Why when I put $F=[0,1,2z]$ in terms of $r(t)=(\cos(t) , \sin(t) , t^2 )$ is the answer $0,1,2t^2?$

Question

I don't understand how the $y$ term becomes $1.$ I assumed it would become $\sin(t)$ $$F=[0,1,2z]. . . . . . r(t)=[\cos(t),\sin(t),t^2], $$ so I thought $$F(r(t))=[0,\sin(t),2t^2].$$ But no, it's apparently $$F(r(t))=[0,1,2t^2].$$ Can you explain why I'm wrong here?

Answer 1

By definition

$$F (x,y,z)=(0,1,2z) $$

thus

$$F (\cos (t),\sin (t),t^2)=(0,1,2t^2) $$