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