June 2022 Paper 2 Q10
10 A gardener has a greenhouse containing 900 tomato plants.
The gardener notices that some of the tomato plants are damaged by insects.
Initially there are 25 damaged tomato plants.
The number of tomato plants damaged by insects is increasing by 32% each day.
(a) The total number of plants damaged by insects, \(x\), is modelled by\[x = A \times B^t\]where \(A\) and \(B\) are constants and \(t\) is the number of days after the gardener first noticed the damaged plants.
(i) Use this model to find the total number of plants damaged by insects 5 days after the gardener noticed the damaged plants. [3 marks]
(ii) Explain why this model is not realistic in the long term. [2 marks]
(b) A refined model assumes the rate of increase of the number of plants damaged by insects is given by\[\frac{\mathrm{d}x}{\mathrm{d}t} = \frac{x(900 - x)}{2700}\]
(i) Show that\[\int \left(\frac{A}{x} + \frac{B}{900 - x}\right)\mathrm{d}x = \int \mathrm{d}t\]where \(A\) and \(B\) are positive integers to be found. [3 marks]
(ii) Hence, find \(t\) in terms of \(x\). [5 marks]
(iii) Hence, find the number of days it takes from when the damage is first noticed until half of the plants are damaged by the insects. [2 marks]
| Scheme | Marks | AO |
|---|---|---|
| (i) Forms correct model Or applies repeated percentage increase 4 times PI by AWRT 75.9 | B1 | 3.3 |
| Substitutes \(t\) = 5 into their model Or Applies repeated percentage increase 5 times | M1 | 3.4 |
| Obtains 101 Condone 100 CAO | A1 | 3.2a |
| (3) | ||
| (ii) Explains that the model grows exponentially Must refer to model and exponential | E1 | 3.5b |
| Refers to 900 plants. eg this can’t be true as there are only 900 tomato plants Condone reference to “tomato(es)” or “plants” in place of “tomato plants” | E1 | 3.5a |
| (2) |
Typical solution
(i)
\[x = 25 \times 1.32^t\]\[t = 5 \Rightarrow x = 25 \times 1.32^5 = 100.18\ldots\]\[x = 101\](ii)
The value predicted by the exponential model will grow without limit.
This can’t be true as there are only 900 tomato plants in the greenhouse.
| Scheme | Marks | AO |
|---|---|---|
| (i) Rearranges to obtain one of the following: \(\dfrac{P}{x(900 - x)}\dfrac{\mathrm{d}x}{\mathrm{d}t} = \dfrac{1}{Q}\) \(\dfrac{P}{x(900 - x)}\,\mathrm{d}x = \dfrac{1}{Q}\,\mathrm{d}t\) \(\dfrac{P}{x(900 - x)} = \dfrac{1}{Q}\dfrac{\mathrm{d}t}{\mathrm{d}x}\) where \(P \times Q = 2700\) If their \(P\) = 2700 no need to see explicit \(\dfrac{1}{Q}\) with \(\mathrm{d}t\), or \(\dfrac{\mathrm{d}t}{\mathrm{d}x}\) May include integral signs | B1 | 3.1a |
| Forms partial fraction equation with correct denominators and uses an appropriate method to find their numerators PI by correct \(A\) and \(B\) without incorrect working | M1 | 3.1a |
| Obtains correct \(A\) and \(B\) and concludes with \(\displaystyle\int \left(\frac{3}{x} + \frac{3}{900 - x}\right)\mathrm{d}x = \int \mathrm{d}t\) Accept \(\displaystyle\int 1\,\mathrm{d}t\) Condone missing brackets | R1 | 2.1 |
| (3) | ||
| (ii) Integrates to obtain \(\ln x\) or \(\pm\ln(900 - x)\) Condone missing brackets for this mark | M1 | 3.1a |
| Integrates to obtain \(\ln x\) and \(\pm\ln(900 - x)\) Condone missing brackets for this mark | M1 | 1.1a |
| Integrates to obtain \(3\big(\ln x - \ln(900 - x)\big) + c = t\) OE Condone missing \(+c\) | A1 | 1.1b |
| Uses \(t\) = 0, \(x\) = 25, to obtain a value for \(c\) | M1 | 3.4 |
| Obtains correct equation for \(t\) in terms of \(x\) ACF If \(c\) given as a decimal accept AWRT 11 eg \(t = 3\big(\ln x - \ln(900 - x)\big) + 3\ln 35\) \(t = 3\ln\left(\dfrac{35x}{900 - x}\right)\) | A1 | 1.1b |
| (5) | ||
| (iii) Substitutes \(x\) = 450 into their model from part (b)(ii) | M1 | 3.4 |
| Obtains 11 CAO | A1 | 3.2a |
| (2) | ||
| (15 marks) |
Typical solution
(i)
\[\frac{\mathrm{d}x}{\mathrm{d}t} = \frac{x(900 - x)}{2700}\]\[\frac{2700}{x(900 - x)}\,\frac{\mathrm{d}x}{\mathrm{d}t} = 1\]\[\int \left(\frac{A}{x} + \frac{B}{900 - x}\right)\mathrm{d}x = \int \mathrm{d}t\]\[\frac{2700}{x(900 - x)} \equiv \frac{A}{x} + \frac{B}{900 - x}\]\[2700 \equiv A(900 - x) + Bx\]\[x = 0 \Rightarrow A = \frac{2700}{900} = 3\]\[x = 900 \Rightarrow B = 3\]\[\therefore \int \left(\frac{3}{x} + \frac{3}{900 - x}\right)\mathrm{d}x = \int \mathrm{d}t\](ii)
\[3\big(\ln x - \ln(900 - x)\big) + c = t\]\[3\big(\ln 25 - \ln(900 - 25)\big) + c = 0\]\[c = 10.67\]\[t = 3\big(\ln x - \ln(900 - x)\big) + 10.67\](iii)
\[3\big(\ln 450 - \ln(450)\big) + 10.67 = 10.67\ldots\]It takes 11 days from when the damage is first noticed until half of the plants are damaged by insects