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Showing two families of curves are orthogonal. 
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#1
Oct207, 06:47 PM

P: 163

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] and [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
Oct207, 10:12 PM

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P: 25,251

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?



#3
Oct207, 10:37 PM

P: 163

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 CauchyRiemann equations (since it is analytic). Thanks for that first step, hopefully the rest of my reasoning is right. 


#4
Oct207, 10:44 PM

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P: 25,251

Showing two families of curves are orthogonal.
Sounds right. Arguing from gradients seems easier, but it does look like they want you to go that way.



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