How do we find the principal unit normal to this curve?

Question

A curve is given in cylindrical coordinates:

$r=r(t)$

$\theta=\theta(t)$

$z=z(t)$

The curve is unit-speed:

$(\frac{dr}{dt})^2+r^2(\frac{d\theta}{dt})^2+(\frac{dz}{dt})^2=1$

How do we find the principal unit normal vector to this curve? Many thanks!

Answer 1

Since we have

$$\vec r'\cdot \vec r'=1 \tag 1$$

then differentiating $(1)$ with respect to $t$ shows that

$$\vec r'' \cdot \vec r'=0. \tag 2$$

Note that $(2)$ implies that $\vec r'$ and $\vec r''$ are orthogonal.

We also know that $\vec r'$ is tangent to the curve spanned by $\vec r$. Thus, $\vec r''$ in a normal to the curve.

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For cylindrical coordinates, we have $\vec r=\hat r r+\hat zz$.

Thus,

$$\vec r'=\hat r r'+\hat \theta r\theta'+\hat zz'$$

and

$$\vec r''=\hat r[r''-r(\theta')^2]+\hat \theta[2r'\theta'+r\theta'']+\hat zz''.$$

To verify that indeed $\vec r'\cdot \vec r''=0$, we form the inner product to find

$$\begin{align} \vec r'\cdot \vec r''&=r'r''-rr'(\theta')^2+2rr'(\theta')^2+r^2\theta' \theta''+z'z''\\\\ &=rr''+rr'(\theta')^2+r^2\theta' \theta''+z'z''\\\\ &=2\frac{d}{dt}\left(r'^2+r^2(\theta')^2+z'^2\right)\\\\ &=0 \end{align}$$

as expected