# Vector function for the curve of intersection of the paraboloid

1. Oct 20, 2004

Original question:

a) Find a vector function for the curve of intersection of the paraboloid z = 3x^2 + 2y^2 and the cylinder y = x^2.
b) Show that this curve passes through (1,1,5) but not (3,3,9).

I really have no idea how to do either parts of this question. Any help would be greatly appreciated!! Thanks!!

2. Oct 21, 2004

### vsage

I am not quite sure how y=x^2 is a cylinder in 3d coordinates but you don't need to have a good sense of visualization to do this question (A TI-89 or other device that does 3D graphs might help you though). Substitute one equation into the other and think about each variable as a function of t. As for part b, all you need to show is that where x(t1)=3, y(t2)=3 and z(t3)=9 that either t1!=t2 or t2!=t3 and show that where x(t)=1 that y(t)=1 and z(t)=5. Hope this helps.

3. Oct 21, 2004

### robphy

b) almost seems too easy... (it seems you don't need the solution to (a) to answer (b) )
If (1,1,5) is an intersection point of two surfaces, then
x=1, y=1, z=5 should yield an equality for both
z = 3x^2 + 2y^2 and y = x^2. Otherwise, it's not an intersection point.
Observe that x=3, y=3, z=9 does not satisfy both equations.

for a),
following up on vsage's comment,
define "t" to be a parameter for the curve, so that ( x(t),y(t),z(t) ) describes a point on that curve. A useful choice is start with is x(t)=t. Then y(t) and then z(t) follow.