# Parametrize intersection of a plane and paraboloid

1. Jul 11, 2009

### re12

1. The problem statement, all variables and given/known data

Parametrize the intersection of
the paraboloid z = x2 + y2
and the plane 3x -7y + z = 4
between 0 $$\leq$$ t $$\geq$$ 2*pi

When t = 0, x will be greatest on the curve.

2. Relevant equations

3. The attempt at a solution

I never really know how to do these kinds of problem. I am more familiar with parametrizing straight lines. Here is what I have done so far

I substitute the z in the plane equation with the paraboloid

3x - 7y + x2 + y2 = 4
x2 + 3x + (3/2)2 + y2 -7y + (7/2)2 = 37/2
(x + 3/2) 2 + (y - 7/2)2 = 37/2

which is a circle centered at (-3/2 , 7/2) with radius 37/2

So to parametrize x, I did

x = $$\sqrt{37/2}$$ - (3/2) at t = 0 so
x = ($$\sqrt{37/2}$$ - 3/2) * cos(t)

This may be wrong, but I am not sure. Please let me know if I am on the right track and how can I continue with this problem. The y and z components seem to be more complicated.

Last edited: Jul 11, 2009
2. Jul 11, 2009

### gabbagabbahey

What happened to the 4?

Close, to find x, you start at the center (-3/2) and add some fraction of the radius....that means you have x=-3/2+f(t)*radius, and so you want x=-3/2+cos(t)*radius not cos(t)*(-3/2+radius)....make sense?

3. Jul 11, 2009

### re12

I added (3/2)2 and (7/2)2 to both side so it will be greater than 4. I think I put in the wrong numbers when I use my calculator. It should be 37/2 instead of 29/2
And that explanation made a lot of sense heh.

So I ended up with
x = -3/2 + ($$\sqrt{37/2}$$*cos(t)
y = 7/2 + ($$\sqrt{37/2}$$*sint(t)

Can anyone lead me on the right track to finding z? Thanks

4. Jul 11, 2009

### gabbagabbahey

z=x^2+y^2

5. Jul 11, 2009

### re12

Love it whenever a problem that looks complicated has simple solution. heh thanks =)