\documentclass[11pt]{article}
\usepackage{amssymb,epsfig,amscd}
\setlength{\textwidth}{6.5in}
\setlength{\topmargin}{-0.5in}
\setlength{\oddsidemargin}{0in}
\setlength{\textheight}{9.0in}
\begin{document}
\centerline{Mathematics 241, section 1 -- Multivariable Calculus}
\centerline{Midterm Exam 2 Solutions}
\centerline{November 1, 2013}
\vskip 10pt
\noindent
I.  All parts of this problem refer to the vector field
\[
{\bf F}(x,y) = (x^2 - 2x, xy - y).
\]
\begin{enumerate}
\item[A.] (10)  Find all critical points of ${\bf F}(x,y)$.
\item[] {\it Solution}:  The critical points are the solutions of the simultaneous
system
\begin{eqnarray*}
x^2 - 2x &=& x(x - 2) = 0\\
xy - y &=& (x - 1)y = 0
\end{eqnarray*}
The solutions are $(0,0)$ and $(2,0)$.   
\item[B.] (5)  There are two vector fields plotted on the back of this sheet.
Say which one shows ${\bf F}(x,y)$ and use that plot to classify each of the 
critical points as a source, sink, saddle, or center.
\item[] {\it Solution:}  This is Vector Field 2 in the plots (the critical points of 
Vector Field 1 are not at the right locations).  From the plot, $(0,0)$ is a {\it sink} and $(2,0)$ is a {\it source}.
\item[C.] (20) Show that $\alpha(t) = (0,4e^{-t})$ and $\beta(t) = \displaystyle{\left(\frac{2}{1+e^{2t}},0\right)}$
are both flow lines of $F$.   What are $\lim_{t\to\infty} \alpha(t)$ and  $\lim_{t\to\infty} \beta(t)$?
\item[] {\it Solution:}  For $\alpha(t)$ we compute $\alpha'(t) = (0,-4e^{-t})$.  On the other hand 
$({\bf F}\circ\alpha)(t) = (0^2 - 0,0\cdot 4e^{-t} - 4e^{-t}) = (0,-4e^{-t})$.  Therefore $\alpha(t)$ is a flow line
of ${\bf F}$.  For $\beta(t)$, similarly, we have 
$$\beta'(t) = \left(\frac{-4e^{2t}}{(1 + e^{2t})^2},0\right)$$
On the other hand, 
\begin{eqnarray*}
({\bf F}\circ\beta)(t) &=& \left(\left(\frac{2}{1+e^{2t}}\right)^2 - \frac{4}{1+e^{2t}}, 0\right)\\
&=& \left(\frac{4 - 4(1 + e^{2t})}{(1 + e^{2t})^2},0\right)\\
&=& \left(\frac{-4e^{2t}}{(1 + e^{2t})^2},0\right).
\end{eqnarray*}
\item[D.] (5) Is there a scalar-valued function $f(x,y)$ such that ${\bf F}(x,y) = \nabla f(x,y)$?  Why or why not? 
\item[] {\it Solution:}  No, there is not.  The reason is that we would need to have 
$\frac{\partial f}{\partial y} = xy - y$, so $f(x,y) = \frac{xy^2}{2} - \frac{y^2}{2} + g(x)$, for 
some function $g(x)$.  
But then $\frac{\partial f}{\partial x} = \frac{y^2}{2} + g'(x)$.  There is no $y^2$ in the first 
component of ${\bf F}$, so this is not possible.  
\end{enumerate}
\vskip 10pt 
\noindent
II.  In the neighborhood of Eagle Pass, the landscape has elevation in feet above sea level given by 
$f(x,y) = \frac{x^2}{4} - y^2 + 1000$.
\begin{enumerate}
\item[A.] (10) Sketch the contours of $f(x,y)$ for $c = 999, 1000, 1001$ on the 
same set of axes.
\item[] {\it Solution:}  The contours for $c = 999$ and $c = 1001$ are hyperbolas, the contour
for $c = 1000$ is the union of the two lines $y = \pm \frac{x}{2}$.  (Those lines are the 
asymptote lines of the hyperbolas.)
\item[B.] (10) Compute the directional derivative $D_u f(2,1)$ for a general 
unit vector.  
\item[] {\it Solution:}  We have $\nabla f(x,y) = \left(\frac{x}{2},-2y\right)$, so $\nabla f(2,1) = (1,-2)$.
The directional derivative $D_u f(2,1) = \nabla f(2,1)\cdot u = u_1 - 2u_2$.
\item[C.] (5) In the direction of which unit vector $u$ should you 
walk from the point with $(x,y) = (2,1)$ in order to decrease your elevation at
the fastest rate?  
\item[] {\it Solution:}  The unit vector in the direction of $-\nabla f(2,1)$, so 
$$u = \left(\frac{-1}{\sqrt{5}},\frac{2}{\sqrt{5}}\right)$$ 
\end{enumerate}
\vskip 10pt
\noindent
III.  All parts of this problem refer to the function 
\[
f(x,y) = 
\frac{x^3 - 3 xy^2}{x^2+y^2} {\rm\ if\ } (x,y) \ne (0,0) {\rm\ and\ } f(0,0) = 0.
\]
\begin{enumerate}
\item[A.]  (15)  Find the tangent plane to $z = f(x,y)$ at $(1,1,f(1,1))$.  
We have, at $(x,y) \ne (0,0)$:
\begin{eqnarray*}
\frac{\partial f}{\partial x} &=& \frac{(x^2 + y^2)(3x^2 - 3y^2) - (x^3 - 3xy^2)(2x)}{(x^2 + y^2)^2}\\
&=& \frac{x^4 + 6x^2 y^2 - 3y^4}{(x^2 + y^2)^2}\\
\therefore \frac{\partial f}{\partial x}(1,1) &=& \frac{4}{4} = 1.
\end{eqnarray*}
Similarly,
\begin{eqnarray*}
\frac{\partial f}{\partial y} &=& \frac{(x^2 + y^2)(-6xy) - (x^3 - 3xy^2)(2y)}{(x^2 + y^2)^2}\\
&=& \frac{-8x^3y}{(x^2 + y^2)^2}\\
\therefore \frac{\partial f}{\partial x}(1,1) &=& \frac{-8}{4} = -2.
\end{eqnarray*}
The tangent plane is 
$z = -1 + (1)(x - 1) + (-2)(y - 1)$, 
or after simplifying:
$z = x - 2y$.  
\item[B.]  (10)  Does $\frac{\partial f}{\partial x}(0,0)$ exist? If so, find it; if not say why not.
\item[] {\it Solution:}  By the limit definition,
\begin{eqnarray*}
\frac{\partial f}{\partial x}(0,0) &=& \lim_{h\to 0} \frac{f(0+h,0) - f(0,0)}{h}\\
&=& \lim_{h\to 0} \frac{h^3/h^2 - 0}{h}\\
&=& \lim_{h\to 0} 1\\
&=& 1.
\end{eqnarray*}
So the answer is yes.
\end{enumerate}
\vskip 10pt
\noindent
IV.  (10)  Can the curve $\alpha(t) = (2\cos(t),\sin(t))$ for $t \in (0,\infty)$ be a
flow line of the vector field $\nabla f(x,y)$ for a differentiable function $f$?  Why or
why not?
\vskip 10pt
\noindent
{\it Solution:}  The answer is NO.  Notice that $\alpha(t)$ is an ellipse with the usual
counterclockwise parametrization and 
$\alpha(0) = \alpha(2\pi) = (2,0)$.  If this was a flow line for the gradient vector field
for some $f(x,y)$, then as we know, $f(x,y)$ would be steadily {\it increasing with $t$} as we move along
the flow line.  However that is not possible since $f(\alpha(0)) = f(\alpha(2\pi))$. 
\vskip 10pt
\noindent
{\it Extra Credit}  (10)  Refer to the function in question III.
 Let $m$ be arbitrary and compute $\lim_{t\to 0} f(t,mt)$ (the limit of the value of $f$ along
the line through the origin in the direction of the vector $(1,m)$).   Is $\lim_{t\to 0} (f\circ\alpha)(t) = 0$
for \emph{every} differentiable curve $\alpha(t)$ with $\alpha(0) = (0,0)$?   Explain. 
\vskip 10pt
\noindent
{\it Solution:}  We have 
$$f(t,mt) = \frac{t^3(1 - 3m^2)}{t^2(1 + m^2)} = \frac{t(1 - 3m^2)}{(1 + m^2)}$$
Hence $\lim_{t\to 0} f(t,mt) = 0$  for all $m$.  It will be true that $\lim_{t\to 0} (f\circ\alpha)(t) = 0$
here because 
$$\frac{x^3 - 3xy^2}{x^2 + y^2} = x \cdot \frac{x^2 - 3y^2}{x^2 + y^2}$$
the second factor takes only values between $1$ and $-3$, while the $x \to 0$ if we 
are moving along any curve $\alpha(t)$ with $\alpha(0) = 0$.  By the squeeze theorem, the limit
must exist and equal zero.  
\vfill\eject
\noindent
Vector Field 1:
\begin{center}
\includegraphics[width=3.0truein,height=3.0truein]{VF1.png}
\end{center}
\noindent
Vector Field 2:
\begin{center}
\includegraphics[width=3.0truein,height=3.0truein]{VF2.png}
\end{center}

 \end{document} 
