Name: Math 241: Calculus & Analytic Geometry A
Exam 3 Solution Guide
November 27 2002

Were it not for Occam's Razor, which always demands simplicity, I'd be tempted to believe that human beings are more influenced by distant causes than immediate ones. This would be especially true of overeducated people, who are capable of thinking past the immediate and becoming obsessed by the remote.
- from ``Straight Man'' by Richard Russo

Instructions: Show all work to receive full or partial credit. All University rules and guidelines for student conduct are applicable.

Beneath each problem is percentage of students who earned 80% or more of the points for that question.

1. [10 pts] Find
   $$\lim_{x \rightarrow \infty} \frac{(2x+1)^2}{3x^2-1}$$
   if it exists.

   
   

   Section 016: 80% Section 017: 92.3% Section 018: 80% All sections: 84.2%

   Simplify and divide the numerator and denominator by $x^2$ which is the greatest power in the denominator, and one obtains

   $$\lim_{x \rightarrow \infty} \frac{(2x+1)^2}{3x^2-1} = \lim_{x \ri... ...\infty} \frac{4 + \frac{4}{x} + \frac{1}{x^2}}{3-\frac{1}{x^2}} = \frac{4}{3}.$$

2. [10 pts] Sketch the curve
   $$y = \frac{(x+4)(x-3)^2}{x^4 (x-1)}$$
   using asymptotes and intercepts, but no derivatives. Label all asymptotes and intercepts.

   
   

   Section 016: 32% Section 017: 69.2% Section 018: 44% All sections: 48.7%

   The rational function is already factored, so it is simply a matter of examining the function. The roots (intercepts) are at $-4$ and $3$. There is no $y$-intercept because $x=0$ is an asymptote as is $x=1$. We also see that $\lim_{x \rightarrow \pm \infty} y = 0$. By inspecting the sign of $y$ on each side of the asymptotes, one can build the following qualitative sketch.

   

   
   

3. [15 pts] Find the area under the curve
   $$f(x) = 4 x^3 + 1$$
   between $x=0$ to $x=1$.

   
   

   Section 016: 80% Section 017: 100% Section 018: 88% All sections: 89.5%

   This is a direct application of the definite integral:

   $$A = \int_0^1 (4 x^3 + 1) dx = \left. x^4 + x \right\vert _0^1 = 2.$$

4. [15 pts] Find the area under the curve
   $$f(x) = \csc^2 x$$
   between $x=\frac{\pi}{4}$ and $x=\frac{3 \pi}{4}$.

   
   

   Section 016: 44% Section 017: 92.3% Section 018: 52% All sections: 63.2%

   Again, this is a direct application of the definite integral, but this time it has a more complicated integrand:

   $$A = \int_{\frac{\pi}{4}}^{\frac{3 \pi}{4}} \csc^2 x dx = \left. - \cot x \right\vert _{\frac{\pi}{4}}^{\frac{3 \pi}{4}} = 2$$

   
   

5. [15 pts] Find $\frac{df}{dx}$ if
   $$f(x) = \int_2^{x^2} (\cos s + 1) ds.$$

   
   

   Section 016: 56% Section 017: 73.1% Section 018: 72% All sections: 67.1%

   This is a direct application of the Fundamental Theorem of Calculus (version 1). In this case, we must apply the chain rule on the upper limit of integration.

   $$\frac{df}{dx} = \left[\cos(x^2) + 1\right] (2 x).$$

6. [15 pts] Approximate the root of
   $$f(x) = x^3 + x +1$$
   using two steps of Newton's Method with an initial guess of $x_0 = 0$.

   
   

   Section 016: 80% Section 017: 96.2% Section 018: 96% All sections: 90.8%

   We recall that the basic algorithm for Newton's method is

   $$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}.$$
   Therefore, we see that
   $$x_{n+1} = x_n - \frac{x_n^3 + x_n +1}{3 x_n^2 + 1}.$$
   If we plug n' chug, we find that
   

   $$x_0 =$$ 0

   $$x_1 = -1$$

   $$x_2 = -\frac{3}{4}$$

   
   

7. [20 pts] A cardboard box is constructed from a sheet of cardboard by folding along the dotted lines and taping along the matching edges (1-7). Maximize the volume enclosed by the box if one has 120 inches of tape, and that the tape must cover each of the matching edges.
   
   .

   
   

   Section 016: 60% Section 017: 80.8% Section 018: 84% All sections: 75%

   Of course, we want to maximize the volume defined to be

   $$V = w^2 h.$$
   The constraint is that we only have a limited amount of tape (but plenty of cardboard apparently). Thus,
   $$4h + 3w = 120$$
   or
   $$h = 30 - \frac{3}{4} w.$$
   Now, if we substitute this expression for $h$ into our equation for $V$, we find that
   $$V = 30 w^2 - \frac{3}{4} w^3.$$
   We maximize volume when $V' = 0$ so
   $$V' = 60 w - \frac{9}{4} w^2 = w (60 - \frac{9}{4} w).$$
   Therefore, we maximize $V$ when $w=\frac{80}{3}$ and $h=10$. Using these values, we see that the greatest possible volume is $\frac{64000}{9}$ cubic inches.