A2 June 2022 Paper 1 Q8
8 A biologist is studying the effect of pesticides on crops. On a certain farm pesticide is regularly applied to a particular crop which grows in soil. Over time, pesticide is transferred between the crop and the soil at a rate which depends on the amount of pesticide in both the crop and the soil. The amount of pesticide in the crop after \(t\) days is \(x\) grams. The amount of pesticide in the soil after \(t\) days is \(y\) grams. Initially, when \(t = 0\), there is no pesticide in either the crop or the soil.
At first it is assumed that no pesticide is lost from the system. The biologist further assumes that pesticide is added to the crop at a constant rate of \(k\) grams per day, where \(k \gt 6\).
After collecting some initial data, the biologist suggests that for \(t \geqslant 0\), this situation can be modelled by the following pair of first order linear differential equations.
\(\dfrac{\mathrm{d}x}{\mathrm{d}t} = -2x + 78y + k\)
\(\dfrac{\mathrm{d}y}{\mathrm{d}t} = 2x - 78y\)
If more than 250 grams of pesticide is found in the crop, then it will fail food safety standards.
Explain why, according to this model, the crop will fail food safety standards as a result of this test. [1]
Further data collection suggests that some pesticide decays in the soil and so is lost from the system. The model is refined in light of this data. The particular solution for \(x\) for this refined model is
\(x = k\left(20 - \mathrm{e}^{-41t}\left(20\cosh\left(\sqrt{1677}\,t\right) + \dfrac{819}{\sqrt{1677}}\sinh\left(\sqrt{1677}\,t\right)\right)\right)\).
In the refined model, it is still assumed that pesticide is added to the crop at a constant rate.
| Scheme | Marks | AO |
|---|---|---|
| (i) \(\dfrac{\mathrm{d}^2x}{\mathrm{d}t^2} = -2\dfrac{\mathrm{d}x}{\mathrm{d}t} + 78(2x - 78y)\) \(= -2\dfrac{\mathrm{d}x}{\mathrm{d}t} + 156x - 78\left(\dfrac{\mathrm{d}x}{\mathrm{d}t} + 2x - k\right)\) | M1 | 3.3 |
| \(\Rightarrow \dfrac{\mathrm{d}^2x}{\mathrm{d}t^2} = -80\dfrac{\mathrm{d}x}{\mathrm{d}t} + 78k\) \(\Rightarrow \dfrac{\mathrm{d}^2x}{\mathrm{d}t^2} + 80\dfrac{\mathrm{d}x}{\mathrm{d}t} = 78k\) | A1 | 1.1 |
| [2] |
Notes
M1: Differentiate \(\dfrac{\mathrm{d}x}{\mathrm{d}t}\) and substitute \(\dfrac{\mathrm{d}y}{\mathrm{d}t}\)
A1: AG convincingly shown after second substitution.
| Scheme | Marks | AO |
|---|---|---|
| (ii) \(\lambda^2 + 80\lambda = 0\) \(\Rightarrow \lambda = 0, -80\) | M1 | 1.1 |
| Complementary function is \(x = A + B\mathrm{e}^{-80t}\) | A1 | 2.2a |
| Trial function is \(x = at\ (+\,b)\) \(\dfrac{\mathrm{d}x}{\mathrm{d}t} = a,\ \dfrac{\mathrm{d}^2x}{\mathrm{d}t^2} = 0\) \(\Rightarrow 80a = 78k \Rightarrow a = \dfrac{39}{40}k\) | M1 | 3.1a |
| \(\Rightarrow\) GS is \(x = A + B\mathrm{e}^{-80t} + \dfrac{39}{40}kt\) | A1 | 1.1 |
| When \(t = 0\), \(x = 0 \Rightarrow A + B = 0\) | M1 | 3.3 |
| \(\dfrac{\mathrm{d}x}{\mathrm{d}t} = -80B\mathrm{e}^{-80t} + \dfrac{39}{40}k\) When \(t = 0, x = 0, y = 0 \Rightarrow \dfrac{\mathrm{d}x}{\mathrm{d}t} = 0 + 0 + k\) \(\Rightarrow k = -80B + \dfrac{39}{40}k\) \(\Rightarrow B = -\dfrac{1}{3200}k,\ A = \dfrac{1}{3200}k\) | M1 | 3.1a |
| Particular solution: \(x = \dfrac{k}{3200}\left(1 - \mathrm{e}^{-80t} + 3120t\right)\) | A1 | 3.3 |
| [7] |
Notes
M1: Attempt to solve auxiliary equation for their DE
M1: Correct (or recovered) trial function for their CF.
A1: For GS
M1: Using \(t = 0\) and \(x = 0\) in their GS to find an equation in \(A\) and \(B\).
M1: Differentiating and using \(t = 0, x = 0, (y = 0), \dot{x} = k\) to find another equation in (\(A\) and) \(B\). Could also use \(\dot{x} + \dot{y} = k\) and \(x + y = kt\).
A1: N.B. \(\dfrac{3120}{3200} = \dfrac{39}{40}\)
Alternative method for first 4 marks
| Scheme | Marks |
|---|---|
| \(\dot{x} + 80x = 78kt + c\) \(\mathrm{e}^{\int 80\,\mathrm{d}t}(\dot{x} + 80x) = (78kt + c)\mathrm{e}^{\int 80\,\mathrm{d}t}\) \(\dfrac{\mathrm{d}}{\mathrm{d}t}\left(\mathrm{e}^{80t}x\right) = (78kt + c)\mathrm{e}^{80t}\) | M1 |
| \(\displaystyle\mathrm{e}^{80t}x = \int(78kt + c)\mathrm{e}^{80t}\,\mathrm{d}t + d\) | M1 |
| \(\displaystyle\mathrm{e}^{80t}x = \frac{(78kt + c)\mathrm{e}^{80t}}{80} - \int\frac{78k}{80}\mathrm{e}^{80t}\,\mathrm{d}t + d\) | M1 |
| \(\mathrm{e}^{80t}x = \dfrac{(78kt + c)\mathrm{e}^{80t}}{80} - \dfrac{39k}{3200}\mathrm{e}^{80t} + d\) \(x = \dfrac{39}{40}kt + \dfrac{c}{80} - \dfrac{39k}{3200} + d\mathrm{e}^{-80t}\) \(= A + B\mathrm{e}^{-80t} + \dfrac{39}{40}kt\) | A1 |
M1: Integrating \(\ddot{x} + 80\dot{x} = 78k\) wrt \(t\) and then using integrating factor
M1: Integrating wrt \(t\)
M1: Integrating by parts (\(78kt + c\) may be separated)
A1: GS of form \(x = A + B\mathrm{e}^{-80t} + \frac{39}{40}kt\)
Alternative method for this M mark
| Scheme | Marks |
|---|---|
| \(78y = \dot{x} + 2x - k\) \(= -78B\mathrm{e}^{-80t} - \dfrac{1}{40}k + 2A + \dfrac{78}{40}kt\) When \(t = 0, x = 0, y = 0\) \(0 = 2A - 78B - \dfrac{1}{40}k\) | M1 |
M1: Finding GS for \(y\) and using \(x = 0, y = 0\) to find another equation for \(A\) and \(B\)
Another alternative method for Q8(a)(ii) for first 4 marks
| Scheme | Marks |
|---|---|
| D.E. is \(\dfrac{\mathrm{d}y}{\mathrm{d}t} + 80y = 78k\) where \(y = \dfrac{\mathrm{d}x}{\mathrm{d}t}\) I.F. \(\mathrm{e}^{80t} \Rightarrow \mathrm{e}^{80t}\dfrac{\mathrm{d}y}{\mathrm{d}t} + 80\mathrm{e}^{80t}y = 78k\mathrm{e}^{80t}\) \(\Rightarrow \dfrac{\mathrm{d}}{\mathrm{d}t}\left(\mathrm{e}^{80t}y\right) = 78k\mathrm{e}^{80t}\) | M1 |
| \(\Rightarrow \mathrm{e}^{80t}y = \dfrac{39k}{40}\mathrm{e}^{80t} + A \Rightarrow y = \dfrac{39k}{40} + A\mathrm{e}^{-80t}\) | A1 |
| \(\Rightarrow \dfrac{\mathrm{d}x}{\mathrm{d}t} = \dfrac{39k}{40} + A\mathrm{e}^{-80t} \Rightarrow x = \dfrac{39kt}{40} - \dfrac{A}{80}\mathrm{e}^{-80t} + B\) | M1 A1 |
Mark scheme for Q8(a)(ii) if candidate solves for \(y\), rather than \(x\)
| Scheme | Marks |
|---|---|
| \(\ddot{y} = 2\dot{x} - 78\dot{y} = 2(-2x + 78y + k) - 78\dot{y}\) \(\ddot{y} = -2(\dot{y} + 78y) + 156y + 2k - 78\dot{y}\) \(\ddot{y} + 80\dot{y} = 2k\) \(\lambda^2 + 80\lambda = 0\) \(\lambda = 0, -80\) | M1 |
| Complementary function is \(y = A + B\mathrm{e}^{-80t}\) | M1 |
| Trial function is \(y = at\) \(\dot{y} = a, \ddot{y} = 0\) So \(80a = 2k \Leftrightarrow a = \frac{1}{40}k\) General solution is \(y = A + B\mathrm{e}^{-80t} + \frac{1}{40}kt\) | M1 |
| \(\dot{y} = -80B\mathrm{e}^{-80t} + \dfrac{1}{40}k\) \(x = \dfrac{1}{2}(\dot{y} + 78y)\) \(= 39A - B\mathrm{e}^{-80t} + \dfrac{39}{40}kt + \dfrac{1}{80}k\) | A1 |
| When \(t = 0, x = 0, y = 0, \dot{y} = 0 + 0 = 0\) \(0 = A + B\) \(0 = 39A - B + \dfrac{1}{80}k\) \(0 = -80B + \dfrac{1}{40}k\) | M1 M1 |
| \(B = \dfrac{1}{3200}k,\ A = -\dfrac{1}{3200}k\) Particular solution for \(x\) is \(x = -\dfrac{39}{3200}k - \dfrac{1}{3200}k\mathrm{e}^{-80t} + \dfrac{39}{40}kt + \dfrac{1}{80}k\) \(= \dfrac{1}{3200}k\left(1 - \mathrm{e}^{-80t} + 3120t\right)\) | A1 |
| [7] |
M1: Auxiliary equation for their DE
M1: Correct trial function for their CF
Note the GS for \(x\) in the main mark scheme achieves A1 but here A1 is not awarded until GS for \(x\) found.
A1: GS of form \(x = C + D\mathrm{e}^{-80t} + \frac{39}{40}kt\)
M1 M1: Using two of \(x = y = \dot{y} = 0\) to find simultaneous equations for \(A\) and \(B\) (M1 for each equation).
Alternative method (solving for \(y\))
| Scheme | Marks |
|---|---|
| \(\dot{y} + 80y = 2kt + c\) \(\mathrm{e}^{\int 80\,\mathrm{d}t}(\dot{y} + 80y) = (2kt + c)\mathrm{e}^{\int 80\,\mathrm{d}t}\) \(\dfrac{\mathrm{d}}{\mathrm{d}t}\left(\mathrm{e}^{80t}y\right) = (2kt + c)\mathrm{e}^{80t}\) | M1 |
| \(\displaystyle\mathrm{e}^{80t}y = \int(2kt + c)\mathrm{e}^{80t}\,\mathrm{d}t + d\) | M1 |
| \(\displaystyle\mathrm{e}^{80t}y = \frac{(2kt + c)\mathrm{e}^{80t}}{80} - \int\frac{2}{80}k\mathrm{e}^{80t}\,\mathrm{d}t + d\) \(\mathrm{e}^{80t}y = \dfrac{(2kt + c)\mathrm{e}^{80t}}{80} - \dfrac{1}{3200}k\mathrm{e}^{80t} + d\) \(y = \dfrac{1}{40}kt + \dfrac{1}{80}c - \dfrac{1}{3200}k + d\mathrm{e}^{-80t}\) \(= A + B\mathrm{e}^{-80t} + \dfrac{1}{40}kt\) | M1 |
M1: Integrating \(\ddot{y} + 80\dot{y} = 2k\) wrt \(t\) and then using integrating factor (corrected from the printed mark scheme, which reads \(\ddot{y} + 78\dot{y} = 2k\))
M1: Integrating wrt \(t\)
M1: Integrating by parts
| Scheme | Marks | AO |
|---|---|---|
| (iii) When \(t = 50\), \(x \gt \dfrac{156000k}{3200}\) \(\left(\text{because } \mathrm{e}^{-4000} \lt 1 \text{ so } 1 - \mathrm{e}^{-4000} \gt 0\right)\) i.e. \(x \gt 48.75k \gt 292.5\) for \(k \gt 6\) i.e. \(x \gt 250\) so fails safety food standards | B1 | 3.4 |
| [1] |
Notes
B1: Use of \(t = 50\) must be seen to give a term in \(k\)
Must reference \(k \gt 6\).
Condone the idea of ignoring exponential term as negligible
Sight of \(292.5 \gt 250\) earns the mark
| Scheme | Marks | AO |
|---|---|---|
| \(\sqrt{1677} = 40.95\ldots \lt 41\) \(\Rightarrow \mathrm{e}^{-41t}\left(a\cosh\left(\sqrt{1677}\,t\right) + b\sinh\left(\sqrt{1677}\,t\right)\right)\) contains only negative exponentials when expanded | M1 | 3.4 |
| so as \(t \to \infty\), the exponential/hyperbolic parts \(\to 0\) so \(x \to 20k \lt 240 \lt 250\) since \(k \lt 12\) so yes, food safety standards are met in the long run. | A1 | 3.2a |
| [2] |
Notes
M1: Turning function into exponentials. Could see eg \(\mathrm{e}^{-0.0488t}\) and \(\mathrm{e}^{-81.95t}\). Condone error(s) in coefficients.
Allow argument such as \(\mathrm{e}^{-41t}\) dominates
A1: Argument must be complete and correct but could be based on sufficiently large values of \(t\) rather than formal limiting process.
SC B1 argument that \(20k = 240 \lt 250\) if 2nd term is assumed to tend to 0
| Scheme | Marks | AO |
|---|---|---|
| Pesticide is likely to be added periodically, eg, during the day; or depending on the weather/time of year; or only when it’s needed; or the amount of pesticide added changes as the amount of crop changes/grows or if it is subject to pest attack | B1 | 3.5a |
| [1] |
Notes
B1: The idea that regularly is not the same as continuously (and may not even mean “at a constant average rate”).
Other sensible answers possible, but must be in context.