The purpose of this note is to entice students to begin the journey from algebra to calculus. This goal is achieved by starting with a physical example in which water is pumped into a tank. Next, the focus gently shifts from an average pumping rate to an instantaneous pumping rate to introduce a basic concept of calculus, the derivative.
Suppose that a technician pumps water into an empty tank. The volume in gallons of water that she pumps is equal to 3 times the square of the time in hours that she pumps. For example, after she has pumped water for 5 hours, the volume of water in the tank will be 3(5²) = 75 gallons. If the technician begins pumping water at 12 noon, she will stop pumping water 5 hours later at 5 pm. The pumping rate is the rate of change of the volume of water in the tank with respect to time. The average pumping rate after pumping water for 5 hours is equal to the volume of water divided by the elapsed time or 75 gal / 5 h = 15 gal / h.
The technician wants to calculate the instantaneous pumping rate at precisely 4 pm after she has pumped water for 4 hours. She reasons that the instantaneous pumping rate at 4 pm is approximately equal to the average pumping rate during a small interval of time that extends beyond 4 pm. At 4 pm, she has pumped 3(4²) = 48 gallons in 4 hours. To numerically test the hypothesis that this approximation will improve as the time interval beyond 4 pm becomes shorter, she initially considers a 6 minute or 0.10 hour time interval extending from 4 pm to 4.10 pm. The average pumping rate during this 0.10 hour interval is equal to [3(4.10²) – 48] gal / 0.10 h = 24.30 gal / h. Next, she reduces the width of the time interval to 0.60 minute or 0.01 hour, so that the time interval extends from 4 pm to 4.01 pm. The average pumping rate during this 0.01 hour interval is [3(4.01²) – 48] gal / 0.01 h = 24.03 gal / h. As the width of the time interval is decreased, the instantaneous pumping rate at 4 pm appears to converge to 24 gal / h.
To demonstrate algebraically that the average pumping rate approaches the instantaneous pumping rate as the width of the time interval shrinks to zero, note that the equation for the volume v of water pumped as a function of the elapsed time t is v =3t². Let w denote the width of a small interval of time extending from time t to time t + w. The average pumping rate in the interval from time t to time t + w equals [3(t + w)² – 3 t²] / w = 6 t + 3 w. To validate the last numerical calculation, let t = 4 hours. When w = 0.01 hour, the average pumping rate is 6(4) + 3(0.01) = 24.03 gal / h. When w goes to zero, the instantaneous pumping rate at time t = 4 hours is 6t = 6(4) = 24 gal / h. In calculus, for v = 3t², the expression 6 t for the instantaneous pumping rate at time t is called the derivative of v with respect to t. The derivative of the volume of water is its instantaneous rate of change.
Theodore Sheskin, P.E. (Retired)