C4 June 2007 Q8
8. A population growth is modelled by the differential equation\[\frac{\mathrm{d}P}{\mathrm{d}t} = kP,\]where \(P\) is the population, \(t\) is the time measured in days and \(k\) is a positive constant.
Given that the initial population is \(P_0\),
Given also that \(k = 2.5\),
In an improved model the differential equation is given as\[\frac{\mathrm{d}P}{\mathrm{d}t} = \lambda P\cos\lambda t,\]where \(P\) is the population, \(t\) is the time measured in days and \(\lambda\) is a positive constant.
Given, again, that the initial population is \(P_0\) and that time is measured in days,
Given also that \(\lambda = 2.5\),
| Scheme | Marks |
|---|---|
| \(\dfrac{\mathrm{d}P}{\mathrm{d}t} = kP\) and \(t = 0,\ P = P_0\) (1) | |
| \(\displaystyle\int\frac{\mathrm{d}P}{P} = \int k\,\mathrm{d}t\) | M1 |
| \(\ln P = kt;\ (+c)\) | A1 |
| When \(t = 0,\ P = P_0 \Rightarrow \ln P_0 = c\) \(\left(\text{or } P = A\mathrm{e}^{kt} \Rightarrow P_0 = A\right)\) | M1 |
| \(\ln P = kt + \ln P_0 \Rightarrow \mathrm{e}^{\ln P} = \mathrm{e}^{kt + \ln P_0} = \mathrm{e}^{kt}.\mathrm{e}^{\ln P_0}\) | |
| Hence, \(\underline{P = P_0\mathrm{e}^{kt}}\) | A1 |
| (4) |
Notes
M1: Separates the variables with \(\displaystyle\int\frac{\mathrm{d}P}{P}\) and \(\displaystyle\int k\,\mathrm{d}t\) on either side with integral signs not necessary.
A1: Must see \(\ln P\) and \(kt\); Correct equation with/without \(+\,c\).
M1: Use of boundary condition (1) to attempt to find the constant of integration.
A1: \(\underline{P = P_0\mathrm{e}^{kt}}\)
\(\underline{P = P_0\mathrm{e}^{kt}}\) written down without the first M1 mark given scores all four marks in part (a).
Aliter 8. (a) Way 2
| Scheme | Marks |
|---|---|
| \(\dfrac{\mathrm{d}P}{\mathrm{d}t} = kP\) and \(t = 0,\ P = P_0\) (1) | |
| \(\displaystyle\int\frac{\mathrm{d}P}{kP} = \int 1\,\mathrm{d}t\) | M1 |
| \(\tfrac{1}{k}\ln P = t;\ (+c)\) | A1 |
| When \(t = 0,\ P = P_0 \Rightarrow \tfrac{1}{k}\ln P_0 = c\) \(\left(\text{or } P = A\mathrm{e}^{kt} \Rightarrow P_0 = A\right)\) | M1 |
| \(\tfrac{1}{k}\ln P = t + \tfrac{1}{k}\ln P_0 \Rightarrow \ln P = kt + \ln P_0\) \(\Rightarrow \mathrm{e}^{\ln P} = \mathrm{e}^{kt + \ln P_0} = \mathrm{e}^{kt}.\mathrm{e}^{\ln P_0}\) | |
| Hence, \(\underline{P = P_0\mathrm{e}^{kt}}\) | A1 |
| (4) |
M1: Separates the variables with \(\displaystyle\int\frac{\mathrm{d}P}{kP}\) and \(\displaystyle\int \mathrm{d}t\) on either side with integral signs not necessary.
A1: Must see \(\tfrac{1}{k}\ln P\) and \(t\); Correct equation with/without \(+\,c\). M1: Use of boundary condition (1) to attempt to find the constant of integration. A1: \(\underline{P = P_0\mathrm{e}^{kt}}\)
Aliter 8. (a) Way 3
| Scheme | Marks |
|---|---|
| \(\displaystyle\int\frac{\mathrm{d}P}{kP} = \int 1\,\mathrm{d}t\) | M1 |
| \(\tfrac{1}{k}\ln(kP) = t;\ (+c)\) | A1 |
| When \(t = 0,\ P = P_0 \Rightarrow \tfrac{1}{k}\ln(kP_0) = c\) \(\left(\text{or } kP = A\mathrm{e}^{kt} \Rightarrow kP_0 = A\right)\) | M1 |
| \(\tfrac{1}{k}\ln(kP) = t + \tfrac{1}{k}\ln(kP_0) \Rightarrow \ln(kP) = kt + \ln(kP_0)\) \(\Rightarrow \mathrm{e}^{\ln(kP)} = \mathrm{e}^{kt + \ln(kP_0)} = \mathrm{e}^{kt}.\mathrm{e}^{\ln(kP_0)}\) \(\Rightarrow kP = \mathrm{e}^{kt}.(kP_0) \Rightarrow kP = kP_0\mathrm{e}^{kt}\) \(\left(\text{or } kP = kP_0\mathrm{e}^{kt}\right)\) | |
| Hence, \(\underline{P = P_0\mathrm{e}^{kt}}\) | A1 |
| (4) |
M1: Separates the variables with \(\displaystyle\int\frac{\mathrm{d}P}{kP}\) and \(\displaystyle\int \mathrm{d}t\) on either side with integral signs not necessary.
A1: Must see \(\tfrac{1}{k}\ln(kP)\) and \(t\); Correct equation with/without \(+\,c\). M1: Use of boundary condition (1) to attempt to find the constant of integration. A1: \(\underline{P = P_0\mathrm{e}^{kt}}\)
\(\underline{P = P_0\mathrm{e}^{kt}}\) written down without the first M1 mark given scores all four marks in part (a).
| Scheme | Marks |
|---|---|
| \(P = 2P_0\) & \(k = 2.5 \Rightarrow \underline{2P_0 = P_0\mathrm{e}^{2.5t}}\) | M1 |
| \(\mathrm{e}^{2.5t} = 2 \Rightarrow \underline{\ln\mathrm{e}^{2.5t} = \ln 2}\) or \(\underline{2.5t = \ln 2}\) …or \(\mathrm{e}^{kt} = 2 \Rightarrow \underline{\ln\mathrm{e}^{kt} = \ln 2}\) or \(\underline{kt = \ln 2}\) | M1 |
| \(\Rightarrow t = \tfrac{1}{2.5}\ln 2 = 0.277258872\ldots\) days | |
| \(t = 0.277258872\ldots \times 24 \times 60 = 399.252776\ldots\) minutes | |
| \(t = \underline{399\,\text{min}}\) or \(t = \underline{6\text{ hr } 39\text{ mins}}\) (to nearest minute) | A1 |
| (3) |
Notes
M1: Substitutes \(P = 2P_0\) into an expression involving \(P\)
M1: Eliminates \(P_0\) and takes ln of both sides
A1: awrt \(t = \underline{399}\) or \(\underline{6\text{ hr } 39\text{ mins}}\)
| Scheme | Marks |
|---|---|
| \(\dfrac{\mathrm{d}P}{\mathrm{d}t} = \lambda P\cos\lambda t\) and \(t = 0,\ P = P_0\) (1) | |
| \(\displaystyle\int\frac{\mathrm{d}P}{P} = \int\lambda\cos\lambda t\,\mathrm{d}t\) | M1 |
| \(\ln P = \sin\lambda t;\ (+c)\) | A1 |
| When \(t = 0,\ P = P_0 \Rightarrow \ln P_0 = c\) \(\left(\text{or } P = A\mathrm{e}^{\sin\lambda t} \Rightarrow P_0 = A\right)\) | M1 |
| \(\ln P = \sin\lambda t + \ln P_0 \Rightarrow \mathrm{e}^{\ln P} = \mathrm{e}^{\sin\lambda t + \ln P_0} = \mathrm{e}^{\sin\lambda t}.\mathrm{e}^{\ln P_0}\) | |
| Hence, \(\underline{P = P_0\mathrm{e}^{\sin\lambda t}}\) | A1 |
| (4) |
Notes
M1: Separates the variables with \(\displaystyle\int\frac{\mathrm{d}P}{P}\) and \(\displaystyle\int \lambda\cos\lambda t\,\mathrm{d}t\) on either side with integral signs not necessary.
A1: Must see \(\ln P\) and \(\sin\lambda t\); Correct equation with/without \(+\,c\).
M1: Use of boundary condition (1) to attempt to find the constant of integration.
A1: \(\underline{P = P_0\mathrm{e}^{\sin\lambda t}}\)
\(\underline{P = P_0\mathrm{e}^{\sin\lambda t}}\) written down without the first M1 mark given scores all four marks in part (c).
Aliter 8. (c) Way 2
| Scheme | Marks |
|---|---|
| \(\dfrac{\mathrm{d}P}{\mathrm{d}t} = \lambda P\cos\lambda t\) and \(t = 0,\ P = P_0\) (1) | |
| \(\displaystyle\int\frac{\mathrm{d}P}{\lambda P} = \int\cos\lambda t\,\mathrm{d}t\) | M1 |
| \(\tfrac{1}{\lambda}\ln P = \tfrac{1}{\lambda}\sin\lambda t;\ (+c)\) | A1 |
| When \(t = 0,\ P = P_0 \Rightarrow \tfrac{1}{\lambda}\ln P_0 = c\) \(\left(\text{or } P = A\mathrm{e}^{\sin\lambda t} \Rightarrow P_0 = A\right)\) | M1 |
| \(\tfrac{1}{\lambda}\ln P = \tfrac{1}{\lambda}\sin\lambda t + \tfrac{1}{\lambda}\ln P_0 \Rightarrow \ln P = \sin\lambda t + \ln P_0\) \(\Rightarrow \mathrm{e}^{\ln P} = \mathrm{e}^{\sin\lambda t + \ln P_0} = \mathrm{e}^{\sin\lambda t}.\mathrm{e}^{\ln P_0}\) | |
| Hence, \(\underline{P = P_0\mathrm{e}^{\sin\lambda t}}\) | A1 |
| (4) |
M1: Separates the variables with \(\displaystyle\int\frac{\mathrm{d}P}{\lambda P}\) and \(\displaystyle\int \cos\lambda t\,\mathrm{d}t\) on either side with integral signs not necessary.
A1: Must see \(\tfrac{1}{\lambda}\ln P\) and \(\tfrac{1}{\lambda}\sin\lambda t\); Correct equation with/without \(+\,c\). M1: Use of boundary condition (1) to attempt to find the constant of integration. A1: \(\underline{P = P_0\mathrm{e}^{\sin\lambda t}}\)
Aliter 8. (c) Way 3
| Scheme | Marks |
|---|---|
| \(\dfrac{\mathrm{d}P}{\mathrm{d}t} = \lambda P\cos\lambda t\) and \(t = 0,\ P = P_0\) (1) | |
| \(\displaystyle\int\frac{\mathrm{d}P}{\lambda P} = \int\cos\lambda t\,\mathrm{d}t\) | M1 |
| \(\tfrac{1}{\lambda}\ln(\lambda P) = \tfrac{1}{\lambda}\sin\lambda t;\ (+c)\) | A1 |
| When \(t = 0,\ P = P_0 \Rightarrow \tfrac{1}{\lambda}\ln(\lambda P_0) = c\) \(\left(\text{or } \lambda P = A\mathrm{e}^{\sin\lambda t} \Rightarrow \lambda P_0 = A\right)\) | M1 |
| \(\tfrac{1}{\lambda}\ln(\lambda P) = \tfrac{1}{\lambda}\sin\lambda t + \tfrac{1}{\lambda}\ln(\lambda P_0)\) \(\Rightarrow \ln(\lambda P) = \sin\lambda t + \ln(\lambda P_0)\) \(\Rightarrow \mathrm{e}^{\ln(\lambda P)} = \mathrm{e}^{\sin\lambda t + \ln(\lambda P_0)} = \mathrm{e}^{\sin\lambda t}.\mathrm{e}^{\ln(\lambda P_0)}\) \(\Rightarrow \lambda P = \mathrm{e}^{\sin\lambda t}.(\lambda P_0)\) \(\left(\text{or } \lambda P = \lambda P_0\mathrm{e}^{\sin\lambda t}\right)\) | |
| Hence, \(\underline{P = P_0\mathrm{e}^{\sin\lambda t}}\) | A1 |
| (4) |
M1: Separates the variables with \(\displaystyle\int\frac{\mathrm{d}P}{\lambda P}\) and \(\displaystyle\int \cos\lambda t\,\mathrm{d}t\) on either side with integral signs not necessary.
A1: Must see \(\tfrac{1}{\lambda}\ln(\lambda P)\) and \(\tfrac{1}{\lambda}\sin\lambda t\); Correct equation with/without \(+\,c\). M1: Use of boundary condition (1) to attempt to find the constant of integration. A1: \(\underline{P = P_0\mathrm{e}^{\sin\lambda t}}\)
\(\underline{P = P_0\mathrm{e}^{\sin\lambda t}}\) written down without the first M1 mark given scores all four marks in part (c).
| Scheme | Marks |
|---|---|
| \(P = 2P_0\) & \(\lambda = 2.5 \Rightarrow 2P_0 = P_0\mathrm{e}^{\sin 2.5t}\) | |
| \(\mathrm{e}^{\sin 2.5t} = 2 \Rightarrow \underline{\sin 2.5t = \ln 2}\) …or… \(\mathrm{e}^{\sin\lambda t} = 2 \Rightarrow \underline{\sin\lambda t = \ln 2}\) | M1 |
| \(\underline{t = \tfrac{1}{2.5}\sin^{-1}(\ln 2)}\) | dM1 |
| \(t = 0.306338477\ldots\) | |
| \(t = 0.306338477\ldots \times 24 \times 60 = 441.1274082\ldots\) minutes | |
| \(t = \underline{441\,\text{min}}\) or \(t = \underline{7\text{ hr } 21\text{ mins}}\) (to nearest minute) | A1 |
| (3) | |
| (14 marks) |
Notes
M1: Eliminates \(P_0\) and makes \(\sin\lambda t\) or \(\sin 2.5t\) the subject by taking ln’s
dM1: Then rearranges to make \(t\) the subject. (must use \(\sin^{-1}\))
A1: awrt \(t = \underline{441}\) or \(\underline{7\text{ hr } 21\text{ mins}}\)
(corrected from the printed mark scheme: the second line is printed “…or… \(\mathrm{e}^{\lambda t} = 2\)”; it should read \(\mathrm{e}^{\sin\lambda t} = 2\))
Note: dM1 denotes a method mark which is dependent upon the award of the previous method mark. ddM1 denotes a method mark which is dependent upon the award of the previous two method marks. depM1* denotes a method mark which is dependent upon the award of M1*. ft denotes “follow through”. cao denotes “correct answer only”. aef denotes “any equivalent form”