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Math Help - Related Rates - Area & Volume

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    Related Rates - Area & Volume

    Ok I have done this problem several times over and my computer is telling me that all of my answers are wrong . Please help!


    We assume that an oil spill is being cleaned up by deploying bacteria that consume the oil at cubic feet per hour. The oil spill itself is modeled in the form of a very thin cylinder as depicted below. Its height is the thickness of the oil slick. Suppose at some moment in time the height is decreasing at foot per hour, the thickness of the slick is foot, and the cylinder is feet in diameter. At what rate is the area covered by the slick changing at that moment? (That is, the area of the base disc of the cylinder).

    I have been finding dr/dt by taking the derivative wrt time of v+pir^2*h and then using dr/dt to find dA/dt by deriving A=pi*r^2. Please Help!
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by rawkstar View Post
    Ok I have done this problem several times over and my computer is telling me that all of my answers are wrong . Please help!


    We assume that an oil spill is being cleaned up by deploying bacteria that consume the oil at cubic feet per hour. The oil spill itself is modeled in the form of a very thin cylinder as depicted below. Its height is the thickness of the oil slick. Suppose at some moment in time the height is decreasing at foot per hour, the thickness of the slick is foot, and the cylinder is feet in diameter. At what rate is the area covered by the slick changing at that moment? (That is, the area of the base disc of the cylinder).

    I have been finding dr/dt by taking the derivative wrt time of v+pir^2*h and then using dr/dt to find dA/dt by deriving A=pi*r^2. Please Help!
    maybe you should post your steps. you're describing the right way to tackle this problem. check your units, and make sure you make the rates negative or positive as appropriate.
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    ok so it is given that dV/dt=-5 and that at a certain point d=300 (so r=150), h=0.001, and dh/dt=-0.0002
    we know that V=pi*r^2*h so dV/dt=2pi*r*h*dr/dt+pi*r^2*dh/dt
    I plugged in dV/dt, h, and dh/dt and got -5=2pi*0.001*r*dr/dt+pi*r^2*-0.0002
    since D=300 r=150
    -5=300pi*0.001*dr/dt+150^2*pi*-0.0002 so dr/dt=(-5+4.5*pi)/(.3pi)
    i also know that A=pi*r^2 so dA/dt=2*pi*r*dr/dt
    so i plugged in dr/dt and 150 for r and got
    dA/dt=300*pi*[(-5+4.5*pi)/(0.3*pi)]

    OK i tried it again and got the right answer using this
    thank you
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