Mathematics 136, section 4 – AP Calculus Solutions for Exam 2

I. From the diagram given with the problem, the distance through the forest along the path is 2\sqrt{x^{2} + 4}, while the distance along the road is 10 − 2x. Using the fact that

time = \frac{distance}{rate}

if we are traveling at a constant speed, the total time the trip will take is

T (x) = \frac{2\sqrt{x^{2} + 4}}{3} + \frac{10 − 2x}{8}

We must find the overall minimum of T (x). To find critical points we differentiate using the chain rule:

T′(x) = \frac{2 · \frac{1}{2} · 2x}{3\sqrt{x^{2} + 4}} − \frac{1}{4} = \frac{2x}{3\sqrt{x^{2} + 4}} − \frac{1}{4}

Set this equal to zero and solve for x:

\begin{aligned}\frac{2x}{3\sqrt{x^{2} + 4}} − \frac{1}{4} & = 0 \\ ⇒ 8x & = 3\sqrt{x^{2} + 4} \\ ⇒ 64x^{2} & = 9(x^{2} + 4) \\ ⇒ 55x^{2} & = 36 \\ ⇒ x & = \frac{6}{\sqrt{55}} ≐ .81 km \end{aligned}

(It is not difficult to see that this is a local minimum for T (x), using the 1st derivative test.)

II. A) We can approximate the total distance traveled: x(5) − x(0) = ∫_{0}^{5} v(t) dt using leftand right-hand Riemann sums for the velocity function (with Δt = 1):

LHS = ∑_{i=1}^{5} v(t_{i−1})Δt = 0 + 10 + 15 + 40 + 45 = 120

and

RHS = ∑_{i=1}^{5} v(t_{i})Δt = 10 + 15 + 40 + 45 + 50 = 170

Since v(t) is apparently increasing on the whole interval, we expect the actual distance traveled to be somewhere between 120 and 170 ft. The average

\frac{120 + 170}{2} = 145

is as good an estimate as we can get from the given information.

B) Since v(t) is apparently increasing the whole time, we expect from class (see also pages 223 - 226 in text) that the true distance traveled is between 120 and 170 and

RHS − LHS = (v(5) − v(0))Δt = 50Δt

To get the average of LHS and RHS to be accurate to within 1 foot we need to take Δt ≤ 1/25. The idea is that if Δt is this small, then the difference between LHS and RHS is at most 2 ft, while the average of LHS and RHS is within 1 ft of every point between them. (I also gave full credit for Δt = 1/50, which says that LHS and RHS are within 1 ft of each other.)

III. A) See class notes.

B) From the given information in the graph of g(t) and the FTC (first part),

G(3) − G(0) = ∫_{0}^{3} g(t) dt = −5 ⇒ G(3) = −5 − 1 = −6
G(4) − G(3) = ∫_{3}^{4} g(t) dt = 1 ⇒ G(4) = −6 + 1 = −5
G(5) − G(4) = ∫_{4}^{5} g(t) dt = −1 ⇒ G(5) = −5 − 1 = −6

The graph of G(t) looks like this:

V. A) Let u = ln(x), so du = 1/x dx. Hence dv = x^{n} dx so v = x^{n+1}/(n + 1). By the integration by parts formula

∫ x^{n} ln(x) dx = \frac{x^{n+1}}{(n + 1)} ln(x) − ∫ \frac{x^{n+1}}{(n + 1)} · \frac{1}{x} dx

= \frac{x^{n+1}}{(n + 1)} ln(x) − \frac{1}{n + 1} ∫ x^{n} dx

= \frac{x^{n+1}}{(n + 1)} ln(x) − \frac{x^{n+1}}{(n + 1)^{2}} + C

(Note – we simplify x^{n+1}/x to x^{n} before integrating on the right.)

B) Let u = sin(2x) (the exponent). Then dx = 2 cos(2x) dx, so the form of the given integral is

\frac{1}{2} ∫ e^{u} du = \frac{1}{2}e^{u} + C = \frac{1}{2}e^{sin(2x)} + C

C) Complete the square in the quadratic in the denominator to see that

x^{2} + 12x + 40 = (x + 6)^{2} + 4

This says that there are no real roots of the quadratic, and hence we want to use # 25 in the table. To set up for this, make the substitution x + 6 = u, so x = u − 6, and du = dx. Then

∫ \frac{5x + 6}{x^{2} + 12x + 40} dx = ∫ \frac{5(u − 6) + 6}{u^{2} + 4} du

= ∫ \frac{5u − 24}{u^{2} + 4} du

= \frac{5}{2} ln |u^{2} + 4| − \frac{24}{2} arctan (\frac{u}{2}) + C

= \frac{5}{2} ln |x^{2} + 6x + 40| − 12 arctan (\frac{x + 6}{2}) + C

D) The most direct way to do this one is to use the trig substitution x = 3 sin(θ), so dx = 3 cos(θ) dθ. Substituting in we get

∫ \frac{x^{2}}{\sqrt{9 − x^{2}}} dx = ∫ \frac{9 sin^{2}(θ) · 3 cos(θ) dθ}{\sqrt{9 − 9 sin^{2}(θ)}}

= ∫ \frac{9 sin^{2}(θ) · 3 cos(θ) dθ}{3 cos(θ)}

= 9 ∫ sin^{2}(θ) dθ

Hence, from # 17 in the table of integrals,

= 9 (−\frac{1}{2} sin(θ) cos(θ) + \frac{1}{2}θ + C)

= \frac{−9}{2} \frac{x}{3} \frac{\sqrt{9 − x^{2}}}{3} + \frac{9}{2} arcsin(x/3) + C

= \frac{−x\sqrt{9 − x^{2}}}{2} + \frac{9}{2} arcsin(x/3) + C

This can also be done using integration by parts if you let u = x, and dv = x/\sqrt{9 − x^{2}} (need the power of x with the radical to be able to integrate it by substitution).