Question: Show that the curve $\alpha(t)=(sint,t,-cost)$ has a constant speed. Is this curve regular curve? Then find a reparametrization of this curve by an arc lenth.
The curve of $\alpha$ has a constant speed and is a regular curve that can be reparametrized by an arc length. To show that it has a constant speed and is a regular curve we must first find the derivative of $\alpha$.
$\alpha'(t)=(cost,1,sint)$
$|\alpha'(t)|= \sqrt{cos^2(t)+1^2+sin^2(t)}= \sqrt{1+1}= \sqrt{2} $
Since $\alpha'(t) \neq 0$ then it is a regular curve by definition. To find arc length,
$s(t)= \int_{0}^{T} \sqrt{2} dt= \sqrt{2}t |^{T}_{0}= \sqrt{2}T- \sqrt{2}(0)= \sqrt{2}T. $
So, $s= \sqrt{2}T$ or $T= \dfrac{s}{\sqrt{2}}$.
Did I finish answering the question? Am I suppose to take a second derivative?
Best Answer
You have indeed completed the necessary computations, but you could definitely present your solution in a nicer manner!
i.e What is the arc-length parametrization? You have found $T(s)$, so you can define $\text{A} (s) = \alpha(T(s))$.
It would probably help (and be educationally re-enforcing) to write things like "By definition, the speed of the parametrization $\alpha(t)$ is defined to be __ which is readily seen to be a constant since (insert computation)."