If a constant number h of fish are harvested from a fishery

In summary, we discussed a conversation about a fish population model and how to solve for the initial value problem and determine the value of P0 such that the fish population becomes extinct in finite time. We also talked about the importance of understanding linear differential equations and their applications in various fields of science.
  • #1
Lin Galido
Hi! Can anyone help me?

If a constant number h of fish are harvested from a fishery per unit time, then a model for the population
P(t) of the fishery at time t is given by:
dP/dt = P(5-P) - h, P(0) = P0.

a. Solve for the IVP if h = 4.
b. Determine the value of P0 such that the fish population becomes extinct in finite time.

My solution so far:
dP/dt = 5p - p^2 -4
dp/(-p^2 +5p - 4) = dt

by integration: ln |p-4| /3 - ln |p-1|/3 = dt
ln |p-4| - ln |p-1| = 3 dt
ln |p-4| - ln |p-1| = 3t + C

At P(0) (when t =0)
ln |P0 -4| - ln |P0 -1| = 3(0) + C
C = ln |P0 -4| - ln |P0 -1|

t = (ln |P-4| - ln |P-1| - ln |P0 -4| - ln |P0 -1|) /3

I end here and I'm not really sure waht to do after.
 
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  • #2
Not checked your working but it looks like the right sort of thing.

When you add or subtract two logs that is the log of what? The inverse of a (natural) log is what?
 
  • #3
epenguin said:
Not checked your working but it looks like the right sort of thing.

When you add or subtract two logs that is the log of what? The inverse of a (natural) log is what?
ln (|P-4|/ |P-1|) ?
I'm not too sure how to answer the IVP part or finding the initial value as well
 
  • #4
I have to go now but you can do some more combinations of logs involving P0.

You must have had some other examples for models involving initial values to base yourself en.
 
  • #5
I actually don't have any other examples for this cause my teacher just surprised us with this. :( If anyone could help, I'd be grateful
 
  • #6
I'm surprised you have never done a simpler example before e.g. dP/dt = -k1P + k2 (P0 - P) which is a basic equation of chemical kinetics which is in all the books of math or physical chemistry and the procedures are exactly the same in principle, your example is just more complicated. Or better in any University level general physical chemistry book you will find the same equation as yours essentially, just without the final -h term - it's called 'second-order kinetics' (some cases).
I'm also surprised there is no constant multiplying your P(5 - P) which seems to me as slightly artificial.

Coming to your case, I think the line 'by integration' and the next line are slightly confused, but that the answer for the integration is OK..
You have already practically done the initial value problem, you just need to complete.
Complete for the rest of the equation what you already started with the log functions of P. Then when you know the natural log of something, what is that something equal to? IOW what is the inverse of a log?
You should be able to get what you get in a whole big class of things like this, that there's a function of P, f(P) say, that equals the same function f(P0) of P0 multiplied by an exponential. In simple cases this function is straightforwardly P itself, but yours is more complicated. You can then do some algebraic manipulation to get the result in the form P = something.
 
  • #7
Lin Galido said:
Hi! Can anyone help me?

If a constant number h of fish are harvested from a fishery per unit time, then a model for the population
P(t) of the fishery at time t is given by:
dP/dt = P(5-P) - h, P(0) = P0.

a. Solve for the IVP if h = 4.
b. Determine the value of P0 such that the fish population becomes extinct in finite time.

My solution so far:
dP/dt = 5p - p^2 -4
dp/(-p^2 +5p - 4) = dt

by integration: ln |p-4| /3 - ln |p-1|/3 = dt
ln |p-4| - ln |p-1| = 3 dt
ln |p-4| - ln |p-1| = 3t + C

At P(0) (when t =0)
ln |P0 -4| - ln |P0 -1| = 3(0) + C
C = ln |P0 -4| - ln |P0 -1|

t = (ln |P-4| - ln |P-1| - ln |P0 -4| - ln |P0 -1|) /3

I end here and I'm not really sure waht to do after.

Because of the absolute-value signs, you need to split up the analysis into cases. For example, if you solve the DE numerically you will see that for some values of ##P_0## the solution ##P(t)## keeps increasing. For other values of ##P_0## the solution decreases to a positive limit and so never reaches 0. For still other values of ##P_0## it decreases and eventually hits zero (after which point the fish are extinct so the differential equation ceases to be valid). You need to look separately at cases such as ##P_0 > 4##, ##1 < P_0 < 4##, and ##0 < P_0 < 1##.
 
  • #8
Well looks like the student has abandoned the thread.
This time I can't, not without saying:
Question (b) is more important than the solution to (a).
If the student did question (b), the observations of Ray in post 7 would come to life, rather than being this rather annoying thing that one learns and forgets or it becomes foggy - although one would recall if one had to calculate some numbers from this, and found that you had written the log of a negative number.
This is a (about the simplest) non-linear differential equation. From what the student says she has not done linear differential equations.
It has been said that the whole of Euclid's Elements is meant really to lead up to the Platonic solids. It could be said that the whole of maths teaching for the mass of sciences students is meant to lead the average student up to linear differential equations (with constant coefficients). There, a variety of stuff learned comes together - calculus, complex numbers, linear algebra, trigonometrical and exponential functions, and a lot of physics and engineering etc. For the majority of scientists linear d.e.'s are something indispensable to grasp,

The trouble is the behaviours they predict are mostly quite boring! Certainly with homogeneous d.e.'s There is just one 'stationary' point in the space of the variables, and they just go there and stay there for all time. Maybe they go there extra-boringly, or else they go there by damped oscillations. OK they never quite arrive at the point but after sufficient time they are as close as you please. Or else the stationary point is unstable, and the variable goes boringly off to infinity. In special cases they might oscillate for ever. Okay with several variables one variable can do one thing, another another. In the case of non-homogeneous equations, I think they mostly end up doing what an imposed force is trying to make them do, though in a foot dragging manner. Well I call them boring - they're not that easy and can be a challenge so maybe they are interesting when you really get into them; this I think would mostly be because of the applications.

Whereas nonlinear differential equations have a more varied repertoire. And this is already seen in this simplest of examples. There are two places the system can end up, depending where it starts. It has two stationary states (not quite same thing as previous sentence), one attracting, one repelling. Fairly typical of nonlinear systems the condition of maximum yield it Is also the most unstable condition. If you try to harvest at the theoretical maximum rate, say you had a contract where you had to deliver this same number of fish all the time, you are at risk of losing your whole resource. (I don't know if this typical feature has name, the something principle.)

You do not need to solve the equation to see this. And in most nonlinear differential equations of interest you cannot solve them in terms of the known elementary analytical etc. functions. But you can analyse them qualitatively. In this case a sketch suffices, usually the qualitative will involve some calculation, in which the linear theory is found useful and necessary. Then not so boring after all.

Possibly because of the risk of boring, the teacher in this case has chosen to stimulate with a nonlinear problem.
Maybe the above has become already obvious to the student having done the problem (wasn't sounding so). I will never understand why students don't complete problems (see also my sig.). But I say just getting the answers is inefficient learning. "When you have got the answer the job is not completed." Students can usually get something additional from helpers on this site when they complete, because, underneath, problems are not isolated the way they tend to be treated in the question-and-answer routine, and helpers can point them to building up connected and empowering knowledge rather than just producing answers to questions asked that end there.
 
Last edited:

Related to If a constant number h of fish are harvested from a fishery

1. How does harvesting a constant number of fish affect the overall fish population?

Harvesting a constant number of fish from a fishery can have a negative impact on the overall fish population. This is because constantly removing the same number of fish can lead to a decrease in the reproductive capacity of the population, making it difficult for the population to replenish itself.

2. What is the ideal number of fish to harvest from a fishery?

The ideal number of fish to harvest from a fishery depends on various factors such as the fish species, the size of the population, and the sustainability of the fishery. It is important to carefully monitor and manage the fishery to determine the appropriate harvest level to maintain a healthy and sustainable fish population.

3. How does harvesting a constant number of fish impact the ecosystem?

Harvesting a constant number of fish can have a ripple effect on the entire ecosystem. Fish play an important role in maintaining the balance of the food chain and their removal can affect the population of other species that depend on them for food. It can also lead to changes in water quality and overall ecosystem health.

4. How can we ensure the sustainability of a fishery if a constant number of fish are being harvested?

To ensure the sustainability of a fishery, it is important to implement effective management strategies. This can include monitoring the fish population, regulating the fishing methods and seasons, and implementing quotas to limit the number of fish that can be harvested. It is also important to consider the impact of other factors such as climate change and pollution on the fishery.

5. What are the potential consequences of not regulating the harvesting of a constant number of fish?

If the harvesting of a constant number of fish is not regulated, it can lead to overfishing and depletion of the fish population. This can have serious consequences not only for the fishery itself but also for the entire ecosystem and the communities that depend on it. It can also result in economic losses and impact the livelihoods of those involved in the fishing industry.

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