Question 3

An indoor swimming pool holds 120 gallons of water. Water is pumped into a swimming pool at 5 gallons per minute and is drained out of the tank at the rate of √(t+1) gallon per minute for when the drainer is on. At time t = 0, the tank contains 2 gallons of water.

a) Suppose that the drain was left on while the water is being pumped. How many gallons of water drained out of the pool from time t=0 to t=3 minutes? And how much water is in the pool at t=3 minutes?

b) The drainer is turned off at t=8 minutes. Find the exact time t, when the pool is at its maximum.

c) Find the average rate of change at which the water fills up the pool to its maximum, under conditions in part b).


Solution

a) Answers are
D(t) = [0 ∫ 3] √(t+1) dt = (14/3) or approx 4.6667
and P(t) = [0 ∫ 3] 5 dt - D(t) + 2 = (37/3) or approx 12.3333

These types of problems involve finding the definite integral.

∙ By integrating a rate gives you the total change in the parent function.
Integrating the rate at which water is drained over an interval will give us the total water drained.

∙ Let D(t) = total water drained in gallons.
By integrating the rate of which water is pumped in gives the total water added to the pool. Therefore by taking the total water pumped in subtracted from total water drained and adding the water you already have in the pool at t=0, gives you the total water you have in the pool in the given time interval. Let P(t)= total water in pool in gallons.

b) Answer t=(1/5)[120 - (1/3)] or approx 23.9333 minutes.

It is given that the maximum of what the pool holds is 120 gallons. Therefore P(t)=120 at its max.

∙ So just by using the equation of P(t) given the time interval of water drains [0,8] we get
P(t) = [0 ∫ t] 5 dt - [0 ∫ 8] √(t+1) dt + 2
120 = [0 ∫ t] 5 dt - [0 ∫ 8] √(t+1) dt + 2

∙ The first Fundamental Theorem of Calculus states that [a ∫ b] f(x) dx = F(b) - F(a).
Which means the integral of a rate change over any given interval is the total change in the parent function over that interval.

∙ And applying the Fundamental Theorem of Calculus we get
120 = [5(t) - 5(0)] - [(1/2)((8)+1)^(-1/2) - (1/2)((0)+1)^(1/2)] + 2
120 = 5t - (1/2)(1/3) - (1/2)
5t = 120 + (1/6) - (1/2)
t = (1/5)[120 - (2/6)]

c) Answer 4.3593 gallons per minute.

When you want to find the average rate of change, you'd multiply the time frame in the interval with the integral in question, using this formula A(x)=[1/(b-a)]([a ∫ b] f(x) dx).

∙ Let A(x)= the average rate of change of total water filling the pool
A(x) = 1/(t-0) ([0 ∫ t] 5 dt) - ([0 ∫ 8] √(t+1) dt) + 2)

note:
Remember you are finding a rate, so don't forget to have the correct units. If it is a rate then there is a value in respect to another, e.g. miles per hour.

∙ Recall that t = (1/5)[120 - (2/6)], so just plugging in the value we get the average rate of change.