Mathematics 133 – Intensive Calculus for Science 1 Final Examination – Solutions December 13, 2005

I. f(x) is approximately exponential. With f(x) = ca^{x}, we have

\begin{aligned}ca^{.3} & = 4.12 \\ ca^{.6} & = 3.74 \end{aligned}

So dividing, a^{.3} ≐ .908 and a ≐ (.908)^{10/3} = .724. Then c ≐ 4.12/(.724)^{.3} = 4.54:

f(x) ≐ 4.54(.724)^{x}.

g(x) is linear since the slope is constant: \frac{g(.6)−g(.3)}{.6−.3} = −.8 so g(x) = (−.8)(x − .3) + 4.31 = −.8x + 4.55:

g(x) = −.8x + 4.55.

II. A) The graph is sinusoidal with period 4, amplitude 5 and vertical shift +3:

f(x) = 5 sin (\frac{2πx}{4}) + 3 = 5 sin (\frac{πx}{2}) + 3

B) The function has an inverse function if we restrict to −1 ≤ x ≤ 1 because that portion of the graph passes the horizontal line test: horizontal lines y = k for −2 ≤ k ≤ 8 intersect that section of the graph exactly once. The domain of the inverse function is −2 ≤ x ≤ 8.

III. A)

f′(x) = lim_{h→0} \frac{5(x + h)^{2} − (x + h) + 3 − 5x^{2} + x − 3}{h}

= lim_{h→0} \frac{5x^{2} + 10xh + 5h^{2} − x − h + 3 − 5x^{2} + x − 3}{h}

= lim_{h→0} 10x − 1 + 5h

= 10x − 1.

B) By the chain rule and the rule for exponential functions:

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

C) Rewrite h(x) using powers: h(x) = 3x^{1/5} + 7x^{−1/3} + 3x^{8}. Then

h′(x) = \frac{3}{5}x^{−4/5} − \frac{7}{3}x^{−4/3} + 24x^{7}

D) By the quotient rule:

i′(x) = \frac{(sin^{2}(x) − 1)(− sin(x)) − cos(x) · 2 sin(x) cos(x)}{(sin^{2}(x) − 1)^{2}}

E)

j′(x) = \frac{1}{1 + (3x + 2)^{2}} · 3

F) Using implicit differentiation:

2y \frac{dy}{dx}\sqrt{x} + y^{2} \frac{1}{2\sqrt{x}} − e^{y} \frac{dy}{dx} = 0

Solving for \frac{dy}{dx}:

\frac{dy}{dx} = \frac{−y^{2} \frac{1}{2\sqrt{x}}}{2y\sqrt{x} − e^{y}} = \frac{−y^{2}}{4xy − 2\sqrt{x}e^{y}}

IV. A) f is increasing on intervals where f′(x) ≥ 0, so here for all x > −.5. f is decreasing

where f′(x) < 0, so x < −.5. B) y = f(x) is concave up on intervals where f′′(x) > 0, or equivalently, where f′(x) is

increasing: x < 0 and x > 1. y = f(x) is concave down on intervals where f′(x) is decreasing: 0 < x < 1.

C) The graph should be parabolic in shape with x-axis intercepts at x = 0, x = 1, opening up. The vertex of the parabola should be at x = .5.

V. A) The temperature at time t = 0 is H(0) = 70 − 50 = 20 degrees. B) The rate of change of the temperature at time t = 10 is H′(10) = 5e^{−10/10} ≐ 1.84

degrees F per minute. C) The temperature reaches 60 degrees when

60 = 70 − 50e^{−t/10} ⇒ \frac{−10}{−50} = e^{\frac{−t}{10}} ⇒ t = −10 ln(1/5) ≐ 16.1

minutes.

VI. By the product and chain rules, f′(x) = −e^{−x} sin(x) + e^{−x} cos(x) = e^{−x}(cos(x) − sin(x)). For 0 ≤ x ≤ π, the only solution of f′(x) = 0 is x = π/4. Then

f(0) = 0  f(π/4) = e^{−π/4} sin(π/4) ≐ .322  f(π) = 0

so the minimum value is 0 and the maximum value is e^{−π/4} sin(π/4).

VII. A) The cost to build the pipeline is 3 times the distance under water, plus 2 times the distance along the shore:

C(x) = 3\sqrt{4 + x^{2}} + 2(6 − x)

B) We find the critical points of C:

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

This is zero when

3x = 2\sqrt{x^{2} + 4} ⇒ 9x^{2} = 4x^{2} + 16 ⇒ 5x^{2} = 16

So x = \frac{4}{\sqrt{5}} ≐ 1.79 (ignore the negative root since it clearly cannot give a minimium of the cost!). This is a minimum by the 1st derivative test: C′(1) = \frac{3}{\sqrt{5}} − 2 ≐ −.66 < 0 while C′(2) = \frac{6}{\sqrt{8}} − 2 ≐ .12 > 0. The pipeline of minimum cost goes from the island to the point with x = 1.79 miles along the shore.

VIII. The volume of the cube is V = x^{3} where x is the length of the side. So

\frac{dV}{dt} = 3x^{2} \frac{dx}{dt}

when V = 216, x = \sqrt[3]{216} = 6, and \frac{dV}{dt} = −10, so

−10 = 3(6)^{2} \frac{dx}{dt} ⇒ \frac{dx}{dt} = \frac{−10}{108} = \frac{−5}{54}

The side of the cube is decreasing at ≐ .093 cm/min.