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{\bf Exam Question

Topic: Tangent Plane}

Find an equation for the plane which is tangent to the surface
whose equation is $z=\ln(x^2+y^2)$ at the point $(1,0,0).$

This tangent plane meets the surface in many other points. Find
all such points for which $x=2.$

 \vspace{0.5in}

{\bf Solution}

$$z=\ln(x^2+y^2);\ \ \frac{\partial z}{\partial
x}=\frac{2x}{x^2+y^2}; \ \ \frac{\partial z}{\partial
y}=\frac{2y}{x^2+y^2}.$$

So when $x=1$ and $y=0$, $\displaystyle\frac{\partial z}{\partial
x}=2$ and $\frac{\partial z}{\partial y}=0.$

The equation of the tangent plane at $(1,0,0)$ is therefore
$z=2(x-1).$

This plane meets the surface where $2(x-1)=\ln(x^2+y^2),$ so if
$x=2$ we have $2=ln(4+y^2).$

Thus $4+y^2=\mathrm{e}^2$ i.e., $y=\pm\sqrt{\mathrm{e}^2-4}.$

There are therefore two such points, namely
$(2,\pm\sqrt{\mathrm{e}^2-4},2).$



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