Arc Length with Parametric Equations — Question 7

Question 7

Problem

A curve is reparametrized from $t$ to $u=t^3$. Explain why its length is unchanged when both parameters increase.

Question 7 – Solution

Solution

1. Let the original curve be $\mathbf r(t)$ for $a\le t\le b$, and suppose $u=t^3$ increases on this interval. Then $$t=u^{1/3}, \qquad \frac{dt}{du}=\frac{1}{3u^{2/3}}$$ wherever this derivative is defined.

2. By the chain rule, the velocity in the new parameter is $$\frac{d\mathbf r}{du} =\frac{d\mathbf r}{dt}\frac{dt}{du}.$$ Taking magnitudes gives $$\left\|\frac{d\mathbf r}{du}\right\| =\left\|\frac{d\mathbf r}{dt}\right\| \left|\frac{dt}{du}\right|.$$

3. Because both parameters increase, $dt/du\ge0$ and $dt=(dt/du)du$. Hence $$\left\|\frac{d\mathbf r}{du}\right\|du =\left\|\frac{d\mathbf r}{dt}\right\|dt.$$ Integrating over corresponding endpoints yields the same value in either parameter.

4. Therefore, reparametrization changes the rate at which the curve is traced but not the geometric arc length: $$\boxed{L_u=L_t}.$$ A possible derivative singularity at the single point $u=0$ does not change the value of the proper or corresponding improper integral.