Find the points on the curve r(t)=(2t)i+(t2+2)j at which r(t) and r′(t) have opposite direction.

r(t)=(2t)i+(t²+2)j and r'(t)=2i+2tj.
r(t)=-ar'(t) where a is a scalar
2t=-2a and t²+2=-2at, equating vector components
So t=-a and a²+2=2a², from which a²=2 and a=±√2, making t=±√2.
r(±√2)=±2√2i+4j, (2√2,4) and (-2√2,4).