| Exam Board | Edexcel |
|---|---|
| Module | C34 (Core Mathematics 3 & 4) |
| Year | 2019 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Implicit equations and differentiation |
| Type | Find stationary points |
| Difficulty | Standard +0.3 This is a straightforward implicit differentiation question requiring product rule and chain rule application, followed by solving a system to find stationary points. While it involves multiple steps, the techniques are standard C3/C4 material with no novel insight required, making it slightly easier than average. |
| Spec | 1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{d(81y^3)}{dx} = 243y^2\frac{dy}{dx}\) | M1 | \(\frac{d(81y^3)}{dx} = ky^2\frac{dy}{dx}\) |
| \(\frac{d(64x^2y)}{dx} = 128xy + 64x^2\frac{dy}{dx}\) | M1 | \(\frac{d(64x^2y)}{dx} = \alpha xy + \beta x^2\frac{dy}{dx}\) |
| \(243y^2\frac{dy}{dx} + 128xy + 64x^2\frac{dy}{dx} + 256 (= 0)\) | A1 | Correct differentiation. The "\(= 0\)" is not required but there should be no extra terms |
| \(243y^2\frac{dy}{dx} + 64x^2\frac{dy}{dx} = -128xy - 256 \Rightarrow \frac{dy}{dx}(243y^2 + 64x^2) = -128xy - 256 \Rightarrow \frac{dy}{dx} = ...\) | M1 | Makes \(\frac{dy}{dx}\) the subject allowing sign errors only with "\(= 0\)" seen or implied. Depends on exactly two \(\frac{dy}{dx}\) terms |
| \(\frac{dy}{dx} = \frac{-128xy - 256}{243y^2 + 64x^2}\) | A1 | Correct expression (oe) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(-128xy - 256 = 0\) | M1 | Sets numerator = 0. May appear from putting \(\frac{dy}{dx} = 0\) into differentiation before making \(\frac{dy}{dx}\) subject |
| \(81y^3 + 64y\left(-\frac{2}{y}\right)^2 + 256\left(-\frac{2}{y}\right) = 0\) or \(81\left(-\frac{2}{x}\right)^3 + 64x^2\left(-\frac{2}{x}\right) + 256x = 0\) | dM1 | Substitutes to obtain equation in one variable. Dependent on first M |
| \(y^4 = \frac{256}{81} \Rightarrow y = \ldots\) or \(x^4 = \frac{81}{16} \Rightarrow x = \ldots\) | dM1 | Solves equation of form \(y^4 = p\) or \(x^4 = q\) \((p,q > 0)\). Depends on previous M |
| \(y = \pm\frac{4}{3}\) or \(x = \pm\frac{3}{2}\) | A1 | 2 correct values for \(x\) or 2 correct values for \(y\). Allow unsimplified |
| \(y = (\pm)\)"\(\frac{4}{3}\)"\(\Rightarrow x = \ldots\) or \(x = (\pm)\)"\(\frac{3}{2}\)"\(\Rightarrow y = \ldots\) | M1 | Attempts at least one value of other variable having previously found and solved equation in one variable |
| \(\left(\pm\frac{3}{2}, \mp\frac{4}{3}\right)\) or \(x = \frac{3}{2}, y = -\frac{4}{3}\) and \(x = -\frac{3}{2}, y = \frac{4}{3}\) | A1 | Correct values, simplified and paired correctly. Do not isw — mark final answer |
## Question 4:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{d(81y^3)}{dx} = 243y^2\frac{dy}{dx}$ | M1 | $\frac{d(81y^3)}{dx} = ky^2\frac{dy}{dx}$ |
| $\frac{d(64x^2y)}{dx} = 128xy + 64x^2\frac{dy}{dx}$ | M1 | $\frac{d(64x^2y)}{dx} = \alpha xy + \beta x^2\frac{dy}{dx}$ |
| $243y^2\frac{dy}{dx} + 128xy + 64x^2\frac{dy}{dx} + 256 (= 0)$ | A1 | Correct differentiation. The "$= 0$" is not required but there should be no extra terms |
| $243y^2\frac{dy}{dx} + 64x^2\frac{dy}{dx} = -128xy - 256 \Rightarrow \frac{dy}{dx}(243y^2 + 64x^2) = -128xy - 256 \Rightarrow \frac{dy}{dx} = ...$ | M1 | Makes $\frac{dy}{dx}$ the subject allowing sign errors only with "$= 0$" seen or implied. Depends on exactly two $\frac{dy}{dx}$ terms |
| $\frac{dy}{dx} = \frac{-128xy - 256}{243y^2 + 64x^2}$ | A1 | Correct expression (oe) |
Note: the final M1A1 in (a) can be recovered in part (b)
## Question (b) [Implicit Differentiation]:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $-128xy - 256 = 0$ | M1 | Sets numerator = 0. May appear from putting $\frac{dy}{dx} = 0$ into differentiation before making $\frac{dy}{dx}$ subject |
| $81y^3 + 64y\left(-\frac{2}{y}\right)^2 + 256\left(-\frac{2}{y}\right) = 0$ or $81\left(-\frac{2}{x}\right)^3 + 64x^2\left(-\frac{2}{x}\right) + 256x = 0$ | dM1 | Substitutes to obtain equation in one variable. Dependent on first M |
| $y^4 = \frac{256}{81} \Rightarrow y = \ldots$ or $x^4 = \frac{81}{16} \Rightarrow x = \ldots$ | dM1 | Solves equation of form $y^4 = p$ or $x^4 = q$ $(p,q > 0)$. Depends on previous M |
| $y = \pm\frac{4}{3}$ or $x = \pm\frac{3}{2}$ | A1 | 2 correct values for $x$ or 2 correct values for $y$. Allow unsimplified |
| $y = (\pm)$"$\frac{4}{3}$"$\Rightarrow x = \ldots$ or $x = (\pm)$"$\frac{3}{2}$"$\Rightarrow y = \ldots$ | M1 | Attempts at least one value of other variable having previously found and solved equation in one variable |
| $\left(\pm\frac{3}{2}, \mp\frac{4}{3}\right)$ or $x = \frac{3}{2}, y = -\frac{4}{3}$ and $x = -\frac{3}{2}, y = \frac{4}{3}$ | A1 | Correct values, simplified and paired correctly. Do not isw — mark final answer |
**Total: (6) marks**
---\begin{enumerate}
\item The curve $C$ has equation
\end{enumerate}
$$81 y ^ { 3 } + 64 x ^ { 2 } y + 256 x = 0$$
(a) Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$ in terms of $x$ and $y$.\\
(b) Hence find the coordinates of the points on $C$ where $\frac { \mathrm { d } y } { \mathrm {~d} x } = 0$\\
\hfill \mbox{\textit{Edexcel C34 2019 Q4 [11]}}