IB Mathematics AHL 5.14 Solving by separation of variables-AI HL Paper 3- Exam Style Questions- New Syllabus

Question

In this question researchers are trying to find the most accurate model to use when modelling a population of wolves.

Historically, a population of wolves in an area had a stable size of \(200\). After some years of disruption, the population was reduced to \(40\) wolves. At this point, the area became a protected space and the population began to grow again.

Researchers in the area wish to model the size of the wolf population, \(x\), as a function of \(t\), where \(t\) is the time, in years, since the area became protected.

(a) Initially, the researchers consider using the logistic model

\(x=\dfrac{L}{1+Ce^{-kt}},\quad \text{where } L,C,k\in\mathbb{R}^{+}.\)

The researchers decide to let \(L=200\).

(i) State the assumption being made in assuming \(L=200\). [1]

At \(t=0\), the population of wolves is \(40\).

(ii) Find the value of \(C\). [2]

At \(t=5\), the population of wolves is found to have increased to \(70\).

(iii) Find the value of \(k\). [2]

(iv) Use your model to predict the size of the wolf population in the area \(10\) years after it became protected. Give your answer correct to the nearest whole number. [2]

(b) An alternative model for population growth is called the Gompertz model. When applied by the researchers to the wolf population, this model satisfies the differential equation

\(\dfrac{dx}{dt}=ax\ln\left(\dfrac{200}{x}\right),\quad a\in\mathbb{R}^{+}.\)

(i) Write down the value of \(\dfrac{dx}{dt}\) when \(x=200\). [1]

(ii) Interpret your answer to part (b)(i) in context. [1]

Consider the function

\(f(x)=\ln(\ln 200-\ln x),\quad \text{where }0<x<200.\)

(iii) Show that

\(f'(x)=\dfrac{-1}{x\ln\left(\dfrac{200}{x}\right)}.\) [2]

(iv) Hence, use separation of variables to show that the general solution of

\(\dfrac{dx}{dt}=ax\ln\left(\dfrac{200}{x}\right),\quad \text{where }0<x<200,\)

can be written as

\(\ln x=\ln 200-Ae^{-at},\)

where \(A\) is an arbitrary positive constant. [5]

(v) Use the size of the wolf population at \(t=0\) to find the value of \(A\). Give your answer in the form \(A=\ln p\), where \(p\in\mathbb{Z}^{+}\). [2]

(vi) Use the size of the wolf population at \(t=5\), given in part (a), to show that \(a=0.0855\), correct to three significant figures. [2]

(vii) Use the Gompertz model to predict the size of the wolf population at \(t=10\). Give your answer correct to the nearest whole number. [3]

After \(10\) years, the wolf population is measured and is found to be \(85\).

(c) Comment on the predictions made by the two models. [1]

By tracking individual wolves, the researchers find that about \(3\%\) of the wolf population emigrate from the protected area each year.

They decide to adapt the Gompertz model to allow for this. The new model will satisfy the differential equation

\(\dfrac{dx}{dt}=0.0855x\ln\left(\dfrac{200}{x}\right)-0.03x.\)

(d)

(i) Use Euler’s method, with a step size of \(0.5\) years and an initial value of \(x_0=70\) when \(t=5\), to find an estimate for the size of the wolf population when \(t=10\). Give your answer correct to the nearest whole number. [4]

(ii) Comment on your answer. [1]

Most-appropriate topic codes (IB DP Mathematics: Applications and Interpretation HL, Paper 3):

• AHL 2.9: Modelling with AHL functions, including logistic models, restricted growth and carrying capacity. (Parts a(i)–a(iv))
• AHL 5.14: Setting up and solving differential equations by separation of variables; general solution; modelling growth from context. (Parts b(i)–b(vii), c)
• AHL 5.16: Euler’s method for approximate solutions of first-order differential equations. (Parts d(i)–d(ii))

▶️ Answer/Explanation

(a) (i)

Since \(L\) is the upper limiting value of the logistic model, assuming \(L=200\) means the researchers are assuming that the long-term stable population will return to \(200\).

✅ Answer: The stable or long-term population will be \(200\) wolves. A1

Marking note: Accept “the carrying capacity is \(200\)” or “the maximum population is \(200\)”.

(a) (ii)

At \(t=0\), \(x=40\), and \(L=200\).

\(40=\dfrac{200}{1+Ce^{0}}\) M1

Since \(e^0=1\),

\(40=\dfrac{200}{1+C}\)

\(40(1+C)=200\)

\(1+C=5\)

\(C=4\)

✅ Answer: \(C=4\) A1

(a) (iii)

Using \(x=70\) when \(t=5\):

\(70=\dfrac{200}{1+4e^{-5k}}\) A1

Rearrange:

\(1+4e^{-5k}=\dfrac{200}{70}=\dfrac{20}{7}\)

\(4e^{-5k}=\dfrac{13}{7}\)

\(e^{-5k}=\dfrac{13}{28}\)

\(-5k=\ln\left(\dfrac{13}{28}\right)\)

\(k=-\dfrac{1}{5}\ln\left(\dfrac{13}{28}\right)=0.153451\ldots\)

✅ Answer: \(k=0.153\) A1

(a) (iv)

Use the logistic model with \(C=4\) and \(k=0.153451\ldots\):

\(x=\dfrac{200}{1+4e^{-10(0.153451\ldots)}}\) A1

\(x=107.397\ldots\)

✅ Answer: \(107\) wolves A1

Explanation: The logistic model predicts the population will continue increasing towards the carrying capacity of \(200\), giving about \(107\) wolves after \(10\) years.

(b) (i)

Substitute \(x=200\) into

\(\dfrac{dx}{dt}=ax\ln\left(\dfrac{200}{x}\right).\)

\(\dfrac{dx}{dt}=a(200)\ln\left(\dfrac{200}{200}\right)\)

\(=200a\ln(1)=0\)

✅ Answer: \(\dfrac{dx}{dt}=0\) A1

(b) (ii)

When the population is \(200\), the rate of change is zero, so the population is not increasing or decreasing at that instant.

✅ Answer: The population is steady or stable at \(200\) wolves. A1

Marking note: Do not just say “\(200\) is the maximum”; the interpretation should refer to the rate of change being zero or the population being stable.

(b) (iii)

We are given

\(f(x)=\ln(\ln 200-\ln x).\)

Differentiate using the chain rule:

\(f'(x)=\dfrac{1}{\ln 200-\ln x}\times\left(-\dfrac{1}{x}\right)\) M1A1

Since

\(\ln 200-\ln x=\ln\left(\dfrac{200}{x}\right),\)

we get

\(f'(x)=\dfrac{-1}{x\ln\left(\dfrac{200}{x}\right)}.\)

✅ Shown AG

(b) (iv)

Start with the differential equation:

\(\dfrac{dx}{dt}=ax\ln\left(\dfrac{200}{x}\right).\)

Separate variables:

\(\dfrac{1}{x\ln\left(\dfrac{200}{x}\right)}\,dx=a\,dt\)

or equivalently, using the derivative found in part (b)(iii),

\(\displaystyle \int \dfrac{-1}{x\ln\left(\dfrac{200}{x}\right)}\,dx=\int -a\,dt\) M1A1

From part (b)(iii), the left side integrates to

\(\ln(\ln 200-\ln x)\).

So,

\(\ln(\ln 200-\ln x)=-at+c\) A1A1

Exponentiate both sides:

\(\ln 200-\ln x=e^{-at+c}=Ae^{-at}\)

where \(A=e^c\), so \(A>0\).

Therefore,

\(\ln x=\ln 200-Ae^{-at}\).

✅ Shown A1AG

Explanation: The derivative in part (b)(iii) helps rewrite the separated integral directly in logarithmic form.

(b) (v)

At \(t=0\), \(x=40\). Substitute into

\(\ln x=\ln 200-Ae^{-at}.\)

\(\ln 40=\ln 200-Ae^{0}\) A1

\(\ln 40=\ln 200-A\)

\(A=\ln 200-\ln 40=\ln\left(\dfrac{200}{40}\right)=\ln5\)

✅ Answer: \(A=\ln5\) A1

(b) (vi)

Using \(x=70\) when \(t=5\), and \(A=\ln5\):

\(\ln70=\ln200-(\ln5)e^{-5a}\) A1

Rearrange:

\((\ln5)e^{-5a}=\ln200-\ln70=\ln\left(\dfrac{200}{70}\right)\)

\(e^{-5a}=\dfrac{\ln\left(\frac{200}{70}\right)}{\ln5}\)

\(-5a=\ln\left(\dfrac{\ln\left(\frac{200}{70}\right)}{\ln5}\right)\)

\(a=-\dfrac{1}{5}\ln\left(\dfrac{\ln\left(\frac{200}{70}\right)}{\ln5}\right)=0.0854528\ldots\)

✅ Answer: \(a=0.0855\) to three significant figures A1AG

(b) (vii)

Use the Gompertz model with \(A=\ln5\) and \(a=0.0854528\ldots\):

\(\ln x=\ln200-(\ln5)e^{-10(0.0854528\ldots)}\) A1

\(x=100.839\ldots\) A1

✅ Answer: \(101\) wolves A1

Explanation: This model predicts a smaller population than the logistic model at \(t=10\).

(c)

The measured population after \(10\) years is \(85\).

The logistic model predicted \(107\), and the Gompertz model predicted \(101\).

Both models overestimate the true population, but the Gompertz model is closer to the measured value.

✅ Answer: Both models overestimate; the Gompertz model is closer to the true value. A1

(d) (i)

The adapted differential equation is

\(\dfrac{dx}{dt}=0.0855x\ln\left(\dfrac{200}{x}\right)-0.03x.\)

Using Euler’s method with step size \(h=0.5\),

\(x_{n+1}=x_n+0.5\left(0.0855x_n\ln\left(\dfrac{200}{x_n}\right)-0.03x_n\right)\) M1A1

Start with

\(x_0=70,\quad t=5.\)

After \(10\) steps of size \(0.5\), the time is \(t=10\). Applying Euler’s method gives

\(x(10)\approx89.4280\ldots\) A1

✅ Answer: \(89\) wolves A1

Explanation: The extra term \(-0.03x\) reduces the growth rate because it models \(3\%\) of the population leaving the protected area each year.

(d) (ii)

The Euler estimate is about \(89\), which is much closer to the measured value of \(85\) than the earlier predictions of \(107\) and \(101\).

✅ Answer: The adapted model is more realistic or more accurate than the previous models. A1

IB Mathematics AHL 5.14 Solving by separation of variables-AI HL Paper 3- Exam Style Questions- New Syllabus

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