Problem 23 - Calculus Early Transcendentals: Pearson New International Edition

Question

A tank of capacity 100 gallons is initially full of pure alcohol. The flow rate of the drain pipe is 5 gallons per minute; the flow rate of the filler pipe can be adjusted to \(c\) gallons per minute. An unlimited amount of \(25 \%\) alcohol solution can be brought in through the filler pipe. Our goal is to reduce the amount of alcohol in the tank so that it will contain 100 gallons of \(50 \%\) solution. Let \(T\) be the number of minutes required to accomplish the desired change. (a) Evaluate \(T\) if \(c=5\) and both pipes are opened. (b) Evaluate \(T\) if \(c=5\) and we first drain away a sufficient amount of the pure alcohol and then close the drain and open the filler pipe. (c) For what values of \(c\) (if any) would strategy (b) give a faster time than (a)? (d) Suppose that \(c=4\). Determine the equation for \(T\) if we initially open both pipes and then close the drain.

Step-by-Step Solution

Answer

(a) 13.33 minutes; (b) 20 minutes; (c) Faster if \(c > 7.5\); (d) Equation for both phases: Net operation, closed drain refilling till condition met.

Step 1: System Setup

The tank initially has 100 gallons of pure alcohol. The goal is to replace this with a 50% alcohol solution using a 25% alcohol solution filler pipe. This means after the operation, there should be 50 gallons of alcohol in 100 gallons of solution.

Step 2: Setting up the Flow Rates

The drain pipe removes liquid at 5 gallons per minute. For part (a) and (b), the flow rate of the filler pipe is also set at 5 gallons per minute. Let's examine the different strategies for achieving the desired concentration.

Step 3: Solve Part (a) - Concurrent Operation

In this scenario, both the drain and the filler pipes are opened simultaneously until the tank is filled with a 50% solution. Since the drain pipe removes 5 gallons of pure alcohol per minute and the filler adds 5 gallons of 25% solution (1.25 gallons of alcohol), the net removal of alcohol is 3.75 gallons per minute.To reach 50 gallons of alcohol from the initial 100, alcohol needs to be reduced by 50 gallons. Thus, the time to achieve this is:\[T = \frac{50}{3.75} = \frac{200}{15} \text{ minutes} \approx 13.33 \text{ minutes.}\]

Step 4: Solve Part (b) - Sequential Operation

First, drain away 50 gallons of pure alcohol from the tank (at 5 gallons per minute). This requires:\[T_1 = \frac{50}{5} = 10 \text{ minutes.}\]Now, fill the tank with the 25% alcohol solution to replace the 50 gallons of drained alcohol. This also requires:\[T_2 = \frac{50}{5} = 10 \text{ minutes.}\]Hence, the total time required is:\[T = T_1 + T_2 = 10 + 10 = 20 \text{ minutes.}\]

Step 5: Answer Part (c) - Comparing Strategies

Strategy (b) takes 20 minutes when \(c = 5\). Strategy (a) took approximately 13.33 minutes. For strategy (b) to be faster, it should take less than 13.33 minutes in total.If \(c\) is increased, more gallons will enter per minute after drainage in strategy (b). To drain 50 gallons of pure alcohol and refill with \(25\%\) solution in less than 13.33 minutes, prioritize maximizing the filler rate \(c > 7.5\) that accounts for refilling time, hence it is possible only when \(c > 7.5 \) gallons per minute refilling is faster.

Step 6: Solve Part (d) - Changing Pipe Rates

If \(c = 4\), the combination of the drain at 5 gallons/minute and filler at 4 gallons/minute is considered. This implies a net decrease in volume by 1 gallon per minute (5-4).Once 1 gallon is left, close the drain pipe, and refill to 100 gallons only with the 25% alcohol solution, adding \(0.25 c = 1\) gallon alcohol per minute.Determine the time until the alcohol content reaches 50 gallons in 100 gallons solution: \[0.75T + 25 = 50 \rightarrow T = \frac{25}{0.75 } = 33.33 \text{ minutes.}\] Therefore total equation combining processes is dissected by period durations in problem requirements for \(c = 4\).

Step 7: Solution Conclusion

Through these calculations, we can evaluate the time \(T\) needed depending on the varying conditions: \(c=5\), full open, sequential strategies, and adjustable filler rates in scenario optimization.

Key Concepts

Flow Rate

The concept of flow rate is pivotal in solving this type of differential equation problem because it defines how quickly liquid leaves or enters the tank. Flow rate can be imagined as how many gallons per minute are flowing.
In this exercise, the flow rate of the drain pipe is set at 5 gallons per minute. This means every minute, 5 gallons of liquid are drained from the tank.
On the other hand, the filler pipe's flow rate is variable—denoted by 'c' gallons per minute. This means the rate at which the solution enters the tank can be adjusted, this adjustment allows flexibility in achieving different concentrations within the tank.
When you work with these parameters, the key is to understand their combined effect on the tank over time. If both pipes are open concurrently, it is the net flow rate, or the difference between the incoming and outgoing flow rates, that determines how quickly the conditions within the tank change.

Alcohol Solution

The problem involves changing the concentration of alcohol in a tank using a weaker alcohol solution. The filler pipe allows an unlimited supply of a 25% alcohol solution.
This means every gallon of this solution contains 0.25 gallons of alcohol. When both drain and filler pipes are open, the challenge is to reach exactly 50% alcohol concentration in the tank.
This translates into having 50 gallons of alcohol in a 100-gallon solution. For every gallon added by the filler, an equivalent amount of alcohol is added commensurate with its 25% concentration.
Understanding how these percentages translate into actual alcohol content helps in determining the appropriate times to open and close the pipes so that the desired concentration is reached more efficiently.

Concurrent and Sequential Operations

To efficiently operate on systems like these, you can choose between concurrent and sequential operation strategies.
Concurrent Operations involve running both the drain and filler simultaneously. This allows continual change in the tank's solution composition, effectively reducing alcohol content at a predictable rate. However, you need to balance the flow rates to reach your desired solution composition exactly.

• For instance, opening both pipes at the same flow rate of 5 gallons per minute means alcohol is being reduced by a net of 3.75 gallons/minute, allowing for efficient reduction to a target concentration.

Sequential Operations, however, act in stages. You might first drain a certain amount of pure alcohol before switching tactics to fill the tank with a diluted solution until the desired concentration is achieved.

• This is seen in Part (b), where after draining the required amount, only the filler pipe was used to reach the target alcohol concentration.

Choosing between these operations requires mindful consideration of time efficiency and accuracy for optimal results.

Tank Capacity Problem

Typically, a tank capacity problem requires you to target a specific level or concentration within a given volume, here exemplified through differential equations.
The fixed capacity of the tank means all operations must result in a total of 100 gallons. The challenge is to manipulate incoming and outgoing solution rates and compositions to hit exactly 50% concentration.
Since the tank initially had 100 gallons of pure alcohol, calculations must ensure that at the end of the process, half of that—50 gallons—should remain as alcohol, mixed with enough 25% solution to total 100 gallons.
Working with tank capacity emphasizes maintaining balance—between what you add, what you remove, and the resultant concentration—to maintain the total while achieving desired properties.