In harvesting of renewable natural resources, it is desirable that policies are formulated to allow maximal harvest of the natural resources, and yet not deplete the resources below a sustainable level. A simple harvesting model devised for the rate of change of the population of wild salmon in a particular region in the Pacific Ocean is given by
dPdt=P(4−P)−h,\frac{\text{d}P}{\text{d}t} = P(4 - P) - h,
where PP is the population of wild salmon in millions at time tt years and hh is the constant harvest rate in millions.

  1. (i)

    Sketch a graph of dPdt\frac{\text{d}P}{\text{d}t} against PP, expressing the turning point in terms of hh.

  2. (ii)

    The Maximum Sustainable Yield (MSY) is the largest harvest rate hh that allows for a sustainable harvest of wild salmon without long-term depletion. State the MSY for the wild salmon.

  3. (iii)

    It is given that the population of wild salmon in that region was 3.2 million in 2015 and the constant harvest rate is 3 million. Find an expression for PP at any time tt.

    Hence find the population of the wild salmon in that region in 2016.

  4. (iv)

    State one assumption you made in your calculation.

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Worked solution

4(i)

1. Express the given differential equation as a quadratic in completed square form to identify the turning point.

dPdt=4P−P2−h\frac{\text{d}P}{\text{d}t} = 4P - P^2 - h
=−(P2−4P)−h= -(P^2 - 4P) - h
=−[(P−2)2−4]−h= -[(P - 2)^2 - 4] - h
=−(P−2)2+4−h= -(P - 2)^2 + 4 - h
Since the coefficient of (P−2)2(P-2)^2 is negative, the graph has a maximum turning point at (2,4−h)(2, 4 - h).

2. Identify the shape and $y$-intercept for the graph sketch.

The equation describes an inverted parabola.
When P=0P = 0, dPdt=−h\frac{\text{d}P}{\text{d}t} = -h.
The required sketch should display this parabolic shape passing through (0,−h)(0, -h) on the vertical axis, reaching a peak at (2,4−h)(2, 4-h).

Answer

A sketch of an inverted (concave down) parabola plotted on dPdt\frac{\text{d}P}{\text{d}t} (vertical axis) against PP (horizontal axis):

  • The maximum turning point is labelled at (2,4−h)(2, 4 - h).
  • The vertical intercept is at (0,−h)(0, -h).

4(ii)

1. Use the maximum value of the rate of change to deduce the condition for sustainable harvest.

For the population to not face long-term depletion, there must be at least one value of PP where the population is not decreasing, meaning dPdt≥0\frac{\text{d}P}{\text{d}t} \ge 0.
From part (i), the expression dPdt=−(P−2)2+4−h\frac{\text{d}P}{\text{d}t} = -(P - 2)^2 + 4 - h has a maximum value of 4−h4 - h.
4−h≥04 - h \ge 0
h≤4h \le 4
The largest sustainable harvest rate hh is therefore 44.

Answer

44 million

4(iii)

1. Formulate the separable differential equation with the given constant harvest rate.

Substitute h=3h = 3 into the given differential equation:
dPdt=P(4−P)−3\frac{\text{d}P}{\text{d}t} = P(4 - P) - 3
dPdt=−P2+4P−3\frac{\text{d}P}{\text{d}t} = -P^2 + 4P - 3
dPdt=−(P2−4P+3)\frac{\text{d}P}{\text{d}t} = -(P^2 - 4P + 3)

2. Separate the variables and prepare for integration by completing the square.

Separating the variables yields:
∫1P2−4P+3 dP=∫−1 dt\int \frac{1}{P^2 - 4P + 3} \, \text{d}P = \int -1 \, \text{d}t
Completing the square for the quadratic denominator:
∫1(P−2)2−12 dP=∫−1 dt\int \frac{1}{(P - 2)^2 - 1^2} \, \text{d}P = \int -1 \, \text{d}t

(Note: Using partial fractions 1P2−4P+3=1/2P−3−1/2P−1\frac{1}{P^2 - 4P + 3} = \frac{1/2}{P-3} - \frac{1/2}{P-1} is an equally valid method to prepare for integration).

3. Integrate both sides using the standard formula.

Using the standard integration result ∫1x2−a2 dx=12aln⁡∣x−ax+a∣\int \frac{1}{x^2 - a^2} \, \text{d}x = \frac{1}{2a}\ln\left|\frac{x-a}{x+a}\right|:
12(1)ln⁡∣(P−2)−1(P−2)+1∣=−t+C\frac{1}{2(1)} \ln \left| \frac{(P - 2) - 1}{(P - 2) + 1} \right| = -t + C
12ln⁡∣P−3P−1∣=−t+C\frac{1}{2} \ln \left| \frac{P - 3}{P - 1} \right| = -t + C

4. Apply the initial condition to determine the arbitrary constant $C$.

Let t=0t = 0 represent the year 2015. We are given that P=3.2P = 3.2 when t=0t = 0:
12ln⁡∣3.2−33.2−1∣=−(0)+C\frac{1}{2} \ln \left| \frac{3.2 - 3}{3.2 - 1} \right| = -(0) + C
C=12ln⁡(0.22.2)=12ln⁡(111)C = \frac{1}{2} \ln \left( \frac{0.2}{2.2} \right) = \frac{1}{2} \ln \left( \frac{1}{11} \right)

5. Rearrange to make $P$ the subject of the formula.

Substitute CC back into the integrated equation:
12ln⁡∣P−3P−1∣=−t+12ln⁡(111)\frac{1}{2} \ln \left| \frac{P - 3}{P - 1} \right| = -t + \frac{1}{2} \ln \left( \frac{1}{11} \right)
Multiply by 22:
ln⁡∣P−3P−1∣=−2t+ln⁡(111)\ln \left| \frac{P - 3}{P - 1} \right| = -2t + \ln \left( \frac{1}{11} \right)
Since P=3.2>3P = 3.2 > 3, the fraction P−3P−1>0\frac{P - 3}{P - 1} > 0. We can thus drop the absolute value signs.
P−3P−1=e−2t+ln⁡(1/11)=111e−2t\frac{P - 3}{P - 1} = e^{-2t + \ln(1/11)} = \frac{1}{11} e^{-2t}
11(P−3)=(P−1)e−2t11(P - 3) = (P - 1)e^{-2t}
11P−33=Pe−2t−e−2t11P - 33 = P e^{-2t} - e^{-2t}
P(11−e−2t)=33−e−2tP(11 - e^{-2t}) = 33 - e^{-2t}
P=33−e−2t11−e−2tP = \frac{33 - e^{-2t}}{11 - e^{-2t}}
(Note: An equivalent and equally valid algebraic form is P=33e2t−111e2t−1P = \frac{33e^{2t} - 1}{11e^{2t} - 1})

6. Calculate the population of the wild salmon in the year 2016.

The year 2016 corresponds to t=1t = 1.
P=33−e−2(1)11−e−2(1)P = \frac{33 - e^{-2(1)}}{11 - e^{-2(1)}}
P=33−e−211−e−2≈3.0249P = \frac{33 - e^{-2}}{11 - e^{-2}} \approx 3.0249
The population in 2016 is approximately 3.023.02 million.

Answer

P=33−e−2t11−e−2tP = \frac{33 - e^{-2t}}{11 - e^{-2t}}

The population of the wild salmon in 2016 is approximately 3.023.02 million.

4(iv)

1. State a valid assumption required for the differential equation model to hold

For the mathematical model dPdt=P(4−P)−h\frac{\text{d}P}{\text{d}t} = P(4 - P) - h to accurately predict the population, we assume that the population changes solely due to the natural birth/death rate modelled and the specified harvesting. This means there are no sudden external environmental changes, disease outbreaks, or migration.

Answer

It is assumed that there is no migration of wild salmon into or out of the region, and no other external factors such as diseases affecting the population.

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