Find a vector function represented by the curve of intersection? I'm struggling with the following problem: 

Given $\, z = \sqrt{x^2 + y^2}\,$ and $\, z = y+1\,$ find the vector function represented by the curve of intersection of the surfaces using the parametrization $\, x =  t$.

I've done problems like this before but this one seems different. The $\,x\,$ is contained in a function with two other variables. My attempt was equating the two functions of $\,z$. Then I solved for $\,y\,$ which got me:
$\,y=\left(t^2-1\right)\big/\,2$. 
Not sure how to approach this problem. 

First, equate $\,z\,$ and square both sides of obtained expression:
$$
z = \sqrt{x^2 + y^2} = y+1 
\implies 
x^2 + y^2 = y^2 + 2y + 1 
\implies 
\boxed{y = \dfrac{x^2 - 1}{2}}
$$
Second, apply parametrization $\, x = t$:
\begin{align}
\begin{cases}
y = \dfrac{x^2 - 1}{2}\\
z = y + 1 \\
z = \sqrt{x^2 + y^2}
\end{cases}
\stackrel{x\, =\, t}{\implies}
\begin{cases}
x = t \\
y  = \dfrac{t^2 - 1}{2} \\
z = y + 1 
\end{cases}
\implies
\begin{cases}
x = t \\
y = \dfrac{t^2 - 1}{2} \\
z = \dfrac{t^2 +1}{2} 
\end{cases}
\end{align}
Therefore your vector function will look like
$$
F\left(t\right) = 
\begin{pmatrix}
t \\ \dfrac{t^2 - 1}{2} \\ \dfrac{t^2 + 1}{2} 
\end{pmatrix}
= \frac 1 2\;
\begin{pmatrix}
2t \\ \ t^2 - 1  \\ t^2  + 1
\end{pmatrix}
$$
