A water container is to be drained for cleansing. If Q represents the number of gallons of the water in the container t minutes after the container has started to drain, and Q(t) = 1/4 (50 − t)^2 .

Question

A water container is to be drained for cleansing. If Q represents the number of gallons of the water in the container t minutes after the container has started to drain, and Q(t) = 1/4 (50 − t)^2 .

Q (Gallons): 625 576 529 484

t (min): 0 2 4 6

a) Estimate at what instant in time is the water draining at a rate of 24 Gallons/min? Show all your work so that it clearly communicates your mathematical thinking.

b) Write a sentence that clearly interprets the physical meaning of your answer in to part (a) in the context of the problem.

Answer 1

a) The question gives us a function Q(t) for the volume of water in the tank, and asks us about the rate at which it's draining; this of course would be the derivative dQ/dt.

Let's express Q(t) as 1/4(u)^2, where u = 50-t.

Applying the power rule and the chain rule,

dQ/dt = 1/2*u*du/dt = (-1/2)*(50-t).

Now let's set that equal to the specified value: (-1/2)*(50-t) = -24.

Why did we choose -24 rather than positive 24? The water is flowing out of the tank, so the volume of water in the tank is decreasing; that is to say, its rate of change must be negative. Once we have the above equation established, we're ready to solve for t.

50-t = 48 and thus t = 2.

b) We've found that the value of t corresponding to dQ/dt = -24 is 2. This means that when two minutes have elapsed, the water is flowing out of the tank at a rate of 24 gallons per minute.