| Exam Board | OCR MEI |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2022 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric curves and Cartesian conversion |
| Type | Convert to Cartesian (polynomial/rational) |
| Difficulty | Moderate -0.3 This is a straightforward parametric equations question requiring standard techniques: finding dy/dx using the chain rule (dy/dt รท dx/dt), substituting a point to verify a gradient, finding when dy/dx equals a specific value, and eliminating the parameter to get a Cartesian equation. All steps are routine A-level procedures with no novel problem-solving required, though the multi-part nature and algebraic manipulation (particularly part d) place it slightly below average difficulty rather than being trivial. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\dfrac{dx}{dt} = 3t^2 - 8\), \(\dfrac{dy}{dt} = 2t\) | M1* | Attempt to differentiate both parametric equations; Only allow complete method for finding \(\frac{dy}{dx}\) in terms of \(t\) using Cartesian equation of curve |
| \(\dfrac{dy}{dx} = \dfrac{2t}{3t^2 - 8}\) | M1(dep) | Combine derivatives to find \(\frac{dy}{dx}\); Do not allow reciprocal |
| cao | A1 | |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| AG: \(t^3 - 8t = 8\) and \(t^2 = 4\) gives \(t = -2\) | M1 | Attempt to establish value of \(t\) at \((8,4)\); Allow \(\pm 2\) or 2 stated; Allow for \(y=4\) used in \(\frac{dy}{dx} = \frac{2\sqrt{y}}{3y-8}\) for M mark only |
| \(\dfrac{dy}{dx} = \dfrac{2(-2)}{3(-2)^2 - 8} = -1\) | E1 | AG; negativity must be clearly established from correct working |
| Alternative: When \(\dfrac{dy}{dx} = \dfrac{2t}{3t^2-8} = -1\) giving \(3t^2 + 2t - 8 = 0\) | M1 | Uses value of derivative to find value of \(t\) at P |
| \(t = \dfrac{4}{3}\) or \(t = -2\); When \(t=-2\) coordinates are \(\left((-2)^3 - 8(-2), (-2)^2\right) = (8,4)\) [which is P] | E1 | Allow without reference to \(t = \frac{4}{3}\) |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\dfrac{dy}{dx} = \dfrac{2t}{3t^2-8} = -1\) giving \(3t^2 + 2t - 8 = 0\) | M1 | Equating their \(\frac{dy}{dx}\) to \(-1\) and rearranging to form quadratic equation |
| \(t = \dfrac{4}{3}\) or \([t = -2\) is the point P\(]\) | A1 | allow www; \(-2\) need not be seen but if seen must be rejected |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Substitute \(t^2 = y\): \(x = t^3 - 8t \Rightarrow x^2 = t^6 - 16t^4 + 64t^2\) | M1, A1 | Allow for \(x^2 = t^2(t^2-8)^2\) |
| \(\Rightarrow x^2 = y^3 - 16y^2 + 64y\) | A1 | |
| Alternative: Substitute \(t = \pm y^{\frac{1}{2}}\): \(x = \pm\left(y^{\frac{3}{2}} - 8y^{\frac{1}{2}}\right)\) | M1 | Substituting for \(t\) in equation for \(x\); allow without \(\pm\) |
| \(x^2 = y(y-8)^2 = \left[y^3 - 16y^2 + 64y\right]\) | A1 | Must be in form \(x^2 = ...\); Need not be simplified; Do not award if \(\pm\) not seen at all |
| [3] |
# Question 8:
## Part (a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\dfrac{dx}{dt} = 3t^2 - 8$, $\dfrac{dy}{dt} = 2t$ | M1* | Attempt to differentiate both parametric equations; Only allow complete method for finding $\frac{dy}{dx}$ in terms of $t$ using Cartesian equation of curve |
| $\dfrac{dy}{dx} = \dfrac{2t}{3t^2 - 8}$ | M1(dep) | Combine derivatives to find $\frac{dy}{dx}$; Do not allow reciprocal |
| cao | A1 | |
| **[3]** | | |
## Part (b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| AG: $t^3 - 8t = 8$ and $t^2 = 4$ gives $t = -2$ | M1 | Attempt to establish value of $t$ at $(8,4)$; Allow $\pm 2$ or 2 stated; Allow for $y=4$ used in $\frac{dy}{dx} = \frac{2\sqrt{y}}{3y-8}$ for M mark only |
| $\dfrac{dy}{dx} = \dfrac{2(-2)}{3(-2)^2 - 8} = -1$ | E1 | AG; negativity must be clearly established from correct working |
| Alternative: When $\dfrac{dy}{dx} = \dfrac{2t}{3t^2-8} = -1$ giving $3t^2 + 2t - 8 = 0$ | M1 | Uses value of derivative to find value of $t$ at P |
| $t = \dfrac{4}{3}$ or $t = -2$; When $t=-2$ coordinates are $\left((-2)^3 - 8(-2), (-2)^2\right) = (8,4)$ [which is P] | E1 | Allow without reference to $t = \frac{4}{3}$ |
| **[2]** | | |
## Part (c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\dfrac{dy}{dx} = \dfrac{2t}{3t^2-8} = -1$ giving $3t^2 + 2t - 8 = 0$ | M1 | Equating their $\frac{dy}{dx}$ to $-1$ and rearranging to form quadratic equation |
| $t = \dfrac{4}{3}$ or $[t = -2$ is the point P$]$ | A1 | allow www; $-2$ need not be seen but if seen must be rejected |
| **[2]** | | |
## Part (d):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Substitute $t^2 = y$: $x = t^3 - 8t \Rightarrow x^2 = t^6 - 16t^4 + 64t^2$ | M1, A1 | Allow for $x^2 = t^2(t^2-8)^2$ |
| $\Rightarrow x^2 = y^3 - 16y^2 + 64y$ | A1 | |
| Alternative: Substitute $t = \pm y^{\frac{1}{2}}$: $x = \pm\left(y^{\frac{3}{2}} - 8y^{\frac{1}{2}}\right)$ | M1 | Substituting for $t$ in equation for $x$; allow without $\pm$ |
| $x^2 = y(y-8)^2 = \left[y^3 - 16y^2 + 64y\right]$ | A1 | Must be in form $x^2 = ...$; Need not be simplified; Do not award if $\pm$ not seen at all |
| **[3]** | | |
---8 A particle moves in the $x - y$ plane so that its position at time $t$ s is given by $x = t ^ { 3 } - 8 t , y = t ^ { 2 }$ for $- 3.5 < t < 3.5$. The units of distance are metres. The graph shows the path of the particle and the direction of travel at the point $\mathrm { P } ( 8,4 )$.\\
\includegraphics[max width=\textwidth, alt={}, center]{9dd6fc6d-b51e-4a73-ace5-d26a7558032c-07_492_924_415_242}
\begin{enumerate}[label=(\alph*)]
\item Find $\frac { \mathrm { dy } } { \mathrm { dx } }$ in terms of $t$.
\item Hence show that the value of $\frac { \mathrm { dy } } { \mathrm { dx } }$ at P is - 1 .
\item Find the time at which the particle is travelling in the direction opposite to that at P .
\item Find the cartesian equation of the path, giving $x ^ { 2 }$ as a function of $y$.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Paper 1 2022 Q8 [10]}}