C4 June 2016 Q4
4. The rate of decay of the mass of a particular substance is modelled by the differential equation \[\frac{\mathrm{d}x}{\mathrm{d}t} = -\frac{5}{2}x, \qquad t \geqslant 0\] where \(x\) is the mass of the substance measured in grams and \(t\) is the time measured in days.
Given that \(x = 60\) when \(t = 0\),
Give your answer to the nearest minute. (3)
| Scheme | Marks |
|---|---|
| \(\dfrac{\mathrm{d}x}{\mathrm{d}t} = -\dfrac{5}{2}x,\quad x \in \mathbb{R}, x \geqslant 0\) | |
| Way 1 | |
| \(\displaystyle\int \dfrac{1}{x}\,\mathrm{d}x = \displaystyle\int -\dfrac{5}{2}\,\mathrm{d}t\) Separates variables as shown. \(\mathrm{d}x\) and \(\mathrm{d}t\) should not be in the wrong positions, though this mark can be implied by later working. Ignore the integral signs. | B1 |
| \(\ln x = -\dfrac{5}{2}t + c\) Integrates both sides to give either \(\pm\dfrac{\alpha}{x} \to \pm\alpha\ln x\) or \(\pm k \to \pm kt\) (with respect to \(t\)); \(k, \alpha \neq 0\) \(\ln x = -\dfrac{5}{2}t + c\), including "\(+c\)" | M1 A1 |
| \(\{t = 0, x = 60 \Rightarrow\}\ \ln 60 = c\) \(\ln x = -\dfrac{5}{2}t + \ln 60 \Rightarrow\) \(\underline{x = 60\mathrm{e}^{-\frac{5}{2}t}}\) or \(x = \underline{\dfrac{60}{\mathrm{e}^{\frac{5}{2}t}}}\) Finds their \(c\) and uses correct algebra to achieve \(x = 60\mathrm{e}^{-\frac{5}{2}t}\) or \(x = \dfrac{60}{\mathrm{e}^{\frac{5}{2}t}}\) with no incorrect working seen | A1 cso |
| (4) |
Notes
B1: For the correct separation of variables. E.g. \(\displaystyle\int \dfrac{1}{5x}\,\mathrm{d}x = \displaystyle\int -\dfrac{1}{2}\,\mathrm{d}t\)
Note: B1 can be implied by seeing either \(\ln x = -\dfrac{5}{2}t + c\) or \(t = -\dfrac{2}{5}\ln x + c\) with or without \(+c\)
Note: B1 can also be implied by seeing \(\big[\ln x\big]_{60}^{x} = \left[-\dfrac{5}{2}t\right]_0^t\)
Note: Allow A1 for \(x = 60\sqrt{\mathrm{e}^{-5t}}\) or \(x = \dfrac{60}{\sqrt{\mathrm{e}^{5t}}}\) with no incorrect working seen
Note: Give final A0 for \(x = \mathrm{e}^{-\frac{5}{2}t} + 60 \to x = 60\mathrm{e}^{-\frac{5}{2}t}\)
Note: Give final A0 for writing \(x = \mathrm{e}^{-\frac{5}{2}t + \ln 60}\) as their final answer (without seeing \(x = 60\mathrm{e}^{-\frac{5}{2}t}\))
Note: Way 1 to Way 5 do not exhaust all the different methods that candidates can give.
Note: Give B0M0A0A0 for writing down \(x = 60\mathrm{e}^{-\frac{5}{2}t}\) or \(x = \dfrac{60}{\mathrm{e}^{\frac{5}{2}t}}\) with no evidence of working or integration seen.
Way 2
| Scheme | Marks |
|---|---|
| \(\dfrac{\mathrm{d}t}{\mathrm{d}x} = -\dfrac{2}{5x}\) or \(t = \displaystyle\int -\dfrac{2}{5x}\,\mathrm{d}x\) Either \(\dfrac{\mathrm{d}t}{\mathrm{d}x} = -\dfrac{2}{5x}\) or \(t = \displaystyle\int -\dfrac{2}{5x}\,\mathrm{d}x\) | B1 |
| \(t = -\dfrac{2}{5}\ln x + c\) Integrates both sides to give either \(t = \ldots\) or \(\pm\alpha\ln px;\ \alpha \neq 0, p > 0\) \(t = -\dfrac{2}{5}\ln x + c\), including "\(+c\)" | M1 A1 |
| \(\{t = 0, x = 60 \Rightarrow\}\ c = \dfrac{2}{5}\ln 60 \Rightarrow t = -\dfrac{2}{5}\ln x + \dfrac{2}{5}\ln 60\) \(\Rightarrow -\dfrac{5}{2}t = \ln x - \ln 60 \Rightarrow\) \(\underline{x = 60\mathrm{e}^{-\frac{5}{2}t}}\) or \(x = \underline{\dfrac{60}{\mathrm{e}^{\frac{5}{2}t}}}\) Finds their \(c\) and uses correct algebra to achieve \(x = 60\mathrm{e}^{-\frac{5}{2}t}\) or \(x = \dfrac{60}{\mathrm{e}^{\frac{5}{2}t}}\) with no incorrect working seen | A1 cso |
| (4) |
Way 3
| Scheme | Marks |
|---|---|
| \(\displaystyle\int_{60}^{x} \dfrac{1}{x}\,\mathrm{d}x = \displaystyle\int_0^t -\dfrac{5}{2}\,\mathrm{d}t\) Ignore limits | B1 |
| \(\big[\ln x\big]_{60}^{x} = \left[-\dfrac{5}{2}t\right]_0^t\) Integrates both sides to give either \(\pm\dfrac{\alpha}{x} \to \pm\alpha\ln x\) or \(\pm k \to \pm kt\) (with respect to \(t\)); \(k, \alpha \neq 0\) \(\big[\ln x\big]_{60}^{x} = \left[-\dfrac{5}{2}t\right]_0^t\) including the correct limits | M1 A1 |
| \(\ln x - \ln 60 = -\dfrac{5}{2}t \Rightarrow\) \(\underline{x = 60\mathrm{e}^{-\frac{5}{2}t}}\) or \(x = \underline{\dfrac{60}{\mathrm{e}^{\frac{5}{2}t}}}\) Correct algebra leading to a correct result | A1 cso |
| (4) |
Way 4
| Scheme | Marks |
|---|---|
| \(\displaystyle\int \dfrac{2}{5x}\,\mathrm{d}x = -\displaystyle\int \mathrm{d}t\) Separates variables as shown. \(\mathrm{d}x\) and \(\mathrm{d}t\) should not be in the wrong positions, though this mark can be implied by later working. Ignore the integral signs. | B1 |
| \(\dfrac{2}{5}\ln(5x) = -t + c\) Integrates both sides to give either \(\pm\alpha\ln(px)\) or \(\pm k \to \pm kt\) (with respect to \(t\)); \(k, \alpha \neq 0;\ p > 0\) \(\dfrac{2}{5}\ln(5x) = -t + c\), including "\(+c\)" | M1 A1 |
| \(\{t = 0, x = 60 \Rightarrow\}\ \dfrac{2}{5}\ln 300 = c\) \(\dfrac{2}{5}\ln(5x) = -t + \dfrac{2}{5}\ln 300 \Rightarrow\) \(\underline{x = 60\mathrm{e}^{-\frac{5}{2}t}}\) or \(x = \underline{\dfrac{60}{\mathrm{e}^{\frac{5}{2}t}}}\) Finds their \(c\) and uses correct algebra to achieve \(x = 60\mathrm{e}^{-\frac{5}{2}t}\) or \(x = \dfrac{60}{\mathrm{e}^{\frac{5}{2}t}}\) with no incorrect working seen | A1 cso |
| (4) |
Way 5
| Scheme | Marks |
|---|---|
| \(\left\{\dfrac{\mathrm{d}t}{\mathrm{d}x} = -\dfrac{2}{5x} \Rightarrow\right\}\ t = \displaystyle\int_{60}^{x} -\dfrac{2}{5x}\,\mathrm{d}x\) Ignore limits | B1 |
| \(t = \left[-\dfrac{2}{5}\ln x\right]_{60}^{x}\) Integrates both sides to give either \(\pm k \to \pm kt\) (with respect to \(t\)) or \(\pm\dfrac{\alpha}{x} \to \pm\alpha\ln x;\ k, \alpha \neq 0\) \(t = \left[-\dfrac{2}{5}\ln x\right]_{60}^{x}\) including the correct limits | M1 A1 |
| \(t = -\dfrac{2}{5}\ln x + \dfrac{2}{5}\ln 60 \Rightarrow -\dfrac{5}{2}t = \ln x - \ln 60\) \(\Rightarrow\) \(\underline{x = 60\mathrm{e}^{-\frac{5}{2}t}}\) or \(x = \underline{\dfrac{60}{\mathrm{e}^{\frac{5}{2}t}}}\) Correct algebra leading to a correct result | A1 cso |
| (4) |
| Scheme | Marks |
|---|---|
| \(20 = 60\mathrm{e}^{-\frac{5}{2}t}\) or \(\ln 20 = -\dfrac{5}{2}t + \ln 60\) Substitutes \(x = 20\) into an equation in the form of either \(x = \pm\lambda\mathrm{e}^{\pm\mu t} \pm \beta\) or \(x = \pm\lambda\mathrm{e}^{\pm\mu t \pm \alpha\ln\delta x}\) or \(\pm\alpha\ln\delta x = \pm\mu t \pm \beta\) or \(t = \pm\lambda\ln\delta x \pm \beta\); \(\alpha, \lambda, \mu, \delta \neq 0\) and \(\beta\) can be 0 | M1 |
| \(t = -\dfrac{2}{5}\ln\left(\dfrac{20}{60}\right)\) \(\{= 0.4394449\ldots \text{ (days)}\}\) Note: \(t\) must be greater than 0 dependent on the previous M mark Uses correct algebra to achieve an equation of the form of either \(t = A\ln\left(\frac{60}{20}\right)\) or \(A\ln\left(\frac{20}{60}\right)\) or \(A\ln 3\) or \(A\ln\left(\frac{1}{3}\right)\) o.e. or \(t = A(\ln 20 - \ln 60)\) or \(A(\ln 60 - \ln 20)\) o.e. \((A \in \mathbb{R}, t > 0)\) | dM1 |
| \(\Rightarrow t = 632.8006\ldots = 633\) (to the nearest minute) awrt 633 or 10 hours and awrt 33 minutes | A1 cso |
| Note: dM1 can be implied by \(t\) = awrt 0.44 from no incorrect working. | |
| (3) | |
| (7 marks) |
Notes
A1: You can apply cso for the work only seen in part (b).
Note: Give dM1(Implied) A1 for \(\dfrac{5}{2}t = \ln 3\) followed by \(t\) = awrt 633 from no incorrect working.
Note: Substitutes \(x = 40\) into their equation from part (a) is M0dM0A0