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Homework Help: Rate of change of a right triangle

  1. Sep 11, 2010 #1
    Imagine a boat on the water being putted
    towards the shore by a winch mounted a height
    h above the water. The winch reels in cable
    (shortening the hypotenuse) at constant rate
    u in m/s . When the boat is a distance d from
    the shore, find its speed and acceleration.
    This is an example of motion at non-uniform
    acceleration.




    im letting hypotenuse here = x, so x^2 = d^2 + h^2



    Im thinking i want to solve for d' to get the velocity at of d at distance d. okay im going to give it a shot, but bear with me, i have little confidence in this answer and i understand that reading these next steps will probably be a little frightening for those who know how to solve this.
    -------------------------------

    x^2 = d^2 + h^2

    d^2 = x^2 - h^2

    2dd' = 2xx' - 2hh'

    d' = [2xx' - 2hh']/(2d) <--- this is what i was thinking to explain the speed of d at distance d . my intuition was the to allow x' be u to explain the constant rate of the shortening of the hypotenuse, but im not sure about it. i want to differentiate again to find d'' to explain acceleration, but i need to establish the correct way to do this first step before moving on.

    can anyone give me a pointer or two.
     
    Last edited: Sep 11, 2010
  2. jcsd
  3. Sep 11, 2010 #2

    rl.bhat

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    Whether the height h is variable or constant?
     
  4. Sep 11, 2010 #3
  5. Sep 12, 2010 #4

    rl.bhat

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  6. Sep 12, 2010 #5
    okay, i understand how you got 2, you multiplied both sides by d then differentiated right?

    okay so now i just simplify..

    (x'*x' - d'*d')/d = d"

    x' here represent the constant rate, u m/s, that the hypotenuse is shortening.
     
    Last edited: Sep 12, 2010
  7. Sep 12, 2010 #6

    rl.bhat

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  8. Sep 12, 2010 #7

    thank you greatly
     
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