18.01 Practice Exam 2, Solutions

1. Recall that the volume of a (right circular) cone is V = \frac{1}{3}πr^{2}h, where r is the radius of the base, and h is the height. Suppose that the distance from the tip to the base’s rim is 1 unit.

(a) (12 points) Show that V = \frac{1}{3}π(h − h^{3}), where h is between 0 and 1.

Solution: From the diagram, we see there is a right triangle with legs of length r and h and hypotenuse of length 1. Therefore r^{2}+h^{2} = 1. Substituting r^{2} = 1 − h^{2} into V = \frac{1}{3}πr^{2}h gives V = \frac{1}{3}π(h − h^{3}), with 0 ≤ h ≤ 1. (We must have h ≥ 0 since it is a length, and h ≤ 1 since the length of the leg of the right triangle is at most that of the hypotenuse.)

(b) (18 points) Determine the maximum volume of such a cone.

Solution: We wish to maximize the function f(x) = \frac{1}{3}π(x − x^{3}), with 0 ≤ x ≤ 1. Setting the derivative,

f′(x) = \frac{1}{3}π(1 − 3x^{2}),

equal to 0, we get

\begin{aligned}\frac{1}{3}π(1 − 3x^{2}) & = 0; \\ 1 − 3x^{2} & = 0; \\ x^{2} & = 1/3; \\ x & = 1/\sqrt{3} \end{aligned}

(ruling out x = −1/\sqrt{3} since x ≥ 0). This value of x does give a maximum since it is the only critical point in the interval [0, 1], and f′′(x) = −2πx is negative at x = 1/\sqrt{3}. The maximum volume therefore occurs when h = 1/\sqrt{3}, and r^{2} = 1 − h^{2} = 2/3, so

V = \frac{1}{3}π · \frac{2}{3} · \frac{1}{\sqrt{3}} = \frac{2π}{9\sqrt{3}}

cubic units.

2. (24 points) Sketch the graph of f(x) = x^{2} + x^{−1}. Indicate the intervals where the function is increasing, decreasing, concave up or concave down. Label all relative maxima and minima and inflection points.

Solution: The function is defined for x ≠ 0. The derivative is f′(x) = 2x − x^{−2}. We find the critical points by solving

\begin{aligned}2x − x^{−2} & = 0; \\ 2x & = x^{−2}; \\ x^{3} & = 1/2. \end{aligned}

Therefore x = 2^{−1/3} is the only critical point. Now we determine whether f(x) is increasing or decreasing on the following intervals:

\begin{matrix}(−∞, 0) & (0, 2^{−1/3}) & (2^{−1/3}, ∞) \\ ↘ & ↘ & ↗ \end{matrix}

(We filled in the arrows by looking at the sign of f′(x): in the first interval both terms are negative, in the second x^{−2} is bigger than 2x, in the third 2x is bigger.)

Therefore there is a relative minimum at x = 2^{−1/3}, where the value is 2^{−2/3} +2^{1/3} = 3·2^{−2/3}. (You could get a rough estimate of the coordinates from ^{3} \sqrt{2} ≈ 5/4.)

Now we analyze the concavity using f′′(x) = 2 + 2x^{−3}. The possible inflection points satisfy 2+2x^{−3} = 0, for which the only solution is x = −1. We then determine whether f(x) is concave up or concave down on the following intervals by looking at the sign of f′′(x) = 2x^{−3}(x^{3} + 1):

\begin{matrix}(−∞, −1) & (−1, 0) & (0, ∞) \\ ∪ & ∩ & ∪ \end{matrix}

Therefore there is a point of inflection at x = −1, y = 0.

3. (12 points) An elastic band of length x and (circular cross-sections) of radius r is being stretched at the rate of 1 cm/sec. Assuming the volume remains constant, at what rate is the radius shrinking when x = 10 cm and r = 0.1 cm.

Solution: The constant volume is given by V = πr^{2}x, which is π · (0.1)^{2}(10) = 0.1π cm^{3}. Solving πr^{2}x = 0.1π for x gives x = 0.1r^{−2}. Therefore

\begin{aligned}\frac{dx}{dt} & = −0.2r^{−3} \frac{dr}{dt} , so \\ \frac{dr}{dt} & = −50r^{3} \frac{dx}{dt}. \end{aligned}

When r = 0.1 cm and \frac{dx}{dt} = 1 cm/sec, we get \frac{dr}{dt} = −0.05 cm/sec.

4. (6 points) State the Mean Value Theorem.

Solution: Suppose that f(x) is continuous on [a, b] and differentiable on (a, b). Then

f(b) − f(a) = f′(c)(b − a)

for some c in (a, b).

5. Approximate \sqrt{4.04} by

(a) (8 points) applying one iteration of Newton’s method to the function f(x) = x^{2} − 4.04;

Solution: Since f′(x) = 2x, the formula from Newton’s Method gives

x_{n+1} = x_{n} − \frac{x_{n}^{2} − 4.04}{2x_{n}}.

Starting with x_{1} = 2, we get x_{2} = 2 + (0.04/4) = 2.01.

(b) (8 points) using a linear approximation of \sqrt{x} at x = 4.

Solution: Since \frac{d}{dx}x^{1/2} = \frac{1}{2}x^{−1/2}, the linear approximation of

f(x) = \sqrt{x} at x = 4 is

\sqrt{x} ≈ 2 + \frac{1}{2} · 4^{−1/2}(x − 4) = 2 + \frac{1}{4}(x − 4).

This gives \sqrt{4.04} ≈ 2.01.

6. Evaluate the following limits (or explain why the limit does not exist):

(a) (6 points) lim_{x→0} \frac{e^{2x} − 1}{sin x};

Solution: This has indeterminate form 0/0. Applying l’Hôpital’s Rule gives

lim_{x→0} \frac{e^{2x} − 1}{sin x} = lim_{x→0} \frac{2e^{2x}}{cos x} = 2.

(b) (6 points) lim_{x→∞} x^{1/x}.

Solution: This has indeterminate form ∞^{0}. The logarithm is ln x^{1/x} =

\frac{ln x}{x} which has the form ∞/∞. Applying l’Hôpital’s Rule gives

L = lim_{x→∞} \frac{ln x}{x} = lim_{x→∞} \frac{1/x}{1} = 0.

Therefore lim_{x→∞} x^{1/x} = e^{L} = 1.