For a fish swimming at a speed v relative to the water, the energy expenditure per unit time is proportional to v3. 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&lt;v), then the time required to swim a distance L is Lv−u and the total energy E required to swim the distance is given by E(v)=av3⋅Lv−u where a is the proportionality constant.

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For a fish swimming at a speed v relative to the water, the energy expenditure per unit time is proportional to v3. 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&amp;lt;v), then the time required to swim a distance L is Lv−u and the total energy E required to swim the distance is given by E(v)=av3⋅Lv−u where a is the proportionality constant.

Terry Ray · Accepted Answer

Step 11) THe domain of v is&amp;nbsp;v&amp;gt;u2)&amp;nbsp;limu⇒v+E(v)=∞&amp;nbsp;and&amp;nbsp;limv⇒+∞E(v)=∞The conclusion is that the minimum requirement and the lowest possible amount are the same.We will differentiate E with respect to v, then solve for the local minimum value.&amp;nbsp;dEdv=0dEdv=ddv[av3Lv−u]The Quotient Rule for differentiationd&amp;nbsp;dx&amp;nbsp;[f(x)g(x)]=f′(x)⋅g(x)−f(x)⋅g′(x)[g(x)]2dEdv=(av3L)′(v−u)−(av3L)(v−u)′(v−u)2dEdv=3av3−1L⋅(v−u)−(av3L)(1−0)(v−u)2dEdv=aL[3v2(v−u)−v3](v−u)2dEdv=aLv2[3(v−u)−v](v−u)2dEdv=aLv2[2v−3u](v−u)2dEdv=0⇒2v−3u=0⇒v=32uE is minimized when&amp;nbsp;v=32u

Medicim6 · Accepted Answer

Step 1Determine the value v that minimizes E.E′(v)=(v−u)⋅3v2aL−av3L⋅1(v−u)2=aLv2(3(v−u)−v)(v−u)2=aLv2(2v−3u)(v−u)2Note that&amp;nbsp;E′(v)&amp;nbsp;is undefined if&amp;nbsp;v=u. SoE′(v)=0when&amp;nbsp;v=0&amp;nbsp;&amp;nbsp;or&amp;nbsp;&amp;nbsp;v=3u2Step 2Here, v = 0 is out of the domian because it implies that the fish is not swimming at all, so the onlycritical number is&amp;nbsp;v=3u2E&amp;nbsp;″(v)=(v−u)2(aLv2⋅2+(2v−3u)⋅2aLv)−aLv2(2v−3u)⋅2(v−u)⋅1((v−u)2)2=(v−u)2(6aLv2−6aLvu)−2aLv2(2u−3v)(v−u)(v−u)4=(v−u)2(v−u)(6aLv)−2aLv2(2u−3v)(v−u)(v−u)4=(v−u)2⋅6aLv−2aLv2(2v−3u)(v−u)3=(v2−2vu+u2)⋅6aLv−4aLv3+6aLv2u(v−u)3&amp;nbsp;

For a fish swimming at a speed v relative to the water, the energy exp

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