MATH 131, section 1 – Calculus for Physical and Life Sciences Solutions for Sample Exam Questions – Exam 3 November 19, 2013

I. Find y′ and simplify.

(a) y = ln(x) (x^{7} − \frac{4}{\sqrt{x}})

Solution: By the product rule,

\begin{aligned}y′ & = ln(x)(7x^{6} + 2x^{−3/2}) + \frac{1}{x} (x^{7} − \frac{4}{x^{1/2}} ) \\ & = ln(x)(7x^{6} + 2x^{−3/2}) + x^{6} − \frac{4}{x^{3/2}} \end{aligned}

(b) y = (e^{2x} + 2)^{3}.

Solution: By the chain rule (twice),

y′ = 3(e^{2x} + 2)^{2} · e^{2x} · 2 = 6e^{2x}(e^{2x} + 2)^{2}.

(c) y = \frac{x+1}{3x^{4}−1}.

Solution: By the quotient rule,

\begin{aligned}y′ & = \frac{(3x^{4} − 1)(1) − (x + 1)(12x^{3})}{(3x^{4} − 1)^{2}} \\ & = \frac{−9x^{4} − 12x^{3} − 1}{(3x^{4} − 1)^{2}} \end{aligned}

(d) y = \frac{sin(x)}{1+cos(x)}

Solution: By the quotient rule,

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

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

= \frac{1}{1 + cos(x)}

(e) y = tan^{−1}(e^{5x}).

Solution: By the derivative rule for inverse tangent and the chain rule,

y′ = \frac{1}{1 + (e^{5x})^{2}} · e^{5x} · 5 = \frac{5e^{5x}}{1 + e^{10x}} .

(f) xy^{2} − 3y^{3} + 2x^{4} = 2.

Solution: Since the equation involves both x and y we use implicit differentiation. Differentiating thinking of y as a function of x,

2xyy′ + y^{2} − 9y^{2}y′ + 8x^{3} = 0

Then solving for y′:

y′ = \frac{−y^{2} − 8x^{3}}{2xy − 9y^{2}} .

(g) y = cos(x)^{x^{3}}.

Solution: For functions of the form u(x)^{v(x)}, we use logarithmic differentiation. First take ln of both sides

ln(y) = ln(cos(x)^{x^{3}}) = x^{3} ln(cos(x)),

then differentiate using the product and chain rules:

\frac{1}{y} y′ = −x^{3} tan(x) + 3x^{2} ln(cos(x)).

So

y′ = y(−x^{3} tan(x) + 3x^{2} ln(cos(x))) = cos(x)^{x^{3}}(−x^{3} tan(x) + 3x^{2} ln(cos(x))).

II. The quantity of a reagent present in a chemical reaction is given by Q(t) = t^{3}−3t^{2}+t+30 grams at time t seconds for all t ≥ 0.

(a) Over which intervals with t ≥ 0 is the amount increasing? decreasing?

Solution: We need to determine the intervals where Q′(t) = 3t^{2} − 6t + 1 is positive and negative. By the quadratic formula, 3t^{2} − 6t + 1 = 0 when

t = \frac{6 ± \sqrt{24}}{6} = \frac{3 ± \sqrt{6}}{3},

which are both positive numbers. By making a sign chart for Q′ we see Q′(t) > 0 for t in [0, \frac{3−\sqrt{6}}{3}) ∪ (\frac{3+\sqrt{6}}{3}, +∞). (Note the problem just said look at t ≥ 0.) Q′(t) < 0

for t in (\frac{3−\sqrt{6}}{3}, \frac{3+\sqrt{6}}{3}).

(b) Over which intervals is the rate of change of Q increasing? decreasing? The rate of change is increasing (decreasing) when Q′′(t) = 6t − 6 is positive (negative). This is increasing for t > 1 and decreasing for 0 ≤ t < 1 (again, the problem said look only at t ≥ 0 so we are ignoring t < 0).

III. A spherical balloon is being inflated at 20 cubic inches per minute. When the radius is 6 inches, at what rate is the radius of the balloon increasing? At what rate is the surface area increasing? (The volume of a sphere of radius r is V = \frac{4πr^{3}}{3} and the surface area is 4πr^{2}.)

Solution: This is a related rates problem. From the volume formula, since V and r are changing with time t,

\frac{dV}{dt} = 4πr^{2} \frac{dr}{dt} .

We are given that \frac{dV}{dt} = 20, so when r = 6:

20 = 4π(6)^{2} \frac{dr}{dt} ,

so

\frac{dr}{dt} = \frac{20}{144π} = \frac{5}{36π}

(inches per minute). The surface area is changing at the rate

\frac{dA}{dt} = 8πr \frac{dr}{dt} = 8π · 6 · \frac{5}{36π} = \frac{20}{3}

(square inches per minute).

IV. All parts of this question refer to f(x) = 4x^{3} − x^{4}.

(a) Find and classify all the critical numbers of f using the First Derivative Test.

Solution: The first derivative is f′(x) = 12x^{2} − 4x^{3} = 4x^{2}(3 − x), so f has critical numbers at x = 0 and x = 3. The first derivative is positive for x < 0, positive for 0 < x < 3, and negative for x > 3. Hence x = 3 is a local maximum, and x = 0 is neither a local max nor a local min.

(b) Over which intervals is the graph y = f(x) concave up? concave down?

Solution: For concavity, we need f′′(x) = 24x − 12x^{2} = 12x(2 − x). This is negative for x < 0 and x > 2, and positive for 0 < x < 2. Hence y = f(x) is concave up on (0, 2) and concave down on (−∞, 0) ∪ (2, +∞). (Since the graph changes concavity at x = 0, 2, these are points of inflection.)

(c) Sketch the graph y = f(x).

Solution: Omitted – Ask Gopal to sketch in the review session.

(d) Find the absolute maximum and minimum of f(x) on the interval [1, 4].

Solution: On this closed interval, we have a critical number at x = 3. The critical value is f(3) = 4 · 27 − 81 = 27. The values at the endpoints are f(1) = 4 − 1 = 3 and f(4) = 4 · 64 − 256 = 0. Hence the maximum value on the interval is f(3) = 27 and the minimum is f(4) = 0.

V. All three parts of this question refer to the function f(x) = x^{2/3} − \frac{1}{5}x^{5/3}.

(a) Find all the critical numbers of f(x).

Solution: The derivative is

f′(x) = \frac{2}{3}x^{−1/3} − \frac{1}{3}x^{2/3} = \frac{1}{3}x^{−1/3}(2 − x)

The function has critical numbers at x = 0 (f′(0) is undefined – the graph has a cusp point there), and x = 2 (f′(2) = 0).

(b) Find all the inflection points of f(x).

Solution: For this we need

f′′(x) = \frac{−2}{9}x^{−4/3} − \frac{2}{9}x^{−1/3} = −\frac{2}{9}x^{−4/3}(1 + x)

There is just one point of inflection at x = −1. Since x^{−4/3} = \frac{1}{(x^{1/3})^{4}} , that term is positive whenever it is defined, so the sign of f′′ changes only at x = −1.

(c) For which of the critical numbers here is the Second Derivative Test applicable? Why? Determine the type of each such critical number using the Second Derivative Test.

Solution: The Second Derivative Test is applicable at critical numbers where the first derivative is equal to zero; here only at x = 2. We have f′′(2) = −\frac{2}{9} · 2^{−4/3}(1 + 2) = −\frac{2}{3} · 2^{−4/3} < 0. Hence f has a local maximum at x = 2 since the graph is concave down there.