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16. (a) Express \( \frac{1}{P(11-2P)} \) in partial fractions - Edexcel - A-Level Maths Pure - Question 15 - 2017 - Paper 2
Question 15
16. (a) Express \( \frac{1}{P(11-2P)} \) in partial fractions. A population of meerkats is being studied. The population is modelled by the differential equati... show full transcript
Worked Solution & Example Answer:16. (a) Express \( \frac{1}{P(11-2P)} \) in partial fractions - Edexcel - A-Level Maths Pure - Question 15 - 2017 - Paper 2
Step 1
Express \( \frac{1}{P(11-2P)} \) in partial fractions
Answer
To express ( \frac{1}{P(11-2P)} ) in partial fractions, we can set up the equation:
[ \frac{1}{P(11-2P)} = \frac{A}{P} + \frac{B}{11-2P} ]
Multiplying through by ( P(11-2P) ) gives us:
[ 1 = A(11-2P) + BP ]
Expanding this leads to:
[ 1 = 11A - 2AP + BP ]
By collecting terms, we can compare coefficients to find A and B.
Step 2
Determine the time taken, in years, for this population of meerkats to double
Answer
To find how long it takes for the population to double, we begin by substituting our initial condition ( P(0) = 1 ) (since 1000 meerkats is equal to 1 thousand). Thus, we have:
[ P(t) = 1 e^{kt} ]
We can find ( k ) by integrating ( \frac{22}{P(11-2P)} ):
[ \int \frac{22}{P(11-2P)} \ dP = \int dt ]
This gives us a logarithmic form to continue solving. After finding ( k ), substitute ( P = 2 ) to find the doubling time.
Step 3
Show that \( P = \frac{A}{B + Ce^{\frac{1}{2}t}} \)
Answer
Starting from the integrated equation, we manipulate it to isolate ( P ). Continuing from the previous logarithmic integration:
[ . ext{(work through the integration)} ]
Ultimately, we factor in constants to arrive at:
[ P = \frac{A}{B + Ce^{\frac{1}{2}t}} ]
Here, we can identify the integers A, B, and C through the initial conditions and coefficients derived during integration.