| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2010 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric integration |
| Type | Parametric loop enclosed area |
| Difficulty | Standard +0.3 This is a standard C4 parametric equations question requiring finding intersection points by setting y=0, then computing area using the formula ∫y(dx/dt)dt. The algebra is straightforward (factoring cubic, differentiating quadratic), and the technique is a direct application of the standard parametric area formula taught in C4. Slightly easier than average due to clean algebraic expressions and routine method application. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| (a) | \(y = 0 \Rightarrow t(9-t^2) = t(3-t)(3+t) = 0\) then \(t = 0, 3, -3\) | Any one correct value B1; Method for finding one value of \(x\) M1 |
| At \(t = 0\), \(x = 5(0)^2 - 4 = -4\); At \(t = 3\), \(x = 5(3)^2 - 4 = 41\); At \(t = -3\), \(x = 5(-3)^2 - 4 = 41\) | Both A1 | (3 marks) |
| (b) | \(\frac{dx}{dt} = 10t\) (Seen or implied) then \(\int y \, dx = \int y \frac{dx}{dt} \, dt = \int t(9-t^2)10t \, dt = \int (90t^2 - 10t^4) \, dt = \frac{90t^3}{3} - \frac{10t^5}{5}(+C)\) (= \(30t^3 - 2t^5(+C)\)) | B1; M1 A1 |
| \(\left[\frac{90t^3}{3} - \frac{10t^5}{5}\right]_0^3 = 30 \times 3^3 - 2 \times 3^5\) (= 324) | M1 | |
| \(A = 2\int y \, dx = 648\) (units\(^2\)) | A1 | (6 marks) |
(a) | $y = 0 \Rightarrow t(9-t^2) = t(3-t)(3+t) = 0$ then $t = 0, 3, -3$ | Any one correct value B1; Method for finding one value of $x$ M1 |
At $t = 0$, $x = 5(0)^2 - 4 = -4$; At $t = 3$, $x = 5(3)^2 - 4 = 41$; At $t = -3$, $x = 5(-3)^2 - 4 = 41$ | Both A1 | (3 marks)
(b) | $\frac{dx}{dt} = 10t$ (Seen or implied) then $\int y \, dx = \int y \frac{dx}{dt} \, dt = \int t(9-t^2)10t \, dt = \int (90t^2 - 10t^4) \, dt = \frac{90t^3}{3} - \frac{10t^5}{5}(+C)$ (= $30t^3 - 2t^5(+C)$) | B1; M1 A1 |
$\left[\frac{90t^3}{3} - \frac{10t^5}{5}\right]_0^3 = 30 \times 3^3 - 2 \times 3^5$ (= 324) | M1 |
$A = 2\int y \, dx = 648$ (units$^2$) | A1 | (6 marks)
**Total: [9]**
---7.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{5ef3ae4a-a06d-48c1-8b79-7d7c3f95d120-12_734_1395_210_249}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
Figure 2 shows a sketch of the curve $C$ with parametric equations
$$x = 5 t ^ { 2 } - 4 , \quad y = t \left( 9 - t ^ { 2 } \right)$$
The curve $C$ cuts the $x$-axis at the points $A$ and $B$.
\begin{enumerate}[label=(\alph*)]
\item Find the $x$-coordinate at the point $A$ and the $x$-coordinate at the point $B$.
The region $R$, as shown shaded in Figure 2, is enclosed by the loop of the curve.
\item Use integration to find the area of $R$.\\
\section*{LU}
\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 2010 Q7 [9]}}