Let $γ\left(t\right)=\pi i\cos\left(t\right),\:-\pi /2\le \:t\le \:\pi /2\:$, how do you find the length of the curve?

Question

I am trying to find curve parametrisation that gives the length of the curve $γ\left(t\right)=\pi i\cos\left(t\right),\:-\pi /2\le \:t\le \:\pi /2\:$, but my answer of $\pi i2$ is rejected. Could anyone explain to me how to solve this problem properly?

Answer 1

Hint: A parametric form of the curve is $p(t)=(0,\pi \cos(t))$ for $-\pi/2 \leq t \leq \pi / 2$. The parametrisation takes on this simple form as the function has no real component. The length of such a curve is $$ \int_{- \frac{\pi}{2}} ^{\frac{\pi}{2}} | p'(s) | ds. $$ Here $|p'(s)|$ is the Euclidean norm of the derivative of $p$. In the current specific case, $|p'(s)|$ is simply $|\frac{\mathrm{d}}{\mathrm{d}s} (\pi \cos(s))|$.

Some intuition: Imagine a particle that is at position $p(t)$ at time $t$. Then the velocity of this particle is $p'(t)$ and its speed is $|p'(t)|$. So the total distance it travels between time $-\pi/2$ and $\pi /2$ is the integral I gave. Thus, if the parametrisation is injective, Integrating the speed gives the total length of the curve.

EDIT: In my original answer I wrote the parametrisation as $(\pi \cos (t), t)$, which is incorrect as noted by JeanMarie.