Arc Length with Vector Functions — Question 2

Question 2 – Solution

Strategy Differentiate first, factor the speed, and use $u=1+t^2$.

Speed $$\vec r'(t)=\left\langle 2t,2t^2\right\rangle,\qquad \|\vec r'(t)\|=2t\sqrt{1+t^2},$$ where $t\ge 0$ allows $\sqrt{t^2}=t$.

Integration $$L=\int_0^2 2t\sqrt{1+t^2}\,dt =\frac 23\left[(1+t^2)^{3/2}\right]_0^2 =\boxed{\frac 23(5\sqrt 5-1)}.$$

Check Both coordinate functions increase. The straight-line distance between the endpoints is $$\sqrt{4^2+(16/3)^2}=\frac{20}{3}.$$ The computed curve length is larger, as it must be.