Showing two families of curves are orthogonal.

[EricVT]

Let the function f(z) = u(x,y) + iv(x,y) be analytic in D, and consider the families of level curves u(x.y)=c1 and v(x,y)=c2 where c1 and c2 are arbitrary constants. Prove that these families are orthogonal. More precisely, show that if zo=(xo,yo) (o is a subscript) is a point in D which is common to two particular curves u(x,y)=c1 and v(x,y)=c2 and if f '(zo) is not equal to zero, then the lines tangent to those curves at (xo,yo) are perpendicular.

I really have absolutely no idea how to show this. It gives the suggestion that

$$\frac{\partial u}{\partial x} + \frac{\partial u}{\partial y}\frac{dy}{dx} = 0$$

and

$$\frac{\partial v}{\partial x} + \frac{\partial v}{\partial y}\frac{dy}{dx} = 0$$

So the total derivatives with respect to x of u and v are both zero. Should I equate these and look for some relationship between the partials? Since the function is analytic we know

$$u_x = v_y$$

$$u_y = -v_x$$

So this can be rewritten in several different ways, but I really just don't know what I am looking for.

Can anyone please offer some advice?

 

[Dick]

Think about it this way. At a given point (x,y) look at the gradient vectors of u and v. grad(u)=(u_x,u_y), grad(v)=(v_x,v_y). The gradient is normal to the slope of the level curve. Compute the dot product of the gradients. What does the tell you about the slopes of the level curves?

 

[EricVT]

So, the slopes should be inverse and opposite?

I think I see how to write this now. I can use the total derivative of u with respect to x and solve for dy/dx and then set the inverse of that to dy/dx for the orthogonal family. Finally, it should work back to the total derivative of v with respect to x using the Cauchy-Riemann equations (since it is analytic).

Thanks for that first step, hopefully the rest of my reasoning is right.

 

[Dick]

Sounds right. Arguing from gradients seems easier, but it does look like they want you to go that way.