| Exam Board | OCR |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2014 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric curves and Cartesian conversion |
| Type | Find stationary/turning points |
| Difficulty | Standard +0.3 This is a standard C4 parametric equations question requiring routine techniques: differentiation using the chain rule, finding stationary points, and converting to Cartesian form using the double angle formula cos(2t) = 1 - 2sin²(t). The algebraic manipulation is straightforward once x = 2sin(t) is substituted. Slightly above average difficulty due to multiple parts and the need to consider the domain restriction, but all techniques are standard textbook exercises. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{dy}{dt}=-2\sin2t+2\cos t\) soi | B1 | NB \(\frac{dx}{dt}=2\cos t\); if B0M0A0; SC3 for \(\frac{dy}{dx}=1-x\) from correct Cartesian equation seen in part (i) or part (ii); B1 for substitution of \(x=2\sin t\) |
| \(\frac{dy}{dx}=\) their \(\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\) oe | M1 | |
| \(\frac{-2\sin2t+2\cos t}{2\cos t}\) soi | A1 | |
| \(\frac{-4\sin t\cos t+2\cos t}{2\cos t}\) or \(\frac{2\cos t(-2\sin t+1)}{2\cos t}\) and completion to \(1-2\sin t\) www | A1 | or equivalent intermediate step |
| \((1,1\frac{1}{2})\) | B1 | NB \(t=\frac{\pi}{6}\); from \(1-2\sin t=0\) |
| [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((y=)\ 1-2\sin^2t+2\sin t\) | B1 | May be awarded after correct substitution for \(x\); eg \((y=)\ 1-\frac{x^2}{4}-\sin^2t+2\sin t\); or \((y=)\ x+\cos2t\) |
| Substitution of \(\sin t=\frac{1}{2}x\) to eliminate \(t\) | M1 | Substitution of \(t=\sin^{-1}(\frac{x}{2})\) to eliminate \(t\) |
| \(y=1+x-\frac{1}{2}x^2\) oe isw | A1 | or B3 www; \(y=x+\cos(2\sin^{-1}(\frac{x}{2}))\) oe isw |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(-2 \leq x \leq 2\) or \(x \geq -2\) (and) \(x \leq 2\) or \( | x | \leq 2\) |
| sketch of negative quadratic with endpoints in 1st and 3rd quadrants | M1 | RH point must be to the right of the maximum |
| positive \(y\)-intercept and one distinguishing feature | A1 | one from: endpoints \((-2,-3)\) and \((2,1)\), vertex at \((1, 1\frac{1}{2})\), \(y\)-intercept is \((0,1)\), \(x\)-intercept is \((1-\sqrt{3}, 0)\) |
| [3] |
# Question 7(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{dy}{dt}=-2\sin2t+2\cos t$ soi | B1 | NB $\frac{dx}{dt}=2\cos t$; if B0M0A0; SC3 for $\frac{dy}{dx}=1-x$ from correct Cartesian equation seen in part (i) or part (ii); B1 for substitution of $x=2\sin t$ |
| $\frac{dy}{dx}=$ their $\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$ oe | M1 | |
| $\frac{-2\sin2t+2\cos t}{2\cos t}$ soi | A1 | |
| $\frac{-4\sin t\cos t+2\cos t}{2\cos t}$ or $\frac{2\cos t(-2\sin t+1)}{2\cos t}$ and completion to $1-2\sin t$ www | A1 | or equivalent intermediate step |
| $(1,1\frac{1}{2})$ | B1 | NB $t=\frac{\pi}{6}$; from $1-2\sin t=0$ |
| **[5]** | | |
# Question 7(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(y=)\ 1-2\sin^2t+2\sin t$ | B1 | May be awarded after correct substitution for $x$; eg $(y=)\ 1-\frac{x^2}{4}-\sin^2t+2\sin t$; or $(y=)\ x+\cos2t$ |
| Substitution of $\sin t=\frac{1}{2}x$ to eliminate $t$ | M1 | Substitution of $t=\sin^{-1}(\frac{x}{2})$ to eliminate $t$ |
| $y=1+x-\frac{1}{2}x^2$ oe isw | A1 | or B3 www; $y=x+\cos(2\sin^{-1}(\frac{x}{2}))$ oe isw |
| **[3]** | | |
# Question 7(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $-2 \leq x \leq 2$ **or** $x \geq -2$ (and) $x \leq 2$ or $|x| \leq 2$ | B1 | cao |
| sketch of negative quadratic with endpoints in 1st and 3rd quadrants | M1 | RH point must be to the right of the maximum |
| positive $y$-intercept and one distinguishing feature | A1 | one from: endpoints $(-2,-3)$ and $(2,1)$, vertex at $(1, 1\frac{1}{2})$, $y$-intercept is $(0,1)$, $x$-intercept is $(1-\sqrt{3}, 0)$ |
| **[3]** | | |
---7 A curve has parametric equations
$$x = 2 \sin t , \quad y = \cos 2 t + 2 \sin t$$
for $- \frac { 1 } { 2 } \pi \leqslant t \leqslant \frac { 1 } { 2 } \pi$.\\
(i) Show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = 1 - 2 \sin t$ and hence find the coordinates of the stationary point.\\
(ii) Find the cartesian equation of the curve.\\
(iii) State the set of values that $x$ can take and hence sketch the curve.
\hfill \mbox{\textit{OCR C4 2014 Q7 [11]}}