| Exam Board | OCR MEI |
|---|---|
| Module | Paper 3 (Paper 3) |
| Year | 2019 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Implicit equations and differentiation |
| Type | Tangent with given gradient |
| Difficulty | Standard +0.8 This question requires implicit differentiation to find dy/dx, then using the relationship between tangent and normal gradients (m_normal = -1/m_tangent), solving the resulting system of equations (one from the curve, one from the gradient condition). It involves multiple algebraic steps and careful manipulation of simultaneous equations, going beyond routine implicit differentiation exercises but using standard A-level techniques throughout. |
| Spec | 1.07m Tangents and normals: gradient and equations1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Gradient of normal \(= 2 \Rightarrow\) gradient of tangent \(= -0.5\) | M1 | 3.1a |
| \(2y\frac{dy}{dx} + y + x\frac{dy}{dx} = 0\) | M1, A1 | 1.1, 1.1 |
| e.g. \(-y + y - 0.5x = 0 \Rightarrow x = 0\) | A1 | 2.2a |
| \(y^2 = 8\) | M1 | 1.1 |
| \(\left(0,\ \sqrt{8}\right)\) and \(\left(0,\ -\sqrt{8}\right)\) | A1 | 1.1 |
| Alternative method: | ||
| \(\frac{dx}{dy} = \frac{-8-y^2}{y^2}\) oe | M1, M1 | |
| \(\frac{8+y^2}{y^2} = 2\) oe | A1 | |
| \(y^2 = 8\) | M1 | |
| \(\left(0,\ \sqrt{8}\right)\) and \(\left(0,\ -\sqrt{8}\right)\) | M1, A1 | |
| [6] |
## Question 9:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Gradient of normal $= 2 \Rightarrow$ gradient of tangent $= -0.5$ | M1 | 3.1a | $\frac{-y}{x+2y} = -\frac{1}{2}$ |
| $2y\frac{dy}{dx} + y + x\frac{dy}{dx} = 0$ | M1, A1 | 1.1, 1.1 | Attempt at implicit differentiation with one algebraic term differentiated correctly; Correct implicit expression; Or M1 for correct expression for $x$ in terms of $y$ and attempt to differentiate |
| e.g. $-y + y - 0.5x = 0 \Rightarrow x = 0$ | A1 | 2.2a | |
| $y^2 = 8$ | M1 | 1.1 | Implied by at least one $y$-value; For one step solving their equation in $x$ and/or $y$ with original equation |
| $\left(0,\ \sqrt{8}\right)$ and $\left(0,\ -\sqrt{8}\right)$ | A1 | 1.1 | |
| **Alternative method:** | | |
| $\frac{dx}{dy} = \frac{-8-y^2}{y^2}$ oe | M1, M1 | | Rearranging to $x = \frac{8-y^2}{y}$ and attempting differentiation; Differentiation by dividing first or quotient rule – allow one error |
| $\frac{8+y^2}{y^2} = 2$ oe | A1 | | Correct differentiation |
| $y^2 = 8$ | M1 | | Using negative reciprocal |
| $\left(0,\ \sqrt{8}\right)$ and $\left(0,\ -\sqrt{8}\right)$ | M1, A1 | | Implied by at least one $y$-value |
| **[6]** | | |
---9 In this question you must show detailed reasoning.\\
The curve $x y + y ^ { 2 } = 8$ is shown in Fig. 9.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{99485c27-9ff8-4bdb-a7e6-49dfcaedc579-7_734_750_397_244}
\captionsetup{labelformat=empty}
\caption{Fig. 9}
\end{center}
\end{figure}
Find the coordinates of the points on the curve at which the normal has gradient 2.
\hfill \mbox{\textit{OCR MEI Paper 3 2019 Q9 [6]}}