The rate of increase of the number, N, of fish in a lake is modelled by the differential equation
$$\frac{dN}{dt} = \frac{(k - t)(5000 - N)}{t} \\ t > 0, \quad 0 < N < 5000$$
In the given equation, the time t is measured in years from the start of January 2000 and k is a positive constant - Edexcel - A-Level Maths: Pure - Question 1 - 2014 - Paper 8
Question 1
The rate of increase of the number, N, of fish in a lake is modelled by the differential equation
$$\frac{dN}{dt} = \frac{(k - t)(5000 - N)}{t} \\ t > 0, \quad 0 < ... show full transcript
Worked Solution & Example Answer:The rate of increase of the number, N, of fish in a lake is modelled by the differential equation
$$\frac{dN}{dt} = \frac{(k - t)(5000 - N)}{t} \\ t > 0, \quad 0 < N < 5000$$
In the given equation, the time t is measured in years from the start of January 2000 and k is a positive constant - Edexcel - A-Level Maths: Pure - Question 1 - 2014 - Paper 8
Step 1
By solving the differential equation, show that $N = 5000 - Ae^{-kt}$
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Answer
To solve the differential equation (\frac{dN}{dt} = \frac{(k - t)(5000 - N)}{t}), we can separate variables.
Rearranging the equation yields:
5000−NdN=t(k−t)dt
Integrating both sides:
∫5000−NdN=∫t(k−t)dt
This results in:
−ln(5000−N)=kln(t)−∫ttdt
Simplifying, we get:
−ln(5000−N)=kln(t)−t+C
Exponentiating both sides leads to:
5000−N=eCt−ke−t
Letting (A = e^{C}) gives us:
N=5000−Ae−kt.
Step 2
Find the exact value of the constant A and the exact value of the constant k.
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Answer
Using the information given:
After one year (January 2001):
N(1)=1200=5000−Ae−k(1)⇒A=5000−1200ek
After two years (January 2002):
N(2)=1800=5000−Ae−k(2)⇒A=5000−1800e2k
Setting the expressions for A equal:
5000−1200ek=5000−1800e2k
Rearranging yields:
1200ek=1800e2k⇒32=ek
Taking natural log gives:
k=ln(32)
Substituting k back into the equation for A, we find:
A=5000−1200∗32=3800.
Step 3
Hence find the number of fish in the lake after five years. Give your answer to the nearest hundred fish.
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