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Length of diagonal rope

  1. Sep 10, 2005 #1
    Hey,
    I need some suggestions to approach the problem:

    A lifeguard standing on a tower throws a buoy to a swinner 5m from the tower. The lifeguard, positioned 3 m above the water, pulls in the rope at a speed of 1m/s.
    How fast is the swinner coming to the shore when he is at 4m from the water edge?

    thanks
     
  2. jcsd
  3. Sep 10, 2005 #2

    quasar987

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    Very cute calc prob.

    If David Haselhoff is pulling the rope at 1m/s, this means that the rate of change of the distance btw the Hasellhoff and the swimmer (lenght of the rope btw swimmer and the Hasellhoff's hairy palms) is of -1m/s. How is the distance btw the Hasellhoff and the swimmer related to the distance btw the swimmer and the tower? hint: old crazy greek cult leader.
     
  4. Sep 10, 2005 #3

    Crazy hint :smile:

    we have a right trangle-swimmer- bottom of tower - Haselhof. (pythagoras )

    but I have still no idea how to solve it since we have two position 5m and 4 m.
     
  5. Sep 10, 2005 #4

    quasar987

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    Say L is the distance Haselhof-swimmer, and D is the distance swimmer-tower. You got

    [tex]L = \sqrt{9+D^2}[/tex]

    And what are you looking for here? You're looking for

    [tex]\frac{dD}{dt}|_{D=4}[/tex]
     
  6. Sep 10, 2005 #5
    Sorry but I do not understand
    [tex]\frac{dD}{dt}|_{D=4}[/tex]

    the derivative of D when D is 4 should not give us 0??
     
  7. Sep 10, 2005 #6

    quasar987

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    This is a little strange to me too... But here's what I think.

    It's like when we have [itex]f(x) = e^{ax}[/itex]. Then [itex]df/dx = ae^{ax} = af(x)[/itex]. df/dx is a function of f and x. Actually it is only a function of x since f is a function of x. But we can still write [itex]df/dx|_{f=c}[/itex] to have is mean [itex]df/dx|_{x^* \ \mbox{such that f(x^*)=c}}[/itex]. And notice that [itex]df/dx|_{f=c} = ac \neq 0[/itex].

    Here we have that

    [tex]\frac{dL}{dt} = \frac{dL}{dD}\frac{dD}{dt} \Leftrightarrow \frac{dD}{dt} = \frac{-1}{\frac{dL}{dD}}[/tex] (since dL/dt = -1)

    So dD/dt is a function of D in the same sense as df/dx above is a function of f. Hence is makes sense to write

    [tex]\frac{dD}{dt}|_{D=4}[/tex]

    SO, to sum up: calculus is a Bitc... That'll be all for today... :zzz:
     
    Last edited: Sep 10, 2005
  8. Sep 10, 2005 #7

    Doc Al

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    The easiest way to solve this is to define the variables L and D just like quasar987 did. Their relationship is:
    [tex]L^2 = 9 + D^2[/tex]

    Now just take the derivative of both sides with respect to time. You'll get an easy expression in terms of L, D, dL/dt, and dD/dt. The only unknown is dD/dt, which is what you want to find. Try it.
     
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