| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric integration |
| Type | Parametric area under curve |
| Difficulty | Standard +0.3 This is a standard C4 parametric equations question covering routine techniques: finding area and volume of revolution using parameter substitution, and converting to Cartesian form. Part (b) is 'show that' which reduces difficulty. The parametric functions are straightforward (polynomial and reciprocal), and the required calculus manipulations follow standard procedures taught in C4 with no novel problem-solving required. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.08e Area between curve and x-axis: using definite integrals4.08d Volumes of revolution: about x and y axes |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(x = 2 \Rightarrow t = 1, \quad x = 9 \Rightarrow t = 2\) | B1 | |
| \(\frac{dx}{dt} = 3t^2\) | M1 | |
| \(\therefore \text{area} = \int_1^2 \frac{2}{t} \times 3t^2 dt = \int_1^2 6t dt\) | A1 | |
| \(= [3t^2]_1^2 = 3(4-1) = 9\) | M1 A1 | |
| (b) \(= \pi \int_1^2 \left(\frac{2}{t}\right)^2 \times 3t^2 dt = \pi \int_1^2 12 dt\) | M1 | |
| \(= \pi[12t]_1^2 = 12\pi(2-1) = 12\pi\) | M1 A1 | |
| (c) \(t = \frac{2}{y} \therefore x = \left(\frac{2}{y}\right)^3 + 1 = \frac{8}{y^3} + 1\) | M1 | |
| \(\therefore y^3 = \frac{8}{x-1}, \quad y = \frac{2}{\sqrt[3]{x-1}}\) | M1 A1 | (11 marks) |
**(a)** $x = 2 \Rightarrow t = 1, \quad x = 9 \Rightarrow t = 2$ | B1 |
$\frac{dx}{dt} = 3t^2$ | M1 |
$\therefore \text{area} = \int_1^2 \frac{2}{t} \times 3t^2 dt = \int_1^2 6t dt$ | A1 |
$= [3t^2]_1^2 = 3(4-1) = 9$ | M1 A1 |
**(b)** $= \pi \int_1^2 \left(\frac{2}{t}\right)^2 \times 3t^2 dt = \pi \int_1^2 12 dt$ | M1 |
$= \pi[12t]_1^2 = 12\pi(2-1) = 12\pi$ | M1 A1 |
**(c)** $t = \frac{2}{y} \therefore x = \left(\frac{2}{y}\right)^3 + 1 = \frac{8}{y^3} + 1$ | M1 |
$\therefore y^3 = \frac{8}{x-1}, \quad y = \frac{2}{\sqrt[3]{x-1}}$ | M1 A1 | (11 marks)7.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{4232f6a1-00ff-4e88-b5f4-1abf3d4742c4-12_560_911_146_456}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Figure 1 shows the curve with parametric equations
$$x = t ^ { 3 } + 1 , \quad y = \frac { 2 } { t } , \quad t > 0 .$$
The shaded region is bounded by the curve, the $x$-axis and the lines $x = 2$ and $x = 9$.
\begin{enumerate}[label=(\alph*)]
\item Find the area of the shaded region.
\item Show that the volume of the solid formed when the shaded region is rotated through $2 \pi$ radians about the $x$-axis is $12 \pi$.
\item Find a cartesian equation for the curve in the form $y = \mathrm { f } ( x )$.\\
7. continued
\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 Q7 [11]}}