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Homework Help: Integration of composite derivative of function

  1. Nov 19, 2012 #1
    1. The problem statement, all variables and given/known data
    While I was in school I thought to see what a function whose derivative is always perpendicular to another functions derivative would look like, so for example for X^2 we have -ln(x)/2.


    2. Relevant equations
    Well all those integration tables I guess


    3. The attempt at a solution
    Obviously their product is -1, so if were looking for g(x),
    g(x) = ∫(-1/f'(x))dx
    However, I don't seem to be able to integrate this.
    I need to have a ln(f'(x)), who's integral is f''(x)/f'(x), but dividing by f''(x) or try to subtract it just makes things much tougher. This isn't homework, but it would be very nice if I was to be helped. Please don't give solution, just clue.

    Thanks, Bonaparte
     
  2. jcsd
  3. Nov 19, 2012 #2

    HallsofIvy

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    It sounds like you are looking for "orthogonal trajectories", a family of curves that are always perpendicular to another, given, family of curves.

    Yes, to find the orthogonal trajectories of [itex]y= x^2+ c[/itex], where c is a constant, we differentiate to get [itex]y'= 2x[/itex] and the constant has disappeared. Now, any curve perpendicular to that has [itex]y'= -1/(2x)[/itex] and the integral of that is -(1/2)ln(x)+ C.

    On the other hand, if the first family were given by [itex]y= cx^2[/itex], so that the constant is multiplied rather than added, [itex]y'= 2cx[/itex] and to get rid of the constant we have to write [itex]y'/y= 2cx/x^2= 2/x[/itex] so that [itex]y'= 2/(xy)[/itex]. Now to find the orthogonal trajectories, we look at [itex]y'= -2/(xy)[/itex] so that [itex]yy'= -2/x[/itex] or [itex]ydy= (-2/x)dx[/itex] and, integrating [itex](1/2)y^2= -2 ln(x)+ C[/itex].

    But there is no general way to solve that problem for all x. It depends to much upon the specific f. "Orthogonal trajectories" is typically a subject for a "Differential Equations" course.
     
  4. Nov 19, 2012 #3
    Could you explain how you got from 2cx/x^2 = 2/x? and then when you had y'/y=2/x, and multiplied both sides by y, how did you get 2/xy? should it not be 2y/x?
    I actually thought you would have 2cx*g'x = -1, g'x=-1/2cx
    gx = (-1/(2c))*ln(x)+C?
    Thanks, Bonaparte
     
  5. Nov 19, 2012 #4

    haruspex

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    HofI meant to write [itex]y'/y= 2cx/(cx^2)= 2/x[/itex], so as you say, [itex]y' = 2y/x[/itex]. For the orthogonals I get [itex]y' = -x/(2y)[/itex], leading to ellipses.
     
  6. Apr 12, 2013 #5
    Why is it not possible to deriviate 2cx and relate to c as if it is a constant? So you have y=2cx^2
    y'=(2cx), g'=-1/(2xc), so g= ln((2xc)/-2c'+C, wouldn't this be the answer?

    Thanks, you are all great :)
    Bonaparte
     
  7. Apr 13, 2013 #6

    haruspex

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    It is only constant in regard to one of the original family of curves. A curve in the orthogonal family runs across these, so the set of intersections treats this as a parameter. Eliminating the constant, as HofI did, gets around this problem by extracting a general truth about the family as a whole.
     
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