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Showing two families of curves are orthogonal.

  1. Oct 2, 2007 #1
    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

    [tex] \frac{\partial u}{\partial x} + \frac{\partial u}{\partial y}\frac{dy}{dx} = 0 [/tex]


    [tex] \frac{\partial v}{\partial x} + \frac{\partial v}{\partial y}\frac{dy}{dx} = 0 [/tex]

    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

    [tex] u_x = v_y [/tex]

    [tex] u_y = -v_x [/tex]

    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?
  2. jcsd
  3. Oct 2, 2007 #2


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    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?
  4. Oct 2, 2007 #3
    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.
  5. Oct 2, 2007 #4


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