IB Mathematics AHL 5.15 Indefinite integrals AA HL Paper 3- Exam Style Questions

(a)(i)
In a logistic model, \(L\) represents the limiting population or carrying capacity.

Therefore, choosing \(L=200\) assumes that the wolf population will eventually stabilize at the same long-term population as it had before the disruption.

✅ Answer: The long-term stable population, or carrying capacity, is assumed to be \(200\) wolves.

(a)(ii)
At \(t=0\), \(x=40\). Substitute these values into the logistic model.

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

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

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

\(1+C=5\)

✅ Answer: \(C=4\)

(a)(iii)
At \(t=5\), \(x=70\), and \(C=4\).

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

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

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

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

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

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

\(k=0.153451\ldots\)

✅ Answer: \(k\approx0.153\)

(a)(iv)
Substitute \(t=10\), \(C=4\) and \(k=0.153451\ldots\) into the logistic model.

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

\(x=107.397\ldots\)

✅ Answer: \(107\) wolves

(b)(i)
When \(x=200\):

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

\(\dfrac{dx}{dt}=200a\ln1\)

✅ Answer: \(\dfrac{dx}{dt}=0\)

(b)(ii)
A rate of change of zero means that the population is not increasing or decreasing.

✅ Answer: A population of \(200\) wolves is an equilibrium or stable population.

(b)(iii)
Given:

\(f(x)=\ln(\ln200-\ln x)\)

Using the chain rule:

\(f'(x)=\dfrac{1}{\ln200-\ln x}\left(-\dfrac{1}{x}\right)\)

Since \(\ln200-\ln x=\ln\left(\dfrac{200}{x}\right)\):

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

(b)(iv)
Start with:

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

Separate the variables.

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

Integrating both sides gives:

\(\ln(\ln200-\ln x)=-at+c\)

Exponentiating:

\(\ln200-\ln x=e^ce^{-at}\)

Let \(A=e^c\). Since an exponential is positive, \(A>0\).

\(\ln200-\ln x=Ae^{-at}\)

Therefore:

✅ \(\ln x=\ln200-Ae^{-at}\)

(b)(v)
At \(t=0\), \(x=40\).

\(\ln40=\ln200-Ae^0\)

\(A=\ln200-\ln40\)

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

✅ Answer: \(A=\ln5\), so \(p=5\)

(b)(vi)
At \(t=5\), \(x=70\), and \(A=\ln5\).

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

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

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

\(e^{-5a}=\dfrac{\ln(20/7)}{\ln5}\)

\(-5a=\ln\left(\dfrac{\ln(20/7)}{\ln5}\right)\)

\(a=\dfrac{1}{5}\ln\left(\dfrac{\ln5}{\ln(20/7)}\right)\)

\(a=0.0854528\ldots\)

✅ Hence, \(a=0.0855\), correct to three significant figures.

(b)(vii)
Using \(A=\ln5\), \(a=0.0854528\ldots\) and \(t=10\):

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

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

\(x=100.839\ldots\)

✅ Answer: \(101\) wolves

(c)
The actual population after \(10\) years is \(85\).

The logistic model predicts \(107\), while the Gompertz model predicts \(101\). Both models overestimate the population.

However, the Gompertz model is closer to the measured value.

✅ Answer: Both models overestimate the wolf population, but the Gompertz model gives the more accurate prediction.

(d)(i)
The modified differential equation is:

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

Using Euler’s formula with step size \(h=0.5\):

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

Starting with \(x_0=70\) when \(t=5\), ten Euler steps are required to reach \(t=10\).

Therefore, when \(t=10\):

✅ Answer: \(89\) wolves, to the nearest whole number

(d)(ii)
The modified model predicts \(89\) wolves, which is much closer to the measured population of \(85\) than the earlier predictions of \(107\) and \(101\).

This suggests that including the emigration of wolves makes the model more realistic and accurate.

✅ Answer: The modified model gives a better prediction because it accounts for wolves leaving the protected area.