Finding a curve in 3 space when two equations intersect

[skate_nerd]

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]

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:

$\displaystyle z=x^2+\frac{1}{4}$

 

[skate_nerd]

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]

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:

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

$2y=1$

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

 

[skate_nerd]

Ahh there we go. Thanks, got it now.