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Math Help - Rate problem

  1. #1
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    Rate problem

    1. A large tank of canola oil ruptures at a time
    t = 0 minutes, and the oil leaks out at a rate of r(t) = 100e^-0.01(t) liters per minute. How much oil leaks out within the first hour?

    I am assuming that you would set u = 100, and dv = e^-0.01(t), so then du = 0 and v = e^t/(-0.01)...right? But I have no idea how to go from there..someone please help?

    Last edited by mr fantastic; March 5th 2010 at 04:56 PM. Reason: Changed post title
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  2. #2
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    Quote Originally Posted by Jgirl689 View Post
    1. A large tank of canola oil ruptures at a time
    t = 0 minutes, and the oil leaks out at a rate of r(t) = 100e^-0.01(t) liters per minute. How much oil leaks out within the first hour?

    I am assuming that you would set u = 100, and dv = e^-0.01(t), so then du = 0 and v = e^t/(-0.01)...right? But I have no idea how to go from there..someone please help?

    liters of leaked oil in the first hour = \int_0^{60} r(t) \, dt
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  3. #3
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    You don't have to use integration by parts to integrate this function. I would recommend reviewing the rules when integrating e^nx. If it still doesn't make sense, come back please
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  4. #4
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    \int_a^b e^{kt} \, dt = \frac{1}{k}\left[e^{kt} \right]_a^b
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  5. #5
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    Quote Originally Posted by Jgirl689 View Post
    1. A large tank of canola oil ruptures at a time
    t = 0 minutes, and the oil leaks out at a rate of r(t) = 100e^-0.01(t) liters per minute. How much oil leaks out within the first hour?

    I am assuming that you would set u = 100, and dv = e^-0.01(t), so then du = 0 and v = e^t/(-0.01)...right? But I have no idea how to go from there..someone please help?

    With dv= e^{-0.01t}, v= -\frac{e^{-0.01t}}{0.01}= -100e^{-0.01t}.

    And, if you are using integration by parts with du=0,

    \int udv= uv- \int vdu= uv.

    In other words, with u a constant, the integral of udv is just uv! The integral of "a constant time f(x)" is that constant times the integral of f(x). I'll bet you learned that before you learned integration by parts!
    Last edited by HallsofIvy; March 8th 2010 at 04:03 AM.
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  6. #6
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    Ok, I came with an approach, I did the integral part:

    u = 100, du = 0, dv = e^(-.01t), v = e^(.01t)/ .01 making the integral of vdu = 0 (because du = 0), so integral of udv = uv = 100 * (e^(.01t)/.01)

    then I just substituted t with 0, and 60 and substracted, since that is the integral....

    100(e^(-0.01*0)/-.01 = 100(e^0)/-.01 = 100/-.01 = -10000
    100(e^-0.01*60)/-.01 = 100(e^-.6)/-.01 = 54.88/-0.01 = -5488

    -5488 - (-10000) = -5488 + 10000 = 4511 liter per minute.. when it is done with a calculator..or

    -10000 e^(-. 6) + 10000 liters per minute

    Is that correct?
    Last edited by Jgirl689; March 7th 2010 at 03:39 PM.
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  7. #7
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    integrating liters per minute w/r to minutes yields liters ...

    \int_0^{60} 100 e^{-.01t} \, dt \approx 4512 liters
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