parametrize intersection of a plane and paraboloidby re12 Tags: intersection, paraboloid, parametrize, plane 

#1
Jul1109, 06:08 PM

P: 3

1. The problem statement, all variables and given/known data
Parametrize the intersection of the paraboloid z = x^{2} + y^{2} and the plane 3x 7y + z = 4 between 0 [tex]\leq[/tex] t [tex]\geq[/tex] 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 + x^{2} + y^{2} = 4 x^{2} + 3x + (3/2)^{2} + y^{2} 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 = [tex]\sqrt{37/2}[/tex]  (3/2) at t = 0 so x = ([tex]\sqrt{37/2}[/tex]  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. 



#2
Jul1109, 06:34 PM

HW Helper
P: 5,004





#3
Jul1109, 07:31 PM

P: 3

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 + ([tex]\sqrt{37/2}[/tex]*cos(t) y = 7/2 + ([tex]\sqrt{37/2}[/tex]*sint(t) Can anyone lead me on the right track to finding z? Thanks 



#4
Jul1109, 07:40 PM

HW Helper
P: 5,004

parametrize intersection of a plane and paraboloid
z=x^2+y^2




#5
Jul1109, 07:52 PM

P: 3

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



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