Name: Calculus III
Exam 1 Solution Notes
4 October 2000



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/graphing calculator. All University rules and guidelines for student conduct are applicable.

1.

[15 pts] If

$$\begin{eqnarray*}\v a & = & \langle5, -1 \rangle, \\ \v b & = & \langle10, 1 \rangle, \\ \v c & = & \langle-1, 2 \rangle, \\ \end{eqnarray*}$$

find scalars k and l such that

$$k \v a + l \v b = \v c.$$

What is the angle between $\v a$ and $\v b$?

The vector equation above is a system of two equations and two unknowns:

$$\begin{eqnarray*}5k + 10l & = & -1 \\ -k + l & = & 2 \end{eqnarray*}$$

The solution is k=-7/5 and l=3/5. Common errors included not understanding vectors and inability to solve linear systems. While solving linear systems is not a topic covered in 92.231, it is a reasonable expectation that students of calculus can solve linear systems. The second question merely applies the relationship

$$\v a \cdot \v b = \vert\v a\vert \vert\v b\vert \cos \theta.$$

Routine calculations yield that

$$\begin{eqnarray*}\v a \cdot \v b & = & 49, \\ \vert\v a\vert & = & \sqrt{26}, \\ \vert\v b\vert & = & \sqrt{101}. \end{eqnarray*}$$

Thus, we find that

$$\theta = \cos^{-1} \left( \frac{49}{\sqrt{26}\sqrt{101}} \right) \approx 0.2971 {\rm\ rad} {\rm\ or \ } 17.02^o.$$

2.

[20 pts] If

$$\v r(t) = e^{-t} \cos(t) \widehat{i} + e^{-t} \sin(t) \widehat{j} + t \widehat{k},$$

find the unit tangent vector, $\v T$, and the curvature, $\kappa$ when t=0. What is the radius of curvature at this point?

One acquires the solution here by computing the following quantities:

$$\begin{eqnarray*}\v T & = & \frac{\v r'(0)}{\vert\v r'(0)\vert}, \\ \kappa & =... ...rac{\vert\v r'(0) \times \v r''(0)\vert}{\vert\v r'(0)\vert^3}. \end{eqnarray*}$$

The radius of curvature is $\frac{1}{\kappa}$. Common problems included a lack of familiarity with the definitions of unit tangent vectors, curvature and radius of curvature as well as a failure to differentiate $\v r(t)$ correctly. To calculate the quantities above, we must differentiate $\v r(t)$ twice.

$$\begin{eqnarray*}\v r'(t) & = & - e^{-t} [\sin t + \cos t] \widehat{i} + e^{-t} ... ... (2 e^{-t} \sin t )\widehat{i} - (2 e^{-t} \cos t )\widehat{j} \end{eqnarray*}$$

Hence,

$$\begin{eqnarray*}\v r'(0) & = & - \widehat{i} + \widehat{j} + \widehat{k}, \\ \v r''(0) & = & - 2 \widehat{j}. \end{eqnarray*}$$

Further computation will reveal that

$$\v T = \frac{1}{\sqrt{3}} (- \widehat{i} + \widehat{j} + \widehat{k})$$

and

$$\kappa = \frac{2 \sqrt{6}}{9}.$$

The radius of curvature simplifies to $\frac{3 \sqrt{6}}{4}$.

3.

[15 pts] Which two of the four parametrized lines below represent the same line?

$$\begin{eqnarray*}\v r_1(t) & = & \langle2, 1, 0 \rangle+ \langle1, -\frac{1}{2},... ...r_4(t) & = & \langle2, 0, 1 \rangle+ \langle-4, 2, 16 \rangle t \end{eqnarray*}$$

Many students correctly determined that $\v r_1$ and $\v r_3$ are parallel. However, two parallel lines are not necessarily the same line, right? In fact, a little more checking would reveal that $\v r_2$ and $\v r_4$ are also parallel. Which pair represents the same line? You must check and see if a point on $\v r_1$ is also on $\v r_3$, or similarly, if a point on $\v r_2$ is on $\v r_4$. For instance, $\v r_1(0) = \langle 2,1,0 \rangle$. Is there a t₀ such that $\v r_3(t_0) = \v r_1(0)$? The answer is yes! You can solve for t₀ and see that

$$\v r_3(-1) = \v r_1(0).$$

You can stop here because the question implies that only two of the four represent the same line. However, if you tried to check on the $\v r_2$ - $\v r_4$ pair, you would find that they have no points in common because you can find no such t₀.

4.

[20 pts] Find the distance from the point (-2, 1, 3) to the plane

x+y+2z=1.

We worked a problem just like this in class, and another one was assigned as a homework problem. The most common problem came when students tried to memorize a formula for solving this problem and put the wrong numbers into the formula. Memorizing formulas for Calc III is a major mistake. The best way to succeed is the learn the material so that you understand how to arrive at the formulas. Essentially, you must set up a four by four linear system

$$\begin{eqnarray*}l \langle 1,1,2 \rangle & = & \langle x_0 + 2, y_0 -1, z_0 -3 \rangle \\ x_0 + y_0 + 2 z_0 & = & 1 \end{eqnarray*}$$

and solve for l. You can also solve for the position of the closest point on the plane, but you do not need it to answer the question. After a little simplification, one finds that l=-2/3. This scalar, l, is the amount you must stretch the normal vector to connect the point to the plane. Thus, the distance is

$$\vert-\frac{2}{3} \langle 1, 1, 2\rangle\vert = \frac{2}{3} \sqrt{6} \approx 1.633.$$

5.

[15 pts] Find the area of the triangle with vertices (-3, -2, -2), (2, 3, -2) and (1,-2,2).

If we were to label the points above as A, B, and C, we would want to find half the area of the parallelogram formed by vectors AB and AC. If we call these vectors $\v v_1$ and $\v v_2$, respectively, we have

$$\begin{eqnarray*}\v v_1 & = & \langle -5, -5, 0 \rangle \\ \v v_2 & = & \langle -4, 0, -4 \rangle. \end{eqnarray*}$$

Note: you could also choose to work with AB and BC. The combination of sides does not matter, as long as you are consistent. The area of the parallelogram is

$$\vert\v v_1 \times \v v_2\vert = \vert\langle -20, -20, -20 \rangle\vert = 20 \sqrt{3}.$$

Thus, the area of the triangle is $10 \sqrt{3}$ or approximately 17.32.

6.

[15 pts] Find the equation of the plane containing the points (1,1,1), (2,1,3) and (-1,4,5).

This problem is straightforward. To find the equation, you must find a point in the plane and a normal vector. You are given three points in the plane, so really, you only need to find a normal. The three points in the plane specify two vectors, so you just need to find a vector perpendicular to both of these vectors. The cross product of two vectors will always yield a mutually perpendicular vector, so the cross product of two non-collinear vectors in a plane will always yield a normal vector. From the points given above, one can see that the vectors

$$\begin{eqnarray*}\v v_1 & = & \langle -1, 0, -2 \rangle, \\ \v v_2 & = & \langle 2, -3, -4 \rangle \end{eqnarray*}$$

lie in the plane. In this case, I constructed these vectors by subtracting the first point from the second point, and the first point from the third point. You could have made other choices to obtain vectors in the plane, but you will arrive at the same solution. The vector,

$$\v v_1 \times \v v_2 = \langle -6, -8, 3\rangle$$

must be a vector normal to the plane. Thus, an equation for the plane is

-6 (x-1) -8(y-1) + 3(z-1) = 0.

If one wanted to do so, one could simplify it to standard form.

-6 x -8 y + 3 z + 11 = 0.