related rates problemThe town of Cougarville monitors the he

(related rates problem)The town of Cougarville monitors the height of water in its(vertical) cylindrical water tank, with an automatic recording device. Water is constantly pumped into the tank at a rate of 3000 ft^3/hour. During a certain 12-hour period(beginning at midnight), the water level rose and fell. If the radius of the tank is 20 feet, at what rate was water being used at 7:00 a.m when a lot of people must\'ve been getting up for their work day? Hint: the rate of change of water volume, V\'(t), is the difference between how fast water Is flowing into the tank, and how fast its flowing out.

Solution

We know that the volume ( V ) of a cylinder with radius r and height h = r² h. Here, r = 20 ft. Therefore, V = 400h…(1)

We know that water is constantly being pumped into the tank at a rate of 3000 ft^3/hour. Therefore      V’(t) = 3000…(2)

We need to relate two quantities to each other. In this case we can relate the volume and the height with the formula for the volume of cylinder. What we really want is a relationship between their derivatives. We can do this by differentiating both sides with respect to t. In other words, we will need to do implicit differentiation on the above formula. Doing this we get V’(t) = 400h’. Now, we know that V’ (t) = 3000. Therefore, 400h’ = 3000, so that h’ = 3000/(400) = 7.5 / . Thus, the height of water in the tank is changing @ 7.5 / ft/hour.