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finding a curve in 3 space when two equations intersect

skatenerd

Active member
Oct 3, 2012
114
So I have two equations that intersect: z=x2+y2 which I know is a paraboloid, and z=x2+(y-1)2 which I know is also a paraboloid shifted one unit in the positive y-direction. However I attempted to find the intersection curve and only way I could think to do that was by setting the two equations equal to each other, x2+y2=x2+(y-1)2 however that ended inconclusively. Any help with this would be appreciated. I have to graph it too.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Equating the two functions gives you $\displaystyle y=\frac{1}{2}$ and then using this value for $y$ in either function gives the curve of intersection as:

$\displaystyle z=x^2+\frac{1}{4}$
 

skatenerd

Active member
Oct 3, 2012
114
I guess I'm overlooking something dumb right now but when I set them equal to each other, the x2's cancel out and then you have y2=(y-1)2, or y=y-1...
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Yes, you do get:

$y^2=(y-1)^2$

Now, at this point you don't want to simply take the positive root of both sides (as this has no solution), you want to expand the right side to get:

$y^2=y^2-2y+1$

$0=-2y+1$

$2y=1$

$\displaystyle y=\frac{1}{2}$

Now, initially you could have equated the positive root on the left with the negative root on the right to get:

$y=-(y-1)=1-y$

$2y=1$

$\displaystyle y=\frac{1}{2}$
 

skatenerd

Active member
Oct 3, 2012
114
Ahh there we go. Thanks, got it now.