| Exam Board | Edexcel |
|---|---|
| Module | F3 (Further Pure Mathematics 3) |
| Year | 2020 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hyperbolic functions |
| Type | Surface area of revolution with hyperbolics |
| Difficulty | Challenging +1.3 This is a structured Further Maths question on surface area of revolution with hyperbolic functions. Part (a) is routine differentiation and algebra using cosh²t - sinh²t = 1. Part (b) applies the standard formula S = 2π∫y√((dx/dt)² + (dy/dt)²)dt. Part (c) requires integration of cosh²t and t·cosh t using standard techniques (double angle formula and integration by parts). While it involves Further Maths content and multiple steps, each part follows predictable methods with clear signposting, making it moderately above average difficulty but not requiring significant novel insight. |
| Spec | 1.07s Parametric and implicit differentiation4.07b Hyperbolic graphs: sketch and properties4.07d Differentiate/integrate: hyperbolic functions8.06b Arc length and surface area: of revolution, cartesian or parametric |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dx}{dt} = \sinh t + 1,\quad \frac{dy}{dt} = \sinh t - 1\) | B1 | Correct derivatives |
| \(\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 = \sinh^2 t + 2\sinh t + 1 + \sinh^2 t - 2\sinh t + 1 = 2\sinh^2 t + 2\) | M1 | Squares correctly, cancels and collects terms |
| \(= 2(1 + \sinh^2 t) = 2\cosh^2 t\) | A1* | Uses \(\cosh^2 t = 1 + \sinh^2 t\) to complete proof with no errors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(S = 2\pi\int y\, ds = 2\pi\int(\cosh t - t)\sqrt{2}\cosh t\, dt\) | M1 | Uses \(S = 2\pi\int y\, ds\) with given \(y\) and result from part (a) |
| \(= 2\sqrt{2}\pi\int_0^{\ln 3}(\cosh^2 t - t\cosh t)\, dt\) | A1* | Correct proof with no errors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\int\cosh^2 t\, dt = \int\pm\frac{1}{2}\pm\frac{1}{2}\cosh 2t\, dt\) | M1 | Uses \(\cosh^2 t = \pm\frac{1}{2}\pm\frac{1}{2}\cosh 2t\) |
| \(\int t\cosh t\, dt = t\sinh t - \int\sinh t\, dt\) | M1 | Attempts integration by parts the right way round on \(t\cosh t\) |
| Correct expression for \(\int t\cosh t\, dt\) | A1 | |
| \(S = (2\sqrt{2}\pi)\int(\cosh^2 t - t\cosh t)\,dt = (2\sqrt{2}\pi)\left[\frac{1}{2}t + \frac{1}{4}\sinh 2t - t\sinh t + \cosh t\right]\) | A1A1 | A1: 2 correct terms; A1: all correct |
| \((S=)\ 2\sqrt{2}\pi\left\{\left(\frac{1}{2}\ln 3 + \frac{10}{9} - \frac{4}{3}\ln 3 + \frac{5}{3}\right) - (1)\right\}\) | dM1 | Correct use of limits 0 and \(\ln 3\); depends on both preceding M marks |
| \(S = \frac{1}{9}\sqrt{2}\pi(32 - 15\ln 3)\) | A1 | cao |
# Question 7(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dx}{dt} = \sinh t + 1,\quad \frac{dy}{dt} = \sinh t - 1$ | B1 | Correct derivatives |
| $\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 = \sinh^2 t + 2\sinh t + 1 + \sinh^2 t - 2\sinh t + 1 = 2\sinh^2 t + 2$ | M1 | Squares correctly, cancels and collects terms |
| $= 2(1 + \sinh^2 t) = 2\cosh^2 t$ | A1* | Uses $\cosh^2 t = 1 + \sinh^2 t$ to complete proof with no errors |
**Total: (3)**
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# Question 7(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $S = 2\pi\int y\, ds = 2\pi\int(\cosh t - t)\sqrt{2}\cosh t\, dt$ | M1 | Uses $S = 2\pi\int y\, ds$ with given $y$ and result from part (a) |
| $= 2\sqrt{2}\pi\int_0^{\ln 3}(\cosh^2 t - t\cosh t)\, dt$ | A1* | Correct proof with no errors |
**Total: (2)**
---
# Question 7(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int\cosh^2 t\, dt = \int\pm\frac{1}{2}\pm\frac{1}{2}\cosh 2t\, dt$ | M1 | Uses $\cosh^2 t = \pm\frac{1}{2}\pm\frac{1}{2}\cosh 2t$ |
| $\int t\cosh t\, dt = t\sinh t - \int\sinh t\, dt$ | M1 | Attempts integration by parts the right way round on $t\cosh t$ |
| Correct expression for $\int t\cosh t\, dt$ | A1 | |
| $S = (2\sqrt{2}\pi)\int(\cosh^2 t - t\cosh t)\,dt = (2\sqrt{2}\pi)\left[\frac{1}{2}t + \frac{1}{4}\sinh 2t - t\sinh t + \cosh t\right]$ | A1A1 | A1: 2 correct terms; A1: all correct |
| $(S=)\ 2\sqrt{2}\pi\left\{\left(\frac{1}{2}\ln 3 + \frac{10}{9} - \frac{4}{3}\ln 3 + \frac{5}{3}\right) - (1)\right\}$ | dM1 | Correct use of limits 0 and $\ln 3$; depends on both preceding M marks |
| $S = \frac{1}{9}\sqrt{2}\pi(32 - 15\ln 3)$ | A1 | cao |
**Total: (7) — Question Total: 12**7. The curve $C$ has parametric equations
$$x = \cosh t + t , \quad y = \cosh t - t \quad 0 \leqslant t \leqslant \ln 3$$
\begin{enumerate}[label=(\alph*)]
\item Show that
$$\left( \frac { \mathrm { d } x } { \mathrm {~d} t } \right) ^ { 2 } + \left( \frac { \mathrm { d } y } { \mathrm {~d} t } \right) ^ { 2 } = 2 \cosh ^ { 2 } t$$
The curve $C$ is rotated through $2 \pi$ radians about the $x$-axis. The area of the curved surface generated is given by $S$.
\item Show that
$$S = 2 \pi \sqrt { 2 } \int _ { 0 } ^ { \ln 3 } \left( \cosh ^ { 2 } t - t \cosh t \right) d t$$
\item Hence find the value of $S$, giving your answer in the form
$$\frac { \pi \sqrt { 2 } } { 9 } ( a + b \ln 3 )$$
where $a$ and $b$ are constants to be determined.
\end{enumerate}
\hfill \mbox{\textit{Edexcel F3 2020 Q7 [12]}}