Problem 3 - Calculus for the Life Sciences: A Modeling Approach Volume I

Question

Exercise 8.3 .3 Some small song birds intermittently flap their wings and glide with wings folded between flapping sessions. Why? R. M. Alexander \(^{3}\) suggests the following analysis. The power required to propel an airplane at a speed \(u\) is $$ P=A u^{3}+B L^{2} / u $$ where \(A\) and \(B\) are constants specific to the airplane and \(L\) is the upward force that lifts the plane. \(A u^{3}\) represents the drag on the airplane due largely to the air striking the front of the craft. a. For what speed is the required power the smallest? b. The energy required to propel the airplane is \(E=P / u\). For what speed is the energy required to propel the plane the smallest. c. How does the speed of minimum energy compare with the speed of minimum power? d. Suppose a bird has drag coefficient \(A_{b}\) with wings folded and \(A_{b}+A_{w}\) with wings extended and flapping, and let \(x\) be the fraction of time the bird spends flapping its wings. Suppose that the speed of the bird while flapping its wings is the same as the speed when the wings are folded and that all of the lift is provided when the wings are flapping. The lift over one complete cycle should be $$(1-x) L_{\text {folded }}+x L_{\text {flapping }}=x L_{\text {flapping }}=m g$$ where \(m\) is the mass of the bird. Then the power required while flapping is $$ P_{\text {flapping }}=\left(A_{b}+A_{w}\right) u^{3}+B\left(\frac{m g}{x}\right)^{2} \frac{1}{u} $$ Write an expression for \(P_{\text {folded }}\). e. The average power over a whole cycle should be $$ \bar{P}=(1-x) P_{\text {folded }}+x P_{\text {flapping }}=A_{b} u^{3}+x A_{w} u^{3}+B \frac{m^{2} g^{2}}{x u} $$ Find the value of \(x\) for which the average power over the whole cycle is minimum. f. The average energy over a whole cycle is \(\bar{E}=\bar{P} / u\). For what value of \(x\) is the average energy a minimum? This problem is continued in Exercise 13.2 .12 Alexander further notes that it may be necessary to consider also the efficiency of muscle contraction at different flapping rates.

Step-by-Step Solution

Answer

The speed for minimum power and energy are different. Optimal x for minimizing average power or energy is \(x = \sqrt{\frac{B m^2 g^2}{A_w u^4}}\).

Step 1: Understanding and Setting Up the Problem

The exercises given involve both calculating and optimizing the power and energy required for flight. We need to analyze the power and energy requirements using given formulas and then find parameters for which these requirements are minimized.

Step 2: Finding Speed of Minimum Power (Part a)

We start by setting up the power equation: \(P = A u^3 + \frac{B L^2}{u}\). To find the speed \(u\) for minimum power, we take the derivative of \(P\) with respect to \(u\), set the derivative equal to zero, and solve for \(u\). This results in \(3 A u^4 = B L^2\), yielding \(u = \left(\frac{B L^2}{3 A}\right)^{1/4}\).

Step 3: Finding Speed of Minimum Energy (Part b)

Energy is given by \(E = \frac{P}{u} = A u^2 + \frac{B L^2}{u^2}\). To find the speed for minimum energy, take the derivative of \(E\) with respect to \(u\), set it to zero, and solve for \(u\). We find \(2 A u^3 = B L^2\), leading to \(u = \left(\frac{B L^2}{2 A}\right)^{1/3}\).

Step 4: Compare Speeds for Minimum Power and Energy (Part c)

The two speeds are \(u_{\text{min power}} = \left(\frac{B L^2}{3 A}\right)^{1/4}\) and \(u_{\text{min energy}} = \left(\frac{B L^2}{2 A}\right)^{1/3}\). These are different due to different powers in the formulas, implying the speed for minimum energy is generally not the same as for minimum power.

Step 5: Expression for Folded Wing Power (Part d)

Given for flapping \(P_{\text{flapping}}\), the power required when folded (\(P_{\text{folded}}\)) is: \(P_{\text{folded}} = A_{b} u^3\). This accounts for the drag coefficient without lifting force as lift is only provided while flapping.

Step 6: Minimizing Average Power over a Cycle (Part e)

Use the expression \(\bar{P} = A_{b} u^3 + x A_{w} u^3 + B \frac{m^2 g^2}{x u}\). To minimize, take the derivative with respect to \(x\), set it to zero, and solve. We derive \(x = \sqrt{\frac{B m^2 g^2}{A_w u^4}}\) for minimum average power.

Step 7: Minimizing Average Energy over a Cycle (Part f)

Average energy \(\bar{E}\) is \(\bar{P}/u\). Minimizing \(\bar{P}\) directly gives the minimization for \(\bar{E}\), for which \(x = \sqrt{\frac{B m^2 g^2}{A_w u^4}}\) based on the power expression discussed above.

Key Concepts

Power Minimization

When flying, both birds and airplanes aim to use energy in the most efficient way possible. This involves minimizing the power required for travel. Power, in simple terms, is the rate at which energy is used. Therefore, less power consumption means more efficient flying. In mathematical terms, the power required to fly at a speed \(u\) can be given by the equation \(P = A u^3 + \frac{B L^2}{u}\). Here, the term \(A u^3\) accounts for the drag force caused mainly by air resistance, and \(\frac{B L^2}{u}\) represents lift-related factors.

To find the speed where this power \(P\) is minimized, a derivative optimization approach is used. This involves taking the derivative of \(P\) with respect to speed \(u\), setting it to zero, and solving for \(u\). Calculations show that the minimum power occurs at a speed given by \(u = \left(\frac{B L^2}{3 A}\right)^{1/4}\). This specific speed is optimal for reducing the energy spent over time.

Understanding this concept is crucial for designing energy-efficient flying patterns and enhances the sustainability of flight operations.

Derivative Optimization

In flight energy optimization, derivatives play a crucial role. Derivative optimization refers to using calculus to find the minimum or maximum points of a function. For instance, in minimizing power, we take the derivative of the power equation \(P = A u^3 + \frac{B L^2}{u}\) with respect to the speed \(u\).

Setting the derivative \(\frac{dP}{du}\) to zero gives us the critical speed where the function reaches a minimum or maximum. Solving this ultimately reveals the optimal speed for minimum power as \(u = \left(\frac{B L^2}{3 A}\right)^{1/4}\).

The process involves:

• Finding the derivative of the function to understand how it changes with speed.
• Setting the derivative to zero to find critical points.
• Determining which of those points minimizes the power.

Using derivatives is a powerful mathematical tool for solving real-world problems like energy optimization in flight.

Drag Coefficient

The drag coefficient is a key factor influencing the efficiency of flight. It quantifies the drag or resistance an object faces as it moves through a fluid, such as air. For birds and planes, minimizing this resistance is essential for energy-efficient travel. In our exercise, the coefficient \(A\) represents this effect in the power equation \(P = A u^3 + \frac{B L^2}{u}\).

A higher drag coefficient means more energy is required to maintain speed, as it indicates greater resistance from the air. Conversely, a lower drag coefficient suggests smoother, more efficient travel. In birds, drag changes when they flap their wings versus when they glide, leading to different energy requirements. The coefficients \(A_b\) and \(A_w\) represent the drag during gliding and flapping phases, respectively.

Reducing drag by designing more aerodynamically efficient shapes and taking advantage of optimal flight conditions can greatly decrease energy consumption.

Minimum Energy Speed

Finding the minimum energy speed is about determining the most efficient speed for flight based on energy instead of power. Energy usage differs in that it is concerned with how much total energy is consumed over a distance, rather than how fast energy is used per time.

The energy required for flying at a speed \(u\) is given by \(E = \frac{P}{u} = A u^2 + \frac{B L^2}{u^2}\). To find the minimum energy speed, we take the derivative of \(E\) with respect to \(u\), set it to zero, and solve for \(u\). This method reveals that the optimal energy speed is \(u = \left(\frac{B L^2}{2 A}\right)^{1/3}\).

This result shows that the speed for minimizing total energy over a cycle of flight is not the same as minimizing power. Instead, it reflects a balance that considers both energy and power use. This distinction is vital for settings where conserving energy over longer distances or times is more relevant than immediate power reduction.