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Thread: Chk Anwser - Draining tank of water

  1. August 28th 2006, 02:30 PM #1
    Yogi_Bear_79
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    Chk Anwser - Draining tank of water

    Q. Suppose that a tank initially contains 2000 gal of water and the rate of change of its volume after the tank drains for t min is `(t) = (0.5)t -30 (in gallons per minute). How much water does the tank contain after it has been draining for 25 minutes?


    A. y(t) = .25t^2 -30t +c.
    C = 2000. Substitute t = 25.
    We get y(25) = .25(25)^2 -30t + 2000 = 1406.25 Gal remaining.
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  2. August 28th 2006, 02:41 PM #2
    topsquark
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    Quote Originally Posted by Yogi_Bear_79
    Q. Suppose that a tank initially contains 2000 gal of water and the rate of change of its volume after the tank drains for t min is `(t) = (0.5)t -30 (in gallons per minute). How much water does the tank contain after it has been draining for 25 minutes?


    A. y(t) = .25t^2 -30t +c.
    C = 2000. Substitute t = 25.
    We get y(25) = .25(25)^2 -30t + 2000 = 1406.25 Gal remaining.
    It looks good to me.

    -Dan
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  3. August 28th 2006, 02:42 PM #3
    ThePerfectHacker
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    Quote Originally Posted by Yogi_Bear_79
    Q. Suppose that a tank initially contains 2000 gal of water and the rate of change of its volume after the tank drains for t min is `(t) = (0.5)t -30 (in gallons per minute). How much water does the tank contain after it has been draining for 25 minutes?


    A. y(t) = .25t^2 -30t +c.
    C = 2000. Substitute t = 25.
    We get y(25) = .25(25)^2 -30t + 2000 = 1406.25 Gal remaining.
    I got the same answer. The only mistake is that,
    y(25)=.25(25)^2-30(25)+2000 but you just probably forgot to write that.
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  4. August 28th 2006, 03:00 PM #4
    Yogi_Bear_79
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    Thanks, this isn't easy for me, so the confirmations and examples this forumn provides are priceless!
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