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8. (a) Express \( \frac{1}{P(5-P)} \) in partial fractions - Edexcel - A-Level Maths Pure - Question 1 - 2012 - Paper 8
Question 1
8. (a) Express \( \frac{1}{P(5-P)} \) in partial fractions. A team of conservationists is studying the population of meerkats on a nature reserve. The population ... 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 - 2012 - 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 assume:
[ \frac{1}{P(5-P)} = \frac{A}{P} + \frac{B}{5-P} ]
Multiplying through by ( P(5-P) ) gives:
[ 1 = A(5-P) + BP ]
When ( P = 0, ) we have ( 1 = 5A ) leading to ( A = \frac{1}{5}. )
When ( P = 5, ) we have ( 1 = 5B ) leading to ( B = \frac{1}{5}. )
Thus, we find:
[ \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) ]
we separate variables:
[ \frac{1}{P(5-P)} dP = \frac{1}{15} dt ]
Substituting our earlier result for ( \frac{1}{P(5-P)} ):
[ \left( \frac{1/5}{P} + \frac{1/5}{5-P} \right) dP = \frac{1}{15} dt ]
Integrating both sides,
[ \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{1}{3}t + C'].
From this, we derive:
[ \frac{P}{5-P} = e^{C'} e^{t/3} ].
Letting ( K = e^{C'} ), we get:
[ P = \frac{5K}{1 + K e^{-t/3}}. ]
Substituting constant definitions leads to the required form:
[ P = \frac{-5}{-5 + 20e^{-t/3}}. ]
Step 3
Hence show that the population cannot exceed 5000.
Answer
Given the form of the solution,
[ P = \frac{5 \cdot 20}{-5 + 20e^{-t/3}} ]
For ( P ) to be positive and bounded, ( -5 + 20e^{-t/3} ) must be positive, leading to the conclusion that ( P < 5000 ) as:
[ P = 20 ] [ \Rightarrow 5 \cdot 20 < 5000. ]
Thus, the population cannot exceed 5000.