C4 January 2013 Q8
8. A bottle of water is put into a refrigerator. The temperature inside the refrigerator remains constant at \(3\,{}^\circ\text{C}\) and \(t\) minutes after the bottle is placed in the refrigerator the temperature of the water in the bottle is \(\theta\,{}^\circ\text{C}\).
The rate of change of the temperature of the water in the bottle is modelled by the differential equation,\[\frac{\mathrm{d}\theta}{\mathrm{d}t} = \frac{(3 - \theta)}{125}\]
Given that the temperature of the water in the bottle when it was put in the refrigerator was \(16\,{}^\circ\text{C}\),
| Scheme | Marks |
|---|---|
| \(\left\{\dfrac{\mathrm{d}\theta}{\mathrm{d}t} = \dfrac{(3 - \theta)}{125}\right\} \Rightarrow \displaystyle\int \frac{1}{3 - \theta}\,\mathrm{d}\theta = \int \frac{1}{125}\,\mathrm{d}t\) or \(\displaystyle\int \frac{125}{3 - \theta}\,\mathrm{d}\theta = \int \mathrm{d}t\) | B1 |
| \(-\ln(\theta - 3) = \dfrac{1}{125}t\ \{+\,c\}\) or \(-\ln(3 - \theta) = \dfrac{1}{125}t\ \{+\,c\}\) See notes. | M1 A1 |
| \(\ln(\theta - 3) = -\dfrac{1}{125}t + c\) \(\theta - 3 = \mathrm{e}^{-\frac{1}{125}t + c}\) or \(\mathrm{e}^{-\frac{1}{125}t}\mathrm{e}^{c}\) \(\theta = A\mathrm{e}^{-0.008t} + 3\ \ *\) Correct completion to \(\theta = A\mathrm{e}^{-0.008t} + 3\). | A1 |
| (4) |
Notes
B1: (M1 on epen) 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.
M1: Both \(\pm\lambda\ln(3 - \theta)\) or \(\pm\lambda\ln(\theta - 3)\) and \(\pm\mu t\) where \(\lambda\) and \(\mu\) are constants.
A1: For \(-\ln(\theta - 3) = \dfrac{1}{125}t\) or \(-\ln(3 - \theta) = \dfrac{1}{125}t\) or \(-125\ln(\theta - 3) = t\) or \(-125\ln(3 - \theta) = t\)
Note: \(+\,c\) is not needed for this mark.
A1: Correct completion to \(\theta = A\mathrm{e}^{-0.008t} + 3\). Note: \(+\,c\) is needed for this mark.
Note: \(\ln(\theta - 3) = -\dfrac{1}{125}t + c\) leading to \(\theta - 3 = \mathrm{e}^{-\frac{1}{125}t} + \mathrm{e}^{c}\) or \(\theta - 3 = \mathrm{e}^{-\frac{1}{125}t} + A\), would be final A0.
Note: From \(-\ln(\theta - 3) = \dfrac{1}{125}t + c\), then \(\ln(\theta - 3) = -\dfrac{1}{125}t + c\)
\(\Rightarrow \theta - 3 = \mathrm{e}^{-\frac{1}{125}t + c}\) or \(\theta - 3 = \mathrm{e}^{-\frac{1}{125}t}\mathrm{e}^{c} \Rightarrow \theta = A\mathrm{e}^{-0.008t} + 3\) is required for A1.
Note: From \(-\ln(3 - \theta) = \dfrac{1}{125}t + c\), then \(\ln(3 - \theta) = -\dfrac{1}{125}t + c\)
\(\Rightarrow 3 - \theta = \mathrm{e}^{-\frac{1}{125}t + c}\) or \(3 - \theta = \mathrm{e}^{-\frac{1}{125}t}\mathrm{e}^{c} \Rightarrow \theta = A\mathrm{e}^{-0.008t} + 3\) is sufficient for A1.
Note: The jump from \(3 - \theta = A\mathrm{e}^{-\frac{1}{125}t}\) to \(\theta = A\mathrm{e}^{-0.008t} + 3\) is fine.
Note: \(\ln(\theta - 3) = -\dfrac{1}{125}t + c \Rightarrow \theta - 3 = Ae^{-\frac{1}{125}t}\), where candidate writes \(A = \mathrm{e}^{c}\) is also acceptable.
| Scheme | Marks |
|---|---|
| \(\{t = 0,\ \theta = 16 \Rightarrow\}\quad 16 = A\mathrm{e}^{-0.008(0)} + 3;\ \Rightarrow \underline{A = 13}\) See notes. | M1; A1 |
| \(10 = 13\mathrm{e}^{-0.008t} + 3\) Substitutes \(\theta = 10\) into an equation of the form \(\theta = A\mathrm{e}^{-0.008t} + 3\), or equivalent. See notes. | M1 |
| \(\mathrm{e}^{-0.008t} = \dfrac{7}{13} \Rightarrow -0.008t = \ln\left(\dfrac{7}{13}\right)\) Correct algebra to \(-0.008t = \ln k\), where \(k\) is a positive value. See notes. | M1 |
| \(\left\{t = \dfrac{\ln\left(\frac{7}{13}\right)}{(-0.008)}\right\} = 77.3799\ldots = 77\ (\text{nearest minute})\) awrt 77 | A1 |
| (5) | |
| (9 marks) |
Notes
M1: (B1 on epen) Substitutes \(\theta = 16,\ t = 0\), into either their equation containing an unknown constant or the printed equation. Note: You can imply this method mark.
A1: (M1 on epen) \(A = 13\). Note: \(\theta = 13\mathrm{e}^{-0.008t} + 3\) without any working implies the first two marks, M1A1.
M1: Substitutes \(\theta = 10\) into an equation of the form \(\theta = A\mathrm{e}^{-0.008t} + 3\), or equivalent.
where \(A\) is a positive or negative numerical value and \(A\) can be equal to 1 or -1.
M1: Uses correct algebra to rearrange their equation into the form \(-0.008t = \ln k\),
where \(k\) is a positive numerical value.
A1: awrt 77 or awrt 1 hour 17 minutes.
Alternative Method 1 for part (b)
| Scheme | Marks |
|---|---|
| \(\displaystyle\int \frac{1}{3 - \theta}\,\mathrm{d}\theta = \int \frac{1}{125}\,\mathrm{d}t \Rightarrow -\ln(\theta - 3) = \frac{1}{125}t + c\) | |
| \(\{t = 0,\ \theta = 16 \Rightarrow\}\ -\ln(16 - 3) = \dfrac{1}{125}(0) + c\) \(\Rightarrow c = -\ln 13\) M1: Substitutes \(t = 0, \theta = 16\), into \(-\ln(\theta - 3) = \dfrac{1}{125}t + c\) A1: \(c = -\ln 13\) | M1 |
| \(-\ln(\theta - 3) = \dfrac{1}{125}t - \ln 13\) or \(\ln(\theta - 3) = -\dfrac{1}{125}t + \ln 13\) \(-\ln(10 - 3) = \dfrac{1}{125}t - \ln 13\) M1: Substitutes \(\theta = 10\) into an equation of the form \(\pm\lambda\ln(\theta - 3) = \pm\dfrac{1}{125}t \pm \mu\) where \(\lambda\), \(\mu\) are numerical values. | M1 |
| \(\ln 13 - \ln 7 = \dfrac{1}{125}t\) M1: Uses correct algebra to rearrange their equation into the form \(\pm 0.008t = \ln C - \ln D\), where \(C\), \(D\) are positive numerical values. | M1 |
| \(t = 77.3799\ldots = 77\ (\text{nearest minute})\) | A1: awrt 77. |
Alternative Method 2 for part (b)
| Scheme | Marks |
|---|---|
| \(\displaystyle\int \frac{1}{3 - \theta}\,\mathrm{d}\theta = \int \frac{1}{125}\,\mathrm{d}t \Rightarrow -\ln|3 - \theta| = \frac{1}{125}t + c\) | |
| \(\{t = 0,\ \theta = 16 \Rightarrow\}\ -\ln|3 - 16| = \dfrac{1}{125}(0) + c\) \(\Rightarrow c = -\ln 13\) M1: Substitutes \(t = 0, \theta = 16\), into \(-\ln(3 - \theta) = \dfrac{1}{125}t + c\) A1: \(c = -\ln 13\) | M1 |
| \(-\ln|3 - \theta| = \dfrac{1}{125}t - \ln 13\) or \(\ln|3 - \theta| = -\dfrac{1}{125}t + \ln 13\) \(-\ln(3 - 10) = \dfrac{1}{125}t - \ln 13\) M1: Substitutes \(\theta = 10\) into an equation of the form \(\pm\lambda\ln(3 - \theta) = \pm\dfrac{1}{125}t \pm \mu\) where \(\lambda\), \(\mu\) are numerical values. | M1 |
| \(\ln 13 - \ln 7 = \dfrac{1}{125}t\) M1: Uses correct algebra to rearrange their equation into the form \(\pm 0.008t = \ln C - \ln D\), where \(C\), \(D\) are positive numerical values. | M1 |
| \(t = 77.3799\ldots = 77\ (\text{nearest minute})\) | A1: awrt 77. |
Alternative Method 3 for part (b)
| Scheme | Marks |
|---|---|
| \(\displaystyle\int_{16}^{10} \frac{1}{3 - \theta}\,\mathrm{d}\theta = \int_0^t \frac{1}{125}\,\mathrm{d}t\) \(= \left[-\ln|3 - \theta|\right]_{16}^{10} = \left[\dfrac{1}{125}t\right]_0^t\) M1A1: \(\ln 13\) M1: Substitutes limit of \(\theta = 10\) correctly. | M1A1 |
| \(-\ln 7 - -\ln 13 = \dfrac{1}{125}t\) M1: Uses correct algebra to rearrange their own equation into the form \(\pm 0.008t = \ln C - \ln D\), where \(C\), \(D\) are positive numerical values. | M1 |
| \(t = 77.3799\ldots = 77\ (\text{nearest minute})\) | A1: awrt 77. |
Alternative Method 4 for part (b)
| Scheme | Marks |
|---|---|
| \(\{\theta = 16 \Rightarrow\}\quad 16 = A\mathrm{e}^{-0.008t} + 3\) \(\{\theta = 10 \Rightarrow\}\quad 10 = A\mathrm{e}^{-0.008t} + 3\) M1*: Writes down a pair of equations in \(A\) and \(t\), for \(\theta = 16\) and \(\theta = 10\) with either \(A\) unknown or \(A\) being a positive or negative value. A1: Two equations with an unknown \(A\). | M1* |
| \(-0.008t = \ln\left(\dfrac{13}{A}\right)\) or \(-0.008t = \ln\left(\dfrac{7}{A}\right)\) \(t_{(1)} = \dfrac{\ln\left(\frac{13}{A}\right)}{-0.008}\) and \(t_{(2)} = \dfrac{\ln\left(\frac{7}{A}\right)}{-0.008}\) M1: Uses correct algebra to solve both of their equations leading to answers of the form \(-0.008t = \ln k\), where \(k\) is a positive numerical value. | M1 |
| \(t = t_{(1)} - t_{(2)} = \dfrac{\ln\left(\frac{13}{A}\right)}{-0.008} - \dfrac{\ln\left(\frac{7}{A}\right)}{-0.008}\) M1: Finds difference between the two times. (either way round). | M1 |
| \(\left\{t = \dfrac{\ln\left(\frac{7}{13}\right)}{(-0.008)}\right\} = 77.3799\ldots = 77\ (\text{nearest minute})\) A1: awrt 77. Correct solution only. | A1 |