| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 8 |
| 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 straightforward implicit differentiation question with standard techniques. Part (a) requires differentiating x with respect to y using the product rule and chain rule, then inverting. Part (b) involves substituting y=-1 to find the point and gradient, then writing the tangent equation. While it requires multiple steps and careful algebra, it follows a predictable pattern with no novel insight required, making it slightly easier than average. |
| Spec | 1.07m Tangents and normals: gradient and equations1.07s Parametric and implicit differentiation |
| Answer | Marks |
|---|---|
| \(\frac{dx}{dy} = 1 \times \sqrt{1 - 2y} + yx \cdot \frac{1}{2}(1 - 2y)^{-\frac{1}{2}} \times (-2)\) | M1 A1 |
| \(= \sqrt{1 - 2y} - \frac{y}{\sqrt{1 - 2y}} = \frac{(1 - 2y) - y}{\sqrt{1 - 2y}} = \frac{1 - 3y}{\sqrt{1 - 2y}}\) | M1 |
| \(\frac{dy}{dx} = 1 + \frac{dx}{dy} = \frac{\sqrt{1 - 2y}}{1 - 3y}\) | M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(y = -1, x = -\sqrt{3}, \text{ grad } = \frac{1}{4}\sqrt{3}\) | B1 | |
| \(\therefore y + 1 = \frac{1}{4}\sqrt{3}(x + \sqrt{3})\) | M1 | |
| \(4y + 4 = \sqrt{3}x + 3\) | ||
| \(\sqrt{3}x - 4y - 1 = 0\) | A1 | \([p = -4, q = -1]\) |
**(a)**
$\frac{dx}{dy} = 1 \times \sqrt{1 - 2y} + yx \cdot \frac{1}{2}(1 - 2y)^{-\frac{1}{2}} \times (-2)$ | M1 A1 |
$= \sqrt{1 - 2y} - \frac{y}{\sqrt{1 - 2y}} = \frac{(1 - 2y) - y}{\sqrt{1 - 2y}} = \frac{1 - 3y}{\sqrt{1 - 2y}}$ | M1 |
$\frac{dy}{dx} = 1 + \frac{dx}{dy} = \frac{\sqrt{1 - 2y}}{1 - 3y}$ | M1 A1 |
**(b)**
$y = -1, x = -\sqrt{3}, \text{ grad } = \frac{1}{4}\sqrt{3}$ | B1 |
$\therefore y + 1 = \frac{1}{4}\sqrt{3}(x + \sqrt{3})$ | M1 |
$4y + 4 = \sqrt{3}x + 3$ |
$\sqrt{3}x - 4y - 1 = 0$ | A1 | $[p = -4, q = -1]$ | (8)4. A curve has the equation $x = y \sqrt { 1 - 2 y }$.
\begin{enumerate}[label=(\alph*)]
\item Show that
$$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { \sqrt { 1 - 2 y } } { 1 - 3 y } .$$
The point $A$ on the curve has $y$-coordinate - 1 .
\item 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.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C3 Q4 [8]}}