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.
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 t and evaluate it at t=10.
Given that the radius of the tank is 7 feet, the volume function is V=πr^2h. We can find the height of the water in the tank at t=10 minutes by substituting t=10 into the height function: h = t/4 = 10/4 = 2.5 feet.
Now we can differentiate the volume function V=πr^2h with respect to t to find the rate of change of volume with respect to time:
dV/dt = d/dt (πr^2h)
= πd(r^2h)/dt
To find d(r^2h)/dt, we consider that r^2 is a constant because the radius of the tank does not change. Therefore, d(r^2)/dt = 0.
Using the product rule, we can differentiate h with respect to t to find dh/dt:
dh/dt = d(t/4)/dt
= (1/4)d(t)/dt
= 1/4
Now we can find the rate of change of volume with respect to time at t=10 minutes by substituting the values:
dV/dt = πd(r^2h)/dt
= π(0)(dh/dt)
= 0
Therefore, the instantaneous rate of change of the volume of water in the tank at t=10 minutes is 0 cubic feet per minute. This means that at t=10 minutes, the volume of water in the tank is not changing.