Water leaks out of a tank through a square hole with 2-inch sides. At time t (in seconds) the velocity of water flowing through the hole is v ft/sec (where v is a function of time, t). Write a definite integral that represents the total amount of water (in cubic feet) lost in the first minute.

V = ∫₀⁶⁰vdt/36 ft³

Step-by-step explanation:

The rate of change of the volume of the tank V is given by dV/dt. Since the volume of the tank is V = Ah where A = cross-sectional area of the square hole and h = height of tank.

Now dV/dt = dV/dh × dh/dt where dh/dt = v = velocity of water flowing through the tank = rate at which the height is changing.

dV/dt = dV/dh × dh/dt

dV/dt = dAh/dh × dh/dt

dV/dt = A × v

dV/dt = Av

Now the area of the dimension of the hole is 2 inches in feet is 2/12 feet. The area of the hole is thus A = (2/12)² = 1/6² = 1/36 ft².

So, dV/dt = v/36

dV = vdt/36

So, for the total volume lost in the first minute, we integrate t from t = 0 to t = 60 s.

V = ∫₀⁶⁰dV

V = ∫₀⁶⁰vdt/36 ft³

So, the definite integral that represents the total amount of water (in cubic feet) lost in the first minute is V = ∫₀⁶⁰vdt/36 ft³

a

step-by-step explanation: