$\newcommand{\Reals}{\mathbf{R}}$Consider the function $r : [1/4, 3/4] → \Reals^2$ defined as $r(t) = (\sqrt{t}, \sqrt{1−t})$. I need to determine whether it is regular and to compute its length. I am given a hint: Computing the length requires the use of the inverse function of the sin, the arcsin.
Well I know that a curve is regular if $r'(t)\neq 0$ $\forall t$ and also the length equals $\int^a_b \|r'(t)\|\, dt$. However I don't know what $t$ to plug in to $r$ in order to show that it is regular and also I cannot find the reason why to use sin and arcsin for the length!
$$ r'(t) = \left(\frac{1}{2\sqrt{t}}, \frac{-1}{2\sqrt{1-t}}\right), \quad\text{and so}\quad \|r'(t)\| = \sqrt{\left(\frac{1}{2\sqrt{t}}\right)^2 + \left(\frac{-1}{2\sqrt{1-t}}\right)^2}. $$