151 MATH EXAM #3A SOLUTIONS FALL 2001 PART 1 1. If = cos1 ( 3/2) , then cos = 3/2 and = /6 . Thus, tan(cos1 ( 3/2)) = tan = tan(/6) = 1/ 3 . The correct answer is (d).
2. Using LHopitals rule 1 1 1 1 ln(1 + x) x = lim = . = lim 1 + x lim 2 x0 x0 x0 2x 2(1 + x) 2 x The correct answer is (d).
3.
d sin1 (2/x) = dx
1 1 (2/x)2
(2/x) =
1 4 1 x2
2 x2
=
1 4 1 x2
2 x4
=
2 x4 4x2
.
The correct answer is (a).
4. If f (x) = 4x + 2 , f (0) = 1 and f (0) = 3 , then f (x) = 2x2 + 2x + 1 f (x) = (2/3)x3 + x2 + x 3 f (3) = (2/3)(3)3 + (3)2 + 3 3 = 27 The correct answer is (e).
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5. The Riemann sum R using four equal subintervals with right endpoints is R = f (0.5)(0.5 0) + f (1)(1.0 0.5) + f (1.5)(1.5 1.0) + f (2)(2.0 1.5) = (0.5)(0.5) + (0.2)(0.5) + (0.1)(0.5) + (0.0)(0.5) = 0.4 The correct answer is (a).
6. If y = (1 + 3x)
cot(x)
, then
3 2 ln(1 + 3x) 1 + 3x = lim 3 cos (x) = 3 = lim lim ln y = lim cot(x) ln(1 + 3x) = lim x0 x0 x0 x0 sec2 (x) x0 1 + 3x tan(x) and
x0
lim (1 + 3x)
cot(x)
lim ln y = e3 . = lim y = lim eln y = e x0
x0 x0
The correct answer is (c).
7. If f (x) = x3 3x2 , then f (x) = 3x2 6x and f (x) = 6x6 . Now f (2) = 0 and f (2) = 6 > 0 . Thus, f has a relative minimum at x = 2 . The correct answer is (d).
8. Dene f (x) = x + 2y = x +
100 100 for x > 0 . Then f (x) = 1 and x = 10 is the only positive x x2 critical number for f (x) . Now f (x) < 0 if 0 < x < 10 and f (x) > 0 if x > 10 . The minimum value 100 = 20 . of x + 2y given that xy = 50 is f (10) = 10 + 10 The correct answer is (c).
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1 9. If 0 1 0 f (x) dx = 7 and
1 g(x) dx = 2 , then 0 [f (x) 4g(x)] dx = 1 0 f (x) dx 4 1 0 g(x) dx = 7 4(2) = 1 .
The correct answer is (c).
10. F (x) =
1 dt 1 + t4 0 1 F (x) = (sin x) 1 + (sin x)4 = F (/3) = = The correct answer is (d). 1 1 + (sin x)4 1 cos x
sin x
cos(/3) 1 + (sin(/3))4 8 1 (1/2) = 4 25 1 + ( 3/2)
11.
y = xcos x ln y = (cos x) ln x 1 y y = (cos x)(1/x) + (ln x)( sin x) y = y (ln x)(sin x) + cos x x
cos x = xcos x (ln x)(sin x) + x The correct answer is (b).
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PART 2
12. If y(t) = number of bacteria at time t , then y (t) = k y(t) y(0) = 108 . The solution is y(t) = y(0)ekt = 108 ekt .
(a) When t = 10 , 108 e10k = y(10) = 1.03 10 8 e10k = 1.03 10k = ln 1.03 1 ln 1.03 k= 10
(b) The population has doubled size its when y(t) = 108 ekt = 2 108 ekt = 2 kt = ln 2 t= ln 2 = 10 k ln 2 ln 1.03 minutes
13. If g(x) = x3 3x2 9x , then g (x) = 3x2 6x 9 . The critical numbers of g(x) solve g (x) = 0 3x2 6x 9 = 0 3(x 3)(x + 1) = 0 . The solutions are x = 1 and x = 3 . Now g(2) = (2)3 3(2)2 9(2) = 2 g(1) = (1)3 3(1)2 9(1) = 5 g(3) = (3)3 3(3)2 9(3) = 27 g(4) = (4)3 3(4)2 9(4) = 20 The function g(x) = x3 3x2 9x has an absolute maximum value of 5 on [2, 4] .
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14. If f (x) = e2x tan1 (ln x) , then f (x) = e2x tan1 (ln x) = e2x 1 1 + (ln x)2 1 = e2x 1 + (ln x)2 + e2x tan1 (ln x)
(ln x) 2e2x tan1 (ln x) 1 2e2x tan1 (ln x) x
15. If g(x) =
1 4 1 6 x x , then 4 6
(a) g (x) = x3 x5 = x3 (1x2 ) and g (x) > 0 on (, 1)(0, 1) and g (x) < 0 on (1, 0)(1, ) . The graph of g(x) is increasing on (, 1) (0, 1) and decreasing on (1, 0) (1, ) .
(b) g (x) = 3x2 5x4 = x2 (3 5x2 ) and g (x) > 0 on ( 3/5, 3/5) and g (x) < 0 on (, 3/5) ( 3/5, ) . The graph of g(x) is concave up on ( 3/5, 3/5) and concave down on (, 3/5) ( 3/5, ) .
16. (a) F (x) = ex is an antiderivative of f (x) = ex . Then ln 9 ln 2 ex dx = F (x) ln 9 ln 2 = F (ln 9) F (ln 2) = eln 9 eln 2 = 9 2 = 7
(b) F (x) = x + ln x is an antiderivative of f (x) =
1 x+1 = 1 + . Then x x
5 5 x+1 dx = F (x) = F (5) F (2) = (5 + ln 5) (2 + ln 2) = 3 + ln(5/2) x 2 2
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17. If a square with side x ft is cut out from each of the four corners, the volume of the box formed by bending up the sides is V (x) = x(6 2x)2 , for 0 < x < 3 . Now V (x) = 4x(6 2x) + (6 2x)2 = (6 2x)(6 6x) = 12(3 x)(1 x) and the critical values for V (x) are x = 1 and x = 3 . Only x = 1 is in the region 0 < x < 3 . Now V (x) > 0 if 0 < x < 1 and V (x) < 0 if 1 < x < 3 . Consequently, the maximum volume occurs when x = 1 and the resulting volume is V (1) = 1(6 2(1)) 2 = 16 .
18. Let s(t) = the position of the car at time t and v(t) = the velocity of the car at time t . Then v (t) = 36 v(t) = 36t + v(0) and the car comes to rest when v(t) = 0 36t + v(0) = 0 v(0) seconds t= 36 Now s (t) = v(t) = 36t + v(0) s(t) = 18t2 + v(0)t + s(0) and the distance traveled by the car during the skid was s(v(0)/36) s(0) = 112.5 ft. Thus, s(v(0)/36) s(0) = 18 112.5 = v(0) 36 2 (v(0)) 2 + v(0) v(0) 36
72 v(0) = 90 ft/sec
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