Remembering the Outward-Pointing Normal Vector

Question

For a closed curve $C$ parameterised anti-clockwise, the outward-pointing normal vector $\overrightarrow{N}$ is described as $(b, -a)$, if the tangent vector $\overrightarrow{r'} = (a, b)$.

I'm seeking an analytical way to remember this outward-pointing normal vector without rote memorisation; therefore, given some tangent vector $(a, b)$ I need to be able to derive it somehow and/or reason about it.

I need to know such a vector for the purpose of calculating flux across a curve. And the reason I want to be able to derive it without rote memorisation is because I don't think rote memorisation is a good strategy for understanding concepts and/or remembering things during examinations.

I was wondering if people could please take the time to provide me with such a methodology so that I may be able to learn the concept more effectively.

Answer 1

One method I just thought of is to use linear algebra.

The map $(a, b) \to (b, -a)$ is a linear transformation with the matrix \begin{matrix} 0 & 1 \\ -1 & 0 \end{matrix}

This is a rotation of angle $\dfrac{-\pi}{2}$.

This makes sense since, if we are dealing with a closed curve that is parameterised anti-clockwise, then the tangent vector is pointing anti-clockwise. Therefore, the outward-pointing normal vector, which by definition is orthogonal to the tangent vector, will be equal to the tangent vector rotated by an angle of $\dfrac{-\pi}{2}$.

I assume that this what the user amd was referring to in his comment.

An easy way to remember this is to reason from an anti-clockwise parameterisation of the unit circle (circle of radius $= 1$):

We have the vector $\overrightarrow{r}(t) = (\cos(t), \sin(t))$.

Since we want the outward-pointing normal vector, we want the vector that is orthogonal to the tangent vector and points in the same direction as a vector representing radius at $t$.

A vector representing the radius at $t = 0$ is $\overrightarrow{r}(0) = (1, 0)$.

The tangent vector to this vector is $\overrightarrow{r}'(t) = (-\sin(t), \cos(t))$.

$\overrightarrow{r}'(0) = (0, 1) = (a, b)$

Therefore, $\overrightarrow{N}(0) = (1, 0) = (b, -a)$ is the outward-pointing normal vector. We could have made the negative apparent by selecting a more appropriate $t$ such as $\pi/4$

Answer 2

As you walk along the curve, in the direction of the curve, your right hand should be outside the enclosed region, and your left hand should be inside the closed region.

If you remember this, it should be easy to figure out the correct $90^\circ$ rotation to apply to the tangent vector to get the outward-pointing normal, which I think of as multiplication by $-i$.