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\begin{center}
  18.022 Recitation Quiz \iftoggle{solutions}{(with solutions)}{} \\
  1 October 2014 \\
  Samuel S.\ Watson
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\itm Find the equation for the plane tangent to the graph of $z=\cos(xy)+2y^2$ at the point whose $x$-coordinate is 0 and whose $y$-coordinate is 2. 

\sol{Let $f(x,y)=\cos(xy)+2y^2$.  The plane tangent to $z=\cos(xy)+2y^2$ at a point $(x_0,y_0,f(\vec{a}))$ (where $\vec{a}=(x_0,y_0)$) is normal to the vector $\left(\pd{f}{x}(\vec{a}),\pd{f}{y}(\vec{a}),-1\right)$. We calculate $\pd{f}{x} = -y\sin(xy)$ and $\pd{f}{y} = -x\sin(xy)+4y$ and substitute to find that $(0,8,-1)$ is a normal vector. Substituting into $\vec{n}\cdot((x,y,z) -P)$ with $P=(0,2,f(0,2))=(0,2,9)$, we find that the tangent plane is $\boxed{8y-z=7}$.}{}


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