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Section 3-2 : Gradient Vector, Tangent Planes and Normal Lines

1. Find the tangent plane and normal line to $${x^2}y = 4z{{\bf{e}}^{x + y}} - 35$$ at $$\left( {3, - 3,2} \right)$$.

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Start Solution

First, we need to do a quick rewrite of the equation as,

${x^2}y - 4z{{\bf{e}}^{x + y}} = - 35$ Show Step 2

Now we need the gradient of the function on the left side of the equation from Step 1 and its value at $$\left( {3, - 3,2} \right)$$. Here are those quantities.

$\nabla f = \left\langle {2xy - 4z{{\bf{e}}^{x + y}},{x^2} - 4z{{\bf{e}}^{x + y}}, - 4{{\bf{e}}^{x + y}}} \right\rangle \hspace{0.25in}\hspace{0.25in}\nabla f\left( {3, - 3,2} \right) = \left\langle { - 26,1, - 4} \right\rangle$ Show Step 3

The tangent plane is then,

$\require{bbox} \bbox[2pt,border:1px solid black]{{ - 26\left( {x - 3} \right) + \left( 1 \right)\left( {y + 3} \right) - 4\left( {z - 2} \right) = 0}}\hspace{0.25in} \Rightarrow \hspace{0.25in}\require{bbox} \bbox[2pt,border:1px solid black]{{ - 26x + y - 4z = - 89}}$

The normal line is,

$\require{bbox} \bbox[2pt,border:1px solid black]{{\vec r\left( t \right) = \left\langle {3, - 3,2} \right\rangle + t\left\langle { - 26,1, - 4} \right\rangle = \left\langle {3 - 26t, - 3 + t,2 - 4t} \right\rangle }}$