# Intersection of multivariate graphs

Sanelma shared this question 1 year ago

Hi, I have two multivariate functions graphed: f(x,y)=x^2 + y^2 and g(x,y) = xy+2. I would like to show the curve of intersection of these two graphs. Is this possible? Couldn't get it to work with regular "intersect" command.

1

The projection of the intersection onto the xOyPlane is:

x² + y² - x y - 2 = 0
Curve(2 / sqrt(2) (sin(t) - cos(t) / sqrt(3)), 2 / sqrt(2) (sin(t) + cos(t) / sqrt(3)), t, 0, 2π)

so the intersection is

Curve(2 / sqrt(2) (sin(t) - cos(t) / sqrt(3)), 2 / sqrt(2) (sin(t) + cos(t) / sqrt(3)),f(2 / sqrt(2) (sin(t) - cos(t) / sqrt(3)), 2 / sqrt(2) (sin(t) + cos(t) / sqrt(3))), t, 0, 2π)
https://www.geogebra.org/m/ebkfuyaj

1

Hi Michael,

I'd like to show the curve of the intersection of x^2+4y^2+9z^2=1 and z = h x + c.

I also want to an animation point on this intersection curve.

Is it possible?

Thank you so much for your help.

1

That's easier, there's a command for that :)

Try:

c:IntersectPath(a, b)
A=Point(c)
StartAnimation(A)

1

Thank you, Michael!

Actually I have a surface.Is there a command for Intersection of a surface and a plane?

I try yours but it didn't work.

1

Hi Sanelma!

One possibility is to solve the equation f(x,y)=g(x,y) and to use the solution for drawing one rather two curves.

Greetings

mire2

1

That's very nice!

I think you need an extra +2 on the z-coordinate

1

Oh - thank you for your kind words and yes, I forgot the +2 on the z-coordinate, so all must be above the x-y-plane.

Good looks!

Kind regards

mire2