A tank, with a rectangular base and having vertical sides of height 4 m, is initially full of water. The water leaks out of a small hole in the horizontal base of the tank at a rate which, at any instant, is proportional to the square root of the depth of the water at that instant.
If x is the depth of water at time t after the leak started, write down a differential equation connecting x and t. If the tank is exactly half empty after one hour, find the further time that elapses before the tank becomes completely empty.
To find a differential equation connecting x and t, we need to consider the rate at which the water is leaking out of the tank at any given time. We are given that the rate of leakage at any instant is proportional to the square root of the depth of the water.
Let's assume that k is the constant of proportionality. Therefore, the rate of leakage can be expressed as k√(x), where x is the depth of the water at time t.
Now, the volume of water in the tank can be calculated as the product of the base area and the depth, which in this case is a rectangle. Since the base area is constant, we only need to consider the depth x.
The rate of change of the volume with respect to time is given by dV/dt = -k√(x), where the negative sign indicates a decrease in volume with time.
We can relate dV/dt and dx/dt by the chain rule. dx/dt represents the rate at which the depth is changing with respect to time.
Since the volume V = base area × depth, we have dV = base area × dx.
Therefore, dV/dt = (base area) × (dx/dt).
Substituting the values, we get (base area) × (dx/dt) = -k√(x).
Dividing both sides by the base area, we obtain dx/dt = -(k/base area) √(x).
Since the base area is a constant, let's represent k/base area as a new constant α.
Therefore, the differential equation connecting x and t is dx/dt = -α √(x).
Now, let's solve this differential equation to find the further time that elapses before the tank becomes completely empty.
We are given that the tank is exactly half empty after one hour, which means x = (1/2) × (base area) × height = (1/2) × (base area) × 4.
To find the solution, let's separate the variables and integrate.
∫ (1/√x) dx = -α ∫ dt.
Integrating the left side gives 2√x = -αt + c1, where c1 is the constant of integration.
Substituting x and t as (1/2) × (base area) × 4 and 1 hour, respectively, we get 2√((1/2) × (base area) × 4) = -α × 1 hour + c1.
Simplifying, we get √((base area) × 4) = -(α/2) + c1.
Since the tank is initially full, we know that x = (base area) × height = (base area) × 4, which gives √((base area) × 4) = 2√((base area)).
Therefore, we have 2√((base area)) = -(α/2) + c1.
We are also given that the tank is completely empty at some further time, let's say t = T.
At t = T, x = 0, which implies 2√((base area)) = -(α/2)T + c1.
Since the tank is completely empty, the depth x is zero, which means √((base area) × 4) = -(α/2)T + c1.
We can solve this equation to find the value of T.
Now, to find the further time that elapses before the tank becomes completely empty, we need to substitute √((base area) × 4) = 2√((base area)) and solve for T.
2√((base area)) = -(α/2)T + c1.
Rearranging the equation, we get -(α/2)T = 2√((base area)) - c1.
Multiplying both sides by -2/α, we obtain T = (-4/α)√((base area)) + (2/α)c1.
Therefore, the further time that elapses before the tank becomes completely empty is given by T = (-4/α)√((base area)) + (2/α)c1.