Arc Length with Vector Functions — Question 7

Question 7 – Solution

Strategy The factor $t\ge 0$ makes the speed easy to integrate; then solve algebraically for $t^2$.

Arc-length function $$\|\vec r'(t)\|=\|\left\langle t,t^2\right\rangle\|=t\sqrt{1+t^2},\qquad s(t)=\frac 13\bigl((1+t^2)^{3/2}-1\bigr).$$

Inverse From $(1+t^2)^{3/2}=3s+1$, $$t(s)=\sqrt{(3s+1)^{2/3}-1}.$$

Unit-speed form With $u=\sqrt{(3s+1)^{2/3}-1}$, $$\boxed{\vec R(s)=\left\langle \frac 12u^2,\frac 13u^3\right\rangle},\qquad s\ge 0.$$ Because $s$ was defined by integrating speed and is strictly increasing for $t>0$, the chain rule gives $\|d\vec R/ds\|=1$ there; the right-hand unit-speed limit also exists at $s=0$.