$\gamma MS$

$\delta S$

show that, at the equilibrium point, the value of the mackerel population density is $\frac{\delta }{\gamma }$;

find the value of the shark population density at the equilibrium point.

Toxic sewage is added to the Mediterranean Sea. Alessia claims this reduces the shark population growth rate and hence the value of $\gamma$ is halved. No other parameter changes.

Global warming increases the temperature of the Mediterranean Sea. Alessia claims that this promotes the mackerel population growth rate and hence the value of $\alpha$ is doubled. No other parameter changes.

Write down the differential equation for $M$ that models this situation.

Show that the expression for the mackerel population density after $t$ years is $M=M_0{\text{e}}^{\alpha t}$

Alessia estimates that the mackerel population density increases by a factor of three every two years. Show that $\alpha =0.549$ to three significant figures.

Write down expressions for $M_{n+1}$ and $S_{n+1}$ in terms of $M_n$ and $S_n$.

Use Euler’s method to find an estimate for the mackerel population density after one year.

Use Euler’s method to sketch the trajectory of the phase portrait, for $4\le M\le 7$ and $1.5\le S\le 3$, over the first nine years.

Using your phase portrait, or otherwise, determine whether the mackerel population density would be sufficient to support sustainable fishing during the first nine years.

State two reasons why Alessia’s conclusion, found in part (f)(ii), might not be valid.

population growth rate / birth rate of sharks (due to eating mackerel)            A1

[1 mark]

(net) death rate of sharks          A1

[1 mark]

$\gamma MS-\delta S=0$          A1

since $S\ne 0$          R1

Note: Accept $S>0$.

getting to given answer without further error by either cancelling or factorizing           A1

$M=\frac{\delta }{\gamma }$           AG

[3 marks]

$\frac{dM}{dt}=0$

$\alpha M-\beta MS=0$          (M1)

(since $M\ne 0$)  $S=\frac{\alpha }{\beta }$          A1

[2 marks]

$M_{eq}=\frac{\delta }{\gamma }\Rightarrow \frac{\delta }{{\displaystyle \frac{1}{2}}\gamma }=2M_{eq}$                      M1

Note: Accept equivalent in words.

Doubles          A1

Note: Do not accept “increases”.

[2 marks]

$M_{eq}=\frac{\delta }{\gamma }$ is not dependent on $\alpha$                      R1

Note: Award R0 for any contextual argument.

no change         A1

Note: Do not award R0A1.

[2 marks]

$\frac{dM}{dt}=\alpha M$                    A1

[1 mark]

$\int \frac{1}{M}dM=\int \alpha \text{d}t$                  M1

Note: Award M1 is for an attempt to separate variables. This means getting to the point $\int f\left(M\right) dM=\int g\left(t\right) \text{d}t$ where the integral can be seen or implied by further work.

$\ln \left|M\right|=\alpha t+c$                  A1

Note: Accept $\ln M$. Condone missing constant of integration for this mark.

$M=k{\text{e}}^{\alpha t}$

when $t=0, M_0=k$                 M1

Note: Award M1 for a clear attempt at using initial conditions to find a constant of integration. Only possible if the constant of integration exists. $t=0$ or “initially” or similar must be seen. Substitution may appear earlier, following the integration.

initial conditions and all other manipulations correct and clearly communicated to get to the final answer                   A1

$M=M_0{\text{e}}^{\alpha t}$                  AG

[4 marks]

$M=3M_0$ seen anywhere                (A1)

substituting $t=2, M=3M_0$ into equation $M=M_0{\text{e}}^{\alpha t}$                (M1)

$3M_0=M_0{\text{e}}^{2\alpha }$

$\alpha =\frac{1}{2}\ln 3$  OR  $0.549306\dots$                  A1

Note: The A1 requires either the exact answer or an answer to at least $4$ sf.

$\approx 0.549$                 AG

[3 marks]

an attempt to set up one recursive equation                (M1)

Note: Must include two given parameters and $M_n$ and $S_n$ and $M_{n+1}$ or $S_{n+1}$ for the (M1) to be awarded.

$M_{n+1}=M_n+0.1\left(0.549M_n-0.236M_n{S}_n\right)$                 A1

$S_{n+1}=S_n+0.1\left(0.244M_n{S}_n-1.39S_n\right)$                 A1

[3 marks]

EITHER
$6.12 (6.11609\dots )$             A2

OR
$6120 (6116.09\dots )$ (mackerel per ${\text{km}}^3$)             A2

[2 marks]

spiral or closed loop shape         A1

approximately $1.25$ rotations (can only be awarded if a spiral)         A1

correct shape, in approximately correct position (centred at approx. $(5.5, 2.5)$)         A1

Note: Award A0A0A0 for any plot of $S$ or $M$ against $t$.

[3 marks]

EITHER

approximate minimum is $(5.07223\dots ) 5.07$ (which is greater than $5$)           A1

OR

the line $M=5$ clearly labelled on their phase portrait           A1

THEN

(the density will not fall below $5000$) hence sufficient for sustainable fishing           A1


Note: Do not award A0A1. Only if the minimum point is labelled on the sketch then a statement here that “the mackerel population is always above $\mathit{5000}$” would be sufficient. Accept the value $5.07$ seen within a table of values.

[2 marks]

Any two from:                           A1A1

• Current values / parameters are only an estimate,

• The Euler method is only an approximate method / choosing $h=0.1$ might be too large.

• There might be random variation / the model has no stochastic component

• Conditions / parameters might change over the nine years,

• A discrete system is being approximated by a continuous system,

Allow any other sensible critique.

If a candidate identifies factors which the model ignores, award A1 per factor identified. These factors could include:

• Other predators

• Seasonality

• Temperature

• The effect of fishing

• Environmental catastrophe

• Migration

Note: Do not allow:
             “You cannot have $5.07$ mackerel”.
             It is only a model (as this is too vague).
             Some factors have been ignored (without specifically identifying the factors).
             Values do not always follow the equation / model. (as this is too vague).

[2 marks]