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

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

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Start Solution

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

$\ln \left( {\frac{x}{{2y}}} \right) - {z^2}\left( {x - 2y} \right) - 3z = 3$ 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 $$\left( {4,2, - 1} \right)$$. Here are those quantities.

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

The tangent plane is then,

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

The normal line is,

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