A water tank is box shaped with a square base of side length 5 ft. The tank is fed by a pipe above and drains from its base. In order to have a well mixed solution, both the feeder pipe and the drain must be opened. The tank is being filled at a rate of 25 ft3/s when the height of the liquid in the tank is 10 ft. If the drains are pulling 30 ft3/s, what is the rate of change in the depth of the solution in the tank

dy/dt = - (1/5) ft/s = - 0.2 ft/s

Step-by-step explanation:

Given

L = 5 ft

Qin = 25 ft³/s

Qout = 30 ft³/s

h = 10 ft

dy/dt = ?

We can apply the relation

ΔQ = Qint - Qout = 25 ft³/s - 30 ft³/s

⇒ ΔQ = - 5 ft³/s

Then we use the formula

Q = v*A

where Q = ΔQ, A = L² is the area of square base and v = dy/dt is the rate of change in the depth of the solution in the tank

⇒ ΔQ = (dy/dt) * L²

⇒ dy/dt = ΔQ/L²

⇒ dy/dt = ( - 5 ft³/s) / (5 ft) ²

⇒ dy/dt = - (1/5) ft/s = - 0.2 ft/s