For a fish swimming at a speed v relative to the water, the energy expenditure per unit time is...

Question:

For a fish swimming at a speed {eq}v {/eq} relative to the water, the energy expenditure per unit time is proportional to {eq}v^{2} {/eq}. 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 with speed {eq}s {/eq} (where {eq}s < v {/eq}), then the time required to swim a distance {eq}L {/eq} is {eq}\frac{L}{v - s} {/eq}, and the total energy {eq}E {/eq} required to swim the distance is given by

$$E(v) = aL \frac{v^{2}}{v - s}, \text{ for } v > s.$$

Here {eq}a, \, s {/eq} and {eq}L {/eq} are positive constants. Find the critical points of {eq}E(v) {/eq} in terms of these constants.

Critical Points:

The critical points on a graph are those places where the derivative is either equal to 0 or undefined. They are critical because something interesting must be happening there: we could possibly have a local extremum or some asymptotic behavior.

We need to know where the derivative is 0 or undefined. We have

{eq}\begin{align*} E' &= \frac{d}{dv} \left( aL\left(\frac{v^{2}}{v-s}\right) \right) \\ &= aL \left( \frac{2v(v-s) - v^2(1)}{(v-s)^2} \right) \\ &= \frac{aLv(v-2s)}{(v-s)^2} \end{align*} {/eq}

When {eq}E' {/eq} is undefined we have

{eq}\begin{align*} (v-s)^2 &= 0 \\ v &= s \end{align*} {/eq}

But since we have {eq}v > s {/eq}, it is outside of our range of acceptable values, and does not correspond to a critical point.

And when {eq}E ' = 0 {/eq}, we have

{eq}\begin{align*} \frac{aLv(v-2s)}{(v-s)^2} &= 0 \\ v-2s &= 0 \\ v &= 2s \end{align*} {/eq}

And then

{eq}\begin{align*} E(2s) &= aL\left(\frac{(2s)^{2}}{2s-s}\right) \\ &= \frac{4aLs^2}s \\ &= 4aLs \end{align*} {/eq}

where we could carry out the last step since {eq}s > 0 {/eq}. Note that this is the minimum energy expenditure and the minimizing speed is {eq}v = 2s {/eq}. The only critical point is

{eq}\begin{align*} (2s, 4aLs) \end{align*} {/eq}