MATH 131, section 1 – Solutions for Practice Questions for Exam 2 October 17, 2007

1. All parts of this question refer to the parametric curve x = 3 cos(2t), y = 5 sin(2t).

a. Eliminate the parameter t and find a Cartesian equation for this curve.

Solution: We have \frac{x}{3} = cos(2t) and \frac{y}{5} = sin(2t), so

(\frac{x}{3})^{2} + (\frac{y}{5})^{2} = cos^{2}(2t) + sin^{2}(2t) = 1.

The equation is \frac{x^{2}}{9} + \frac{y^{2}}{25} = 1, which we recognize as the equation of an ellipse centered at (0, 0), with semimajor axis 5 (along the y-axis), and semiminor axis 3 (along the x-axis).

b. What portion of the curve is traced out for 0 ≤ t ≤ \frac{π}{2}, and in which direction is the curve being traced?

Solution: As t increases from 0 to \frac{π}{2}, cos(2t) is decreasing from 1 to −1, and sin(2t) is increasing from 0 to 1 at t = \frac{π}{4}, then decreasing back to 0 at t = \frac{π}{2}. Therefore, x is decreasing from 3 to −3 and y is increasing up to 5, then decreasing back to 0. The portion of the curve traced out is the top half of the ellipse and we are moving counterclockwise.

c. What would change in your answer to part b if the curve above was replaced by x = 3 cos(−2t), y = 5 sin(−2t)?

Solution: We would be moving clockwise around the bottom half of the ellipse rather than counterclockwise around the top half.

2. Use the sum, product, and/or quotient rules to compute the following derivatives. You may use any correct method, but must show work and simplify your answers for full credit.

a. \frac{d}{dx}(5x\sqrt{x} − \frac{2}{x^{3}} + 11x − 4)

Solution: By the power and sum rules,

\frac{d}{dx}(5x\sqrt{x} − \frac{2}{x^{3}} + 11x − 4) = \frac{15}{2}x^{1/2} + 6x^{−4} + 11.

b. \frac{d}{dt}(t^{2}e^{t})

Solution: By the product rule,

\frac{d}{dt}(t^{2}e^{t}) = t^{2}e^{t} + 2te^{t} = (t^{2} + 2t)e^{t}.

c. \frac{d}{dz} \frac{z^{2} − 2z + 4}{z^{2} + 1}

Solution: By the quotient rule,

\frac{d}{dz} \frac{z^{2} − 2z + 4}{z^{2} + 1} = \frac{(z^{2} + 1)(2z − 2) − (z^{2} − 2z + 4)(2z)}{(z^{2} + 1)^{2}}
= \frac{2z^{2} − 6z − 2}{(z^{2} + 1)^{2}}.

d. \frac{d}{dx}(\frac{π^{2} + tan(e^{π}) − 2x^{e}}{4} )

Solution: Don’t be fooled – most of this function is constant. The only part that is not is the −\frac{x^{e}}{2} , so the derivative is

\frac{d}{dx}(\frac{π^{2} + tan(e^{π}) − 2x^{e}}{4} ) = \frac{d}{dx} (\frac{π^{2}}{4} + \frac{tan(e^{π})}{4} − \frac{x^{e}}{2} ) = 0+0− \frac{ex^{e−1}}{2} = −\frac{ex^{e−1}}{2} .

3. The graph of a function f is shown below with several points marked. Find all the marked points at which the following are true, and give explanations for your answers.

A  B  C D E  F  G

a. f is discontinuous.

Solution: Points B, C, and G. Note that both one-sided limits exist and are equal at x = B, but the limit is not the same as the value of the function there (a removable discontinuity). At C, f has an infinite discontinuity (vertical asymptote). At G, f has a jump discontinuity (both one-sided limits exist but they are unequal). b. f is continuous, but the graph of f has a vertical tangent line.

Solution: This happens at point A only. c. f is continuous, but the graph of f has no tangent line.

Solution: This happens at points E (a “corner”) and F (a cusp).

4. Compute the indicated limits. Show all work for full credit.

a. lim_{x→1} \frac{3x^{2} − 5x − 2}{x^{2} − 4x + 4}

Solution: We have

lim_{x→1} 3x^{2} − 5x − 2 = 3(lim_{x→1} x)^{2} − 5 lim_{x→1} x − 2 = 3 − 5 − 2 = −4

and

lim_{x→1} x^{2} − 4x + 4 = (lim_{x→1} x)^{2} − 4 lim_{x→1} x + 4 = 1 − 4 + 4 = 1

by the limit sum and product rules. Since the limit of the denominator is not zero, the limit quotient rule then says

lim_{x→1} \frac{3x^{2} − 5x − 2}{x^{2} − 4x + 4} = \frac{lim_{x→1} 3x^{2} − 5x − 2}{lim_{x→1} x^{2} − 4x + 4} = \frac{−4}{1} = −4.

b. lim_{x→2} \frac{3x^{2} − 5x − 2}{x^{2} − 4x + 4}

Solution: We cannot apply the technique used in part a here since now

lim_{x→2} 3x^{2} − 5x − 2 = 3(lim_{x→2} x)^{2} − 5 lim_{x→2} x − 2 = 12 − 10 + 2 = 0.

and

lim_{x→2} x^{2} − 4x + 4 = (lim_{x→2} x)^{2} − 4 lim_{x→2} x + 4 = 4 − 8 + 4 = 0.

For x ≠ 2, we have after factoring and cancellation:

\frac{3x^{2} − 5x − 2}{x^{2} − 4x + 4} = \frac{(3x + 1)(x − 2)}{(x − 2)(x − 2)} = \frac{3x + 1}{x − 2}.

Since the denominator is still going to 0 as x → 2 while the numerator is approaching 7, this limit does not exist. In terms of the one-sided limits, in fact,

lim_{x→2^{−}} \frac{3x + 1}{x − 2} = −∞, while lim_{x→2^{+}} \frac{3x + 1}{x − 2} = +∞.

c. lim_{x→∞} \frac{3x^{2} − 5x − 2}{x^{2} − 4x + 4}

Solution: Divide the top and bottom by the fastest growing power of x that appears and then take the limit:

lim_{x→∞} \frac{3x^{2} − 5x − 2}{x^{2} − 4x + 4} = lim_{x→∞} \frac{(3x^{2} − 5x − 2)\frac{1}{x^{2}}}{(x^{2} − 4x + 4)\frac{1}{x^{2}}}
= lim_{x→∞} \frac{3 − \frac{5}{x} − \frac{2}{x^{2}}}{1 − \frac{4}{x} + \frac{4}{x^{2}}}
= 3/1
= 3.

(Recall, this computation would show that the graph y = \frac{3x^{2} − 5x − 2}{x^{2} − 4x + 4} has a

horizontal asymptote at y = 3 as x → +∞ and also as x → −∞.)

d. lim_{x→2^{+}} \frac{|x − 2|}{x^{2} − 5x + 6}

Solution: Note x^{2} − 5x + 6 = (x − 2)(x − 3). Hence the function is

\frac{|x − 2|}{(x − 2)(x − 3)} = \frac{|x − 2|}{x − 2} · \frac{1}{x − 3} = \begin{cases}\frac{1}{x−3} & if x > 2 \\ \frac{−1}{x−3} & if x < 2 \end{cases} .

This shows

lim_{x→2^{+}} \frac{|x − 2|}{x^{2} − 5x + 6} = lim_{x→2^{+}} \frac{1}{x − 3} = −1.

5. Let f(x) = x^{3} − x^{2}.

a. Find all intervals on which f is decreasing.

Solution: To answer this, we need f′(x), since f is decreasing on an interval when f′(x) < 0 on that interval. Here f′(x) = 3x^{2} − 2x = x(3x − 2). The graph y = f′(x) is a parabola opening up. We see f′(x) = 0 at x = 0 and x = 2/3, so f′(x) is negative for 0 < x < 2/3 and positive for x < 0, x > 2/3. Answer: In interval notation, f is decreasing for x ∈ (0, 2/3).

b. Find all intervals on which f is concave up.

Solution: To answer this, we need f′′(x), since f is concave up on an interval when f′′(x) > 0 on that interval. We have f′′(x) = 6x − 2, so f′′(x) < 0 for x < 1/3 and f′′(x) > 0 for x > 1/3. Answer: f is concave up for x ∈ (1/3, +∞) c. Find all intervals on which f is both increasing and concave down.

Solution: We want the intervals where f′(x) > 0 and f′′(x) < 0. From the solutions for parts a, b, this is true only when x < 0. Answer: f is increasing and concave down for x ∈ (−∞, 0).

6. Do not use the differentiation rules from Chapter 3 in this question.

a. State the limit definition of the derivative f′(x).

Solution:

f′(x) = lim_{h→0} \frac{f(x + h) − f(x)}{h},

if the limit exists. b. Use the definition to compute the derivative function of f(x) = \frac{1}{3x}.

Solution: Using the definition from part a,

f′(x) = lim_{h→0} \frac{\frac{1}{3(x + h)} − \frac{1}{3x}}{h}
= lim_{h→0} \frac{3x − 3(x + h)}{9hx(x + h)} (simplify fraction)
= lim_{h→0} \frac{−3h}{9hx(x + h)} (cancel like terms on top)
= lim_{h→0} \frac{−1}{3x(x + h)} (cancel the 3h top and bottom)
= \frac{−1}{3x^{2}} .

c. Find the equation of the line tangent to the graph y = \frac{1}{3x} at x = 2.

Solution: When x = 2, f(2) = \frac{1}{6}, and f′(2) = \frac{−1}{12} from part b. So we calculate the point-slope form of the equation of the tangent line, which is y − \frac{1}{6} = \frac{−1}{12}(x − 2), or after simplifying,

y = \frac{−1}{12}x + \frac{1}{3}.

7. The total cost (in $) of repaying a car loan at interest rate of r% per year is C = f(r). a. What is the meaning of the statement f(7) = 20000?

Solution: This means that if the interest rate is 7% per year, then the cost of repaying the loan is $20000.

(a) What is the meaning of the statement f′(7) = 3000? What are the units of f′(7)?

Solution: This means that when r = 7% per year, the rate of change of the cost of repaying the loan is $3000 dollars per % per year (the units are dollars per % per year).