Chapter 14 - Calculus of Vector-Valued Functions - 14.1 Vector-Valued Functions - Exercises - Page 709: 3

$$ r(2)= \langle 0, 4,\frac{1}{5} \rangle,$$ $$ r(-1)= \langle -1, 1,\frac{1}{2} \rangle.$$

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