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bii17 Group Title

A weight is attached to one end of a 33 foot rope which passes over a pulley 18 feet above the ground. the other end is attached to a truck at a point 3 feet above the ground. if the truck moves away at a rate of 2 feet per second, how fast is the weight rising when the truck is 8 feet from the spot directly under the pulley?

  • 2 years ago
  • 2 years ago

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  1. perl Group Title
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    |dw:1349004422200:dw|

    • 2 years ago
  2. goformit100 Group Title
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    |dw:1349004500723:dw|

    • 2 years ago
  3. bii17 Group Title
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    I need a solution :)

    • 2 years ago
  4. perl Group Title
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    which direction is the trick moving, it is not clear

    • 2 years ago
  5. sauravshakya Group Title
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    |dw:1349004602933:dw|

    • 2 years ago
  6. goformit100 Group Title
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    |dw:1349004739213:dw|

    • 2 years ago
  7. sauravshakya Group Title
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    |dw:1349004667017:dw|

    • 2 years ago
  8. sauravshakya Group Title
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    |dw:1349004778629:dw|

    • 2 years ago
  9. sauravshakya Group Title
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    dx/dt =2 feet per second y is constant..... now, can u find dh/dt

    • 2 years ago
  10. bii17 Group Title
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    how?

    • 2 years ago
  11. sauravshakya Group Title
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    |dw:1349004927462:dw|

    • 2 years ago
  12. sauravshakya Group Title
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    Since y is constant dy/dt=0 right?

    • 2 years ago
  13. bii17 Group Title
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    yes. then?

    • 2 years ago
  14. sauravshakya Group Title
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    |dw:1349005061647:dw|

    • 2 years ago
  15. sauravshakya Group Title
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    Now, given x=8

    • 2 years ago
  16. sauravshakya Group Title
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    y=18-3=15

    • 2 years ago
  17. sauravshakya Group Title
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    So, |dw:1349005186713:dw|

    • 2 years ago
  18. sauravshakya Group Title
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    right?

    • 2 years ago
  19. bii17 Group Title
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    @sauravshakya the answer is 16/ 17

    • 2 years ago
  20. goformit100 Group Title
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    Use of calculus is good

    • 2 years ago
  21. sauravshakya Group Title
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    |dw:1349005396208:dw|

    • 2 years ago
  22. sauravshakya Group Title
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    Now we need to find dl/dt

    • 2 years ago
  23. bii17 Group Title
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    how to fnd it @sauravshakya

    • 2 years ago
  24. perl Group Title
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    you need a better diagram

    • 2 years ago
  25. perl Group Title
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    A weight is attached to one end of a 33 foot rope which passes over a pulley 18 feet above the ground. the other end is attached to a truck at a point 3 feet above the ground. if the truck moves away at a rate of 2 feet per second, how fast is the weight rising when the truck is 8 feet from the spot directly under the pulley?

    • 2 years ago
  26. sauravshakya Group Title
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    |dw:1349005616181:dw||dw:1349005671316:dw|

    • 2 years ago
  27. sauravshakya Group Title
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    Did I did some algebra wrong or what???????

    • 2 years ago
  28. bii17 Group Title
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    hmm i dont know.. the answer should be 16/ 17

    • 2 years ago
  29. perl Group Title
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    |dw:1349005592570:dw|

    • 2 years ago
  30. perl Group Title
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    x is the distance the weight has been raised, and y is the horizontal distance, where the rope is attached, to the vertical line passing through the pulley.

    • 2 years ago
  31. goformit100 Group Title
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    its ok

    • 2 years ago
  32. sauravshakya Group Title
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    @bii17 u sure the answer is 16/17

    • 2 years ago
  33. goformit100 Group Title
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    its correct

    • 2 years ago
  34. bii17 Group Title
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    @sauravshakya yes im sure. that is the answer given in the book

    • 2 years ago
  35. sauravshakya Group Title
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    But I got 16/15 @goformit100

    • 2 years ago
  36. goformit100 Group Title
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    i did the calculation @sauravshakya is correct now

    • 2 years ago
  37. sauravshakya Group Title
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    @ganeshie8 if u r free PLZ see this

    • 2 years ago
  38. goformit100 Group Title
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    @Callisto check it

    • 2 years ago
  39. goformit100 Group Title
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    plz

    • 2 years ago
  40. sauravshakya Group Title
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    @hartnn can U plz check it

    • 2 years ago
  41. perl Group Title
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    oh i will do it again

    • 2 years ago
  42. perl Group Title
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    |dw:1349007585718:dw|

    • 2 years ago
  43. perl Group Title
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    15 ^2 + x^2 = (33- y)^2

    • 2 years ago
  44. perl Group Title
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    y is the height that the weight rises from the ground. , x is the distance the truck has moved to the left

    • 2 years ago
  45. hartnn Group Title
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    |dw:1349007886322:dw| @sauravshakya

    • 2 years ago
  46. perl Group Title
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    did you get 16/17?

    • 2 years ago
  47. sauravshakya Group Title
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    Oh ya....... @hartnn

    • 2 years ago
  48. hartnn Group Title
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    that still doesn't give answer.....

    • 2 years ago
  49. sauravshakya Group Title
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    Yep

    • 2 years ago
  50. perl Group Title
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    15 ^2 + x^2 = (33- y)^2 2x *dx/dt = 2 ( 33 - y) * (-dy/dt) x * dx/dt = (33-y) * (-dy/dt) x/ (33-y) * dx/dt = -dy/dt , this works

    • 2 years ago
  51. perl Group Title
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    8 / ( 33 - 16 ) * 2 = 16/17 , x = 8 , that is given, y = 16 by plugging in 15^2 + x^2 = (33-y)^2

    • 2 years ago
  52. perl Group Title
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    the only weird part is the negative sign

    • 2 years ago
  53. sauravshakya Group Title
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    @hartnn that gives the answer

    • 2 years ago
  54. perl Group Title
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    oh nevermind, it should be plus

    • 2 years ago
  55. perl Group Title
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    15^2 + x^2 = (33+y)^2

    • 2 years ago
  56. sauravshakya Group Title
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    |dw:1349008213306:dw|

    • 2 years ago
  57. perl Group Title
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    |dw:1349008265370:dw|

    • 2 years ago
  58. sauravshakya Group Title
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    So,|dw:1349008256578:dw|

    • 2 years ago
  59. hartnn Group Title
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    yes it does, u did it @sauravshakya ! :)

    • 2 years ago
  60. perl Group Title
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    yes that is simpler. the trick is to see that y is constant

    • 2 years ago
  61. perl Group Title
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    |dw:1349008430327:dw|

    • 2 years ago
  62. perl Group Title
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    y = 15,

    • 2 years ago
  63. perl Group Title
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    wait, dh/dt is the rate at which the rope is being pulled

    • 2 years ago
  64. perl Group Title
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    so dh/dt = the rate at which the weight is moving up ? interesting

    • 2 years ago
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