TITLE: Arc length with trig vectors
QUESTION [0 upvotes]: A particle moves with speed  $s'(t)=|t\sin t|$ in direction $(\sin t \tanh t,\cos t \tanh t,\operatorname{sech} t)$  at time $t$. What is the length of the arc that it travels from $t=0$ to $t=\pi$?
Normally I would normalize the direction vector and multiply by $t\sin t$.  However in this case, I don't know how to normalize the vector. Am I missing something?

REPLY [1 votes]: The arc length is simply
$$s(\pi)=\int_{0}^{\pi} |t\sin t| \, dt$$
Since $t\in [0,\pi] \implies \sin t\ge 0$, and using integration by parts
\begin{align}
  s(\pi) &= \int_{0}^{\pi} t\sin t \, dt \\
  &= [-t\cos t]_{0}^{\pi}+\int_{0}^{\pi} \cos t \, dt \\
  &= \pi
\end{align}
Note that $\mathbf{T}(t)=(\sin t \tanh t,\cos t \tanh t,\operatorname{sech} t)$ is already a unit tangent vector.  So the velocity is simply
$$\mathbf{v}(t)=|t\sin t|(\sin t \tanh t,\cos t \tanh t,\operatorname{sech} t)$$
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