Normal Vectors

Normal Vectors to 2D curves

Consider some curve parametrized by its length $s$:

$$C:x=x(s),y=y(s)$$

We know that the Unit Tangent Vector to this curve at $s$ is:

$$\hat{T}(t)=\frac{d\vec{r}}{ds}=\frac{dx}{ds}\vec{i}+\frac{dy}{ds}\vec{j}$$

The unit normal vector to the curve $C$ at $s$ can then simply be found by swapping the vector components and negating one of them:

$$\hat{N}=-\frac{dy}{ds}\vec{i}+\frac{dx}{ds}\vec{j}$$

All other normal vectors to the curve are scalar multiples of $\hat{N}$.

Normal Vectors to 3D curves

For a 3D curve parametrized by its length $s$:

$$C:x=x(s),y=y(s),z=z(s),0\le s\le L$$

We already know how to compute the unit tangent vector, $\hat{T}$, which satisfies:

$$1=\hat{T}\cdot \hat{T}$$

Differentiating with respect to $s$ gives:

$$\begin{array}{rl}0& =\frac{d\hat{T}}{ds}\cdot \hat{T}+\hat{T}\cdot \frac{d\hat{T}}{ds}\\ & =2\left(\hat{T}\cdot \frac{d\hat{T}}{ds}\right)\end{array}$$

If $\vec{T}$ and $\frac{d\vec{T}}{ds}$ are non-zero, then $\frac{d\vec{T}}{ds}$ is orthogonal to $\hat{T}$ and can be used as a normal vector: $\vec{N}=\frac{d\vec{T}}{ds}$ Unit normal vector:

$$\hat{N}=\frac{\vec{N}}{\mid N\mid }$$

This is also called the principal normal vector to $C$. Unlike the 2D case, there are normal vectors $C$ at $s$ that are not a scalar multiple of $\hat{N}$.

What if $C$ is not parametrized by $s$ but by some other $t$?

$$C:x=x(t),y=y(t),z=z(t),\alpha \le t\le \beta$$

An application to $\hat{T}$ yields this

$$\frac{d\hat{T}}{ds}=\frac{d\hat{T}}{dt}\frac{dt}{ds}$$

• Note: $\frac{dt}{ds}$ is non-negative since both $s$ and $t$ increase along the direction of $C$

We hence define a unit normal vector by:

$$\begin{array}{rl}\vec{N}& =\frac{d\hat{T}/ds}{\mid d\vec{T}/ds\mid }\\ & =\frac{(d\hat{T}/ds)(dt/ds)}{\mid (d\vec{T}/ds)(dt/ds)\mid }\\ & =\frac{(d\hat{T}/ds)(dt/ds)}{\mid d\hat{T}/dt\mid dt/ds}\\ & =\frac{d\hat{T}/dt}{\mid d\hat{T}/dt\mid }\end{array}$$