| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2007 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric differentiation |
| Type | Find normal equation at parameter |
| Difficulty | Standard +0.3 This is a straightforward parametric differentiation question requiring standard techniques: find dx/dt and dy/dt, compute dy/dx = (dy/dt)/(dx/dt), then find the normal gradient and equation at a specific parameter value. The trigonometric expressions are slightly more complex than basic examples (involving 7t terms), but the method is entirely routine for C4 students with no novel problem-solving required. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.07m Tangents and normals: gradient and equations1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dx}{dt} = -7\sin t + 7\sin 7t,\quad \frac{dy}{dt} = 7\cos t - 7\cos 7t\) | M1 | Attempt to differentiate \(x\) and \(y\) w.r.t. \(t\); \(\frac{dx}{dt}\) in form \(\pm A\sin t \pm B\sin 7t\); \(\frac{dy}{dt}\) in form \(\pm C\cos t \pm D\cos 7t\) |
| Correct \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\) | A1 | |
| \(\therefore \dfrac{dy}{dx} = \dfrac{7\cos t - 7\cos 7t}{-7\sin t + 7\sin 7t}\) | B1\(\checkmark\) | Candidate's \(\dfrac{\frac{dy}{dt}}{\frac{dx}{dt}}\) |
| Total: 3 marks | [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| When \(t=\frac{\pi}{6}\): \(m(T) = \dfrac{7\cos\frac{\pi}{6}-7\cos\frac{7\pi}{6}}{-7\sin\frac{\pi}{6}+7\sin\frac{7\pi}{6}}\) | M1 | Substitutes \(t=\frac{\pi}{6}\) or \(30°\) into their \(\frac{dy}{dx}\) expression |
| \(= \dfrac{\frac{7\sqrt{3}}{2}-\left(-\frac{7\sqrt{3}}{2}\right)}{-\frac{7}{2}-\frac{7}{2}} = \dfrac{7\sqrt{3}}{-7} = -\sqrt{3}\) (awrt \(-1.73\)) | A1 cso | To give any of the four underlined expressions; (must be correct solution only) |
| Hence \(m(N) = \dfrac{-1}{-\sqrt{3}} = \dfrac{1}{\sqrt{3}}\) (awrt \(0.58\)) | A1\(\checkmark\) oe | Uses \(m(T)\) to 'correctly' find \(m(N)\). Can be ft from "their tangent gradient". |
| When \(t=\frac{\pi}{6}\): \(x = 7\cos\frac{\pi}{6}-\cos\frac{7\pi}{6} = \frac{7\sqrt{3}}{2}-\left(-\frac{\sqrt{3}}{2}\right) = \frac{8\sqrt{3}}{2} = 4\sqrt{3}\) | B1 | The point \(\left(4\sqrt{3},\,4\right)\) or (awrt 6.9, 4) |
| \(y = 7\sin\frac{\pi}{6}-\sin\frac{7\pi}{6} = \frac{7}{2}-\left(-\frac{1}{2}\right) = \frac{8}{2} = 4\) | ||
| \(\mathbf{N}:\ y-4 = \dfrac{1}{\sqrt{3}}\left(x-4\sqrt{3}\right)\) | M1 | Finding equation of normal with their point and their normal gradient or finds \(c\) using \(y=(\text{their gradient})x + \text{"c"}\) |
| \(\mathbf{N}:\ y = \dfrac{1}{\sqrt{3}}x\quad\) or \(\quad y=\dfrac{\sqrt{3}}{3}x\quad\) or \(\quad 3y=\sqrt{3}x\) | A1 oe | Correct simplified EXACT equation of normal. Dependent on candidate using correct \(\left(4\sqrt{3},\,4\right)\) |
| Total: 6 marks | [6] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x = 7\cos t - \cos 7t\), \(y = 7\sin t - \sin 7t\) | ||
| \(\frac{dx}{dt} = -7\sin t + 7\sin 7t\), \(\frac{dy}{dt} = 7\cos t - 7\cos 7t\) | M1 | Attempt to differentiate x and y w.r.t. \(t\) to give \(\frac{dx}{dt}\) in form \(\pm A\sin t \pm B\sin 7t\); \(\frac{dy}{dt}\) in form \(\pm C\cos t \pm D\cos 7t\) |
| Correct \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\) | A1 | |
| \(\frac{dy}{dx} = \frac{7\cos t - 7\cos 7t}{-7\sin t + 7\sin 7t} = \frac{-7(-2\sin 4t\sin 3t)}{-7(2\cos 4t\sin 3t)} = \tan 4t\) | B1 \(\checkmark\) | Candidate's \(\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| When \(t = \frac{\pi}{6}\), \(m(\mathbf{T}) = \frac{dy}{dx} = \tan\frac{4\pi}{6}\) | M1 | Substitutes \(t = \frac{\pi}{6}\) or \(30°\) into their \(\frac{dy}{dx}\) expression |
| \(= \frac{2\left(\frac{\sqrt{3}}{2}\right)(1)}{2\left(-\frac{1}{2}\right)(1)} = \frac{-\sqrt{3}}{1} = -\sqrt{3}\) (awrt \(-1.73\)) | A1 cso | To give any of the three underlined expressions (must be correct solution only) |
| \(m(\mathbf{N}) = \frac{-1}{-\sqrt{3}}\) or \(\frac{1}{\sqrt{3}}\) = awrt \(0.58\) | A1\(\checkmark\) oe | Uses \(m(\mathbf{T})\) to 'correctly' find \(m(\mathbf{N})\). Can be ft from "their tangent gradient" |
| \(x = 7\cos\frac{\pi}{6} - \cos\frac{7\pi}{6} = \frac{7\sqrt{3}}{2} - \left(-\frac{\sqrt{3}}{2}\right) = \frac{8\sqrt{3}}{2} = 4\sqrt{3}\) | ||
| \(y = 7\sin\frac{\pi}{6} - \sin\frac{7\pi}{6} = \frac{7}{2} - \left(-\frac{1}{2}\right) = \frac{8}{2} = 4\) | B1 | The point \((4\sqrt{3}, 4)\) or (awrt \(6.9, 4\)) |
| \(\mathbf{N}\): \(y - 4 = \frac{1}{\sqrt{3}}\left(x - 4\sqrt{3}\right)\) | M1 | Finding equation of normal with their point and gradient, or finds \(c\) using \(y = (\text{their gradient})x + \text{"c"}\) |
| \(\mathbf{N}\): \(y = \frac{1}{\sqrt{3}}x\) or \(y = \frac{\sqrt{3}}{3}x\) or \(3y = \sqrt{3}x\) | A1 oe | Correct simplified EXACT equation of normal. Dependent on candidate using correct \((4\sqrt{3}, 4)\) |
# Question 3(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dx}{dt} = -7\sin t + 7\sin 7t,\quad \frac{dy}{dt} = 7\cos t - 7\cos 7t$ | M1 | Attempt to differentiate $x$ and $y$ w.r.t. $t$; $\frac{dx}{dt}$ in form $\pm A\sin t \pm B\sin 7t$; $\frac{dy}{dt}$ in form $\pm C\cos t \pm D\cos 7t$ |
| Correct $\frac{dx}{dt}$ and $\frac{dy}{dt}$ | A1 | |
| $\therefore \dfrac{dy}{dx} = \dfrac{7\cos t - 7\cos 7t}{-7\sin t + 7\sin 7t}$ | B1$\checkmark$ | Candidate's $\dfrac{\frac{dy}{dt}}{\frac{dx}{dt}}$ |
| **Total: 3 marks** | **[3]** | |
---
# Question 3(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| When $t=\frac{\pi}{6}$: $m(T) = \dfrac{7\cos\frac{\pi}{6}-7\cos\frac{7\pi}{6}}{-7\sin\frac{\pi}{6}+7\sin\frac{7\pi}{6}}$ | M1 | Substitutes $t=\frac{\pi}{6}$ or $30°$ into their $\frac{dy}{dx}$ expression |
| $= \dfrac{\frac{7\sqrt{3}}{2}-\left(-\frac{7\sqrt{3}}{2}\right)}{-\frac{7}{2}-\frac{7}{2}} = \dfrac{7\sqrt{3}}{-7} = -\sqrt{3}$ (awrt $-1.73$) | A1 cso | To give any of the four underlined expressions; **(must be correct solution only)** |
| Hence $m(N) = \dfrac{-1}{-\sqrt{3}} = \dfrac{1}{\sqrt{3}}$ (awrt $0.58$) | A1$\checkmark$ oe | Uses $m(T)$ to 'correctly' find $m(N)$. Can be ft from "their tangent gradient". |
| When $t=\frac{\pi}{6}$: $x = 7\cos\frac{\pi}{6}-\cos\frac{7\pi}{6} = \frac{7\sqrt{3}}{2}-\left(-\frac{\sqrt{3}}{2}\right) = \frac{8\sqrt{3}}{2} = 4\sqrt{3}$ | B1 | The point $\left(4\sqrt{3},\,4\right)$ or (awrt 6.9, 4) |
| $y = 7\sin\frac{\pi}{6}-\sin\frac{7\pi}{6} = \frac{7}{2}-\left(-\frac{1}{2}\right) = \frac{8}{2} = 4$ | | |
| $\mathbf{N}:\ y-4 = \dfrac{1}{\sqrt{3}}\left(x-4\sqrt{3}\right)$ | M1 | Finding equation of normal with their point and their normal gradient or finds $c$ using $y=(\text{their gradient})x + \text{"c"}$ |
| $\mathbf{N}:\ y = \dfrac{1}{\sqrt{3}}x\quad$ or $\quad y=\dfrac{\sqrt{3}}{3}x\quad$ or $\quad 3y=\sqrt{3}x$ | A1 oe | Correct simplified EXACT equation of normal. Dependent on candidate using correct $\left(4\sqrt{3},\,4\right)$ |
| **Total: 6 marks** | **[6]** | |
## Question 3(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x = 7\cos t - \cos 7t$, $y = 7\sin t - \sin 7t$ | | |
| $\frac{dx}{dt} = -7\sin t + 7\sin 7t$, $\frac{dy}{dt} = 7\cos t - 7\cos 7t$ | M1 | Attempt to differentiate x **and** y w.r.t. $t$ to give $\frac{dx}{dt}$ in form $\pm A\sin t \pm B\sin 7t$; $\frac{dy}{dt}$ in form $\pm C\cos t \pm D\cos 7t$ |
| Correct $\frac{dx}{dt}$ and $\frac{dy}{dt}$ | A1 | |
| $\frac{dy}{dx} = \frac{7\cos t - 7\cos 7t}{-7\sin t + 7\sin 7t} = \frac{-7(-2\sin 4t\sin 3t)}{-7(2\cos 4t\sin 3t)} = \tan 4t$ | B1 $\checkmark$ | Candidate's $\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$ |
**[3 marks]**
---
## Question 3(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| When $t = \frac{\pi}{6}$, $m(\mathbf{T}) = \frac{dy}{dx} = \tan\frac{4\pi}{6}$ | M1 | Substitutes $t = \frac{\pi}{6}$ or $30°$ into their $\frac{dy}{dx}$ expression |
| $= \frac{2\left(\frac{\sqrt{3}}{2}\right)(1)}{2\left(-\frac{1}{2}\right)(1)} = \frac{-\sqrt{3}}{1} = -\sqrt{3}$ (awrt $-1.73$) | A1 **cso** | To give any of the three underlined expressions **(must be correct solution only)** |
| $m(\mathbf{N}) = \frac{-1}{-\sqrt{3}}$ or $\frac{1}{\sqrt{3}}$ = awrt $0.58$ | A1$\checkmark$ oe | Uses $m(\mathbf{T})$ to 'correctly' find $m(\mathbf{N})$. Can be ft from "their tangent gradient" |
| $x = 7\cos\frac{\pi}{6} - \cos\frac{7\pi}{6} = \frac{7\sqrt{3}}{2} - \left(-\frac{\sqrt{3}}{2}\right) = \frac{8\sqrt{3}}{2} = 4\sqrt{3}$ | | |
| $y = 7\sin\frac{\pi}{6} - \sin\frac{7\pi}{6} = \frac{7}{2} - \left(-\frac{1}{2}\right) = \frac{8}{2} = 4$ | B1 | The point $(4\sqrt{3}, 4)$ or (awrt $6.9, 4$) |
| $\mathbf{N}$: $y - 4 = \frac{1}{\sqrt{3}}\left(x - 4\sqrt{3}\right)$ | M1 | Finding equation of normal with their point and gradient, or finds $c$ using $y = (\text{their gradient})x + \text{"c"}$ |
| $\mathbf{N}$: $y = \frac{1}{\sqrt{3}}x$ or $y = \frac{\sqrt{3}}{3}x$ or $3y = \sqrt{3}x$ | A1 oe | Correct simplified EXACT equation of normal. Dependent on candidate using correct $(4\sqrt{3}, 4)$ |
**[6 marks] — Total: 9 marks**
---\begin{enumerate}
\item A curve has parametric equations
\end{enumerate}
$$x = 7 \cos t - \cos 7 t , y = 7 \sin t - \sin 7 t , \quad \frac { \pi } { 8 } < t < \frac { \pi } { 3 }$$
(a) Find an expression for $\frac { \mathrm { d } y } { \mathrm {~d} x }$ in terms of $t$. You need not simplify your answer.\\
(b) Find an equation of the normal to the curve at the point where $t = \frac { \pi } { 6 }$.
Give your answer in its simplest exact form.\\
\hfill \mbox{\textit{Edexcel C4 2007 Q3 [9]}}