| Exam Board | OCR |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Implicit equations and differentiation |
| Type | Find tangent equation at point |
| Difficulty | Standard +0.3 This is a standard implicit differentiation question with straightforward algebra. Part (i) requires routine application of the product and chain rules to differentiate implicitly, then rearranging. Part (ii) involves substituting a given y-value to find the point and gradient, then forming a tangent equation. While it requires careful algebraic manipulation, it follows a well-practiced procedure with no novel problem-solving required, making it slightly easier than the average A-level question. |
| Spec | 1.07m Tangents and normals: gradient and equations1.07s Parametric and implicit differentiation |
\begin{enumerate}
\item A curve has the equation $x = y \sqrt { 1 - 2 y }$.\\
(i) Show that
\end{enumerate}
$$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { \sqrt { 1 - 2 y } } { 1 - 3 y } .$$
The point $A$ on the curve has $y$-coordinate - 1 .\\
(ii) Show that the equation of tangent to the curve at $A$ can be written in the form
$$\sqrt { 3 } x + p y + q = 0$$
where $p$ and $q$ are integers to be found.\\
\hfill \mbox{\textit{OCR C3 Q4 [7]}}