| Exam Board | OCR |
|---|---|
| Module | Further Additional Pure (Further Additional Pure) |
| Year | 2018 |
| Session | September |
| Marks | 10 |
| Topic | Vectors: Cross Product & Distances |
| Type | Find parameter value for geometric condition |
| Difficulty | Challenging +1.8 This is a Further Maths question requiring parametric arc length and surface of revolution formulas with integration. While the formulas are standard, the algebraic manipulation (differentiating parametric equations, simplifying under square roots, and evaluating definite integrals in exact form) requires careful multi-step work. The parametric expressions lead to manageable but non-trivial integrals, making this significantly harder than typical A-level questions but within reach for well-prepared Further Maths students. |
| Spec | 4.08d Volumes of revolution: about x and y axes8.06b Arc length and surface area: of revolution, cartesian or parametric |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dx}{dt} = 3t^2 - 3\) and \(\frac{dy}{dt} = 6t\) | B1, M1, A1 | DR Both correct; Attempted; Must be in the form of a perfect square (here or later) |
| \(\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 = 9t^4 - 18t^2 + 9 + 36t^2\) | M1, A1 | Attempted; Correct arc-length formula statement (FT the integrand) |
| \(L = \int_0^5 (3t^2 + 3) dt\) | M1, A1 | Correct arc-length formula statement (FT the integrand); DR Correct integration of (at least) a quadratic; ignore limits |
| \(= \left[t^3 + 3t\right]_0^5\) | A1, A1 | DR Correct integration of (at least) a quadratic; ignore limits; (blank) |
| \(= 140\) | A1 | |
| (ii) \(A = 2\pi \int_0^5 (3t^2 - 1)(3t^2 + 3) dt\) | M1, A1 | DR Correct surface area formula statement (FT the integrand) |
| \(= 6\pi \int_0^5 (3t^4 + 2t^2 - 1) dt\) | A1 | Correct, in an integrable form |
| \(= 6\pi \left[\frac{3}{5}t^5 + \frac{2}{3}t^3 - t\right]\) | A1 | Correct integration of (at least) a quartic; ignore limits |
| \(= 11720\pi\) | A1 |
**(i)**
$\frac{dx}{dt} = 3t^2 - 3$ and $\frac{dy}{dt} = 6t$ | B1, M1, A1 | DR Both correct; Attempted; Must be in the form of a perfect square (here or later)
$\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 = 9t^4 - 18t^2 + 9 + 36t^2$ | M1, A1 | Attempted; Correct arc-length formula statement (FT the integrand)
$L = \int_0^5 (3t^2 + 3) dt$ | M1, A1 | Correct arc-length formula statement (FT the integrand); DR Correct integration of (at least) a quadratic; ignore limits
$= \left[t^3 + 3t\right]_0^5$ | A1, A1 | DR Correct integration of (at least) a quadratic; ignore limits; (blank)
$= 140$ | A1 |
**(ii)** $A = 2\pi \int_0^5 (3t^2 - 1)(3t^2 + 3) dt$ | M1, A1 | DR Correct surface area formula statement (FT the integrand)
$= 6\pi \int_0^5 (3t^4 + 2t^2 - 1) dt$ | A1 | Correct, in an integrable form
$= 6\pi \left[\frac{3}{5}t^5 + \frac{2}{3}t^3 - t\right]$ | A1 | Correct integration of (at least) a quartic; ignore limits
$= 11720\pi$ | A1 |In this question, you must show detailed reasoning.
A curve is defined parametrically by $x = t^3 - 3t + 1$, $y = 3t^2 - 1$, for $0 \leq t \leq 5$. Find, in exact form,
\begin{enumerate}[label=(\roman*)]
\item the length of the curve, [6]
\item the area of the surface generated when the curve is rotated completely about the $x$-axis. [4]
\end{enumerate}
\hfill \mbox{\textit{OCR Further Additional Pure 2018 Q2 [10]}}