Problem 66 - Algebra and Trigonometry

Question

Migrating Fish A fish swims at a speed \(v\) relative to the water, against a current of 5 mi/h. 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 fish minimizes energy consumption by swimming at 7.5 mi/h.

Step 1: Understand the Problem

We are given the energy function \( E(v) = 2.73 v^{3} \frac{10}{v-5} \) where \( v \) is the speed of the fish relative to the water in miles per hour. We need to find the value of \( v \) that minimizes this energy function.

Step 2: Differentiate the Energy Function

To find the minimum energy expenditure \( E(v) \), we need to take the derivative of \( E(v) \) with respect to \( v \). Use the quotient rule and the chain rule to find \( E'(v) \).

Step 3: Apply the Quotient Rule

The quotient rule states that if you have a function \( f(x) = \frac{g(x)}{h(x)} \), its derivative is \( f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{[h(x)]^2} \). Apply this rule to \( E(v) \) where \( g(v) = 2.73v^3 \) and \( h(v) = v - 5 \).

Step 4: Calculate Derivatives

Compute \( g'(v) = 8.19v^2 \) and \( h'(v) = 1 \). Substitute these derivatives into the quotient rule formula to find \( E'(v) \).

Step 5: Set the Derivative to Zero

Set \( E'(v) = 0 \) and solve for \( v \) to find the critical points. This will help identify potential minimum points of the function.

Step 6: Solve the Equation

Solve the equation formed in Step 5 to find the critical point that minimizes the energy function \( E(v) \).

Step 7: Verify the Result

Check the critical point to ensure it is indeed a minimum. Use the second derivative test or analyze the sign changes of \( E'(v) \) around the critical point.

Step 8: Relate Result to Problem Context

The critical point found where energy is minimized should align with the problem's context, which is 50% greater than the current's speed of 5 mi/h. Calculate \( v = 1.5 \times 5 = 7.5 \) mi/h to verify.

Key Concepts

Energy Minimization

When analyzing problems like the migrating fish scenario, energy minimization is a key concept. The goal is to find the speed that allows the fish to use the least amount of energy while swimming against a current. For this problem, the energy function is given as \( E(v) = 2.73 v^{3} \frac{10}{v-5} \). This function represents the total energy required to swim a specific distance. The idea of energy minimization involves calculus techniques to find the value of \( v \) that makes this expression as small as possible.
By achieving energy minimization, organisms like fish can travel efficiently, which is crucial for survival during processes like migration. This optimization is why understanding how to minimize energy effectively is an important application of calculus.

Quotient Rule

To find where the energy function is minimized, we use the derivative, and to calculate this derivative, the quotient rule comes into play. The quotient rule is vital when dealing with ratios of functions, specifically when a function is expressed as a fraction. It states that if you have two functions, \( g(v) \) and \( h(v) \), then:

• \( f'(v) = \frac{g'(v)h(v) - g(v)h'(v)}{[h(v)]^2} \)

Apply this formula to our energy equation where \( g(v) = 2.73v^3 \) and \( h(v) = v - 5 \). First, compute the derivative of each part: \( g'(v) \) and \( h'(v) \). Then, plug these back into the quotient rule to find the derivative of the energy function.
This helps in setting up the process for identifying critical points where the function might reach a minimum.

Derivative Calculation

Once you have the formula set up with the quotient rule, the next step is to actually compute the derivatives. This involves calculating \( g'(v) \), which is the derivative of \( 2.73v^3 \). This evaluates to \( 8.19v^2 \). Similarly, \( h'(v) = 1 \) since the derivative of \( v - 5 \) is just 1. With these derivatives calculated, substitute back into the quotient rule equation from the previous concept.
The computation yields the expression for \( E'(v) \), which will allow us to determine where potential minima or maxima occur by setting it to zero and solving for the variable \( v \).
This step crucially sets the stage for finding critical points, which is our next focus area.

Critical Points

Critical points are the values where the derivative \( E'(v) \) equals zero or is undefined, and they play a central role in finding minima or maxima. Set the derivative of the energy function to zero to find these points. This model represents possible speeds where the fish might use minimum energy to swim. Solve the equation \( E'(v) = 0 \) to pinpoint these positions on the energy curve.
Once these points are identified, they must be analyzed further—these critical points are candidates for the minima or maxima, but not all are guaranteed to result in the desired minimum. Understanding these points helps in optimizing the fish's energy expenditure.

Second Derivative Test

After identifying the critical points, the next step is to decide which of these points actually represent minima. This is where the second derivative test comes in. The second derivative test tells us about the concavity of the function at that critical point:

• If \( E''(v) > 0 \), the function has a local minimum at that point.
• If \( E''(v) < 0 \), the function has a local maximum.
• If \( E''(v) = 0 \), the test is inconclusive.

Calculate the second derivative of the energy function, \( E''(v) \), to apply this test. For our fish problem, once the second derivative confirms a minimum, we find that the optimal speed of the fish coincides with this minimum energy point.
In our analysis, the critical point aligns with swimming at a speed that is 50% greater than the current, ensuring the fish uses as little energy as possible. This verification step ensures the calculus models match observations in biological studies.