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The solutions to practice exam iii for math 112, which covers topics such as parametric curves, taylor series, and convergence. The exam includes questions on identifying curves described by given parametric equations, finding derivatives and areas, and determining intervals of convergence for power series.
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April 19, 2004 Math 112 — Practice Exam III
Show all work clearly; an answer with no justifying computations may not receive credit (except for the “write down but do not evaluate” questions).
(i) x = t + sin(2t) , y =
t 2 + 0.3 cos(3t)^ (ii)^ x^ =^
3 t t^2 + 1 , y^ =
∣∣ ∣∣^ t 2
∣∣ ∣∣
x = cos^3 θ , y = sin^3 θ :
(a) Find in terms of t the first and second derivatives dy/dx and d^2 y/dx^2 ; (b) At which points (x, y) is the tangent to the curve horizontal or vertical? (It’s enough to consider θ-values in [0, 2 π].) (c) On which interval or intervals (θ-values this time, in [0, 2 π]) is the curve concave down? (d) Find the area in the first quadrant under the curve.
(a) xe−x^2 at x = 0 (b) cos x at x = π/ 2
(−1)n^
4 n+ 3 n+1(2n + 1) (x^ −^ 3)
n
Some possibly useful formulas: ∑ an converges if (^) nlim→∞^ n
√ |an| < 1 or (^) nlim→∞
∣∣ ∣∣^ an+ an
∣∣ ∣∣ < 1.
an, bn > 0 , (^) nlim→∞^ an bn
∑ an converges ⇐⇒
∑ bn converges
Solutions to Practice Exam III:
y dx =
∫ (^0) π/ 2
(sin^3 θ)(3 cos^2 θ(− sin θ))dθ = − 3
∫ (^0) π/ 2
sin^4 θ cos^2 θ dθ
∫ (^0)
π/ 2
(1 − cos 2θ)^2 (1 + cos 2θ)dθ
∫ (^0)
π/ 2
(1 − cos 2θ − cos^2 2 θ + cos^3 2 θ)dθ
= − 3 8
∫ (^0)
π/ 2
(1 − cos 2θ − 1 2
(1 + cos 4θ) + (1 − sin^2 2 θ) cos 2θ)dθ
[ θ −
2 sin 2θ^ −^
2 (θ^ +
4 sin 4θ) +
2 sin 2θ^ −^
6 sin
(^3 2) θ
] 0
π/ 2 = 0 +^3 8
( (^) π 2
( (^) π 2
)
) =^3 π 32
cos x = 0 −
1 (x^ −^
π 2 )^ −^ 0 +
3! (x^ −^
π 2 )
5! (x^ −^
π 2 )
= −(x − π 2
(x − π 2
(x − π 2
∑^ ∞ n=
(−1)n (2n − 1)!
(x − π 2
)^2 n−^1
(−1)n+^
4 n+ 3 n+2(2n + 3) (x^ −^ 3)
n+
(−1)n^4
n+ 3 n+1(2n + 1)
(x − 3)n
∣∣ ∣∣ ∣∣ ∣∣ ∣
= 4(23(2nn^ + 1)+ 3) |x − 3 | n −→→∞ 43 |x − 3 | < 1
when |x − 3 | < 3 /4, i.e., 2 14 < x < 3 34 , so on this interval the series converges, and for x < (^2 ) or x > 3 34 it diverges. At x = 2 14 and x = 3 34 , the series becomes
∑^ ∞ n=
3(2n + 1)
and
∑^ ∞ n=
(−1)n^16 3(2n + 1)
respectively. The first of these diverges, by Limit Comparison to the harmonic series (i.e., the p-series with p = 1), and the second converges by the Alternating Series Test; but the convergence is not absolute, because the sum of the absolute values of the terms is the first series, which, as we have just noted, diverges.