Differential Calculus: From Practice to Theory

Suppose we are running an industrial trout fishery. There are a number of questions we’d really like to be able to answer at any given time. How large is the current stock? Do we have too few? Too many? Is our stock of trout in danger of dying off due to under or over population? How frequently can we harvest our stock without endangering it? Even if precise numbers aren’t possible, it would still be useful to have a good, qualitative understanding of the life cycle of our stock of trout so we could make reliable estimates.

To gain such an understanding we will build a model of our trout population as a function of time. As a first approximation assume that we have complete control of the lake where we will hold the trout. We can supply all of the food and space that our trout need to continue reproducing.

Suppose that our lake is stocked with an initial purchase of \(10\) tons of trout. Let \(P(t)\) represent the amount of fish we have, in tons, at time, \(t\text{,}\) in years, (so \(P(0)=10\) tons). We can estimate the rate of growth of our population by casting a net into the same point of the lake each year and weighing the catch. Comparing the weight of the catch from year to year will give a reasonable estimate of the rate at which the trout population grows in one year.

This IVP should look familiar. When we were modeling the growth of a bacteria colony in Example 8.6.1 we got IVP (8.18), which is almost the same equation.‘ The only difference between them is that in IVP (11.1) the coefficient of \(P\) is \(0.1\) and in IVP (8.18) the coefficient of \(P\) is \(0.3\text{.}\)

The graph of \(P(t)\) is shown below. As you can see our trout population is growing exponentially, so it appears that we’re going to get very rich as soon as we sell of all of those fish.

These assumptions are unrealistic in the long run. But in the short run they are not completely crazy.

Recall that in Section 8.6 we described a situation in Lake Erie near Toledo, Ohio when, for a few days, the algae population grew exponentially in much the same manner as that suggested by our trout model. Naturally such growth cannot be sustained. Eventually the algae will consume all of the available nutrients, or will simply run out of space. They will then begin to die off. But our model, simple as it is, gives pretty accurate predictions of the population for small values of \(t\text{,}\) be they a population of bacteria, algae, trout, or even humans.

It would be nice if we could find a formula for \(P(t)\) in terms of \(t\) but this will have to wait for a course in Integral Calculus. Still we would like to provide a reasonable sketch of the graph of \(P = P(t)\text{.}\) Aside from the initial condition, \(P(0) = 10\text{,}\) we don’t know any specific values of \(P(t)\) but we can at least capture the qualitative behavior of the population.

We will obtain a sketch of the graph of \(P(t)\) in the same systematic way as before. We can use the first derivative test to determine when the trout population is increasing or decreasing. But keep in mind that we have \(\dfdx{P}{t}\) in terms of \(P\text{,}\) not \(t\text{,}\) so we won’t be able to say that the population is increasing or decreasing at a particular time. This will change how we interpret the results. For example we can say that when we have \(25\) tons of trout the population is increasing even though we do not know when this happens.

It should be straightforward to see that the POTPs are \(P = 0, 100\text{.}\) Since the context of our problem prohibits \(P \lt 0\) (why?), we will look at intervals \([0,100)\) and \((100, \infty)\) to determine for which values of \(P\) the population is increasing or decreasing. We summarize this analysis in the following table:

We now have enough data on to plot a reasonable sketch of the graph of \(P=P(t)\) but because all of our formulas are in terms of the dependent variable \(P\text{,}\) our transition points and intervals are will all be shown on the vertical axis, not the horizontal as seen below.

Notice that we drew solid horizontal lines at \(P=0\) and \(P= 100\text{.}\) At these population levels \(\dfdx{P}{t}=0\) so, theoretically at least, once the population has achieved either of these levels it will never change again. This is correct when \(P(0)=0\text{.}\) Once all of our fish have died the only way we can get more is to buy them and start again. But for other values of \(P\) it is still not entirely clear what’s happening to our fish population.

It is not possible to maintain a population of trout at a constant value of \(100\) tons. There will always be a little fluctuation above and below. All models are inaccurate in some way. This is one way in which this model is inaccurate.

In the sketch above the dashed line represents the population level where the graph of the population transitions from concave up to concave down. Physically this means that the growth rate of the population transitions from accelerating to decelerating. We can visualize this (below) using slope field technique we saw in Section 10.3.

Starting with an initial value of \(P(0)=10\text{,}\) we can plot the graph of the population and make a prediction of what will happened to the population in the long term. We just can’t tell what “long term” means. That would have to be determined experimentally.

Notice that on the vertical interval \((0,50)\) the population is accelerating until it reaches \(P=50\) tons. After that, it still grows, but is starting to decelerate. Over the long run it will approach (but never quite reach) a level of \(P=100\) tons.

When we modeled the populations of trout and bass in an industrial farm we did not have to account for competition between species because we assumed that the two species of fish were kept in separate lakes. For fish living in the wild we cannot make that assumption, but we can tweak our original model to account for competition between species in a natural setting. To do this we’ll have to use two equations, one for each species.

Notice that we took the growth rates from before and subtracted an identical term \(0.0001BW\text{.}\) This represents the detrimental effect of each species on the other’s growth rate. For simplicity, we assumed that the effect of each on the other is the same. The fact that the new term, \(-0.0001BW\text{,}\) has the product \(BW\) in it reflects the idea that to have an appreciable effect on each other, both species must have a sufficient number to produce the required interactions. Again, this model is overly simplified, but let’s see what long term effect this tweak has on both species.

To see that this is truly just a tweak of our previous model notice that if \(W=0\text{,}\) then we would recover the previous model and over time \(B\) would approach its carrying capacity of \(500\) tons. Likewise, if \(B=0\text{,}\) then, again we recover the previous model and over time \(W\) would approach its carrying capacity of \(400\) tons.

Given that \(B,W\ge0\text{,}\) we want to determine what this model will predict happens to both species in the long term. Could they coexist or would one species drive the other to extinction? Take your best guess.

Next we form the (\(B\))ass—(\(W\))alleye plane, where each point \((B,W)\) identifies a particular pair of possible populations. For example the point \((200, 500)\) represents the situation when there are \(200\) tons of bass and \(500\) tons of walleye in the lake.

In the sketch below we see from equation (11.10) and equation (11.11) that on each of the red lines we have \(\dfdx{B}{t}=0\text{.}\)

Finally, notice that when \(W=0\) the bass population will increase until it reaches the carrying capacity of the lake for bass alone (\(500\)), after which it will start to decrease. But if there are walleye in the lake the bass population will not reach the carrying capacity before it begins to decrease. This is consistent with the notion that the number of walleye adversely affect the growth rate of the bass.

Next we need to perform a similar analysis for the walleye population.

If we superimpose the sketch you (should have) obtained from Problem 11.1.11 onto the Figure 11.1.10 we get the \(4\) regions seen in the sketch below.

The points where the green and red lines intersect are called equilibrium points. Theoretically, if the two population levels are at an equilibrium point then neither population will ever change again because the growth rates are both zero.

But what happens if we don’t start at an equilibrium point? Would the levels tend toward an equilibrium over time as the name suggests? Take your best guess.

We can figure this out by looking at the growth rates in each region. For example, suppose the initial populations of bass and walleye, \((B,W)\) is in the blue region in Figure 11.1.14. The bass population is increasing. Think of this as an arrow pointing to the right. In the same region the walleye population is decreasing which we can think of as an arrow pointing downward. Combining these arrows gives us an arrow pointing down and to the right.

Similarly, in the purple area the trajectory of the populations will be up and to the right since both populations are increasing. In the red area it will be up and to the left, while in the white area it will be down and to the left.

Figure 11.1.14 gives us an rough, qualitative idea of the population trajectories that this model predicts for the bass and walleye populations starting from various initial values.

Notice that these trajectories tend toward the equilibrium point, \(E=\left(E_B=437.5, E_W=312.5\right)\text{.}\) This says that given these initial levels of bass and walleye, the levels will approach \(E_B=437.5\) tons of bass and \(E_W=312.5\) tons of walleye. This kind of problem is called a dynamical system and the path that the point \((B,W)\) follows for any given set of initial conditions is called its trajectory.