[RESOLVED] Relation between gradients of two perpendicular graphs.

[eranga262154]

Guys and gals,

Say I have two graphs (linear on non-linear) on two planes. As an example on XYZ 3D orbit, one graph on XY plane and the other one is on YZ plane. How can I formulate the combination of gradients of two graphs.

Thanks,

[jemidiah]

I'm a little confused I think.

These graphs are 2d, so the gradient at any point is a 2d vector. Could you simply add a 3rd component to each vector that is 0 at all times, and add the two gradient vectors to combine them? So the graph on the XY plane would have 3d gradient <g1x(x), g1y(y), 0> and the graph on the YZ plane would have 3d gradient <0, g2y(y), g2z(z)>, and their "combined" gradient would be <g1x(x), g1y(y) + g2y(y), g2z(z)>.

How exactly do you want to combine the two gradients?

[eranga262154]

Ya, I agreed with you. But no need to go such a way.

This is one way to do that I found from a book. First find a common point on the common axis, that is Y according to my example. Then it says that,

grd(XY plane graph) = -1 / grd(YZ plane graph)

I'm confusing how it is happened. Try to prove by using elementary of two graphs and fail. Any idea?

[jemidiah]

I remember from years ago that two perpendicular lines have slopes related by slope1 = -1/slope2. Maybe that's just what's happening here?

I'm still a little confused about this setup (and dividing by a vector, but I assume it can be done by dot products). Like you said, the two graphs should meet only on the y-axis, in which case they have perpendicular gradients only if the dot product of grad1 and grad2 are 0--but, this seems wayy too arbitrary to be useful. (The dot product in this case would have to have a 3rd dimension added like we've talked about.)

Are you saying that grad1 [dot] grad 2 = 0 is equivalent to grad1 = -1/grad2?


Perhaps the grad1 = -1/grad2 is a special case or alternate formulation of the dot product rule? Checkin': (* will mean dot product instead of regular product)

grad1 * grad2 = 0 'condition for perpendicularity
grad1 = -1/grad2 = -1 * grad2 / (grad2 * grad2) = -grad2 / |grad2| -- no...

Well, I'm more confused now than when I started. Any thoughts?

[eranga262154]

That's what happened to me also. Well I'll try it with your explanation and let see what we can do.