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\centerline{Mathematics 241, section 1 -- Multivariable Calculus}
\centerline{Solutions for Midterm Exam 2}
\centerline{November 5, 2010}
\vskip 10pt
\noindent
I.  All parts of this problem refer to the vector field
\[
{\bf F}(x,y) = (x^2 - 2x, y^2 + 2y).
\]
\begin{enumerate}
\item[A.] (5)  Find all critical points of ${\bf F}(x,y)$.
\item[] {\it Solution:}  The critical points are the solutions of 
$x^2 - 2x = 0$ and $y^2 + 2y = 0$, so $x = 0, 2$ and $y = 0, -2$.
This gives four critical points all together:  $(0,0)$, $(2,0)$, 
$(0,-2)$ and $(2,-2)$.   
\item[B.] (10)  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:}  The correct plot is Vector Field Two, since
Vector Field One apparently has only two critical points.  Consulting 
the plot, we see that $(0,0)$ and $(2,-2)$ are saddle points, $(2,0)$ is a
source, and $(0,-2)$ is a sink.
\item[C.] (10)  Show that $\alpha(t) = \left(\frac{2}{1-2e^{2t}},\frac{-2}{1 - 3e^{-2t}}\right)$ is a flow line of ${\bf F}$.  
\item[] {\it Solution:}  We must show that $\alpha'(t) = {\bf F}(\alpha(t))$.
The left side is 
\[
\alpha'(t) = \left(\frac{8 e^{2t}}{(1 - 2e^{2t})^2},
\frac{12 e^{-2t}}{(1 - 3e^{-2t})^2} \right).
\]
The right side is 
\begin{eqnarray*}
{\bf F}(\alpha(t)) &=& \left(\left(\frac{2}{1-2e^{2t}}\right)^2 - 2\left(\frac{2}{1-2e^{2t}}\right), \left(\frac{-2}{1-3e^{-2t}}\right)^2 + 2\left(\frac{-2}{1-3e^{-2t}}\right)\right)\\
&=& \left(\frac{4 - 4(1 - 2e^{2t})}{(1 - 2e^{2t})^2}, \frac{4 - 4(1 - 3e^{-2t})}{(1 - 3e^{-2t})^2}\right)\\
&=& \left(\frac{8 e^{2t}}{(1 - 2e^{2t})^2},
\frac{12 e^{-2t}}{(1 - 3e^{-2t})^2}\right).
\end{eqnarray*}
This shows what we wanted.
\item[D.] (10) Show that any $f(x,y) = \frac{x^3}{3} - x^2 + \frac{y^3}{3} + y^2 + c$
satisfies $\nabla f = {\bf F}$.  What are the critical points of such an $f$?
Classify each of them as a local maximum, local minimum, or saddle point.  
(Don't ``start over'' here; use previous parts of this question as appropriate.) 
\item[] {\it Solutions:}  For any such $f$, we have 
$\frac{\partial f}{\partial x} = x^2 - 2x$ and $\frac{\partial f}{\partial y} = y^2 + 2y$.
So $\nabla f(x,y) = (x^2 - 2x, y^2 + 2y) = {\bf F}(x,y)$.  By properties
of gradient vector fields, we know that $f(x,y)$ has a local minimum at 
$(2,0)$, a local maximum at $(0,-2)$, and saddle points at $(0,0)$ and $(2,-2)$.  
\end{enumerate}
\vskip 10pt
\noindent
II.  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.]  (10)  Compute $\frac{\partial f}{\partial x}(x,y)$ and $\frac{\partial f}{\partial y}(x,y)$
for $(x,y) \ne (0,0)$.
\item[] {\it Solution:}  By the quotient rule
\begin{eqnarray*}
\frac{\partial f}{\partial x} &=& \frac{(x^2 + y^2)(3x^2 - 3y^2) - (x^3 - 3x y^2)(2x)}{(x^2 + y^2)^2}\\
&=& \frac{x^4 + 6x^2y^2 - 3y^4}{(x^2 + y^2)^2}.
\end{eqnarray*}
and 
\begin{eqnarray*}
\frac{\partial f}{\partial y} &=& \frac{(x^2 + y^2)(-6xy) - (x^3 - 3x y^2)(2y)}{(x^2 + y^2)^2}\\
&=& \frac{-8x^3 y}{(x^2 + y^2)^2}.
\end{eqnarray*}

\item[B.]  (10)  Do $\frac{\partial f}{\partial x}(0,0)$ and $\frac{\partial f}{\partial y}(0,0)$ exist?
If so, find them; if not say why not.
\item[] {\it Solution:}  The formulas from part A are not defined at $(0,0)$. 
However, this just says we need to use the limit definition to see whether
the partial derivatives exist.  We find
\begin{eqnarray*}
\frac{\partial f}{\partial x}(0,0) &=& \lim_{h\to 0} \frac{f(h,0) - f(0,0)}{h}\\
&=& \lim_{h\to 0} \frac{h^3/h^2 - 0}{h}\\
&=& 1
\end{eqnarray*}
and 
\begin{eqnarray*}
\frac{\partial f}{\partial y}(0,0) &=& \lim_{h\to 0} \frac{f(0,h) - f(0,0)}{h}\\
&=& \lim_{h\to 0} \frac{0/h^2 - 0}{h}\\
&=& 0.
\end{eqnarray*}
\item[C.]  (10)  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 $(1,m)$).  
Does this {\it prove} anything about the existence
of the limit $\lim_{(x,y) \to (0,0)} f(x,y)$?  Explain your 
answer using the definition of the statement
$\lim_{(x,y) \to (0,0)} f(x,y) = L$.  
\item[] {\it Solution:}  We have 
\[ 
f(t,mt) = \frac{(1 - 3m^2)t^3}{(1+m^2)t^2} = \left(\frac{1 - 3m^2}{1+m^2}\right) t.
\]
So as $t \to 0$, the limit is zero.  This {\it does not show} that
$\lim_{(x,y) \to (0,0)} f(x,y) = 0$, though.  For that, we would have to 
show that for all $\varepsilon > 0$, there exists $r > 0$ such that 
$|f(x,y)| < \varepsilon$ for all $(x,y)$ in the open ball $B_r(0,0)$,
except possibly $(0,0)$.   
\item[D.]  (10 Extra Credit)  Is $f(x,y)$ {\it differentiable} at $(0,0)$?  Explain 
your answer.
\item[] {\it Solution:}  The answer is no.  To show differentiability, we
would need to show that 
\[
\lim_{(x,y) \to (0,0)} \frac{f(x,y) - \ell(x,y)}{\sqrt{x^2 + y^2}} = 0, 
\]
where $\ell(x,y)$ is the linear approximation at $(0,0)$.  
Here from part B we know $\ell(x,y) = x$, so 
\[
\frac{f(x,y) - \ell(x,y)}{\sqrt{x^2 + y^2}} =  \frac{-4xy^2}{(x^2 + y^2)^{3/2}}.
\]
If you let $(x,y) \to (0,0)$ along the $x$ or $y$ axes, 
this is going to $0$.  However, along the line $y = x$ this goes to the 
value $-4/2^{3/2} \ne 0$.   Hence the required limit does not exist.
Since it is not equal to zero, $f(x,y)$ is not differentiable.
\end{enumerate}
\vskip 10pt
\noindent
III.  All parts of this problem refer to $f(x,y) = 4x^2 - y^2$.
\begin{enumerate}
\item[A.] (15) Sketch the contours of $f(x,y)$ for $c = -1, 0, 1$ on the 
same set of axes.
\item[] {\it Solution:}  
\begin{center}
\includegraphics[width=3.0truein,height=3.0truein]{P3.ps}
\end{center}
(These are hyperbolas; 
the $c = -1$ contour opens up and down; the $c = 0$ contour
is the union of the two lines $y = \pm 2x$; the $c = 1$ contour opens left and right.)
\item[B.] (10) Consider 
\[
S = \{(x,y) \in \mathbb{R}^2\mid -1 < f(x,y) < 0\}.
\]
Is $S$ open, closed, or neither?  Why?  (An intuitive explanation referring
to the definition of open or closed is OK here.)
\item[] {\it Solution:}  $S$ is open since none of the boundary points of $S$
(the points on the two contours for $c = -1$ and $c = 0$) are included
in $S$.  This means that for any $(x_0,y_0) \in S$, there exists
a ball $B_r(x_0,y_0) \subset S$ for some $r > 0$.  
\item[C.]  (10) For which unit vectors $u$ is the directional derivative 
$D_u f(1,2) = 0$?
\item[] {\it Solution:}  The best way is to recognize that  $f(1,2) = 4 - 4 = 0$.
So $(1,2)$ is on the contour consisting of the two lines $y = \pm 2 x$. 
We want $u$ in the direction of the line with the plus sign to get the 
directional derivative equal to zero, so 
\[
u = \pm \left(\frac{1}{\sqrt{5}},\frac{2}{\sqrt{5}}\right).
\]
\end{enumerate}




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