Problem 52 - Algebra and Trigonometry

Question

Migrating Fish A fish swims at a speed \(v\) relative to the water, against a current of 5 mifh. Using a mathematical model of energy expenditure, it can be shown that the total energy \(E\) required to swim a distance of 10 \(\mathrm{mi}\) is given by $$E(v)=2.73 v^{3} \frac{10}{v-5}$$ Biologists believe that migrating fish try to minimize the total energy required to swim a fixed distance. Find the value of \(v\) that minimizes energy required. NOTE: This result has been verified; migrating fish swim against a current at a speed 50\(\%\) greater than the speed of the current.

Step-by-Step Solution

Answer

The minimum energy is achieved at a swimming speed of 7.5 mifh, which aligns with the biological observation.

Step 1: Understanding the Problem

We are given a formula for the energy expenditure of a fish swimming 10 miles against a current. We need to find the speed, \(v\), that minimizes the energy expenditure function \(E(v) = 2.73 v^3 \frac{10}{v-5}\). Biologists suggest that the optimal speed should be 50% greater than the current speed, which is 5 mifh.

Step 2: Setting Up the Derivative

To find the minimum energy, we need to take the derivative of \(E(v)\) with respect to \(v\). The formula for \(E(v)\) is \(E(v) = 2.73 v^3 \frac{10}{v-5}\). Use the quotient rule, given by \(\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}\), to find \(E'(v)\). Let \(u = 2.73 v^3 \times 10\) and \(w = v - 5\).

Step 3: Calculating the Derivative

Calculate the derivative using the quotient rule:- Find \(u' = \frac{d}{dv}(27.3 v^3) = 81.9 v^2\).- Find \(w' = \frac{d}{dv}(v - 5) = 1\).Using the quotient rule:\[ E'(v) = \frac{(81.9 v^2)(v-5) - (27.3 v^3)(1)}{(v-5)^2} \].

Step 4: Setting the Derivative to Zero

To find the minimum energy, set \(E'(v) = 0\):\[ (81.9 v^2)(v-5) - 27.3 v^3 = 0 \].Factor out \(27.3 v^2\):\[ 27.3 v^2 (3(v-5) - v) = 0 \].Simplify inside the parentheses:\[ 27.3 v^2 (2v - 15) = 0 \].This gives us \(v^2 = 0\) or \(2v - 15 = 0\).

Step 5: Solving for v

Ignore the solution \(v^2 = 0\) because \(v = 0\) is not feasible for swimming. Solve the equation \(2v - 15 = 0\):\[ 2v = 15 \Rightarrow v = 7.5 \].

Step 6: Verifying the Result

Substitute \(v = 7.5\) back into the original function to verify if it minimizes \(E(v)\). Also note that biologically, fish swim at a speed 50% greater than the current of 5 mifh, which means \(v = 1.5 \times 5 = 7.5\). This confirms our solution.

Key Concepts

Mathematical Modeling

Mathematical modeling is a powerful tool used to represent real-world situations with mathematical expressions. This method allows us to predict and analyze complex phenomena in fields like physics, engineering, and biology.
One primary goal of mathematical modeling is to simplify and quantify complex systems. In the case of migrating fish, we use a model to estimate the energy they expend while swimming against a current.
The energy expenditure for a fish swimming 10 miles is modeled by the function \(E(v) = 2.73 v^3 \frac{10}{v-5}\). This expression helps biologists and mathematicians understand how speed affects the fish's energy usage.
A good model captures the essential features of the phenomenon being studied while remaining simple enough for analysis. This fish example incorporates both speed-related energy costs and the impact of the current.

Derivative Calculus

Derivative calculus plays a crucial role in finding optimal solutions by analyzing how functions change. In this exercise, understanding the derivative of the energy function \(E(v)\) with respect to speed \(v\) is essential.
The derivative \(E'(v)\) provides information on the rate of change of energy with respect to speed. Calculating \(E'(v)\) involves using the quotient rule for derivatives, crucial for functions presented as fractions of other functions.
For \(E(v) = 2.73 v^3 \frac{10}{v-5}\), the quotient rule is applied with distinct components: \(u = 2.73 v^3 \times 10\) and \(w = v - 5\).
This derivative helps determine the speed \(v\) at which energy reaches its minimal point by setting \(E'(v) = 0\). This is a typical approach to finding minima or maxima in calculus.

Optimal Speed Calculation

Calculating the optimal speed for energy efficiency involves setting the derivative of the energy function to zero. This step finds the critical points that tell us where energy expenditure is minimized.
We solve the equation \[(81.9 v^2)(v-5) - 27.3 v^3 = 0\] by factoring, resulting in \[27.3 v^2 (2v - 15) = 0\]. Here, \(v^2 = 0\) is discarded as it leads to no motion and does not fulfill the physical requirements.
The other factor, \(2v - 15 = 0\), reveals the solution \(v = 7.5\). This speed is 50% more than the current, aligning with biologists' expectations. Thus, 7.5 mifh is identified as the optimal speed for migrating fish against a 5 mifh current, minimizing energy use.

Biological Application of Mathematics

This exercise is an excellent example of mathematical concepts being applied to solve biological problems. Mathematical modeling and calculus are used to understand and predict the behavior of migrating fish.
By identifying an optimal swimming speed of 7.5 mifh, biologists gain insights into natural behaviors and energy efficiency strategies used by fish.

• Optimal speed is an essential feature of animal behavior, critical for survival and energy conservation.


• Mathematics helps scientists create models that can be tested and verified - like the fish's speed being 50% greater than the current.


• This integration showcases the synergy between mathematics and biology, highlighting how mathematical tools aid in biological investigation.

Understanding such interactions assists in conservation efforts and exploring how animals adapt to their environments.