C4 January 2007 Q3
3. A curve has parametric equations
\[x = 7\cos t - \cos 7t, \quad y = 7\sin t - \sin 7t, \qquad \frac{\pi}{8} \lt t \lt \frac{\pi}{3}.\]
Give your answer in its simplest exact form. (6)
| Scheme | Marks |
|---|---|
| \(x = 7\cos t - \cos 7t,\quad y = 7\sin t - \sin 7t,\) | |
| \(\dfrac{\mathrm{d}x}{\mathrm{d}t} = -7\sin t + 7\sin 7t,\quad \dfrac{\mathrm{d}y}{\mathrm{d}t} = 7\cos t - 7\cos 7t\) | M1 A1 |
| \(\therefore \dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{7\cos t - 7\cos 7t}{-7\sin t + 7\sin 7t}\) | B1ft |
| (3) |
Notes
M1 Attempt to differentiate \(x\) and \(y\) with respect to \(t\) to give \(\frac{\mathrm{d}x}{\mathrm{d}t}\) in the form \(\pm A\sin t \pm B\sin 7t\), \(\frac{\mathrm{d}y}{\mathrm{d}t}\) in the form \(\pm C\cos t \pm D\cos 7t\)
A1 Correct \(\frac{\mathrm{d}x}{\mathrm{d}t}\) and \(\frac{\mathrm{d}y}{\mathrm{d}t}\)
B1ft Candidate’s \(\dfrac{\frac{\mathrm{d}y}{\mathrm{d}t}}{\frac{\mathrm{d}x}{\mathrm{d}t}}\)
Aliter (a) Way 2
| \(x = 7\cos t - \cos 7t,\quad y = 7\sin t - \sin 7t,\) | |
| \(\dfrac{\mathrm{d}x}{\mathrm{d}t} = -7\sin t + 7\sin 7t,\quad \dfrac{\mathrm{d}y}{\mathrm{d}t} = 7\cos t - 7\cos 7t\) | M1 A1 |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{7\cos t - 7\cos 7t}{-7\sin t + 7\sin 7t} = \dfrac{-7(-2\sin 4t\sin 3t)}{7(2\cos 4t\sin 3t)} = \tan 4t\) | B1ft |
| [3] |
M1 Attempt to differentiate \(x\) and \(y\) with respect to \(t\) to give \(\frac{\mathrm{d}x}{\mathrm{d}t}\) in the form \(\pm A\sin t \pm B\sin 7t\), \(\frac{\mathrm{d}y}{\mathrm{d}t}\) in the form \(\pm C\cos t \pm D\cos 7t\)
A1 Correct \(\frac{\mathrm{d}x}{\mathrm{d}t}\) and \(\frac{\mathrm{d}y}{\mathrm{d}t}\)
B1ft Candidate’s \(\dfrac{\frac{\mathrm{d}y}{\mathrm{d}t}}{\frac{\mathrm{d}x}{\mathrm{d}t}}\)
CHECK (corrected from the printed mark scheme: in Way 2 the denominator is printed as \(-7(2\cos 4t\sin 3t)\); since \(-7\sin t + 7\sin 7t = 7(2\cos 4t\sin 3t)\), it is \(7(2\cos 4t\sin 3t)\), which gives \(\tan 4t\).)
| Scheme | Marks |
|---|---|
| When \(t = \dfrac{\pi}{6}\), \(m(\mathbf{T}) = \dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{7\cos\frac{\pi}{6} - 7\cos\frac{7\pi}{6}}{-7\sin\frac{\pi}{6} + 7\sin\frac{7\pi}{6}}\); | M1 |
| \(= \underline{\dfrac{\frac{7\sqrt{3}}{2} - \left(-\frac{7\sqrt{3}}{2}\right)}{-\frac{7}{2} - \frac{7}{2}}} = \underline{\dfrac{7\sqrt{3}}{-7}} = \underline{-\sqrt{3}} = \underline{\text{awrt } -1.73}\) | A1 cso |
| Hence \(m(\mathbf{N}) = \dfrac{-1}{-\sqrt{3}}\) or \(\dfrac{1}{\sqrt{3}} = \) awrt 0.58 | A1ft oe. |
| When \(t = \frac{\pi}{6}\), \(x = 7\cos\frac{\pi}{6} - \cos\frac{7\pi}{6} = \frac{7\sqrt{3}}{2} - \left(-\frac{\sqrt{3}}{2}\right) = \frac{8\sqrt{3}}{2} = 4\sqrt{3}\) \(y = 7\sin\frac{\pi}{6} - \sin\frac{7\pi}{6} = \frac{7}{2} - \left(-\frac{1}{2}\right) = \frac{8}{2} = 4\) | B1 |
| \(\mathbf{N}\): \(y - 4 = \frac{1}{\sqrt{3}}\left(x - 4\sqrt{3}\right)\) | M1 |
| \(\mathbf{N}\): \(\underline{y = \frac{1}{\sqrt{3}}x}\) or \(\underline{y = \frac{\sqrt{3}}{3}x}\) or \(\underline{3y = \sqrt{3}x}\) | A1 oe |
| or \(4 = \frac{1}{\sqrt{3}}\left(4\sqrt{3}\right) + c \Rightarrow c = 4 - 4 = 0\) Hence \(\mathbf{N}\): \(\underline{y = \frac{1}{\sqrt{3}}x}\) or \(\underline{y = \frac{\sqrt{3}}{3}x}\) or \(\underline{3y = \sqrt{3}x}\) | |
| (6) | |
| (9 marks) |
Notes
M1 Substitutes \(t = \frac{\pi}{6}\) or \(30^\circ\) into their \(\frac{\mathrm{d}y}{\mathrm{d}x}\) expression;
A1 cso to give any of the four underlined expressions oe (must be correct solution only)
A1ft oe. Uses \(m(\mathbf{T})\) to ‘correctly’ find \(m(\mathbf{N})\). Can be ft from “their tangent gradient”.
B1 The point \(\underline{\left(4\sqrt{3}, 4\right)}\) or \(\underline{(\text{awrt } 6.9, 4)}\)
M1 Finding an equation of a normal with their point and their normal gradient or finds \(c\) by using \(y = (\text{their gradient})x + \text{“}c\text{”}\).
A1 oe Correct simplified EXACT equation of normal. This is dependent on candidate using correct \(\left(4\sqrt{3}, 4\right)\)
Aliter (b) Way 2
| When \(t = \dfrac{\pi}{6}\), \(m(\mathbf{T}) = \dfrac{\mathrm{d}y}{\mathrm{d}x} = \tan\frac{4\pi}{6}\); | M1 |
| \(= \underline{\dfrac{2\left(\frac{\sqrt{3}}{2}\right)(1)}{2\left(-\frac{1}{2}\right)(1)}} = \underline{-\sqrt{3}} = \underline{\text{awrt } -1.73}\) | A1 cso |
| Hence \(m(\mathbf{N}) = \dfrac{-1}{-\sqrt{3}}\) or \(\dfrac{1}{\sqrt{3}} = \) awrt 0.58 | A1ft oe. |
| When \(t = \frac{\pi}{6}\), \(x = 7\cos\frac{\pi}{6} - \cos\frac{7\pi}{6} = \frac{7\sqrt{3}}{2} - \left(-\frac{\sqrt{3}}{2}\right) = \frac{8\sqrt{3}}{2} = 4\sqrt{3}\) \(y = 7\sin\frac{\pi}{6} - \sin\frac{7\pi}{6} = \frac{7}{2} - \left(-\frac{1}{2}\right) = \frac{8}{2} = 4\) | B1 |
| \(\mathbf{N}\): \(y - 4 = \frac{1}{\sqrt{3}}\left(x - 4\sqrt{3}\right)\) | M1 |
| \(\mathbf{N}\): \(\underline{y = \frac{1}{\sqrt{3}}x}\) or \(\underline{y = \frac{\sqrt{3}}{3}x}\) or \(\underline{3y = \sqrt{3}x}\) | A1 oe |
| or \(4 = \frac{1}{\sqrt{3}}\left(4\sqrt{3}\right) + c \Rightarrow c = 4 - 4 = 0\) Hence \(\mathbf{N}\): \(\underline{y = \frac{1}{\sqrt{3}}x}\) or \(\underline{y = \frac{\sqrt{3}}{3}x}\) or \(\underline{3y = \sqrt{3}x}\) | |
| [6] |
M1 Substitutes \(t = \frac{\pi}{6}\) or \(30^\circ\) into their \(\frac{\mathrm{d}y}{\mathrm{d}x}\) expression;
A1 cso to give any of the three underlined expressions oe (must be correct solution only)
A1ft oe. Uses \(m(\mathbf{T})\) to ‘correctly’ find \(m(\mathbf{N})\). Can be ft from “their tangent gradient”.
B1 The point \(\underline{\left(4\sqrt{3}, 4\right)}\) or \(\underline{(\text{awrt } 6.9, 4)}\)
M1 Finding an equation of a normal with their point and their normal gradient or finds \(c\) by using \(y = (\text{their gradient})x + \text{“}c\text{”}\).
A1 oe Correct simplified EXACT equation of normal. This is dependent on candidate using correct \(\left(4\sqrt{3}, 4\right)\)
Beware: A candidate finding an \(m(\mathbf{T}) = 0\) can obtain A1ft for \(m(\mathbf{N}) \to \infty\), but obtains M0 if they write \(y - 4 = \infty(x - 4\sqrt{3})\). If they write, however, \(\mathbf{N}\): \(x = 4\sqrt{3}\), then they can score M1.
Beware: A candidate finding an \(m(\mathbf{T}) = \infty\) can obtain A1ft for \(m(\mathbf{N}) = 0\), and also obtains M1 if they write \(y - 4 = 0(x - 4\sqrt{3})\) or \(y = 4\).