bii17
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?
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perl
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|dw:1349004422200:dw|
goformit100
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|dw:1349004500723:dw|
bii17
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I need a solution :)
perl
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which direction is the trick moving, it is not clear
sauravshakya
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|dw:1349004602933:dw|
goformit100
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|dw:1349004739213:dw|
sauravshakya
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|dw:1349004667017:dw|
sauravshakya
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|dw:1349004778629:dw|
sauravshakya
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dx/dt =2 feet per second
y is constant..... now, can u find dh/dt
bii17
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how?
sauravshakya
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|dw:1349004927462:dw|
sauravshakya
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Since y is constant dy/dt=0 right?
bii17
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yes. then?
sauravshakya
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|dw:1349005061647:dw|
sauravshakya
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Now, given x=8
sauravshakya
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y=18-3=15
sauravshakya
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So, |dw:1349005186713:dw|
sauravshakya
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right?
bii17
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@sauravshakya the answer is 16/ 17
goformit100
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Use of calculus is good
sauravshakya
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|dw:1349005396208:dw|
sauravshakya
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Now we need to find dl/dt
bii17
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how to fnd it @sauravshakya
perl
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you need a better diagram
perl
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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?
sauravshakya
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|dw:1349005616181:dw||dw:1349005671316:dw|
sauravshakya
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Did I did some algebra wrong or what???????
bii17
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hmm i dont know.. the answer should be 16/ 17
perl
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|dw:1349005592570:dw|
perl
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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.
goformit100
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its ok
sauravshakya
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@bii17 u sure the answer is 16/17
goformit100
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its correct
bii17
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@sauravshakya yes im sure. that is the answer given in the book
sauravshakya
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But I got 16/15 @goformit100
goformit100
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i did the calculation @sauravshakya is correct now
sauravshakya
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@ganeshie8 if u r free PLZ see this
goformit100
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@Callisto check it
goformit100
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plz
sauravshakya
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@hartnn can U plz check it
perl
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oh i will do it again
perl
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|dw:1349007585718:dw|
perl
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15 ^2 + x^2 = (33- y)^2
perl
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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
hartnn
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|dw:1349007886322:dw|
@sauravshakya
perl
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did you get 16/17?
sauravshakya
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Oh ya....... @hartnn
hartnn
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that still doesn't give answer.....
sauravshakya
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Yep
perl
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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
perl
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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
perl
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the only weird part is the negative sign
sauravshakya
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@hartnn that gives the answer
perl
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oh nevermind, it should be plus
perl
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15^2 + x^2 = (33+y)^2
sauravshakya
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|dw:1349008213306:dw|
perl
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|dw:1349008265370:dw|
sauravshakya
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So,|dw:1349008256578:dw|
hartnn
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yes it does, u did it @sauravshakya ! :)
perl
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yes that is simpler. the trick is to see that y is constant
perl
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|dw:1349008430327:dw|
perl
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y = 15,
perl
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wait, dh/dt is the rate at which the rope is being pulled
perl
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so dh/dt = the rate at which the weight is moving up ? interesting