| Exam Board | OCR MEI |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric differentiation |
| Type | Find stationary points of parametric curve |
| Difficulty | Standard +0.3 This is a straightforward parametric differentiation question requiring verification of parametric equations (simple substitution), finding dy/dx using the chain rule (dy/dt รท dx/dt), and identifying stationary points by setting dy/dx = 0. All techniques are standard C4 material with no novel insight required, making it slightly easier than average. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(y^2 - x^2 = (t+1/t)^2 - (t-1/t)^2 = t^2 + 2 + 1/t^2 - t^2 + 2 - 1/t^2 = 4\) | M1, E1 [2] | Substituting for \(x\) and \(y\) in terms of \(t\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| EITHER: \(dx/dt = 1 + 1/t^2\), \(dy/dt = 1 - 1/t^2\) | B1 | For both results |
| \(\Rightarrow \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{1-1/t^2}{1+1/t^2} = \frac{t^2-1}{t^2+1} = \frac{(t-1)(t+1)}{t^2+1}\) | M1, E1 | |
| OR: \(2y\frac{dy}{dx} - 2x = 0 \Rightarrow \frac{dy}{dx} = \frac{x}{y} = \frac{t-1/t}{t+1/t} = \frac{t^2-1}{t^2+1} = \frac{(t-1)(t+1)}{t^2+1}\) | B1, M1, E1 | |
| OR: \(y = \sqrt{(4+x^2)} \Rightarrow \frac{dy}{dx} = \frac{x}{\sqrt{4+x^2}} = \frac{t^2-1}{t^2+1} = \frac{(t-1)(t+1)}{t^2+1}\) | B1, M1, E1 | |
| \(dy/dx = 0\) when \(t = 1\) or \(t = -1\) | M1 | |
| \(t = 1 \Rightarrow (0, 2)\) | A1 | |
| \(t = -1 \Rightarrow (0, -2)\) | A1 [6] |
## Question 3:
### Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $y^2 - x^2 = (t+1/t)^2 - (t-1/t)^2 = t^2 + 2 + 1/t^2 - t^2 + 2 - 1/t^2 = 4$ | M1, E1 [2] | Substituting for $x$ and $y$ in terms of $t$ |
### Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| **EITHER:** $dx/dt = 1 + 1/t^2$, $dy/dt = 1 - 1/t^2$ | B1 | For both results |
| $\Rightarrow \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{1-1/t^2}{1+1/t^2} = \frac{t^2-1}{t^2+1} = \frac{(t-1)(t+1)}{t^2+1}$ | M1, E1 | |
| **OR:** $2y\frac{dy}{dx} - 2x = 0 \Rightarrow \frac{dy}{dx} = \frac{x}{y} = \frac{t-1/t}{t+1/t} = \frac{t^2-1}{t^2+1} = \frac{(t-1)(t+1)}{t^2+1}$ | B1, M1, E1 | |
| **OR:** $y = \sqrt{(4+x^2)} \Rightarrow \frac{dy}{dx} = \frac{x}{\sqrt{4+x^2}} = \frac{t^2-1}{t^2+1} = \frac{(t-1)(t+1)}{t^2+1}$ | B1, M1, E1 | |
| $dy/dx = 0$ when $t = 1$ or $t = -1$ | M1 | |
| $t = 1 \Rightarrow (0, 2)$ | A1 | |
| $t = -1 \Rightarrow (0, -2)$ | A1 [6] | |
---3 A curve has carlesian equation $\mathrm { y } ^ { 2 } - \mathrm { x } _ { 2 } = 4$.\\
(i) Verify that
$$\boldsymbol { x } = \boldsymbol { t } - - ^ { 1 } \quad \boldsymbol { t ^ { \prime } } \quad \boldsymbol { y } = \boldsymbol { t } + \frac { 1 } { \boldsymbol { t } ^ { \prime } }$$
are parametric equations of the curve.\\
(u) Show lhat $\left. \underset { d x } { \mathbf { d y } } = \frac { ( t - I ) ( r } { 12 + 1 } + 1 \right)$. Hence find the coordinates of the staionary points of the curve.
\hfill \mbox{\textit{OCR MEI C4 Q3 [8]}}