## Question 8:

### Part (a):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\dfrac{d\theta}{dt} = \dfrac{3-\theta}{125} \Rightarrow \int\dfrac{1}{3-\theta}\,d\theta = \int\dfrac{1}{125}\,dt$ | B1 | Separates variables correctly; $d\theta$ and $dt$ in correct positions |
| $-\ln(\theta - 3) = \dfrac{1}{125}t \{+c\}$ or $-\ln(3-\theta) = \dfrac{1}{125}t \{+c\}$ | M1 A1 | M1: both $\pm\lambda\ln(3-\theta)$ and $\pm\mu t$; A1: correct equation, $+c$ not needed here |
| $\theta = Ae^{-0.008t} + 3$ | A1 | Correct completion; $+c$ needed for this mark |

### Part (b):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $t=0, \theta=16 \Rightarrow 16 = Ae^0 + 3 \Rightarrow A = 13$ | M1 A1 | Substitutes $\theta=16, t=0$; finds $A=13$ |
| $10 = 13e^{-0.008t} + 3$ | M1 | Substitutes $\theta=10$ into equation of form $\theta = Ae^{-0.008t}+3$ |
| $e^{-0.008t} = \dfrac{7}{13} \Rightarrow -0.008t = \ln\!\left(\dfrac{7}{13}\right)$ | M1 | Correct algebra to $-0.008t = \ln k$ where $k$ is positive |
| $t = \dfrac{\ln\!\left(\frac{7}{13}\right)}{(-0.008)} = 77.3799... \approx 77$ (nearest minute) | A1 | awrt 77 |

## Question 8(b):

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**Part (b) – Main Method:**

| Answer/Working | Mark | Guidance |
|---|---|---|
| Substitutes $\theta = 16, t = 0$ into equation containing unknown constant | M1 | B1 on open; can imply this method mark |
| $A = 13$; note $\theta = 13e^{-0.008t} + 3$ without working implies M1A1 | A1 | |
| Substitutes $\theta = 10$ into equation **of the form** $\theta = Ae^{-0.008t} + 3$ | M1 | $A$ is positive or negative numerical value; $A$ can equal $\pm 1$ |
| Rearranges **their equation** into form $-0.008t = \ln k$ | M1 | $k$ is a **positive numerical value** |
| $t =$ awrt 77 or awrt 1 hour 17 minutes | A1 | |

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**Alternative Method 1:**

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int \frac{1}{3-\theta}\,d\theta = \int \frac{1}{125}\,dt \Rightarrow -\ln(\theta-3) = \frac{1}{125}t + c$ | | |
| Sub $t=0, \theta=16$: $-\ln(16-3) = \frac{1}{125}(0)+c \Rightarrow c = -\ln 13$ | M1/A1 | Substitutes $t=0, \theta=16$ into $-\ln(\theta-3)=\frac{1}{125}t+c$ |
| $c = -\ln 13$ | A1 | |
| Substitutes $\theta=10$ into equation **of the form** $\pm\lambda\ln(\theta-3)=\pm\frac{1}{125}t \pm\mu$ | M1 | $\lambda, \mu$ are numerical values |
| $-\ln(10-3)=\frac{1}{125}t - \ln 13$ | | |
| Rearranges **their equation** into form $\pm 0.008t = \ln C - \ln D$ | M1 | $C, D$ are **positive numerical values** |
| $\ln 13 - \ln 7 = \frac{1}{125}t$ | | |
| $t = 77.3799... = 77$ (nearest minute) | A1 | awrt 77 |

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**Alternative Method 2:**

| Answer/Working | Mark | Guidance |
|---|---|---|
| $-\ln|3-\theta| = \frac{1}{125}t + c$ | | |
| Sub $t=0, \theta=16$: $c = -\ln 13$ | M1/A1 | |
| Substitutes $\theta=10$ into equation of the form $\pm\lambda\ln(3-\theta)=\pm\frac{1}{125}t\pm\mu$ | M1 | |
| Rearranges into form $\pm 0.008t = \ln C - \ln D$, $C,D$ positive | M1 | |
| $\ln 13 - \ln 7 = \frac{1}{125}t$; $t=77$ (nearest minute) | A1 | awrt 77 |

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**Alternative Method 3:**

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int_{16}^{10}\frac{1}{3-\theta}\,d\theta = \int_0^t \frac{1}{125}\,dt$ | | |
| $\left[-\ln|3-\theta|\right]_{16}^{10} = \left[\frac{1}{125}t\right]_0^t$ | M1A1 | $\ln 13$ |
| Substitutes limit $\theta=10$ correctly | M1 | |
| Rearranges into form $\pm 0.008t = \ln C - \ln D$ | M1 | $C,D$ positive numerical values |
| $t = 77.3799... = 77$ (nearest minute) | A1 | awrt 77 |

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**Alternative Method 4:**

| Answer/Working | Mark | Guidance |
|---|---|---|
| Writes pair of equations in $A$ and $t$ for $\theta=16$ and $\theta=10$: $16=Ae^{-0.008t}+3$ and $10=Ae^{-0.008t}+3$ | M1* | $A$ unknown or positive/negative value |
| Two equations with unknown $A$ | A1 | |
| Uses correct algebra to solve both equations leading to $-0.008t = \ln k$, $k$ **positive numerical value** | M1 | |
| Finds difference between the two times (either way round) | M1 | |
| $t = \dfrac{\ln\!\left(\frac{7}{13}\right)}{(-0.008)} = 77.3799... = 77$ (nearest minute) | A1 | awrt 77; correct solution only |

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**Note (General):**
$\ln(\theta-3) = -\frac{1}{125}t + c \Rightarrow \theta - 3 = Ae^{-\frac{1}{125}t}$, where candidate writes $A = e^c$ is also acceptable.