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{\bf NOTES:  Lab 3, Clairaut}
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Here are some notes to help you with the algebra in the lab:

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\item Let
$$
f(x,y)=\frac{xy(x^2-y^2)}{x^2+y^2}
$$
Notice that this could be written as:
$$
f(x,y)=xy\cdot
\left(\frac{x^2}{x^2+y^2}-\frac{y^2}{x^2+y^2}\right)
$$

To show that $-|xy|\leq f(x,y)\leq |xy|$, show that the quantity
in parenthesis is between $\pm 1$:

We note that the two numbers in parenthesis must sum to 1.  We
could rephrase that:  Let $a+b=1$, where $0\leq a, b\leq 1$.  
Find the max and min of $a-b$.
We could do a substitution, so we would find the max and min of
$a-(1-a)=2a-1$, where $a$ is between $0$ and $1$.  The maximum is
attained at $a=1$ ($b=0$) and the minimum is where $a=0$ ($b=1$).
Therefore,
$$
-1\leq \frac{x^2-y^2}{x^2+y^2}\leq 1
$$


\item Show that $f_x$ should be $0$ at $(x,y)=(0,0)$ by seeing that
it is trapped between $2|y|$ and $-2|y|$.  To show this, note
that:
$$
\frac{x^4-y^4+4x^2y^2}{(x^2+y^2)^2}=\frac{x^4+2x^2y^2+y^4+2x^2y^2-2y^4}{(x^2+y^2)^2}
$$
We wrote the fraction in this way to simplify things a bit:
$$
\frac{x^4+2x^2y^2+y^4+2x^2y^2-2y^4}{(x^2+y^2)^2}=\frac{(x^2+y^2)^2+2y^2(x^2-y^2)}{(x^2+y^2)^2}
$$
Now simplify:
$$
\frac{(x^2+y^2)^2+2y^2(x^2-y^2)}{(x^2+y^2)^2}=1+2\frac{y^2}{x^2+y^2}\left(
\frac{x^2}{x^2+y^2}-\frac{y^2}{x^2+y^2}\right)
$$
As before, let $a=x^2/(x^2+y^2)$, and $b=y^2/(x^2+y^2)$.  Then $a,
b \geq 0$,  $a+b=1$, and we want to find the minimum and maximum
of:
$$
1+2b(a-b)
$$
Substituting $a=1-b$, we find the min and max of
$$
1+2b((1-b)-b)=1+2b(1-2b)=1+2b-4b^2, \qquad 0\leq b \leq 1
$$
Using Calculus, you should find the the maximum occurs at $b=1/4$,
and the minimum occurs at $b=1$.  Put these back into the
expression to see that:
$$
-1 \leq \frac{x^4-y^4+4x^2y^2}{(x^2+y^2)^2}\leq \frac{5}{4}
$$
Therefore,
$$
-|y|\leq f_x(x,y) \leq \frac{5}{4}|y|
$$

Now see if you can do something similar for $f_y(x,y)$.

\item  For the second mixed partials, try plotting.  Does
the graph look familiar (like something from our practice 
Maple sheet)? 

Algebraically, take note of $f_x(0,y)$ and $f_y(x,0)$.  Then
compute:
$$
f_{xy}(0,0)=\lim_{h\rightarrow 0}\frac{f_x(0,0+h)-f_x(0,0)}{h}
$$
and
$$
f_{yx}(0,0)=\lim_{h\rightarrow 0}\frac{f_y(0+h,0)-f_y(0,0)}{h}
$$
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Some things I want you to get from doing Lab 3:
\begin{itemize}
\item Maple is a very powerful visualization and computational tool,
especially in three dimensions.
\item We should never trust Maple completely- Always do a ``reality check''
on what Maple is giving you to see if you believe it.
\item From our algebra in these notes, it's clear that while
Maple is a great tool, nothing beats old fashioned mathematical
reasoning and skill!
\end{itemize}
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