Showing that a Unit Speed Curve is a Circle.

Question

In my recent differential geometry tutorial, we were given the question:

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Given the unit speed curve,

$$\boldsymbol{r}(s)=\left(\frac{4}{5}\cos(s),1-\sin(s),-\frac{3}{5}\cos(s)\right)$$

show that this represents a circle with centre $(0,1,0)$ with radius 1.

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My first intuitive thought is to simply find the distance between $(0,1,0)$ and $\boldsymbol{r}(s)$ and show that it is 1 for all $s$ - this is the definition of a circle, correct? However, my tutor advised that we had to look into the torsion of the curve and use the fact that it is 0.

Any help/advice would be greatly appreciated. Thank you in advanced.

Answer 1

Notice that in three dimension, a curve of constant distant to a fixed point is a curve on the sphere. So not only you shall show what you mentioned, but that torsion vanishes to ensure the curve is restricted to a plane, hence a circle in 2 dimensional subspace.

Answer 2

By looking at x,y and z coordinates, notice that the curve is intersection of an elliptic cylinder and plane through y-axis at a certain angle to get equal axes for the ellipse.. as circle.Directly use Frenet-Serret to find that torsion vanishes, curvature is constant.

Answer 3

With the rotation matrix around axis $Oy$,

$$R_\theta=\left(\begin{matrix} \cos \theta & 0 & -\sin \theta \\ 0 & 1 & 0 \\ \sin \theta & 0 & \cos \theta \\ \end{matrix}\right)$$

And

$$r=\left(\begin{matrix} \frac45\cos s\\ 1-\sin s\\ -\frac35\cos s \end{matrix}\right)$$

Then

$$R_\theta \cdot r=\left(\begin{matrix} \frac{(3\sin \theta + 4\cos \theta)\cos s}5\\ 1-\sin s \\ \frac{(4\sin \theta - 3\cos\theta)\cos s}5 \\ \end{matrix}\right)$$

Thus, with $\theta=\mathrm{Arctan} \frac34$, the third component vanishes, and (you've got a $3-4-5$ triangle):

$$\cos \theta=\frac45, \ \sin\theta=\frac35$$

So

$$R_\theta \cdot r=\left(\begin{matrix} \frac{(3\sin \theta + 4\cos \theta)\cos s}5\\ 1-\sin s \\ 0 \\ \end{matrix}\right)=\left(\begin{matrix} \cos s\\ 1-\sin s \\ 0 \\ \end{matrix}\right)$$

So the rotated curve has a simpler equation.