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

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!

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.

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

• Dr. MV, thank you for taking time to answer this question!. I hear your reasoning; it is the established way to finding the normal in Cartesian coordinates. My issue is with the cylindrical coordinate system, because even though the curve is unit-speed in Cartesian coordinates via $|r^{\prime}|=1$, this same expression is no longer true in cylindrical coordinates. Also in cylindrical coordinates, $r^{\prime\prime}\cdot r^{\prime}$ is not zero. Where is my mistake? – Jay May 26 '15 at 17:04
• @Jay You're welcome. My pleasure. Note that the development is independent of a coordinate system and thus applies to cylindrical coordinates as well as Cartesian coordinates. – Mark Viola May 26 '15 at 17:06
• Then my mistake is in writing ${\mathbf r}$ improperly. I wrote ${\mathbf r} = r \hat{r} + \theta \hat{\theta} + z \hat{z}$. – Jay May 26 '15 at 17:18
• If there is a good textbook that explains the position vector in cylindrical coordinates and its derivatives I would be happy to read more. – Jay May 26 '15 at 17:22
• @Jay I understand the reason why it is tempting to add the $\hat \theta \theta$ term to the position vector. But, if one simply takes $\vec r=\hat xx+\hat yy+\hat zz$ and uses $\hat x=\hat r \cos \theta-\hat \theta \sin \theta$ and $\hat y=\hat r \sin \theta+\hat \theta \cos \theta$ along with $x=r\cos \theta$ and $y=r\sin \theta$, one finds that $\vec r=\hat r r+\hat zz$. Alternatively, one can intuit this by drawing a picture and identifying that for the position vector, there is no $\hat \theta$ component. – Mark Viola May 26 '15 at 17:30