Problem 15 - Calculus for the Life Sciences: A Modeling Approach Volume I

Question

What is the optimum radius of the trachea when coughing? The objective is for the flow of air to create a strong force outward in the throat to clear it. For this problem you should perform the following experiment. Hold your hand about \(10 \mathrm{~cm}\) from your mouth and blow on it (a) with your lips compressed almost closed but with a small stream of air escaping, (b) with your mouth wide open, and (c) with your lips adjusted to create the largest force on your hand. With (a) your lips almost closed there is a high pressure causing rapid air flow but a small stream of air and little force. With (b) your mouth wide open there is a large stream of air but with little pressure so that air flow is slow. The largest force (c) is created with an intermediate opening of your lips where there is a notable pressure and rapid flow of substantial volume of air. Let \(R\) be the normal radius of the trachea and \(r

Step-by-Step Solution

Answer

The optimal tracheal radius for coughing is \( r = \frac{2R}{3} \).

Step 1: Understanding the Problem

We need to determine the radius \( r \) of the trachea which maximizes the momentum \( M = k r^2 (R - r) \) during coughing. Given: air velocity is proportional to pressure difference (\( R-r \)), and air mass is proportional to tracheal area (\( \pi r^2 \)). We aim to maximize \( M \).

Step 2: Express Momentum Function

The momentum \( M \) is given as a function of \( r \): \[ M = k r^2 (R - r) \]. Here, \( k \) and \( R \) are constants, and \( r \) is a variable we need to optimize.

Step 3: Simplify the Function

We know that \( M = k r^2 (R - r) \). To focus on the optimization process, let's consider the function without the constant \( k \) since it won't affect the location of the maximum. Simplified, it becomes: \[ f(r) = r^2 (R - r) \].

Step 4: Differentiate the Momentum Function

Differentiate \( f(r) = r^2 (R - r) \) with respect to \( r \). Apply the product rule: \( f'(r) = \frac{d}{dr} (r^2) \cdot (R-r) + r^2 \cdot \frac{d}{dr}(R-r) \). This gives \( f'(r) = 2r(R-r) - r^2 \).

Step 5: Find Critical Points

Set the derivative \( f'(r) = 2r(R-r) - r^2 = 0 \) to find the critical points. Simplifying the expression gives: \( 2rR - 2r^2 - r^2 = 0 \), which leads to \( 3r^2 = 2rR \). Therefore, critical points occur when \( r = 0 \) and \( r = \frac{2R}{3} \).

Step 6: Determine Optimal Radius

Test the critical points and any boundaries for the maximum value of \( M \). Since \( r \) must be greater than 0 and less than \( R \), it follows that \( r = \frac{2R}{3} \) is feasible. After checking, this value indeed maximizes \( f(r) \).

Step 7: Final Conclusion

The optimal radius of the trachea when coughing, that maximizes the momentum \( M \), occurs when \( r = \frac{2R}{3} \).

Key Concepts

Optimization

In calculus, optimization refers to the process of finding the best solution to a problem within a defined set of constraints. This concept is crucial when one wants to maximize or minimize a function. In the context of the trachea exercise, the main goal is to maximize the momentum of air expelled during coughing. The optimization process involves working with the function representing momentum and determining the radius, \( r \), that gives the highest value.

To achieve this, we consider the momentum function \( M = k r^2 (R - r) \). Here, \( k \) and \( R \) are constants, while \( r \) is the variable to be optimized. By simplifying this to \( f(r) = r^2 (R - r) \), we ignore \( k \) because it does not impact the position of the maximum value. Then, through differentiation and finding critical points (places where the slope of the function is zero), we identify the optimal \( r \). This process is a practical application of optimization in calculus.

Differentiation

Differentiation is a key mathematical tool in calculus used to determine the rate of change of a function with respect to a variable. In simple terms, differentiation tells us how a function is changing at any given point. It's especially useful in optimization problems since it helps find maximum or minimum values by identifying critical points.

In the trachea radius problem, we differentiate the momentum function \( f(r) = r^2 (R - r) \) to find \( f'(r) \). This derivative is computed using the product rule, resulting in \( f'(r) = 2r(R-r) - r^2 \). By setting \( f'(r) = 0 \), we find the critical points which are potential candidates for maximizing or minimizing the function. Differentiation enables us to analyze the behavior of the momentum function, leading us to conclude that \( r = \frac{2R}{3} \) maximizes \( M \).

Understanding differentiation provides insight into how the function varies and is essential for successfully executing optimization.

Momentum Maximization

Momentum maximization involves adjusting variables within physical constraints to achieve the highest momentum outcome. This concept is particularly important in contexts where force and movement are key factors, such as fluid dynamics or mechanical systems.

In the problem at hand, momentum is defined as the product of mass and velocity. For the trachea, the air's momentum is influenced by its radius during coughing. The momentum function \( M = k r^2 (R - r) \) encapsulates this relationship, where \( r^2 \) relates to the cross-sectional area (or mass flow), and \( (R-r) \) is linked to the pressure difference driving the velocity.

• **Mass of air flow**: Proportional to the area of the trachea, \( \pi r^2 \).
• **Velocity of air flow**: Proportional to the pressure difference, \( R - r \).

By maximizing \( M \), we find the radius \( r = \frac{2R}{3} \), which optimally balances these two factors. This specific radius permits a favorable combination of pressure and air mass, resulting in the most efficient expulsion of air and thus maximal force. This maximization is key for effective coughing.