Math 181: FINAL EXAM NAME: (Combined Sections): May 5, 1992 SS#

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(Scoring:)

1,2,3.  (30: 10 each)
4,5,6.  (30: 10 each)
7,8,9.  (30: 10 each)

Circle your TA and discussion time:

(Prof. Fong) 10,11,12. (30: 10 each)

Boris: 8 9 10

Rasoulian: 8 9 11 13. (20 pts)

(Prof. Radford)

Shi: 11 12 1 14. (20 pts)

Xiong: 10 11 12

(Prof. Friedlander) 15. (20 pts)

Chai: 12 1 2

Yang: 12 1 2 16. (20 pts)

(Prof. Smith)

Tan: 11  TOTAL.  (200 pts)

Multiple-choice Problems

In Problems 1 through 12, CIRCLE the correct answer. (There is no partial credit, but you must show work to receive credit).

Problem 1 A curve has parametric equations x = t + t^{2}, y = t + e^{t}. Find the slope of the tangent line at the point (0, 1).

a) −2

b) −1

c) 0

d) 1

e) 2

Answer e): \frac{dx}{dt} = 1 + 2t, \frac{dy}{dt} = 1 + e^{t}, \frac{dy}{dx} = \frac{1+e^{t}}{1+2t}. At (0, 1) we have t = 0, so the slope is 2.

Problem 2 Find the length of the polar curve r = e^{−θ} from θ = 0 to θ = ln 2.

a) 2\sqrt{2}

b) \sqrt{2}

c) 2π\sqrt{2}

d) π(2^{\frac{3}{2}} − 1)

e) \frac{\sqrt{2}}{2}

Answer e): \frac{dr}{dθ} = −e^{−θ}, so length is ∫_{0}^{ln 2} \sqrt{(e^{−θ})^{2} + (−e^{−θ})^{2}} dθ = ∫_{0}^{ln 2} \sqrt{2e^{−2θ}} dθ = \sqrt{2} ∫_{0}^{ln 2} e^{−θ} dθ = \sqrt{2} · −e^{−θ}|_{0}^{ln 2} = \sqrt{2}(−\frac{1}{2} + 1) = \frac{\sqrt{2}}{2}.

Problem 3 Find the area of the surface of revolution obtained by rotating y = x^{2}/2 for 0 ≤ x ≤ 1 about the y-axis.

a) (4\sqrt{2} − 2)π/3

b) (2\sqrt{2} − 2)π/3

c) (4\sqrt{2} + 1)π/3

d) (4\sqrt{2} + 2)π/3

e) (2\sqrt{2} + 2)π/3

f) (4\sqrt{2} − 1)π/3

Answer a): y = x^{2}/2, y′ = x, so ds = \sqrt{1 + x^{2}} dx. The area of the surface is S = ∫_{0}^{1} 2πx ds = 2π ∫_{0}^{1} x\sqrt{1 + x^{2}} dx = \frac{2π}{3}(1 + x^{2})^{3/2}]_{0}^{1} = \frac{π}{3}(4\sqrt{2} − 2).

Problem 4 Evaluate ∫_{1}^{∞} \frac{ln x}{x^{3}} dx. a) 1/4

b) 1/3

c) 1/2

e) ln 2

f) ln 3

g) ln 4

h) divergent Answer a): (by parts) u = ln x,dv = \frac{dx}{x^{3}} gives ∫ \frac{ln x}{x^{3}} dx = −\frac{ln x}{2x^{2}} + ∫ \frac{dx}{2x^{3}} = −\frac{ln x}{2x^{2}} − \frac{1}{4x^{2}} . Then ∫_{1}^{∞} \frac{ln x}{x^{3}} dx = lim_{t→∞} ∫_{1}^{t} \frac{ln x}{x^{3}} dx = lim_{t→∞}[(−\frac{ln t}{2t^{2}} − 0) + (−\frac{1}{4t^{2}} + \frac{1}{4})] = \frac{1}{4}.

Problem 5 Find ∫ sin^{3} x dx. a) cos^{3} x/3

b) sin^{3} x/3

c) − cos x + cos^{3} x/3

d) sin x − sin^{3} x/3

e) cos x − sin^{3} x/3

f) sin x − cos^{3} x/3 Answer c): ∫ sin^{3} x dx = ∫ (1 − cos^{2} x) sin x dx = − cos x + \frac{1}{3} cos^{3} x.

Problem 6 Find ∫ \frac{dx}{x^{2}\sqrt{x^{2} + 4}} a) \sqrt{x^{2} + 4}

b) x\sqrt{x^{2} + 4}

c) −\sqrt{x^{2} + 4}/4x

d) −\sqrt{x^{2} + 4}/x^{2}

e) x/\sqrt{x^{2} + 4}

f) 1/\sqrt{x^{2} + 4} Answer c): x = 2 tan θ, dx = 2 sec^{2} θ dθ converts ∫ \frac{dx}{x^{2}\sqrt{x^{2}+4}} into ∫ \frac{2 sec^{2} θ dθ}{4 tan^{2} θ·2 sec θ} = ∫ \frac{sec θ dθ}{4 tan^{2} θ} = \frac{1}{4} ∫ \frac{cos θ dθ}{sin^{2} θ} = −\frac{1}{4} \frac{1}{sin θ} = −\frac{\sqrt{x^{2}+4}}{4x}.

Problem 7 Find ∑_{n=0}^{∞} 2^{2n}5^{1−n}.

a) 10

b) 25

c) 2

d) 5

e) divergent Answer b): ∑_{n=0}^{∞} 2^{2n}5^{1−n} = 5 ∑_{n=0}^{∞}(\frac{4}{5})^{n} = 5[\frac{1}{1−4/5}] = 25.

Problem 8 Which of the following series is convergent?

a) ∑_{n=1}^{∞} \frac{ln n}{n}

b) ∑_{n=1}^{∞} \frac{3n}{4n^{2} − 1}

c) ∑_{n=1}^{∞} \frac{(−1)^{n}n^{2}}{n^{3} + 1}

d) ∑_{n=1}^{∞} \frac{n^{2}}{n^{3} + 1}

e) ∑_{n=1}^{∞} \frac{n^{n}}{n!}

Answer c): converges since it is an alternating series with a_{n+1} ≤ a_{n} and a_{n} ⟶ 0. [ a), b), d) diverge by limit comparison with harmonic; e) since n^{n}/n! ⟶ ∞ ].

Problem 9 Which of the following series is divergent?

a) ∑_{n=1}^{∞}(\frac{n + 4}{2n − 1})^{n}

b) ∑_{n=1}^{∞} \frac{(−1)^{n}n^{3}}{3^{n}}

c) ∑_{n=1}^{∞} \frac{(−1)^{n}}{n}

d) ∑_{n=1}^{∞}(\frac{2n + 1}{n + 1})^{n}

e) ∑_{n=1}^{∞} \frac{2^{3n}}{3^{2n}}

Answer d): diverges by the root test. [ The others converge: a) by the root test,

b) by the ratio test, c) by the alternating test, e) is a geometric series with r = 8/9 ].

Problem 10 Find the radius of convergence of ∑_{n=0}^{∞} \frac{n(x − 1)^{n}}{4^{n+1}} .

a) 1

b) 2

c) 3

d) 4

e) 5 Answer d): The ratio \frac{(n+1)|x−1|^{n+1}}{4^{n+2}} · \frac{4^{n+1}}{n|x−1|^{n}} ⟶ \frac{|x−1|}{4}. The series converges when |x − 1| < 4, diverges when |x − 1| > 4, so R = 4.

Problem 11 Find lim_{x→∞} x^{3/x^{2}}.

a) 3/2

b) e

c) e^{3/2}

d) ln(3/2)

e) ∞

f) 1 Answer f): x^{3/x^{2}} = e^{\frac{3 ln x}{x^{2}}} ∼ e^{3 \frac{1}{x} /2x} = e^{\frac{3}{2x^{2}}} ⟶ 1.

Problem 12 A bacteria colony in a culture has been growing exponentially. If there were 80 bacteria at the end of 3 minutes and 160 at the end of 5 minutes, how many were present initially?

a) 40

b) 160/2^{3/2}

c) 80/2^{3/2}

d) 40/2^{3/2}

e) 80/ ln 2

f) 160/ ln 2 Answer c): P (t) = P (0)e^{kt}, 80 = P (0)e^{3k}, 160 = P (0)e^{5k}. Thus 2 = e^{2k}, k = ln \sqrt{2}, and P (0) = 80/e^{3 ln \sqrt{2}} = 80/2^{3/2}.

In the remaining problems 13-16, show all your work. (There is partial credit).

Problem 13 Evaluate ∫ \frac{3x^{2} + x + 4}{x^{3} + x} dx.

Answer: \frac{3x^{2}+x+4}{x^{3}+x} = \frac{A}{x} + \frac{Bx+C}{x^{2}+1} gives 3x^{2} + x + 4 = A(x^{2} + 1) + (Bx + C)x. So (const) A = 4; (x) C = 1; and then (x^{2}) 3 = A + B = 4 + B gives B = −1. So integrate: ∫ \frac{4}{x} dx − ∫ \frac{x}{x^{2}+1} dx + ∫ \frac{1}{x^{2}+1} dx = 4 ln |x| − \frac{1}{2} ln |x^{2} + 1| + tan^{−1}(x) + C.

Problem 14 Find the area inside the polar graph of r = 4 sin θ but outside the polar graph of r = 2 sin θ.

Answer: The circles touch only at the origin. So the desired area is A = 2 · \frac{1}{2} ∫_{0}^{π/2}(16 sin^{2} θ − 4 sin^{2} θ) dθ = ∫_{0}^{π/2} 6(1 − cos 2θ) dθ = (6θ − 3 sin 2θ)]_{0}^{π/2} = 3π.

Problem 15

a) Solve the differential equation \frac{dQ}{dt} = 2Q(5 − Q) for Q(t), with initial condition Q(0) = 1.

b) Find lim_{t→∞} Q(t).

Answer: Separating variables gives \frac{dQ}{Q(5−Q)} = dt.

Now \frac{1}{Q(5−Q)} = \frac{1}{5}(\frac{1}{Q} + \frac{1}{5−Q}), so ln(\frac{Q}{5−Q}) = 10t + C (note Q(0) = 1 > 0, 5 − Q(0) = 4 > 0, so no absolute values are needed). The initial condition gives C = ln \frac{1}{4}. For a): ln(\frac{4Q}{5−Q}) = 10t, \frac{4Q}{5−Q} = e^{10t}, Q(t) = \frac{5e^{10t}}{4+e^{10t}}. Finally for (b), lim_{t→∞} Q(t) = 5.

Problem 16

a) Find the Taylor series for f(x) = (1 + x)^{−1/2} expanded about x = 0.

b) Let R_{3}(x) be the third remainder term (that is, beginning with the x^{4}-term) of the series in a). Find M > 0 such that |R_{3}(x)| ≤ M for 0 ≤ x ≤ 1.

Answer: f^{(0)}(x) = (1 + x)^{−1/2}, f^{(1)}(x) = (−\frac{1}{2})(1 + x)^{−3/2}, f^{(2)}(x) = (−\frac{1}{2})(−\frac{3}{2})(1 + x)^{−5/2},

f^{(3)}(x) = (−\frac{1}{2})(−\frac{3}{2})(−\frac{5}{2})(1 + x)^{−7/2}, f^{(4)}(x) = (−\frac{1}{2})(−\frac{3}{2})(−\frac{5}{2})(−\frac{7}{2})(1 + x)^{−9/2}. So f^{(n)}(x) = (−1)^{n} \frac{1·3···(2n−1)}{2^{n}}(1 + x)^{\frac{2n+1}{2}} , and the Taylor series is ∑_{0}^{∞}(−1)^{n} \frac{1·3···(2n−1)}{2^{n}n!}x^{n}. The remainder term |R_{3}(x)| = \frac{1·3·5·7}{16·24}|1 + z|^{−9/2} ≤ \frac{105}{384} for 0 ≤ x ≤ 1.