Question

Find a unit vectot normal to the surface x^3-xyz-z^3 at the point ( 1,1,1)

Answer 1

Given surface is,

$\small\displaystyle\longrightarrow x^3-xyz-z^3+1=0$

We need to find a unit vector normal to this surface at the point (1, 1, 1).

Let us define the function,

$\small\displaystyle\longrightarrow f(x,\ y,\ z)=x^3-xyz-z^3+1$

The partial derivative of f wrt x is,

$\small\displaystyle\longrightarrow \dfrac{\partial f}{\partial x}(x,\ y,\ z)=3x^2-yz$

At $\small(x,\ y,\ z)=(1,\ 1,\ 1),$

$\small\displaystyle\longrightarrow \dfrac{\partial f}{\partial x}(1,\ 1,\ 1)=2$

The partial derivative of f wrt y is,

$\small\displaystyle\longrightarrow \dfrac{\partial f}{\partial y}(x,\ y,\ z)=-xz$

At $\small(x,\ y,\ z)=(1,\ 1,\ 1),$

$\small\displaystyle\longrightarrow \dfrac{\partial f}{\partial y}(1,\ 1,\ 1)=-1$

The partial derivative of f wrt z is,

$\small\displaystyle\longrightarrow \dfrac{\partial f}{\partial z}(x,\ y,\ z)=-xy-3z^2$

At $\small(x,\ y,\ z)=(1,\ 1,\ 1),$

$\small\displaystyle\longrightarrow \dfrac{\partial f}{\partial z}(1,\ 1,\ 1)=-4$

Then a normal vector to the surface is given by,

$\small\displaystyle\longrightarrow\vec{n}=\left<\dfrac{\partial f}{\partial x},\ \dfrac{\partial f}{\partial y},\ \dfrac{\partial f}{\partial z}\right>$

$\small\displaystyle\longrightarrow\vec{n}=\left<2,\ -1,\ -4\right>$

Then a unit vector normal to the surface is,

$\small\text{ \displaystyle\longrightarrow\underline{\underline{\hat n=\dfrac{1}{\sqrt{21}}\left<2,\ -1,\ -4\right>}} }$