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{\bf Exam Question

Topic: Tangent Plane}

Find the equation of the tangent plane at the point $(-1,1,0)$ to
the surface given by the implicit equation $$xy+yz+\cos(zx)=0.$$

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{\bf Solution}
 Let $F(x,y,z)=xy+yz+\cos(zx).$ Differentiating gives
 \begin{eqnarray*}
 \frac{\partial F}{\partial x}&=&y-z\sin(zx)=1\ \ \mathrm{at}\ \
 (-1,1,0)\\
\frac{\partial F}{\partial y}&=&x+z=-1\ \ \mathrm{at}\ \
 (-1,1,0)\\
 \frac{\partial F}{\partial z}&=&y+x\sin(zx)=1\ \ \mathrm{at}\ \
 (-1,1,0)
 \end{eqnarray*}
 So the equation of the tangent plane is $(x+1)-(y-1)+z=0;\ \
 x-y+z=-2.$


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