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 square inches. The water level is initially 2 feet above the level of the hole. Find the time required for the water level to drop 1 foot.
Tom gave us a nice little formula to use for these leaky tank problems, as long as we get our setup down right, it’s simply a matter of evaluating the integrals.
Or as I like to set it up
With y being the height of the water, A(y) being the cross-sectional area of the tank at height y, and A_0 being the area of the hole in the bottom of the tank. We are given the areas A(y) and A_0, along with the values of y_0, y_1, and t_0 as follows
And the problem asks us to find the value of t_1 that satisfies the equation.
But, before we set up the integral, we need to do a unit conversion to match the unit of area on both sides. As such, I changed the units of A_0 from square inches to square feet
Now, to set up the integrals
Some rearranging and simplifying gives us
Evaluating the integrals on both sides gives us
Plugging these numbers into a calculator gives us