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# In Exercise, find an equation for the tangent plane and parametric equations for the normal line to the surface at the point $P .$$z = 4 x ^ { 3 } y ^ { 2 } + 2 y : P ( 1 , - 2,12 )$

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The equation of the tangent plane at $P \left( x _ { 0 } , y _ { 0 } , z _ { 0 } \right)$ to the surface $z = f ( x , y )$ as

$$$f _ { x } \left( x _ { 0 } , y _ { 0 } \right) \left( x - x _ { 0 } \right) + f _ { y } \left( x _ { 0 } , y _ { 0 } \right) \left( y - y _ { 0 } \right) - \left( z - z _ { 0 } \right) = 0$$$

and the normal line at $P _ { 0 }$ has parametric equation:

$$$x = x _ { 0 } + f x \left( x _ { 0 } , y _ { 0 } \right) t , \quad y = y _ { 0 } + f _ { y } \left( x _ { 0 } , y _ { 0 } \right) t , \quad z = z _ { 0 } - t$$$

Hence for $z = 4 x ^ { 3 } y ^ { 2 } + 2 y$ at $P ( 1 , - 2,12 )$

$f x = \frac { \partial z } { \partial x } = 12 x ^ { 2 } y ^ { 2 } \quad , \Rightarrow f _ { x } \left( x _ { 0 } , y _ { 0 } \right) = 12 ( 1 ) ^ { 2 } ( - 2 ) ^ { 2 } = 4 ( 12 ) = 48$

$f_{y}= \frac { \partial z } { \partial y } = 8 x ^ { 3 } y + 2 \Rightarrow f y \left( x _ { 0 } , y _ { 0 } \right) = 8 ( 1 ) ^ { 3 } ( - 2 ) + 2 = - 16 + 2 = - 14$

and the equation of the tangent plane results:

$48 ( x - 1 ) - 14 ( y - ( - 2 ) ) - ( z - 12 ) = 0$

$\Rightarrow 48 x - 14 y - z = 48 + 28 - 12 = 64$

$48 x - 14 y - z = 64$

Now, the equation of the normal line (2) results:

$x = 1 + 48 t , \quad y = - 2 - 14 t \quad , \quad z = 12 - t$

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