| Exam Board | OCR |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2013 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric differentiation |
| Type | Find stationary points of parametric curve |
| Difficulty | Standard +0.3 This is a standard C4 parametric equations question requiring routine application of the chain rule (dy/dx = (dy/dt)/(dx/dt)), finding stationary points by setting dy/dx = 0, and converting to Cartesian form. All techniques are textbook exercises with straightforward algebra, making it slightly easier than average for C4 level. |
| Spec | 1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{dy}{dt}=2(+)-\frac{2}{t^3}\); \(\frac{dx}{dt}=-\frac{1}{t^2}\) oe soi ISW | B1, B1 | |
| \(\frac{2}{t}-2t^2\) or \(-2\left(t^2-\frac{1}{t}\right)\), \(\frac{2t^3-2}{-t}\), \(-t^2\left(2-\frac{2}{t^3}\right)\) oe | B1 | ISW. Must not involve (implied)'triple-deckers' e.g. fractions with neg powers... e.g. \(\frac{2-2t^{-3}}{-t^{-2}}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| (Any of their expressions for \(\frac{dy}{dx}\)) \(=0\) or their \(\frac{dy}{dt}=0\) | M1 | |
| \(t=1\rightarrow\) (stationary point) \(=(0,3)\) | A1 | Not awarded if \(\frac{dy}{dx}\) is wrong in (i) and used here BUT allow recovery if only explicitly considering \(\frac{dy}{dt}=0\) |
| Consider values of \(x\) on each side of their critical value of \(x\) which lead to finite values of \(\frac{dy}{dx}\) | M1 | |
| Hence \((0,3)\) is a minimum point www | A1 | Totally satis; values of \(x\) must be close to 0 & not going below or equal to \(x=-1\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Attempt to find \(t\) from \(x=\frac{1}{t}-1\) and substitute into the equation for \(y\) | M1 | |
| \(y=\frac{2}{x+1}+(x+1)^2\) oe (can be unsimplified) ISW | A1 |
## Question 9(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{dy}{dt}=2(+)-\frac{2}{t^3}$; $\frac{dx}{dt}=-\frac{1}{t^2}$ oe soi ISW | B1, B1 | |
| $\frac{2}{t}-2t^2$ or $-2\left(t^2-\frac{1}{t}\right)$, $\frac{2t^3-2}{-t}$, $-t^2\left(2-\frac{2}{t^3}\right)$ oe | B1 | ISW. Must not involve (implied)'triple-deckers' e.g. fractions with neg powers... e.g. $\frac{2-2t^{-3}}{-t^{-2}}$ |
---
## Question 9(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| (Any of their expressions for $\frac{dy}{dx}$) $=0$ or their $\frac{dy}{dt}=0$ | M1 | |
| $t=1\rightarrow$ (stationary point) $=(0,3)$ | A1 | Not awarded if $\frac{dy}{dx}$ is wrong in (i) and used here BUT allow recovery if only explicitly considering $\frac{dy}{dt}=0$ |
| Consider values of $x$ on each side of their critical value of $x$ which lead to finite values of $\frac{dy}{dx}$ | M1 | |
| Hence $(0,3)$ is a minimum point www | A1 | Totally satis; values of $x$ must be close to 0 & not going below or equal to $x=-1$ |
---
## Question 9(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Attempt to find $t$ from $x=\frac{1}{t}-1$ and substitute into the equation for $y$ | M1 | |
| $y=\frac{2}{x+1}+(x+1)^2$ oe (can be unsimplified) ISW | A1 | |
---9 A curve has parametric equations $x = \frac { 1 } { t } - 1$ and $y = 2 t + \frac { 1 } { t ^ { 2 } }$.\\
(i) Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$ in terms of $t$, simplifying your answer.\\
(ii) Find the coordinates of the stationary point and, by considering the gradient of the curve on either side of this point, determine its nature.\\
(iii) Find a cartesian equation of the curve.
\hfill \mbox{\textit{OCR C4 2013 Q9 [9]}}