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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?

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    $\begingroup$ Yes, your answer is fine, there is no need for second derivative $\endgroup$ – Semsem Mar 29 '14 at 3:43
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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)."

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  • $\begingroup$ So it would sound better if I end it by saying: By definition, the reparametrization of $\alpha(t)$ is defined to be $T= \dfrac{s}{\sqrt{2}}$ which is a constant since $ |\alpha'(t)|= \sqrt{2}$. $\endgroup$ – Ruth Gutierrez Mar 29 '14 at 4:14
  • $\begingroup$ $T=\frac{s}{\sqrt 2}$ is not a reparametrization per-se. It's an equation expressing the relationship between the arc-length traversed, starting from $t=0$, at time $T$. You really just need one more line saying something like "I can plug $T=\frac{s}{\sqrt 2}$ into the above parametrization to obtain a new parametrization in terms of arc-length." $\endgroup$ – BlueBuck Mar 29 '14 at 4:17
  • $\begingroup$ Oh ok yea so that's why you use parametrization instead? $\endgroup$ – Ruth Gutierrez Mar 29 '14 at 4:19
  • $\begingroup$ Oh like saying $\alpha'(t)= (cos\dfrac{s}{\sqrt{2}},1, sin \dfrac{s}{\sqrt{2}})$ $\endgroup$ – Ruth Gutierrez Mar 29 '14 at 4:21
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    $\begingroup$ ALMOST!!!! Notice that the vector on the right has no t's!!!!! Define a new parametrization $A(s)$ (note this is in terms of arc length which is what we wanteD!!) by $A(s) = \alpha (t(s)) = (sin(\frac{s}{\sqrt 2}, \frac{s}{\sqrt 2}, -cos\frac{s}{\sqrt 2})$ $\endgroup$ – BlueBuck Mar 29 '14 at 4:29

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