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\centerline{MATH 241 -- Multivariable Calculus}
\centerline{Sample Exam Questions for Exam 1 -- Solutions}
\centerline{September 23, 2013}
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\noindent
I.  Consider the set $Q = \{(x,y,z) \in {\bf R}^3 \mid x^2 - y^2/4 - z^2 = 1\}$.
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\item{A)} Identify the slices of $Q$ in planes parallel to the three coordinate planes. 
Take $x,y,z = -2,-1,0,1,2$ (at least).
\item{} {\it Solution:}  The slices by the planes $x = c$ have equations
$y^2/4 + z^2 = c^2 - 1$.  These are ellipses if $c > 1$ or $c < -1$, single points if
$c = \pm 1$ and empty if $-1 < c < 1$.  The slices by planes $y = c$ have equations
$x^2 - z^2 = 1 + c^2/4$.  These are hyperbolas for all $c$ 
in the planes parallel to the $xz$-plane).  The slices by planes $z = c$ have
equations $x^2 - y^2/4 = 1 + c^2$.  These are also hyperbolas.
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\item{B)} Use that information to generate a rough sketch of Q or a verbal description.
\item{} {\it Solution:}
See the Maple plot posted with these solutions.  A verbal description would go something 
like this:  The surface consists of two sheets: one with $x \ge 1$ and the other with $x \le -1$.
There are no points on the surface with $-1 < x < 1$.  
The two sheets both have elliptical cross-sections by planes $x = c$.   
This is called a hyperboloid of two sheets because the sections by planes $y = c$ and $z = c$
are both hyperbolas.  
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\item{C)} Show that the image of the parametric curve 
$\alpha(t) = ((e^t + e^{-t})/2, 0, (e^t - e^{-t})/2)$ for $t$ in ${\bf R}$ lies entirely in $Q$.
\item{} {\it Solution:}  Substituting $x = (e^t + e^{-t})/2$, $y = 0$, and $z = (e^t - e^{-t})/2$
into the equation of $Q$ we have 
$$\eqalign{
x^2 - y^2/4 - z^2 &= \left((e^t + e^{-t})/2\right)^2 - 0 - \left((e^t - e^{-t})/2\right)^2\cr
&= (e^{2t} + 2 + e^{-2t})/4 - (e^{2t} - 2 + e^{-2t})/4\cr
&= 1.\cr}$$
Therefore all points on the image of $\alpha$ satisfy the equation of $Q$, so the 
curve lies entirely in the surface.  
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\noindent
II.
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\item{A)} Find the equation of the plane in ${\bf R}^3$
 containing the point $P = (1,1,1)$ and the line through $Q = (0,1,-1)$ with direction vector $v = (-4,1,2)$.
\item{} {\it Solution:}  The plane contains the vector $(-4,1,2)$ and $P - Q = (1,0,2)$.  Hence a 
normal vector can be found by computing the cross product
$$N = (-4,1,2) \times (1,0,2) = (2,10,-1)$$
The plane has equation 
$$N\cdot (x - 1, y - 1, z - 1) = 2x + 10y - z - 11 = 0.$$
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\item{B)} The points $P,Q,Q + v$ from part A are three corners of a parallelogram in the plane you found. 
Find the fourth corner and the area of that parallelogram.
\item{} {\it Solution:}  The fourth corner is $P + v = (-3,2,3)$.  The area of the parallelogram is
$$\| (-4,1,2) \times (1,0,2) \| = \|(2,10,-1)\| = \sqrt{4 + 100 + 1} = \sqrt{105}.$$
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\item{C)} Two lines with direction vectors $v = (0,1,2)$ and $w = (4,3,2)$ meet at $(0,0,0)$. 
Find the acute angle between the two lines.
\item{} {\it Solution:}  The angle $\theta$ satisfies:
$$\cos\theta = {v\cdot w\over \|v\| \|w\|} = {7\over \sqrt{5} \sqrt{29}}.$$
Hence $\theta = \cos^{-1}\left({7\over \sqrt{5} \sqrt{29}}\right) \doteq 1.31$ radians.
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\item{D)} Show that if $u = (u_1,u_2,u_3)$, $v = (v_1,v_2,v_3)$, and 
$w = (w_1,w_2,w_3)$ are general vectors in space, then $u \times (v + w) = u \times v + u \times w$.
\item{} {\it Solution:}  We have, after applying the distributive law and splitting up the 
vector into a sum of two other vectors:
$$\eqalign{
u \times (v + w) &= (u_1,u_2,u_3) \times (v_1+w_1,v_2+w_2,v_3+w_3)\cr
&= (u_2(v_3 + w_3) - u_3(v_2 + w_2), -(u_1(v_3 + w_3) - u_3(v_1 + w_1)),\cr
&\qquad\qquad u_1(v_2 + w_2) - u_2(v_1 + w_1))\cr
&= (u_2v_3 - u_3v_2 + u_2w_3 - u_3w_2, -(u_1v_3 - u_3v_1) - (u_1w_3 - u_3w_1),\cr 
&\qquad\qquad u_1v_2 - u_2v_1 + u_1w_2 - u_2w_1)\cr
&= (u_2v_3 - u_3v_2,-(u_1v_3 - u_3v_1), u_1v_2 - u_2v_1 ) + \cr
&\qquad\qquad (u_2w_3 - u_3w_2,-(u_1w_3 - u_3w_1),u_1w_2 - u_2w_1) \cr
&= u \times v + u \times w.\cr
}$$
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\noindent
III.
\item{A)} Sketch the parametric curve 
$\alpha(t) = (4 \cos(\pi t), 3 \sin(\pi t))$ for $0 \le t \le 1$. Show the starting and final points and 
indicate the direction the curve is traced with an arrow.
\item{} {\it Solution:}  This is the top half of the ellipse ${x^2 \over 16} + {y^2 \over 9} = 1$, starting at
$\alpha(0) = (4,0)$ and ending at $\alpha(1) = (-4,0)$.  The curve is traced in the counter-clockwise direction.
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\item{B)} Give a parametrization of the circle with center $(1,3)$ and radius $r = 4$ so 
that one full circuit of the circle is traced out clockwise for $0 \le t \le 4\pi$. 
\item{} {\it Solution:}  $\beta(t) = (4\cos(t/2) + 1, -4\sin(t/2) + 3)$ is one such parametrization.
starting from $\beta(0) = (5,3)$.)  Another curve satisfying the given conditions is
$\gamma(t) = (4\sin(t/2) + 1, 4\cos(t/2) + 3)$.  This one starts from $\gamma(0) = (1,7)$.
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\noindent 
IV. All parts of this problem refer to the parametric curve
$$\alpha(t) = ((1+ \cos(t))\sin(t),-(1 + \cos(t))\cos(t)),$$
called a cardioid (see the Maple plot).  
\bigskip
\item{A)} Show that for all $t$, the distance from the origin to the point $\alpha(t)$ is given by $1 + \cos(t)$.
\item{} {\it Solution:}  The distance is the same as the magnitude $\|\alpha(t)\|$.  Computing this, by means
of the basic trig identity, we find:
$$\eqalign{
\|\alpha(t)\| &= \sqrt{(1 + \cos(t))^2 \sin^2(t) + (1 + \cos(t))^2 \cos^2(t)}\cr
&= \sqrt{(1 + \cos(t))^2 (\sin^2(t) + \cos^2(t))}\cr
&= \sqrt{(1 + \cos(t))^2 }\cr
&= |1 + \cos(t)| \cr
&= 1 + \cos(t),\cr
}$$
since $\cos(t) \ge -1$ for all $t$, so $1+ \cos(t) \ge 0$.  
\bigskip
\item{B)} At how many different times $t$ in $[0,2\pi]$ is the $y$-coordinate of $\alpha(t)$ equal to zero? Find them.
\item{} {\it Solution:}  The $y$-coordinate is $-(1 + \cos(t))\cos(t)$.  This is only $= 0$ if either $\cos(t) = -1$
or $\cos(t) = 0$.  There are three such $t$ values in the given interval:  $t = \pi/2, \pi, 3\pi/2$.
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\item{C)} Find a parametrization of the tangent line to the cardioid at the point $\alpha(\pi/4)$.
\item{} {\it Solution:}  The point on the cardioid is
$$\alpha(\pi/4) = ((1 + \sqrt{2}/2)\sqrt{2}/2, -(1 + \sqrt{2}/2)\sqrt{2}/2) 
= \left({1 + \sqrt{2}\over 2}, -{1 + \sqrt{2}\over 2}\right).$$
The tangent vector at a general $\alpha(t)$ is 
$$\alpha'(t) = (-\sin^2(t) + \cos^2(t) + \cos(t), 2\sin(t)\cos(t) + \sin(t)).$$
So $\alpha'(\pi/4) = (\sqrt{2}/2, 1 + \sqrt{2}/2)$.  The tangent line is 
$$\lambda(u) = \alpha(\pi/4) + u \alpha'(\pi/4) 
= \left({1 + \sqrt{2}\over 2}, -{1 + \sqrt{2}\over 2}\right) + u \left({\sqrt{2}\over 2}, 1 + {\sqrt{2}\over 2}\right).$$
\item{D)} Give a parametrization for any one circle whose interior region completely contains the cardioid. 
Explain how you determined your center and radius.
\item{} {\it Solution:}  By part A, every point on the cardioid is distance at most $2$ from the origin.  
So if we take any radius $r > 2$ and center $(0,0)$, the cardioid will be contained in the interior
region of the circle: $\delta(t) = (3\cos(t),3\sin(t))$ gives one such circle if we take $t \in [0,2\pi]$.   
\bye 
