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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!

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

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