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Orthogonal Trajectories

  1. Nov 14, 2007 #1
    1. The problem statement, all variables and given/known data

    Find the orthogonal trajectories of the family of curves. Use a graphing device to draw several members of each family on a common screen.
    y = x/(1+kx)

    2. The attempt at a solution

    I have been trying this problem for hours, and I get a different answer every time. I have a big exam over this material tomorrow, and I just discovered that someone in this forum may be able to help me. Any help will be GREATLY appreciated!

    The following is my most recent attempt. This was after I got help from a friend. I know that it must be wrong, but it's as far as I have gotten.

    y = x/(1+kx)
    y' = dy/dx = x/[(x+kx)^2]
    slope of tangent line = x/[(x+kx)^2]
    slope of orthogonal line = -(1+kx)^2
    dy = [-(1+kx)^2]dx

    After that, I am completely stuck. I do not think I am anywhere close to the answer. Also, I have no idea how to go about graphing it. Please help!
     
    Last edited: Nov 14, 2007
  2. jcsd
  3. Nov 14, 2007 #2

    Dick

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    I'm ignoring what I think are typos, like the correct y' is 1/(1+kx)^2. Your final line looks good though. Now just integrate both sides. Use a substitution u=1+kx. Or just square it out if you have to.
     
  4. Nov 14, 2007 #3
    Do you really think I am on the right track? I thought I was supposed to take K out early on, but I do not know when, or what to replace it with. I just think that my final answer is not supposed to have K in it.
     
  5. Nov 14, 2007 #4

    Dick

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    Yes, I think you are on the right track. And I don't see why you think the final answer shouldn't have a k in it.
     
  6. Nov 14, 2007 #5
    My final answer is y=-[(1+KX)^3]/3K + C. I think that should be correct. To graph it, will I just graph the original equation y = x/(1+kx) and then add y=-[(1+KX)^3]/3K + C for different values of K and C?

    In my math book, it says that K is an arbitrary constant and that you have to do it for all values of K, so you find K as follows:
    y = x/(1+kx)
    1+kx = x/y
    k=(x-1)/y

    I just do not know when I would add it into the problem. Perhaps this is correct:
    y = x/(1+kx)
    y' = dy/dx = x/[(x+kx)^2]
    And replacing k with (x-1)/y...
    slope of tangent line = x/[(x+[(x-1)/y]x)^2] = (xy^2)/(x^4)

    Does that look right? I'm so confused...
     
  7. Nov 14, 2007 #6

    Dick

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    For someone who has done all the right stuff, you are mighty confused. Let f(x)=x/(1+kx) and g(x)=c+(1+kx)^3/(-3k). You have shown that f'(x)=-1/g'(x). So if you put in any values of c and k, then if f(x) and g(x) intersect, they will intersect at a right angle. They are orthogonal.
     
  8. Nov 14, 2007 #7
    LOL, I think I'm just getting stressed out because it's almost the end of the semester, and I want to be prepared for Calculus III next semester. Thank you so much for helping me; I will just go to bed and hope I pass the exam tomorrow.

    Thanks again!
    (:
     
  9. Nov 14, 2007 #8

    Dick

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    Good strategy, you probably will if you get some sleep. You are doing a lot of stuff right.
     
  10. Nov 15, 2007 #9

    HallsofIvy

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    Dick, the differential equation defining the family y= x/(1+kx) should NOT have a "k" in it. The k determines which member of the family you have. Once you have determined y'= f(x,y), write y'= -1/f(x,y) and integrate that to find the orthogonal trajectories. You will get a constant of integration in that, of course.

    I would be inclined to do it this way: write y= x/(1+kx) as y(1+ kx)= x. That allows you to do two things. First, differentiating both sides, y'(1+ kx)+ ky= 1. Also y+ kxy= x so kxy= x- y and you have both kx= (x-y)/y and ky= (x-y)/x. y'(1+ (x-y)/y)+ (x-y)/x= 1.
    1+ (x-y)/y= x/y and 1- (x-y)/x= y/x. The equation of the original family is (x/y)y'= (y/x) or y'= y2/x2.
     
  11. Nov 15, 2007 #10

    Dick

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    Thanks, Halls. Or just write (1+kx)=x/y and put that into the derivative the OP already has y'=1/(1+kx)=y^2/x^2. I had a feeling something was going awry...
     
  12. Nov 15, 2007 #11
    Wow, my answer was really wrong. I guess it does not matter since my exam has already passed, but thanks for the help, guys.
     
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