Find the points of the trajectory in which the velocity is 9

Question

I have a parametric curve given by $M:[0,2]\rightarrow\mathbb{R}; x(t) = 3t^2, y(t) = t^3,z(t)=6t$. I need to find the points of its trajectory in which the velocity is 9. I also need to represent those points in matlab.
I know that the velocity is given by the first derivative, so I found the derivative, which is $x'(t) = 6t, y'(t) = 3t^2, z'(t) = 6$. When I calculate the magnitude of $|M(t)| = 9$, I get the values of $t$ as being $i-2$ and $-i-2$.
Now, how do I plot those points in matlab or is my solution for this problem incorect?

Answer 1

$$v^2=(6t)^2+(3t^2)^2+6^2=9t^4+36t^2+36=9^2=81$$ You can simplify this to $$t^4+4t^2-5=0$$ This yields $$t^2=\frac{-4\pm\sqrt{4^2-4(-5)}}{2}=\frac{-4\pm6}{2}$$ Since we want $t^2>0$, we have only $t^2=1$ so $t=\pm1$.