## Question 13:

$$N = \frac{240}{1 + ke^{-\frac{t}{16}}}$$

### Part (a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{240}{1 + ke^{(0)}} = 50 \Rightarrow k = \ldots$ | M1 | Substitutes $t = 0$ and $N = 50$ and solves for $k$ |
| $k = 3.8 \left(= \frac{19}{5}\right)$ | A1 | cao |

**(2 marks total)**

---

### Part (b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $100 = \frac{240}{1 + 3.8e^{-\frac{t}{16}}} \Rightarrow 3.8e^{-\frac{t}{16}} = 1.4$ | M1 | Puts $N = 100$ and solves as far as $pe^{-\frac{t}{16}} = q$ using correct processing (allow sign/copying/arithmetic slips) |
| $e^{-\frac{t}{16}} = \frac{7}{19} \Rightarrow -\frac{t}{16} = \ln\left(\frac{7}{19}\right)$ | dM1 | Takes ln's correctly to reach $\pm\frac{t}{16} = \ln(\alpha)$, $\alpha > 0$. Dependent on previous M |
| $t = 16\ln\left(\frac{19}{7}\right)$ or $-16\ln\left(\frac{7}{19}\right)$ or $8\ln\left(\frac{361}{49}\right)$ or $4\ln\left(\frac{130321}{2401}\right)$ etc. | A1 | cao (accept equivalents) or awrt 16 |

**(3 marks total)**

---

### Part (c) — Way 1:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $N = 240\left(1 + ke^{-\frac{t}{16}}\right)^{-1} \Rightarrow \frac{dN}{dt} = Ae^{-\frac{t}{16}}\left(1 + Be^{-\frac{t}{16}}\right)^{-2}$ | M1 | Attempt at differentiating |
| $\frac{dN}{dt} = -240\left(1 + 3.8e^{-\frac{t}{16}}\right)^{-2} \times -\frac{3.8}{16}e^{-\frac{t}{16}}$ | A1ft | Correct derivative. Follow through their $k$ or the letter $k$ |
| $N = \frac{240}{1+ke^{-\frac{t}{16}}} \Rightarrow 1 + ke^{-\frac{t}{16}} = \frac{240}{N}$ | M1 | Attempt to find $e^{-\frac{t}{16}}$ or $ke^{-\frac{t}{16}}$ or $1 + ke^{-\frac{t}{16}}$ in terms of $N$ |
| $\frac{dN}{dt} = \frac{57\left(\frac{240-N}{3.8N}\right)}{\left(\frac{240}{N}\right)^2}$ | M1 | Obtains $\frac{dN}{dt}$ in terms of $N$ only (may include $k$'s) |
| $\frac{dN}{dt} = \frac{1}{16}N - \frac{1}{3840}N^2$ | A1 | cao (Allow $p = 16$, $q = 3840$) |

**(5 marks total)**

# Question (c) – Differential Equation Derivation

## Way 2

| Working | Mark | Guidance |
|---------|------|----------|
| $N = \frac{240}{1+ke^{-t/16}} \Rightarrow ke^{-t/16} = \frac{240}{N}-1$ | | Starting point |
| $-\frac{k}{16}e^{-t/16}\frac{dt}{dN} = -\frac{240}{N^2}$ or $-\frac{k}{16}e^{-t/16} = -\frac{240}{N^2}\frac{dN}{dt}$ | M1 | Differentiates to obtain $Ae^{-t/16}\frac{dt}{dN} = \frac{B}{N^2}$ or $Ae^{-t/16} = \frac{B}{N^2}\frac{dN}{dt}$ |
| Correct differentiation following through their $k$ or the letter $k$ | A1ft | Correct differentiation, follow through their $k$ or the letter $k$ |
| $ke^{-t/16} = \frac{240}{N}-1$, attempt to find $e^{-t/16}$ or $ke^{-t/16}$ or $1+ke^{-t/16}$ in terms of $N$ | M1 | Note: may be scored by replacing $1+ke^{-t/16}$ by $\frac{240}{N}$ |
| $\frac{dN}{dt} = \frac{\frac{1}{16}\left(\frac{240}{N}-1\right)}{\frac{240}{N^2}}$ | M1 | Obtains $\frac{dN}{dt}$ in terms of $N$ only (may include $k$'s) |
| $\frac{dN}{dt} = \frac{1}{16}N - \frac{1}{3840}N^2$ | A1 | Cao (Allow $p=16$, $q=3840$) |

**Total: (5)**

---

## Way 2 mis-read $N = \frac{240}{1+ke^{+t/16}}$ (Max 4/5)

| Working | Mark | Guidance |
|---------|------|----------|
| $\frac{k}{16}e^{t/16}\frac{dt}{dN} = -\frac{240}{N^2}$ or equivalent | M1 | Differentiates to obtain $Ae^{t/16}\frac{dt}{dN} = \frac{B}{N^2}$ or $Ae^{t/16} = \frac{B}{N^2}\frac{dN}{dt}$ |
| Correct differentiation following through their $k$ | A1ft | Follow through their $k$ or the letter $k$ |
| Attempt to find $e^{t/16}$ or $ke^{t/16}$ or $1+ke^{t/16}$ in terms of $N$ | M1 | Note: may be scored by replacing $1+ke^{t/16}$ by $\frac{240}{N}$ |
| $\frac{dN}{dt}$ in terms of $N$ only | M1 | May include $k$'s |
| $\frac{dN}{dt} = -\frac{1}{16}N + \frac{1}{3840}N^2$ | A0 | |

---

## Way 3

| Working | Mark | Guidance |
|---------|------|----------|
| $-\frac{t}{16} = \ln\frac{1}{k}\left(\frac{240}{N}-1\right)$, makes $t$ subject, takes ln's and differentiates using chain rule | M1 | |
| $\frac{dt}{dN} = -16\left(\frac{N}{240-N}\right)\left(-\frac{240}{N^2}\right)$ | A1ft | Correct differentiation, follow through their $k$ or the letter $k$ |
| Attempt to find $e^{-t/16}$ or $ke^{-t/16}$ or $1+ke^{-t/16}$ in terms of $N$ | M1 | Note: may be scored by replacing $1+ke^{-t/16}$ by $\frac{240}{N}$ |
| $= \frac{3840}{N(240-N)} \Rightarrow \frac{dN}{dt} = \frac{N(240-N)}{3840}$ | M1 | Obtains $\frac{dN}{dt}$ in terms of $N$ only (may include $k$'s) |
| $\frac{dN}{dt} = \frac{1}{16}N - \frac{1}{3840}N^2$ | A1 | Cao (Allow $p=16$, $q=3840$) |

---

## Way 3 mis-read $N = \frac{240}{1+ke^{+t/16}}$ (Max 4/5)

| Working | Mark | Guidance |
|---------|------|----------|
| $\frac{t}{16} = \ln\frac{1}{k}\left(\frac{240}{N}-1\right)$, makes $t$ subject | M1 | Takes ln's and differentiates using chain rule |
| $\frac{dt}{dN} = 16\left(\frac{N}{240-N}\right)\left(-\frac{240}{N^2}\right)$ | A1ft | Correct differentiation, follow through their $k$ or the letter $k$ |
| Attempt to find $e^{t/16}$ or $ke^{t/16}$ or $1+ke^{t/16}$ in terms of $N$ | M1 | May be scored by replacing $1+ke^{t/16}$ by $\frac{240}{N}$ |
| $= \frac{3840}{N(N-240)} \Rightarrow \frac{dN}{dt} = \frac{N(N-240)}{3840}$ | M1 | Obtains $\frac{dN}{dt}$ in terms of $N$ only (may include $k$'s) |
| $\frac{dN}{dt} = -\frac{1}{16}N + \frac{1}{3840}N^2$ | A0 | |

---

## Way 4

| Working | Mark | Guidance |
|---------|------|----------|
| $\left(1+ke^{-t/16}\right)N = 240 \Rightarrow N\times\left(-\frac{k}{16}e^{-t/16}\right) + \left(1+ke^{-t/16}\right)\frac{dN}{dt} = 0$ | M1 | Multiplies by $\left(1+ke^{-t/16}\right)$ and differentiates with respect to $t$ or $N$ using product rule |
| Correct differentiation following through their $k$ | A1ft | Follow through their $k$ or the letter $k$ |
| Attempt to find $e^{-t/16}$ or $ke^{-t/16}$ or $1+ke^{-t/16}$ in terms of $N$ | M1 | May be scored by replacing $1+ke^{-t/16}$ by $\frac{240}{N}$ |
| $\frac{dN}{dt} = \frac{N(240-N)}{3840}$ | M1 | Obtains $\frac{dN}{dt}$ in terms of $N$ only (may include $k$'s) |
| $\frac{dN}{dt} = \frac{1}{16}N - \frac{1}{3840}N^2$ | A1 | Cao (Allow $p=16$, $q=3840$) |

---

## Way 4 mis-read $N = \frac{240}{1+ke^{+t/16}}$ (Max 4/5)

| Working | Mark | Guidance |
|---------|------|----------|
| $\left(1+ke^{t/16}\right)N = 240 \Rightarrow N\times\frac{k}{16}e^{t/16} + \left(1+ke^{t/16}\right)\frac{dN}{dt} = 0$ | M1 | Multiplies by $\left(1+ke^{t/16}\right)$ and differentiates with respect to $t$ or $N$ using product rule |
| Correct differentiation following through their $k$ | A1ft | Follow through their $k$ or the letter $k$ |
| Attempt to find $e^{t/16}$ or $ke^{t/16}$ or $1+ke^{t/16}$ in terms of $N$ | M1 | May be scored by replacing $1+ke^{t/16}$ by $\frac{240}{N}$ |
| $\frac{dN}{dt} = \frac{N(N-240)}{3840}$ | M1 | Obtains $\frac{dN}{dt}$ in terms of $N$ only (may include $k$'s) |
| $\frac{dN}{dt} = -\frac{1}{16}N + \frac{1}{3840}N^2$ | A0 | |