# Question 8:

## Part (a):

| Answer/Working | Marks | Guidance |
|---|---|---|
| At 6 hours $t = 0.25$ so "$h$" is $0.25$ | **B1** | Identifies correct step length; 6 hours is a quarter of a day, so $h = 0.25$ |
| At $t=0$: $\frac{\mathrm{d}x}{\mathrm{d}t} = \frac{3+\cosh 0}{3\times 3^2 \cosh 0} - \frac{1}{3}(3)\tanh 0 = \ldots \left(=\frac{4}{27}\right)$ | **M1** | Uses $y_0 = x(0) = 3$ and $t=0$ to find $\left(\frac{\mathrm{d}x}{\mathrm{d}t}\right)_0$; accept whichever notation used |
| So $x_1 \approx 3 + \text{"0.25"} \times \frac{\text{"4"}}{27} = \ldots$ | **M1** | Applies approximation formula with their $h$ and their $\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)_0$ |
| After 6 hours concentration is approximately $3.04$ ppm (3 s.f.) or $\frac{82}{27}$ ppm | **A1** | awrt $3.04$ ppm; accept $\frac{82}{27}$ ppm |

---

## Part (b):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{\mathrm{d}u}{\mathrm{d}t} = 3x^2 \times \frac{\mathrm{d}x}{\mathrm{d}t}$ or $\frac{1}{3x^2}\frac{\mathrm{d}u}{\mathrm{d}t} = \frac{\mathrm{d}x}{\mathrm{d}t}$ or $\frac{\mathrm{d}u}{\mathrm{d}t} = \frac{\mathrm{d}u}{\mathrm{d}x}\times\frac{\mathrm{d}x}{\mathrm{d}t} = 3x^2\!\left(\frac{3+\cosh t}{3x^2\cosh t}-\frac{1}{3}x\tanh t\right)$ | **B1** | A correct equation relating $\frac{\mathrm{d}u}{\mathrm{d}t}$ and $\frac{\mathrm{d}x}{\mathrm{d}t}$ from the chain rule |
| $\frac{\mathrm{d}x}{\mathrm{d}t}=\frac{3+\cosh t}{3x^2\cosh t}-\frac{1}{3}x\tanh t \rightarrow \frac{1}{3x^2}\frac{\mathrm{d}u}{\mathrm{d}t}=\frac{3+\cosh t}{3x^2\cosh t}-\frac{1}{3}x\tanh t$ $\rightarrow \frac{\mathrm{d}u}{\mathrm{d}t}=\frac{3}{\cosh t}+1-u\tanh t$ | **M1** | Makes complete substitution for $x$ and $\frac{\mathrm{d}x}{\mathrm{d}t}$ in equation (I), or complete substitution for $u$ and $\frac{\mathrm{d}u}{\mathrm{d}t}$ in equation (II) |
| $\frac{\mathrm{d}u}{\mathrm{d}t} + u\tanh t = 1 + \frac{3}{\cosh t}$ * | **A1*** | Simplifies correctly to achieve the given result |

---

## Part (c):

| Answer/Working | Marks | Guidance |
|---|---|---|
| I.F. $= \exp\!\left(\int \tanh t\, \mathrm{d}t\right) = \exp(\ln\cosh t) = \cosh t$ | **B1** | Correct integrating factor found or spotted; allow $e^{\ln\cosh t}$ |
| $\Rightarrow u'\cosh t' = \int \!\cosh t'\!\left(1+\frac{3}{\cosh t}\right)\mathrm{d}t = \left\{\int \cosh t + 3\, \mathrm{d}t\right\}$ | **M1** | Applies IF to achieve $u\,\text{"}\cosh t\text{"} = \int\text{"}\cosh t\text{"}\!\left(1+\frac{3}{\cosh t}\right)\mathrm{d}t$ |
| $\Rightarrow u\cosh t = \sinh t + 3t\,(+c)$ | **M1** | Reasonable attempt to integrate RHS; need not include constant if IF correct; allow $\pm\sinh t + 3t\,(+c)$ |
| $u\cosh t = \sinh t + 3t + c$ or $u = \tanh t + \frac{3t}{\cosh t} + \frac{c}{\cosh t}$ oe | **A1** | Correct general solution, implicit or explicit, including constant of integration |

---

## Part (d):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $t=0 \Rightarrow x=3,\; u=27 \Rightarrow c = 27\cosh 0 - \sinh 0 - 3(0) = 27$ | **M1** | Uses initial conditions in appropriate equation to find constant; either $t=0$, $u=27$ in answer to (c), or $t=0$, $x=3$ if substitution for $x$ occurs first |
| $\Rightarrow x = \left(\tanh t + \frac{3t+\text{"27"}}{\cosh t}\right)^{\!\frac{1}{3}}$ | **M1** | Reverses substitution and rearranges to find equation for $x$, with evaluated constant included |
| $x = \left(\tanh t + \frac{3t+27}{\cosh t}\right)^{\!\frac{1}{3}}$ (oe) | **A1** | Correct equation, any equivalent form, but must be $x=\ldots$ |

---

## Part (e):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $x(0.25) = \left(\tanh 0.25 + \frac{(0.75+27)}{\cosh 0.25}\right)^{\!\frac{1}{3}} = \ldots\,(= 3.0055\ldots)$ | **M1** | Uses their model solution to find value at $t=0.25$ |
| $\%$ error is $\frac{3.0055\ldots - 3.037\ldots}{3.0055\ldots}\times 100 = \ldots$ | **M1** | Applies $\frac{\text{actual value}-\text{estimate}}{\text{actual value}}\times 100$ with their values |
| Estimate in (a) is an overestimate by $1.05\%$ (3 s.f.) | **A1** | States part (a) is overestimate by $1.05\%$ |