1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Orthogonal Families

  1. Mar 19, 2013 #1
    1. The problem statement, all variables and given/known data
    Naajjmloos.jpg


    2. Relevant equations
    The product of the slopes has to be equal to -1.

    3. The attempt at a solution
    Well, as the function, call it u, is equal to a constant, the derivative of u with respect to x is
    the partial derivative with respect to x + (the partial derivative of u with respect to y * dy/dx) = 0

    So dy/dx = - (∂u/∂x) / (∂u/∂y)

    But from here on I don't know how to continue. Could anyone provide a hint?
     
  2. jcsd
  3. Mar 20, 2013 #2
    I apologize, I had not realized that it was in the wrong section before.

    To be precise, I found that
    dy/dx = [tex]-\frac{x \left(B^2+L\right)}{y \left(A^2+L^2\right)}[/tex]

    Before imposing anything on a, lamda and b.

    And thus, If I call the lamba of the first family L and the lamba of the second family L2:

    [tex]\frac{x^2 \left(B^2+L\right) \left(B^2+\text{L2}\right)}{y^2 \left(A^2+L^2\right) \left(A^2+\text{L2}^2\right)}=-1[/tex]

    Is what I have to show, if they are indeed orthogonal.

    Should I express Y in terms of X and plug that in?
     
    Last edited: Mar 20, 2013
  4. Mar 20, 2013 #3

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    I've been looking at this problem since shortly after you posted it. It does look quite difficult.

    Here's my take on it. -- nowhere near to getting a solution.

    For the case [itex]\displaystyle \ -A^2<\lambda<-B^2\ :[/itex] (Yes, I prefer using the upper case for a & b, but like to stick with lower case Greek for λ.)
    λ < -B2 so then B2 + λ < 0 .

    So the equation for the family of curves becomes:

    [itex]\displaystyle \frac{x^2}{A^2+\lambda^2}-\frac{y^2}{\left|B^2+\lambda\right|}=1\ .[/itex]

    Also, A2>B2 and λ2 > 0 so of course, A2 + λ2 > | B2 + λ | .

    This equation describes a hyperbola, centered at (0, 0) with semi-major axis of [itex] \displaystyle \ \sqrt{A^2+\lambda^2\ } [/itex] along the x-axis and semi-minor axis of [itex] \displaystyle \ \sqrt{\left|B^2+\lambda\right|\ } [/itex] along the y-axis.

    Similarly you can show that for λ > -B2 > -A2, B2 + λ > 0, so that the family of curves described are ellipses.

    So, it seems to me that the problem makes most sense, if we consider A and B to be each fixed at some value and that the "families" are to be parametrized by λ .


    I realize that this is very far from a solution. -- It's even far from beginning a solution. -- But, hopefully it helps understanding what's being asked for.

    SammyS
     
  5. Mar 20, 2013 #4
    Edit 2: Indeed, the first labda is NOT supposed to be squared, that is a typo! In that case I know of a method on how to solve it, it basically just follows the website I posted below.

    Dear SammyS,

    Thank you for your response. Indeed, I also thought that it would be most natural to think of λ to be the changing factor, while keeping A and B constant. Interpreting what it meant has proven to be a bit problematic for me, so your input is very useful.

    The 'book' I use in my course - complex analysis - is Schaum's Outline of Complex Variables, which is basically just a summary type of book. It does have 1 worked out exercise dealing with orthogonal families, and that is the approach I followed. It uses implicit differentiation to find dy/dx, for both of the equations, and then you can see that the product is -1 without too much hassle.
    This problem is clearly not in the same ballpark, as it is very nontrivial (if it can even be seen?) that the product is -1 in this case. I'm going to work out a few examples with values of a and b, and see if that gets me anywhere..

    Alright, so what this shows me is that the approach does not work, as far as I can tell. I don't get that the product is -1, for specific values of a, b and the lamda's. Guess I'll have to think of something else..

    Making a few contour plots in mathematica, for different values of lamda, the first family is indeed a family of hyperbola's and the second one a family of ellipses.

    Now, I don't understand the approach, but http://planetmath.org/hyperbolasorthogonaltoellipses treats how to show that a certain family of ellipses, not too much unlike the one I have here, is perpendicular to a family of hyperbola's.

    Looking at it, it does show some promise, but I can't reproduce the line after (3). Applying the same method to my system seems to be working, up to that point, where I extract a quantity that is not equal to zero.

    Edit: Alright, so I used the method posted on that website. It ALMOST works! The only issue is that, in the second denominator, lamda is not squared, while the first one is. So if the first one is also not supposed to be squared, then it works.


    Edit 2:
    Indeed, the first labda is NOT supposed to be squared, that is a typo!
     
    Last edited: Mar 20, 2013
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted