Name: Math 243: Calculus C
Exam 1 solution guide
28 February 2002

Yes, mathematics has two faces; it is the rigorous science of Euclid but it is also something else. Mathematics presented in the Euclidean way appears as a systematic, deductive science; but mathematics in the making appears as an experimental, inductive science.
- from ``How to Solve It'' by G. Polya

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.

[20 pts] Find the equation of the plane passing through the points (-1,1,2), (2,1,3) and (0,3,-1).

We must find a normal vector to the plane and a point in the plane. Since we already have three points in the plane, the only work we have to do is to find a normal vector. Subtracting the second point from the first, and the third from the first, we obtain two vectors lying along the plane: $\langle-3,0,1\rangle$ and $\langle-1,-2,3\rangle$. To find a normal vector, we need a vector that is perpendicular to both of these vectors. Thus,

$$\v n = \langle-3,0,1\rangle \times \langle-1,-2,3\rangle = \langle-2,10,6\rangle$$

Now that we have a normal, we can choose any point on the plane to find its equation:

-2 (x+1) + 10 (y-1) + 6(z-2) = 0,

or if you want to simplify it,

-2x+10y+6z = 24.

2.

[15 pts] For what values of parameter s are the vectors $\langle 1,1,-1 \rangle$ and $\langle 2, s^2, 3 \rangle$ orthogonal?

Two vectors are orthogonal if and only if their dot product is zero. So,

$$\langle 1,1,-1 \rangle \cdot \langle 2,s^2,3 \rangle = s^2 -1.$$

Therefore, we see that the two vectors are orthogonal when s=1 or when s=-1.

3.

[20 pts] Find the equation of the line tangent to the spacecurve

$${\v r}(t) = \langle e^{\cos t}, t^2 -t, \sin t \rangle$$

at $t = \pi$.

To find the equation of a line, we need a point on the line and a direction vector. To find the point on the line, we will use

$${\v r}(\pi) =\langle e^{-1},\pi^2-\pi,0 \rangle.$$

To find a direction vector, we must differentiate once

$${\v r}'(t) =\langle -\sin t e^{\cos t},2t -1,\cos t \rangle,$$

and evaluate it at $t = \pi$,

$${\v r}'(\pi) =\langle 0,2\pi -1,-1 \rangle.$$

Putting all this together, the equation for the tangent line is

$$\begin{eqnarray*}{\v l}(t) & = & \langle e^{-1},\pi^2-\pi,0 \rangle + \langle 0,... ...le t \\ & = & \langle e^{-1},\pi^2-\pi + (2\pi-1) t,-t \rangle. \end{eqnarray*}$$

4.

[15 pts] Sketch the surface

x² - 8x + y² + 4y + z² + 16z + 80 = 0.

Be sure to label your axes.

Seeing the quadratic equation, we can put this expression into a recognizable form by completing the square:

(x-4)² + (y+2)² + (z+8)² = 4.

Thus, we see that it is a sphere of radius 2, centered at the point (4,-2,-8). To make the sketch, I used a figure drawing package so that I could provide the class with an online solution guide. Honestly, a hand sketch would have been quicker, easier and probably would have looked better.

5.

[15 pts] Sketch the surface

$$x^2 + y^2 - 4 z^2 = 0, \ \ \ \ z \geq 0.$$

Be sure to label your axes.

This is the top half of a cone, scaled as shown. (Again, I used a figure drawing package.)

6.

[15 pts] Express the surface

$$x^2 + y^2 - 4 z^2 = 0, \ \ \ \ z \geq 0,$$

from the previous problem in cylindrical coordinates AND in spherical coordinates.

For cylindrical coordinates, we can substitute r² for x² + y²to obtain the simple relation r² = 4z². Since we know that zmust be positive, we can take the square root of both sides to get r = 2z, and we know that $\theta$ has no restrictions because the surface is symmetric about the z axis.

For spherical coordinates, we can see by inspection that $\rho$ and $\theta$ are unrestricted. The cone is generated by fixing an angle $\phi$ and then allowing $\rho$ and $\theta$ to vary over their full range. To find the angle, $\phi$, we can apply some trigonometry. For instance, we know that in the yz-plane, the edge of the cone passes through the point (0,2,1). Thus,

$$\phi = \arctan(2)$$

is the equation for the surface in spherical coordinates.