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Solving for Orthogonal Trajectories

  1. Sep 15, 2009 #1
    1. The problem statement, all variables and given/known data

    Which of the following is the set of orthogonal trajectories for the family indicated by

    (x-c)^2 + y^2 = c^2

    a). (x-c)^2 + y^2 = c^2

    b). (x-c)^2 - y^2 = c^2

    c). x^2 + (y-c)^2 = c^2

    d). x^2 - (y-c)^2 = c^2

    e). None of the above




    2. Relevant equations

    I think this is a homogenous equation after playing with the variables and substituting for C. I came up with:

    y' = (y^2 - x^2)/xy

    I then substitute y=xv after setting y' = 1/y' for the orthogonal trajectories and tried to solve the equation, but ended up with answers no where close to the choices.



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 15, 2009 #2

    tiny-tim

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    Hi vorse! :smile:

    Let's do it the easy way :wink:

    draw a few curves, and that should show you the most obvious orthogonal trajectory …

    which of a) to e) does that one lie on? :smile:
     
  4. Sep 15, 2009 #3
    I was afraid you'd said I have to graph the equation; I can never make heads to tail of what I draw out.
     
  5. Sep 16, 2009 #4

    tiny-tim

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    ok … (x-c)2 + y2 = c2 is obviously a circle.

    So, for each value of c, what is its centre, what is its radius, and what are the endpoints of its "horizontal" diameter?
     
  6. Sep 16, 2009 #5
    Ok, I figured out the answer. I think it's answer choice C. It's just my answer has something like 2c as one of the constant, and I guess the answer choice reduce it down to just C. Btw, I didn't graph the equation; it is a homogenous equation and can be solve that way; just the algebra manipulation was confusing. Thanks for all the help PF. Thanks for trying to help me tiny tim.
     
    Last edited: Sep 16, 2009
  7. Sep 17, 2009 #6

    tiny-tim

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    Yup! :smile:

    But a quick sketch always helps …

    in this case, the original curves are the circles with centres on the x-axis, all touching each other (and the y-axis) at the origin.

    Quickly hovering over this with a pencil shows that the orthogonal trajectories must approach the y-axis horizontally, and must all go through the origin …

    which choice C does, since it's the circles with centres on the y-axis. :wink:

    hmm … are those all of the orthogonal trajectories?

    (btw, the wiki page http://en.wikipedia.org/wiki/Orthogonal_trajectories" [Broken])
     
    Last edited by a moderator: May 4, 2017
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