So I have a curve(in $\mathbb{R}^3$) descripted as an intersection of two surfaces: $$z^2=x^2+y^2, z=1-y$$
so I get
$$x=\sqrt{1-2y}$$ $$y=y$$ $$z=1-y$$
Is this a good parametrization. But what does this Curve represent? How can I draw it from this? And how can I bound y? is it just $y\in{\mathbb{R}}$ or should I bound it differently.
Any help would be apreciated. Thank you in advance.
On
The first surface is a cone. The second surface is a plane parallel to the angle of the cone, so the curve is a parabola lying in the plane $z=1-y.$
You have
$$(1-y)^2 = x^2+y^2$$
So
$$y=\frac{1-x^2}{2}$$
is the equation of the projection of the parabola onto the $xy$-plane.
So let $x$ be your parameter.
No, since you have lost those points of the curve whose first coordinate is negative. Note that, since $x^2=1-2y$, you have $y=\frac{1-x^2}2$ and $z=1-y=\frac{1+x^2}2$. So, take the parametrization$$\left\{\begin{array}{l}x=t\\y=\frac{1-t^2}2\\z=\frac{1+t^2}2,\end{array}\right.$$with $t\in\mathbb R$.