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:

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

Cross-multiplying gives:

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

This simplifies to:

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

Rearranging the equation:

$$250 = 800e^{0.1t} - 750e^{0.1t} = 50e^{0.1t}.$$

Dividing through by 50 gives:

$$5 = e^{0.1t}.$$

Taking the natural logarithm:

$$\ln(5) = 0.1t \quad \Rightarrow \quad t = \frac{\ln(5)}{0.1} = 10 \ln(5).$$

This can be expressed as:

10 \ln(5), ext{ where } a = 10, b = 5.$$

To find \frac{dP}{dt}$, we first need to differentiate P with respect to t. Using the quotient rule:

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

let u = 800e^{0.1t} and v = 1 + 3e^{0.1t}.

Using the quotient rule:

$$\frac{dP}{dt} = \frac{v \frac{du}{dt} - u \frac{dv}{dt}}{v^2}.$$

Where:

\frac{du}{dt} = 800 \cdot 0.1 e^{0.1t}, \quad \frac{dv}{dt} = 3 \cdot 0.1 e^{0.1t} \cdot 3,$$ At t = 10: First, evaluate: $$e^{0.1 \cdot 10} = e^{1}.$$ Thus: $$rac{du}{dt} = 80 e^{1},\frac{dv}{dt} = 0.3 e^{1}.$$ Substituting back: $$\frac{dP}{dt} = \frac{(1+3e^1)(80e^1)-(800e^1)(0.3e^1)}{(1+3e^1)^2}.$$ Now simplifying: This will yield a specific numerical value that can be computed further.

To explain why the population cannot reach 270, we observe the equation for P:

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

The maximum population occurs as t approaches infinity, which is:

$$\lim_{t \to \infty} P = \frac{800}{3} \approx 266.67.$$

Since the maximum population approaches 266, it can never reach 270, making it impossible for the population of primroses to be 270.