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Orthogonal Trojectories

  1. Feb 8, 2007 #1
    1. The problem statement, all variables and given/known data

    Write the equation to find the orthogonal trojectory for [tex]x^2 + y^2 = r^2[/tex]

    2. The attempt at a solution

    Well.. you solve this by just integrating.. but, I don't know what to do with the r^2 ?

    [tex]x^2 + y^2 = r^2[/tex]
    [tex]x^2 + y^2 \frac{dy}{dx} = r^2 \frac{dy}{dx}[/tex]

    is that even right? then... integrate r^2 to become [tex]\frac{r^3}{3}[/tex] ?
     
  2. jcsd
  3. Feb 8, 2007 #2

    Dick

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    I believe you are supposed to consider x^2+y^2=r^2 as a family of curves (one curve for each value of r). Now you are supposed to find another family of curves whose intersections with these are always orthogonal. Hint: derive a differential equation satisfied by the first set. How would you need to change the differential equation to force orthogonality? Hint 2): check your differentiation, r is a CONSTANT.
     
  4. Feb 8, 2007 #3

    HallsofIvy

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    You might also note that the family of curves is a family of cocentric circles! What curves are perpendicular to all of them?

    And your differentiation is awful! The derivative of x2 is NOT x2, the derivative of y2 is NOT y2 dy/dx and the derivative of r2 is NOT r2 dy/dx!
     
    Last edited: Feb 8, 2007
  5. Feb 8, 2007 #4
    ohh??

    so let me give this a shot...
    [tex]x^2 + y^2 = r^2[/tex]

    [tex]2x + 2y \frac{dy}{dx} = 2r[/tex]

    [tex]r = x + y \frac{dy}{dx}[/tex]

    [tex]\sqrt{x^2 + y^2} = x + y \frac{dy}{dx}[/tex]

    is this right so far...?
     
  6. Feb 9, 2007 #5

    Dick

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    This is the last time I'm saying this. r is a CONSTANT. What does this say about it's derivative?
     
  7. Feb 9, 2007 #6
    dy/dx is the tangent of the angle of the curve right? So you could find the tangent of the angle perpendicular to that and derive a new DE, which you could solve for your answer... I guess there's a geometric way to do this as well (without using calculus), like HallsofIvy said, what family of curves is perpendicular to that family of circles?
     
  8. Feb 9, 2007 #7

    Dick

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    Yes, as Halls said - it's pretty easy to just see the answer in this case. But I'm guessing this a HW for a calculus course so you might want to do it that way. Actually, doing it that way is not that much harder than guessing...
     
  9. Feb 9, 2007 #8

    HallsofIvy

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    No, it's not right. If you want to be really technical, the derivative of r2 with respect to x is [itex]2r\frac{dr}{dx}[/itex], just as you did with y- but since r is a constant, what is [itex]\frac{dr}{dx}[/itex]?
    Once you have found [itex]\frac{dy}{dx}[/itex] remember that a line is orthogonal to a line with slope m if and only if it has slope -1/m.
     
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