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The Product Rule is a fundamental concept in calculus used to differentiate expressions where two functions are multiplied together. In our exercise, the equation for air velocity during coughing, \(v = k(R-r)r^2\), involves the product of two functions: \(u = R-r\) and \(v = r^2\).
To differentiate such a function, we apply the Product Rule, which states that:

• The derivative of a product \((uv)'\) is given by \(u'v + uv'\).

Let's break it down:

• Differentiate \(u = R - r\) to get \(u' = -1\), since \(R\) is a constant.
• Differentiate \(v = r^2\) to get \(v' = 2r\) using the power rule.
• Apply the product rule to get \(\frac{dv}{dr} = (-1) \cdot r^2 + (R-r) \cdot 2r\).

This yields the derived expression: \(\frac{dv}{dr} = -2kr + 2kr(R - r)\).
Understanding the Product Rule is crucial for proceeding to find the critical points, which helps determine the maximal velocity.

Critical Points are values of the variable where a function's derivative is either zero or undefined. Finding these points is essential for identifying potential maxima or minima within a function.
In the context of our problem, we need to locate the value of \(r\) that maximizes the air velocity \(v\).
To find critical points, we set the derivative \(\frac{dv}{dr}\) equal to zero and solve for \(r\):

• Set the equation \(-2kr + 2kr(R - r) = 0\).
• Solve for \(r\) to find \(r = \frac{1}{3}R\).

This value, \(r = \frac{1}{3}R\), is our critical point. It is an essential value as it indicates where the air velocity could be at either a maximum or minimum. To confirm this, further analysis such as the second derivative test is needed.

The Second Derivative Test helps determine whether a critical point is a maximum, minimum, or a point of inflection. It involves taking the second derivative of a function and evaluating it at the critical point.
For our problem, we calculate the second derivative of \(v\) with respect to \(r\):

• We find that \(\frac{d^2v}{dr^2} = -2k + 2k = 0\).

Since the second derivative evaluates to zero, the traditional second derivative test does not give us conclusive evidence if \(r = \frac{1}{3}R\) is a maximum, minimum, or neither.
In these cases, alternative methods such as analyzing the behavior of the first derivative around \(r = \frac{1}{3}R\) or using physical reasoning can be employed. Given the problem’s context related to coughing, \(r = \frac{1}{3}R\) is established as a logical point for the maximum air velocity based on the constraints imposed by the cough mechanics.

Coughing Mechanics addresses the physiological aspects that influence air movement in the trachea during a cough. The trachea undergoes temporary narrowing, affecting air velocity, modeled by the equation \(v=k(R-r)r^{2}\).
This model demonstrates:

• The impact of tracheal radius \(r\) on air velocity \(v\), which increases until it reaches a maximum when \(r = \frac{1}{3}R\).
• The balance between the residual radius \(R-r\) and the area through which air is forced \(r^2\).

Understanding these mechanics is crucial in comprehending why certain tracheal radii optimize airflow. This model's predictions align with clinical knowledge that suggests an optimal degree of tracheal constriction to maintain efficient airflow during a cough.
Coughing effectively clears airways by maximizing air velocity, thus explaining why \(r = \frac{1}{3}R\) would provide the greatest expelling force to clear obstructions.