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Orthogonal Curves

  1. Mar 9, 2005 #1
    How do I prove that two curves are orthogonal when they interest each other at a specific point?

    Do I just take the derivative of both and compare the slopes?
    The slopes should be negative reciprocals of each other, correct?
  2. jcsd
  3. Mar 9, 2005 #2


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    well since you have good ideas, what would convince you they are right?
  4. Mar 9, 2005 #3
    I believe I am correct; however, I do wish to prove it.
    I need two equations that are orthogonal.
    I can't find any in my Calculus book.

  5. Mar 9, 2005 #4


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    do you mean you want the definition of orthogonal curves? to me it would be as you assume, i.e. at least for smooth curves, that their tangent lines are orthogonal.
  6. Mar 9, 2005 #5
    Yea that works, but do you have any equations I could play with?

    Thanks :)
  7. Mar 10, 2005 #6


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    You've got the basic idea, so what's stopping you from developing that idea into an equation?
  8. Mar 10, 2005 #7


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    Do you mean you want examples?

    y= x is orthogonal to y= -x!

    Can you find the equation of the line orthogonal to y= x2 at (1,1)?

    (Find the derivative of x2 at x= 1. Yes, the slope of a line orthogonal to that is the negative of the reciprocal of the derivative.)

    A common problem in differential equations is to find the family of curves orthogonal to a given family: To find the set of all curves orthogonal to the family of curves
    xy= a, differentiate with respect to x: y+ xy'= 0 (thus eliminating the constant a).
    Then y'= -y/x at each point (except x= 0). Any curve orthogonal to that must have
    y'= x/y or yy'= x so (1/2)y2= (1/2)x2+ C or
    x2- y2= c (c= -2C). The family of curves orthogonal to the hyperbolas having the axes as asymptotes is the family of hyperbolas having y= x and y= -x as asymptotes.
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