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I need a solution :)
which direction is the trick moving, it is not clear
dx/dt =2 feet per second y is constant..... now, can u find dh/dt
Since y is constant dy/dt=0 right?
Now, given x=8
@sauravshakya the answer is 16/ 17
Use of calculus is good
Now we need to find dl/dt
how to fnd it @sauravshakya
you need a better diagram
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?
Did I did some algebra wrong or what???????
hmm i dont know.. the answer should be 16/ 17
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.
@bii17 u sure the answer is 16/17
@sauravshakya yes im sure. that is the answer given in the book
But I got 16/15 @goformit100
i did the calculation @sauravshakya is correct now
@ganeshie8 if u r free PLZ see this
@Callisto check it
@hartnn can U plz check it
oh i will do it again
15 ^2 + x^2 = (33- y)^2
y is the height that the weight rises from the ground. , x is the distance the truck has moved to the left
did you get 16/17?
Oh ya....... @hartnn
that still doesn't give answer.....
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
8 / ( 33 - 16 ) * 2 = 16/17 , x = 8 , that is given, y = 16 by plugging in 15^2 + x^2 = (33-y)^2
the only weird part is the negative sign
@hartnn that gives the answer
oh nevermind, it should be plus
15^2 + x^2 = (33+y)^2
yes it does, u did it @sauravshakya ! :)
yes that is simpler. the trick is to see that y is constant
y = 15,
wait, dh/dt is the rate at which the rope is being pulled
so dh/dt = the rate at which the weight is moving up ? interesting