June 2024 Paper 1 Q10
10 Zac is measuring the growth of a culture of bacteria in a laboratory. The initial area of the culture is \(8\ \text{cm}^2\). The area one day later is \(8.8\ \text{cm}^2\).
At first, Zac uses a model of the form \(A = a + bt\), where \(A\ \text{cm}^2\) is the area \(t\) days after he begins measuring and \(a\) and \(b\) are constants.
Explain why this model may not be suitable after the first day. [1]
Zac decides to use a different model for \(A\). His new model is \(A = P\mathrm{e}^{kt}\), where \(P\) and \(k\) are constants.
| Scheme | Marks | AO |
|---|---|---|
| When \(t = 0, A = 8\) so \(a = 8\) | B1 | 3.3 |
| When \(t = 1, A = 8.8\) so \(b = 0.8\) | B1 | 3.3 |
| [2] |
Notes
B1: Allow embedded in \(A = 8 + 0.8t\)
| Scheme | Marks | AO |
|---|---|---|
| \(15 = 8 + 0.8t\) so \(t = 8.75\) | B1 | 3.4 |
| [1] |
Notes
B1: Allow for “after 9 days” oe FT their values in (a)
| Scheme | Marks | AO |
|---|---|---|
| The model is linear so gives the same increase each day. 10% increase each day gives bigger increases as the size of the culture increases. | B1 | 3.5b |
| [1] |
Notes
B1: Must indicate the mis-match between the linear model and the exponential observed results either in general terms or for a particular day
(Amounts are 8, 8.8 and 9.6 for the model and 8, 8.8 9.68 for the 10% increase)
(see appendix)
Appendix: exemplar responses for Q10(c)
| Response | Mark |
|---|---|
| Larger values of \(t\) give inaccurate results | B0 |
| Exponential doesn’t give the same increase each day | B0 |
| 10% of the new area is not 10% of the original area | B0 |
| Original model predicts an increase of 0.8 each day but the increase 10% each day so the model is an underestimate | B1 |
| This model doesn’t grow exponentially but increasing by 10% each day would be modelled by that | B1 |
| The model is linear whereas 10% each day is exponential | B1 |
| Scheme | Marks | AO |
|---|---|---|
| Using \(A = P\mathrm{e}^{kt}\) when \(t = 0\) gives \(P = 8\) | B1 | 3.3 |
| When \(t = 1,\quad 8.8 = 8\mathrm{e}^{1k}\) | M1 | 3.3 |
| So \(k = \ln 1.1 = [0.0953\ldots]\) | A1 | 3.3 |
| [3] |
Notes
B1: cao
M1: Forming an equation for \(k\) using \(t = 1\) and \(A = 8.8\) oe
FT their value for \(P\)
A1: Allow for \(\ln 1.1\) or a decimal answer to at least 2sf
| Scheme | Marks | AO |
|---|---|---|
| Area \(15\ \text{cm}^2\) when \(15 = 8\mathrm{e}^{(\ln 1.1)t}\) \(t = \dfrac{\ln\left(\frac{15}{8}\right)}{\ln 1.1} = 6.60\) | M1 A1 | 3.4 1.1 |
| [2] |
Notes
M1: Correct use of logs in an attempt to solve indicial equation
Similarly for \(15 = 8 \times 1.1^t\)
A1: FT their \(P\) and \(k\)
| Scheme | Marks | AO |
|---|---|---|
| The model predicts unlimited growth which is not possible in the laboratory | B1 | 3.5b |
| [1] |
Notes
B1: Must describe what the model predicts and compare with the situation being modelled (see appendix)
Appendix: exemplar responses for Q10(f)
| Response | Mark |
|---|---|
| It would get too big | B0 |
| It would get impossibly big | B1 |
| There is no limit to the size of the area, but growth will eventually reduce or stop | B0 |
| The area would be too large to be plausible and have to outgrow the lab (would be implies the model) | B1 BOD |
| For large values of \(t\), the area would be too large (large values of \(t\) implies the model being used) | B0 |
| For large values of \(t\), the area would be impossibly large | B1 |
| It suggests that the culture grows to a size that is unrealistic (“it” refers to the model) | B1 |
| The model predicts unlimited growth but this is not possible | B1 |