Suppose water is leaking from a tank through a circular hole of area $A_h$ at its bottom. When water leaks through a hole, friction and contraction of the stream near the hole reduce the volume of the water leaving the tank per second to $c A_h \sqrt{2gh}$, where $c$ (with $0 < c < 1$) is an empirical constant. Determine a differential equation for the height $h$ of water at time $t$ for the cubical tank in the figure below. The radius of the hole is 4 in., g = 32 ft/$s^2$

Neetesh Kumar | November 08, 2024

Differential Equation Homework Help

This is the solution to Math 2A, section 13Z, Fall 2023 | WebAssign
Math002ACh1Sec03 (Homework) Question - 5
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Step-by-Step-Solution:

Let $h(t)$ be the water height in the tank at time $t$. The volume of water in a cubical tank with a square base of side length $L$ and height $h$ is given by:

$V = L^2 h$

The area of the circular hole can be calculated as follows:

1. Convert the radius of the hole from inches to feet:

   $r_h = \frac{4 \text{ in}}{12} = \frac{1}{3} \text{ ft}$

2. Calculate the area $A_h$ of the hole:

   $A_h = \pi r_h^2 = \pi \left(\frac{1}{3}\right)^2 = \frac{\pi}{9} \text{ ft}^2$

Step 1: Determine the Rate of Change of Volume

The rate of water leaving the tank per second through the hole is given by:

$\frac{dV}{dt} = -c A_h \sqrt{2gh}$

Step 2: Relate Volume to Height

Since the volume of the tank is given by $V = L^2 h$, the rate of change of volume can also be expressed as:

$\frac{dV}{dt} = L^2 \frac{dh}{dt}$

Step 3: Set the Two Expressions for Volume Change Equal

Setting the two expressions for $\frac{dV}{dt}$ equal gives us:

$L^2 \frac{dh}{dt} = -c A_h \sqrt{2gh}$

Step 4: Substitute for $A_h$

Now substituting $A_h$ into the equation gives:

$L^2 \frac{dh}{dt} = -c \left( \frac{\pi}{9} \right) \sqrt{2gh}$

Step 5: Rearrange the Equation

Rearranging gives us the differential equation:

$\frac{dh}{dt} = -\frac{c \pi}{9 L^2} \sqrt{2gh}$

Conclusion

The differential equation governing the height $h(t)$ of water in the tank at time $t$ is:

$\frac{dh}{dt} = -k \sqrt{h}$

where

$k = \frac{c \pi \sqrt{2g}}{9 L^2}$

Here, $g = 32 \text{ ft/s}^2$.

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