Chapter 4.6, Problem 32E

i.

To determine

To calculate: The value of $v$ that minimizes of $E$ 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 the current $u\left(u<v\right)$ , then the time required to swim a distance $L$ is $\raisebox{1ex}{$L$}\!\left/ \!\raisebox{-1ex}{$\left(v-u\right)$}\right.$ and the total energy $E$ required to swim the distance is given by

  $E\left(v\right)=a{v}^3·\frac{L}{v-u}$

Where $a$ is proportionality constant

Answer to Problem 32E

The value of $v$ that minimizes of $E$ for a fish swimming at a speed $v$ relative to the water is $\frac{3u}{2}$

Explanation of Solution

Given information:

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 the current $u\left(u<v\right)$ , then the time required to swim a distance $L$ is $\raisebox{1ex}{$L$}\!\left/ \!\raisebox{-1ex}{$\left(v-u\right)$}\right.$ and the total energy $E$ required to swim the distance is given by

  $E\left(v\right)=a{v}^3·\frac{L}{v-u}$

Where $a$ is proportionality constant

.

.

Formula used:

Division Rule: If the two functions $f\left(x\right)\text{and }g\left(x\right)$ are differentiable then the quotient is differentiable and,

  ${\left(\frac{f}{g}\right)}^{\prime }=\frac{g{f}^1-f{g}^{\prime }}{g^2}$

Product rule: If the two function $f\left(x\right)\text{ and }g\left(x\right)$ are differentiable then the product is differentiable and ,

  ${\left(fg\right)}^{\prime }=f{g}^{\prime }+g{f}^{\prime }$

And

Let $f$ be a differentiable function defined on an interval $I$ and let $a\in I$ .

Then

1. $x=a$ is a point of local maximum value of $f$, if
1. $f^{\prime }(a)=0$ and
2. $f^{\prime }(x)$ changes sign from positive to negative as $x$ increases through $a$ , i.e. if $f^{\prime }(x)>0$ at every point sufficiently close to and to the left of $a$ , and $f^{\prime }(x)<0$ at every point sufficiently close to and to the right of $a$ , then $a$ is a point of local maxima
2. $x=a$ is a point of local maximum value of $f$, if
1. $f^{\prime }(a)=0$ and
2. $f^{\prime }(x)$ changes sign from negative to positive as $x$ increases through $a$ , i.e. if $f^{\prime }(x)<0$ at every point sufficiently close to and to the left of $a$ , and at $f^{\prime }(x)>0$ every point sufficiently close to and to the right of $a$ , then $a$ is a point of local minima.
3. $f^{\prime }(a)=0$ and If $f^{\prime }(x)$ does not change sign as $x$ increases through $a$ , then $a$ is neither a point of local maxima nor a point of local minima.
4. • If $f^{″}(a)>0$ then $f$ has a local minimum at $x=a$
   • If $f^{″}(a)<0$ then $f$ has a local maximum at $x=a$

Calculation:

As per the given problem

Draw the diagram of 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 the current $u\left(u<v\right)$ , then the time required to swim a distance $L$ is $\raisebox{1ex}{$L$}\!\left/ \!\raisebox{-1ex}{$\left(v-u\right)$}\right.$ and the total energy $E$ required to swim the distance is given by

  $E\left(v\right)=a{v}^3·\frac{L}{v-u}$

Where $a$ is proportionality constant

The total energy $E$ required to swim the distance is $E\left(v\right)=a{v}^3·\frac{L}{v-u}$

Recall that,

Let $f$ be a differentiable function defined on an interval $I$ and let $a\in I$ .

Then

1. $x=a$ is a point of local maximum value of $f$, if
1. $f^{\prime }(a)=0$ and
2. $f^{\prime }(x)$ changes sign from positive to negative as $x$ increases through $a$ , i.e. if $f^{\prime }(x)>0$ at every point sufficiently close to and to the left of $a$ , and $f^{\prime }(x)<0$ at every point sufficiently close to and to the right of $a$ , then $a$ is a point of local maxima
2. $x=a$ is a point of local maximum value of $f$, if
1. $f^{\prime }(a)=0$ and
2. $f^{\prime }(x)$ changes sign from negative to positive as $x$ increases through $a$ , i.e. if $f^{\prime }(x)<0$ at every point sufficiently close to and to the left of $a$ , and at every point sufficiently close to and to the right of $a$ , then $a$ is a point of local minima.
3. $f^{\prime }(a)=0$ and If $f^{\prime }(x)$ does not change sign as $x$ increases through $a$ , then $a$ is neither a point of local maxima nor a point of local minima.
4. • If $f^{″}(a)>0$ then $f$ has a local minimum at $x=a$
   • If $f^{″}(a)<0$ then $f$ has a local maximum at $x=a$

Differentiate on both sides,

  $\frac{d}{dv}E\left(v\right)=\frac{d}{dv}\left\{a{v}^3·\frac{L}{v-u}\right\}$

Recall that,

Division Rule: If the two functions $f\left(x\right)\text{and }g\left(x\right)$ are differentiable then the quotient is differentiable and,

  ${\left(\frac{f}{g}\right)}^{\prime }=\frac{g{f}^1-f{g}^{\prime }}{g^2}$

  $\begin{array}{l}\frac{d}{dv}E\left(v\right)=\frac{d}{dv}\left\{a{v}^3·\frac{L}{v-u}\right\}\\ =\frac{\left(v-u\right)\frac{d}{dv}a{v}^{3}L-a{v}^{3}L\frac{d}{dv}\left(v-u\right)}{{\left(v-u\right)}^2}\\ =\frac{3\left(v-u\right)a{v}^{2}L-a{v}^{3}L}{{\left(v-u\right)}^2}\\ =\frac{3a{v}^{3}L-3au{v}^{2}L-a{v}^{3}L}{{\left(v-u\right)}^2}\\ =\frac{2a{v}^{3}L-3au{v}^{2}L}{{\left(v-u\right)}^2}\\ =\frac{a{v}^{2}L\left(2v-3u\right)}{{\left(v-u\right)}^2}\mathrm{......}\left(2\right)\end{array}$

For maximum and minimum $\frac{d}{dv}E\left(v\right)=0$ , and simplified

  $\begin{array}{l}\frac{2a{v}^{3}L-3au{v}^{2}L}{{\left(v-u\right)}^2}=0\\ \Rightarrow 2a{v}^{3}L-3au{v}^{2}L=0\\ \Rightarrow 2a{v}^{3}L=3au{v}^{2}L\\ \Rightarrow v=\frac{3u}{2}\end{array}$

Differentiate equation $\left(2\right)$ with respect to $v$

  $\frac{d^2}{d{v}^2}E\left(v\right)=\frac{d}{dv}\left\{\frac{a{v}^{2}L\left(2v-3u\right)}{{\left(v-u\right)}^2}\right\}$

Recall that, Division Rule: If the two functions $f\left(x\right)\text{and }g\left(x\right)$ are differentiable then the quotient is differentiable and,

  ${\left(\frac{f}{g}\right)}^{\prime }=\frac{g{f}^1-f{g}^{\prime }}{g^2}$

  $=\frac{{\left(v-u\right)}^2\frac{d}{dv}\left\{a{v}^{2}L\left(2v-3u\right)\right\}-\left\{a{v}^{2}L\left(2v-3u\right)\right\}\frac{d}{dv}{\left(v-u\right)}^2}{{\left\{\left({\left(v-u\right)}^2\right)\right\}}^2}$

Recall that, Product rule: If the two function $f\left(x\right)\text{ and }g\left(x\right)$ are differentiable then the product is differentiable and ,

  ${\left(fg\right)}^{\prime }=f{g}^{\prime }+g{f}^{\prime }$

  $\begin{array}{l}=\frac{{\left(v-u\right)}^2\frac{d}{dv}\left\{a{v}^{2}L\left(2v-3u\right)\right\}-\left\{a{v}^{2}L\left(2v-3u\right)\right\}\frac{d}{dv}{\left(v-u\right)}^2}{{\left\{{\left(v-u\right)}^2\right\}}^2}\\ =\frac{{\left(v-u\right)}^2\left\{a{v}^{2}L\frac{d}{dv}\left(2v-3u\right)+\left(2v-3u\right)\frac{d}{dv}a{v}^{2}L\right\}-2\left\{a{v}^{2}L\left(2v-3u\right)\right\}\left(v-u\right)}{{\left(v-u\right)}^4}\\ =\frac{{\left(v-u\right)}^2\left\{2a{v}^{2}L+2avL\left(2v-3u\right)\right\}-2\left\{a{v}^{2}L\left(2v-3u\right)\right\}\left(v-u\right)}{{\left(v-u\right)}^4}\\ =\frac{\left(v-u\right)\left\{2a{v}^{2}L+2avL\left(2v-3u\right)\right\}-2\left\{a{v}^{2}L\left(2v-3u\right)\right\}}{{\left(v-u\right)}^3}\end{array}$

Solve further to get,

  $=\frac{2avL\left(v^2-3vu+3u^2\right)}{{\left(v-u\right)}^3}$

Substitute $v=\frac{3}{2}u$ and solve,

  $\begin{array}{l}\frac{d^2}{d{v}^2}E\left(\frac{3}{2}u\right)=\frac{2avL\left\{{\left(\frac{3}{2}u\right)}^2-3\times \frac{3}{2}{u}^2+3u^2\right\}}{{\left(\frac{3}{2}u-u\right)}^3}\\ =\frac{3aLu\left(\frac{3}{4}{u}^2\right)}{\frac{u^3}{8}}\\ =18aL\end{array}$

Therefore,

  $\frac{d^2}{d{v}^2}E\left(\frac{3}{2}u\right)=18aL\ge 0$

Because $L>0\text{and }a>0$

So by the second derivatives test $v=\frac{3}{2}u$ gives a local minimum for the energy.

Conclusion:

Thus the value of $v$ that minimizes of $E$ for a fish swimming at a speed $v$ relative to the water is $\frac{3u}{2}$

ii.

To determine

Sketch the graph of $E$ .

Answer to Problem 32E

The graph of $E$

  

Explanation of Solution

Given information:

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 the current $u\left(u<v\right)$ , then the time required to swim a distance $L$ is $\raisebox{1ex}{$L$}\!\left/ \!\raisebox{-1ex}{$\left(v-u\right)$}\right.$ and the total energy $E$ required to swim the distance is given by

  $E\left(v\right)=a{v}^3·\frac{L}{v-u}$

Where $a$ is proportionality constant

Formula Used:

Division Rule: If the two functions $f\left(x\right)\text{and }g\left(x\right)$ are differentiable then the quotient is differentiable and,

  ${\left(\frac{f}{g}\right)}^{\prime }=\frac{g{f}^1-f{g}^{\prime }}{g^2}$

Product rule: If the two function $f\left(x\right)\text{ and }g\left(x\right)$ are differentiable then the product is differentiable and ,

  ${\left(fg\right)}^{\prime }=f{g}^{\prime }+g{f}^{\prime }$

And

Let $f$ be a differentiable function defined on an interval $I$ and let $a\in I$ .

Then

1. $x=a$ is a point of local maximum value of $f$, if
1. $f^{\prime }(a)=0$ and
2. $f^{\prime }(x)$ changes sign from positive to negative as $x$ increases through $a$ , i.e. if $f^{\prime }(x)>0$ at every point sufficiently close to and to the left of $a$ , and $f^{\prime }(x)<0$ at every point sufficiently close to and to the right of $a$ , then $a$ is a point of local maxima
2. $x=a$ is a point of local maximum value of $f$, if
1. $f^{\prime }(a)=0$ and
2. $f^{\prime }(x)$ changes sign from negative to positive as $x$ increases through $a$ , i.e. if $f^{\prime }(x)<0$ at every point sufficiently close to and to the left of $a$ , and at $f^{\prime }(x)>0$ every point sufficiently close to and to the right of $a$ , then $a$ is a point of local minima.
3. $f^{\prime }(a)=0$ and If $f^{\prime }(x)$ does not change sign as $x$ increases through $a$ , then $a$ is neither a point of local maxima nor a point of local minima.

Calculation:

As per the given problem

Draw the diagram of 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 the current $u\left(u<v\right)$ , then the time required to swim a distance $L$ is $\raisebox{1ex}{$L$}\!\left/ \!\raisebox{-1ex}{$\left(v-u\right)$}\right.$ and the total energy $E$ required to swim the distance is given by

  $E\left(v\right)=a{v}^3·\frac{L}{v-u}$

Where $a$ is proportionality constant

The total energy $E$ required to swim the distance is $E\left(v\right)=a{v}^3·\frac{L}{v-u}$

Recall that,

Let $f$ be a differentiable function defined on an interval $I$ and let $a\in I$ .

Then

1. $x=a$ is a point of local maximum value of $f$, if
1. $f^{\prime }(a)=0$ and
2. $f^{\prime }(x)$ changes sign from positive to negative as $x$ increases through $a$ , i.e. if $f^{\prime }(x)>0$ at every point sufficiently close to and to the left of $a$ , and $f^{\prime }(x)<0$ at every point sufficiently close to and to the right of $a$ , then $a$ is a point of local maxima
2. $x=a$ is a point of local maximum value of $f$, if
1. $f^{\prime }(a)=0$ and
2. $f^{\prime }(x)$ changes sign from negative to positive as $x$ increases through $a$ , i.e. if $f^{\prime }(x)<0$ at every point sufficiently close to and to the left of $a$ , and at every point sufficiently close to and to the right of $a$ , then $a$ is a point of local minima.
3. $f^{\prime }(a)=0$ and If $f^{\prime }(x)$ does not change sign as $x$ increases through $a$ , then $a$ is neither a point of local maxima nor a point of local minima.

Differentiate on both sides,

  $\frac{d}{dv}E\left(v\right)=\frac{d}{dv}\left\{a{v}^3·\frac{L}{v-u}\right\}$

Recall that,

Division Rule: If the two functions $f\left(x\right)\text{and }g\left(x\right)$ are differentiable then the quotient is differentiable and,

  ${\left(\frac{f}{g}\right)}^{\prime }=\frac{g{f}^1-f{g}^{\prime }}{g^2}$

  $\begin{array}{l}\frac{d}{dv}E\left(v\right)=\frac{d}{dv}\left\{a{v}^3·\frac{L}{v-u}\right\}\\ =\frac{\left(v-u\right)\frac{d}{dv}a{v}^{3}L-a{v}^{3}L\frac{d}{dv}\left(v-u\right)}{{\left(v-u\right)}^2}\\ =\frac{3\left(v-u\right)a{v}^{2}L-a{v}^{3}L}{{\left(v-u\right)}^2}\\ =\frac{3a{v}^{3}L-3au{v}^{2}L-a{v}^{3}L}{{\left(v-u\right)}^2}\\ =\frac{2a{v}^{3}L-3au{v}^{2}L}{{\left(v-u\right)}^2}\\ =\frac{a{v}^{2}L\left(2v-3u\right)}{{\left(v-u\right)}^2}\mathrm{......}\left(2\right)\end{array}$

For maximum and minimum $\frac{d}{dv}E\left(v\right)=0$ , and simplified

  $\begin{array}{l}\frac{2a{v}^{3}L-3au{v}^{2}L}{{\left(v-u\right)}^2}=0\\ \Rightarrow 2a{v}^{3}L-3au{v}^{2}L=0\\ \Rightarrow 2a{v}^{3}L=3au{v}^{2}L\\ \Rightarrow v=\frac{3u}{2}\end{array}$

Differentiate equation $\left(2\right)$ with respect to $v$

  $\frac{d^2}{d{v}^2}E\left(v\right)=\frac{d}{dv}\left\{\frac{a{v}^{2}L\left(2v-3u\right)}{{\left(v-u\right)}^2}\right\}$

Recall that, Division Rule: If the two functions $f\left(x\right)\text{and }g\left(x\right)$ are differentiable then the quotient is differentiable and,

  ${\left(\frac{f}{g}\right)}^{\prime }=\frac{g{f}^1-f{g}^{\prime }}{g^2}$

  $=\frac{{\left(v-u\right)}^2\frac{d}{dv}\left\{a{v}^{2}L\left(2v-3u\right)\right\}-\left\{a{v}^{2}L\left(2v-3u\right)\right\}\frac{d}{dv}{\left(v-u\right)}^2}{{\left\{\left({\left(v-u\right)}^2\right)\right\}}^2}$

Recall that, Product rule: If the two function $f\left(x\right)\text{ and }g\left(x\right)$ are differentiable then the product is differentiable and ,

  ${\left(fg\right)}^{\prime }=f{g}^{\prime }+g{f}^{\prime }$

  $\begin{array}{l}=\frac{{\left(v-u\right)}^2\frac{d}{dv}\left\{a{v}^{2}L\left(2v-3u\right)\right\}-\left\{a{v}^{2}L\left(2v-3u\right)\right\}\frac{d}{dv}{\left(v-u\right)}^2}{{\left\{{\left(v-u\right)}^2\right\}}^2}\\ =\frac{{\left(v-u\right)}^2\left\{a{v}^{2}L\frac{d}{dv}\left(2v-3u\right)+\left(2v-3u\right)\frac{d}{dv}a{v}^{2}L\right\}-2\left\{a{v}^{2}L\left(2v-3u\right)\right\}\left(v-u\right)}{{\left(v-u\right)}^4}\\ =\frac{{\left(v-u\right)}^2\left\{2a{v}^{2}L+2avL\left(2v-3u\right)\right\}-2\left\{a{v}^{2}L\left(2v-3u\right)\right\}\left(v-u\right)}{{\left(v-u\right)}^4}\\ =\frac{\left(v-u\right)\left\{2a{v}^{2}L+2avL\left(2v-3u\right)\right\}-2\left\{a{v}^{2}L\left(2v-3u\right)\right\}}{{\left(v-u\right)}^3}\end{array}$

Solve further to get,

  $=\frac{2avL\left(v^2-3vu+3u^2\right)}{{\left(v-u\right)}^3}$

Substitute $v=\frac{3}{2}u$ and solve,

  $\begin{array}{l}\frac{d^2}{d{v}^2}E\left(\frac{3}{2}u\right)=\frac{2avL\left\{{\left(\frac{3}{2}u\right)}^2-3\times \frac{3}{2}{u}^2+3u^2\right\}}{{\left(\frac{3}{2}u-u\right)}^3}\\ =\frac{3aLu\left(\frac{3}{4}{u}^2\right)}{\frac{u^3}{8}}\\ =18aL\end{array}$

Therefore,

  $\frac{d^2}{d{v}^2}E\left(\frac{3}{2}u\right)=18aL\ge 0$

Because $L>0\text{and }a>0$

So by the second derivatives test $v=\frac{3}{2}u$ gives a local minimum for the energy.

Conclusion:

Thus the graph of $E$