| Exam Board | Edexcel |
|---|---|
| Module | C34 (Core Mathematics 3 & 4) |
| Year | 2018 |
| Session | October |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric integration |
| Type | Parametric loop enclosed area |
| Difficulty | Standard +0.3 This is a standard parametric equations question covering routine techniques: finding tangent equations (differentiation using chain rule), finding axis intersections (setting y=0), and computing area under a parametric curve. All methods are textbook exercises requiring careful algebra but no novel insight. Slightly easier than average due to straightforward parametric forms and standard integration. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.07s Parametric and implicit differentiation1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| At \(t=1 \Rightarrow x=2, y=8\) | B1 | Score if \((2,8)\) is used in tangent equation |
| \(\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{9-3t^2}{14t}\) | M1A1 | M1: Attempts \(\frac{dy}{dx}=\frac{dy/dt}{dx/dt}\). Condone slips in multiplying out \(t(9-t^2)\) before differentiating or slips in product rule, but do not allow \(\frac{dy}{dt}=1\times-2t\). A1: \(\frac{dy}{dx}=\frac{9-3t^2}{14t}\) or exact equivalent |
| Equation of tangent: \(y-8=\frac{6}{14}(x-2)\) | M1 | Valid attempt at tangent to \(C\) at \(t=1\). Allow \(y-\)"\(8\)"\(=\)"\(\frac{6}{14}\)"\((x-\)"\(2\)"\()\) |
| \(3x-7y+50=0\) | A1 | Allow \(k(3x-7y+50=0)\) where \(k\) is an integer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| At \(A\), \(x=-5\) | B1 | \(A=(-5,0)\). Allow \(x=-5\) and \(A=-5\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y=0 \Rightarrow t(9-t^2)=t(3-t)(3+t)=0\), so \(t=0,3,-3\) | M1 | For attempting to solve \(t(9-t^2)=0\) to produce a non-zero value for \(t\) and substitute into \(x=7t^2-5\) |
| At \(t=3\), \(x=7(3)^2-5=58\) or at \(t=-3\), \(x=7(-3)^2-5=58\), so at \(B\), \(x=58\) | A1 | \(B=(58,0)\). Allow \(x=58\) and \(B=58\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\int y\,dx = \int y\frac{dx}{dt}\,dt = \int t(9-t^2)14t\,dt\) | M1 | Attempts area \(= \int y\frac{dx}{dt}\,dt\). Do not be concerned with limits |
| \(= \int(126t^2-14t^4)\,dt\) | A1 | Must be multiplied out. Alternatively accept \(\int 14t^2(9-t^2)\,dt\) by parts with \(u\) being one of \(14t^2\) or \((9-t^2)\) and \(v\) the other |
| \(= \frac{126t^3}{3}-\frac{14t^5}{5}\ (+C)\) \(\left(=42t^3-2.8t^5\ (+C)\right)\) | M1 | Integrates to a form \(At^3+Bt^5\ (+c)\). Condone slips on coefficients only |
| Either \(2\times\left[\frac{126t^3}{3}-\frac{14t^5}{5}\right]_0^3 = 2\times(42\times3^3-2.8\times3^5)=907.2\) | ddM1A1 | ddM1: Full method to find area of \(R\) using limits. Accept either \(2\times[\ldots]_0^3\) or \([\ldots]_{-3}^3\). Dependent on both M marks. A1: \(907.2\) (units\(^2\)) or equivalent \(4536/5\) |
| Or \(\left[\frac{126t^3}{3}-\frac{14t^5}{5}\right]_{-3}^3 = (42\times3^3-2.8\times3^5)-(42\times(-3)^3-2.8\times(-3)^5)=907.2\) |
# Question 12:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| At $t=1 \Rightarrow x=2, y=8$ | B1 | Score if $(2,8)$ is used in tangent equation |
| $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{9-3t^2}{14t}$ | M1A1 | M1: Attempts $\frac{dy}{dx}=\frac{dy/dt}{dx/dt}$. Condone slips in multiplying out $t(9-t^2)$ before differentiating or slips in product rule, but do not allow $\frac{dy}{dt}=1\times-2t$. A1: $\frac{dy}{dx}=\frac{9-3t^2}{14t}$ or exact equivalent |
| Equation of tangent: $y-8=\frac{6}{14}(x-2)$ | M1 | Valid attempt at tangent to $C$ at $t=1$. Allow $y-$"$8$"$=$"$\frac{6}{14}$"$(x-$"$2$"$)$ |
| $3x-7y+50=0$ | A1 | Allow $k(3x-7y+50=0)$ where $k$ is an integer |
## Part (b)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| At $A$, $x=-5$ | B1 | $A=(-5,0)$. Allow $x=-5$ and $A=-5$ |
## Part (b)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y=0 \Rightarrow t(9-t^2)=t(3-t)(3+t)=0$, so $t=0,3,-3$ | M1 | For attempting to solve $t(9-t^2)=0$ to produce a non-zero value for $t$ and substitute into $x=7t^2-5$ |
| At $t=3$, $x=7(3)^2-5=58$ or at $t=-3$, $x=7(-3)^2-5=58$, so at $B$, $x=58$ | A1 | $B=(58,0)$. Allow $x=58$ and $B=58$ |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int y\,dx = \int y\frac{dx}{dt}\,dt = \int t(9-t^2)14t\,dt$ | M1 | Attempts area $= \int y\frac{dx}{dt}\,dt$. Do not be concerned with limits |
| $= \int(126t^2-14t^4)\,dt$ | A1 | Must be multiplied out. Alternatively accept $\int 14t^2(9-t^2)\,dt$ by parts with $u$ being one of $14t^2$ or $(9-t^2)$ and $v$ the other |
| $= \frac{126t^3}{3}-\frac{14t^5}{5}\ (+C)$ $\left(=42t^3-2.8t^5\ (+C)\right)$ | M1 | Integrates to a form $At^3+Bt^5\ (+c)$. Condone slips on coefficients only |
| Either $2\times\left[\frac{126t^3}{3}-\frac{14t^5}{5}\right]_0^3 = 2\times(42\times3^3-2.8\times3^5)=907.2$ | ddM1A1 | ddM1: Full method to find area of $R$ using limits. Accept either $2\times[\ldots]_0^3$ or $[\ldots]_{-3}^3$. Dependent on both M marks. A1: $907.2$ (units$^2$) or equivalent $4536/5$ |
| Or $\left[\frac{126t^3}{3}-\frac{14t^5}{5}\right]_{-3}^3 = (42\times3^3-2.8\times3^5)-(42\times(-3)^3-2.8\times(-3)^5)=907.2$ | | |
---12.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{c6bde466-61ec-437d-a3b4-84511a98d788-40_520_663_255_644}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
Figure 3 shows a sketch of part of the curve $C$ with parametric equations
$$x = 7 t ^ { 2 } - 5 , \quad y = t \left( 9 - t ^ { 2 } \right) , \quad t \in \mathbb { R }$$
\begin{enumerate}[label=(\alph*)]
\item Find an equation of the tangent to $C$ at the point where $t = 1$
Write your answer in the form $a x + b y + c = 0$, where $a , b$ and $c$ are integers.
The curve $C$ cuts the $x$-axis at the points $A$ and $B$, as shown in Figure 3
\item \begin{enumerate}[label=(\roman*)]
\item Find the $x$ coordinate of the point $A$.
\item Find the $x$ coordinate of the point $B$.
The region $R$, shown shaded in Figure 3, is enclosed by the loop of the curve $C$.
\end{enumerate}\item Use integration to find the area of $R$.
\begin{center}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{Edexcel C34 2018 Q12 [13]}}