October 2020 Paper 3 Q7
7
The population of fish in a lake is modelled by the differential equation
\(\dfrac{\mathrm{d}x}{\mathrm{d}t} = \dfrac{x(400 - x)}{400}\)
where \(x\) is the number of fish and \(t\) is the time in years.
When \(t = 0\), \(x = 100\).
Find the number of fish in the lake when \(t = 10\), as predicted by the model. [8]
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{A}{x(A - x)}\) | B1 | 1.1 |
| [1] |
Notes
B1: Denominator may be multiplied out
| Scheme | Marks | AO |
|---|---|---|
| DR \(\displaystyle\int \frac{400}{x(400 - x)}\,\mathrm{d}x = \int \mathrm{d}t\) | M1 | 1.1a |
| \(\displaystyle\int \left(\frac{1}{x} + \frac{1}{400 - x}\right)\mathrm{d}x = \int \mathrm{d}t\) | M1 | 3.1a |
| \(\ln x - \ln(400 - x) = t + c\) | M1 A1 | 1.1 1.1 |
| \(\ln\left(\dfrac{x}{400 - x}\right) = t + c\) | ||
| When \(t = 0\), \(x = 100\) so \(c = \ln\left(\dfrac{1}{3}\right)\) | M1 | 1.1 |
| \(t = \ln\left(\dfrac{3x}{400 - x}\right)\) When \(t = 10\), \(10 = \ln\left(\dfrac{3x}{400 - x}\right)\) | M1 | 3.4 |
| \(\dfrac{3x}{400 - x} = \mathrm{e}^{10} \Rightarrow 3x = \mathrm{e}^{10}(400 - x)\) | M1 | 2.1 |
| 400 | A1 | 3.2a |
| [8] |
Notes
M1: Separates variables (inc \(\mathrm{d}x\) and \(\mathrm{d}t\))
400 either side
M1: Use of result from (a) or method for partial fractions
M1: Attempt to integrate (at least one term correct)
A1: All integration correct
Or \(\ln\left|\dfrac{Ax}{400 - x}\right| = t\)
M1: Finding constant ( or \(A = 3\))
M1: Using log laws to combine the \(c\) value
M1: Attempt to rearrange to find \(x\)
A1: Must be rounded to nearest whole number
Alternative method (OR)
| Scheme | Marks |
|---|---|
| DR \(\displaystyle\int_{100}^{X} \frac{400}{x(400 - x)}\,\mathrm{d}x = \int_0^{10} \mathrm{d}t\) | M1 |
| \(\displaystyle\int_{100}^{X} \left(\frac{1}{x} + \frac{1}{400 - x}\right)\mathrm{d}x = \int_0^{10} \mathrm{d}t\) | M1 M1 |
| \(\big[\ln x - \ln(400 - x)\big]_{100}^{X} = [t]_0^{10}\) \(\left[\ln\dfrac{x}{(400 - x)}\right]_{100}^{X} = [t]_0^{10}\) | A1 |
| \(\ln\dfrac{X}{(400 - X)} - \ln\dfrac{100}{300} = 10 - 0\) \(\ln\dfrac{X}{(400 - X)} - \ln\dfrac{1}{3} = 10\) | M1 |
| \(\ln\dfrac{3X}{(400 - X)} = 10\) | M1 |
| \(\dfrac{3X}{(400 - X)} = \mathrm{e}^{10}\) \(3X = \mathrm{e}^{10}(400 - X)\) | M1 |
| \(X = 400\) | A1 |
M1: Separates variables
M1 M1: Use result from (a) or method for partial fractions
Attempt to integrate (at least one term correct)
A1: All integration correct
M1: Limits applied (condone one error)
M1: Combining log terms together
M1: Attempt to rearrange to find \(X\)
A1: Must be rounded to nearest whole number