I'm working on the following problem from a practice midterm:
Let C be the intersection of the sphere of radius 2 centered at the origin and the plane y + z = 0.
(a) Write parametric equations for C.
(b) Choose your favorite point on C (any point will do) and write parametric equations for the tangent line to C at that point.
For (a), I've gotten up to the point where I found the equation of the sphere ($x^2+y^2+z^2=4$). I also wrote $z$ as $z = -y$.
Now I'm not entirely sure how I'm supposed to parametrize this equation. My friend told me to 'say y = t' and to 'find z in terms of y' and 'x in terms of y.'
I interpreted this as plugging in z = -y into the equation for a sphere and solving accordingly. I got x = -2t+2, z = t, and y = t, though I'm not sure if I computed that correctly.
My question is this: is the reason why were are getting the parametric equation of C so that we can construct an equation we can take the derivative of so we can get the tangent line (what part b of the question is asking for, basically)? I'm confused as to why we're allowed to 'say y = t' when trying to compute the parametric equations - is there a generalized approach for finding the parametric equation of an intersection like C?
The intersection of a sphere and a plane is a circle, and the projection of this circle in the $xy$ plane is the ellipse $$ x^2+y^2+(-y)^2 = x^2+2y^2 = 4 $$ This information we can use to find a suitable parametrization. Indeed, you can parametrize the ellipse as follows $$ x= 2 \cos t\quad y = \sqrt{2} \sin t $$ with $t\in [0,2\pi] $. And since you know that the curve belongs to the plane $z=-y$, you end up with \begin{cases} x= 2 \cos t\\y = \sqrt{2} \sin t \qquad t\in [0,2\pi]\\ z=-\sqrt{2} \sin t \end{cases}