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Constructing Basic Differential Equation

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    Here's the question;

    Fluid flows out of a cylindrical tank with constant cross section. At time t minutes, t>0, the volume of fluid remaining in the tank is V m3. The rate at which the fluid flows in m3 min-1 is proportional to the square root of V.
    Show that the depth h metres of fluid in the tank satisfies the differential equation:


    \dfrac{dh}{dt}=-kh^{\frac{1}{2}}

    Could someone please point me in the right direction with this. Thanks.
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    I suggest you try using

    \frac{dV}{dt}=\frac{dV}{dh} \times \frac{dh}{dt}

    You'll need the formula for the volume of a cylinder, remember that the cross-sectional area is constant.

    The volume of fluid in the cylinder, V, as well as the height of fluid in the cylinder, h, will decrease as time passes which is why the negative is present.

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    (Original post by JeremyB)
    Here's the question;

    Fluid flows out of a cylindrical tank with constant cross section. At time t minutes, t>0, the volume of fluid remaining in the tank is V m3. The rate at which the fluid flows in m3 min-1 is proportional to the square root of V.
    Show that the depth h metres of fluid in the tank satisfies the differential equation:


    \dfrac{dh}{dt}=-kh^{\frac{1}{2}}

    Could someone please point me in the right direction with this. Thanks.
    Havent read the whole question but if you ask me to solve the differential equation given her is what I do
    1)multiply both sides by dt
    2)divide both sides by h1/2
    3)leave k as where it is
    then just integrate both sides and remember to put +C on the right hand side

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