So I've been working on a way to prove rigorously that a parametric curve is the intersection of two surfaces but I'm unsure of how to show it
Question
Show that the curve with parametric equations
$$x(t) = cos(t)$$ $$y(t) = sin(t)$$ $$z(t) = sin^2(t)$$
is the equation of the intersection curve of the surfaces
$$z = x^2$$ $$x^2 + y^2 =1$$
Solution Attempt
The intersection curve of the two surfaces is all the points such that
$$\{(x,y,z)\in \mathbb{R}^3 | z = x^2, x^2 + y^2 = 1\}$$
On the other and, we know that
$$z(t) = sin^2(t) = (sin(t)) ^2 = x^2(t) \iff z(t) = x^2(t), $$
and $$x^2(t) + y^2(t) = cos^2(t) + sin^2(t) = 1 \iff x^2(t) + y^2(t) = 1$$
So we know that the parametric curve, whatever $t \in \mathbb{R}$ is, respect the constraints.
Therefore, if C is the parametic curve, then
$$C = \{ (x,y,z)\in \mathbb{R}^3 | x^2(t) + y^2(t) = 1, z(t) = x^2(t)\}$$
Here is where I dont know how to continue How do I know that it is exactly the intersection? In other words, how do I use $t$ to make a conclusion and show that it is exactly the intersection?
Thank you!
Plug the curve $\langle\sin t, \cos t, (\sin t)^2\rangle$ into the equations for the surfaces. Once you do that, you'll see that the equations are true, proving the curve indeed lies on the intersection: For example, you have: $z=x^2 \implies (\sin t)^2 = (\sin t)^2 $
Then do the same thing for the other surface equation!