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8. (a) Express $\frac{1}{P(5 - P)}$ in partial fractions - Edexcel - A-Level Maths Pure - Question 1 - 2011 - Paper 8
Question 1
8. (a) Express $\frac{1}{P(5 - P)}$ in partial fractions. (3) A team of conservationists is studying the population of meerkats on a nature reserve. The popula... show full transcript
Worked Solution & Example Answer:8. (a) Express $\frac{1}{P(5 - P)}$ in partial fractions - Edexcel - A-Level Maths Pure - Question 1 - 2011 - Paper 8
Step 1
Express $\frac{1}{P(5 - P)}$ in partial fractions.
Answer
To express ( \frac{1}{P(5 - P)} ) in partial fractions, we can assume:
[
\frac{1}{P(5 - P)} = \frac{A}{P} + \frac{B}{5 - P}
]
Multiplying through by the denominator gives:
[
1 = A(5 - P) + BP
]
Setting coefficients for (P) and the constant term:
- For the constant term: ( A \cdot 5 = 1 \Rightarrow A = \frac{1}{5} )
- For the coefficient of (P): ( -A + B = 0 \quad \Rightarrow \quad -\frac{1}{5} + B = 0 \Rightarrow B = \frac{1}{5} )
Thus, the partial fraction decomposition is:
[
\frac{1}{P(5 - P)} = \frac{1/5}{P} + \frac{1/5}{5 - P}
]
Step 2
solve the differential equation, giving your answer in the form, $P = \frac{a}{b + ce^{tf}}$.
Answer
To solve the differential equation ( \frac{dP}{dt} = \frac{1}{15} P(5 - P) ), separate variables:
[
\frac{dP}{P(5 - P)} = \frac{1}{15} dt
]
Integrating both sides:
[
\int \frac{1}{P(5 - P)} dP = \frac{1}{15} \int dt
]
Using the partial fraction result:
[
\frac{1}{5} \ln |P| - \frac{1}{5} \ln |5 - P| = \frac{t}{15} + C
]
This simplifies to:
[
\ln \left| \frac{P}{5 - P} \right| = \frac{t}{3} + C'
]
Exponentiating gives:
[
\frac{P}{5 - P} = e^{C'}e^{\frac{t}{3}} \Rightarrow P = 5 \cdot \frac{e^{C'}e^{\frac{t}{3}}}{1 + e^{C'}e^{\frac{t}{3}}}
]
Letting ( a = 5 ), ( b = 1 ), and ( c = e^{C'} ), we have ( P = \frac{5}{1 + ce^{\frac{t}{3}}} ).
Step 3
Hence show that the population cannot exceed 5000.
Answer
From the solution ( P = \frac{5}{1 + ce^{\frac{t}{3}}} ), as ( t ) approaches infinity, ( e^{\frac{t}{3}} ) approaches infinity. Therefore, the denominator tends towards infinity, leading to:
[
P = \frac{5}{\infty} \to 0
]
This demonstrates that the maximum population is 5 when ( c ) is minimized, and thus:
[
P < 5000
]
Hence, the population of meerkats cannot exceed 5000.