Question 14 - A-Level Maths

Edexcel PMT Mocks — Question 14 14 marks

Exam Board	Edexcel
Module	PMT Mocks (PMT Mocks)
Marks	14
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Differential equations
Type	Logistic/bounded growth
Difficulty	Standard +0.3 This is a structured logistic growth question with clear signposting through multiple parts. Parts (a)-(d) involve routine substitution, algebraic manipulation, and solving exponential equations—all standard A-level techniques. Part (e) requires differentiation of a quotient and finding a maximum, but the 'show that' format provides the target answer. While it covers several skills, each step is straightforward with no novel insight required, making it slightly easier than average.
Spec	1.06i Exponential growth/decay: in modelling context 1.07j Differentiate exponentials: e^(kx) and a^(kx)1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates

14. A population of ants being studied on an island. The number of ants, $P$, in the population, is modelled by the equation. $$P = \frac { 900 k e ^ { 0.2 t } } { 1 + k e ^ { 0.2 t } } , \text { where } k \text { is a constant. }$$ Given that there were 360 ants when the study started,
a. show that $k = \frac { 2 } { 3 }$.
b. Show that $P = \frac { 1800 } { 2 + 3 e ^ { - 0.2 t } }$. The model predicts an upper limit to the number of ants on the island.
c. State the value of this limit.
d. Find the value of $t$ when $P = 520$. Give your answer to one decimal place.
e. i. Show that the rate of growth, $\frac { \mathrm { d } P } { d t } = \frac { P ( 900 - P ) } { 4500 }$ ii. Hence state the value of $P$ at which the rate of growth is a maximum.

Show mark scheme Show mark scheme source

Question 14:

Part a: Show that $k = \frac{2}{3}$

Answer	Marks	Guidance
Working/Answer	Mark	Guidance
At $t = 0$, $P = 360 \Rightarrow 360 = \frac{900k}{1+k}$	M1	Setting $P = 360$ at $t = 0$
$360(1+k) = 900k \Rightarrow 360 + 360k = 900k$	M1	Obtains equation in form $360 = 540k$
$k = \frac{2}{3}$	A1	Correct solution only

(3 marks)

Part b: Show that $P = \frac{1800}{2+3e^{-0.2t}}$

Answer	Marks	Guidance
Working/Answer	Mark	Guidance
$P = \frac{900 \times \frac{2}{3}e^{0.2t}}{1 + \frac{2}{3}e^{0.2t}} = \frac{1800e^{0.2t}}{3 + 2e^{0.2t}} = \frac{1800}{3e^{-0.2t}+2}$	B1	Substitutes $k = \frac{2}{3}$ and divides numerator and denominator by $e^{0.2t}$ to get $P = \frac{1800}{2+3e^{-0.2t}}$

(1 mark)

Part c: State the upper limit

Answer	Marks	Guidance
Working/Answer	Mark	Guidance
As $t \to \infty$, $P = 900$	B1	Sets $t \to \infty$

(1 mark)

Part d: Find $t$ when $P = 520$

Answer	Marks	Guidance
Working/Answer	Mark	Guidance
$520 = \frac{1800}{2+3e^{-0.2t}} \Rightarrow 2 + 3e^{-0.2t} = \frac{1800}{520}$	M1	Substitutes $P = 520$ and proceeds to equation of form $Ae^{-0.2t} = B$
$3e^{-0.2t} = \frac{19}{13} \Rightarrow e^{-0.2t} = \frac{19}{39}$	A1	Correct equation: $1560e^{-0.2t} = 760$ or equivalent
$-0.2t = \ln\frac{19}{39}$	M1	Correct order of operations using ln to make $t$ the subject
$t = 3.6$	A1	Correct answer only

(4 marks)

Part e(i): Show $\frac{dP}{dt} = \frac{P(900-P)}{4500}$

Answer	Marks	Guidance
Working/Answer	Mark	Guidance
$\frac{dP}{dt} = \frac{1800 \cdot (-0.6)e^{-0.2t}}{(2+3e^{-0.2t})^2}$	M1	Attempts to differentiate using quotient rule
$= \frac{1080e^{-0.2t}}{(2+3e^{-0.2t})^2}$	A1	Correct differentiation: $\frac{1080e^{-0.2t}}{(2+3e^{-0.2t})^2}$
Substitutes $2 + 3e^{-0.2t} = \frac{1800}{P}$ and $e^{-0.2t} = \frac{\frac{1800}{P}-2}{3}$ into $\frac{dP}{dt}$	M1	Substitutes to form equation linking $\frac{dP}{dt}$ and $P$
$\frac{dP}{dt} = \frac{1800P - 2P^2}{9000} = \frac{P(900-P)}{4500}$	A1	Correct algebra leading to $\frac{dP}{dt} = \frac{P(900-P)}{4500}$

Part e(ii): Value of P at maximum rate of growth

Answer	Marks	Guidance
Working/Answer	Mark	Guidance
$\frac{d^2P}{dt^2} = 900 - 2P = 0 \Rightarrow P = 450$	B1	$P = 450$

(5 marks total for part e)

(Total for Question 14: 14 marks)

## Question 14:

### Part a: Show that $k = \frac{2}{3}$

| Working/Answer | Mark | Guidance |
|---|---|---|
| At $t = 0$, $P = 360 \Rightarrow 360 = \frac{900k}{1+k}$ | M1 | Setting $P = 360$ at $t = 0$ |
| $360(1+k) = 900k \Rightarrow 360 + 360k = 900k$ | M1 | Obtains equation in form $360 = 540k$ |
| $k = \frac{2}{3}$ | A1 | Correct solution only |

**(3 marks)**

### Part b: Show that $P = \frac{1800}{2+3e^{-0.2t}}$

| Working/Answer | Mark | Guidance |
|---|---|---|
| $P = \frac{900 \times \frac{2}{3}e^{0.2t}}{1 + \frac{2}{3}e^{0.2t}} = \frac{1800e^{0.2t}}{3 + 2e^{0.2t}} = \frac{1800}{3e^{-0.2t}+2}$ | B1 | Substitutes $k = \frac{2}{3}$ and divides numerator and denominator by $e^{0.2t}$ to get $P = \frac{1800}{2+3e^{-0.2t}}$ |

**(1 mark)**

### Part c: State the upper limit

| Working/Answer | Mark | Guidance |
|---|---|---|
| As $t \to \infty$, $P = 900$ | B1 | Sets $t \to \infty$ |

**(1 mark)**

### Part d: Find $t$ when $P = 520$

| Working/Answer | Mark | Guidance |
|---|---|---|
| $520 = \frac{1800}{2+3e^{-0.2t}} \Rightarrow 2 + 3e^{-0.2t} = \frac{1800}{520}$ | M1 | Substitutes $P = 520$ and proceeds to equation of form $Ae^{-0.2t} = B$ |
| $3e^{-0.2t} = \frac{19}{13} \Rightarrow e^{-0.2t} = \frac{19}{39}$ | A1 | Correct equation: $1560e^{-0.2t} = 760$ or equivalent |
| $-0.2t = \ln\frac{19}{39}$ | M1 | Correct order of operations using ln to make $t$ the subject |
| $t = 3.6$ | A1 | Correct answer only |

**(4 marks)**

### Part e(i): Show $\frac{dP}{dt} = \frac{P(900-P)}{4500}$

| Working/Answer | Mark | Guidance |
|---|---|---|
| $\frac{dP}{dt} = \frac{1800 \cdot (-0.6)e^{-0.2t}}{(2+3e^{-0.2t})^2}$ | M1 | Attempts to differentiate using quotient rule |
| $= \frac{1080e^{-0.2t}}{(2+3e^{-0.2t})^2}$ | A1 | Correct differentiation: $\frac{1080e^{-0.2t}}{(2+3e^{-0.2t})^2}$ |
| Substitutes $2 + 3e^{-0.2t} = \frac{1800}{P}$ and $e^{-0.2t} = \frac{\frac{1800}{P}-2}{3}$ into $\frac{dP}{dt}$ | M1 | Substitutes to form equation linking $\frac{dP}{dt}$ and $P$ |
| $\frac{dP}{dt} = \frac{1800P - 2P^2}{9000} = \frac{P(900-P)}{4500}$ | A1 | Correct algebra leading to $\frac{dP}{dt} = \frac{P(900-P)}{4500}$ |

### Part e(ii): Value of P at maximum rate of growth

| Working/Answer | Mark | Guidance |
|---|---|---|
| $\frac{d^2P}{dt^2} = 900 - 2P = 0 \Rightarrow P = 450$ | B1 | $P = 450$ |

**(5 marks total for part e)**

**(Total for Question 14: 14 marks)**

Show LaTeX source

14. A population of ants being studied on an island. The number of ants, $P$, in the population, is modelled by the equation.

$$P = \frac { 900 k e ^ { 0.2 t } } { 1 + k e ^ { 0.2 t } } , \text { where } k \text { is a constant. }$$

Given that there were 360 ants when the study started,\\
a. show that $k = \frac { 2 } { 3 }$.\\
b. Show that $P = \frac { 1800 } { 2 + 3 e ^ { - 0.2 t } }$.

The model predicts an upper limit to the number of ants on the island.\\
c. State the value of this limit.\\
d. Find the value of $t$ when $P = 520$. Give your answer to one decimal place.\\
e. i. Show that the rate of growth, $\frac { \mathrm { d } P } { d t } = \frac { P ( 900 - P ) } { 4500 }$\\
ii. Hence state the value of $P$ at which the rate of growth is a maximum.

\hfill \mbox{\textit{Edexcel PMT Mocks  Q14 [14]}}

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Edexcel PMT Mocks — Question 14 14 marks

Question 14:

Part a: Show that \(k = \frac{2}{3}\)

(3 marks)

Part b: Show that \(P = \frac{1800}{2+3e^{-0.2t}}\)

(1 mark)

Part c: State the upper limit

(1 mark)

Part d: Find \(t\) when \(P = 520\)

(4 marks)

Part e(i): Show \(\frac{dP}{dt} = \frac{P(900-P)}{4500}\)

Part e(ii): Value of P at maximum rate of growth

(5 marks total for part e)

(Total for Question 14: 14 marks)

This paper (14 questions)