t/2.9 Division by 0.1 gives = 2t/2.9 Take the base two logarithm of both sides, then multiply by 2.9 to obtain t = 2.9 log2 ≈ 6.73 days 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 logarithm 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 second: ln 100 − ln 28 = m(ln 0.6 − ln 0.4), so m ≈ 3.14; then c = 28/(0.4)3.14 ≈ 497.4 We obtain W = 497L3.14 42 a We solve the equation Q0 /2 = Q0 (0.85)t Taking natural logarithms of Test Bank for Calculus for The Life Sciences 1st Edition by Schreiber Full file at https://TestbankDirect.eu/ 29 both sides (after dividing them by Q0 ) gives t = ln(1/2)/ ln 0.85 ≈ 4.27 years 45 a b We have to solve the equation Q0 /2 = Q0 r100 , so r = 100 1/2 ≈ 0.993, which means the depletion rate is 0.7% 43 a b Technology gives that the best fitting line is ln y = 1.6314 + 0.00296x 46 The figure shows the plot of (t, ln x) The best fitting line is −2.42 + 0.0012t Because x = cert , we obtain that ln x = ln c + rt, and then r = 0.0012 b Technology gives that the best fitting line is ln y = 0.9366 + 0.7488 ln x 44 a 47 The figure shows the plot of (t, ln x) The best fitting line is −9.13 + 0.005t Because x = cert , we obtain that ln x = ln c + rt, and then r = 0.005 b Technology gives that the best fitting line is ln y = −0.6575 + 0.5261 ln x Full file at https://TestbankDirect.eu/ Test Bank for Calculus for The Life Sciences 1st Edition by Schreiber Full file at https://TestbankDirect.eu/ 30 48 The figure shows the plot of (t, ln x) The best fitting line is −28.92 + 0.015t Because x = cert , we obtain that ln x = ln c + rt, and then r = 0.015 a1 = cos(π/2) = 0, a2 = cos(π) = −1, a3 = cos(3π/2) = 0, a4 = cos(2π) = 1, a5 = cos(5π/2) = Problem Set 1.7 - Sequences and Difference Equations a1 = − 1/1 = 0, a2 = − 1/2 = 1/2, a3 = − 1/3 = 2/3, a4 = − 1/4 = 3/4, a5 = − 1/5 = 4/5 a1 = cos(2π)/1 = 1, a2 = cos(4π)/2 = 1/2, a3 = cos(6π)/3 = 1/3, a4 = cos(8π)/4 = 1/4, a5 = cos(10π)/5 = 1/5 a1 = (−1)2 = 1, a2 = (−1)3 = −1, a3 = (−1)4 = 1, a4 = (−1)5 = −1, a5 = (−1)6 = Full file at https://TestbankDirect.eu/ a1 = 1, a2 = 4, a3 = 2, a4 = 8, a5 = Test Bank for Calculus for The Life Sciences 1st Edition by Schreiber Full file at https://TestbankDirect.eu/ 31 a1 = 2, a2 = 7, a3 = 1, a4 = 8, a5 = a1 = −4, a2 = 6, a3 = − = 2, a4 = + = 8, a5 = + = 10 10 a1 = 1, a2 = 2, a3 = · = 2, a4 = · = 4, a5 = · = √ √ a1 = √ 256, a2 = 256 = 16, a = 16 = √ 4, a4 = = 2, a5 = 11 a1 = 0, a2 = + = 8, a3 = + = 16, a4 = 16 + = 24, a5 = 24 + = 32 12 a1 = 1, a2 = · = 3, a3 = · = 9, a4 = · = 27, a5 = · 27 = 81 a1 = 2, a2 = 22 = 4, a3 = 42 = 16, a4 = 162 = 256, a5 = 2562 = 65536 Full file at https://TestbankDirect.eu/ 13 a1 = 100, a2 = 100/2 + = 52, a3 = 52/2 + = 28, a4 = 28/2 + = 16, Test Bank for Calculus for The Life Sciences 1st Edition by Schreiber Full file at https://TestbankDirect.eu/ 32 a5 = 16/2 + = 10 14 a1 = 1000, a2 = 1000/10+2 = 102, a3 = 102/10 + = 12.2, a4 = 12.2/10 + = 3.22, a5 = 3.22/10 + = 2.322 15 a1 = 0, a2 = · + = 2, a3 = · + = 12, a4 = · 12 + = 62, a5 = · 62 + = 312 16 a1 = 0, a2 = − · = 1, a3 = − · = −1, a4 = 1−2·(−1) = 3, a5 = 1−2·3 = −5 17 a1 = 8, a2 = 2·8+1 = 17, a3 = 2·17+1 = 35, a4 = · 35 + = 71, a5 = · 71 + = 143 23 The equilibria can be found by solving x = 3x/(1 + x); subtracting x from both sides and factoring gives = x(3/(1+x)−1) This product is zero when x = or x = 18 a1 = 0, a2 = − 0/2 = 1, a3 = − 1/2 = 1/2, a4 = − 1/4 = 3/4, a5 = − 3/8 = 5/8 19 a1 = 1, a2 = 2/2 = 1, a3 = 2/2 = 1, a4 = 2/2 = 1, a5 = 2/2 = 20 a1 = 1, a2 = · · = 0, a3 = · · = 0, a4 = · · = 0, a5 = · · = 21 The equilibria can be found by solving x = 2x(1 − x); subtracting x from both sides and factoring gives = x(1−2x) This product is zero when x = or x = 1/2 22 The equilibria can be found by solving x = x(2 − x); subtracting x from both sides and factoring gives = x(1 − x) This product is zero when x = or x = Full file at https://TestbankDirect.eu/ 24 The equilibria can be found by solving x = 3x/(1 + x); subtracting x from both sides and factoring gives = x(3/(1+x)−1) This product is zero when x = or x = 25 The equilibria can be found by solving x = + x/2; we get x = Test Bank for Calculus for The Life Sciences 1st Edition by Schreiber Full file at https://TestbankDirect.eu/ 33 29 Equilibria are 0, ≈ 1, and ≈ −1 26 The equilibria can be found by solving x = 1/(1 + x); multiplication and rearrangement gives x2 +x−1 √ = Thus the equilibria √ are x = (−1 + 5)/2 or x = (−1 − 5)/2 30 Equilibria are 0, ≈ 3, and ≈ −3 27 Equilibria are and 31 Let the original amount be A At the end of the first hour, the amount is A/2 At the end of the second hour, the amount is A/2/2 = A/4 At the end of the third hour, the amount is A/8 At the end of hours, we get A/16 This is 6.25% At the end of n hours, the drug present is A/2n 28 Equilibria are and ≈ Full file at https://TestbankDirect.eu/ 32 a a1 = 500, an = 0.2an−1 + 500 Test Bank for Calculus for The Life Sciences 1st Edition by Schreiber Full file at https://TestbankDirect.eu/ 34 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 c This means the difference equation is xn+1 = axn − 150 We choose a1 = 1249 and iterate We obtain the values 1219, 1186, 1150, 1110, 1067, 1019, 967, 910, 847, 779, 703, 621, 530, 431, 322, 203, 73, and then we get negative values The model predicts extinction after 17 years 33 a First, a1 = 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 A when (1 − c)/c > 1; i.e when − 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 that an = (1 − c)an−1 + A b The equilibrium is the solution of x = (1 − c)x + A, which is x = A/c c The equilibrium value is bigger than 2A when A/c > 2A; i.e when c < 1/2 35 a We need that x18 = a17 x1 , i.e 1249 = a17 263, which gives a = 17 1249/263 ≈ 1.096 b The model gives a very good fit Full file at https://TestbankDirect.eu/ 36 The equilibria are given by the equation x = bxe−cx , which is the same as x(1 − be−cx ) = The product is zero if either x = or − be−cx = The second equation gives x = ln b/c This is positive when b > The figures show the cobweb diagrams for b = 0.9, b = 2.0, b = 8.0 and b = 20.0, if a1 = 2, c = 1.0 Test Bank for Calculus for The Life Sciences 1st Edition by Schreiber Full file at https://TestbankDirect.eu/ 35 generation will consist of the previous generation, plus the addition coming from one cycle before Thus an+1 = an + an−1 b r1 = 1, r2 = 2, r3 = 3/2, r4 = 5/3, r5 = 8/5, r6 = 13/8, r7 = 21/13, r8 = 34/21, r9 = 55/34, r10 = 89/55 c rn+1 = an+2 /an+1 = (an +an+1 )/an+1 = + 1/rn 39 a x2 = x1 /(1 + x1 ), x3 = x2 /(1 + x2 ) = x1 /((1 + x1 )(1 + x1 /(1 + x1 )) = x1 /(1 + 2x1 ), x4 = x1 /(1 + 3x1 ), x5 = 1/(1 + 4x1 ) b We guess xn = x1 /(1 + (n − 1)x1 ) c We check that xn+1 = xn /(1 + xn ) = (x1 /(1+(n−1)x1 ))/(1+x1 /(1+(n−1)x1 )) = x1 /(1 + nx1 ) 37 a a1 = 1, a2 = + 1/1 = 2, a3 = + 1/2 = 3/2, a4 = + 1/(3/2) = 5/3, a5 = + 1/(5/3) = 8/5 b The equilibria are the solutions of x = + 1/x; multiplication by x gives the quadratic x2 − x − = with solutions √ (1 ± 5)/2 c The iterates converge toward the positive equilibrium with the initial condition a1 = 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 )2 N ) = (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 = 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 )2 N ) = (xn /3)/(2xn /3 + (1 − xn )) = xn /(3 − xn ) 38 a See figure in book; generally, the next Full file at https://TestbankDirect.eu/ 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 Test Bank for Calculus for The Life Sciences 1st Edition by Schreiber Full file at https://TestbankDirect.eu/ 36 c Equilibria are given by x = x/(3 − x), i.e x = and x = d a disappears faster in this case 42 a With the assumption given, xn+1 = number of a alleles/total number of alleles = (2xn (1 − xn )N/3)/(4xn (1 − xn )N/3+ (1 − xn )2 N ) = (2xn /3)/(4xn /3 + (1 − xn )) = 2xn /(3 + xn ) b x1 = 1/2, x2 = 2/7, x3 = 4/23, x4 = 8/73, x5 = 16/227, x6 = 32/697, x7 = 64/2123, x8 = 128/6433, x9 = 256/19427, x10 = 512/58537 c Equilibria are given by x = 2x/(3 + x), i.e x = and x = −1 d a disappears faster in this case 43 As expected, a disappears more rapidly when it kills a bigger proportion of Aa types For y = x, the residuals are e1 = 0.7, e2 = 0.1, e3 = 0.2, so the sum-of-squares is 0.49 + 0.01 + 0.04 = 0.54; for y = x/2 + 1, the residuals are e1 = 0.3, e2 = 0.4, e3 = 0.2, so the sum-of-squares is 0.09 + 0.16 + 0.04 = 0.29 The second line is a better fit a T ∝ A ∝ l2 , W ∝ V ∝ l3 , so T ∝ W 2/3 Review Questions b According to the previous part, T2 /T1 = (W2 /W1 )2/3 = 22/3 ≈ 1.59 a We need that log x + ≥ 0, which means log x ≥ −1, so x ≥ 1/10 is the domain The range is [0, ∞) The best fitting line for the pairs (ln x, ln y) is 2.35 − 0.373x, thus the m value is −0.373 If we need an integer, m = −1 is the best estimate y = √ b For the inverse, we have to ysolve −1 log x + for x We get x = 10 The domain is [0, ∞), the range is [1/10, ∞) The amplitude B is (92 − 52)/2 = 20; the average temperature A is (92 + 52)/2 = 72 The period is t = 24 hours, so C = 2π/24 Finally, the maximum is at t = 17, so D = 17 The function approximating the temperature is T (t) = 72 + 20 cos(2π(t − 17)/24) The equilibria are at x = −1.4 and x = 1.4 Full file at https://TestbankDirect.eu/ a Interest compounded once a year gives 1000(1 + 0.1) = 1100; twice a year: Test Bank for Calculus for The Life Sciences 1st Edition by Schreiber Full file at https://TestbankDirect.eu/ 37 1000(1 + 0.1/2)2 = 1102.5; three times a year: 1000(1 + 0.1/3)3 = 1103.37 b It approaches 1000e0.1 ≈ 1105.17 y = log a + x log b, thus a = 100 and b = 10−1 = 0.1 10 a First, a1 = Then we obtain a2 = (1−c)A+(1−c)a1 = (1−c)(A+a1 ), where c is the proportion disappearing; in this case, c = 1/2, and A = 100 (mcg) Continuing, a3 = (1 − c)A + (1 − c)a2 = (1 − c)(A + a2 ) We obtain that an = (1 − c)(A + an−1 ) = (1/2)(100 + an−1 ) = an−1 /2 + 50 b The equilibrium is the solution of x = x/2 + 50, so x = 100 This is the amount of T4 in the individual’s body on the long run right before taking the next dose c The greatest amount in the body will be right after taking the replacement hormone, about 100 + 100 = 200 (mcg) 11 a With the assumption given, xn+1 = number of a alleles/total number of alleles = (9xn (1 − xn )N/10)/(9xn (1 − xn )N/5+ (1 − xn )2 N ) = (9xn /10)/(9xn /5 + (1 − xn )) = 9xn /(10 + 8xn ) b x1 = 0.9, x2 = 0.47, x3 = 0.31, x4 = 0.22 12 a Logarithmic b Logarithmic c Exponential d Exponential 13 a f (x/2) is B - consider the x-axis b 2f (x) is D - consider the y-axis Full file at https://TestbankDirect.eu/ c f (−x) is C - the graph is reflected on the y-axis d −f (x) is A - the graph is reflected on the x-axis 14 a N (0) = a = 20; N (2007 − 1941) = N (66) = ae66r = 20e66r = 518 Thus r = ln(518/20)/66 ≈ 0.049 b The doubling time is T = ln 2/r ≈ 14 years 15 a a1 = 2−(1/2) = 3/2, a2 = 2−(2/3) = 4/3, a3 = − (3/4) = 5/4, a4 = − (4/5) = 6/5, a5 = − (5/6) = 7/6 b a1 = (1/2)0 = 1, a2 = (1/2)1 = 1/2, a3 = (1/2)2 = 1/4, a4 = (1/2)3 = 1/8, a5 = (1/2)4 = 1/16 c a1 = 2, a2 = 3, a3 = 5, a4 = 7, a5 = 11 16 The pollution level is 0.5P after the treatment, so P (t) = 25 − 15 cos(2πt/365) We have to solve P = 40, which gives t = 365/2 This is the only time we get 40 17 n(d) is not a linear function w(n) is a linear function a We graph w ◦ n = w(n(d)) = 70 − 50d/(6 + d) The mathematical domain is d = −6; the physical meaning implies that d > Test Bank for Calculus for The Life Sciences 1st Edition by Schreiber Full file at https://TestbankDirect.eu/ 38 b The weight decreases, toward the minimum weight of 20 (grams) 18 a 2002 is represented by t = 10, and f (10) = 1.25(10)2 − 26 · 10 + 161 = 26 1996 is represented by t = 4, and f (4) = 100 So the answer is 74 tons /day less b 19 The data implies that one female produces 150 · (1/3) · 0.5 female larvae Thus after one cycle, we have 25 female larvae, after two, we have 252 , and so on So the total number of descendants in one year is 25 + 252 + 253 + 254 + 255 , and these eat a total of 203.5 kg of wool 20 a The slope of the line is given by (0.11 − 0.17)/(10 − 0) = −0.006 The equation is P (t) = −0.006t + 0.17, because P (0) = 0.17 b We have to solve P (t) = −0.006t + 0.17 = 0.05 This gives t = 20, which corresponds to the year 2020 Problem Set 2.1 - Rates of Change and Tangent Lines By definition, the average rate of change is −2 − 13 f (2) − f (−3) = = −3 − (−3) By definition, the average rate of change is f (3) − f (−3) 5−5 = = − (−3) By definition, the average rate of change is 27 − f (3) − f (1) = = 12 3−1 By definition, the average rate of change is f (4) − f (1) −24 − = = −9 4−1 By definition, the average rate of change is f (9) − f (4) 3−2 = = 9−4 5 By definition, the average rate of change is f (5) − f (1) (−1/3) − (−1) = = 5−1 By definition, the instantaneous rate of change at this point is − 3b − 13 −3(b + 3) f (b) − f (−3) = lim = lim = −3 b→−3 b→−3 b − (−3) b→−3 b − (−3) b+3 lim Full file at https://TestbankDirect.eu/ ... https://TestbankDirect.eu/ Test Bank for Calculus for The Life Sciences 1st Edition by Schreiber Full file at https://TestbankDirect.eu/ 22 10 11 12 Full file at https://TestbankDirect.eu/ Test Bank for. .. for Calculus for The Life Sciences 1st Edition by Schreiber Full file at https://TestbankDirect.eu/ 23 13 16 14 17 15 18 Full file at https://TestbankDirect.eu/ Test Bank for Calculus for The Life. .. https://TestbankDirect.eu/ Test Bank for Calculus for The Life Sciences 1st Edition by Schreiber Full file at https://TestbankDirect.eu/ 30 48 The figure shows the plot of (t, ln x) The best