| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2012 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Implicit equations and differentiation |
| Type | Find stationary points |
| Difficulty | Standard +0.3 This is a standard implicit differentiation question requiring product rule and algebraic manipulation to find dy/dx, followed by solving simultaneous equations (setting dy/dx = 0 with the original curve equation). While it involves multiple steps, the techniques are routine C4 content with no novel insight required, making it slightly easier than average. |
| Spec | 1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Differentiating implicitly to obtain \(\pm ay^2\frac{dy}{dx}\) and/or \(\pm bx^2\frac{dy}{dx}\) | M1 | |
| \(48y^2\frac{dy}{dx} + \ldots - 54\ldots\) | A1 | |
| \(9x^2y \rightarrow 9x^2\frac{dy}{dx} + 18xy\) | B1 | or equivalent |
| \(\left(48y^2 + 9x^2\right)\frac{dy}{dx} + 18xy - 54 = 0\) | M1 | |
| \(\frac{dy}{dx} = \frac{54 - 18xy}{48y^2 + 9x^2}\left(= \frac{18 - 6xy}{16y^2 + 3x^2}\right)\) | A1 (5) |
| Answer | Marks | Guidance |
|---|---|---|
| \(18 - 6xy = 0\) | M1 | |
| Using \(x = \frac{3}{y}\) or \(y = \frac{3}{x}\): | M1 | |
| \(16y^3 + 9\left(\frac{3}{y}\right)^2 y - 54\left(\frac{3}{y}\right) = 0\) or \(16\left(\frac{3}{x}\right)^3 + 9x^2\left(\frac{3}{x}\right) - 54x = 0\) | M1 | |
| \(16y^4 + 81 - 162 = 0\) or \(16 + x^4 - 2x^4 = 0\) | ||
| \(y^4 = \frac{81}{16}\) or \(x^4 = 16\) | ||
| \(y = \frac{3}{2}, -\frac{3}{2}\) or \(x = 2, -2\) | A1 A1 | |
| Substituting into \(xy = 3\) to obtain other variable | M1 | |
| \(\left(2, \frac{3}{2}\right), \left(-2, -\frac{3}{2}\right)\) | A1 (7) | both |
## Question 5:
### Part (a)
| Differentiating implicitly to obtain $\pm ay^2\frac{dy}{dx}$ and/or $\pm bx^2\frac{dy}{dx}$ | M1 | |
| $48y^2\frac{dy}{dx} + \ldots - 54\ldots$ | A1 | |
| $9x^2y \rightarrow 9x^2\frac{dy}{dx} + 18xy$ | B1 | or equivalent |
| $\left(48y^2 + 9x^2\right)\frac{dy}{dx} + 18xy - 54 = 0$ | M1 | |
| $\frac{dy}{dx} = \frac{54 - 18xy}{48y^2 + 9x^2}\left(= \frac{18 - 6xy}{16y^2 + 3x^2}\right)$ | A1 (5) | |
### Part (b)
| $18 - 6xy = 0$ | M1 | |
| Using $x = \frac{3}{y}$ or $y = \frac{3}{x}$: | M1 | |
| $16y^3 + 9\left(\frac{3}{y}\right)^2 y - 54\left(\frac{3}{y}\right) = 0$ or $16\left(\frac{3}{x}\right)^3 + 9x^2\left(\frac{3}{x}\right) - 54x = 0$ | M1 | |
| $16y^4 + 81 - 162 = 0$ or $16 + x^4 - 2x^4 = 0$ | | |
| $y^4 = \frac{81}{16}$ or $x^4 = 16$ | | |
| $y = \frac{3}{2}, -\frac{3}{2}$ or $x = 2, -2$ | A1 A1 | |
| Substituting into $xy = 3$ to obtain other variable | M1 | |
| $\left(2, \frac{3}{2}\right), \left(-2, -\frac{3}{2}\right)$ | A1 (7) | both |
---\begin{enumerate}
\item The curve $C$ has equation
\end{enumerate}
$$16 y ^ { 3 } + 9 x ^ { 2 } y - 54 x = 0$$
(a) Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$ in terms of $x$ and $y$.\\
(b) Find the coordinates of the points on $C$ where $\frac { \mathrm { d } y } { \mathrm {~d} x } = 0$.\\
\hfill \mbox{\textit{Edexcel C4 2012 Q5 [12]}}