# What is the unit vector with positive z component which is normal to z = xy + x/y at (1,1,2)?

Dec 3, 2015

This is really a calculus question, but the answer is $\vec{n} = < - \frac{2}{\sqrt{5}} , 0 , \frac{1}{\sqrt{5}} > = - \frac{2}{\sqrt{5}} \hat{i} + \frac{1}{\sqrt{5}} \hat{k}$.

#### Explanation:

Rewrite the equation $z = x y + \frac{x}{y}$ as $x y + \frac{x}{y} - z = 0$ and let $F \left(x , y , z\right) = x y + \frac{x}{y} - z$ so that the equation is $F \left(x , y , z\right) = 0$.

The gradient vector of $F$ is normal to the surface at any given point on the surface. That gradient vector is

$\nabla F = < \frac{\partial F}{\partial x} , \frac{\partial F}{\partial y} , \frac{\partial F}{\partial z} > = < y + \frac{1}{y} , x - \frac{x}{y} ^ 2 , - 1 >$.

At the point $\left(x , y , z\right) = \left(1 , 1 , 2\right)$ (which you should check is on the original surface), this gradient vector is $\nabla F = < 2 , 0 , - 1 >$.

Since $| | \nabla F | | = \sqrt{{2}^{2} + {0}^{2} + {\left(- 1\right)}^{2}} = \sqrt{5}$, it follows that $\vec{n} = - \frac{\nabla F}{|} | \nabla F | | = < - \frac{2}{\sqrt{5}} , 0 , \frac{1}{\sqrt{5}} > = - \frac{2}{\sqrt{5}} \hat{i} + \frac{1}{\sqrt{5}} \hat{k}$ is a unit vector normal to this surface with a positive $z$-component.

Here's a picture of this situation: