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\begin{document}


%%%(change to appropriate class and semester)
MATH 2222HJ Spring 2016

%%%(change to appropriate quiz type and date)
Quiz 3 \hspace{1.9in} SOLUTION
\vspace{2pc}

%%%(modify rules, time, points as appropriate)
Show all work clearly and in order, and circle your final answers.  

Justify your answers algebraically whenever possible. Unjustified work may not receive full credit.
\vspace{2pc}

\begin{problem}{2}
If the following limit exists, compute it. If it does not exist, explain why not:
$$\displaystyle\lim_{(x,y) \rightarrow (4,5)} \dfrac{3-\sqrt{x+y}}{9-x-y}.$$
\end{problem}
\textit{Solution}: Trying to plug in $x=4$ and $y=5$ into the function yields an indeterminate form $\dfrac{0}{0}$, so we must find the limit (if it exists) another way. We claim that the limit exists. To see it, first notice that we may rationalize the numerator of the function by multiplying by a ``convenient form of $1$" (this shows that the point $(4,5)$ is a \href{https://en.wikipedia.org/wiki/Removable_singularity}{removable singularity} of $\dfrac{3-\sqrt{x+y}}{9-x-y}$ ):
$$\dfrac{3-\sqrt{x+y}}{9-x-y} = \dfrac{3-\sqrt{x+y}}{9-x-y} \left( \dfrac{3+\sqrt{x+y}}{3+\sqrt{x+y}} \right) = \dfrac{9-x-y}{(9-x-y)(3+\sqrt{x+y})} = \dfrac{1}{3+\sqrt{x+y}},$$
showing that for $(x,y) \neq (4,5)$, $\dfrac{3-\sqrt{x+y}}{9-x-y} = \dfrac{1}{3+\sqrt{x+y}}$. Moreover, the function $\dfrac{1}{3+\sqrt{x+y}}$ is \href{https://en.wikipedia.org/wiki/Continuous_function}{continuous} at $(4,5)$ because it is a rational function (such functions are known to be continuous on their domain) and so computing the limit there is easy (i.e. we simply plug $(4,5)$ into it). Therefore we may \href{http://www.wolframalpha.com/input/?i=limit+as+%28x,y%29-%3E%284,5%29+of+%283-sqrt%28x%2By%29%29%2F%289-x-y%29}{compute}
$$\displaystyle\lim_{(x,y) \rightarrow (4,5)} \dfrac{3-\sqrt{x+y}}{9-x-y} = \displaystyle\lim_{(x,y) \rightarrow (4,5)} \dfrac{1}{3+\sqrt{x+y}}=\dfrac{1}{3+\sqrt{4+5}}=\dfrac{1}{6}.$$
\begin{problem}{3}
Let $f(x,y,z,w,t)=[\sin(wt)]^{xyz}$.  Compute $\dfrac{\partial^2 f}{\partial t\partial x}$ and $f_{yw}$. \\
\end{problem}
\textit{Solution:} Recall that the notation $\dfrac{\partial^2 f}{\partial t \partial x}$ means to first differentiate with respect to $x$ and then differentiate with respect to $t$ and t he notation $f_{yw}$ means to first differentiate with respect to $y$ and then differentiate with respect to $w$. So \href{http://www.wolframalpha.com/input/?i=d%2Fdx+[+[sin%28wt%29]^%28xyz%29]}{compute}
$$\dfrac{\partial f}{\partial x} = \dfrac{\partial}{\partial x} \left[ e^{xyz \log(\sin(wt))} \right]=[\sin(wt)]^{xyz} yz \log(\sin(wt))$$
and so we \href{http://www.wolframalpha.com/input/?i=d%2Fdt+[+yz*log(sin(wt))*+[sin(wt)]^(xyz)]}{see}
$$\begin{array}{ll}
\dfrac{\partial^2 f}{\partial t \partial x} &= \dfrac{\partial}{\partial t} \dfrac{\partial f}{\partial x} \\
&= yz \log(\sin(wt)) \dfrac{\partial}{\partial t} \left( [\sin(wt)]^{xyz} \right) + yz [\sin(wt)^{xyz}] \dfrac{\partial}{\partial t} \left( \log(\sin(wt)) \right) \\
&= yz \log(\sin(wt))(xyz) [\sin(wt)]^{xyz-1} (w \cos(wt)) + yz [\sin(wt)^{xyz}] \dfrac{1}{\sin(wt)} (w \cos(wt)) \\
&= xw y^2 z^2 \log(\sin(wt)) [\sin(wt)]^{xyz-1} \cos(wt)+\dfrac{yzw [\sin(wt)]^{xyz} \cos(wt)}{\sin(wt)} \\
&= xw y^2 z^2 \log(\sin(wt)) [\sin(wt)]^{xyz-1} \cos(wt)+yzw [\sin(wt)]^{xyz-1} \cos(wt) \\
\end{array}$$
For the other derivative first \href{http://www.wolframalpha.com/input/?i=d%2Fdy+[sin%28wt%29^%28xyz%29]}{compute}
$$f_y = \left[ e^{xyz \log(\sin(wt))} \right]_y=xz[\sin(wt)]^{xyz}  \log(\sin(wt)),$$
and then \href{http://www.wolframalpha.com/input/?i=d%2Fdw+xz[sin%28wt%29]^%28xyz%29log%28sin%28wt%29%29}{compute}
$$\begin{array}{ll}
f_{yw} &= \dfrac{\partial}{\partial w} \left[ xz[\sin(wt)]^{xyz}  \log(\sin(wt)) \right] \\
&= xz \log(\sin(wt)) \dfrac{\partial}{\partial w} [(\sin(wt))^{xyz}] + xz (\sin(wt))^{xyz} \dfrac{\partial}{\partial w} [\log(\sin(wt)] \\
&=xz \log(\sin(wt)) (xyz) [\sin(wt)]^{xyz-1} t \cos(wt) + \dfrac{xz [\sin(wt)]^{xyz}]}{\sin(wt)} \cos(wt)t \\
&= x^2 z^2 yt \log(\sin(wt)) [\sin(wt)]^{xyz-1} \cos(wt) + txz [\sin(wt)]^{xyz-1} \cos(wt).
\end{array}$$
\vfill
\showpoints
\end{document}