C4 June 2006 Q4
4.
The curve shown in Figure 2 has parametric equations
\[x = \sin t, \quad y = \sin\left(t + \tfrac{\pi}{6}\right), \qquad -\tfrac{\pi}{2} \lt t \lt \tfrac{\pi}{2}.\]
\[y = \frac{\surd 3}{2}x + \frac{1}{2}\surd(1 - x^2), \qquad -1 \lt x \lt 1.\]
(3)| Scheme | Marks |
|---|---|
| \(x = \sin t,\quad y = \sin\left(t + \frac{\pi}{6}\right)\) | |
| \(\dfrac{\mathrm{d}x}{\mathrm{d}t} = \cos t,\quad \dfrac{\mathrm{d}y}{\mathrm{d}t} = \cos\left(t + \frac{\pi}{6}\right)\) | M1 A1 |
| When \(t = \dfrac{\pi}{6}\), \(\underline{\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{\cos\left(\frac{\pi}{6} + \frac{\pi}{6}\right)}{\cos\left(\frac{\pi}{6}\right)} = \dfrac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} = \dfrac{1}{\sqrt{3}} = \text{awrt } 0.58}\) | A1 |
| When \(t = \dfrac{\pi}{6}\), \(x = \dfrac{1}{2},\ y = \dfrac{\sqrt{3}}{2}\) | B1 |
| \(\mathbf{T}\): \(\underline{y - \frac{\sqrt{3}}{2} = \frac{1}{\sqrt{3}}\left(x - \frac{1}{2}\right)}\) | dM1 A1 oe |
| or \(\frac{\sqrt{3}}{2} = \frac{1}{\sqrt{3}}\left(\frac{1}{2}\right) + c \Rightarrow c = \frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{6} = \frac{\sqrt{3}}{3}\) or \(\mathbf{T}\): \(\left[\underline{y = \frac{\sqrt{3}}{3}x + \frac{\sqrt{3}}{3}}\right]\) | |
| (6) |
Notes
M1 Attempt to differentiate both \(x\) and \(y\) wrt \(t\) to give two terms in cos
A1 Correct \(\frac{\mathrm{d}x}{\mathrm{d}t}\) and \(\frac{\mathrm{d}y}{\mathrm{d}t}\)
A1 Divides in correct way and substitutes for \(t\) to give any of the four underlined oe: Ignore the double negative if candidate has differentiated \(\sin \to -\cos\)
B1 The point \(\underline{\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)}\) or \(\underline{\left(\frac{1}{2}, \text{awrt } 0.87\right)}\)
dM1 Finding an equation of a tangent with their point and their tangent gradient or finds \(c\) and uses \(y = (\text{their gradient})x + \text{“}c\text{”}\).
A1 oe Correct EXACT equation of tangent oe.
Aliter (a) Way 2
| \(x = \sin t,\quad y = \sin\left(t + \frac{\pi}{6}\right) = \sin t\cos\frac{\pi}{6} + \cos t\sin\frac{\pi}{6}\) (Do not give this for part (b)) | |
| \(\dfrac{\mathrm{d}x}{\mathrm{d}t} = \cos t,\quad \dfrac{\mathrm{d}y}{\mathrm{d}t} = \cos t\cos\frac{\pi}{6} - \sin t\sin\frac{\pi}{6}\) | M1 A1 |
| When \(t = \dfrac{\pi}{6}\), \(\underline{\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{\cos\frac{\pi}{6}\cos\frac{\pi}{6} - \sin\frac{\pi}{6}\sin\frac{\pi}{6}}{\cos\left(\frac{\pi}{6}\right)}}\) \(\underline{= \dfrac{\frac{3}{4} - \frac{1}{4}}{\frac{\sqrt{3}}{2}} = \dfrac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} = \dfrac{1}{\sqrt{3}} = \text{awrt } 0.58}\) | A1 |
| When \(t = \dfrac{\pi}{6}\), \(x = \dfrac{1}{2},\ y = \dfrac{\sqrt{3}}{2}\) | B1 |
| \(\mathbf{T}\): \(\underline{y - \frac{\sqrt{3}}{2} = \frac{1}{\sqrt{3}}\left(x - \frac{1}{2}\right)}\) | dM1 A1 oe |
| or \(\frac{\sqrt{3}}{2} = \frac{1}{\sqrt{3}}\left(\frac{1}{2}\right) + c \Rightarrow c = \frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{6} = \frac{\sqrt{3}}{3}\) or \(\mathbf{T}\): \(\left[\underline{y = \frac{\sqrt{3}}{3}x + \frac{\sqrt{3}}{3}}\right]\) | |
| [6] |
M1 Attempt to differentiate \(x\) and \(y\) wrt \(t\) to give \(\frac{\mathrm{d}x}{\mathrm{d}t}\) in terms of cos and \(\frac{\mathrm{d}y}{\mathrm{d}t}\) in the form \(\pm a\cos t \pm b\sin t\)
A1 Correct \(\frac{\mathrm{d}x}{\mathrm{d}t}\) and \(\frac{\mathrm{d}y}{\mathrm{d}t}\)
A1 Divides in correct way and substitutes for \(t\) to give any of the four underlined oe:
B1 The point \(\underline{\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)}\) or \(\underline{\left(\frac{1}{2}, \text{awrt } 0.87\right)}\)
dM1 Finding an equation of a tangent with their point and their tangent gradient or finds \(c\) and uses \(y = (\text{their gradient})x + \text{“}c\text{”}\).
A1 oe Correct EXACT equation of tangent oe.
Aliter (a) Way 3
| \(y = \frac{\sqrt{3}}{2}x + \frac{1}{2}\sqrt{(1 - x^2)}\) | |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{\sqrt{3}}{2} + \left(\dfrac{1}{2}\right)\left(\dfrac{1}{2}\right)(1 - x^2)^{-\frac{1}{2}}(-2x)\) | M1 A1 |
| \(\underline{\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{\sqrt{3}}{2} + \left(\dfrac{1}{2}\right)\left(\dfrac{1}{2}\right)(1 - (0.5)^2)^{-\frac{1}{2}}(-2(0.5)) = \dfrac{1}{\sqrt{3}}}\) | A1 |
| When \(t = \dfrac{\pi}{6}\), \(x = \dfrac{1}{2},\ y = \dfrac{\sqrt{3}}{2}\) | B1 |
| \(\mathbf{T}\): \(\underline{y - \frac{\sqrt{3}}{2} = \frac{1}{\sqrt{3}}\left(x - \frac{1}{2}\right)}\) | dM1 A1 oe |
| or \(\frac{\sqrt{3}}{2} = \frac{1}{\sqrt{3}}\left(\frac{1}{2}\right) + c \Rightarrow c = \frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{6} = \frac{\sqrt{3}}{3}\) or \(\mathbf{T}\): \(\left[\underline{y = \frac{\sqrt{3}}{3}x + \frac{\sqrt{3}}{3}}\right]\) | |
| [6] |
M1 Attempt to differentiate two terms using the chain rule for the second term.
A1 Correct \(\frac{\mathrm{d}y}{\mathrm{d}x}\)
A1 Correct substitution of \(x = \frac{1}{2}\) into a correct \(\frac{\mathrm{d}y}{\mathrm{d}x}\)
B1 The point \(\underline{\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)}\) or \(\underline{\left(\frac{1}{2}, \text{awrt } 0.87\right)}\)
dM1 Finding an equation of a tangent with their point and their tangent gradient or finds \(c\) and uses \(y = (\text{their gradient})x + \text{“}c\text{”}\).
A1 oe Correct EXACT equation of tangent oe.
| Scheme | Marks |
|---|---|
| \(y = \sin\left(t + \frac{\pi}{6}\right) = \sin t\cos\frac{\pi}{6} + \cos t\sin\frac{\pi}{6}\) | M1 |
| Nb: \(\sin^2 t + \cos^2 t \equiv 1 \Rightarrow \cos^2 t \equiv 1 - \sin^2 t\) | |
| \(\therefore x = \sin t\) gives \(\cos t = \sqrt{(1 - x^2)}\) | M1 |
| \(\therefore y = \frac{\sqrt{3}}{2}\sin t + \frac{1}{2}\cos t\) | |
| gives \(y = \frac{\sqrt{3}}{2}x + \frac{1}{2}\sqrt{(1 - x^2)}\) AG | A1 cso |
| (3) | |
| (9 marks) |
Notes
M1 Use of compound angle formula for sine.
M1 Use of trig identity to find \(\cos t\) in terms of \(x\) or \(\cos^2 t\) in terms of \(x\).
A1 cso Substitutes for \(\sin t\), \(\cos\frac{\pi}{6}\), \(\cos t\) and \(\sin\frac{\pi}{6}\) to give \(y\) in terms of \(x\).
Aliter (b) Way 2
| \(x = \sin t\) gives \(y = \frac{\sqrt{3}}{2}\sin t + \frac{1}{2}\sqrt{(1 - \sin^2 t)}\) | M1 |
| Nb: \(\sin^2 t + \cos^2 t \equiv 1 \Rightarrow \cos^2 t \equiv 1 - \sin^2 t\) | |
| \(\cos t = \sqrt{(1 - \sin^2 t)}\) | M1 |
| gives \(y = \frac{\sqrt{3}}{2}\sin t + \frac{1}{2}\cos t\) | |
| Hence \(y = \sin t\cos\frac{\pi}{6} + \cos t\sin\frac{\pi}{6} = \sin\left(t + \frac{\pi}{6}\right)\) | A1 cso |
| [3] |
M1 Substitutes \(x = \sin t\) into the equation give in \(y\).
M1 Use of trig identity to deduce that \(\cos t = \sqrt{(1 - \sin^2 t)}\).
A1 cso Using the compound angle formula to prove \(y = \sin\left(t + \frac{\pi}{6}\right)\)