You didn't include the function.
If the function is, V = 20,000 (1 - t/20)^2, as you can see, for t = 20, V = 0
Then I am going to show you the work with that function, and you can use the same procedure with your own fucntion.
The rate of change of the volume is the derivative of the function with respect to the time.
[dV/dt] = 20,000 2 [1 -t/20] [-1/20] = - 2000 [1 - t/20] = 2000[t/20 - 1]
At t = 3
dV/dt = 2000[3/20 - 1] = -1700 gallons / min
The negative sign mean decreasing.
The answer is - 1700 gallons / min
Now, for the same function, if the time is t = 2, you would obtain one of the results of the list of options.
dV/dt = 20,000 [2/20 - 1] = -1800 gallons/minute, which is the option D of the list.
