A rare species of primrose is being studied - Edexcel - A-Level Maths Pure - Question 9 - 2014 - Paper 5

To find ( t ) when ( P = 250 ), we set up the equation:

$250 = \frac{800e^{0.1t}}{1 + 3e^{0.1t}}.\n$ Cross-multiplying gives:

$250(1 + 3e^{0.1t}) = 800e^{0.1t}.$ Expanding this, we have:

$250 + 750e^{0.1t} = 800e^{0.1t}.$ Bringing terms involving ( e^{0.1t} ) together results in:

$250 = 800e^{0.1t} - 750e^{0.1t} = 50e^{0.1t}.$ Solving for ( e^{0.1t} ):
$e^{0.1t} = \frac{250}{50} = 5.$ Taking natural logarithms gives:

$0.1t = \ln(5), \quad\text{thus } t = 10 \ln(5).$

To find ( \frac{dP}{dt} ), we differentiate the population equation using the quotient rule:

$P = \frac{800e^{0.1t}}{1 + 3e^{0.1t}}.$ Using the quotient rule:

$\frac{dP}{dt} = \frac{(1 + 3e^{0.1t}) \cdot (800 \cdot 0.1 e^{0.1t}) - (800e^{0.1t}) \cdot (3 \cdot 0.1 e^{0.1t})}{(1 + 3e^{0.1t})^2}$ At ( t = 10 ):

First, we find ( P ):

$P = \frac{800e^{1}}{1 + 3e^{1}}.$
Now substituting back into the derivative we calculate:

$\frac{dP}{dt} = \frac{800 \cdot 0.1 e^{1} \cdot (1 + 3e^{1}) - (400e^{1}) \cdot (3e^{1})}{(1 + 3e^{1})^2}.$
This simplifies to yield a final answer for ( \frac{dP}{dt} ). Calculating this gives approximately 266.

To determine why the population of primroses can never reach 270, we analyze the behavior of the function:

$P = \frac{800e^{0.1t}}{1 + 3e^{0.1t}}.$
As ( t ) approaches infinity, the limit of ( P ) is:

$P = \frac{800}{3} \approx 266.67.$
This means that the population approaches, but never exceeds, this value. Thus, the population can never be 270.