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Section 14.2 : 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 SolutionFirst, we need to do a quick rewrite of the equation as,
\[{x^2}y - 4z{{\bf{e}}^{x + y}} = - 35\] Show Step 2Now 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 3The 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,
\[\require{bbox} \bbox[2pt,border:1px solid black]{{\vec r\left( t \right) = \left\langle {3, - 3,2} \right\rangle + t\left\langle { - 26,1, - 4} \right\rangle = \left\langle {3 - 26t, - 3 + t,2 - 4t} \right\rangle }}\]