C4 June 2013 Q6
6. Water is being heated in a kettle. At time \(t\) seconds, the temperature of the water is \(\theta\) °C.
The rate of increase of the temperature of the water at any time \(t\) is modelled by the differential equation \[\frac{\mathrm{d}\theta}{\mathrm{d}t} = \lambda(120 - \theta), \qquad \theta \leqslant 100\] where \(\lambda\) is a positive constant.
Given that \(\theta = 20\) when \(t = 0\),
When the temperature of the water reaches 100 °C, the kettle switches off.
| Scheme | Marks | ||||||||
|---|---|---|---|---|---|---|---|---|---|
| \(\dfrac{\mathrm{d}\theta}{\mathrm{d}t} = \lambda(120 - \theta), \quad \theta \leqslant 100\) | |||||||||
| \(\displaystyle\int \dfrac{1}{120 - \theta}\,\mathrm{d}\theta = \displaystyle\int \lambda\,\mathrm{d}t\) or \(\displaystyle\int \dfrac{1}{\lambda(120 - \theta)}\,\mathrm{d}\theta = \displaystyle\int \mathrm{d}t\) | B1 | ||||||||
| \(-\ln(120 - \theta);\ = \lambda t + c\) or \(-\dfrac{1}{\lambda}\ln(120 - \theta);\ = t + c\) See notes | M1 A1; M1 A1 | ||||||||
| \(\{t = 0,\ \theta = 20 \Rightarrow\}\ -\ln(120 - 20) = \lambda(0) + c\) See notes | M1 | ||||||||
| \(c = -\ln 100 \Rightarrow -\ln(120 - \theta) = \lambda t - \ln 100\) | |||||||||
| dddM1 A1 * | ||||||||
| (8) | |||||||||
Notes
B1: Separates variables as shown. \(\mathrm{d}\theta\) and \(\mathrm{d}t\) should be in the correct positions, though this mark can be implied by later working. Ignore the integral signs.
| Either | or |
| M1: \(\displaystyle\int \dfrac{1}{120 - \theta}\,\mathrm{d}\theta \to \pm A\ln(120 - \theta)\) | \(\displaystyle\int \dfrac{1}{\lambda(120 - \theta)}\,\mathrm{d}\theta \to \pm A\ln(120 - \theta)\), \(A\) is a constant. |
| A1: \(\displaystyle\int \dfrac{1}{120 - \theta}\,\mathrm{d}\theta \to -\ln(120 - \theta)\) | \(\displaystyle\int \dfrac{1}{\lambda(120 - \theta)}\,\mathrm{d}\theta \to -\dfrac{1}{\lambda}\ln(120 - \theta)\) or \(-\dfrac{1}{\lambda}\ln(120\lambda - \lambda\theta)\), |
| M1: \(\displaystyle\int \lambda\,\mathrm{d}t \to \lambda t\) | \(\displaystyle\int 1\,\mathrm{d}t \to t\) |
| A1: \(\displaystyle\int \lambda\,\mathrm{d}t \to \lambda t + c\) | or \(\displaystyle\int 1\,\mathrm{d}t \to t + c\) The \(+\,c\) can appear on either side of the equation. |
IMPORTANT: \(+\,c\) can be on either side of their equation for the 2nd A1 mark.
M1: Substitutes \(t = 0\) AND \(\theta = 20\) in an integrated or changed equation containing \(c\) (or \(A\) or \(\ln A\)).
Note that this mark can be implied by the correct value of \(c\). { Note that \(-\ln 100 = -4.60517...\) }.
dddM1: Uses their value of \(c\) which must be a \(\ln\) term, and uses fully correct method to eliminate their logarithms. Note: This mark is dependent on all three previous method marks being awarded.
A1*: This is a given answer. All previous marks must have been scored and there must not be any errors in the candidate’s working. Do not accept huge leaps in working at the end. So a minimum of either:
(1): \(\mathrm{e}^{-\lambda t} = \dfrac{120 - \theta}{100} \Rightarrow 100\mathrm{e}^{-\lambda t} = 120 - \theta \Rightarrow \theta = 120 - 100\mathrm{e}^{-\lambda t}\)
or (2): \(\mathrm{e}^{\lambda t} = \dfrac{100}{120 - \theta} \Rightarrow (120 - \theta)\mathrm{e}^{\lambda t} = 100 \Rightarrow 120 - \theta = 100\mathrm{e}^{-\lambda t} \Rightarrow \theta = 120 - 100\mathrm{e}^{-\lambda t}\)
is required for A1.
Note: \(\displaystyle\int \dfrac{1}{(120\lambda - \lambda\theta)}\,\mathrm{d}\theta \to -\dfrac{1}{\lambda}\ln(120\lambda - \lambda\theta)\) is ok for the first M1A1 in part (a).
Aliter 6. (a) Way 2
| Scheme | Marks |
|---|---|
| \(\displaystyle\int \dfrac{1}{120 - \theta}\,\mathrm{d}\theta = \displaystyle\int \lambda\,\mathrm{d}t\) | B1 |
| \(-\ln(120 - \theta) = \lambda t + c\) See notes | M1 A1; M1 A1 |
| \(-\ln(120 - \theta) = \lambda t + c\) \(\ln(120 - \theta) = -\lambda t + c\) \(120 - \theta = A\mathrm{e}^{-\lambda t}\) \(\theta = 120 - A\mathrm{e}^{-\lambda t}\) | |
| \(\{t = 0,\ \theta = 20 \Rightarrow\}\ 20 = 120 - A\mathrm{e}^0\) | M1 |
| \(A = 120 - 20 = 100\) So, \(\theta = 120 - 100\mathrm{e}^{-\lambda t}\) | dddM1 A1 * |
| (8) |
B1M1A1M1A1: Mark as in the original scheme.
M1: Substitutes \(t = 0\) AND \(\theta = 20\) in an integrated equation containing their constant of integration which could be \(c\) or \(A\). Note that this mark can be implied by the correct value of \(c\) or \(A\).
dddM1: Uses a fully correct method to eliminate their logarithms and writes down an equation containing their evaluated constant of integration.
Note: This mark is dependent on all three previous method marks being awarded.
Note: \(\ln(120 - \theta) = -\lambda t + c\) leading to \(120 - \theta = \mathrm{e}^{-\lambda t} + \mathrm{e}^c\) or \(120 - \theta = \mathrm{e}^{-\lambda t} + A\), would be dddM0.
A1*: Same as the original scheme.
Note: The jump from \(\ln(120 - \theta) = -\lambda t + c\) to \(120 - \theta = A\mathrm{e}^{-\lambda t}\) with no incorrect working is condoned in part (a).
Aliter 6. (a) Way 3
| Scheme | Marks | ||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| \(\displaystyle\int \dfrac{1}{120 - \theta}\,\mathrm{d}\theta = \displaystyle\int \lambda\,\mathrm{d}t\ \left\{\Rightarrow \displaystyle\int \dfrac{-1}{\theta - 120}\,\mathrm{d}\theta = \displaystyle\int \lambda\,\mathrm{d}t\right\}\) | B1 | ||||||||||||
| \(-\ln|\theta - 120| = \lambda t + c\) Modulus required for 1st A1. | M1 A1 M1 A1 | ||||||||||||
| \(\{t = 0,\ \theta = 20 \Rightarrow\}\ -\ln|20 - 120| = \lambda(0) + c\) Modulus not required here! | M1 | ||||||||||||
| \(\Rightarrow c = -\ln 100 \Rightarrow -\ln|\theta - 120| = \lambda t - \ln 100\) | |||||||||||||
Understanding of modulus is required here! | dddM1 A1 * | ||||||||||||
| (8) | |||||||||||||
B1: Mark as in the original scheme.
M1: Mark as in the original scheme ignoring the modulus.
A1: \(\displaystyle\int \dfrac{1}{120 - \theta}\,\mathrm{d}\theta \to -\ln|\theta - 120|\). (The modulus is required here).
M1A1: Mark as in the original scheme.
M1: Substitutes \(t = 0\) AND \(\theta = 20\) in an integrated equation containing their constant of integration which could be \(c\) or \(A\). Mark as in the original scheme ignoring the modulus.
dddM1: Mark as in the original scheme AND the candidate must demonstrate that they have converted \(\ln|\theta - 120|\) to \(\ln(120 - \theta)\) in their working. Note: This mark is dependent on all three previous method marks being awarded.
A1: Mark as in the original scheme.
Aliter 6. (a) Way 4
Use of an integrating factor
| \(\dfrac{\mathrm{d}\theta}{\mathrm{d}t} = \lambda(120 - \theta) \Rightarrow \dfrac{\mathrm{d}\theta}{\mathrm{d}t} + \lambda\theta = 120\lambda\) | |
| \(\text{IF} = \mathrm{e}^{\lambda t}\) | B1 |
| \(\dfrac{\mathrm{d}}{\mathrm{d}t}\left(\mathrm{e}^{\lambda t}\theta\right) = 120\lambda\mathrm{e}^{\lambda t}\), | M1A1 |
| \(\mathrm{e}^{\lambda t}\theta = 120\mathrm{e}^{\lambda t} + k\) | M1A1 |
| \(\theta = 120 + K\mathrm{e}^{-\lambda t}\) | M1 |
| \(\{t = 0,\ \theta = 20 \Rightarrow\}\ -100 = K\) | |
| \(\theta = 120 - 100\mathrm{e}^{-\lambda t}\) | M1A1 |
(corrected from the printed mark scheme: the line \(\mathrm{e}^{\lambda t}\theta = 120\mathrm{e}^{\lambda t} + k\) is printed as \(\mathrm{e}^{\lambda t}\theta = 120\lambda\mathrm{e}^{\lambda t} + k\); integrating \(120\lambda\mathrm{e}^{\lambda t}\) gives \(120\mathrm{e}^{\lambda t}\), as the next line \(\theta = 120 + K\mathrm{e}^{-\lambda t}\) shows.)
| Scheme | Marks |
|---|---|
| \(\{\lambda = 0.01,\ \theta = 100 \Rightarrow\}\quad 100 = 120 - 100\mathrm{e}^{-0.01t}\) | M1 |
| \(\Rightarrow 100\mathrm{e}^{-0.01t} = 120 - 100 \Rightarrow -0.01t = \ln\left(\dfrac{120 - 100}{100}\right)\) \(t = \dfrac{1}{-0.01}\ln\left(\dfrac{120 - 100}{100}\right)\) Uses correct order of operations by moving from \(100 = 120 - 100\mathrm{e}^{-0.01t}\) to give \(t = \ldots\) and \(t = A\ln B\), where \(B > 0\) | dM1 |
| \(\left\{t = \dfrac{1}{-0.01}\ln\left(\dfrac{1}{5}\right) = 100\ln 5\right\}\) | |
| \(t = 160.94379... = 161\) (s) (nearest second) awrt 161 | A1 |
| (3) | |
| (11 marks) |
Notes
M1: Substitutes \(\lambda = 0.01\) and \(\theta = 100\) into the printed equation or one of their earlier equations connecting \(\theta\) and \(t\). This mark can be implied by subsequent working.
dM1: Candidate uses correct order of operations by moving from \(100 = 120 - 100\mathrm{e}^{-0.01t}\) to \(t = \ldots\)
Note: that the 2nd Method mark is dependent on the 1st Method mark being awarded in part (b).
A1: awrt 161 or “awrt” 2 minutes 41 seconds. (Ignore incorrect units).