AS June 2024 Q2

EdexcelAS paperCurrent spec6 marksNumerical Methods

2. An area of woodland contains a mixture of blue and yellow flowers.

A study found that the proportion, \(x\), of blue flowers in the woodland area satisfies the differential equation

\[\frac{\mathrm{d}x}{\mathrm{d}t} = \frac{xt(0.8 - x)}{x^2 + 5t} \qquad t \gt 0\]

where \(t\) is the number of years since the start of the study.

Given that exactly 3 years after the start of the study half of the flowers in the woodland area were blue,

(a) use one application of the approximation formula \(\left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)_n \approx \dfrac{y_{n+1} - y_n}{h}\) to estimate the proportion of blue flowers in the woodland area half a year later. (5)

(b) Deduce from the differential equation the proportion of flowers that will be blue in the long term. (1)

Mark scheme (a)

Scheme	Marks	AO
\(t_0 = 3\) and step is half a year, so \(h = \dfrac{1}{2}\) and \(x_0 = \dfrac{1}{2}\)	B1	3.3
\(\left(\dfrac{\mathrm{d}x}{\mathrm{d}t}\right)_0 = \dfrac{0.5 \times 3 \times (0.8 - 0.5)}{0.5^2 + 5 \times 3} = \ldots\left(\dfrac{9}{305} = 0.0295\ldots\right)\)	M1 A1	3.4 1.1b
So when \(t = 3.5\), \(x \approx \dfrac{1}{2} + \dfrac{1}{2} \times \text{'}\dfrac{9}{305}\text{'} = \ldots\)	M1	1.1b
\(= \dfrac{157}{305} \approx 0.51475\ldots\)	A1	3.2a
	(5)

Notes

B1: Uses the given information to set up correct parameters for the model, \(t_0 = 3, x_0 = \dfrac{1}{2}\), and \(h = \dfrac{1}{2}\) seen or implied. Look for e.g. \(x = \dfrac{1}{2}\) and \(t = 3\) used in the d.e. and \(h = \dfrac{1}{2}\) used in the approximation formula.

M1: Uses their values for \(x\) and \(t\) in the given equation to find \(\left(\dfrac{\mathrm{d}x}{\mathrm{d}t}\right)_0\)
Condone one slip when substituting their \(x\) and their \(t\) into the d.e.
This may be implied by \(\left(\dfrac{\mathrm{d}x}{\mathrm{d}t}\right)_0 = \text{awrt } 0.03\).
May be seen embedded e.g. \(x = \dfrac{1}{2} + \dfrac{1}{2}\left(\dfrac{0.5 \times 3 \times (0.8 - 0.5)}{0.5^2 + 5 \times 3}\right)\)
or may be seen in a table e.g.

\(n\)	\(x\)	\(t\)	d\(y\)/d\(x\)
0	½	3	0.0295…
1

A1: For \(\left(\dfrac{\mathrm{d}x}{\mathrm{d}t}\right)_0 = \dfrac{9}{305}\) or awrt 0.03. May be implied by subsequent work or seen as \(\dfrac{0.45}{15.25}\)

M1: Applies the approximation formula with their \(x\), their \(h\) and their \(\left(\dfrac{\mathrm{d}x}{\mathrm{d}t}\right)_0\) to find a value for \(x\).

A1: Correct proportion found, accept as fraction or awrt 0.515 or e.g. 51.5%

Scheme	Marks	AO
Long term proportion is \(\dfrac{4}{5}\)	B1	3.4
	(1)
	(6 marks)