## Calculus (3rd Edition)

$$r(2)= \langle 0, 4,\frac{1}{5} \rangle,$$ $$r(-1)= \langle -1, 1,\frac{1}{2} \rangle.$$
Since $$r(t)=\langle \sin \frac{\pi}{2}t, t^2,(t^2+1)^{-1} \rangle,$$ then $$r(2)=\langle \sin \pi, 4,(4+1)^{-1} \rangle=\langle 0, 4,\frac{1}{5} \rangle,$$ $$r(-1)=\langle \sin \frac{-\pi}{2}t, 1,(1+1)^{-1} \rangle=\langle -1, 1,\frac{1}{2} \rangle.$$