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Related Rates, formulating the problem

  1. Jan 6, 2009 #1
    1. The problem statement, all variables and given/known data
    A wight W is attached to a rope 50 ft long that passes over a pulley at point P, 20 ft above the ground. The other end of the rope is attached to a truck at a point A, 2 ft above the round. If th truck moves away at the rate of 9 ft/s, how fast is the weight rising when it is 6 ft above the ground.




    3. The attempt at a solution

    I got the solution and it says:
    y2 = (30 + x)2 - (18)2

    and I don't understand why it is not

    y2 = (30+x)2 - (20-x)2

    since the rope is 50 ft long
     
  2. jcsd
  3. Jan 7, 2009 #2

    HallsofIvy

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    Actually, neither of those is a "solution" to the problem stated because neither is a "rate" in ft per second. Also no equation in x and y will make sense until you say what "x" and "y" represent. I assume you mean that "y" is the height of the object above the ground but then "x" cannot the horizontal distance to the truck because you appear to be using the Pythagorean theorem but x and y do not form a right triangle.

    If "x" is in fact, the horizontal distance of the truck from the point directly under the object being raised, then the rope, from the pulley to the truck, is the hypotenuse of a right triangle with one leg of length x and the other leg of length 20: that does NOT depend on the height of the lifted object. By the Pythagorean theorem, the length of that hypotenuse is [itex]\sqrt{x^2+ 400}[/itex]. If the length of the rope is 50 feet and y is the height of the lifted object, [itex]20- y+ \sqrt{x^2+ 400}= 50[/itex] so [itex]y= 20+ \sqrt{x^2+ 400}[/itex]
     
  4. Jan 7, 2009 #3
    I am sorry, x is the distance the weight has been raised, and y is the horizontal distance from point A, where the rope is attached, to the vertical line passing through the pulley.

    From the textbook solution

    y2 = (30 + x)2 - (18)2

    dy/dt = [tex]\frac{30 + x}{y}[/tex][tex]\frac{dx}{dt}[/tex]

    x = 6
    y = 18[tex]\sqrt{3}[/tex]
    dy/dt = 9

    then [tex]\frac{dx}{dt}[/tex] = [tex]\frac{9}{2}[/tex][tex]\sqrt{3}[/tex] ft/sec

    Edit: Solved,
    "x is the distance the weight has been raised, and y is the horizontal distance from point A, where the rope is attached, to the vertical line passing through the pulley.
    "
     
    Last edited: Jan 8, 2009
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