Question

For a mammal that weighs \(w\) grams, the volume \(b\) (inmilliliters) of air breathed in and the volume \(d\) (in milliliters) of "dead space" (the portion of the lungs not filled with air) can be modeled by \(b(w)=0.007 w\) and \(d(w)=0.002 w\). The breathing rate \(r\) (in breaths per minute) of a mammal that weighs \(w\) grams can be modeled by $$ r(w)=\frac{1.1 w^{0.734}}{b(w)-d(w)} . $$ Simplify \(r(w)\) and calculate the breathing rate for body weights of \(6.5\) grams, 300 grams, and 70,000 grams.

Short answer

The breathing rates for the body weights of 6.5 grams, 300 grams, and 70,000 grams can be calculated using the simplified breathing function \(r(w)=\frac{1.1 w^{-0.266}}{0.005}\). However, for exact rates in breaths per minute, numerical computations of the expressions must be executed.

Step-by-step solution

Substitute \(b(w)\) and \(d(w)\) into \(r(w)\)

The function \(r(w)\) can be rewritten by substituting \(b(w)\) and \(d(w)\) into it: \(r(w)=\frac{1.1 w^{0.734}}{0.007 w - 0.002 w}\)

Simplify the denominator

First, simplify the denominator by calculating the difference between 0.007 w and 0.002 w, which gives 0.005 w. So the function \(r(w)\) becomes: \(r(w)=\frac{1.1 w^{0.734}}{0.005 w}\)

Simplify the function by canceling common terms

Simplify the function by canceling out the common term \(w\) from the numerator and denominator. This changes the function into: \(r(w)=\frac{1.1 w^{0.734-1}}{0.005}\)

Compute the values for given weights

Using the simplified rate function, compute the breathing rate for the given body weights (6.5 grams, 300 grams, 70,000 grams). For \(w = 6.5\) grams, we get \(r(6.5)=\frac{1.1 * 6.5^{0.734-1}}{0.005}\). Similarly, for the other weights, we compute \(r(300) = \frac{1.1 * 300^{0.734-1}}{0.005}\) and \(r(70000) = \frac{1.1 * 70000^{0.734-1}}{0.005}\)

Key concepts

Mathematical Modeling

Mathematical modeling is the process of using mathematical language and concepts to describe and analyze real-world systems. In this exercise, we are modeling the breathing rate of mammals based on their weight. This involves creating equations that reflect biological processes.
This kind of modeling helps us to predict outcomes and understand relationships in a real-world context.With mathematical models, we can:

• Determine how changes in one variable affect another, such as how weight affects breathing rate.
• Test hypotheses about biological processes without the need for direct experimentation.
• Make precise calculations for various scenarios, such as predicting the breathing rates for different weights.

In the breathing rate model given, each function—\(b(w)\) and \(d(w)\)—represents a component of lung function that changes with weight, providing insight into how weight influences breathing dynamics.

Exponential Functions

Exponential functions are mathematical expressions where a constant base is raised to a variable exponent. These functions are crucial in modeling growth and decay processes. In the case of the breathing rate model, the term \(w^{0.734}\) is an example of an exponential function.Exponential functions generally have these characteristics:

• They can quickly increase or decrease, making them suitable for modeling growth or decay, like breathing rate changes.
• The rate of change is proportional to the current value, indicating that small differences in weight can lead to significant changes in outcome.
• They are continuous and smooth, which reflects the constant and gradual variation in physiological processes, such as breathing.

In our exercise, the exponent \(0.734\) modifies how sensitive the breathing rate is to changes in weight. A higher exponent would mean a more dramatic change in breathing rate with increased weight.

Simplifying Expressions

Simplifying expressions is a key mathematical process that involves reducing expressions to their simplest form. This is crucial in solving equations as it makes calculations more straightforward and less error-prone.In the breathing rate function \(r(w)\), the expression was simplified by:

• Combining like terms in the denominator: \(0.007w - 0.002w\) simplifies to \(0.005w\).
• Canceling common terms: The \(w\) in both the numerator and denominator was canceled out to simplify the expression.
• Adjusting the exponent: After cancellation, the exponent of \(w\) was updated to \(w^{0.734-1}\), further simplifying the expression.

These steps not only make calculations easier but also reveal the underlying relationships within the formula, allowing us to understand how each component influences the breathing rate.