| Exam Board | OCR |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric differentiation |
| Type | Tangent parallel to axis condition |
| Difficulty | Standard +0.3 This is a standard C4 parametric equations question requiring routine techniques: chain rule for dy/dx, finding stationary points by setting dy/dx=0, finding a tangent equation, and eliminating the parameter. All methods are textbook exercises with no novel insight required, making it slightly easier than average. |
| 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 |
|---|---|
| \(\frac{dx}{dt} = 6\cos t \times (-\sin t)\) | M1 |
| \(\frac{dy}{dt} = 2\cos 2t\) | |
| \(\frac{dy}{dx} = \frac{2\cos 2t}{-6\cos t \sin t} = \frac{2\cos 2t}{-3\sin 2t} = -\frac{2}{3}\cot 2t\) | M1 A1 |
| Answer | Marks |
|---|---|
| \(-\frac{2}{3}\cot 2t = 0 \Rightarrow 2t = \frac{\pi}{2}, \frac{3\pi}{2} \Rightarrow t = \frac{\pi}{4}, \frac{3\pi}{4}\) | M1 A1 |
| \(\therefore \left(\frac{3}{2}, 1\right), \left(\frac{3}{2}, -1\right)\) | A1 |
| Answer | Marks |
|---|---|
| \(t = \frac{\pi}{6}\), \(x = \frac{9}{4}\), \(y = \frac{\sqrt{3}}{2}\), \(\text{grad} = -\frac{2}{3\sqrt{3}}\) | B1 |
| \(\therefore y - \frac{\sqrt{3}}{2} = -\frac{2}{3\sqrt{3}}\left(x - \frac{9}{4}\right)\) | M1 |
| Answer | Marks |
|---|---|
| \(2x + 3\sqrt{3}\,y = 9\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(y^2 = \sin^2 2t = 4\sin^2 t\cos^2 t = 4(1-\cos^2 t)\cos^2 t\) | M1 | |
| \(\cos^2 t = \frac{x}{3}\ \therefore y^2 = 4\left(1-\frac{x}{3}\right)\frac{x}{3},\ y^2 = \frac{4}{9}x(3-x)\) | M1 A1 | (12) |
# Question 6:
## Part (i):
$\frac{dx}{dt} = 6\cos t \times (-\sin t)$ | M1 |
$\frac{dy}{dt} = 2\cos 2t$ | |
$\frac{dy}{dx} = \frac{2\cos 2t}{-6\cos t \sin t} = \frac{2\cos 2t}{-3\sin 2t} = -\frac{2}{3}\cot 2t$ | M1 A1 |
## Part (ii):
$-\frac{2}{3}\cot 2t = 0 \Rightarrow 2t = \frac{\pi}{2}, \frac{3\pi}{2} \Rightarrow t = \frac{\pi}{4}, \frac{3\pi}{4}$ | M1 A1 |
$\therefore \left(\frac{3}{2}, 1\right), \left(\frac{3}{2}, -1\right)$ | A1 |
## Part (iii):
$t = \frac{\pi}{6}$, $x = \frac{9}{4}$, $y = \frac{\sqrt{3}}{2}$, $\text{grad} = -\frac{2}{3\sqrt{3}}$ | B1 |
$\therefore y - \frac{\sqrt{3}}{2} = -\frac{2}{3\sqrt{3}}\left(x - \frac{9}{4}\right)$ | M1 |
$6\sqrt{3}\,y - 9 = -4x + 9$
$2x + 3\sqrt{3}\,y = 9$ | A1 |
## Part (iv):
$y^2 = \sin^2 2t = 4\sin^2 t\cos^2 t = 4(1-\cos^2 t)\cos^2 t$ | M1 |
$\cos^2 t = \frac{x}{3}\ \therefore y^2 = 4\left(1-\frac{x}{3}\right)\frac{x}{3},\ y^2 = \frac{4}{9}x(3-x)$ | M1 A1 | **(12)**
---6. A curve has parametric equations
$$x = 3 \cos ^ { 2 } t , \quad y = \sin 2 t , \quad 0 \leq t < \pi$$
(i) Show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = - \frac { 2 } { 3 } \cot 2 t$.\\
(ii) Find the coordinates of the points where the tangent to the curve is parallel to the $x$-axis.\\
(iii) Show that the tangent to the curve at the point where $t = \frac { \pi } { 6 }$ has the equation
$$2 x + 3 \sqrt { 3 } y = 9$$
(iv) Find a cartesian equation for the curve in the form $y ^ { 2 } = \mathrm { f } ( x )$.\\
\hfill \mbox{\textit{OCR C4 Q6 [12]}}