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{\bf Question}

The height of the ground in kilometers near an extinct volcano is
given by the formula : $$h = \exp\left( -(x^2 + y^2 -
0.25)^2\right)$$ where $x$ and $y$ are the distances in kilometers
from the centre of the crater in the north and east directions
respectively.

Let $x=r\cos\theta$ and $y=r\sin\theta.$

\begin{description}
\item[(a)] Derive a formula for $\displaystyle\frac{\partial h}{\partial
\theta}$, and show that $\displaystyle\frac{\partial h}{\partial
\theta} = 0$. What is the physical meaning of this result?
\item[(b)] Find a general formula for $\displaystyle\frac{\partial h}{\partial r}$,
and show that $\displaystyle\frac{\partial h}{\partial r} = 0$ for
$r = 0$ and $r = 0.5$. What is the physical meaning of this
result?
\end{description}

\vspace{.25in}

{\bf Answer}

\begin{description}
\item[(a)]

$\displaystyle x = r \cos \theta \Rightarrow \frac{\partial
x}{\partial \theta} = -r \sin \theta = -y$

$\displaystyle  y = r \sin \theta \Rightarrow \frac{\partial
y}{\partial \theta} = r \cos \theta = x$

Chain rule $\displaystyle \Rightarrow \frac{\partial h}{\partial
\theta} = \frac{\partial h}{\partial x}\frac{\partial x}{\partial
\theta} + \frac{\partial h}{\partial y}\frac{\partial y}{\partial
\theta}$

with \begin{eqnarray*} h  & = & \exp\left( -(x^2 + y^2 -
0.25)^2\right)
\\ \frac{\partial h}{\partial x} & = & -4x(x^2 + y^2 -0.25) \exp\left( -(x^2 + y^2 -
0.25)^2\right) \\ \frac{\partial h}{\partial y} & = & -4y(x^2 +
y^2 -0.25) \exp\left( -(x^2 + y^2 - 0.25)^2\right) \\ {\rm
substituting:}
\\ \frac{\partial h}{\partial \theta} & = & -4(x^2 + y^2 -0.25)
\exp\left( -(x^2 + y^2 - 0.25)^2\right) \times (-xy + xy)
\\ & \equiv & 0 \end{eqnarray*}

Physical meaning: $(r, \theta)$ are polar coordinates.

We can rewrite $ h =\exp\left( -(r^2 - 0.25)^2\right)$. This is
independent of the angle $\theta$, so we expect the height to be
independent of the angle $\theta$, and the derivative with respect
to $\theta$ to be zero.


\item[(b)]


Chain rule $\displaystyle \Rightarrow \frac{\partial h}{\partial
r} = \frac{\partial h}{\partial x}\frac{\partial x}{\partial r} +
\frac{\partial h}{\partial y}\frac{\partial y}{\partial r}$

$\displaystyle  x = r \cos \theta \Rightarrow \frac{\partial
x}{\partial r} =  \cos \theta = \frac{x}{r}$

$\displaystyle  y = r \sin \theta \Rightarrow \frac{\partial
y}{\partial r} = \sin \theta = \frac{y}{r}$

Substituting:  \begin{eqnarray*}  \frac{\partial h}{\partial r} &
= & -4(x^2 + y^2 -0.25) \exp\left( -(x^2 + y^2 - 0.25)^2\right)
\times \left( x\frac{x}{r} + y \frac{y}{r}\right) \\ & = &
\frac{-4(x^2 + y^2)}{r} (x^2 + y^2 -0.25)\exp\left( -(x^2 + y^2 -
0.25)^2\right)
\end{eqnarray*}

Now $x= r \cos \theta, y = r \sin \theta \Rightarrow x^2 + y^2 =
r^2(\cos^2 \theta + \sin^2 \theta) = r^2$

Hence $$ \frac{\partial h}{\partial r} =  -4r(r^2 -0.25)
\exp\left( -(r^2 - 0.25)^2\right)$$

From the formula $\displaystyle\frac{\partial h}{\partial r} = 0$
if $r = 0$ or $r^2 = 0.25 \Rightarrow r = 0.5$

The height  has a minimum at the centre of crater $(r = 0)$

The height has maxima at all points on the rim of the crater $(r =
0.5)$

\end{description}


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