June 2023 Paper 1 Q9
9 Conservationists are studying how the number of bees in a wildflower meadow varies according to the number of wildflower plants. The study takes place over a series of weeks in the summer. A model is suggested for the number of bees, \(B\), and the number of wildflower plants, \(F\), at time \(t\) weeks after the start of the study.
In the model \(B = 20 + 2t + \cos 3t\) and \(F = 50\mathrm{e}^{0.1t}\).
The model assumes that \(B\) and \(F\) can be treated as continuous variables.
| Scheme | Marks | AO |
|---|---|---|
| The rate at which the number of bees changes relative to the rate at which the number of flowers changes or Rate of increase of bees per increase in wildflower plants | B1 | 3.3 |
| [1] |
Notes
B1: Must mention ‘rate of change’ of bees, or equiv, and how this relates to wildflower plants
B0 for ‘change’ not ‘rate of change’
B1 for rate of change of bees with respect to plants
B1 for rate of change of bees as the plants change
B1 BOD for rate of change of bees compared to the number of plants
State or imply that it is bees compared to flowers, so B0 if other way around
Must relate to bees and plants, and not just \(B\) and \(F\)
See appendix for further examples (below)
Appendix: exemplar responses for Q9(a)
| Response | Mark | Comment |
|---|---|---|
| The rate of increase in the number of bees regarding the increase in the number of plants | B1 | BOD ‘regarding’ |
| Rate of bees with respect to flowers | B0 | no ‘change’ |
| Rate of change in number of bees in terms of the number of flowers | B1 | BOD ‘in terms of’ |
| The rate of growth in the number of bees when the number of plants increases over time | B1 | |
| The rate of change of flowers according to the number of bees | B0 | wrong way around |
| The rate of increase of bees over the rate of increase in wildflowers | B0 | not ‘over’ – suggests fraction |
| The rate of change of number of bees compared to number of plants | B1 | BOD ‘number’ |
| Rate at which the number of bees increases as the number of plants increase | B1 | Includes ‘rate’ and ‘increases’ |
| The change in number of bees in respect to the number of flowers | B0 | no ‘rate’ |
| The rate of increase in number of bees in accordance to the number of plants | B1 | BOD ‘in accordance’ |
| How the number of bees vary with the number of flowers | B0 | not rate of change |
| The rate at which the bees to flowers ratio is changing | B1 | BOD ‘ratio’ |
| Rate of growth of bees depending on the number of flowers | B1 | |
| Rate of change of bees per wildflower plant | B1 | |
| Rate of change of the number of bees in terms of flowers | B1 | |
| The rate of change between the bees and the plants | B0 | no dependency implied |
| The rate of change in number of bees against the change in plants | B1 | |
| The rate of change of bees compared to flowers | B1 | |
| Rate of change of the number of bees as the number of flowers vary | B1 | |
| The rate at which the number of bees increase with the number of plants | B1 | BOD ‘with’ |
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{\mathrm{d}F}{\mathrm{d}t} = 5\mathrm{e}^{0.1t}\) | B1 | 1.1 |
| \(\dfrac{\mathrm{d}B}{\mathrm{d}t} = 2 - 3\sin 3t\) | B1 | 1.1 |
| \(\dfrac{\mathrm{d}B}{\mathrm{d}F} = \dfrac{\mathrm{d}B}{\mathrm{d}t} \times \dfrac{\mathrm{d}t}{\mathrm{d}F} = \dfrac{2 - 3\sin 3t}{5\mathrm{e}^{0.1t}}\) | M1 | 3.4 |
| \(\dfrac{\mathrm{d}B}{\mathrm{d}F} = \dfrac{2 - 3\sin 12}{5\mathrm{e}^{0.4}} = 0.484\) | A1 | 3.4 |
| [4] |
Notes
B1: Correct derivative
No need to see \(\dfrac{\mathrm{d}F}{\mathrm{d}t}\) notation
Could be unsimplified
B1: Correct derivative
No need to see \(\dfrac{\mathrm{d}B}{\mathrm{d}t}\) notation
NB watch out for \(2 - 3\sin t\)
M1: Correct method to combine their two derivatives – algebraic or numerical
B0B0M1 is possible
Must have \(\mathrm{e}^{0.1t}\) in denominator, or \(\mathrm{e}^{-0.1t}\) in the numerator, but allow muddles with the placing of the 5
A1: Substitute \(t = 4\) to obtain 0.484, or better
Alternative method
| Scheme | Marks |
|---|---|
| \(t = 10\ln\left(\dfrac{F}{50}\right)\) \(B = 20 + 20\ln\left(\dfrac{F}{50}\right) + \cos\left(30\ln\left(\dfrac{F}{50}\right)\right)\) | B1 |
| \(\dfrac{\mathrm{d}B}{\mathrm{d}F} = \dfrac{20}{F} - \dfrac{30}{F}\sin\left(30\ln\left(\dfrac{F}{50}\right)\right)\) | M1 |
| A1 | |
| \(\dfrac{\mathrm{d}B}{\mathrm{d}F} = \dfrac{20}{50\mathrm{e}^{0.4}} - \dfrac{30}{50\mathrm{e}^{0.4}}\sin\left(30\ln\left(\dfrac{50\mathrm{e}^{0.4}}{50}\right)\right)\) \(= 0.484\) | A1 |
B1: Correct expression for \(B\) as a function of \(F\)
M1: Attempt differentiation
May see ln terms split first (possibly even including use of \(\cos(A - B)\))
A1: Obtain correct derivative aef
A1: Substitute \(F = 50\mathrm{e}^{0.4}\) to obtain 0.484, or better
Could use \(t = 4\) if derivative now in terms of \(t\)
| Scheme | Marks | AO |
|---|---|---|
| The data comes from the summer, so taking it beyond 12 weeks is unlikely to be reliable | B1 | 3.5b |
| [1] |
Notes
B1: Summer will be over so pattern may not continue
Summer is not greater than 12 weeks
Fewer bees and/or flowers in autumn/winter
Any reason referring to a change in season having an effect on bees and/or flowers
B0 for just considering long-term behaviour eg flowers will not continue to increase exponentially
Reasons must reference seasons / different time of year (could be implied by ‘weather getting colder’)