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Section 3-1 : Tangent Planes and Linear Approximations

1. Find the equation of the tangent plane to \(\displaystyle z = {x^2}\cos \left( {\pi y} \right) - \frac{6}{{x{y^2}}}\) at \(\left( {2, - 1} \right)\) .

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First, we know we’ll need the two 1st order partial derivatives. Here they are,

\[{f_x} = 2x\cos \left( {\pi y} \right) + \frac{6}{{{x^2}{y^2}}}\hspace{0.5in}{f_y} = - \pi {x^2}\sin \left( {\pi y} \right) + \frac{{12}}{{x{y^3}}}\] Show Step 2

Now we also need the two derivatives from the first step and the function evaluated at \(\left( {2, - 1} \right)\) . Here are those evaluations,

\[f\left( {2, - 1} \right) = - 7\hspace{0.5in}{f_x}\left( {2, - 1} \right) = - \frac{5}{2}\hspace{0.5in}{f_y}\left( {2, - 1} \right) = - 6\] Show Step 3

The tangent plane is then,

\[\require{bbox} \bbox[2pt,border:1px solid black]{{z = - 7 - \frac{5}{2}\left( {x - 2} \right) - 6\left( {y + 1} \right) = - \frac{5}{2}x - 6y - 8}}\]