Name: Math 243: Calculus C
Exam 2 Solution Guide
28 March 2002

My mind rebels at stagnation. Give me problems, give me work ... and I am in my own proper atmosphere.
-Sherlock Holmes from The Sign of the Four by A. C. Doyle.

Instructions: Show all work to receive full or partial credit. You may use a scientific calculator on this exam. All University rules and guidelines for student conduct are applicable.

1.

[15 pts] Find the arc length of the curve

$$\v r = \langle t \cos t - \sin t, t \sin t + \cos t \rangle,$$

from t=0 to t=3.

This problem involves a straightforward computation of the arclength of a parametric curve. We know that

$$L = \int_0^3 \vert r'(t)\vert dt.$$

We know that

$$r'(t) = \langle -t \sin t, t \cos t \rangle,$$

so that

|r'(t)| = t,

when t is positive. Thus,

$$L = \int_0^3 t dt = \frac{9}{2}.$$

2.

[15 pts] Find the curvature of

$$\v r = \langle e^t \cos t, e^t \sin t, t \rangle$$

at t=0.

There are several ways to compute $\kappa$, but since we are given the parametric curve, the most direct is by computing

$$\kappa(0)=\frac{\vert r'(0) \times r''(0)\vert}{\vert r'(0)\vert^3}$$

So, we must differential r(t) twice.

$$\begin{eqnarray*}r'(t) & = & \langle e^t(\cos t - \sin t), e^t(\sin t + \cos t)\rangle \\ r''(t) & = & \langle -2 e^t \sin t, 2 e^t \cos t\rangle \end{eqnarray*}$$

Now, you can substitute t=0, and compute the curvature:

$$\kappa(0) = \left(\frac{2}{3}\right)^{\frac{3}{2}}$$

3.

[15 pts] Prove that

$$\lim_{(x,y) \rightarrow (0,0)} \frac{xy}{x^2+3y^2}$$

does not exist.

We seek to find two (or more) paths, r₁ and r₂, which pass through the origin at t=0 such that $\lim_{t \rightarrow 0} f[r_1(t)] \neq \lim_{t \rightarrow 0} f[r_2(t)]$. There are many ways to do this. One example would be

$$\begin{eqnarray*}r_1(t) & = & \langle t, 0\rangle \\ r_2(t) & = & \langle t, t\rangle \end{eqnarray*}$$

In the first case, $\lim_{t \rightarrow 0} f[r_1(t)] = 0$. In the second case, $\lim_{t \rightarrow 0} f[r_2(t)] = \frac{1}{4}$.

4.

[20 pts] Find an equation for the tangent plane of the surface

f(x,y) = 3 - 2 xy + y² - x²

at (1,3,5).

To find the equation of the tangent place, we first find the gradient because we know that

$$\left\langle \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, -1 \right\rangle$$

is normal to the tangent plane. Thus, $\langle -8, 4, -1\rangle$ is normal to the tangent plane. Using the standard form for a plane knowing a normal vector and a point on the plane

-8(x-1) + 4(y-3) - (z-5) = 0.

5.

[15 pts] Find the absolute maximum and minimum of

z = 7 + x + 3 y

on the disk $x^2 + y^2 \leq 4$.

To solve this problem, you must find the critical points in the interior and also check the boundary using Lagrange multipliers or some other technique. Since

$$\nabla f = \langle 1,3 \rangle,$$

we see that there are no critical points in the interior. If we use

g(x,y) = x² + y² = 4

as a constraint for the boundary, we must solve

$$\nabla f = \lambda \nabla g.$$

Solving these equations, we see that y=3x. Applying the constraint, we find that the two candidates are $(\sqrt{10}/5,3\sqrt{10}/5)$ and $(-\sqrt{10}/5,-3\sqrt{10}/5)$. The former corresponds to a maximum at $7 + 2\sqrt{10}$ and the latter corresponds to the minimum at $7-2\sqrt{10}$.

6.

[20 pts] Find all critical points, the absolute maximum and absolute minimum of

$$f(x,y) = \sin(\pi x) \sin(\pi y),$$

on the domain $0 \leq x \leq 2, 0 \leq y \leq 2$. Sketch a contour plot of f(x,y).

This one looks harder than it is. You must check the interior and the boundary. However, if you look at f, you can see that f(x,y)=0 on the boundary. Now, we only need to check the interior for critical points.

$$\nabla f = \langle \pi \cos(\pi x) \sin(\pi y),\pi \sin(\pi x) \cos(\pi y) \rangle.$$

Both components must be zero at the same time. For instance, for the first component to the be zero, either $x=\frac{1}{2}$ $x=\frac{3}{2}$or y = 1. Remember, we are only interested in the interior of the domain, so we do not think about y=0 or y=2, or many other possibilities for x and y that are completely outside the domain. However, notice also that if $x=\frac{1}{2}$ than y must be either $\frac{1}{2}$ or $\frac{3}{2}$ to force the second component of the gradient to be zero. Below, I will tabulate all the interior critical points together with D and the points' classification. As usual, we can distinguish between a max and a min by checking the second derivative in either the x or y direction.

Point	D	Type
$\left(\frac{1}{2},\frac{1}{2}\right)$	1	max
$\left(\frac{1}{2},\frac{3}{2}\right)$	1	min
$\left(\frac{3}{2},\frac{1}{2}\right)$	1	min
$\left(\frac{3}{2},\frac{3}{2}\right)$	1	max
$\left(1,1\right)$	-1	saddle

A rough sketch might look like