MATH 2260 FALL 2013 PRACTICE MIDTERM EXAM 2: SOLUTIONS. PETE L.
CLARK. I. Consider the three-dimensional region with bottom face the unit disk
x2 ...
MATH 2260 FALL 2013 PRACTICE MIDTERM EXAM 2: SOLUTIONS PETE L. CLARK
I. Consider the three-dimensional region with bottom face the unit disk x2 + y 2 ≤ 1 in the plane, and whose cross-section at x is an equilateral triangle whose side lies in the plane and is perpendicular to the x-axis. Find the volume of the region. ∫1 Solution: The volume is −1 A(x)dx, where A(x) is the area of the equilateral tri√ angle giving the cross-section at x. This triangle has side length s(x) = 2 1 − x2 , √ √ 2 so its area is s(x)4 3 = 3(1 − x2 ). Thus √ ( ) √ ∫ 1 √ x3 1 4 3 V = 3 (1 − x2 )dx = 3 x − |−1 = . 3 3 −1 II. Let R be the bounded region enclosed by x + y = 4 and x = −y 2 + 2y + 2. a) Sketch the region R. Solution: A verbal sketch: the first curve is a line with slope −1, so cuts out an isosceles right triangle in the first quadrant. The second curve is a parabola which opens to the left. We expect two intersection points; let us find them, using x = 4 − y: x = (4 − y) = −y 2 + 2y + 2, y 2 − 3y + 2 = 0, (y − 1)(y − 2) = 0, so the intersection points are (3, 1) and (2, 2). So the region in question has y ranging from 1 to 2 and is bounded on the left by the curve x = 4 − y and on the right by x = −y 2 + 2y + 2. b) Find the area of R. Solution: Using part a) we get ∫ 2 ∫ 2 (−y 2 + 3y − 2)dy = (−y 2 + 2y + 2) − (4 − y)dy = A= 1
1
−y 3 3y 2 1 + − 2y|21 = . 3 2 6 c) Find the volume of the region obtained by revolving R about the line y = −5. Solution: Since we are revolving around a horizontal line, our first instinct perhaps is to use disks/washers, but to do this we would have to integrate with respect to x. Rather, it is easier to use shells and integrate with respect to y. The top function 1
2
PETE L. CLARK
is (still) −y 2 + 2y + 2 and the bottom function is (still) 4 − y; the radius is y + 5. So we get ∫ 2 ∫ 2 V = 2πr(ht − hb )dy = 2π (y + 5)(−y 2 + 2y + 2 − (4 − y))dy = 1
1
2π ·
13 13π = . 12 6
Solution: We integrate with respect to y and use washers. The outer radius is R = 4 − x = 4 − (4 − y) = y. The inner radius is r = 4 − x = 4 − (−y 2 + 2y + 2) = y 2 − 2y + 2. So ∫ 2 ∫ 2 7π 2 2 V =π (R(y) − r(y) )dy = π (y 2 − (y 2 − 2y + 2)2 )dy = . 15 1 1 III. Find the arclength of the curve y = log sec x on the interval 0 ≤ x ≤ Solution: We have
∫
π 4
√
L=
(
1+ 0
dy dx
π 4.
)2 dx.
dy 1 = (sec x tan x) = tan x, dx sec x so
∫
π 4
L=
∫ √ 1 + tan2 x =
0
π 4
√ π sec x = log(sec x + tan x) |04 = log( 2 + 1).
0
IV. Find the surface area √ of the surface of revolution obtained by revolving the portion of the curve y = x from x = 4 to x = 9 about the x-axis. Solution: We have
√
)2 dy dx dx 4 √ ∫ 9 ∫ 9 √ √ 1 = 2π x 1+ dx = π 4x + 1dx. 4x 4 4 3 1 1 π π = (4x + 1) 2 |94 = (37 3 − 17 3 ). 6 6 ∫
A = 2π
9
y
(
1+
V. Determine whether each of the following series converges or diverges. If it converges, ∑∞ find the sum. a) n=1 log( n+3 n ). First Solution: This is a telescoping sum: SN =
N ∑ n=1
= log(N + 3) + log(N + 2) + log(N + 1) − (log 1 + log 2 + log 3).
MATH 2260 FALL 2013 PRACTICE MIDTERM EXAM 2: SOLUTIONS
3
Since limx→∞ log x = ∞, SN → ∞: the series diverges. Second Solution: We have limx→0 lim
log( n+3 n ) 3 n
n→∞
Since b)
∑∞
3 n=1 n
∑∞ n=5
log(1+x) x
= 3lim
= 1 (by L’Hopital’s Rule). Thus log(1 + n3 ) 3 n
n →0
= 1.
= ∞, the given series diverges by the Limit Comparison Test. 2n+3
4 2013 32n+10 2n .
Solution: This is a convergent geometric series. ∞ ∑
2013
n=5
( =
2013 · 43 310
∞ ∑ 42n+3 43 16n = 2013 32n+10 2n 310 9n 2n n=5
)∑ ∞ ( n=5
16 18
(
)n =
2013 · 43 310
) ( 8 )5 9
1−
8 9
.
VI. Find all values of x for which the following series converges, and if it converges find the sum: ∞ ∑ 2 en(x −4) . n=1
Solution: Since en(x
2
−4)
( 2 )n 2 = ex −4 , this is a geometric series with r = ex −4 .
It is convergent if and only if |r| < 1; since exponentials are always positive, this 2 occurs if and only if ex −4 < 1 if and only if x2 − 4 < 0 if and only if −2 < x < 2. For such x the sum is
r 1−r
2
=
ex −4 . 1−ex2 −4
VII. True or false: Suppose {a∑ of real numbers n } and {bn } are two sequences ∑∞ ∞ with an ≤ bn for all n. Then if n=0 bn converges, then n=0 an converges. Solution: False. This looks like the Comparison Test, but the hypothesis that an ≥ 0 for all sufficiently large n is missing. Without it nothing stops the partial sums from being e.g. unbounded below. A simple example is bn = 0 and an = −1 for all n. Much more complicated things can happen if bn is allowed to be both positive and negative. VIII. Determine whether each of the following series converges or diverges: 1.2 ∑∞ (a) 2 (log nn) . Solution: Since limn→∞ (log n)1.2 = ∞, for sufficiently large n the numerator is greater than 1 one and hence ∑ the terms of the series are eventually larger than those of the harmonic series n n1 . Since the harmonic series is divergent, the given series diverges by comparison. (b)
∑∞
n7 +100n4 +6 n=1 1,000,000n6 +50n5 +39n2 +10n+4 .
4
PETE L. CLARK
Solution: The series diverges by Limit Comparison to (c)
∑∞ n=1
∑
1 n n.
sin( n12 ).
Solution: Since limx→0 ∑ 1 n n2 .
sin x x
= 1, the series converges by Limit Comparison to
IX. State the (Direct) Comparison Test and the Limit Comparison Test. Solution: Theorem 1. (Comparison Test) Let {an } and {bn } be two sequences of real numbers. Suppose that for 0 ≤ an ≤ bn for all∑n ≥ N . ∑∞ ∑∞some positive integer N we ∑have ∞ ∞ Then n=N an ≤ n=N b∑ b n . It follows that if n=1 ∑∞ n converges then n=1 an ∞ converges. Equivalently, if n=1 an diverges then n=1 bn diverges. Theorem 2. (Limit Comparison Test) Let {an } and {bn } be two sequences of real numbers. Suppose that for some positive integer N we have an ≥ 0 for all n ≥ N and bn > 0 for all n ≥ N . Suppose moreover that L = limn→∞ abnn exists and is ∑∞ ∑∞ nonzero. Then either both series n=1 an and n=1 bn are convergent, or both series are divergent.
