Let $C$ be the curve of intersection of the parabolic cylinder $x^2 = 2y$ and the surface $3z = xy$. Find the length of the part of $C$ from $(0, 0, 0)$ to $(6, 18, 36)$. (Hint: It may be useful to note the identity a $2 + 4a + 4 = (a + 2)^2$ in the middle of computation.)
We were given the solution to this problem. I am given the two equations $x^2=xy$ and $3z=xy$ which are then parameterized into $\langle t,.5t^2,(1/6)t^3\rangle$. I was wondering if someone could help me with this parameterization step.
A start: If $x=t$ then from the first equation we get $y=\frac{t^2}{2}$ and from the second we then get $z=\frac{t^3}{6}$.