Math Is Fun Forum

cadjeff

Member

Registered: 2007-05-08

Posts: 26

Population growth and effects of hunting

Hello,
i'm having trouble with the following task.

Basically there's an island inhabited with buffalo.
For the last 12 months they've been the target of hunters.

The present population is 3920 heads.
The population 12 months ago was 3790.
The population 10 years ago was 580.

The number of hunters during the last 12 months is expected to have varied continuously but smoothly with the time of year, being the minimum at around the new year and peaking towards the end of june. At the beginning of january/end of december, six 'hard core' hunters were known to be operating on the island, but it is not known how many were around at the height of the season.

1) Is there any evidence to suggest that hunters managed to kill significant numbers (>10) of buffalo during the last twelve months?
2) Roughly how many hunters were active in mid-summer?

THIS DATA is also supplied:
The rate of change in the size of a herd of bison constrained to the area of a particular nature reserve, during the 3 month hunting seasons every year in the period 1990-2005, was invariably found to be directly proportional to the number of hunters. During the 2001 season, for example, the population decreased from 4734 to 4531 in the space of 3 months with 87 hunters constantly in attendance.

Ok, i got the population growth formula from dP/dt= KP

i ended up with N = No e^rt

(where N = present pop.   No = initial pop.   r= growth rate   t=length of time)

So by substituting the 10 year and 12 month old populations and t=9 (time between the different populations) i obtained the 'r' (rate of growth without hunters)
I then used 'r' to find what the present population would be had there been no hunters.

I got the approx population of 4670 heads.

Clearly this suggested that there had been significant numbers of buffalo killed by hunters.

So thats part 1 done really.

Part 2: I've been told the following:
          - i need to model the number of hunters around at any time (does not require differential equations at all)
          - I need to consider how the population might vary as a function of the number of hunters present. (the extra data could be of use here)

I've managed to work out that the varying number of hunters can be represented by a sin wave (or cos) and that the minimum point represents the hunter no. at New year and the maximum point represents the no. of hunters at the end of june.
ASSUMPTION: The minimum and maximum no. of hunters are 6 months apart.

This is as far as i've got - i'm not sure how to do the rest of part 2.

Some help would be really appreciated!

Thank you

Caddy

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Population growth and effects of hunting

I initially assumed the graph of number of hunters (H) against time of year (T) to be a cosine curve, but I got some untidy answers. (I’ve deleted them.) Instead, I am going to assume that the graph of H against T is the arch of an inverted parabola. Let H[sub]max[/sub] be the peak number of hunters, which occurs at T = 6. Then the equation between H and T can be written as

When T = 0, H = 6. Therefore

So roughly 42 hunters were active in midsummer.

So the number of hunters around at any time is given by

Let B be the population of the buffalo. Then

where k is a constant. ∴

where B[sub]0[/sub] is another constant. When T = 0, B = 3790; ∴ B[sub]0[/sub] = 3790. When T = 12, B = 3920; thus

If we express T in terms of H:

Substituting for T into

will enable B to be expressed directly as a function of H before and after midsummer.

Last edited by JaneFairfax (2007-05-09 04:24:12)

cadjeff

Member

Registered: 2007-05-08

Posts: 26

Re: Population growth and effects of hunting

Thank you so much, that was a great help.
Could you briefly explain how you got the second to last part equation that says:
'If we express T in terms of H:' just above it?

thank you very very much!

caddy

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Population growth and effects of hunting

cadjeff

Member

Registered: 2007-05-08

Posts: 26

Re: Population growth and effects of hunting

Sorry, i've been looking at what you started with,

H = Hmax - (T - 6)^2

but i'm still a little puzzled, it looks right but where does this equation come from? or what stages did you go through to get it?

thank you,

caddy

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Population growth and effects of hunting

It's just an assumption.

Last edited by JaneFairfax (2007-05-11 01:07:05)

cadjeff

Member

Registered: 2007-05-08

Posts: 26

Re: Population growth and effects of hunting

By assumption do you mean you made this equation up from nothing? If it doesn't come from some known formula or rule how would you know if it was right?

cheers,

caddy

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Population growth and effects of hunting

How do I know if it’s right? Well, I don’t. The question is not specific enough about the exact type of graph required – all it says is that the graph must be smooth and continuous and have a maximum peak. I thought I was allowed to choose any reasonable graph fitting this description to model the problem.

cadjeff

Member

Registered: 2007-05-08

Posts: 26

Re: Population growth and effects of hunting

Ok, i see. It is supposed to be a sin wave according to my teacher. I'm trawling through 'Advanced engineering mathematics' by Kreyszig to see if any new equations/methods come to light...not going to well at the moment, haha!

I'm looking at this other part of the task now but i'm not sure where to start.

This is the basics of it anyway...Waste is being dumped by a factory into a river and it's contaminating the lake that the buffalo drink from.
1) The rate of change in the concentration, C, of the pollutant in the lake is directly proportional to the amount P of pollutant present at any time. The proportionality constant is K1.
2) When the amount of pollutant in the lake is constant the concentration appears to decay exponentially with time, from its starting value to 1/e of its starting value in 2 days.
   However, if there is a change in the amount of pollutant in the lake, the concentration seems to become directly proportional to the rate of change in the amount of pollutant. The proportionality constant is K2. In other words, the concentration of pollutant in the lake seems to be affected by two phenomena independently ( a natural decay and any new waste arriving through the river).

I'm to provide a formula for the amount of pollutant in the lake as a function of time and the various proportionality constants above.

Bed time me tinks!

cheers,

caddy

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Population growth and effects of hunting

Case 1: P constant

Let the initial concentration be C[sub]0[/sub]

From #1,

We are also told that C decays exponentially with t when P is constant; let

Combining (1) and (2),

(NB: C and t vary in such a way that P remains constant.)

I have to go now; I'll come back and constant the other case later.

Last edited by JaneFairfax (2007-05-17 00:13:43)

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Population growth and effects of hunting

Case 2: P variable

This is a standard homogeneous second-order linear ordinary differential equation. The general solution depends on whether K[sub]1[/sub]⁄K[sub]2[/sub] is positive or negative. If it’s positive, the general solution is

If K[sub]1[/sub]⁄K[sub]2[/sub] is negative, the general solution is

In both cases, we can eliminate one of the “arbitrary” constants P[sub]1[/sub] and P[sub]2[/sub] (since we started with a first-order differential equation). Unfortunately there is not enough information to enable the other arbitrary constant to be determined. At best, the other arbitrary constant can be expressed in terms of P[sub]0[/sub], the amount of pollutant in the lake at time t = 0.

Last edited by JaneFairfax (2007-05-17 03:49:03)

cadjeff

Member

Registered: 2007-05-08

Posts: 26

Re: Population growth and effects of hunting

How do you go from:  Co/e = 2k1P + Co

To:    Co  = 2ek1 p / 1 – e      ?   
   
I can’t seem to rearrange so as to get that answer.

Co/e = Coe^-2a  ,  a = ½  , how do you know what ‘a’ is?

Therefore Co = Ce1/2t    How do get to this from the previous equation?

Hopefully if that’s cleared the combined (1) and (2) equations will be clear to me.

Next post:

How do you get  1/K2 .dC/dt = K1/K2.P    ?

Sorry for the mass of questions.
As you may have already noticed (haha!) but I’ve been finding differentials difficult.
I covered 1st order/ 2nd order / Homogeneous differentials a month or so ago and I've struggled a lot with them.

Cheers,
caddy

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Population growth and effects of hunting

Co/e = Coe^-2a  ,  a = ½  , how do you know what ‘a’ is?

Just cancel the C[sub]0[/sub] from both sides and solve for a.

Therefore Co = Ce1/2t    How do get to this from the previous equation?

How do you get  1/K2 .dC/dt = K1/K2.P    ?

From the first part.

Last edited by JaneFairfax (2007-05-17 21:08:22)