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 t4 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 V with respect to time t and evaluate it at t=10.
The volume function is given by V = πr^2h, where r is the radius of the tank and h is the height of the water in the tank.
In this case, the radius r is 7 feet. The height h at time t is t^4 feet.
So, we can write the volume function V as V = π(7^2)(t^4) = 49πt^4.
To find the derivative of V with respect to t, we use the power rule for differentiation:
dV/dt = d(49πt^4)/dt = 196πt^3.
Now, we evaluate the derivative at t=10:
dV/dt = 196π(10^3) = 196π(1000) = 196000π.
Therefore, the instantaneous rate of change of the volume of water in the tank at t=10 minutes is 196000π ft^3/min.
However, this answer is not listed among the options provided. So, the correct answer is none of the given options.