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A man, a boat, and the tractrix formed

  1. Mar 3, 2004 #1
    I really need help solving question #2 from the following webpage: Applied Homework 2. I've attempted to solve this, but looking at my answers, I'm not overly confident. In part a, I'm suppose to prove that tractrix somehow, and in part b, I believe I am to just integrate that given equation...

    But I'm not 100% sure (especially with part a)...


    I'd be eternally grateful to any help given. :smile:
     
  2. jcsd
  3. Mar 3, 2004 #2

    HallsofIvy

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    "Applied Homework"? Now that's a peculiar name. The only good reason for doing homework is to learn something that you can later apply!

    In any case, you have a problem about a "tractrix", the curve that has the property that "the rope is always tangent to the curve".

    Okay, call the curve "y= f(x)". The man walks along the "y-axis" (I get that from the picture) so let's designate his postion at any time by (0, Y). The boat is at (x,y)= (x,f(x)) and the distance between him and the boat is &sqrt((0-x)2+ (Y-f(x))2)= L. The slope of the line from the boat to the man is (Y-f(x))/(0-x)= f(x)/x- Y/x and, by definition of a "tractrix" that is the tangent of the curve: df/dx= f(x)x- Y/x. "Y" varies from moment to moment but we can solve the "L" equation for Y: x2+(Y-f(x))2= L2 so Y= f(x)+ L2- x2.

    From that you should be able to get the equation in part (a).
    That looks to me like a straight forward first order, separable differential equation. Try to solve it and if you have trouble with it come back here.
     
  4. Mar 5, 2004 #3
    Thank you for the help, HallsofIvy.

    I was able to follow just about everything you mentioned. I understand that solving for Y would yield Y= f(x)+ sqrt(L2- x2), and then when that Y value is placed into the equation of the slope f(x)/x- Y/x and simplify, I will get that original equation.

    There is only one thing I am confused about-- When you said "by definition of a "tractrix" that is the tangent of the curve: df/dx= f(x)x- Y/x," were you referring to df/dx= dY/dx= f'(x)? I just want to make sure I understand perfectly. Thanks.
    :smile:
     
  5. Mar 5, 2004 #4

    HallsofIvy

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