Is it possible to convert vector valued function in this form: $r(t)=\langle3,t,t^2-1\rangle$ into non parametric form? For instance with $r(t)=\langle t+1, t^2-2\rangle$, we can substitute it with x values and y values to get $y=t^2-2=(x-1)^2-2$ form that is easy to graph. How can we do the same for $r(t)=\langle 3,t,t^2-1\rangle$?
I managed to sub $z=y^2-1$, but not sure where to place the $x=3$ in the equation.
 A: Actually, you have it ! What you need to plot is the function of equation $z=y^2-1$, plotted in the $yz$ plane of equation $x=3$. Here is a drawing (using online Geogebra 3D, equation : $(3,y,y^2-1)$).

A: You've got a parabola $z=y^2-1$ in the $y-z$ plane, translated three units parallel to that plane.  So it isn't a surface,  but rather $1$-dimensional.  There's only one free variable.
A: The usual way to represent that curve in a non-parametric way is, as already observed in other answers, as a pair of equations: $z=y^2-1, x=3$.
If one is been stubborn, a rather convoluted way to obtain a single equation could be
$$(y^2-z-1)^2+(x-3)^2=0$$
or may be
$$|y^2-z-1|+|x-3|=0$$
but those are just artificial and more complicated ways to join the two equations and should be avoided.
A: Your curve $r(t)$ (dimension 1) is immersed in $\mathbb{R}^3$. It is said that the codimension is 3-1=2. Then, you need two implicit expressions to properly describe your curve. In this case, is the intersection of two surfaces:
$$
z=y^2-1
$$
and the plane
$$
x=3
$$
By the way, a surface (dimension 2) has codimension 3-2=1, so you only need 1 equation. The intersection of surfaces is a curve (in general).
