Mathematics 136 – AP Calculus Makeup Exam 2 Solutions – November 14, 2003

I. (20) In an alternate universe, the “super-attractive” force exerted by two objects of mass m_{1}, m_{2} is directly proportional to the square of the distance between them: F = Sm_{1}m_{2}r^{2}, where S is a positive constant and r is the distance between the masses. Four 1kg masses are fixed at points x = −12, −4, 2, 7 respectively along a straight line in this alternate universe. At what location x along that line should another 2kg mass be placed to minimize the sum of the super-attractive forces exerted on it by the four unit masses?

Solution: The distance between a and b along the number line is |b − a| = |a − b|. When we square, the absolute value can be disregarded, since the sign does not matter. Hence the total force exerted by the four masses on the 2kg mass at x is

\begin{aligned}F (x) & = 2S((x − (−12))^{2} + (x − (−4))^{2} + (x − 2)^{2} + (x − 7)^{2}) \\ & = 2S((x + 12)^{2} + (x + 4)^{2} + (x − 2)^{2} + (x − 7)^{2}) \end{aligned}

To minimize F we differentiate and set to zero:

0 = F′(x) = 2S(2(x + 12) + 2(x + 4) + 2(x − 2) + 2(x − 7)) = 4S(4x + 7)

This is zero if and only if x = −7/4. (This is clearly a local minimum for F, since F′′ = 16S > 0.)

II. In a car moving at 90 ft/sec, the driver suddenly saw an obstacle 400 feet ahead and braked to a stop in 10 seconds. The car’s velocity was recorded by a sensor in its onboard computer every two seconds:

\begin{matrix}t (sec) & 0 & 2 & 4 & 6 & 8 & 10 \\ v(t) (ft/sec) & 90 & 75 & 50 & 25 & 5 & 0 \end{matrix}

A) (10) Can you say for sure from the information here whether the car hit the obstacle or not? Explain, using the left- and right-hand Riemann sums for v(t).

Solution: The left-hand Riemann sum gives

LHS = 2 · 90 + 2 · 75 + 2 · 50 + 2 · 25 + 2 · 5 = 490

The right-hand Riemann sum gives

RHS = 2 · 75 + 2 · 50 + 2 · 25 + 2 · 5 + 2 · 0 = 310

Since the velocity is apparently decreasing on the whole interval, we expect the LHS is an overestimate of the actual distance travelled and the RHS is an underestimate of the actual distance. This says 310 ≤ actual distance ≤ 490. But it is not possible to say for sure from this information whether a collision occurred. B) (10) From the given information, what is your best estimate about whether the car

hit the obstacle?

Solution: The average (RHS + LHS)/2 = 400 exactly. There is a good chance the car did hit or just brush the obstacle.

III. A) (10) State the first and second parts of the Fundamental Theorem of Calculus.

Solution: See text or class notes.

B) (10) The following graph shows y = f(x) for some function. Sketch the graph of the antiderivative F (x) of g(x) with F (x) = 0. What is F (4)?

Solution: The graph of F should have a linear segment from x = 0 to x = 1 of slope 1, so F (1) = 1. Then between x = 1 and x = 2, the graph is part of a parabola opening down. The areas above and below the x-axis between x = 1 and x = 2 exactly balance out so F (2) = F (1) = 1. Between x = 2 and x = 3, the graph is part of a parabola opening up; the area below the axis is −1/2 (a triangle of base 1 and height 1). So F (3)−F (2) = −1/2 and F (3) = 1/2. Then the rest of the graph to x = 4 is part of a parabola opening down and F (4) − F (3) = −1/2, so F (4) = 0.

IV. Methods of integration. In B,C,D you may use the table of integrals. A) (10) Using integration by parts, show that

∫ x^{n} cos(ax) dx = \frac{x^{n}}{a} sin(ax) − \frac{n}{a} ∫ x^{n−1} sin(ax) dx.

Solution: Let u = x^{n}, dv = cos(ax) dx. Then du = nx^{n−1} and v = \frac{1}{a} sin(ax), so using the integration by parts formula,

∫ x^{n} cos(ax) dx = ∫ u dv = uv − ∫ v du = \frac{x^{n}}{a} sin(ax) − \frac{n}{a} ∫ x^{n−1} sin(ax) dx

which is what we had to show.

B) (10) ∫ tan(\sqrt{x})/\sqrt{x} dx

Solution: (This is a u-substitution form.) Let u = \sqrt{x} = x^{1/2}. Then du = \frac{1}{2\sqrt{x}} so the \sqrt{x} in the denominator of the integrand is part of du. We have, using # 7 in the table (or by another substitution)

∫ tan(\sqrt{x})/\sqrt{x} dx = 2 ∫ tan(u) du
= −2 ln(cos(u)) + C
= −2 ln(cos(\sqrt{x})) + C

C) (10) ∫ (3x + 2)/(x^{2} + 8x + 7) dx

Solution: The quadratic in the denominator factors as x^{2} + 8x + 7 = (x + 1)(x + 7) (or, completing square, get (x + 4)^{2} − 9 and the negative sign says there are real roots). Hence we want to use #27 in the table with a = −1, b = −7, c = 3, d = 2. The result is

\frac{1}{6} (((−1)(3) + 2) ln |x + 1| − ((−7)(3) + 2) ln |x + 7|) + C

which simplifies to

\frac{1}{6} (− ln |x + 1| + 19 ln |x + 7|) + C

D) (10) ∫ (4 − x^{2})^{−3/2} dx

Solution: (This is a trigonometric substitution form.) Let x = 2 sin(θ), so dx = 2 cos(θ)dθ. Then the integral becomes

∫ (4 − x^{2})^{−3/2} dx = 2 ∫ (4 cos^{2}(θ))^{−3/2} cos(θ) dθ

= \frac{1}{4} ∫ cos^{−3}(θ) cos(θ) dθ

= \frac{1}{4} ∫ cos^{−2}(θ) dθ

= \frac{1}{4} ∫ sec^{2}(θ) dθ

= \frac{1}{4} tan(θ) + C

To convert back to x, we use sin(θ) = x/2, so cos(θ) = \sqrt{4 − x^{2}}/2, and tan(θ) = sin(θ)/ cos(θ) = x/\sqrt{4 − x^{2}}. The final answer is

∫ (4 − x^{2})^{−3/2} dx = \frac{1}{4} \frac{x}{\sqrt{4 − x^{2}}} + C