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7. (a) Express \( \frac{2}{P(P - 2)} \) in partial fractions - Edexcel - A-Level Maths Pure - Question 1 - 2015 - Paper 4
Question 1
7. (a) Express \( \frac{2}{P(P - 2)} \) in partial fractions. A team of biologists is studying a population of a particular species of animal. The population is mo... show full transcript
Worked Solution & Example Answer:7. (a) Express \( \frac{2}{P(P - 2)} \) in partial fractions - Edexcel - A-Level Maths Pure - Question 1 - 2015 - Paper 4
Step 1
Express \( \frac{2}{P(P - 2)} \) in partial fractions
Answer
To express ( \frac{2}{P(P - 2)} ) in partial fractions, we assume a form:
[ \frac{2}{P(P - 2)} = \frac{A}{P} + \frac{B}{P - 2} ]
Multiplying through by the denominator ( P(P - 2) ) gives:
[ 2 = A(P - 2) + BP ]
Expanding and rearranging leads to:
[ 2 = AP - 2A + BP ] [ 2 = (A + B)P - 2A ]
Setting coefficients equal leads to the system of equations:
[ A + B = 0 ] [ -2A = 2 ]
From ( -2A = 2 ), we have ( A = -1 ) and substituting back gives ( B = 1 ). Thus, the partial fraction decomposition is:
[ \frac{2}{P(P - 2)} = \frac{-1}{P} + \frac{1}{P - 2} ]
Step 2
solve this differential equation to show that \( P = \frac{6}{3 - e^{-2 \sin 2t}} \)
Answer
To solve the differential equation:
[ \frac{dP}{dt} = \frac{1}{2} P(P - 2) \cos 2t ]
First, we can separate the variables:
[ \frac{1}{P(P - 2)} dP = \frac{1}{2} \cos 2t , dt ]
Integrating both sides gives us:
[ \int \frac{1}{P(P - 2)} dP = \frac{1}{2} \int \cos 2t , dt ]
The left-hand side can be integrated using the previous result:
[ , \ln |P - 2| - \ln |P| = \sin 2t + c ]
This can be rearranged to:
[ \ln \left( \frac{P - 2}{P} \right) = \sin 2t + c ]
Exponentiating gives:
[ \frac{P - 2}{P} = e^{\sin 2t + c} ] [ P - 2 = P e^{\sin 2t + c} ]
By simplifying, we find:
[ P(1 - e^{\sin 2t + c}) = 2 ] [ P = \frac{2}{1 - e^{\sin 2t + c}} ]
Substituting the initial condition ( P = 3 ) when ( t = 0 ), we can find the constant:
[ 3 = \frac{2}{1 - e^{c}} \implies 1 - e^{c} = \frac{2}{3} \implies e^{c} = \frac{1}{3} ]
This leads to the formula:
[ P = \frac{6}{3 - e^{-2 \sin 2t}} ]
Step 3
find the time taken for the population to reach 4000 for the first time
Answer
To find the time for which ( P = 4000 ) (or ( P = 4 ) in thousands):
Using the derived formula:
[ 4 = \frac{6}{3 - e^{-2 \sin 2t}} ]
Cross-multiplying gives:
[ 4(3 - e^{-2 \sin 2t}) = 6 ]
Solving for ( e^{-2 \sin 2t} ) leads to:
[ 12 - 4e^{-2 \sin 2t} = 6 \implies 6 = 4e^{-2 \sin 2t} ] [ e^{-2 \sin 2t} = \frac{3}{2} ]
This implicates that we will later need to compute:
[ -2 \sin 2t = \ln(\frac{3}{2}) ] [ \sin 2t = -\frac{1}{2} \ln(\frac{3}{2}) ]
Looking for the first occurrence where ( 2t = \frac{7\pi}{6} ), it gives:
[ t = \frac{7\pi}{12} \approx 1.83 \text{ years} ]
Rounding to three significant figures, the final answer is:
( t \approx 1.83 \text{ years} )