A city work fills the cylindrical tank of a new water tower. The tank has a radius of 7 feet. At time t minutes, the height of the water in the tank is t/4 feet and the volume of water in the water tower is V=πr2h cubic feet. Find the instantaneous rate of change of the volume of water in the tank at t=10 minutes. (1 point) Responses 49π40 ft.3min. 49 π 40 ft. 3 min. 49π10 ft.3min. 49 π 10 ft. 3 min. 49π4 ft.3min. 49 π 4 ft. 3 min. 189π4 ft.3min. 189 π 4 ft. 3 min. 2452 ft.3min.
To find the instantaneous rate of change of the volume of water in the tank at t=10 minutes, we need to find the derivative of the volume function with respect to time, and then evaluate it at t=10.
The volume of water in the tank is given by V=πr^2h, where r is the radius of the tank (7 feet) and h is the height of the water at time t.
At time t minutes, the height of the water is given as h = t/4 feet.
Substituting the given values, the volume function becomes:
V = π(7^2)(t/4) = 49π(t/4) = 49πt/4
Now, we differentiate the volume function with respect to time:
dV/dt = 49π/4
To find the instantaneous rate of change at t=10 minutes, we evaluate the derivative at t=10:
dV/dt = 49π/4 = 49π/4 = 49π/4 ft^3/min
Therefore, the correct answer is 49π/4 ft^3/min.