Question

Coughing forces the trachea (windpipe) to contract, which in turn affects the velocity of the air through the trachea. The velocity of the air during coughing can be modeled by \(v=k(R-r) r^{2}, 0 \leq r

Short answer

The radius that will produce the maximum air velocity during coughing is \(\frac{2R}{3}\), with \(R\) being the normal radius of the trachea.

Step-by-step solution

Differentiate the function with respect to \(r\)

To find the maximum of the function, you first need to find its derivative with respect to \(r\). The derivative of \(v\) with respect to \(r\) is \(\frac{dv}{dr} = k(2rR - 3r^{2})\).

Solve the first derivative for \(r\)

To find the maximum, set the derivative equal to zero and solve for \(r\): \(k(2rR - 3r^{2}) = 0\). This simplifies to \(2rR - 3r^{2} = 0\), and by factoring \(r\) you get \(r(2R - 3r) = 0\). This gives two solutions: \(r = 0\) and \(r = \frac{2R}{3}\).

Determine which value of \(r\) gives a maximum

To figure out which of these solutions gives a maximum, you need to consider the second derivative of \(v\) with respect to \(r\), which is \(\frac{d^2v}{dr^2} = -6kr\). For \(r = 0\), \(\frac{d^2v}{dr^2} = 0\), and for \(r = \frac{2R}{3}\), \(\frac{d^2v}{dr^2} < 0\). Therefore, the velocity is maximized when \(r = \frac{2R}{3}\).

Key concepts

Differential Calculus

Differential calculus, a branch of calculus, involves the study of how functions change when their inputs change. The core tool of differential calculus is the derivative. It is defined as the rate at which a function value changes as its input changes. Essentially, it represents the slope of the tangent to the graph of the function at any point.

When solving the max velocity in the trachea problem, differential calculus is applied by first finding the derivative of the velocity function. The function is given as a product of a constant, the difference between two radii, and the square of the radius during coughing. By differentiating this function, we obtain the rate of change of velocity with respect to radius. This step is crucial as it sets up the opportunity to find the exact value of the radius that maximizes velocity through the critical point analysis.

Optimization Problem

An optimization problem in mathematics seeks to find the maximum or minimum value of a function given a set of constraints. The function we are looking to optimize represents some quantity we want to maximize or minimize, such as profit, distance, area, or, as in our case, the air velocity in the trachea.

In the given problem, the radius that maximizes the velocity is sought. By setting the first derivative of the velocity function to zero, we find potential candidates for maximum values. These critical points are where the velocity's rate of change switches signs, indicating a possible maximum or minimum. The second derivative test then aids in discerning whether these points correspond to a maximum. In this exercise, we're using the optimization principles of differential calculus to determine the best possible outcome given the biological constraints of the trachea.

Applied Mathematics

Applied mathematics involves using mathematical methods to solve real-world problems. It often requires the translation of a physical situation into a mathematical model, which is then analyzed and solved using various mathematical tools.

In the context of finding maximum air velocity in the trachea, applied mathematics comes to life. We start with a physiological event—coughing—which is represented by a mathematical model showing how air velocity changes with the contraction of the trachea. Through the optimization techniques of differential calculus, applied mathematics enables solving for the unknown radius that maximizes air velocity—a result that can have direct implications in medical diagnostics or treatment strategies.