Please checking to find an arc-length reparametrization

Question

Find an arc-length reparametrization of

$$c(t)=\langle \cos t+t\sin t, \sin t-t\cos t\rangle$$ for $t\in [\pi, 3\pi/2]$

solution trial:

$$c'(t)=\langle -\sin t+\sin t+t\cos t, \cos t-\cos t+t\sin t\rangle= \langle t\cos t, t\sin t\rangle$$

Note that For $t\in [\pi, 3\pi/2]$, $c'(t)\not=0$ so, $c(t)$ is regular paramtrized curve.

$$\lVert c'(t)\rVert =\sqrt{(t\cos t)^2+(t\sin t)^2}=|t|=t$$

$$s=\int^{t_0}_{t}\lVert c'(t)\rVert du =\int_{\pi}^{t} u du=\frac{1}{2}(t^2 -\pi^2)$$

So, $t=\sqrt{2s+\pi^2}$

$$\tilde{c}(s)=\langle \cos(\sqrt{2s}+\pi^2)+\sqrt{2s+\pi^2}\sin(\sqrt{2s+\pi^2}), \sin(\sqrt{2s+\pi^2})-\sqrt{2s+\pi^2}\cos(\sqrt{2s+\pi^2})\rangle$$ For $s\in [\pi/\sqrt{2},3\pi/2\sqrt{2}]$

Additionally, $$\lVert \tilde{c}'(s)\rVert \not =1$$ so not unit speed.

Is this solution enough and true?

Answer 1

Choosing $\sqrt{2t}$ gives you:

$$c\left(\sqrt{2t}\right) = \{\sqrt{2} \sqrt{t} \sin \left(\sqrt{2} \sqrt{t}\right)+\cos \left(\sqrt{2} \sqrt{t}\right),\sin \left(\sqrt{2} \sqrt{t}\right)-\sqrt{2} \sqrt{t} \cos \left(\sqrt{2} \sqrt{t}\right)\}$$

And: $$\frac{d}{dt}c\left(\sqrt{2t}\right)=\left\{\text{Cos}\left[\sqrt{2} \sqrt{t}\right],\text{Sin}\left[\sqrt{2} \sqrt{t}\right]\right\}$$

Which is obviously a unit sized injection. Substitute it to t to get an arc-parametrication.

Answer 2

No. Pick some $t_0\in [\ pi,3/2\pi]$ and evaluate your integral again: $$s(t)=\int_{t_0}^t\|c'(t)\|\,dt,$$