| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2008 |
| Session | June |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric integration |
| Type | Parametric normal then bounded area |
| Difficulty | Standard +0.3 This is a standard C4 parametric equations question covering routine techniques: finding parameter values from coordinates, finding normals using dy/dx from the chain rule, setting up area integrals with the parametric formula, and evaluating using trigonometric identities and substitution. All parts follow textbook methods with no novel insight required, making it slightly easier than average. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.07s Parametric and implicit differentiation1.08d Evaluate definite integrals: between limits1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| At \(P(4, 2\sqrt{3})\) either \(4 = 8\cos t\) or \(2\sqrt{3} = 4\sin 2t\) | M1 | \(4=8\cos t\) or \(2\sqrt{3}=4\sin 2t\) |
| \(\Rightarrow\) only solution is \(t = \frac{\pi}{3}\) where \(0 \leq t \leq \frac{\pi}{2}\) | A1 | \(t=\frac{\pi}{3}\) or awrt 1.05 (radians) only, stated in the range \(0 \leq t \leq \frac{\pi}{2}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x = 8\cos t\), \(y = 4\sin 2t\); \(\frac{dx}{dt} = -8\sin t\), \(\frac{dy}{dt} = 8\cos 2t\) | M1 | Attempt to differentiate both \(x\) and \(y\) wrt \(t\) to give \(\pm p\sin t\) and \(\pm q\cos 2t\) respectively |
| A1 | Correct \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\) | |
| At \(P\): \(\frac{dy}{dx} = \frac{8\cos\!\left(\frac{2\pi}{3}\right)}{-8\sin\!\left(\frac{\pi}{3}\right)}\) | M1* | Divides in correct way and substitutes their value of \(t\) into their \(\frac{dy}{dx}\) expression |
| \(= \frac{8(-\frac{1}{2})}{(-8)(\frac{\sqrt{3}}{2})} = \frac{1}{\sqrt{3}}\) awrt 0.58 | You may need to check candidate's substitutions for M1* | |
| Hence \(m(\mathbf{N}) = -\sqrt{3}\) or \(\frac{-1}{\frac{1}{\sqrt{3}}}\) | dM1* | Uses \(m(\mathbf{N}) = -\frac{1}{\text{their } m(\mathbf{T})}\) |
| N: \(y - 2\sqrt{3} = -\sqrt{3}(x-4)\) | dM1* | Uses \(y - 2\sqrt{3} = (\text{their } m_N)(x-4)\) or finds \(c\) using \(x=4\) and \(y=2\sqrt{3}\) and uses \(y=(\text{their } m_N)x + \text{"c"}\) |
| N: \(y = -\sqrt{3}x + 6\sqrt{3}\) AG | A1 cso AG | \(y = -\sqrt{3}x + 6\sqrt{3}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(A = \int_0^4 y\, dx = \int 4\sin 2t \cdot (-8\sin t)\, dt\) | M1 | attempt at \(A = \int y \frac{dx}{dt}\, dt\) |
| Correct expression | A1 | ignore limits and \(dt\) |
| \(A = \int -32\sin 2t \cdot \sin t\, dt = \int -32(2\sin t \cos t)\cdot \sin t\, dt\) | M1 | Seeing \(\sin 2t = 2\sin t \cos t\) anywhere in PART (c) |
| \(A = \int -64\sin^2 t \cos t\, dt\) | A1 AG | Correct proof. Appreciation of how the negative sign affects the limits. Note that the answer is given in the question. |
| \(A = \int 64\sin^2 t \cos t\, dt\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Using substitution \(u = \sin t \Rightarrow \frac{du}{dt} = \cos t\) | ||
| Change limits: when \(t = \frac{\pi}{3}\), \(u = \frac{\sqrt{3}}{2}\) and when \(t = \frac{\pi}{2}\), \(u = 1\) | ||
| \(A = 64\left[\frac{\sin^3 t}{3}\right]_{\frac{\pi}{3}}^{\frac{\pi}{2}}\) or \(A = 64\left[\frac{u^3}{3}\right]_{\frac{\sqrt{3}}{2}}^{1}\) | M1 | \(k\sin^3 t\) or \(ku^3\) with \(u = \sin t\) |
| Correct integration ignoring limits | A1 | |
| \(A = 64\left[\frac{1}{3} - \left(\frac{1}{3}\cdot\frac{\sqrt{3}}{2}\cdot\frac{\sqrt{3}}{2}\cdot\frac{\sqrt{3}}{2}\right)\right]\) | dM1 | Substitutes limits of either \(\left(t = \frac{\pi}{2}\text{ and }t = \frac{\pi}{3}\right)\) or \(\left(u=1\text{ and }u=\frac{\sqrt{3}}{2}\right)\) and subtracts the correct way round |
| \(A = 64\left(\frac{1}{3} - \frac{1}{8}\sqrt{3}\right) = \dfrac{64}{3} - 8\sqrt{3}\) | A1 aef isw | \(\dfrac{64}{3} - 8\sqrt{3}\); Aef in the form \(a + b\sqrt{3}\), with awrt 21.3 and anything that cancels to \(a = \frac{64}{3}\) and \(b = -8\) |
| (Note that \(a = \frac{64}{3}\), \(b = -8\)) |
## Question 8:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| At $P(4, 2\sqrt{3})$ either $4 = 8\cos t$ or $2\sqrt{3} = 4\sin 2t$ | M1 | $4=8\cos t$ or $2\sqrt{3}=4\sin 2t$ |
| $\Rightarrow$ only solution is $t = \frac{\pi}{3}$ where $0 \leq t \leq \frac{\pi}{2}$ | A1 | $t=\frac{\pi}{3}$ or awrt 1.05 (radians) only, stated in the range $0 \leq t \leq \frac{\pi}{2}$ |
**[2 marks]**
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x = 8\cos t$, $y = 4\sin 2t$; $\frac{dx}{dt} = -8\sin t$, $\frac{dy}{dt} = 8\cos 2t$ | M1 | Attempt to differentiate both $x$ and $y$ wrt $t$ to give $\pm p\sin t$ and $\pm q\cos 2t$ respectively |
| | A1 | Correct $\frac{dx}{dt}$ and $\frac{dy}{dt}$ |
| At $P$: $\frac{dy}{dx} = \frac{8\cos\!\left(\frac{2\pi}{3}\right)}{-8\sin\!\left(\frac{\pi}{3}\right)}$ | M1* | Divides in correct way and substitutes their value of $t$ into their $\frac{dy}{dx}$ expression |
| $= \frac{8(-\frac{1}{2})}{(-8)(\frac{\sqrt{3}}{2})} = \frac{1}{\sqrt{3}}$ awrt 0.58 | | You may need to check candidate's substitutions for M1* |
| Hence $m(\mathbf{N}) = -\sqrt{3}$ or $\frac{-1}{\frac{1}{\sqrt{3}}}$ | dM1* | Uses $m(\mathbf{N}) = -\frac{1}{\text{their } m(\mathbf{T})}$ |
| **N**: $y - 2\sqrt{3} = -\sqrt{3}(x-4)$ | dM1* | Uses $y - 2\sqrt{3} = (\text{their } m_N)(x-4)$ or finds $c$ using $x=4$ and $y=2\sqrt{3}$ and uses $y=(\text{their } m_N)x + \text{"c"}$ |
| **N**: $y = -\sqrt{3}x + 6\sqrt{3}$ **AG** | A1 cso AG | $y = -\sqrt{3}x + 6\sqrt{3}$ |
**[6 marks]**
## Question 8:
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $A = \int_0^4 y\, dx = \int 4\sin 2t \cdot (-8\sin t)\, dt$ | M1 | attempt at $A = \int y \frac{dx}{dt}\, dt$ |
| Correct expression | A1 | ignore limits and $dt$ |
| $A = \int -32\sin 2t \cdot \sin t\, dt = \int -32(2\sin t \cos t)\cdot \sin t\, dt$ | M1 | Seeing $\sin 2t = 2\sin t \cos t$ anywhere in PART (c) |
| $A = \int -64\sin^2 t \cos t\, dt$ | A1 **AG** | Correct proof. Appreciation of how the negative sign affects the limits. **Note that the answer is given in the question.** |
| $A = \int 64\sin^2 t \cos t\, dt$ | | |
**[4]**
### Part (d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Using substitution $u = \sin t \Rightarrow \frac{du}{dt} = \cos t$ | | |
| Change limits: when $t = \frac{\pi}{3}$, $u = \frac{\sqrt{3}}{2}$ and when $t = \frac{\pi}{2}$, $u = 1$ | | |
| $A = 64\left[\frac{\sin^3 t}{3}\right]_{\frac{\pi}{3}}^{\frac{\pi}{2}}$ or $A = 64\left[\frac{u^3}{3}\right]_{\frac{\sqrt{3}}{2}}^{1}$ | M1 | $k\sin^3 t$ or $ku^3$ with $u = \sin t$ |
| Correct integration ignoring limits | A1 | |
| $A = 64\left[\frac{1}{3} - \left(\frac{1}{3}\cdot\frac{\sqrt{3}}{2}\cdot\frac{\sqrt{3}}{2}\cdot\frac{\sqrt{3}}{2}\right)\right]$ | dM1 | Substitutes limits of either $\left(t = \frac{\pi}{2}\text{ and }t = \frac{\pi}{3}\right)$ or $\left(u=1\text{ and }u=\frac{\sqrt{3}}{2}\right)$ and subtracts the correct way round |
| $A = 64\left(\frac{1}{3} - \frac{1}{8}\sqrt{3}\right) = \dfrac{64}{3} - 8\sqrt{3}$ | A1 aef isw | $\dfrac{64}{3} - 8\sqrt{3}$; Aef in the form $a + b\sqrt{3}$, with awrt 21.3 and anything that cancels to $a = \frac{64}{3}$ and $b = -8$ |
| (Note that $a = \frac{64}{3}$, $b = -8$) | | |
**[4]**
**Total: 16 marks**8.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{fb1924cc-9fa3-4fde-ba4d-6fb095f7f70b-11_639_972_228_484}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
Figure 3 shows the curve $C$ with parametric equations
$$x = 8 \cos t , \quad y = 4 \sin 2 t , \quad 0 \leqslant t \leqslant \frac { \pi } { 2 } .$$
The point $P$ lies on $C$ and has coordinates $( 4,2 \sqrt { } 3 )$.
\begin{enumerate}[label=(\alph*)]
\item Find the value of $t$ at the point $P$.
The line $l$ is a normal to $C$ at $P$.
\item Show that an equation for $l$ is $y = - x \sqrt { 3 } + 6 \sqrt { 3 }$.
The finite region $R$ is enclosed by the curve $C$, the $x$-axis and the line $x = 4$, as shown shaded in Figure 3.
\item Show that the area of $R$ is given by the integral $\int _ { \frac { \pi } { 3 } } ^ { \frac { \pi } { 2 } } 64 \sin ^ { 2 } t \cos t \mathrm {~d} t$.
\item Use this integral to find the area of $R$, giving your answer in the form $a + b \sqrt { } 3$, where $a$ and $b$ are constants to be determined.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 2008 Q8 [16]}}