Mathematics 133 – Intensive Calculus for Science 1 Solutions for Midterm Exam 3 – December 2, 2005

I. f(x) = xe^{−x^{2}}. A) By the product and chain rules:

f′(x) = e^{−x^{2}} + xe^{−x^{2}}(−2x) = (1 − 2x^{2})e^{−x^{2}}.

B) The critical points of f occur when 1 − 2x^{2} = 0, so x = \frac{±1}{\sqrt{2}} = \frac{±\sqrt{2}}{2} ≐ ±.707. Of these, only −\sqrt{2}/2 ≐ −.707 is in our interval. Evaluating f at the critical point and at the endpoints:

f(−2) = −2e^{−4} ≐ −.0366
f(−\sqrt{2}/2) = −\sqrt{2}/2 · e^{−1/2} ≐ −.4289
f(0) = 0

So the maximum value is f(0) = 0 and the minimum is f(−\sqrt{2}/2) = −\sqrt{2}/2 · e^{−1/2} ≐ −.4289.

C) Yes it does. From the form of f′ we can see that f′ changes from negative to positive at −\sqrt{2}/2, so by the First Derivative Test, this is a local minimum.

II. A) By the quotient rule:

f′(x) = \frac{(cos(x) + sin(x)) cos(x) − sin(x)(− sin(x) + cos(x))}{(sin(x) + cos(x))^{2}}

= \frac{cos^{2}(x) + sin(x) cos(x) + sin^{2}(x) − sin(x) cos(x)}{(sin(x) + cos(x))^{2}}

= \frac{1}{(sin(x) + cos(x))^{2}}

B) By the rule for natural log plus the chain rule:

g′(x) = \frac{1}{3x + \frac{1}{x}} (3 − \frac{1}{x^{2}} ) = \frac{3x^{2} − 1}{3x^{3} + x}

C) By the rule for arcsin and the chain rule:

h′(x) = \frac{1}{\sqrt{1 − (4x)^{2}}} · 4 = \frac{4}{\sqrt{1 − 16x^{2}}} .

D) By implicit differentiation (using product rule):

y^{2} + 2xy \frac{dy}{dx} + 2xy + x^{2} \frac{dy}{dx} = 0

Solving for \frac{dy}{dx}:

(2xy + x^{2})\frac{dy}{dx} = −2xy − y^{2},

so

\frac{dy}{dx} = \frac{−2xy − y^{2}}{2xy + x^{2}} .

III. Call the dimensions of the rectangle x, y (in yards). Say x is the side parallel to the existing fence. Then we have xy = 500 and we want to minimize L = x + 2y = x + \frac{1000}{x}. Then L′ = 1 − \frac{1000}{x^{2}} = 0 when x = \sqrt{1000} = 10\sqrt{10}. This is a minimum, since it is the only critical point with positive x, and L′′ = \frac{1000}{x^{3}}, so L′′ > 0 for all x > 0. The dimensions of the lot with minimum fence are x = 10\sqrt{10} ≐ 31.6 and y = 5\sqrt{10} ≐ 15.8 (from xy = 5000).

IV. For the sphere, V = \frac{4π}{3} r^{3}. So taking time derivatives:

\frac{dV}{dt} = 4πr^{2} \frac{dr}{dt}

We are given \frac{dr}{dt} = 3, so when r = 6,

\frac{dV}{dt} = 4π · (6)^{2} · 3 = 432π

(units are mm^{3}/min).