Normal Vector for Sphere

Objective: Find the unit outward normal vector for the sphere $x^2 + y^2 +z^2 =1$ at point $P\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}},0\right)$.

Gradient of the Surface Function

Define the surface function $F(x, y, z) = x^2 + y^2 + z^2$. The gradient vector $\nabla F$ is normal to the surface.

$\nabla F = \frac{\partial F}{\partial x}\hat{i} + \frac{\partial F}{\partial y}\hat{j} + \frac{\partial F}{\partial z}\hat{k} = 2x\hat{i} + 2y\hat{j} + 2z\hat{k}$.

Evaluate Gradient at Point P

At $P\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}},0\right)$, the gradient is:

$\nabla F \bigg|_P = 2\left(\frac{1}{\sqrt{2}}\right)\hat{i} + 2\left(\frac{1}{\sqrt{2}}\right)\hat{j} + 2(0)\hat{k} = \sqrt{2}\hat{i} + \sqrt{2}\hat{j}$.

This vector points outward from the sphere's center (origin).

Calculate Unit Normal Vector

Find the magnitude of $\nabla F \bigg|_P$:

$|\nabla F \bigg|_P| = \sqrt{(\sqrt{2})^2 + (\sqrt{2})^2 + 0^2} = \sqrt{2+2} = 2$.

Normalize the vector:

Unit Normal Vector $\vec{N} = \frac{\nabla F \bigg|_P}{|\nabla F \bigg|_P|} = \frac{\sqrt{2}\hat{i} + \sqrt{2}\hat{j}}{2} = \frac{1}{\sqrt{2}}\hat{i} + \frac{1}{\sqrt{2}}\hat{j}$.

Result

The unit outward normal vector is $\frac{1}{\sqrt{2}}\hat{i} + \frac{1}{\sqrt{2}}\hat{j}$.