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{\bf Question}

The height of the ground in kilometers near an extinct volcano is
given by the formula : $$h = \exp-(x^2 + y^2 - 0.25)^2$$ where $x$
and $y$ are the distances in kilometers from the centre of the
crater in the north and east directions respectively.
\begin{description}
\item[(a)] Sketch the shape of the volcano in section; find the
height of the centre of the crater, of the rim around the crater,
and at a large distance from the mountain.

Now find the slope of the paths:
\item[(b)] at $(0.5,0)$ in the $(1,0)$ direction
\item[(c)] at $(1,1)$ in the $(1,1)$ direction
\item[(d)] at $(1,1)$ in the $(2,-1)$ direction
\end{description}

\vspace{.25in}

{\bf Answer}

$$h = \exp-(x^2 + y^2 - 0.25)^2 = \exp -f(xy)$$
\begin{description}
\item[(a)]
Shape:

\begin{center}
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\epsfig{file=175-8-1.eps, width=50mm}
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\rm{cross-section}
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$$\\$$
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\end{center}


At the centre of the crater $x = y =0$ so $h = \exp -0.25^2
\approx 0.9394$km

On the rim  $x^2 + y^2 = 0.25$; rim has radius $\frac{1}{2}$km and
$h=1$

(the crater is approximately 60.6 metres below the rim)

Far from volcano $x,y \rightarrow \infty \Rightarrow h \rightarrow
0$

\item[ (b), (c), (d)] all need $\nabla h$

$$\nabla h = \frac{\partial h}{\partial x}{\bf i} + \frac{\partial
h}{\partial y}{\bf j} = \frac{\partial h}{\partial
f}\frac{\partial f}{\partial x}{\bf i} + \frac{\partial
h}{\partial f}\frac{\partial f}{\partial y}{\bf j}$$

\begin{eqnarray*}
f(x,y) &amp; = &amp; (x^2 + y^2 - 0.25)^2 \\ \frac{\partial f}{\partial x}
&amp; = &amp; 2(x^2 + y^2 - 0.25) \times 2x = 4x(x^2 + y^2 - 0.25) \\
\frac{\partial f}{\partial y} &amp; = &amp; 2(x^2 + y^2 - 0.25) \times 2y
= 4y (x^2 + y^2 - 0.25) \\ h &amp; = &amp; \exp(-f) \Rightarrow
\frac{dh}{df} = -\exp(-f) = -h\end{eqnarray*}

$ \nabla h  =  -\exp-(x^2 + y^2 - 0.25)^2 (4x(x^2 + y^2 -
0.25){\bf i}+ 4y(x^2 + y^2 - 0.25){\bf j})$

\item[(b)]
At $(0.5,0)$ $\nabla h = -(1)(0{\bf i} + 0{\bf j}) = 0$

$\displaystyle \frac{\partial f}{\partial n} = \nabla f \cdot {\bf
\hat n} \hspace{.2in} {\bf \hat n} = (1,0)
\hspace{.2in}\Rightarrow \frac{\partial h}{\partial n} = \nabla f
\cdot {\bf \hat n} = 0$

all paths at (1,0) are locally flat  [(1,0) is on the rim]

\item[(c)]

At (1,1) in the {\bf n} = (1,1)  direction

Unit vector in the direction (1,1) is ${\bf \hat n} = \left(
\frac{1}{\sqrt2}, \frac{1}{\sqrt2} \right)$

$\displaystyle\frac{\partial h}{\partial n} = \nabla h \cdot {\bf
\hat n}
$

$(x^2 + y^2 -.25) = f^{\frac{1}{2}} = 1.75$

$f=(x^2 + y^2 -.25)^2 = 3.0625 \Rightarrow e^{-f} = h =
e^{-3.0625} \approx 0.0468$

\begin{eqnarray*} \nabla h &amp; = &amp; -0.0468 \times \{ 4 \times 1.75 {\bf i}
+ 4 \times 1.75 {\bf j} \} \\ &amp; = &amp; -0.327 ({\bf i} + {\bf j}) \\
\nabla h \cdot {\bf \hat n} &amp; = &amp; -0.327 ({\bf i} + {\bf j}) \cdot
\frac{1}{\sqrt2} ({\bf i} + {\bf j}) \\ &amp; = &amp; \frac{-2}{\sqrt2}
\times 0.327 \\ &amp; \approx &amp; -0.463
\end{eqnarray*}

\item[(d)]

At $(1,1)$ in the ${\bf n} = (2,-1)$  direction

$\displaystyle {\bf \hat n} = \frac{1}{\sqrt{2^2 +(-1)^2}} (2,-1)
= \left( \frac{2}{\sqrt5}{\bf i} - \frac{1}{\sqrt5}{\bf j} \right)
\approx 0.894{\bf i} - 0.447{\bf j}$


\begin{eqnarray*}\frac{\partial h}{\partial n} = \nabla h
 \cdot {\bf \hat n} &amp; = &amp; -0.327 ({\bf i} + {\bf
j}) \times 0.447 (2{\bf i} + {\bf j}) \\ &amp; = &amp; -0.327 \times 0.447
\\ &amp; \approx &amp; -0.146
\end{eqnarray*}

\end{description}



\end{document}
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