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Find the depth of water in a leaky tank as time goes to infinity

Consider a water tank with vertical sides and with cross-section a square of area 4 feet. Water exits the tank through a hole which as area equal to \frac{5}{3} square inches, and water is added to the tank at a rate of 100 cubic inches per second. Show that the water level approaches the value of \left( \frac{25}{24} \right)^2 feet above the hole no matter what the initial water level was.


Incomplete.

One comment

  1. Evangelos says:

    Can’t say I didn’t try on this one, but it’s got me stumped…

    We know from the given information that the rate of change of the volume of the water in the tank can be written as:

        \begin{align*} \frac{dV}{dt} &= A(y) \frac{dy}{dt} + \frac{100}{1728} \frac{ft^3}{s} \\ &= -4.8 A_{0} \sqrt{y} + \frac{100}{1728} \frac{ft^3}{s} \\ &= \frac{100 - 96 \sqrt{y}}{1728} \frac{ft^3}{s} \\ \end{align*}

    Dividing by the cross-sectional area A(y) = 4 ft^2 gives us the change in y over t

        \begin{align*} \frac{dy}{dt} &= \frac{25 - 24 \sqrt{y}}{1728} \frac{ft}{s} \\ \end{align*}

    Which is a separable equation. Rearranging terms and integrating gives us

        \begin{align*} \int \frac{dy}{25 - 24 \sqrt{y}} &= \frac{t}{1728} + C \end{align*}

    And this is where I got stuck. I guess if you send t to +infinity, the only y that would satisfy the equation would be a number that sends 25 – 24 sqrt(y) to 0, which is

        \begin{align*} y = (\frac{25}{24})^2 \end{align*}

    Not sure if that holds though…

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