Problem 52 - Calculus with Concepts in Calculus

Question

According to one model of coughing, the flow \(F\) (volume per unit time) of air through the windpipe during a cough is a function of the radius \(r\) of the windpipe, given by $$ F=k\left(r_{0}-r\right) r^{4} \quad \text { for } \frac{1}{2} r_{0} \leq r \leq r_{0} $$ where \(k\) is a positive constant and \(r_{0}\) is the normal (noncoughing) radius. a. Find the value of \(r\) that maximizes the flow \(F\). b. According to the same model, the velocity \(v\) of air through the windpipe during a cough is given by $$ v=\frac{k}{\pi}\left(r_{0}-r\right) r^{2} \quad \text { for } \frac{1}{2} r_{0} \leq r \leq r_{0} $$ Find the value of \(r\) that maximizes the velocity \(v\). c. During a cough the windpipe is constricted. According to parts (a) and (b), is that likely to assist or hinder the cough?

Step-by-Step Solution

Answer

The maximum airflow occurs at \( r = \frac{4}{5}r_0 \), and max velocity at \( r = \frac{r_0}{2} \). Constriction favors velocity.

Step 1: Set up the derivative for maximizing F

Given the flow equation \( F = k(r_0 - r)r^4 \), we need to find the critical points by taking the derivative of \( F \) with respect to \( r \) and setting it to zero. First, expand the function: \( F = kr_0r^4 - kr^5 \). Now, differentiate with respect to \( r \): \( \frac{dF}{dr} = kr_0(4r^3) - k(5r^4) = 4kr_0r^3 - 5kr^4 \). Set the derivative equal to zero for critical points: \( 4kr_0r^3 - 5kr^4 = 0 \).

Step 2: Solve for the critical points of F

From the equation \( 4kr_0r^3 - 5kr^4 = 0 \), factor out \( kr^3 \): \( kr^3(4r_0 - 5r) = 0 \). Since \( kr^3 eq 0 \) (as \( k \) and \( r eq 0 \)), solve \( 4r_0 - 5r = 0 \) to find \( r = \frac{4}{5}r_0 \).

Step 3: Verify the maximum for F

To confirm \( r = \frac{4}{5}r_0 \) provides a maximum, check the second derivative. Differentiate \( \frac{dF}{dr} = 4kr_0r^3 - 5kr^4 \) again: \( \frac{d^2F}{dr^2} = 12kr_0r^2 - 20kr^3 \). Evaluate \( \frac{d^2F}{dr^2} \) at \( r = \frac{4}{5}r_0 \): it results in \( -k \times 20 \left(\frac{4}{5}r_0\right)^3 < 0 \), indicating a local maximum.

Step 4: Set up the derivative for maximizing v

For the velocity function \( v = \frac{k}{\pi}(r_0 - r)r^2 \), differentiate it with respect to \( r \): \( \frac{dv}{dr} = \frac{k}{\pi}[(r_0 - 2r)r^2 + (r_0 - r)2r] = \frac{k}{\pi}(r_0r^2 - 2r^3 + 2r_0r - 2r^2) \). Simplify it: \( \frac{dv}{dr} = \frac{k}{\pi}(r_0r^2 + 2r_0r - 4r^3) \). Set it to zero: \( r_0r^2 + 2r_0r - 4r^3 = 0 \).

Step 5: Solve for the critical points of v

Factor the expression \( r_0r^2 + 2r_0r - 4r^3 = 0 \): \( r(r^2r_0 + 2r_0 - 4r^2) = 0 \). Ignoring the trivial solution \( r = 0 \), solve \( r_0 + 2 - 4r = 0 \) to find \( r = \frac{r_0}{2} \).

Step 6: Verify the maximum for v

Take the second derivative \( \frac{d^2v}{dr^2} = \frac{k}{\pi} [2r_0 - 12r] \). Evaluate it at \( r = \frac{r_0}{2} \): \( \frac{d^2v}{dr^2} = \frac{k}{\pi}(2r_0 - 12\times \frac{r_0}{2}) = -4\frac{kr_0}{\pi} < 0 \), confirming the maximum.

Step 7: Conclusion regarding the effects on cough

According to the model, for airflow (volume) to be maximum, \( r \) should be \( \frac{4}{5}r_0 \), and for maximum velocity, \( r \) should be \( \frac{r_0}{2} \). During a cough, the windpipe constricts toward \( \frac{r_0}{2} \), favoring maximum velocity rather than maximum volume of airflow.

Key Concepts

Coughing Model

The coughing model provides a mathematical representation of how air flows through the windpipe during a cough. This model helps us understand the dynamics of air movement based on changes in the windpipe's radius. The volume flow rate of air during a cough, denoted by \( F \), depends on the radius \( r \) of the windpipe in a special way. The equation \( F = k(r_0 - r)r^4 \) helps demonstrate how the air flow is affected when the windpipe's radius changes. Here, \( r_0 \) represents the windpipe's normal radius when not coughing, and \( k \) is a constant. Adjustments in \( r \) influence the flow rate \( F \), requiring calculus techniques to find the conditions for optimizing this flow.

Maximizing Flow

In calculus, to maximize the air flow \( F \) through the windpipe, we need to locate the value of \( r \) which gives us the highest flow. This process involves finding critical points where the derivative of the flow function equals zero, a foundational step in optimization problems. For \( F = k(r_0 - r)r^4 \), the details come alive as we apply differentiation. We start with the expansion and differentiation, arriving at \( \frac{dF}{dr} = 4kr_0r^3 - 5kr^4 \). By setting this derivative to zero, we determine critical points where \( r \) can potentially yield maximum flow. Solving \( 4kr_0r^3 - 5kr^4 = 0 \) provides a critical value at \( r = \frac{4}{5}r_0 \). To ensure this is a maximum, we use the second derivative test to verify it shows a concave down curve at this point.

Critical Points

Critical points in calculus are where a function's derivative is zero or undefined, indicating potential highs, lows, or inflection points. Identifying critical points is crucial when analyzing functions for optimization. In our coughing model, critical points help us find the optimal windpipe radius to maximize air flow and velocity during a cough. We used the first derivative \( \frac{dF}{dr} = 4kr_0r^3 - 5kr^4 \) to locate critical points where the air flow could be highest. Once identified, we analyze these points to determine their behavior, ensuring they represent maximum or minimum values as required in the problem scenario.

Second Derivative Test

The second derivative test is a powerful tool to determine whether critical points identified via the first derivative lead to a local maximum, minimum, or an inflection point. After obtaining critical points through \( \frac{dF}{dr} = 4kr_0r^3 - 5kr^4 \), we take the second derivative \( \frac{d^2F}{dr^2} = 12kr_0r^2 - 20kr^3 \). For a positive test, this second derivative helps check the concavity of the function at critical points. When \( \frac{d^2F}{dr^2} \) evaluated at \( r = \frac{4}{5}r_0 \) returns a negative value, it indicates that the function curves down at this point, confirming a local maximum. Similarly, this test is crucial for the velocity equation too, verifying the maximal velocity configuration in the airflow dynamics.