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Thread: Accumulators

  1. #1
    Dec 2007


    Ok the question reads:
    A long lake is used as a reservoir to supply a hydro dam. The lake is approximately the shape of a rectangular prism with a length of 80 km, a width of 2km and a depth of 500m. On January 1st, the amound of water in the reservoir is 80 billion m^3.

    The flow of water into the lake is E(t) = 90/ 2 + cos((2pi/365) t)
    measured in m^3/day
    The flow of water out of the lake is L(t) = 146/ 3 + cos((2pi/365) t)
    measured in m^3/day

    What is the rate of depth increase on the 182nd day of the year? (in cm/hour)

    The answer in my text book says it's 0.4427 cm/hour

    Could you please give a step by step guide, Thanks a ton

    Last edited by the-G; Apr 6th 2008 at 11:51 AM.
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  2. #2
    MHF Contributor Mathstud28's Avatar
    Mar 2008

    Unless I misread it

    which i often do...this is jsut a derivative at a point problem...ok $\displaystyle f(t)=\frac{90}{2}+cos\bigg(\frac{2\pi{t}}{365}\big g)$ the derivative is $\displaystyle f'(t)=-sin\bigg(\frac{2\pi{t}}{35}\bigg)\cdot\frac{2\pi}{ 365}$ evaluate it at $\displaystyle 182$ and you get $\displaystyle f'(182)=-sin\bigg(\frac{2\pi}{182}{236}\bigg)\cdot\frac{2\p i}{365}$=-.000142....o I messed up... you have to add the two functions together and differntiate then plug in 182...use same concept as above...the answer should be -3.6666
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