Problem 70 - Single Variable Calculus

Question

Velocity of Airflow During a Cough When a person coughs, the trachea (windpipe) contracts, allowing air to be expelled at a maximum velocity. It can be shown that the velocity \(v\) of airflow during a cough is given by $$ v=f(r)=k r^{2}(R-r) \quad 0 \leq r \leq R $$ where \(r\) is the radius of the trachea in centimeters during a cough, \(R\) is the normal radius of the trachea in centimeters, and \(k\) is a constant that depends on the length of the trachea. Find the radius for which the velocity of airflow is greatest.

Step-by-Step Solution

Answer

The velocity of airflow during a cough is greatest when the radius of the trachea is \(r = \frac{2R}{3}\), where \(R\) is the normal radius of the trachea.

Step 1: Differentiation of the velocity function

To find critical points of the velocity function \(v(r)=kr^2(R-r)\), we need to find its derivative with respect to \(r\). Using the power rule and the product rule, the derivative of \(v(r)\) can be found: \( \frac{dv}{dr} = \frac{d}{dr}\left(kr^2(R-r)\right) \) Differentiating \(kr^2\) and \((R-r)\) with respect to \(r\) and applying the product rule, we get: \( \frac{dv}{dr} = k(2r)(R-r) + kr^2(-1) \)

Step 2: Finding critical points

To find critical points, we need to find values of \(r\) that satisfy \(\frac{dv}{dr} = 0\). So, we set the derivative equal to zero: \( 0 = k(2r)(R-r) + kr^2(-1) \) Solving for r: \( 0 = 2kr(R-r) - kr^2 \) Divide both sides by \(k\) (assuming \(k \neq 0\)): \( 0 = 2r(R-r) - r^2 \)

Step 3: Solve the equation for r

To solve this equation, first, simplify it: \( 0 = 2rR - 2r^2 - r^2 \) Combine the terms with \(r^2\): \( 0 = 2rR - 3r^2 \) Factor out an r: \( 0 = r(2R - 3r) \) This equation has two solutions: \(r=0\) and \(r=\frac{2R}{3}\). Since the trachea always has some radius, we discard the solution \(r=0\). Therefore, we have: \( r = \frac{2R}{3} \)

Step 4: Conclusion

Based on our analysis, the velocity of airflow during a cough is greatest when the radius of the trachea is \(\frac{2R}{3}\), where \(R\) is the normal radius of the trachea.

Key Concepts

Derivative

The concept of a derivative in calculus is fundamental for understanding how a function changes as its input changes. In simpler terms, when you take the derivative of a function, you are finding out how the function is increasing or decreasing at any given point. For the velocity function in the given problem, the derivative helps determine the rate of change of airflow velocity concerning the change in the radius of the trachea.
Finding the derivative is the first step in identifying points where the function may have a maximum or minimum value, known as critical points. By doing this, we can determine where the velocity of airflow during a cough is at its peak—and thus, where it is optimized.

Critical Points

Critical points are where the derivative of the function equals zero or does not exist. They are incredibly useful for optimization problems, such as the one we are considering with the airflow during a cough.
By setting the derivative of the velocity function to zero, we find out specific values of the radius that may lead to maximum airflow. These critical points give us insight into where the function changes its increasing or decreasing behavior, and can thus reveal points of optimization or extremum, which are essential for solving real-world problems linked to calculus optimization.

Power Rule

The power rule is a simple yet powerful tool used in calculus to find the derivative of functions that are in the form of a power. In mathematical terms, if you have a function of the form \(x^n\), its derivative is \(nx^{n-1}\). Applying the power rule helps us quickly calculate the derivative of parts of functions that have powers, like \(kr^2\) in our velocity function.
In the exercise, the power rule was applied to the term \(kr^2\) to obtain its derivative with respect to \(r\), resulting in \(2kr\). This is a crucial step in simplifying the overall derivative and helps in finding the critical points.

Product Rule

The product rule is used when differentiating functions that are the product of two or more functions. It's another essential derivative technique and is defined mathematically as the derivative of \(u \, v\) is given by \(u'v + uv'\), where \(u\) and \(v\) are functions of the same variable.
In our problem, \(v(r) = kr^2(R-r)\) involves two functions multiplied together: \(u = kr^2\) and \(v = R-r\). The product rule is used to correctly differentiate the entire expression. This involves taking the derivative of each part, multiplying them appropriately, and then combining the results to find the full derivative. These steps lead to correctly establishing the derivative needed to determine the critical points, making the product rule indispensable in solving complex expressions in calculus optimization.