Name: Calculus III
Exam 2 Solutions
1 November 2000



``... I never went to Cambridge as my brothers did. I had the chance, but I refused it. I wanted to get out into the world. I've always regretted it. I think it would have saved me a lot of mistakes. You learn more quickly under the guidance of experienced teachers. You waste a lot of time going down blind alleys if you have no one to lead you.''

``You may be right. I don't mind if I make mistakes. It may be that in one of the blind alleys I may find something to my purpose.''

from The Razor's Edge by W. Somerset Maugham


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

1.

[20 pts] If

$$f(x,y) = \cos(x^2 + 3xy + 2y^2),$$

calculate $\frac{\partial f}{\partial x}$, $\frac{\partial f}{\partial y}$, $\frac{\partial^2 f}{\partial x^2}$ and $\frac{\partial^2 f}{\partial x \partial y}$.

These calculations are straightforward applications of the chain rule and the product rule.

$$\begin{eqnarray*}\frac{\partial f}{\partial x} & = & -(2x+3y)\sin(x^2 + 3xy + 2... ...-(2x+3y)(3x+4y)\cos(x^2 + 3xy + 2y^2) - 3\sin(x^2 + 3xy + 2y^2) \end{eqnarray*}$$

2.

[20 pts] Prove that

$$\frac{x^2-y^2}{x^2+2y^2}$$

is not continuous at (0,0).

Here it suffices to find two parametric paths to the origin which approach different limits. It is not enough to observe that the limit of the expression above is an indeterminant form. Many continuous limits are indeterminant forms. (For instance, all derivatives are indeterminant forms.) The easiest limits to study follow the coordinate axes. So you could try

$$x_1 = t, \ \ \ \ \ y_1 = 0$$

for the x-axis. Substituting this parametrization into the limit, you obtain

$$\lim_{t \rightarrow 0} \frac{t^2-0}{t^2+0} = \lim_{t \rightarrow 0} 1 = 1.$$

For the y-axis, you could try

$$x_2 = 0, \ \ \ \ \ y_2 = t$$

for the x-axis. Substituting this parametrization into the limit, you obtain

$$\lim_{t \rightarrow 0} \frac{0-t^2}{0-2t^2} = \lim_{t \rightarrow 0} -\frac{1}{2} = -\frac{1}{2}.$$

Since the limit approaching the origin varies with the path, the function cannot be continuous at the origin.

3.

[20 pts] Find the equation of plane tangent to the surface

x² - y² - z² = 1

at the point $(\sqrt{3},1,1)$. Sketch the surface and the tangent plane. (Be sure to clearly label the x, y and z axes.)

Since a point on the plane is already supplied in the question, one just needs to find the normal vector to the plane. Since the surface is expressed as a level set, the gradient will be a normal vector.

$$\nabla(x^2 - y^2 - z^2) = 2 x \widehat{i} - 2 y \widehat{j} - 2 z \widehat{k}$$

At the point in question, the vector $2 \sqrt{3} \widehat{i} - 2 \widehat{j} - 2 \widehat{k}$ is a normal vector. Notice that any scalar multiple of this vector is also a normal vector to the plane. Thus, an equation for the tangent plane would be

$$2 \sqrt{3} (x-\sqrt{3}) - 2(y-1) - 2(z-1) =0.$$

If you so choose, you can simplify it further:

$$\sqrt{3} x - y - z = 1.$$

As for the sketch, I am no artist, but I can do much better by hand than with a drawing program.

$\parbox{1.5in}{\epsfxsize=1.5in \epsfbox{solnA.eps}}$

4.

[20 pts] Find the critical points of the function

f(x,y) = x² + 5xy + 2y².

This problem was relatively simple. Some students found it deceptively so. I offer no apologies for this, and I ask that you have confidence in your knowledge of mathematics so that you can find a solution and know that it is true. To find the critical points, one must find all points (x,y) such that

$$\nabla f = \v 0.$$

Thus, the following two equations must be true.

$$2x + 5 y = 0 \ \ \ \ \ 5 x + 4 y = 0$$

The only solution that satisfies both of these equations is (0,0).

5.

[20 pts] Find the maximum and minimum of

f(x,y) = xy

subject to the constraint

x² + y² = 4.

Here, we use Lagrange multipliers.

$$\begin{eqnarray*}\nabla f & = & \lambda \nabla g \\ \langle y,x \rangle& = & \lambda \langle2 x, 2 y \rangle \end{eqnarray*}$$

Thus,

$$\frac{y}{2x} = \frac{x}{2y}$$

or

x² = y².

A common error in this problem was to assume that y=x. In fact, $y=\pm x$. Substituting this back into the constraint, one finds four solutions:

$$(\sqrt{2},\sqrt{2}), (\sqrt{2},-\sqrt{2}), (-\sqrt{2},\sqrt{2}), (-\sqrt{2},-\sqrt{2}).$$

If you check each of these points, you find that the minimum is -2 and the maximum is 2.