Problem 17 - Algebra 1

Question

In the Chapter Opener you learned that there is a relationship between the breathing rate of a cyclist and the bicycle speed. $$\begin{array}{|l|c|c|c|c|c|}\hline \text { Bicycle speed, } x & 0 & 5 & 10 & 15 & 20 \\\\\hline \text { Breathing rate, } y & 6.4 & 10.7 & 18.1 & 30.5 & 51.4 \\\\\hline\end{array}$$ Let \(x\) represent the speed of the bike in miles per hour, and let \(y\) represent the cyclist's breathing rate in liters of air taken into the lungs per minute. The breathing rate of a cyclist can be modeled by \(y=6.37(1.11)^{x} .\) What is the cyclist's breathing rate if the bike is traveling 19 miles per hour? 25 miles per hour?

Step-by-Step Solution

Answer

The cyclist's breathing rate is approximately 44.32 liters of air per minute when the bike is traveling at 19 miles per hour and approximately 99.24 liters of air per minute when the bike is traveling at 25 miles per hour.

Step 1: Substitute the given speeds into the function

The given function is \(y=6.37(1.11)^{x}\). Firstly, substitute \(x=19\) into the function. This yields: \(y=6.37(1.11)^{19}\).

Step 2: Calculate the breathing rate at 19 mph

Use a calculator to simplify \(6.37(1.11)^{19}\). This results in approximately 44.32, meaning the cyclist's breathing rate is approximately 44.32 liters of air per minute when the bike is traveling at a speed of 19 miles per hour.

Step 3: Substitute the second given speed into the function

Now substitute \(x=25\) into the function. This yields: \(y=6.37(1.11)^{25}\).

Step 4: Calculate the breathing rate at 25 mph

Use a calculator to simplify \(6.37(1.11)^{25}\). This results in approximately 99.24, meaning the cyclist's breathing rate is approximately 99.24 liters of air per minute when the bike is traveling at a speed of 25 miles per hour.

Key Concepts

Modeling Real-World Problems

When solving real-world problems, we often create mathematical models to represent complex situations. This model gives us the ability to predict outcomes and understand relationships in real-world phenomena.
In our exercise, we predicted the cyclist's breathing rate using an exponential function. Exponential functions are chosen when relationships grow rapidly between variables. In this case, the breathing rate increases substantially as the cycling speed rises.

• The speed of the bicycle was the independent variable, represented by \(x\).
• The breathing rate was the dependent variable, represented by \(y\).
• Despite complex underlying physiological processes, our model simplifies the relationship between these variables using a single formula.

The primary goal of modeling real-world problems is to create easy-to-use equations that closely mirror actual data. Such models can then offer insights and make predictions.

Substitution Method

The substitution method is a powerful tool in mathematics to find specific values within an equation. Here, we used a given speed and substituted it into our exponential equation to find the corresponding breathing rate.
The steps are simple:

• Identify the value to substitute. In this exercise, the speeds given were 19 mph and 25 mph.
• Replace the variable in the equation with this value. Such as \(y = 6.37(1.11)^{19}\) for a speed of 19 mph.
• Solve the equation with the substituted value to find your result.

This method helps solve equations step-by-step by simplifying complex problems into straightforward calculations.

Function Calculation

Function calculation involves substituting values into a function and performing arithmetic operations to solve for an unknown variable.
For this problem:

• Start with the provided function, \(y=6.37(1.11)^{x}\).
• Substitute the given speed to find \(y\), for instance, \(x=19\) or \(x=25\).
• Calculate using tools like a calculator to simplify expressions such as \(6.37(1.11)^{19}\).

The challenge in function calculation is managing large numbers and exponents, especially when dealing with exponential functions. Calculators or computer software are typically used for precision.
This ensures accurate calculations, which is critical for verifying the reliability of a model.

Mathematical Expressions

Mathematical expressions consist of numbers, variables, and operations that help us solve problems and express calculations clearly.
Our focus in this solution was the use of the expression:\[y=6.37(1.11)^{x}\]
In this context:

• \(6.37\) is the initial value that engineers or scientists determined from experimental data.
• \(1.11\) represents a growth factor. When raised to the power of \(x\), it describes how breathing rates increase exponentially with speed.
• The exponent \(x\) showcases the rapid increase in breathing rate as speed grows.

Understanding mathematical expressions helps manipulate and reformulate equations to reflect real-life aspects accurately. They serve as the building blocks for constructing and deconstructing formulas to assess relationships, trends, and outcomes.