A tank of liquid has both an inlet pipe allowing liquid to be added to the tank and a drain allowing liquid to be drained from the tank.

The rate at which liquid is entering the tank through the inlet pipe is modeled by the function i(x)=3x2+2, where the rate is measured in gallons per hour. The rate at which liquid is being drained from the tank is modeled by the function d(x)=4x−1, where the rate is measured in gallons per hour.

What does (i−d)(3) mean in this situation?

The rate at which the amount of liquid in the tank is changing at t = 3 hours is 40 gallons per hour.

There are 40 gallons of liquid in the tank at t = 3 hours.

The rate at which the amount of liquid in the tank is changing at t = 3 hours is 18 gallons per hour.

There are 18 gallons of liquid in the tank at t = 3 hours.

Guest Mar 14, 2019

#1**+1 **

(i - d) = [ 3x^2 + 2 ] - [ 4x - 1 ] = [3x^2 - 4x + 3]

(i - d)(3) = 3(3)^3 - 4(3) + 3 = 27 - 12 + 3 = 18 means that

The rate at which the amount of liquid in the tank is changing at t = 3 hours is 18 gallons per hour

[At 3 hours.... It is entering at 3(3)^2 + 2 = 29 gallons/hr and leaving at 4(3) - 1 = 11 gal/hr....so...the rate difference is 29 - 11 = 18 gal/hr ]

CPhill Mar 14, 2019