# Finding an ODE to satisfy the family of curves orthogonal to another family of curves

## Homework Statement

Consider the family F of circles in the xy-plane (x-c)2+y2=c2 that are tangent to the y-axis at the origin. What is a differential equation that is satisfied by the family of curves orthogonal to F?

∇f(x,y)=<fx,fy>

## The Attempt at a Solution

My general strategy: I know that for a function of two variables, f(x,y), it's gradient ∇f(x,y) is orthogonal to every level curve of f(x,y) on the xy plane. So I first wanted to find a function of two variables, f(x,y) for which F is its level curves, then take ∇f because it would be orthogonal to all the curves in the family F.

So I started as follows. I plug z in for c, then solve for z

(x-z)2+y2=z2
x2-2xz+z2+y2=z2
x2-2xz+y2=0
(x2+y2)/(2x) = z = f(x,y)

(Originally I just left z implicitally in the equation and when I took the gradient I did implicit partial differentiation. But it just seemed like this was simpler.)

So now I realize that the family of level curves for f(x,y) is F that we started with. Now, to find curves that are orthogonal to F, I take the gradient of f.

∇f(x,y)=<fx,fy> = <1/2-y2/2x2, y/x>

Now I know that for a curve in the family of F, at any point on the curve, ∇f(x,y) specifies a vector that is orthogonal to the curve at that point. Now this is where I get hazy. I want to find now an ODE that will satisfy this. So what I thought was just going:

0=M(x,y)dx + N(x,y)dy where M(x,y) = fx, N(x,y) = fy (I thought of doing this because I felt, this would make an "exact" equation)

0 = (1/2-y2/2x2)dx + (y/x)dy

0 = dy/dx + (1/2-y2/2x2)/(y/x)

0 = dy/dx + (x2-y2)/2xy

dy/dx = (y2-x2)/2xy

OK... now the weird part is that the answer in the book is the negative reciprocal of this.... they say dy/dx = 2xy/((x2-y2)

!

I don't understand why. I know that if two lines have slopes that are negative reciprocals then they are perpendicular (orthogonal), so I'm assuming it has something to do with that. But I'm not sure what. It seems since my answer is the negative reciprocal, have I just found the ODE that satisfies my original F, not the curves orthogonal to it? If so, why? I thought the gradient was orthogonal to the level curves of f(x,y). I don't know where my logic is wrong but I assume it's in going from having found ∇f to finding the ODE that satisfies it. Because I am a little confusd. ∇f just specifies a vector at a certain point. But I'm looking for a function. NOTE: I'm not looking for someone to solve the problem for me. I just really want to understand where my logic is wrong so I can understand the concepts.

Now I know that for a curve in the family of F, at any point on the curve, ∇f(x,y) specifies a vector that is orthogonal to the curve at that point. Now this is where I get hazy. I want to find now an ODE that will satisfy this.

∇f(x,y) is the vector orthogonal to the F curves. The curves you are looking for must also be orthogonal to F curves, which means they must be tangential to ∇f(x,y). You, however, look for curves orthogonal to ∇f(x,y) - which means the F curves you started with. That's why you get the equation for the F curves in the end.

HallsofIvy
I don't know why you "plug z in for c". Rather, just differenting $(x- c)^2+ y^2= c[itex], with respect to x, you get [itex]2(x- c)+ 2y (dy/dx)= 0$. From that dy/dx= -(x- c)/y so that a curve that is orthogonal to that has dy/dx= y/(x- c).