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Question

In Exercise, find an equation for the tangent plane and parametric equations for the normal line to the surface at the point P.P .

z=4x3y2+2y:P(1,2,12)z = 4 x ^ { 3 } y ^ { 2 } + 2 y : P ( 1 , - 2,12 )

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The equation of the tangent plane at P(x0,y0,z0)P \left( x _ { 0 } , y _ { 0 } , z _ { 0 } \right) to the surface z=f(x,y)z = f ( x , y ) as

fx(x0,y0)(xx0)+fy(x0,y0)(yy0)(zz0)=0\begin{equation} 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 \end{equation}

and the normal line at P0P _ { 0 } has parametric equation:

x=x0+fx(x0,y0)t,y=y0+fy(x0,y0)t,z=z0t\begin{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 \end{equation}

Hence for z=4x3y2+2yz = 4 x ^ { 3 } y ^ { 2 } + 2 y at P(1,2,12)P ( 1 , - 2,12 )

fx=zx=12x2y2,fx(x0,y0)=12(1)2(2)2=4(12)=48f 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

fy=zy=8x3y+2fy(x0,y0)=8(1)3(2)+2=16+2=14f_{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(x1)14(y(2))(z12)=048 ( x - 1 ) - 14 ( y - ( - 2 ) ) - ( z - 12 ) = 0

48x14yz=48+2812=64\Rightarrow 48 x - 14 y - z = 48 + 28 - 12 = 64

48x14yz=6448 x - 14 y - z = 64

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

x=1+48t,y=214t,z=12tx = 1 + 48 t , \quad y = - 2 - 14 t \quad , \quad z = 12 - t

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