Problem 75 - Precalculus Mathematics for Calculus

Question

A truck radiator holds 5 gal and is filled with water. A gallon of water is removed from the radiator and replaced with a gallon of antifreeze; then a gallon of the mixture is removed from the radiator and again replaced by a gallon of antifreeze. This process is repeated indefinitely. How much water remains in the tank after this process is repeated 3 times? 5 times? \(n\) times?

Step-by-Step Solution

Answer

2.56 gallons after 3 times, 1.6384 gallons after 5 times, \( \left( \frac{4}{5} \right)^n \times 5 \) after \( n \) times.

Step 1: Understand the Process

To solve this problem, first understand that each cycle involves removing 1 gallon of the current mixture and replacing it with 1 gallon of pure antifreeze. This changes the proportion of water in the radiator.

Step 2: Calculate Fractional Representation

Initially, the radiator is full of water, so it has 5 gallons of water. After the first gallon is replaced, we are left with 4 parts water and 1 part antifreeze out of a total of 5 parts. So, the fraction of water remaining after one cycle is \( \frac{4}{5} \).

Step 3: General Pattern Recognition

For each subsequent cycle, the same fractional \(rac{4}{5}\) process applies to what remains. Thus after two cycles, it is \( \left( \frac{4}{5} \right)^2 \) of the original 5 gallons.

Step 4: Apply the Process for 3 Times

After 3 cycles, the amount of water remaining is \( \left( \frac{4}{5} \right)^3 \times 5 \). Calculate this: \( \left( \frac{4}{5} \right)^3 \times 5 = \frac{64}{125} \times 5 = 2.56 \) gallons.

Step 5: Apply the Process for 5 Times

After 5 cycles, apply the pattern: \( \left( \frac{4}{5} \right)^5 \times 5 = \frac{1024}{3125} \times 5 = 1.6384 \) gallons. This is the amount of water remaining after 5 times.

Step 6: Generalize for n Times

For \( n \) cycles, the formula becomes: \( \left( \frac{4}{5} \right)^n \times 5 \). This formula will give you the amount of water remaining in the radiator after \( n \) repetitions of the process.

Key Concepts

Fractional Representation

Understanding how to express quantities as fractions is key to solving this type of problem accurately. Initially, the radiator is entirely filled with 5 gallons of water. The moment you substitute 1 gallon with antifreeze, you're essentially shifting the ratios of components in the radiator.
After the first gallon is swapped, 4 parts remain water, and 1 part becomes antifreeze. The fractional representation of water in this mix then becomes \( \frac{4}{5} \).
This fraction quickly becomes your go-to expression for each following step.This idea of breaking down a physical process into fractional parts allows you to easily calculate how much of a substance remains after each cycle. You just multiply the fraction of water remaining after each cycle to understand how much water is left overall.
Applying this fraction to successive cycles quickly creates a pattern that forms the foundation of the solution.

Pattern Recognition

Recognizing patterns is essential in complex calculations as it simplifies repetition concepts into easily understandable steps. Once you complete the initial cycle of removing and replacing 1 gallon, you’ll notice a remarkable pattern.
After one cycle, the water remaining is \( \frac{4}{5} \) of the original 5 gallons.
For each new cycle, you apply this \( \frac{4}{5} \) fraction further to what already remains in the radiator.This repetition forms an exponential pattern: after two cycles, it becomes \( \left( \frac{4}{5} \right)^2 \), and after three cycles \( \left( \frac{4}{5} \right)^3 \), continuing this way for any number of cycles \( n \).
Recognizing this predictive pattern enhances your ability to anticipate how much water remains, making it a crucial tool when dealing with repetitive processes like this one.

Sequence and Series

Understanding exponential decay through sequences and series is very informative in this problem. You begin with a fixed amount of water and notice how it decreases exponentially or geometrically with each replacement cycle.
The formula \( \left( \frac{4}{5} \right)^n \times 5 \) determines how much water remains after \( n \) cycles.This transformation into a geometric series provides an elegant way to compute results for multiple cycles without going through each individual cycle manually.
The beauty of sequences and series lies in their ability to model such exponential decay and allow easy calculation for large \( n \) values.This method efficiently handles repetitive calculations, adding to your mathematical toolbox for solving similar problems in the future. With this understanding, it's possible to predict volume changes even as the cycles continue hypothetically to infinity, offering deep insights into the underlying exponential decay.