A declining species and an increasing logistic species · source reference

Printed question

On a mangrove swamp, scientists are investigating the population of mudskippers and crabs. Initially, there are 800 mudskippers and 800 crabs on the mangrove swamp. At time $t$ years, the number of mudskippers and crabs are $M,C$ respectively.

(a) For the mudskippers, the scientists discovers that every year, the growth rate is 0.4% of the population size and the death rate is 5 mudskippers per year.

(i) Write down a differential equation relating $M,t$. [1]

(ii) Solve the differential equation in part (i) and determine what happens to the mudskipper population in the future. [4]

(b) For the crabs, the scientists propose that $C,t$ are related by the differential equation $\frac{dC}{dt}=8C-0.005C^2$.

(i) Find the number of crabs in the mangrove swamp when the rate of change of $C$ is a maximum. [2]

(ii) Find $C$ in terms of $t$. Sketch a graph of $C$ against $t$. [5]

[2021/TJC/Prelim/I/10]

Printed working

Mudskippers (a)

Growth rate $(0.4/100)M=0.004M$; death rate 5. $M'=0.004M-5=0.004(M-1250)$.

$M'/(M-1250)=0.004$; $\int dM/(M-1250)=\int0.004dt$; $\ln|M-1250|=0.004t+c$.

$|M-1250|=e^{0.004t+c}$; $M-1250=\pm e^{0.004t+c}=Ae^{0.004t}$, $A=\pm e^c$.

At 0, $M=800$, $A=-450$; $M=1250-450e^{0.004t}$.

As $t\to\infty$, this formal expression tends to $-\infty$; the population becomes extinct. $M=0\Rightarrow e^{0.004t}=25/9\Rightarrow t=255.4$: extinction in the 256th year; physical model stops at zero.

Crabs maximum growth (b)(i): three printed methods

1. $R=C'=8C-0.005C^2$. $dR/dC=8-0.01C=0\Rightarrow C=800$. $d^2R/dC^2=-0.01<0$, so maximum.
2. Complete square: $C'=-0.005(C^2-1600C)=-0.005(C-800)^2+3200\le3200$. Maximum when $C=800$.
3. $C'=C(8-0.005C)=0$ at $C=0,1600$. The downward quadratic attains its maximum at the midpoint 800; graph vertex $(800,3200)$.

Crabs solution (b)(ii): printed partial-fraction method

$C'=0.005C(1600-C)$; $\int dC/[C(1600-C)]=\int0.005dt$.

$\frac1{1600}\int[1/C+1/(1600-C)]dC=\int0.005dt$.

$[\ln|C|-\ln|1600-C|]/1600=0.005t+d$; $\ln|C/(1600-C)|=8t+1600d$.

$C/(1600-C)=\pm e^{8t+1600d}=Be^{8t}$, $B=\pm e^{1600d}$.

At 0, $C=800$, $B=1$; $C=1600e^{8t}/(1+e^{8t})=1600/(1+e^{-8t})$.

Printed complete-square method

$C'=-0.005[(C-800)^2-800^2]$; $\int dC/[(C-800)^2-800^2]=\int-0.005dt$.

$\frac1{1600}\ln\left|\frac{(C-800)-800}{(C-800)+800}\right|=-0.005t+d$.

$(C-1600)/C=\pm e^{-8t+1600d}=Be^{-8t}$; at 0, $B=-1$.

$1-1600/C=-e^{-8t}\Rightarrow1600/C=1+e^{-8t}\Rightarrow C=1600/(1+e^{-8t})$.

As $t\to\infty$, $e^{-8t}\to0$ so $C\to1600$.

Printed answer key

(a)(i) $dM/dt=0.004M-5$. (ii) $M=1250-450e^{0.004t}$; Mudskipper population will become extinct in the future. (b)(i) 800. (ii) $C=1600/(1+e^{-8t})$; native sketch below.

Teacher note 1

$dM/dt$ is the overall net change in the number of mudskippers $M$.

Teacher note 2

Given $dM/dt=f(M)$, $(1/f(M))dM/dt=1$. Integrate $\int dM/f(M)=\int1dt$.

Need to put in the modulus sign when integration results in a logarithm, since its argument may be negative. In removing the modulus sign, remember to put in “$\pm$”. Here $M-1250$ may be negative.

Teacher note 3

Complete the square: $(C-800)^2\ge0$, so $-0.005(C-800)^2\le0$ and $dC/dt\le3200$.

Teacher note 4

Since coefficient of $C^2$ is negative, this quadratic curve has a maximum. For a quadratic, max/min occurs at the midpoint of the $x$-intercepts.

Teacher note 5

$1/[C(1600-C)]=A/C+B/(1600-C)$; $1=A(1600-C)+BC$. At $C=0$, $A=1/1600$; at $C=1600$, $B=1/1600$. Cover-up rule gives the same values.

Need to put in the modulus sign when integration results in a logarithm, since its argument may be negative. In removing the modulus sign, remember to put in “$\pm$”.

Teacher note 6

As $t\to\infty$, $e^{-8t}\to0$, so $C\to1600$.

Diagram 10-16-growth

Diagram 10-16-population

Mathematical review

(a)(i) $M'=0.004M-5$. (ii) $M=1250+Ae^{0.004t}$; initial M=800 gives $M=1250-450e^{0.004t}$. It reaches zero at $t=250\ln(25/9)\approx255.41$ years. Interpret this as modeled extinction and stop the population model there.

(b)(i) $R(C)=8C-0.005C^2=3200-0.005(C-800)^2$, maximal at C=800, with rate 3200/year. (ii) Separation of $C'=0.005C(1600-C)$ gives $\ln[C/(1600-C)]=8t+A$. Initial C=800 gives A=0, so $C=1600/(1+e^{-8t})$. It starts at the logistic inflection (0,800), is increasing and concave down for t>0, and approaches C=1600. Substitution into the DE and both initial conditions verifies the formulas.

Provenance

TJC JC2 2025 H2 Maths (9758) MSM. Native transcription checked against original visible question, working and sidebars. Question page(s): 6. Working page(s): 23, 24, 25, 26. Original records remain private for QA; this native article has no PDF or raster worksheet dependency.