## 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

1. perl

|dw:1349004422200:dw|

2. goformit100

|dw:1349004500723:dw|

3. bii17

I need a solution :)

4. perl

which direction is the trick moving, it is not clear

5. sauravshakya

|dw:1349004602933:dw|

6. goformit100

|dw:1349004739213:dw|

7. sauravshakya

|dw:1349004667017:dw|

8. sauravshakya

|dw:1349004778629:dw|

9. sauravshakya

dx/dt =2 feet per second y is constant..... now, can u find dh/dt

10. bii17

how?

11. sauravshakya

|dw:1349004927462:dw|

12. sauravshakya

Since y is constant dy/dt=0 right?

13. bii17

yes. then?

14. sauravshakya

|dw:1349005061647:dw|

15. sauravshakya

Now, given x=8

16. sauravshakya

y=18-3=15

17. sauravshakya

So, |dw:1349005186713:dw|

18. sauravshakya

right?

19. bii17

@sauravshakya the answer is 16/ 17

20. goformit100

Use of calculus is good

21. sauravshakya

|dw:1349005396208:dw|

22. sauravshakya

Now we need to find dl/dt

23. bii17

how to fnd it @sauravshakya

24. perl

you need a better diagram

25. perl

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?

26. sauravshakya

|dw:1349005616181:dw||dw:1349005671316:dw|

27. sauravshakya

Did I did some algebra wrong or what???????

28. bii17

hmm i dont know.. the answer should be 16/ 17

29. perl

|dw:1349005592570:dw|

30. perl

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.

31. goformit100

its ok

32. sauravshakya

@bii17 u sure the answer is 16/17

33. goformit100

its correct

34. bii17

@sauravshakya yes im sure. that is the answer given in the book

35. sauravshakya

But I got 16/15 @goformit100

36. goformit100

i did the calculation @sauravshakya is correct now

37. sauravshakya

@ganeshie8 if u r free PLZ see this

38. goformit100

@Callisto check it

39. goformit100

plz

40. sauravshakya

@hartnn can U plz check it

41. perl

oh i will do it again

42. perl

|dw:1349007585718:dw|

43. perl

15 ^2 + x^2 = (33- y)^2

44. perl

y is the height that the weight rises from the ground. , x is the distance the truck has moved to the left

45. hartnn

|dw:1349007886322:dw| @sauravshakya

46. perl

did you get 16/17?

47. sauravshakya

Oh ya....... @hartnn

48. hartnn

49. sauravshakya

Yep

50. perl

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

51. perl

8 / ( 33 - 16 ) * 2 = 16/17 , x = 8 , that is given, y = 16 by plugging in 15^2 + x^2 = (33-y)^2

52. perl

the only weird part is the negative sign

53. sauravshakya

54. perl

oh nevermind, it should be plus

55. perl

15^2 + x^2 = (33+y)^2

56. sauravshakya

|dw:1349008213306:dw|

57. perl

|dw:1349008265370:dw|

58. sauravshakya

So,|dw:1349008256578:dw|

59. hartnn

yes it does, u did it @sauravshakya ! :)

60. perl

yes that is simpler. the trick is to see that y is constant

61. perl

|dw:1349008430327:dw|

62. perl

y = 15,

63. perl

wait, dh/dt is the rate at which the rope is being pulled

64. perl

so dh/dt = the rate at which the weight is moving up ? interesting