Finding Tangent Line to Graph Find a vector equation for the tangent line to the curve of intersection
of the cylinders

$\ x^2 + y^2 = 25$
and
$\ y^2 + z^2 = 25$

at the point (3,4,2).


I don't understand the answer key. I've explained my interpretations below. Could someone clarify:



Why is the projection of C onto the xy plane:$\ x^2 + y^2 = 25$.  I understand that a projection in 3D space is the "shadow" of the graph onto one plane, but I don't quite see that C's projection can be modeled as given.

Why is z >= 0?
The intersection is a subset (or, if you prefer, contained within) the cylinder given by $x^2+y^2=25$, right? That equation is not sufficient to determine the points on the curve, but they all do fulfill it, because one thing the intersection is is exactly being part of that cylinder.

Now, if you take any point on the cylinder and project it onto the $x/y$ plane, what do you get? You get a point with $z=0$ and $x^2+y^2=25$. Try with the given point, $3^2+4^2=25$.

As for $z&gt;0$, you are in the vicinity of $(3,4,2)$, and for that point, $z=2&gt;0$.
The projection of the cylinder $x^2 + y^2 = 25$ onto the $xy$ plane is the circle $x^2 + y^2 = 25,\,z = 0$. Any intersection with the cylinder will lie on the cylinder. Thus the projection of the intersection will be some part of the projection of the cylinder.

$z$ is close to 2 near the point $(3,\,4,\,2)$ and is thus more than 0.