| Exam Board | Edexcel |
|---|---|
| Module | F3 (Further Pure Mathematics 3) |
| Year | 2014 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric integration |
| Type | Arc length with substitution |
| Difficulty | Challenging +1.8 This is a Further Maths F3 question requiring surface area of revolution and arc length with parametric equations, plus a hyperbolic substitution. While the techniques are advanced (F3 level), the execution is relatively straightforward: standard formula application, routine integration, and a guided substitution. The multi-part structure and hyperbolic functions place it well above average difficulty, but the step-by-step guidance and standard methods prevent it from reaching the highest difficulty tier. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian4.08d Volumes of revolution: about x and y axes |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dx}{dt} = 6t\), \(\frac{dy}{dt} = 12\) | B1 | Both derivatives correct |
| \(S = (2\pi)\int 12t\sqrt{(6t)^2 + 12^2}\, dt\) | M1A1 | M1: Use of a correct surface area formula with their derivatives (\(2\pi\) not needed for this mark). A1: Correct expression including \(2\pi\) which may be implied by later work |
| \(= \frac{2\pi}{9}\left[(36t^2+144)^{\frac{3}{2}}\right]\) | dM1 | Recognisable attempt at integration e.g. \(t = 2\tan\theta\). Dependent on the first M |
| \(= \frac{2\pi}{9}\left\{720^{\frac{3}{2}} - 144^{\frac{3}{2}}\right\}\) | dM1 | Uses the limits 0 and 4 and subtracts. Dependent on the first M |
| \(= \pi(1920\sqrt{5} - 384)\) | A1 | cao (Allow equivalent fractions for 1920 and or 384) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(L = \int_0^4 \sqrt{(6t)^2 + 12^2}\, dt = 6\int_0^4 \sqrt{t^2+4}\, dt\) | B1 | Use of a correct arc length formula and obtains \(k=6\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(t = 2\sinh\theta \Rightarrow \frac{dt}{d\theta} = 2\cosh\theta\) | B1 | Correct derivative |
| \(L = 6\int\sqrt{4\sinh^2\theta + 4} \times 2\cosh\theta\, d\theta\) | M1 | Complete substitution |
| \(= 24\int\cosh^2\theta\, d\theta = 12\int(\cosh 2\theta + 1)\, d\theta\) | M1 | Uses \(\cosh^2\theta = \pm\frac{1}{2} \pm \frac{1}{2}\cosh 2\theta\) |
| \(6\sinh 2\theta + 12\theta\) | A1 | Correct integration |
| \(L = 6\sinh 2(\text{arsinh}\, 2) + 12\,\text{arsinh}\,2 - (0)\) | M1 | Use limits \(\text{arsinh}\,2\) (and 0) |
| \(= 24\sqrt{5} + 12\ln(2+\sqrt{5})\)* | A1* | Correct solution with no errors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(24\int\cosh^2\theta\, d\theta = 12\int\left(\frac{e^\theta + e^{-\theta}}{2}\right)^2 d\theta = 6\int(e^{2\theta} + e^{-2\theta} + 2)\, d\theta\) | M1 | Substitutes the correct exponential form of \(\cosh\theta\) and squares |
| \(3e^{2\theta} - 3e^{-2\theta} + 12\theta\) | A1 | Correct integration |
| \(L = 3e^{2\,\text{arsinh}\,2} - 3e^{-2\,\text{arsinh}\,2} + 12\,\text{arsinh}\,2 - (0)\) | M1 | Use limits \(\text{arsinh}\,2\) (and 0) |
| \(= 24\sqrt{5} + 12\ln(2+\sqrt{5})\)* | A1* | Correct solution with no errors |
# Question 7:
## Part 7(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dx}{dt} = 6t$, $\frac{dy}{dt} = 12$ | B1 | Both derivatives correct |
| $S = (2\pi)\int 12t\sqrt{(6t)^2 + 12^2}\, dt$ | M1A1 | M1: Use of a correct surface area formula with their derivatives ($2\pi$ **not needed for this mark**). A1: Correct expression **including** $2\pi$ which may be implied by later work |
| $= \frac{2\pi}{9}\left[(36t^2+144)^{\frac{3}{2}}\right]$ | dM1 | Recognisable attempt at integration e.g. $t = 2\tan\theta$. **Dependent on the first M** |
| $= \frac{2\pi}{9}\left\{720^{\frac{3}{2}} - 144^{\frac{3}{2}}\right\}$ | dM1 | Uses the limits 0 and 4 and subtracts. **Dependent on the first M** |
| $= \pi(1920\sqrt{5} - 384)$ | A1 | cao (Allow equivalent fractions for 1920 and or 384) |
## Part 7(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $L = \int_0^4 \sqrt{(6t)^2 + 12^2}\, dt = 6\int_0^4 \sqrt{t^2+4}\, dt$ | B1 | Use of a correct arc length formula and obtains $k=6$ |
## Part 7(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $t = 2\sinh\theta \Rightarrow \frac{dt}{d\theta} = 2\cosh\theta$ | B1 | Correct derivative |
| $L = 6\int\sqrt{4\sinh^2\theta + 4} \times 2\cosh\theta\, d\theta$ | M1 | Complete substitution |
| $= 24\int\cosh^2\theta\, d\theta = 12\int(\cosh 2\theta + 1)\, d\theta$ | M1 | Uses $\cosh^2\theta = \pm\frac{1}{2} \pm \frac{1}{2}\cosh 2\theta$ |
| $6\sinh 2\theta + 12\theta$ | A1 | Correct integration |
| $L = 6\sinh 2(\text{arsinh}\, 2) + 12\,\text{arsinh}\,2 - (0)$ | M1 | Use limits $\text{arsinh}\,2$ (and 0) |
| $= 24\sqrt{5} + 12\ln(2+\sqrt{5})$* | A1* | Correct solution with no errors |
### Alternative (exponentials, last 4 marks):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $24\int\cosh^2\theta\, d\theta = 12\int\left(\frac{e^\theta + e^{-\theta}}{2}\right)^2 d\theta = 6\int(e^{2\theta} + e^{-2\theta} + 2)\, d\theta$ | M1 | Substitutes the correct exponential form of $\cosh\theta$ and squares |
| $3e^{2\theta} - 3e^{-2\theta} + 12\theta$ | A1 | Correct integration |
| $L = 3e^{2\,\text{arsinh}\,2} - 3e^{-2\,\text{arsinh}\,2} + 12\,\text{arsinh}\,2 - (0)$ | M1 | Use limits $\text{arsinh}\,2$ (and 0) |
| $= 24\sqrt{5} + 12\ln(2+\sqrt{5})$* | A1* | Correct solution with no errors |
---7. The curve $C$ has parametric equations
$$x = 3 t ^ { 2 } , \quad y = 12 t , \quad 0 \leqslant t \leqslant 4$$
The curve $C$ is rotated through $2 \pi$ radians about the $x$-axis.
\begin{enumerate}[label=(\alph*)]
\item Show that the area of the surface generated is
$$\pi ( a \sqrt { 5 } + b )$$
where $a$ and $b$ are constants to be found.
\item Show that the length of the curve $C$ is given by
$$k \int _ { 0 } ^ { 4 } \sqrt { \left( t ^ { 2 } + 4 \right) } \mathrm { d } t$$
where $k$ is a constant to be found.
\item Use the substitution $t = 2 \sinh \theta$ to show that the exact value of the length of the curve $C$ is
$$24 \sqrt { 5 } + 12 \ln ( 2 + \sqrt { 5 } )$$
\end{enumerate}
\hfill \mbox{\textit{Edexcel F3 2014 Q7 [13]}}