Suppose that a curve $\gamma$ is parametrically defined by two continuously differentiable functions $f(t),$ $a \le t \le b$ and $g(u),$ $\alpha \le u \le \beta$.
These functions are called equivalent parameterizations of $\gamma$ if there is a continuously differential function $\phi$ such that:

  1. $a = \phi(\alpha$) and $b = \phi(\beta)$.
  2. $f(\phi(u)) = g(u),$ $\alpha \le u \le \beta$
  3. $\phi'(u) > 0,$ $\alpha < u <\beta$.

Show that

$$f(t) = (cos(t), sin(t)), 0 \le t \le \frac{\pi}{2}$$ and $$g(u) = \left(\frac{1-u^2}{1+u^2}, \frac{2u}{1+u^2} \right), 0 \le u \le 1$$

are equivalent parameterizations of a quarter-circle

How do I approach this? I'm unfamiliar with equivalent paramterizations and my book touches on it once with an example which isn't helpful.

I thought of placing $\cos(t)$ in replacement for $u$ in $\frac{1-u^2}{1+u^2}$ and $\sin(t)$ for $\frac{2u}{1+u^2}$, then taking the derivative of $g(u)$, though I'm unsure how to proceed with setting up the integrals as I'm given two bounds?


1 Answer 1


I read the problem and quickly the solution $\phi(u)=2\arctan(u)$ came to mind (I'll justify this intuition a little at the end$^*$). You can prove this works by means of the double angle formula.

To find a solution on a more systematic manner, assume that there is a function such that $f\circ\phi=g$. That is $$\begin{bmatrix}\cos(\phi(u))\\ \sin(\phi(u))\end{bmatrix}=\begin{bmatrix}\frac{1-u^2}{1+u^2}\\ \frac{2u}{1+u^2} \end{bmatrix}$$ Which leads to the system of equations $$\begin{cases}\phi(u)=\arccos\left(\frac{1-u^2}{1+u^2}\right)\\ \phi(u)=\arcsin\left(\frac{2u}{1+u^2}\right)\end{cases}$$

So now you have two candidates for a solution. Just show that either of these satisfies. (Btw, we have that $\forall u\in [0,1],\;\arccos\left(\frac{1-u^2}{1+u^2}\right)=\arcsin\left(\frac{2u}{1+u^2}\right)$).

$^*$ $f$ has trigonometric components and $g$ has rational components, a good guess is that $\phi$ will be similar to the inverse of a trig function. $\phi=\arctan$ seemed like a good candidate since both $\sin(\arctan(u))$ and $\cos(\arctan(u))$ have $\sqrt{1+u^2}$ on the denominator. I just felt like adding a $2$ before $\arctan$ so that $\phi(1)=2\arctan(1)=\frac{\pi}{2}$ (which was required) and all the other conditions sort of just worked.


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