Problem 69 - Algebra and Trigonometry

Question

Coughing When a foreign object lodged in the trachea (windpipe) forces a person to cough, the diaphragm thrusts upward causing an increase in pressure in the lungs. At the same time, the trachea contracts, causing the expelled air to move faster and increasing the pressure on the foreign object. According to a mathematical model coughing, the velocity \(v\) of the airstream through an average-sized person's trachea is related to the radius \(r\) of the trachea (in centimeters) by the function $$ v(r)=3.2(1-r) r^{2}, \quad \frac{1}{2} \leq r \leq 1 $$ Determine the value of \(r\) for which \(v\) is a maximum.

Step-by-Step Solution

Answer

The value of \( r \) for which \( v \) is maximum is \( r = \frac{2}{3} \).

Step 1: Understanding the Problem

We are given a function \( v(r) = 3.2(1-r)r^2 \) that models the velocity \( v \) of the airstream in a person's trachea. We need to find the value of \( r \) in the interval \( \frac{1}{2} \leq r \leq 1 \) that maximizes this function.

Step 2: Differentiate the Function

To find the value of \( r \) that maximizes the function, we first need to find its derivative. The function given is \( v(r) = 3.2(1-r)r^{2} \). We will apply the product rule and the constant multiple rule to differentiate it:\[ v'(r) = \frac{d}{dr}[3.2(1-r)r^2] = 3.2 \left( \frac{d}{dr}[(1-r)r^2] \right) \]Using the product rule, \( \frac{d}{dr}[(1-r)r^2] = (1-r) \cdot \frac{d}{dr}[r^2] + r^2 \cdot \frac{d}{dr}[1-r] \).Thus:\[ v'(r) = 3.2 \left( (1-r) \cdot 2r + r^2 \cdot (-1) \right) = 3.2 (2r(1-r) - r^2) \] \[ v'(r) = 3.2 (2r - 2r^2 - r^2) = 3.2 (2r - 3r^2) \]\[ v'(r) = 3.2 (2r - 3r^2) = 6.4r - 9.6 r^2 \]

Step 3: Set the Derivative to Zero

To find the critical points where velocity is maximized, solve \( v'(r) = 0 \):\[ 6.4r - 9.6r^2 = 0 \]\[ r(6.4 - 9.6r) = 0 \]The solutions of the equation are \( r = 0 \) or \( 6.4 - 9.6r = 0 \). Solving \( 6.4 - 9.6r = 0 \):\[ 9.6r = 6.4 \]\[ r = \frac{6.4}{9.6} = \frac{2}{3} \]

Step 4: Check the Interval Boundaries

Verify that the critical point found lies within the given interval \( \frac{1}{2} \leq r \leq 1 \). The critical point \( r = \frac{2}{3} \) is within the interval.Also, consider the endpoints of the interval, \( r = \frac{1}{2} \) and \( r = 1 \):- For \( r = \frac{1}{2} \), \( v\left(\frac{1}{2}\right) = 3.2(1-\frac{1}{2})\left(\frac{1}{2}\right)^2 = 3.2(\frac{1}{2})(\frac{1}{4}) = 0.4 \).- For \( r = 1 \), \( v(1) = 3.2(1-1)(1)^2 = 0 \).- For \( r = \frac{2}{3} \), \( v\left(\frac{2}{3}\right) = 3.2\left(1-\frac{2}{3}\right)\left(\frac{2}{3}\right)^2 = 3.2 \times \frac{1}{3} \times \left(\frac{4}{9}\right) = \frac{4}{9} \times 3.2 \times \frac{1}{3} = \frac{12.8}{27} \approx 0.474 \).

Step 5: Determine the Maximum Value

Compare the values of \( v(r) \) at the critical point and the interval endpoints. We found:- \( v\left(\frac{1}{2}\right) = 0.4 \)- \( v\left(\frac{2}{3}\right) \approx 0.474 \)- \( v(1) = 0 \)The maximum velocity occurs at \( r = \frac{2}{3} \).

Key Concepts

Derivative

In calculus, the derivative of a function measures how a function's output value changes as the input changes. It is a fundamental tool for analyzing the behavior of functions. When you take the derivative, you get a new function, called the derivative, which tells you the rate of change at any given point.
For the function given in the exercise, which models the velocity of airflow in the trachea, we take the derivative to understand how changes in the trachea's radius impact airflow rate. The form of the function is gentle enough to allow us to use basic rules of differentiation. When analyzing motion or flow like in this problem, the derivative lets us identify points of maximal or minimal velocity, which are crucial for understanding optimum performance and avoiding errors.

Critical Points

Critical points of a function are the values of the input where the first derivative is zero or undefined. These points are essential because they can indicate where a function reaches its maximum or minimum values.
In the air velocity problem, we found the critical points by setting the derivative equal to zero. Solving this equation allows us to find where the slope of the function becomes zero, indicating a potential maximum or minimum. It's helpful to remember that while critical points are important, they are not guaranteed to be maximum or minimum. Further analysis, such as evaluations at endpoints or applying the second derivative test, is needed to confirm their nature.
For this exercise, finding the critical points within the given interval helps to determine the airflow's behavior within the natural limits defined.

Product Rule

The product rule is a fundamental technique in calculus for differentiating products of two functions. If you have a function that is the product of two separate functions, like the function in the air velocity problem, you use this rule to find the derivative.
Here's how it works: if you have two functions, say, \( u(x) \) and \( v(x) \), and you want the derivative of their product \( u(x) \cdot v(x) \), you need to find the derivative of each function separately and then apply the product rule formula:

• \( \frac{d}{dx}[u(x) \cdot v(x)] = u(x) \cdot v'(x) + v(x) \cdot u'(x) \)

In our exercise, we have used the product rule to differentiate \((1-r)r^2\), focusing on ensuring that the rule correctly multiplies parts of the expression by their respective derivatives.

Interval Analysis

Interval analysis is a useful technique for studying functions over a specified range of values. This concept involves checking the behavior of the function at critical points as well as at the boundaries of the defined interval.
In the exercise, we are tasked to determine where the maximum velocity occurs within the interval \( \frac{1}{2} \leq r \leq 1 \). After finding the critical points using the derivative, we perform interval analysis to confirm whether they lie within the interval and how they compare to endpoint values.

• Begin by checking the critical points discovered.
• Next, evaluate the function at the boundaries of the interval.

By assessing the function values at these points, we can ascertain the true maximum or minimum within the segment of interest. It highlights any differences or peculiarities about performance in restricted domains, offering critical insights in applications like optimizing the velocity in medical scenarios.