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8. (a) Express \( \frac{1}{P(5-P)} \) in partial fractions - Edexcel - A-Level Maths Pure - Question 2 - 2011 - Paper 5
Question 2
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 populati... show full transcript
Worked Solution & Example Answer:8. (a) Express \( \frac{1}{P(5-P)} \) in partial fractions - Edexcel - A-Level Maths Pure - Question 2 - 2011 - Paper 5
Step 1
Express \( \frac{1}{P(5-P)} \) in partial fractions.
Answer
To express ( \frac{1}{P(5-P)} ) in partial fractions, we assume it can be written as: [ \frac{1}{P(5-P)} = \frac{A}{P} + \frac{B}{5-P} ] Multiplying through by the common denominator ( P(5-P) ) gives: [ 1 = A(5-P) + BP ] Expanding this, we have: [ 1 = 5A - AP + BP ] Rearranging and equating coefficients, we get two equations:
- For the constant term: ( 5A = 1 ) leads to ( A = \frac{1}{5} )
- For the coefficients of ( P ): ( -A + B = 0 ) gives ( B = A = \frac{1}{5} )
Thus, we have: [ \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 ] Integrating both sides, we get: [ \int \frac{1}{P(5-P)} dP = \frac{1}{15} \int dt ] Using partial fractions from part (a), we find: [ \int \left( \frac{1/5}{P} + \frac{1/5}{5-P} \right) dP = \frac{1}{15} t + C ] The integrals yield: [ \frac{1}{5} \ln|P| - \frac{1}{5} \ln|5-P| = \frac{1}{15} t + C ] This simplifies to: [ \ln \left( \frac{P}{5-P} \right) = \frac{5}{15} t + C' ] Exponentiating gives: [ \frac{P}{5-P} = e^{C'} e^{\frac{1}{3}t} ] Letting ( k = e^{C'} ): [ P = \frac{5k}{1+k} ] and solving for ( P ) yields: [ P = \frac{25}{5 + 20e^{-\frac{1}{3}t}} ]
Step 3
Hence show that the population cannot exceed 5000.
Answer
From part (b), observing the solution ( P = \frac{25}{5 + 20e^{-\frac{1}{3}t}} ), we analyze the limit as ( t \to \infty ):
- As ( t o \infty ), ( e^{-\frac{1}{3}t} \to 0 ): therefore, ( P = \frac{25}{5 + 0} = 5 ) (in thousands). This implies that the maximum population ( P = 5 ) translates to 5000 meerkats, confirming that the population indeed cannot exceed 5000.