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Related Rates calculation

  1. May 8, 2007 #1
    1) (Related Rates) One end of a rope 20 meters long is attached to a box resting on the floor. The other end is passed over a pulley directly above the box, 5 meters above the floor, and attached to the back of a truck at a point 1 meters above the ground. The truck then drives in a straight line away from the pulley at a speed of 0.5 m/s. At what speed is the box rising when the top of the box is 2 meters above the ground?


    I really don't get this problem. I can't even start doing the calcultions because I am not sure how to set up the variables in the problem...can someone please give me some guidelines/hints?

    Thanks a lot!
     
  2. jcsd
  3. May 8, 2007 #2
    It's a triangle: (not to scale)
    Code (Text):
    Pulley
     |\
    4| \15
     |__\Truck
    1| X
    Box
    With dx/dt=0.5
     
  4. May 8, 2007 #3
    But this is a "related rates" problem, so I believe that there should be more than 1 variable....but I can't figure out where and how to put in the second (or third) variable...
     
  5. May 8, 2007 #4
    Let y be the distance from pulley to the angle opposite 15. Then you have x^2 + y^2 = 225, 2xdx + 2ydy = 0. Then plug in? Will that work?
     
  6. May 8, 2007 #5
    But it says "At what speed is the box rising when the top of the box is 2 meters above the ground?", does it matter that it's 2 meters above? It is higher than the baseline of the triangle...
     
  7. May 8, 2007 #6
    Solve for dy/dt
     
  8. May 9, 2007 #7
    I am quite lost...

    If I let y to be the distance from pulley to the angle opposite 15, the box would be somewhere within the line, not at the bottom end of the line...

    When the box is 2 meters above, it is higher than the baseline of the triangle...how can I use Pythagoras when the triangle is not fixed? This is the part that I really don't get...

    Can anyone please help me? Thanks!
     
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