I am stuck at a problem that seems quite easy, but I just couldn't sort it out.
Determine a unit-speed parametrisation for the curves:
(i) $\gamma(t) = $(cos$^2(t)$, sin$^2(t))$ for $t∈(0,\frac{\pi}{2})$
(ii) $\gamma(t) = (t$, cosh$t)$ for $t∈R$
I know that these two curves are regular since $||\gamma \dot (t)||$ never vanishes within their domain, and we need it to be $1$ to make it unit-speed.
In (i), I found $\gamma \dot (t) = (-2$ sin$t$ cos$t$, $2$ sin$t$ cos$t)$, and $||\gamma \dot (t)|| =2 \sqrt(2)$ sin$t$ cos$t = \sqrt(2)$sin$2t$. Then $\Gamma \dot(t) = \frac{\gamma \dot (t)}{||\gamma \dot (t)||} = (-\frac{\sqrt(2)}{2}, \frac{\sqrt(2)}{2})$, which makes $\Gamma(t) = (-\frac{\sqrt(2)}{2}t, \frac{\sqrt(2)}{2}t)$.
$\Gamma(t)$ satisfies the conditions of unit speed, but the method and result do seem weird. This couldn't be applied to (ii), either, due to the difficulties in integrating the division result.
Could someone please guide me to the right track? Thanks!!
Best Answer
The result is not correct. Note that $\cos^2 t + \sin^2 t = 1$, so the curve is (part of) the line $x+y=1$. When you integrated to obtain $\Gamma(t)$, you forgot the integration constants. That's why you obtained a line that is parallel to the correct line.
This method will not work in general. In the simple case of a line it works, but here's the general approach:
Step 1. Find the arclength parameter: $$ s(t) = \int_0^t \|\dot\gamma(t)\|\,dt = \int_0^t \sqrt{2}\sin 2t\,dt = \frac{\sqrt{2}}{2}\left(1-\cos 2t\right). $$ Step 2. Write the parameter $t=t(s)$ in function of the arclength $s$. This means you have to find the inverse of the function $s(t)$: $$ t=\frac{1}{2}\arccos(1-\sqrt{2}s).$$ Step 3. Substitute this expression in the parametrisation $\gamma(t)$. You will obtain $$ \gamma(s) = \left(1-\frac{s}{\sqrt{2}},\frac{s}{\sqrt{2}}\right) \quad s\in(0,\sqrt{2}). $$ Now you can tackle the curve in (ii).