Find the unit vector normal to the curve y = $x^{2}$ at the point (2, 4, 1).

Problem – Find the unit vector normal to the curve y =  at the point (2, 4, 1).

Solution

We have given Equation of curve surface is

y = $x^{2}$

$x^{2}$- y = 0

A constant scalar function $\varPhi$ on this surface is given by

$\varPhi(x,y,z)$ = $x^{2}$- y

we know normal unit vector is given by

$\hat{n}$ = $\pm\frac{\overrightarrow{\nabla}\varPhi}{\left|\overrightarrow{\nabla}\varPhi\right|}$

taking gradient of $\varPhi(x,y,z)$ = $x^{2}$- y

$\overrightarrow{\nabla}\varPhi$ =  $\overrightarrow{\nabla(}x^{2}- y)$

= $(\frac{\partial}{\partial x}\hat{x}+\frac{\partial}{\partial y}\hat{y}+\frac{\partial}{\partial z}\hat{z}).(x^{2}- y)$

=$\frac{\partial}{\partial x}(x^{2}-y)\hat{x}+\frac{\partial}{\partial y}(x^{2}-y)\hat{y}+\frac{\partial}{\partial z}(x^{2}-y)\hat{z}$

=$2x\hat{x}-\hat{y}$

now gradient of surface $\varPhi$ at point (2,4,1) is

$\overrightarrow{\nabla}\varPhi$ = $4\hat{x}-\hat{y}$

magnitude of gradient

$\left|\overrightarrow{\nabla}\varPhi\right|$ = $\sqrt{4^{2}+(-1)^{2}}$ = $\sqrt{17}$

now,

$\hat{n}=\pm\frac{\overrightarrow{\nabla}\varPhi}{\left|\overrightarrow{\nabla}\varPhi\right|}$ =

$\hat{n}$ = $\pm\frac{4\hat{x}-\hat{y}}{\left|4\hat{x}-\hat{y}\right|}$

$\hat{n}$ = $\pm\frac{4\hat{x}-\hat{y}}{\sqrt{17}}$