Problem 91 - 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

After 3 times: 2.56 gallons of water. After 5 times: 1.6384 gallons. For n times: \( 5 \times \left( \frac{4}{5} \right)^n \) gallons of water.

Step 1: Initial Setup

The radiator can hold 5 gallons of liquid in total and initially contains 5 gallons of water. We will define the problem such that first we remove 1 gallon of water, replacing it with 1 gallon of antifreeze.

Step 2: First Repetition

After replacing the first gallon of water with antifreeze, the radiator now contains 4 gallons of water and 1 gallon of antifreeze.

Step 3: Second Repetition

We remove a gallon of the mixture (4/5 of water and 1/5 of antifreeze), leaving behind 3.2 gallons of water and 0.8 gallons of antifreeze. We then add 1 gallon of antifreeze, for a total of 4.2 gallons of water and 0.8 gallons of antifreeze.

Step 4: Third Repetition

Next, we remove another gallon of the mixture (4.2/5 of water and 0.8/5 of antifreeze), leaving behind approximately 3.36 gallons of water. Then, we add another gallon of antifreeze, leading to approximately 3.36 gallons of water and 1.64 gallons of antifreeze.

Step 5: General Formula

To find how much water is left after n repetitions, we observe that each repetition reduces the amount of water by a factor of \( \frac{4}{5} \). Thus, the formula for the amount of water remaining after n repetitions is given by \( 5 \times \left( \frac{4}{5} \right)^n \).

Step 6: Calculations for Specific Cases

- For 3 iterations: Calculate \( 5 \times \left( \frac{4}{5} \right)^3 = 5 \times \frac{64}{125} = 2.56 \) gallons.- For 5 iterations: Calculate \( 5 \times \left( \frac{4}{5} \right)^5 = 5 \times \frac{1024}{3125} = 1.6384 \) gallons.

Key Concepts

Recursion

Recursion is a method of solving problems where the solution involves solving smaller instances of the same problem. It is a fundamental concept in mathematics and computer science. In this exercise, recursion helps us understand how the amount of water in the radiator changes with each iteration of the process. The essence of recursion lies in the repeated action, here being the removal and replacement of the liquid in the radiator. This recurrent process gradually reduces the water content in a predictable way.

• Every time a gallon of the mixture is removed, the model recalculates the remaining water, starting from the previous amount.
• The new state is dependent upon the earlier state with the effect of the action applied.

Thus, each step becomes the starting point for the next calculation, embodying the principle of recursion.

Exponential Decay

Exponential decay refers to a decrease at a continuously diminishing rate. This concept is crucial for understanding how the water in the radiator decreases over repeated iterations. In this problem, the proportion of water is progressively less, following an exponential decay pattern as given by the formula. Each time a gallon of the mixture is removed, the remaining water content is reduced by a constant fraction, which in this case is \( \frac{4}{5} \).

Key aspects:

• This process forms a geometric sequence where each term is \( \frac{4}{5} \) times the previous term.
• The formula for the sequence is \( 5 \times \left(\frac{4}{5}\right)^n \), illustrating how the water decreases over time.

The sequence decays towards zero, showing the continual decrease in water as the number of iterations increases.

Algebraic Modeling

Algebraic modeling involves using mathematical expressions to represent real-world scenarios. Here, we express the situation with a formula that predicts the amount of water remaining after a certain number of iterations. Algebra is employed to create the model that works for this recursive process.

To better understand the model:

• The scenario is represented by the formula \( 5 \times \left( \frac{4}{5} \right)^n \), which enables predictions for any given \( n \).
• This uses basic algebraic concepts to form a model that, while simple, captures the essential dynamics of the process.

By applying algebraic modeling to this exercise, we gain a clear, concise formula that can calculate the remaining amount of water for any specified iteration, offering a mathematical explanation of the process.