How do I find the slope of a curve of intersection at a given point? I am asked to find the slope of the curve of intersection between the upper half of the unit sphere and the plane $y=\frac12$ at the point $P(\frac12, \frac12,\frac {1} {\sqrt2})$
I made a drawing to illustrate how I visualize the problem.

I'm confused about how to parametrize the curve. Yeah, I see that it's a circle. I can probably use a trigonometric parametrization. But which one? And how do I say why that's the one I chose to use?
I understand that the partial derivatives will give me the slope of the curve at a point. So in order to find the slope at $P$ I just plug it in to the first partials. But if I have parametrized the curve, does that mean I have to change the coordinates of $P$?
I need help understanding this problem. The more I search for resources on the internet the more confused I become.

Assuming that you mean the unit sphere given by $x^2+y^2+z^2=1$, you have for $y=\frac{1}{2}$ the equation $x^2+z^2=\frac{3}{4}$. Solving this wrt $z$ we have $z=\pm\sqrt{\frac{3}{4}-x^2}$, and since you are looking for the upper half dome of your sphere we use the $+$-solution, that is, $z(x)=\sqrt{\frac{3}{4}-x^2}$. Now you can differentiate this wrt $x$ to obtain the slope. Is this what you are looking for?
