I apologize for the generic title. By type I mean that it belongs to a class of curves, such as helixes, circles, lines, etc.

To the point however, I stumbled into an exercise where asked to describe which type of curve that has the following parameterization:

$$\vec{r} = (a \cos t \sin t)\hat{i} + (a \sin^2 t)\hat{j} + bt\hat{k}, $$

What I did was that I tried to express the $x$- and $y$-components in the form $C_1 \sin f(t)$ and $C_2 \cos f(t)$ respectively. So:

$$x=a \cos t \sin t=\frac{a}{2} \sin 2t$$ $$y = a \sin^2 t=a\frac{1-\cos^2 2t}{2} \iff 2y - a = -a\cos 2t$$

Using this, I confirmed that: $$x^2 + \left(y - \frac{a}{2}\right)^2 = \left( \frac{a}{2} \right) ^2$$

This implies that the curve lies on the cylinder with the above equation. However, this is where I get a bit confused as to how to proceed. It is seemingly clear that the equation represents a helix, considering that $z$ is increasing and both $x$ and $y$ draws out projections that are circles in the $xy$-plane. But how do I actually prove this? Technically, it could be the line $x = y = 0$ as well, if I don't proceed to somehow that $x$ and $y$ are "behaving circular" (you get what I mean, hopefully). Worth mentioning is that the book doesn't seem to prove this either.

Question: Can I in this particular case prove that the curve is a helix? My initial guess was that it had to do with $x,y \in C^0$, along with some other constraint, but I can't really get any further.

Any help would be gladly appreciated!


Let $u=2t-\dfrac{\pi}{2}$

$$\begin{pmatrix} x \\ y \\ z \end{pmatrix}= \begin{pmatrix} \dfrac{a}{2} \sin 2t \\ \dfrac{a}{2}-\dfrac{a}{2} \cos 2t \\ bt \end{pmatrix}= \begin{pmatrix} 0 \\ \dfrac{a}{2} \\ \dfrac{b\pi}{4} \end{pmatrix}+ \begin{pmatrix} \dfrac{a}{2} \cos u \\ \dfrac{a}{2} \sin u \\ \dfrac{bu}{2} \end{pmatrix}$$

which is a helix with axis $(x,y,z)=\left( 0, \dfrac{a}{2}, z \right)$, curvature $\dfrac{2}{\sqrt{a^2+b^2}}$ and torsion $\dfrac{2b}{\sqrt{a^2+b^2}}$

  • $\begingroup$ I can't believe how I didn't think about how there may be an equation for a helix, rather than going on about continuity. So just to clarify, it's enough to just refer to that equation then? Of course, with the steps that you showed, to be clear. $\endgroup$ – Max Jan 30 '17 at 16:44
  • 1
    $\begingroup$ I think so, it's just a translated one. $\endgroup$ – Ng Chung Tak Jan 30 '17 at 16:47

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