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\begin{center}
{\Large Key for TEST 2}
\end{center}

\begin{enumerate}

\item  
\begin{enumerate}
\item  Let $(x,y,z)=(-2,-2,-3)$, find the coordinates of this point in cylindrical coordinates.
\answer: We need to find $(r, \theta, z)$.
	Converting to cylindrical coordinate, $z$ stays the same so we don't need to worry about that.
	$r^2=x^2+y^2 = 4+4=8$.  So, we have $r=\sqrt{8}=2\sqrt{2}$.
	Next, $\tan \theta=y/x=1$.  We just have to make sure $\theta$ in the the correct quadrant.
	Notice that $x$ and $y$ are both negative.  Therefore, $\theta$ need to be $5\pi/4$ 
	(and not $\pi/4$ which is in quadrant I).
	So, $(r, \theta, z)=(2\sqrt{2}, 5\pi/4, -3)$.

	Another correct answer would be $(r, \theta, z)=(-2\sqrt{2}, \pi/4, -3)$.

\item  Let $(\rho,\theta,\phi)=(4,5\pi/3,3\pi/4)$, find the Cartesian coordinates of this point.
\answer: Here we just use the formulas (you could have either remembered them, or derived them geometrically):
	\begin{eqnarray*}
	r & = & \rho \sin \phi = 4 \sin(3\pi/4) = 4 \left( \frac{\sqrt{2}}{2} \right) = 2\sqrt{2} \\
	x & = & r \cos \theta = 2\sqrt{2} \cos(5\pi/3) = 2\sqrt{2}(1/2) = \sqrt{2} \\
	y & = & r \sin \theta = 2\sqrt{2} \sin(5\pi/3) = 2\sqrt{2}\left(- \frac{\sqrt{3}}{2} \right) = -\sqrt{6} \\
	z & = & \rho \cos \phi = 4 \cos(5\pi/3) = 4 \left(- \frac{\sqrt{2}}{2} \right) = -2\sqrt{2}
	\end{eqnarray*}
	So we get:
	$(x,y,z) = (\sqrt{2}, -\sqrt{6}, -2\sqrt{2})$.
\end{enumerate}

\item  Consider the function:
\[
f(x,y)=xe^y + \cos(xy)
\]
\begin{enumerate}
\item  Find all first partial derivatives ($f_x, f_y$).
\answer: Remember to use the chain rule:
	\begin{eqnarray*}
	f_x & = & e^y - y \sin(xy) \\
	f_y & = & xe^y - x \sin(xy)
	\end{eqnarray*}

\item  Find all second partial derivatives ($f_{xx}, f_{xy}, f_{yx}, f_{yy}$)
\answer: to find $f_{xx}=(f_x)_x)$ and $f_{yy}=(f_y)_y$, you just use the chain rule again.
	To find $f_{xy}$ and $f_{yx}$, you need to use the product rule.
	Remember that these mixed partials should be equal.
	\begin{eqnarray*}
	f_{xx} & = & -y^2\cos(xy) \\
	f_{yy} & = & xe^y-x^2\cos(xy) \\
	f_{xy} & = & e^y-\sin(xy)-xy\cos(xy) \\
	f_{yx} & = & e^y-\sin(xy)-xy\cos(xy)
	\end{eqnarray*}
\end{enumerate}

\item  Graph the level curves $z=k$ for $k=-2,0,2$:
\[
z=-x^2-4y^2+1
\]
\answer:  You need to graph the three equations in the $xy$-plane (because level curves are in the $xy$-plane):
(I rearranged a bit too)
\begin{eqnarray*}
k=-2: && x^2+4y^2 = 3 \\
k=0:  && x^2+4y^2 = 1 \\
k=2:  && x^2+4y^2 = -1
\end{eqnarray*}
Notice that that last equation will not have a graph ($x^2+4y^2$ is always positive).
The other two graphs will be ellipses in the $xy$-plane centered at the origin.

\item  Graph the following function:
\[
z = 1-x^2
\]
\answer:  There is a $y$ ``missing'' from the equation.  
Therefore this is a ``cylinder'' in the sense of section~14.6
Thus we graph $z=1-x^2$ in the $xz$-plane and extend this out in the $y$ directions.

\item  Consider the limit:
\[
\lim_{(x,y) \ra (0,0)} \frac{x^2}{x^2-y^2}
\]
Consider the paths to the origin along the $x$-axis and then along the $y$-axis.
Show that the limit does not exist.
\answer:  A path to the origin along the $x$ axis is obtained by making $y=0$.
	Similarly, a path to the origin along the $y$-axis is obtained by making $x=0$.
	We look at the limits along these paths:
	\begin{eqnarray*}
	\lim_{(x,0) \ra (0,0)} \frac{x^2}{x^2-(0)^2} & = & \lim_{(x,0) \ra (0,0)} \frac{x^2}{x^2} = 1 \\
	\lim_{(0,y) \ra (0,0)} \frac{(0)^2}{(0)^2-y^2} & = & 0
	\end{eqnarray*}
	We get different limits along different paths, so therefore the limit in question does not exist.
	This is similar to example~1 in section~15.3.


\item  Describe the following surfaces where ``$\mathrm{constant}$'' could be any constant number. 
	(You might want to graph them to help me understand your description).
\begin{enumerate}
\item  $\theta=\mbox{constant}$, in cylindrical coordinates.
	\answer:  These will be planes that contain the $z$-axis.  These plane will make an angle of $\theta$
	with the $xz$-plane.
\item  $\theta=\mbox{constant}$, in spherical coordinates.
	\answer:  The $\theta$ in cylindrical coordinates measures the same thing as $\theta$ in spherical coordinates.
	Therefore, this answer is the same as the previous one.
\item  $\phi=\mbox{constant}$, in spherical coordinates.
	\answer: $\phi$ is the angle from the $z$-axis to a point.
	All the points with this value the same form a cone with ``cone axis'' the $z$-axis.
	The ``cone angle'' is $\phi$.
	The best picture of this in the picture of the elliptical cone in section~14.6.  
	Notice how the cone really consists of two pieces.

	Also, there are two exceptional cases that can happen.
	If $\phi=\pi/2$, then the graph is the $xy$-plane.
	If $\phi=0$, then the graph is the $z$-axis.
\item  $r=\mbox{constant}$, in cylindrical coordinates.
	\answer: These are cylinders (in the sense we usually think of cylinders).
	See the picture in section~14.7 (figure~2).

	There is also an exceptional case here.
	The graph of $r=0$ is the $z$-axis.
\item  $\rho=\mbox{constant}$, in spherical coordinates.
	\answer:  These are spheres, see figure~4 in section~14.7.

	Again, there is an exceptional case.
	The graph of $\rho=0$ is just the origin
	(and $\rho=(\mbox{a negative number})$ is nothing).
\end{enumerate}



\end{enumerate}

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