A rare species of primrose is being studied - Edexcel - A-Level Maths Pure - Question 1 - 2014 - Paper 5

To find t when P = 250, we substitute 250 into the population equation:

$250 = \frac{800 e^{0.1t}}{1 + 3 e^{0.1t}}.$

Cross-multiplying gives:

$250(1 + 3 e^{0.1t}) = 800 e^{0.1t}$

This simplifies to:

$250 + 750 e^{0.1t} = 800 e^{0.1t}$

Rearranging leads to:

$250 = 50 e^{0.1t} \Rightarrow e^{0.1t} = \frac{250}{50} = 5.$

Taking the natural logarithm:

$0.1t = \ln(5) \Rightarrow t = 10 \ln(5).$

To find ( \frac{dP}{dt} ), we first differentiate the population equation. We use the quotient rule:

$P = \frac{800 e^{0.1t}}{1 + 3 e^{0.1t}}.$

Applying the quotient rule:

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

Evaluating this at t = 10:

First calculate e^{0.1 \cdot 10} = e^1 \

Substituting this in:

$\frac{dP}{dt} = \frac{(1 + 3e)(800 \cdot 0.1 e) - 800e \cdot (3 \cdot 0.1 e)}{(1 + 3e)^2}$
This simplifies to:

$\frac{800(0.1) e (1 + 3e) - 800(0.3)e^2}{(1 + 3e)^2}.$

To understand why the population can never reach 270, we analyze the limit of the population equation:

As t approaches infinity, we find:

$P = \frac{800 e^{0.1t}}{1 + 3 e^{0.1t}} \approx \frac{800 e^{0.1t}}{3 e^{0.1t}} = \frac{800}{3} \approx 266.67.$
Thus, as time goes on, the population nears but never reaches 270, since the maximum population tends towards approximately 266.67.