Answer to Question #223671 in Chemical Engineering for Lokika

Question #223671
Find unit normal vector to the surface xy^{2}+2yz=8 at the point (3,-2,1)
1
Expert's answer
2021-08-05T18:05:47-0400

Given:

F(x,y,z)=xy2+2yz8,M(3,2,1)F(x,y,z)=xy^2+2yz-8,\quad M(3,-2,1)

The unit normal vector to the surface F(x,y,z)=0F(x,y,z)=0 is given by

N=FF{\bf N}=\frac{\nabla F}{|\nabla F|}

F=(ix+jy+kz)(xy2+2yz8)\nabla F=\left({\bf i}\frac{\partial}{\partial x}+{\bf j}\frac{\partial}{\partial y}+{\bf k}\frac{\partial}{\partial z}\right)(xy^2+2yz-8)

=(iy2+j(2xy+2z)+k2y)=4i10j4k=\left({\bf i}y^2+{\bf j}(2xy+2z)+{\bf k}2y\right)=4{\bf i}-10{\bf j}-4{\bf k}

F=42+(10)2+(4)2=233|\nabla F|=\sqrt{4^2+(-10)^2+(-4)^2}=2\sqrt{33}

Finally

N=2i5j2k33{\bf N}=\frac{2{\bf i}-5{\bf j}-2{\bf k}}{\sqrt{33}}


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