Unit normal vector of a surface

Background

• Partial derivatives of parametric surfaces
  • In particular, make sure you have a strong intuition for the partial derivatives of a function parameterizing a surface, and what they represent.
• Cross product (video)

What we're building to

• If a surface is parameterized by a function $\overrightarrow{\mathbf{\text{v}}}(t,s)$‍, the unit normal vector to this surface is given by the expression

  $\begin{array}{r}\pm {\displaystyle \frac{({\displaystyle \frac{\partial \overrightarrow{\mathbf{\text{v}}}}{\partial t}}(t,s))\times ({\displaystyle \frac{\partial \overrightarrow{\mathbf{\text{v}}}}{\partial s}}(t,s))}{|({\displaystyle \frac{\partial \overrightarrow{\mathbf{\text{v}}}}{\partial t}}(t,s))\times ({\displaystyle \frac{\partial \overrightarrow{\mathbf{\text{v}}}}{\partial s}}(t,s))|}}\end{array}$‍

• You always have two choices for a unit vector function. If a surface is closed, like a sphere or a torus, those choices can be interpreted as outward-facing and inward-facing vectors.
• This is useful for the idea of flux in three-dimensions, covered in the next article.

Unit normal vector

Example: How to compute a unit normal vector

Step 1: Find a (not necessarily unit) normal vector

Choose 1 answer:

Choose 1 answer:

• (Choice A)  

  $\begin{array}{r}({\displaystyle \frac{\partial \overrightarrow{\mathbf{\text{v}}}}{\partial t}}(1,-2))\times ({\displaystyle \frac{\partial \overrightarrow{\mathbf{\text{v}}}}{\partial s}}(1,-2))\end{array}$‍

  A

  $\begin{array}{r}({\displaystyle \frac{\partial \overrightarrow{\mathbf{\text{v}}}}{\partial t}}(1,-2))\times ({\displaystyle \frac{\partial \overrightarrow{\mathbf{\text{v}}}}{\partial s}}(1,-2))\end{array}$‍
• (Choice B)  

  $\begin{array}{r}({\displaystyle \frac{\partial \overrightarrow{\mathbf{\text{v}}}}{\partial t}}(1,-2))\cdot ({\displaystyle \frac{\partial \overrightarrow{\mathbf{\text{v}}}}{\partial s}}(1,-2))\end{array}$‍

  B

  $\begin{array}{r}({\displaystyle \frac{\partial \overrightarrow{\mathbf{\text{v}}}}{\partial t}}(1,-2))\cdot ({\displaystyle \frac{\partial \overrightarrow{\mathbf{\text{v}}}}{\partial s}}(1,-2))\end{array}$‍
• (Choice C)  

  $\begin{array}{r}({\displaystyle \frac{\partial \overrightarrow{\mathbf{\text{v}}}}{\partial t}}(1,-2))+({\displaystyle \frac{\partial \overrightarrow{\mathbf{\text{v}}}}{\partial s}}(1,-2))\end{array}$‍

  C

  $\begin{array}{r}({\displaystyle \frac{\partial \overrightarrow{\mathbf{\text{v}}}}{\partial t}}(1,-2))+({\displaystyle \frac{\partial \overrightarrow{\mathbf{\text{v}}}}{\partial s}}(1,-2))\end{array}$‍

Check

Answer

The first answer choice is correct:

$\begin{array}{r}({\displaystyle \frac{\partial \overrightarrow{\mathbf{\text{v}}}}{\partial t}}(1,-2))\times ({\displaystyle \frac{\partial \overrightarrow{\mathbf{\text{v}}}}{\partial s}}(1,-2))\end{array}$‍

If this seems unclear, consider reviewing the video on cross products.

Both partial derivative vectors, ${\displaystyle \frac{\partial \overrightarrow{\mathbf{\text{v}}}}{\partial t}}(1,-2)$‍ and ${\displaystyle \frac{\partial \overrightarrow{\mathbf{\text{v}}}}{\partial s}}(1,-2)$‍, are tangent to the surface at the point $\overrightarrow{\mathbf{\text{v}}}(1,-2)$‍. Their cross product gives a vector that is perpendicular to both of them, and which is therefore normal to the surface at that point.

$\begin{array}{r}({\displaystyle \frac{\partial \overrightarrow{\mathbf{\text{v}}}}{\partial t}}(t,s))\times ({\displaystyle \frac{\partial \overrightarrow{\mathbf{\text{v}}}}{\partial s}}(t,s))=\end{array}$‍

$\widehat{\mathbf{\text{i}}}+$‍

$\widehat{\mathbf{\text{j}}}+$‍

$\widehat{\mathbf{\text{k}}}$‍

Check

Answer

Start off by computing each partial derivative. First, let's do it with respect to $t$‍:

$\begin{array}{r}{\displaystyle \frac{\partial \overrightarrow{\mathbf{\text{v}}}}{\partial t}}=\left[\begin{array}{c}{\displaystyle \frac{\partial }{\partial t}}(t+1)\\ \\ {\displaystyle \frac{\partial }{\partial t}}(s)\\ \\ {\displaystyle \frac{\partial }{\partial t}}(s^2-t^2+1)\end{array}\right]=\left[\begin{array}{c}1\\ 0\\ -2t\end{array}\right]\end{array}$‍

Next up, with respect to $s$‍:

$\begin{array}{r}{\displaystyle \frac{\partial \overrightarrow{\mathbf{\text{v}}}}{\partial t}}=\left[\begin{array}{c}{\displaystyle \frac{\partial }{\partial s}}(t+1)\\ \\ {\displaystyle \frac{\partial }{\partial s}}(s)\\ \\ {\displaystyle \frac{\partial }{\partial s}}(s^2-t^2+1)\end{array}\right]=\left[\begin{array}{c}0\\ 1\\ 2s\end{array}\right]\end{array}$‍

Now, take the cross product:

$\begin{array}{rl}\left[\begin{array}{c}1\\ 0\\ -2t\end{array}\right]\times \left[\begin{array}{c}0\\ 1\\ 2s\end{array}\right]& =\text{det}\left(\begin{array}{ccc}\widehat{\mathbf{\text{i}}}& \widehat{\mathbf{\text{j}}}& \widehat{\mathbf{\text{k}}}\\ 1& 0& -2t\\ 0& 1& 2s\end{array}\right)\\ \\ & =(0-(-2t))\widehat{\mathbf{\text{i}}}+(0-2s)\widehat{\mathbf{\text{j}}}+(1-0)\widehat{\mathbf{\text{k}}}\\ \\ & =2t\widehat{\mathbf{\text{i}}}+-2s\widehat{\mathbf{\text{j}}}+1\widehat{\mathbf{\text{k}}}\\ \end{array}$‍

Or, written as an upright vector, here's what we get:

$\begin{array}{r}\left[\begin{array}{c}2t\\ -2s\\ 1\end{array}\right]\end{array}$‍

Step 2: Make that a unit normal vector

$\widehat{\mathbf{\text{i}}}+$‍

$\widehat{\mathbf{\text{j}}}+$‍

$\widehat{\mathbf{\text{k}}}$‍

Check

Answer

As I said in the previous section, when we plug in $(t,s)=(1,-2)$‍ to our general expression for a normal vector, we get:

$\begin{array}{r}\left[\begin{array}{c}2(1)\\ -2(-2)\\ 1\end{array}\right]=\left[\begin{array}{c}2\\ 4\\ 1\end{array}\right]\end{array}$‍

And this vector has the following magnitude:

$\sqrt{2^2+4^2+1^2}=\sqrt{4+16+1}=\sqrt{21}$‍

To make this a unit vector, then, we will scale it from a vector with length $\sqrt{21}$‍ to one with length $1$‍. To do this, divide each component by $\sqrt{21}$‍:

$\begin{array}{r}\left[\begin{array}{c}2/\sqrt{21}\\ 4/\sqrt{21}\\ 1/\sqrt{21}\end{array}\right]\leftarrow \text{Unit vector}\end{array}$‍

$\widehat{\mathbf{\text{i}}}+$‍

$\widehat{\mathbf{\text{j}}}+$‍

$\widehat{\mathbf{\text{k}}}$‍

Check

Answer

Our general expression for a (not necessarily unit) normal vector is

$\begin{array}{r}\left[\begin{array}{c}2t\\ -2s\\ 1\end{array}\right]\end{array}$‍

As a function of $t$‍ and $s$‍, this vector has magnitude

$\sqrt{(2t{)}^2+(-2s{)}^2+1}=\sqrt{4t^2+4s^2+1}$‍

To turn our normal vector expression into a unit normal vector expression, divide each term by this magnitude:

$\begin{array}{r}\left[\begin{array}{c}2t/\sqrt{4t^2+4s^2+1}\\ -2s/\sqrt{4t^2+4s^2+1}\\ 1/\sqrt{4t^2+4s^2+1}\end{array}\right]\end{array}$‍

Choosing orientation

Summary

• Given a surface parameterized by a function $\overrightarrow{\mathbf{\text{v}}}(t,s)$‍, to find an expression for the unit normal vector to this surface, take the following steps:
• Step 1: Get a (non necessarily unit) normal vector by taking the cross product of both partial derivatives of $\overrightarrow{\mathbf{\text{v}}}(t,s)$‍:

  $\begin{array}{r}({\displaystyle \frac{\partial \overrightarrow{\mathbf{\text{v}}}}{\partial t}}(t,s))\times ({\displaystyle \frac{\partial \overrightarrow{\mathbf{\text{v}}}}{\partial s}}(t,s))\end{array}$‍

• Step 2: Turn this vector-expression into a unit vector by dividing it by its own magnitude:

• You can also multiply this expression by $-1$‍, and it will still give unit normal vectors.
• The main reason for learning this skill is to compute three-dimensional flux.