Question

for a fish swimming at a speed v relative to the water, the energy expenditure per unit time is proportional to ( v^{3} ). it is believed that migrating fish try to minimize the total energy required to swim a fixed distance. if the fish are swimming against a current u (( u < v )), then the time required to swim a distance l is ( \frac{l}{(v - u)} ), and the total energy e required to swim the distance is given by ( e(v)=\frac{a v^{3} cdot l}{v - u} ), where a is the proportionality constant. (a) determine the value of v that minimizes e. note: this result has been verified experimentally; migrating fish swim against a current at a speed 50% greater than the current speed. v = (b) sketch the graph of e.

Explanation:

Step1: Find the derivative of \(E(v)\)

Given \(E(v)=a v^{3} \cdot \frac{L}{v - u}\), where \(a\) and \(L\) are constants. Using the quotient rule \((\frac{f}{g})^\prime=\frac{f^\prime g - fg^\prime}{g^{2}}\), with \(f = aLv^{3}\), \(f^\prime=3aLv^{2}\), and \(g = v - u\), \(g^\prime = 1\).

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Step2: Set \(E^\prime(v)=0\)

Set \(\frac{aLv^{2}(2v - 3u)}{(v - u)^{2}} = 0\). Since \(a
eq0\), \(L
eq0\), and \(v>u\) (because \(v - u>0\) for the time formula \(\frac{L}{v - u}\) to be meaningful in the context of the problem), \(v^{2}>0\) and \((v - u)^{2}>0\). Then \(2v-3u = 0\).

Solve \(2v-3u = 0\) for \(v\):

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Step3: Analyze the sign of \(E^\prime(v)\)

• When \(u < v<\frac{3}{2}u\), \(2v-3u<0\), so \(E^\prime(v)<0\), and \(E(v)\) is decreasing.
• When \(v>\frac{3}{2}u\), \(2v - 3u>0\), so \(E^\prime(v)>0\), and \(E(v)\) is increasing.

By the first - derivative test, \(E(v)\) has a minimum at \(v = \frac{3}{2}u\).

Answer:

\(v=\frac{3}{2}u\)