Problem 67 - Precalculus: Mathematical 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 operations: 2.56 gallons; after 5 operations: 1.6384 gallons; after \( n \) operations: \( 5 \times \left( \frac{4}{5} \right)^n \) gallons.

Step 1: Understand the Problem

The radiator initially holds 5 gallons of pure water. In each step, we remove 1 gallon of the mixture from the radiator and replace it with 1 gallon of antifreeze.

Step 2: Analyze the First Operation

Initially, there are 5 gallons of water in the radiator. After removing 1 gallon of water, 4 gallons of water remain. Since we add 1 gallon of antifreeze to the radiator, it now contains 4 gallons of water and 1 gallon of antifreeze.

Step 3: Analyze the Second Operation

The mixture now consists of 4 gallons of water out of 5 total gallons. The fraction of water in the radiator is therefore \( \frac{4}{5} \). In the second operation, we remove 1 gallon of this mixture, leaving \( \frac{4}{5} \times 5 = 4 \) gallons of water in the radiator before the removal. After removing \( \frac{4}{5} \) gallon of water, the remaining water is \( 4 - \frac{4}{5} = \frac{16}{5} \). After adding another gallon of antifreeze, the radiator now contains \( \frac{16}{5} \) gallons of water and an increased amount of antifreeze.

Step 4: Determine the Pattern

For each subsequent operation, the amount of water remaining is \( \frac{4}{5} \) of what it was before the removal. After each operation, if originally there were 5 gallons, after the \( n \)-th operation, there will be \( 5 \times \left( \frac{4}{5} \right)^n \) gallons of water.

Step 5: Calculate Water After 3 Operations

Using the pattern, after 3 operations, the amount of water remaining is \( 5 \times \left( \frac{4}{5} \right)^3 = 5 \times \frac{64}{125} = \frac{320}{125} = 2.56 \) gallons.

Step 6: Calculate Water After 5 Operations

Similarly, after 5 operations, the amount of water remaining is \( 5 \times \left( \frac{4}{5} \right)^5 = 5 \times \frac{1024}{3125} = \frac{5120}{3125} = 1.6384 \) gallons.

Step 7: Generalize for n Operations

For the \( n \)-th operation, the amount of water remaining in the radiator is given by the formula: \( 5 \times \left( \frac{4}{5} \right)^n \) gallons.

Key Concepts

Geometric Sequences

A geometric sequence is a series of numbers where each term after the first is found by multiplying the previous term by a fixed, non-zero number called the common ratio. In our exercise involving the truck radiator, we have a similar situation. Initially, the radiator is filled completely with water. As each step involves removing and replacing some of the liquid, the portion of water in the radiator diminishes geometrically.

• The first operation changes the amount of water from 5 gallons to 4 gallons.
• Each subsequent operation multiplies the remaining water by the fraction \( \frac{4}{5} \).
• This is the common ratio: \( \frac{4}{5} \).

Thus, after each operation, the remaining amount of water can be calculated using a geometric sequence formula for the \( n \)-th term: \( a_n = a_1 \times r^{(n-1)} \), but in this problem, it's more convenient to adjust and start with 5 gallons of water, giving us: \( 5 \times \left( \frac{4}{5} \right)^n \). As you can see, this is our geometric sequence formula applied to find how much water remains.

Recursion

Recursion in mathematics often refers to defining a sequence, set, or functions in terms of other elements in its standard. In the context of our radiator mixture problem, recursion helps to establish a pattern through iterative processes.
Each step of our problem influences the next. In simpler terms:

• At each step, the amount of water is computed from the results of the prior step.
• Start with 5 gallons of water.
• At every iteration, compute \( \frac{4}{5} \) of the remaining water in the radiator.

Recursion is a robust way to find solutions incrementally and is useful, especially in complex calculations like those involving series and sequences, such as our pattern of decreasing water content.

Exponential Decay

Exponential decay describes a process where the quantity decreases at a rate proportional to its current value. Here, the amount of water in the radiator dwindles each step because it is replaced by antifreeze, leading to a reduction proportional to how much water was there before.
The formula for exponential decay is \( y = y_0 \times e^{-kt} \), where:

• \( y_0 \) is the initial quantity.
• \( k \) is the decay constant.
• \( t \) is the time or iteration step.

In our scenario, traditional exponential decay needs a minor adaptation. We don’t use \( e^{-kt} \), since our decay factor is simply \( \frac{4}{5} \). Thus, the decay becomes:

• Initial amount: 5 gallons.
• Decay factor: \( \frac{4}{5} \).
• Formula: \( 5 \times \left( \frac{4}{5} \right)^n \).

This concept is key as it helps to model how quickly the water content decreases with each repetition, making it a valuable tool for understanding transformations in various scientific and mathematical models.