| Exam Board | CAIE |
|---|---|
| Module | Further Paper 2 (Further Paper 2) |
| Year | 2021 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hyperbolic functions |
| Type | Surface area of revolution with hyperbolics |
| Difficulty | Challenging +1.8 This is a Further Maths question combining hyperbolic functions with surface area of revolution. Part (a) requires standard differentiation and using the identity cosh²t - sinh²t = 1. Part (b) involves setting up the surface area formula with parametric equations, then integrating products of hyperbolic functions using identities and integration by parts. While technically demanding with multiple steps, the techniques are standard for Further Maths students who have studied both topics, making it challenging but not exceptional. |
| Spec | 4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials4.07d Differentiate/integrate: hyperbolic functions4.08f Integrate using partial fractions4.08h Integration: inverse trig/hyperbolic substitutions8.06b Arc length and surface area: of revolution, cartesian or parametric |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{\mathrm{d}x}{\mathrm{d}t} = 2\sinh t\) | B1 | |
| \(\frac{\mathrm{d}y}{\mathrm{d}t} = \frac{3}{2} - \frac{1}{2}\cosh 2t = 1 - \left(\frac{1}{2}\cosh 2t - \frac{1}{2}\right) = 1 - \sinh^2 t\) | M1 A1 | Applies \(2\sinh^2 t = \cosh 2t - 1\), AG |
| 3 (total) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\left(\frac{\mathrm{d}x}{\mathrm{d}t}\right)^2 + \left(\frac{\mathrm{d}y}{\mathrm{d}t}\right)^2 = 4s^2 + (s^2-1)^2 = 4s^2 + s^4 - 2s^2 + 1 = (s^2+1)^2\) | M1 | Factorises \(\left(\frac{\mathrm{d}x}{\mathrm{d}t}\right)^2 + \left(\frac{\mathrm{d}y}{\mathrm{d}t}\right)^2\) |
| \(\cosh^4 t\) | A1 | |
| \(2\pi\int_0^1\left(\frac{3}{2}t - \frac{1}{4}\sinh 2t\right)\cosh^2 t\,\mathrm{d}t = \pi\int_0^1\left(\frac{3}{2}t - \frac{1}{4}\sinh 2t\right)(\cosh 2t + 1)\,\mathrm{d}t\) | M1 A1 | Correct formula for surface area, AG. A0 if limits missing. \(\left(\frac{\mathrm{d}x}{\mathrm{d}t}\right)^2 + \left(\frac{\mathrm{d}y}{\mathrm{d}t}\right)^2\) does not need to be simplified for M1 |
| 4 (total) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\int_0^1 \frac{1}{4}\sinh 2t(\cosh 2t + 1)\,\mathrm{d}t = \left[\frac{1}{16}(1+\cosh 2t)^2\right]_0^1 = \frac{1}{16}(1+\cosh 2)^2 - \frac{1}{4}\) | M1 A1 | Integrates |
| \(\frac{3}{2}\int_0^1 t(\cosh 2t+1)\,\mathrm{d}t = \frac{3}{2}\left[t\left(\frac{1}{2}\sinh 2t + t\right)\right]_0^1 - \frac{3}{2}\int_0^1 \frac{1}{2}\sinh 2t + t\,\mathrm{d}t\) | M1 A1 | Integrates by parts |
| \(\frac{3}{2}\left[t\left(\frac{1}{2}\sinh 2t + t\right) - \frac{1}{4}\cosh 2t - \frac{1}{2}t^2\right]_0^1 = \frac{3}{2}\left(\frac{1}{2}\sinh 2 - \frac{1}{4}\cosh 2 + \frac{3}{4}\right)\) | A1 | |
| \(\pi\left(\frac{3}{4}\sinh 2 - \frac{3}{8}\cosh 2 - \frac{1}{16}(1+\cosh 2)^2 + \frac{11}{8}\right)\) | A1 | OE. Must be exact. (Decimal answer is \(3.980131435\ldots\)) |
| 6 (total) |
## Question 8(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{\mathrm{d}x}{\mathrm{d}t} = 2\sinh t$ | **B1** | |
| $\frac{\mathrm{d}y}{\mathrm{d}t} = \frac{3}{2} - \frac{1}{2}\cosh 2t = 1 - \left(\frac{1}{2}\cosh 2t - \frac{1}{2}\right) = 1 - \sinh^2 t$ | **M1 A1** | Applies $2\sinh^2 t = \cosh 2t - 1$, AG |
| | **3** (total) | |
---
## Question 8(b)(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\left(\frac{\mathrm{d}x}{\mathrm{d}t}\right)^2 + \left(\frac{\mathrm{d}y}{\mathrm{d}t}\right)^2 = 4s^2 + (s^2-1)^2 = 4s^2 + s^4 - 2s^2 + 1 = (s^2+1)^2$ | **M1** | Factorises $\left(\frac{\mathrm{d}x}{\mathrm{d}t}\right)^2 + \left(\frac{\mathrm{d}y}{\mathrm{d}t}\right)^2$ |
| $\cosh^4 t$ | **A1** | |
| $2\pi\int_0^1\left(\frac{3}{2}t - \frac{1}{4}\sinh 2t\right)\cosh^2 t\,\mathrm{d}t = \pi\int_0^1\left(\frac{3}{2}t - \frac{1}{4}\sinh 2t\right)(\cosh 2t + 1)\,\mathrm{d}t$ | **M1 A1** | Correct formula for surface area, AG. A0 if limits missing. $\left(\frac{\mathrm{d}x}{\mathrm{d}t}\right)^2 + \left(\frac{\mathrm{d}y}{\mathrm{d}t}\right)^2$ does not need to be simplified for M1 |
| | **4** (total) | |
---
## Question 8(b)(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\int_0^1 \frac{1}{4}\sinh 2t(\cosh 2t + 1)\,\mathrm{d}t = \left[\frac{1}{16}(1+\cosh 2t)^2\right]_0^1 = \frac{1}{16}(1+\cosh 2)^2 - \frac{1}{4}$ | **M1 A1** | Integrates |
| $\frac{3}{2}\int_0^1 t(\cosh 2t+1)\,\mathrm{d}t = \frac{3}{2}\left[t\left(\frac{1}{2}\sinh 2t + t\right)\right]_0^1 - \frac{3}{2}\int_0^1 \frac{1}{2}\sinh 2t + t\,\mathrm{d}t$ | **M1 A1** | Integrates by parts |
| $\frac{3}{2}\left[t\left(\frac{1}{2}\sinh 2t + t\right) - \frac{1}{4}\cosh 2t - \frac{1}{2}t^2\right]_0^1 = \frac{3}{2}\left(\frac{1}{2}\sinh 2 - \frac{1}{4}\cosh 2 + \frac{3}{4}\right)$ | **A1** | |
| $\pi\left(\frac{3}{4}\sinh 2 - \frac{3}{8}\cosh 2 - \frac{1}{16}(1+\cosh 2)^2 + \frac{11}{8}\right)$ | **A1** | OE. Must be exact. (Decimal answer is $3.980131435\ldots$) |
| | **6** (total) | |8 The curve $C$ has parametric equations
$$\mathbf { x } = 2 \cosh t , \quad \mathbf { y } = \frac { 3 } { 2 } \mathbf { t } - \frac { 1 } { 4 } \sinh 2 \mathbf { t } , \text { for } 0 \leqslant t \leqslant 1$$
\begin{enumerate}[label=(\alph*)]
\item Find $\frac { \mathrm { dx } } { \mathrm { dt } }$ and show that $\frac { \mathrm { dy } } { \mathrm { dt } } = 1 - \sinh ^ { 2 } \mathrm { t }$.\\
The area of the surface generated when $C$ is rotated through $2 \pi$ radians about the $x$-axis is denoted by $A$.
\item \begin{enumerate}[label=(\roman*)]
\item Show that $\mathrm { A } = \pi \int _ { 0 } ^ { 1 } \left( \frac { 3 } { 2 } \mathrm { t } - \frac { 1 } { 4 } \sinh 2 \mathrm { t } \right) ( 1 + \cosh 2 \mathrm { t } ) \mathrm { dt }$.
\item Hence find $A$ in terms of $\pi , \sinh 2$ and $\cosh 2$.\\
If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 2 2021 Q8 [13]}}