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Math Help - one last application integral

  1. #1
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    one last application integral

    OK,skeeter, if you read this, I need your help one last time.

    A 5 lb bucket that is leaking is lifted from the ground into the air by a worker pulling 20 ft of rope at a constant speed. The rope weighs 0.08 lb/ft.
    The bucket starts with 2 gallons of liquid. Each gallon of liquid weighs 16 pounds. The liquid leaks at a constant rate. It finishes leaking (becomes empty) just as it reaches the top. Find the work done.


    Tell me what parts I have correct and correct me where I need correcting.


    The interval that is sliced is [0,20] because it is the distance the bucket travels?


    Since the weight of the rope changes as it is pulled up, we would have
    .08(20-yi)

    How do I also factor in the liquid in the bucket?
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  2. #2
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    Quote Originally Posted by gammaman View Post
    OK,skeeter, if you read this, I need your help one last time.

    A 5 lb bucket that is leaking is lifted from the ground into the air by a worker pulling 20 ft of rope at a constant speed. The rope weighs 0.08 lb/ft.
    The bucket starts with 2 gallons of liquid. Each gallon of liquid weighs 16 pounds. The liquid leaks at a constant rate. It finishes leaking (becomes empty) just as it reaches the top. Find the work done.


    Tell me what parts I have correct and correct me where I need correcting.



    The interval that is sliced is [0,20] because it is the distance the bucket travels?


    Since the weight of the rope changes as it is pulled up, we would have
    .08(20-yi)

    How do I also factor in the liquid in the bucket?
    So the total weight at the bottom is

    5+20(.08)+2(16)=38.6

    The weight at the top is 5

    This gives two ordered pairt (h,w) height and weight

    (0,38.6) and (20,5) this gives us the equation of the line( we can do this becuase they both change linearly or at a constant rate)

    w=-\frac{33.6}{20}h+38.6=-1.68h+38.6

    \int_{0}^{20}w dh=\int_{0}^{20}(-1.68h+38.6)dh=...
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  3. #3
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    Well actually according to my notes.

    16(-.1y+2).

    Can you tell me where this comes from?
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    Quote Originally Posted by gammaman View Post
    Well actually according to my notes.

    16(-.1y+2).

    Can you tell me where this comes from?

    This is only the weight of the liquid

    16 is the weight per pound

    at the bottom you have two pounds y=0 and at the top you have y=20 you have zero pounds.

    If you add this two the weight of the buck and rope you will get the equation that I gave you above

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  5. #5
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    Can I look at it this way.

    At the ground the bucket has 2 gallons of water therefore (0,2)
    When it reaches the top at height 20 the water is gone (20,0)

    So, 2-0/0-20 = -.1.

    So 16, the weight of each gallon and 2 the number of gallons we started with, becomes

    16(-.1y+2).

    So the final integral would be

    W=\int_{0}^{10}[5+.08(20-y)+16(-.1y+2)]
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    Quote Originally Posted by gammaman View Post
    Can I look at it this way.

    At the ground the bucket has 2 gallons of water therefore (0,2)
    When it reaches the top at height 20 the water is gone (20,0)

    So, 2-0/0-20 = -.1.

    So 16, the weight of each gallon and 2 the number of gallons we started with, becomes

    16(-.1y+2).

    So the final integral would be

    W=\int_{0}^{10}[5+.08(20-y)+16(-.1y+2)]

    Yes but I think you have a typo in the upper limit of integration it should be 20.

    also,(again) if you collect all of the like terms in your integrand you will end up with the same integral I gave you above.
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    Woops. My typo. The only thing I am unsure of is the +2 at the end. Is that really just the number of gallons we started with? Or is it found another way?

    I feel a little uncomfortable doing it the way you did, only because the what you did it feels more physics based. I am trying to look at it purely from a calculus perspective where W=Fi*Di, and a limit is used to find the work done.
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    Quote Originally Posted by gammaman View Post
    Woops. My typo. The only thing I am unsure of is the +2 at the end. Is that really just the number of gallons we started with? Or is it found another way?

    I feel a little uncomfortable doing it the way you did, only because the what you did it feels more physics based. I am trying to look at it purely from a calculus perspective where W=Fi*Di, and a limit is used to find the work done.

    The two and the end is the volume when y = 0 as y increased the term -.1y is how much fluid is lost per foot. so when y=20 you have no fluid left.
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  9. #9
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    Sorry to be such a bother, but would anyone know where I could practice more word problems like this? My Calculus book only has 2 questions.
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