| Exam Board | OCR MEI |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Implicit equations and differentiation |
| Type | Find normal equation at point |
| Difficulty | Moderate -0.8 This is a straightforward implicit differentiation question with a simple circle equation. Part (i) is routine application of implicit differentiation rules, and part (ii) requires finding the normal (negative reciprocal of tangent gradient) and using point-slope form. Both parts are standard textbook exercises requiring no problem-solving insight, making this easier than average. |
| Spec | 1.07m Tangents and normals: gradient and equations1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(x^2 + y^2 = 25 \Rightarrow 2x + 2y\frac{dy}{dx} = 0\) | M1A1 | |
| \(\Rightarrow \frac{dy}{dx} = -\frac{x}{y}\) | E1 | |
| Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Gradient of normal \(= \frac{4}{3}\) | B1 | |
| \(\Rightarrow y - 4 = \frac{4}{3}(x-3)\) | M1 | |
| \(\Rightarrow 3y = 4x\) or equivalent | A1 | |
| Total: 3 |
## Question 5:
### Part (i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $x^2 + y^2 = 25 \Rightarrow 2x + 2y\frac{dy}{dx} = 0$ | M1A1 | |
| $\Rightarrow \frac{dy}{dx} = -\frac{x}{y}$ | E1 | |
| | **Total: 3** | |
### Part (ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| Gradient of normal $= \frac{4}{3}$ | B1 | |
| $\Rightarrow y - 4 = \frac{4}{3}(x-3)$ | M1 | |
| $\Rightarrow 3y = 4x$ or equivalent | A1 | |
| | **Total: 3** | |
---5 The equation of a circle is $x ^ { 2 } + y ^ { 2 } = 25$.\\
(i) Show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = - \frac { x } { y }$.\\
(ii) Hence find the equation of the normal to the circle at the point ( 3,4 ).
\hfill \mbox{\textit{OCR MEI C3 Q5 [6]}}