Question

Find a vector function that represents the curve of intersection of the two surfaces: The cone z=sqrt(x^2 + y^2) and the plane z =1 + y.

Answer 1

Try to express all of x, y, z in terms of just one of the three variables. It is a bit tricky to do, and for some intersections of surfaces not possible.But this time you can do it. Here is one way.

Squaring first equation, then substituting for z from the second:

x^2+y^2 = z^2 = (1+y)^2 = 1+2y+y^2.

Subtracting y^2 from both sides:
x^2 = 1 + 2y.

So, starting with x you can use the preceding equation to find y= (x^2-1)/2, then from that you can use z=1+y to get z. So x is a good parameter.

Parameterization:
x=t; y = (t^2-1)/2; z = 1+(t^2-1)/2 = (t^2+1)/2.

All values of t.