I have the vector-valued function $\vec{r}(t) = t \hat{i}+2t\cos(t) \hat{j} + 2t \sin(t) \hat{k}$, and the cone $4x^2=y^2+z^2$.
I figured, that by showing I could re-write the equation for the cone as the vector-valued function, it would be sufficient proof that the vector-valued function lies on the cone.
So, I let $y = 2x\cos(t)$ and $z=2x\sin(t)$, and, finally, let $x=t$
This shows that the cone can be written as $\vec{r}(t) = t \hat{i}+2t\cos(t) \hat{j} + 2t \sin(t) \hat{k}$
However, the answer is: 
Did I correctly answer this question? I don't think the answer wrote out enough steps.
Best Answer
It's worth noting that the cone, itself, cannot be written in the form $$\vec r(t)=t\hat i+2t\cos(t)\hat j+2t\sin(t)\hat k.$$
The question at hand is, does that curve lie on the cone? In particular, all we need to show is that for $\langle x,y,z\rangle$ on that curve--that is, $x=t,y=2t\cos(t),z=2t\sin(t)$ for some $t$--we have $y^2+z^2=4x^2$, and the book wrote out precisely enough steps to show that.