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Section 3-1 : Tangent Planes and Linear Approximations

3. Find the linear approximation to \(z = 4{x^2} - y{{\bf{e}}^{2x + y}}\) at \(\left( { - 2,4} \right)\) .

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Recall that the linear approximation to a function at a point is really nothing more than the tangent plane to that function at the point.

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

\[{f_x} = 8x - 2y{{\bf{e}}^{2x + y}}\hspace{0.5in}\hspace{0.25in}{f_y} = - {{\bf{e}}^{2x + y}} - y{{\bf{e}}^{2x + y}}\] Show Step 2

Now we also need the two derivatives from the first step and the function evaluated at \(\left( { - 2,4} \right)\) . Here are those evaluations,

\[f\left( { - 2,4} \right) = 12\hspace{0.5in}{f_x}\left( { - 2,4} \right) = - 24\hspace{0.5in}{f_y}\left( { - 2,4} \right) = - 5\] Show Step 3

The linear approximation is then,

\[\require{bbox} \bbox[2pt,border:1px solid black]{{L\left( {x,y} \right) = 12 - 24\left( {x + 2} \right) - 5\left( {y - 4} \right) = - 24x - 5y - 16}}\]