College of the Holy Cross, Spring 2014 Solutions for Math 136, Midterm Exam 3 May 2

I. Let R be the region bounded by y = sin(x), the x-axis, and 0 ≤ x ≤ π/3.

A. [10 points] Write down (but do not try to evaluate) the integral that would compute the arc-length of the top edge of R.

Solution: The top is the graph y = sin(x), so \frac{dy}{dx} = cos(x), and the arclength is

∫_{0}^{π/3} \sqrt{1 + (\frac{dy}{dx})^{2}} dx = ∫_{0}^{π/3} \sqrt{1 + cos^{2}(x)} dx.

B. [15 points] A thin metal plate of constant density has the shape of R. Find the xcoordinate of the center of mass of the plate.

Solution: The x-coordinate of the center of mass is

\frac{∫_{0}^{π/3} x sin(x) dx}{∫_{0}^{π/3} sin(x) dx}

The denominator integral is simpler here:

∫_{0}^{π/3} sin(x) dx = − cos(x)|_{0}^{π/3} = 1 − \frac{1}{2} = \frac{1}{2}.

For the numerator, we need to integrate by parts (with u = x, dv = sin(x) dx, so du = dx and v = − cos(x)):

∫_{0}^{π/3} x sin(x) dx = −x cos(x)|_{0}^{π/3} + ∫_{0}^{π/3} cos(x) dx
= −\frac{π}{6} + sin(x)|_{0}^{π/3}
= \frac{\sqrt{3}}{2} − \frac{π}{6} .

So the x-coordinate of the center of mass is

\overline{x} = \sqrt{3} − \frac{π}{3} ≐ .6849

II. Both parts of this problem deal with the differential equation \frac{dy}{dx} = xy.

A. [15 points] Find the general solution y(x) of the equation by separating variables and integrating.

Solution: This is separable:

\frac{dy}{dx} = xy
∫ \frac{dy}{y} = ∫ x dx
ln |y| = \frac{x^{2}}{2} + c
y = ke^{x^{2}/2} where k = ±e^{c}.

B. [5 points] Find the particular solution y(x) satisfying the initial condition y(0) = 4 and compute the exact value of y(2).

Solution: With y(0) = 4, we get 4 = ke^{0}, so k = 4. The particular solution is y = 4e^{x^{2}/2}. With x = 2, we have y = 4e^{2} ≐ 29.56.

III. [10 points] A drug is administered to a patient intravenously at a constant rate of 10mg per hour. The patient’s body breaks down the drug and removes it from the bloodstream at a rate proportional to the amount present, with some proportionality constant k. Write a differential equation for the function Q(t) = amount of the drug present (in mg) in the bloodstream at time t (in hours) that describes this situation. Note: You do not need to solve the equation.

Solution: The differential equation comes from analyzing both contributions to the change of the amount of drug in the bloodstream:

\frac{dQ}{dt} = kQ + 10.

Here k < 0 is the proportionality constant giving the rate at which the drug is broken down and removed from the bloodstream. The term +10 represents the constant intravenous dose of 10mg per hour.

IV.

A. [10 points] Does the geometric series ∑_{n=0}^{∞} ·\frac{2^{n}}{π^{n}} converge or diverge? If it is convergent,

say why and find the sum; if it is not convergent say why not.

Solution: This is a geometric series with a = 1 and r = \frac{2}{π}. Since |r| < 1, it converges to the sum

∑_{n=0}^{∞} \frac{2^{n}}{π^{n}} = \frac{1}{1 − \frac{2}{π}} = \frac{π}{π − 2} ≐ 2.752

B. [10 points] Explain why the Integral Test can be applied to the series ∑_{n=1}^{∞} \frac{1}{n^{2} + 4} and

use it to determine if the series converges or diverges.

Solution: The nth term of the series is \frac{1}{n^{2}+4} = f(n), for the function f(x) = \frac{1}{x^{2}+4}. This function is continuous at all x > 1, it is decreasing (since f′(x) = \frac{−2x}{(x^{2}+4)^{2}} < 0 for all x > 1), and lim_{x→∞} \frac{1}{x^{2}+4} = 0. Hence the Integral Test applies and

∫_{1}^{∞} \frac{dx}{x^{2} + 4} = lim_{b→∞} ∫_{1}^{b} \frac{dx}{x^{2} + 4}
= lim_{b→∞} \frac{1}{2} tan^{−1} (\frac{x}{2})|_{1}^{b}
= lim_{b→∞} \frac{1}{2} tan^{−1} (\frac{b}{2}) − \frac{1}{2} tan^{−1} (\frac{1}{2})
= \frac{π}{4} − \frac{tan^{−1} (\frac{1}{2})}{2}.

Since the improper integral converges (is finite), the same is true for the infinite series.

V. All parts of this question refer to the power series

∑_{n=1}^{∞} \frac{(x − 1)^{n}}{n3^{n}} .

A. [15 points] Use the Ratio Test to determine the radius of convergence.

Solution: Applying the Ratio Test,

lim_{n→∞} |\frac{(x − 1)^{n+1}}{(n + 1)3^{n+1}} · \frac{n3^{n}}{(x − 1)^{n}} | = lim_{n→∞} \frac{n}{3(n + 1)}|x − 1|
= \frac{1}{3}|x − 1|.

For absolute convergence, we need \frac{1}{3}|x − 1| < 1, so |x − 1| < 3, or −2 < x < 4. The series is centered at a = 1, so the radius of convergence is 3.

B. [10 points] Test convergence at the endpoints of the interval from part A to determine the interval of convergence. Explain your conclusions. Solution: The inequality |x − 1| < 3 is equivalent to −3 < x − 1 < 3, or −2 < x < 4. At x = 4, we substitute and obtain

∑_{n=1}^{∞} \frac{1}{n}.

This is the harmonic series which diverges. At x = −2, we have

∑_{n=1}^{∞} \frac{(−1)^{n}}{n} .

The Alternating Series Test implies that this converges. So the interval of convergence is [−2, 4), or all x with −2 ≤ x < 4.