Arc Length with Vector Functions — Question 10

Question 10 – Solution

Strategy The speed contains $\sqrt{t^2}=|t|$; use symmetry across the cusp.

Speed $$\vec r'(t)=\left\langle 2t,3t^2\right\rangle,\qquad \|\vec r'(t)\|=|t|\sqrt{4+9t^2}.$$

Length $$L=2\int_0^1t\sqrt{4+9t^2}\,dt =\frac{2}{27}\left[(4+9t^2)^{3/2}\right]_0^1 =\boxed{\frac{2}{27}(13\sqrt{13}-8)}.$$

Why care about $|t|$? Replacing $|t|$ by $t$ on the full symmetric interval would make the integrand odd and falsely give zero. Although velocity vanishes at the cusp, each half is smooth and their nonnegative lengths add.