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Math 112 Practice Exam III: Parametric Curves, Taylor Series, and Convergence, Exams of Calculus

Colgate UniversityCalculus

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.

Typology: Exams

Pre 2010

Uploaded on 08/18/2009

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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).
1. (13 points) Which of the curves below are described by each of the given sets of parametric
equations?
(i) x=t+ sin(2t), y =t
2+ 0.3 cos(3t) (ii) x=3t
t2+ 1 , y =
t
2
pf3
pf4
pf5

Partial preview of the text

Download Math 112 Practice Exam III: Parametric Curves, Taylor Series, and Convergence and more Exams Calculus in PDF only on Docsity!

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).

  1. (13 points) Which of the curves below are described by each of the given sets of parametric equations?

(i) x = t + sin(2t) , y =

t 2 + 0.3 cos(3t)^ (ii)^ x^ =^

3 t t^2 + 1 , y^ =

∣∣ ∣∣^ t 2

∣∣ ∣∣

  1. (32 points) Given the parametric equations

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.

  1. (15 points) Find the following Taylor series, centered at the points specified, as simply as possible:

(a) xe−x^2 at x = 0 (b) cos x at x = π/ 2

  1. (30 points) For which values of x does the following power series converge absolutely? (If it is a finite interval, tell what it does at the endpoints.) ∑^ ∞ n=

(−1)n^

4 n+ 3 n+1(2n + 1) (x^ −^ 3)

n

  1. (10 points) Find, in terms of the angle 6 ACB = t the coordinates of the point P. The two circles centered at the origin have radii 1 and 3, and the angles at A and P are right angles.

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:

  1. (i) Both the x- and y-values are unbounded, so this must be (c). (ii) The y-values are bounded below by 0, and the x-values are bounded between −3 and 3, so this must be (d).
  2. (a) dy/dx = (3 sin^2 θ cos θ)/(3 cos^2 θ(− sin θ)) = − tan θ, and d^2 y/dx^2 = (− sec^2 θ)/(−3 cos^2 θ sin θ) = 1 /(3 cos^4 θ sin θ). (b) The tangent is horizontal when dy/dx = 0, i.e., when sin θ = 0, which occurs at θ = 0 and π; the corresponding points are (± 1 , 0). It is vertical when dy/dx is undefined, i.e., when cos θ = 0, which occurs at θ = π/2 and 3π/2; the corresponding points are (0, ±1). (c) The curve is concave down when d^2 y/dx^2 < 0, which is true when sin θ < 0, which is true when π < θ < 2 π (except at θ = − 3 π/2, where the second derivative is undefined, because the cosine is 0). (d) As the x-values go from 0 to 1 (the largest x can be), the t-values go from π/2 to 0; so the desired area is ∫ (^1) 0

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

)

  • 0 − 0

) =^3 π 32

  1. (a) Because ex^ = ∑∞ n=0 xn/n!, we have xe−x^2 = x ∑∞ n=0(−x^2 )n/n! = ∑∞ n=0(−1)nx^2 n+1/n!. (b) The derivatives cycle through cos x, − sin x, − cos x, sin x, so their values at π/2 cycle through 0, − 1 , 0 , 1. Dividing the n-th value by n! and putting the results into the Taylor series formula gives

cos x = 0 −

1 (x^ −^

π 2 )^ −^ 0 +

3! (x^ −^

π 2 )

5! (x^ −^

π 2 )

= −(x − π 2

) +^1

(x − π 2

)^3 − 1

(x − π 2

)^5 +... =

∑^ ∞ n=

(−1)n (2n − 1)!

(x − π 2

)^2 n−^1

  1. Start with the Ratio Test: ∣∣ ∣∣ ∣∣ ∣∣ ∣

(−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.

  1. The y-coordinate of P is the same as the y-coordinate of D, i.e., y = sin t. And the x- coordinate of P is equal to the x-coordinate of B, i.e., the hypotenuse of the right triangle with side CA = 3 adjacent to angle t; so x = 3 sec t.