Create curve function from intersection of two surfaces

[Addez123]

Homework Statement

Create the curve r(u) from
$$4x - y^2 = 0$$
and
$$x^2+y^2-z = 0$$

Relevant Equations

Vector calculus

What I do is set the two equations equal to one another and solve for z.
This gives:
$$z = \sqrt{x^2+2y^2-4x}$$
which is a surface and not a curve.

What am I doing wrong?

 

[Mark44]

Create the curve r(u) from
$$4x - y^2 = 0$$
and
$$x^2+y^2-z = 0$$

What I do is set the two equations equal to one another and solve for z.
This gives:
$$z = \sqrt{x^2+2y^2-4x}$$
which is a surface and not a curve.

What am I doing wrong?

I don't see how you got the equation you show.

The two equations can be rewritten as
$y^2 = 4x$ and
$z = x^2 + y^2$
Note that from the equations above, that $x \ge 0$ and $z \ge 0$.
Substituting, we get $z = x^2 + 4x$.

 

[Orodruin]

Homework Statement:: Create the curve r(u) from
$$4x - y^2 = 0$$
and
$$x^2+y^2-z = 0$$
Relevant Equations:: Vector calculus

What I do is set the two equations equal to one another and solve for z.

What do you even mean by this? The two equations are independent equations and putting two independent equations equal to each other makes no sense whatsoever.

If you mean that you put the non-zero sides of the equations equal to each other then that is where you went wrong because you just threw away the information that both expressions are equal to zero independently and you will end up with the surface along which those expressions take the same value regardless of whether that value is zero or not.

 

[Addez123]

Ahh it's true!
I eliminated the information when i set them equal. I should've done like @Mark44 suggested and used
$$y^2 = 4x$$
into the second equation. Then I can easily get a one variable function/curve.

 

[Orodruin]

Note: I would use $y$ as the curve parameter instead.