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Thread: Minimum volume of empty space in container with water

  1. #1
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    Minimum volume of empty space in container with water

    Volume of a container is $\frac{4 \pi}{3}$. Water can flow in and out of container.The volume of water in container is given by: $g(t), 0 \leq t \leq 4$, where $t$ is time in hours and $g(t)$ is measured in $m^3$. The rate of change of the volume of water in the container is: $g'(t)= 0.9 - 2.5 cos(0.4 t^2)$.
    a) The volume of water is increasing when $p < t < q$. Find $p$ and $q$. During that interval the volume of water increases by k $ m^3$. Find k.

    When $t =0$, the volume of water in the container is $2.3 m^3$. The container is never completely full of water during 4 hour period.
    b) Find the minimum volume of empty space in the container during the 4 hour period.

    I did part a) and found that p= 1.733874 and q = 3.56393, so volume increased by $ g(q) - g(p)= \int_{p}^{q} g'(t) dt = 3.7454$ $m^3$ during that period.

    I am not sure how to do part b). I know that I have to find maximum of g(t). I couldn't find integral of g'(t), because Fresnel's integral appears.

    I found that integral of g'(t) from t=0 to t=1.733874 is - 2.187, what means that volume of water decreased, so volume is 2.3 - 2.187 = 0.113. Then, from t=1.733874 to t=3.56393 volume increases by 3.7454 so it will be 0.113+3.7454=3.8584. Finally, integral from t=3.56393 to t=4 is - 0.4678, so volume is 3.8584 - 0.4678= 3.3906. Maximum volume is 3.8584, and therefore minimum empty space is 0.33.

    Could someone check whether this is correct? Thanks in advance.
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  2. #2
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    Re: Minimum volume of empty space in container with water

    I get a min empty space of 0.33m^3 occurring at t =3.56394 as well.
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