| Exam Board | OCR MEI |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2020 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric differentiation |
| Type | Find parameter value given gradient condition |
| Difficulty | Standard +0.3 This is a standard parametric differentiation question requiring chain rule (dy/dx = (dy/dt)/(dx/dt)), finding a parameter value from a gradient condition, and eliminating the parameter to find a Cartesian equation. All techniques are routine A-level procedures with straightforward algebra, 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 | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{dx}{dt} = -2t^{-3}\) and \(\frac{dy}{dt} = -3t^{-4} + t^{-2}\) | M1 (2.1) | Attempt to differentiate both equations |
| \(\frac{dy}{dx} = \frac{-3t^{-4} + t^{-2}}{-2t^{-3}}\) | M1 (2.1) | Combining derivatives for \(\frac{dy}{dx}\) |
| Multiply top and bottom by \(t^4\): \(\frac{dy}{dx} = \frac{-3+t^2}{-2t} = \frac{3-t^2}{2t}\) | A1 (2.1) [3] | AG Correct derivative in required form. Note \(\frac{dy}{dx} = \left(-\frac{3}{t^4}-\frac{1}{t^2}\right)\times\left(-\frac{t^3}{2}\right)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Tangent parallel when \(\frac{dy}{dx} = -\frac{1}{4}\) | B1 (3.1a) | Establishing gradient \(-\frac{1}{4}\). Note: \(y = -\frac{1}{4}x + \frac{1}{4}\) not sufficient on its own |
| \(\frac{3-t^2}{2t} = -\frac{1}{4}\), giving \(4t^2 - 2t - 12 = 0\) | M1 (1.1a) | Forming and solving quadratic equation |
| Roots \(2, \left[-\frac{3}{2}\right]\) [but since \(t > 0\), \(t = 2\)] | ||
| When \(t=2\): \(x = \frac{1}{4}\), \(y = \frac{1}{8} - \frac{1}{2} = -\frac{3}{8}\). Coordinates are \(\left(\frac{1}{4}, -\frac{3}{8}\right)\) | A1 (1.1) [3] | Using the value of \(t\) for both coordinates. Ignore any point based on \(t = -\frac{3}{2}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Rearrange \(t = x^{-\frac{1}{2}}\) | B1 (3.1a) | Or equivalent e.g. \(\frac{1}{t} = \sqrt{x}\) |
| Substitute \(y = \left(x^{-\frac{1}{2}}\right)^{-3} - \left(x^{-\frac{1}{2}}\right)^{-1} = x^{\frac{3}{2}} - x^{\frac{1}{2}}\) | M1 (1.1) | Attempt to eliminate \(t\) |
| \(= x^{\frac{1}{2}}(x-1) = (x-1)\sqrt{x}\) | A1 (1.1) [3] | Factorised form. Allow surd or index form. Do not allow \(\pm(x-1)\sqrt{x}\) |
# Question 10:
## Part (a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{dx}{dt} = -2t^{-3}$ and $\frac{dy}{dt} = -3t^{-4} + t^{-2}$ | M1 (2.1) | Attempt to differentiate both equations |
| $\frac{dy}{dx} = \frac{-3t^{-4} + t^{-2}}{-2t^{-3}}$ | M1 (2.1) | Combining derivatives for $\frac{dy}{dx}$ |
| Multiply top and bottom by $t^4$: $\frac{dy}{dx} = \frac{-3+t^2}{-2t} = \frac{3-t^2}{2t}$ | A1 (2.1) [3] | AG Correct derivative in required form. Note $\frac{dy}{dx} = \left(-\frac{3}{t^4}-\frac{1}{t^2}\right)\times\left(-\frac{t^3}{2}\right)$ |
## Part (b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Tangent parallel when $\frac{dy}{dx} = -\frac{1}{4}$ | B1 (3.1a) | Establishing gradient $-\frac{1}{4}$. Note: $y = -\frac{1}{4}x + \frac{1}{4}$ not sufficient on its own |
| $\frac{3-t^2}{2t} = -\frac{1}{4}$, giving $4t^2 - 2t - 12 = 0$ | M1 (1.1a) | Forming and solving quadratic equation |
| Roots $2, \left[-\frac{3}{2}\right]$ [but since $t > 0$, $t = 2$] | | |
| When $t=2$: $x = \frac{1}{4}$, $y = \frac{1}{8} - \frac{1}{2} = -\frac{3}{8}$. Coordinates are $\left(\frac{1}{4}, -\frac{3}{8}\right)$ | A1 (1.1) [3] | Using the value of $t$ for both coordinates. Ignore any point based on $t = -\frac{3}{2}$ |
## Part (c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Rearrange $t = x^{-\frac{1}{2}}$ | B1 (3.1a) | Or equivalent e.g. $\frac{1}{t} = \sqrt{x}$ |
| Substitute $y = \left(x^{-\frac{1}{2}}\right)^{-3} - \left(x^{-\frac{1}{2}}\right)^{-1} = x^{\frac{3}{2}} - x^{\frac{1}{2}}$ | M1 (1.1) | Attempt to eliminate $t$ |
| $= x^{\frac{1}{2}}(x-1) = (x-1)\sqrt{x}$ | A1 (1.1) [3] | Factorised form. Allow surd or index form. Do not allow $\pm(x-1)\sqrt{x}$ |
---10 In this question you must show detailed reasoning.
Fig. 10 shows the curve given parametrically by the equations $\mathrm { x } = \frac { 1 } { \mathrm { t } ^ { 2 } } , \mathrm { y } = \frac { 1 } { \mathrm { t } ^ { 3 } } - \frac { 1 } { \mathrm { t } }$, for $t > 0$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{7de77679-59c0-4431-a9cb-6ab11d2f9062-07_611_595_708_260}
\captionsetup{labelformat=empty}
\caption{Fig. 10}
\end{center}
\end{figure}
\begin{enumerate}[label=(\alph*)]
\item Show that $\frac { d y } { d x } = \frac { 3 - t ^ { 2 } } { 2 t }$.
\item Find the coordinates of the point on the curve at which the tangent to the curve is parallel to the line $4 \mathrm { y } + \mathrm { x } = 1$.
\item Find the cartesian equation of the curve. Give your answer in factorised form.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Paper 1 2020 Q10 [9]}}