APPM 1360 Final Exam Fall 2013 - Applied...

APPM 1360 Final Exam Fall 2013

On the front of your bluebook, please write: a grading key, your name, student ID, and section andinstructor. This exam is worth 150 points and has 9 questions.

• Show all work! Answers with no justification will receive no points.

• Please begin each problem on a new page.

• No notes, calculators, or electronic devices are permitted.

1. (20 points) Evaluate the following:

(a)∫

3

(x− 1)(x2 + 2)dx

(b) Solve for y in the differential equation:dy

dx= xy sinx with initial value y(π) = 1.

Solution:

(a) Use partial fractions.∫3

(x− 1)(x2 + 2)dx =

∫ (A

x− 1+Bx+ C

x2 + 2

)dx

A(x2 + 2) + (Bx+ C)(x− 1) = 3

(A+B)x2 + (−B + C)x+ (2A− C) = 3

A+B = 0, −B + C = 0, 2A− C = 3

A = 1, B = −1, C = −1

∫ (1

x− 1− x+ 1

x2 + 2

)dx =

∫ (1

x− 1− x

x2 + 2︸ ︷︷ ︸u=x2+2

− 1

x2 + 2

)dx

= ln |x− 1| − 1

2ln∣∣x2 + 2

∣∣− 1√2arctan

(x√2

)+ C

(b) This is a separable equation.

dy

dx= xy sinx ⇒

∫dy

y=

∫x︸︷︷︸u

sinx dx︸ ︷︷ ︸dv

Use integration by parts with u = x, du = dx, and dv = sinx dx, v = − cosx.

ln |y| = −x cosx+

∫cosx dx = −x cosx+ sinx+ C

|y| = e−x cosx+sinx+C

Plug in the initial value (π, 1).

1 = eπ+C ⇒ C = −π

y = e−x cosx+sinx−π

2. (14 points) Consider the region shown at right. Set up, butdo not evaluate, the integrals to calculate:

(a) The volume generated by rotating this region about theline x = 3.

(b) Consider the upper curve y = x2/2, defined on theinterval shown on the graph. Find the surface area ofthe solid generated by rotating the curve about they-axis. (The lower curve is not relevant for this part.)

y �

x2

2

y �

x4

4-

x2

2

x

y

Solution:

(a) The curves intersect at (0, 0) and (2, 2). By the Shell Method,

V =

∫2πrh dx =

∫ 2

02π(3− x)

(x2

2−(x4

4− x2

2

))dx

(b)

S =

∫2πr

√1 + (dy/dx)2 dx =

∫ 2

02πx

√1 + x2 dx

3. (24 points) Do the following converge or diverge? Explain your work and name any test or theoremyou use.

(a)∫ ∞0

dx

1 + ex(b){(lnm)2

m

}∞m=15

(c)∞∑n=3

√n+ 2

n− 2

Solution:

(a) Since 0 <1

1 + ex<

1

exand

∫ ∞0

dx

exis convergent, by the Comparison Theorem

∫ ∞0

dx

1 + ex

also is convergent .

Alternate solution:∫ ∞0

dx

1 + ex= lim

t→∞

∫ t

0

dx

1 + ex

Let u = 1 + ex, du = ex dx.

= limt→∞

∫ 1+et

2

du

u(u− 1)= lim

t→∞

∫ 1+et

2

(1

u− 1− 1

u

)du = lim

t→∞

[ln |u− 1| − ln |u|

]1+et2

= limt→∞

[ln

∣∣∣∣u− 1

u

∣∣∣∣]1+et2

= limt→∞

(ln

et

1 + et− ln

1

2

)= ln 1 + ln 2 = ln 2

The integral is convergent .

(b)

limm→∞

am = limm→∞

(lnm)2

m

L′H= lim

m→∞

2 lnm

m

L′H= lim

m→∞

2

m= 0

The sequence converges to 0.

(c) Since√n+ 2

n− 2>

√n

n=

1√n

and∞∑n=1

1√n

is a divergent p-series, by the Comparison Test

∞∑n=3

√n+ 2

n− 2also is divergent .

Alternate solution: Use the Limit Comparison Test with bn = 1/√n.

limn→∞

∣∣∣∣anbn∣∣∣∣ = lim

n→∞

∣∣∣∣√n+ 2

n− 2·√n

1

∣∣∣∣ = limn→∞

√n2 + 2n

n2 − 2n+ 4= 1

Since∞∑n=1

1√n

is a divergent p-series,∞∑n=3

√n+ 2

n− 2also is divergent .

4. (16 points) Find the values of x for which the following series converge absolutely, converge condi-

tionally, or diverge: (a)∞∑n=2

(x+ 3)n

n2 − 1(b)

∞∑n=0

3n

n!x2n

Solution:

(a) Use the Ratio Test.

limn→∞

∣∣∣∣an+1

an

∣∣∣∣ = limn→∞

∣∣∣∣ (x+ 3)n+1

(n+ 1)2 − 1· n

2 − 1

(x+ 3)n

∣∣∣∣ = |x+ 3| < 1

The series is absolutely convergent on −4 < x < −2. Now check the endpoints. When

x = −2, the series∞∑n=2

1

n2 − 1converges by the Limit Comparison Test with the convergent

p-series∞∑n=1

1

n2. When x = −4, the series

∞∑n=2

(−1)n

n2 − 1converges by the Alternating Series Test

since1

(n+ 1)2 − 1<

1

n2 − 1and lim

n→∞

1

n2 − 1= 0.

Therefore the series∞∑n=2

(x+ 3)n

n2 − 1is absolutely convergent for −4 ≤ x ≤ −2 , is not condi-

tionally convergent for any x, and is divergent for x < −4, x > −2 .

(b) Use the Ratio Test.

limn→∞

∣∣∣∣an+1

an

∣∣∣∣ = limn→∞

∣∣∣∣3n+1x2n+2

(n+ 1)!· n!

3nx2n

∣∣∣∣ = limn→∞

∣∣∣∣ 3x2

n+ 1

∣∣∣∣ = 0

Since the ratio is 0, the series is absolutely convergent for all x , and not conditionally conver-gent or divergent for any x.

5. (14 points) Let a1 = 1, a2 =1 + 1/4

1 + 1/2, a3 =

1 + 1/4 + 1/42

1 + 1/2 + 1/22, a4 =

1 + 1/4 + 1/42 + 1/43

1 + 1/2 + 1/22 + 1/23, . . ..

(a) Does the sequence {an} converge? If yes, find its limit. Explain.

(b) Does∞∑n=1

an converge? Explain your work and state any test or theorem you use.

Solution:

(a) an is the quotient of geometric series. The numerator has ratio 1/4 and the denominator hasratio 1/2. The sum of an infinite geometric series is S = a/(1− r).

limn→∞

an = limn→∞

1 + 14 + 1

42+ · · ·+ 1

4n−1

1 + 12 + 1

22+ · · ·+ 1

2n−1

=limn→∞

(1 + 1

4 + 142

+ · · ·+ 14n−1

)limn→∞

(1 + 1

2 + 122

+ · · ·+ 12n−1

) =1/(1− 1

4)

1/(1− 12)

=2

3

(b) By the Test for Divergence, since limn→∞

an 6= 0, the series diverges .

6. (20 points)

(a) Find a Maclaurin series for x sin(x2).

(b) Use your answer from part (a) to find a series for∫x sin(x2) dx.

(c) Use your answer from part (b) to find the 8th order Taylor polynomial T8(x) for the integral.

(d) Use T8(x) to approximate the value of∫ 1

0x sin(x2) dx.

(e) Is your approximation an overestimate or underestimate? Explain your reasoning.

Solution:

(a) sinx =∞∑n=0

(−1)n x2n+1

(2n+ 1)!⇒ x sin(x2) =

∞∑n=0

(−1)n x(x2)2n+1

(2n+ 1)!=

∞∑n=0

(−1)nx4n+3

(2n+ 1)!

(b)∫x sin(x2) dx = C +

∞∑n=0

(−1)nx4n+4

(2n+ 1)!(4n+ 4)

(c) T8(x) = C +x4

4− x8

48

(d)∫ 1

0x sin(x2) dx ≈ T8(1)− T8(0) =

1

4− 1

48=

11

48

(e) Because the series is alternating, the approximation is an underestimate .

7. (14 points)

(a) Sketch the graphs of r = −2 cos θ and r = 1.

(b) Find the area of the region inside r = −2 cos θ and outside r = 1.

Solution:

(a)

r@t_D := -2 Cos@tD;s@t_D := 1;Show@ParametricPlot@88r Cos@tD r@tD, r Sin@tD r@tD<, 8r Cos@tD s@tD, r Sin@tD s@tD<<,

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8Arrowheads@AutomaticD, Thickness@.003D<<,LabelStyle Ø 24,PlotStyle Ø 88Thickness@.01D, Black<, 8Thickness@.01D, Black<<D,

PolarPlot@8-2 Cos@tD, 1<, 8t, 0, 2 Pi<,PlotStyle Ø 88Thickness@.006D, Black<, 8Thickness@.006D, Black<<,Mesh Ø NoneD

D

68 1360-exams-fall13.nb

(b) We wish to find the area of the shaded region. The curve r = −2 cos θ (for θ = 0 to π) andcurve r = 1 (for θ = 0 to 2π) intersect at −2 cos θ = 1 ⇒ θ = 2π/3. We double the area ofthe shaded region above the x-axis.

A = 2

∫1

2

(r21 − r22

)dθ = 2

∫ π

2π/3

1

2

((−2 cos θ)2 − 12

)dθ =

∫ π

2π/3

(4 cos2 θ − 1

)dθ

=

∫ π

2π/3(2(1 + cos 2θ)− 1) dθ =

∫ π

2π/3(1 + 2 cos 2θ) dθ

=

[θ + sin 2θ

]π2π/3

=

(π −

(2π

3−√3

2

))=

π

3+

√3

2

8. (16 points) Write the word TRUE (if always true) or FALSE. For this problem, no explanation isnecessary.

(a) If∞∑n=1

an diverges, then∞∑n=1

|an| diverges.

(b) The length of the curve r2 = 2 cos θ is given by L =

∫ π/2

04

√2 cos θ +

sin2 θ

2 cos θdθ.

(c) If the parametric equations x = f(t) and y = g(t) are twice differentiable, thend2y

dx2=d2y/dt2

d2x/dt2.

(d) The second order Taylor polynomial T2(x) centered at a = 1 is used to approximate f(x) =√x

on the interval [1, 1.1]. An estimate for the error of the approximation using Taylor’s Formula is1

16(0.1)3.

Solution:

(a) True because if∞∑n=1

|an| converges, then so does∞∑n=1

an.

(b) True. Note that r is not defined for cos θ < 0. If r =√2 cos θ, then

dr

dθ=− sin θ√2 cos θ

. The length

of the entire curve is four times the length for θ = 0 to π/2.

L =

∫ √r2 +

(dr

dθ

)2

dθ = 4

∫ π/2

0

√2 cos θ +

sin2 θ

2 cos θdθ.

(c) False.d2y

dx2=

ddt(dy/dx)

dx/dt

(d) True. R2(x) =f ′′′(z)

3!(x − 1)3 for z between x and 1. On the interval [1, 1.1], |f ′′′(z)| =∣∣∣ 3

8z5/2

∣∣∣ ≤ 38 and

∣∣(x− 1)3∣∣ ≤ (0.1)3, so |R2(x)| ≤ 3

8 ·13! · (0.1)

3 = 116(0.1)

3.

9. (12 points) Match the graphs shown below to the following equations. No explanation is necessary.

(a) x2 = 1 + y2/2 (d) r2 = 4/θ(b) x2 = 1− y2/4 (e) r2 = (cos θ)/4(c) x = 2 cos t, y = sin t (f) r = 1/2 + cos(2θ)

-2 2x

-2

2

y

H1L

-2 2x

-2

2

y

H2L

-2 2x

-2

2

y

H3L

-2 2x

-2

2

y

H4L

Solution: (1) d (2) f (3) c (4) a