## Question 8(a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dx}{dy}=\frac{y}{4}$ or $2y\frac{dy}{dx}=8$ or $\frac{dy}{dx}=\frac{1}{2}(2\sqrt{2})x^{-\frac{1}{2}}$ or $\frac{1}{2}(2\sqrt{2})\left(\frac{2\sqrt{2}}{y}\right)$ oe | B1 | Any correct equation in $\frac{dx}{dy}$ or $\frac{dy}{dx}$ in terms of $y$ or $x$ |
| $\int\sqrt{1+\left(\frac{dx}{dy}\right)^2}\,dy = \int\sqrt{1+\left(\frac{y}{4}\right)^2}\,dy$ or $\int\sqrt{1+\left(\frac{dy}{dx}\right)^2}\cdot\frac{dx}{dy}\,dy = \int\sqrt{1+\left(\frac{4}{y}\right)^2}\cdot\frac{y}{4}\,dy$ | M1 | Forms $\int\sqrt{1+\left(\frac{dx}{dy}\right)^2}\,dy$ or $\int\sqrt{1+\left(\frac{dy}{dx}\right)^2}\cdot\frac{dx}{dy}\,dy$ correctly with their derivative |
| $x=18 \Rightarrow y^2=144 \Rightarrow \beta=12,\ \alpha=24$; $\Rightarrow$ perimeter of $R = 24 + 2\int_0^{12}\sqrt{1+\frac{y^2}{16}}\,dy$ | A1 | Correct expression |

**(3 marks)**

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## Question 8(b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = 4\sinh u \Rightarrow \frac{dy}{du}=4\cosh u$ | B1 | Correct derivative; condone $\frac{dy}{dx}=4\cosh u$ |
| $\int\sqrt{1+\frac{y^2}{16}}\,dy = \int\sqrt{1+\frac{(4\sinh u)^2}{16}}(4\cosh u)\,du = 4\int\cosh^2 u\,du$ | M1 | Full substitution, correct for their $\frac{dy}{du}$ |
| $\int\cosh^2 u\,du = \int\!\left(\frac{1}{2}\cosh 2u+\frac{1}{2}\right)du = \frac{1}{4}\sinh 2u + \frac{1}{2}u$ **or** $\int\!\left(\frac{e^u+e^{-u}}{2}\right)^2du = \int\!\left(\frac{e^{2u}}{4}+\frac{1}{2}+\frac{e^{-2u}}{4}\right)du = \frac{e^{2u}}{8}+\frac{1}{2}u-\frac{e^{-2u}}{8}$ | dM1 A1 | dM1: Uses $\cosh^2 u = \pm\frac{1}{2}\cosh 2u \pm \frac{1}{2}$ and integrates to get $a\sinh 2u + bu$, **or** uses $k(e^u+e^{-u})$ for $\cosh u$, expands and integrates to get $ae^{2u}+bu+ce^{-2u}$. A1: Correct integration |
| $= 24+(2)(4)\left[\frac{1}{4}\sinh 2u+\frac{1}{2}u\right]_0^{\text{arsinh }3=\ln(3+\sqrt{10})}$ $= 24+2\left[2\sinh u\sqrt{1+\sinh^2 u}+2u\right]_0^{\text{arsinh }3=\ln(3+\sqrt{10})}$ $= 24+2\left[(2)(3)\sqrt{1+3^2}+2\ln(3+\sqrt{10})\right]$ | ddM1 | Substitutes $\text{arsinh}\,3$ and/or $\ln(3+\sqrt{3^2+1})$ into their expression using correct identities, **or** correctly removes exponentials to obtain numerical expression in constants and lns only; accept calculator use e.g. $\sinh(2\,\text{arsinh}\,3)=6\sqrt{10}$ |
| $24+12\sqrt{10}+4\ln(3+\sqrt{10})$ or e.g. $4\!\left(6+3\sqrt{10}+\ln(3+\sqrt{10})\right)$ | A1 | Correct answer — any exact simplified equivalent |

**(6 marks) | Total 9**

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