Exercise
Let [math]P:[a,b] \goesto \R^2[/math] and [math]Q:[c,d]\goesto R^2[/math] be two parametrizations of the same curve [math]C[/math] such that all four coordinate functions are continuously differentiable. (A function is continuously differentiable if its derivative exists and is continuous at every number in its domain.) Then [math]P[/math] and [math]Q[/math] are called equivalent parametrizations of [math]C[/math] if there exists a continuously differentiable function [math]f[/math] with domain [math][a,b][/math] and range [math][c,d][/math] which has a continuously differentiable inverse function, and in addition satisfies (i) [math]f(a) = c[/math] and [math]f(b) = d[/math], (ii) [math]P(t) = Q(f(t)),[/math] for every [math]t[/math] in [math][a,b][/math].
- Using the Chain Rule and the Change of Variable Theorem for Definite Integrals (for the latter, see Theorem \ref{thm 4.6.6}), prove that equivalent parametrizations assign the same arc length to [math]C[/math].
- Show that
[[math]] P(t) = (\cos t, \sin t), \quad 0 \leq t \leq \frac{\pi}2 , [[/math]][[math]] Q(s) = \left( \frac{1-s^2}{1+s^2}, \frac{2s}{1+s^2}\right), \quad 0 \leq s \leq 1 , [[/math]]are equivalent parametrizations of the same curve [math]C[/math], and identify the curve.
- Show that
[[math]] P(t) = (\cos t, \sin t), \quad 0 \leq t \leq 2\pi , [[/math]]and[[math]] Q(s) = (\cos 5t, \sin 5t), \quad 0 \leq t \leq 2\pi , [[/math]]are nonequivalent parametrizations of the circle.