# Normal vector to a polar curve

I'm struggling with working through a proof. Suppose I have a polar curve of the form $$r = f(\phi)$$.

How do I find the $$\textbf{normal vector}$$ to this curve? The end result I need should be in terms of $$\hat{r}$$ and $$\hat{\phi}$$ but I'm unsure of what I should do.

All I know so far is how to find the tangent slope and normal slope at any given $$\phi$$. Any help would be very appreciated. Thanks!

• If you know the slope of the normal, then you can make a normal vector out of it. – Berci Nov 29 '18 at 18:06
• I added the "differential-geometry" tag to your post. Cheers! – Robert Lewis Nov 29 '18 at 18:27

You have the vector of position:

$$\vec{r}=\hat{e}_r r=\hat{e}_r f(\phi)$$

differentiate over $$\phi$$ to get the tangent:

$$\vec{t}=\frac{d}{d\phi}\vec{r}=\frac{d\hat{e}_r}{d\phi}f(\phi)+\hat{e}_rf'(\phi)$$ But in polar coordinates, we know how unit vector in the radial direction changes with angle: $$\frac{d\hat{e}_r}{d\phi}=\hat{e}_\phi$$ (if unsure, write it as $$\hat{e}_r=(\cos\phi,\sin\phi)$$ and take the derivative.

So you have

$$\vec{t}=f(\phi)\hat{e}_\phi+f'(\phi)\hat{e}_r$$

You can normalize it you want.

A normal is just a $$90^\circ$$ rotation of this vector, and because $$\hat{e}_r$$ and $$\hat{e}_\phi$$ are orthogonal and unit sized, you can do the same as in the cartesian coordinates: $$90^\circ$$ positive rotation is just exchanging components and negating the one that gets the "y" (in this case, $$\phi$$) component.

$$\vec{n}=-f(\phi)\hat{e}_r+f'(\phi)\hat{e}_\phi$$

Orthogonality is easily verified with dot product.