| Exam Board | Edexcel |
|---|---|
| Module | P4 (Pure Mathematics 4) |
| Year | 2022 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Implicit equations and differentiation |
| Type | Find constant from gradient condition |
| Difficulty | Standard +0.3 This is a straightforward implicit differentiation problem requiring standard application of the product rule and chain rule, followed by algebraic manipulation to find a constant. The working is methodical rather than requiring insight, and finding k from the turning point condition is a routine substitution into the curve equation. Slightly easier than average due to clear structure and standard techniques. |
| Spec | 1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \((8)y^3\rightarrow(8\times)3y^2\frac{dy}{dx}\) | B1 | For \(y^3\rightarrow 3y^2\frac{dy}{dx}\); allow if seen in aside working without the 8 |
| \(-9kx^2y\rightarrow...kxy\pm...-9kx^2\frac{dy}{dx}\) | M1 | Correct attempt at implicit differentiation on \(-9kx^2y\) |
| \(48x^2-18kxy-9kx^2\frac{dy}{dx}+24y^2\frac{dy}{dx}=0\Rightarrow\frac{dy}{dx}(24y^2-9kx^2)=18kxy-48x^2\Rightarrow\frac{dy}{dx}=...\) | M1 | Collects both \(\frac{dy}{dx}\) terms, collects non-\(\frac{dy}{dx}\) terms other side, factorises and divides; must have two \(\frac{dy}{dx}\) terms |
| \(\frac{dy}{dx}=\frac{6kxy-16x^2}{8y^2-3kx^2}\) * | A1* | Achieves given answer with no errors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{dy}{dx}=0, x=\frac{5}{2}\Rightarrow\frac{6k(\frac{5}{2})y-16(\frac{5}{2})^2}{8y^2-3k(\frac{5}{2})^2}=0\) or \(x=\frac{5}{2}\Rightarrow 16(\frac{5}{2})^3-9k(\frac{5}{2})^2y+8y^3=875\) | M1 | Uses information to produce one equation in \(k\) and \(y\); sets \(\frac{dy}{dx}=0\) and substitutes \(x=\frac{5}{2}\), or substitutes \(x=\frac{5}{2}\) into given equation; allow one slip |
| \(15ky-100=0\) or \(250-\frac{225}{4}ky+8y^3=875\) | A1 | Correct equation without fraction and with simplified coefficients |
| E.g. \(16(\frac{5}{2})^3-9k(\frac{5}{2})^2(\frac{20}{3k})+8(\frac{20}{3k})^3=875\Rightarrow k^3=...\left(=\frac{64}{27}\right)\Rightarrow k=...\) | M1 | Complete method to find \(k\); solves equations simultaneously to achieve value for \(k\); may find \(y\) first |
| \(k=\frac{4}{3}\) | A1 |
# Question 4:
## Part (a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $(8)y^3\rightarrow(8\times)3y^2\frac{dy}{dx}$ | B1 | For $y^3\rightarrow 3y^2\frac{dy}{dx}$; allow if seen in aside working without the 8 |
| $-9kx^2y\rightarrow...kxy\pm...-9kx^2\frac{dy}{dx}$ | M1 | Correct attempt at implicit differentiation on $-9kx^2y$ |
| $48x^2-18kxy-9kx^2\frac{dy}{dx}+24y^2\frac{dy}{dx}=0\Rightarrow\frac{dy}{dx}(24y^2-9kx^2)=18kxy-48x^2\Rightarrow\frac{dy}{dx}=...$ | M1 | Collects both $\frac{dy}{dx}$ terms, collects non-$\frac{dy}{dx}$ terms other side, factorises and divides; must have two $\frac{dy}{dx}$ terms |
| $\frac{dy}{dx}=\frac{6kxy-16x^2}{8y^2-3kx^2}$ * | A1* | Achieves given answer with no errors |
## Part (b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{dy}{dx}=0, x=\frac{5}{2}\Rightarrow\frac{6k(\frac{5}{2})y-16(\frac{5}{2})^2}{8y^2-3k(\frac{5}{2})^2}=0$ or $x=\frac{5}{2}\Rightarrow 16(\frac{5}{2})^3-9k(\frac{5}{2})^2y+8y^3=875$ | M1 | Uses information to produce one equation in $k$ and $y$; sets $\frac{dy}{dx}=0$ and substitutes $x=\frac{5}{2}$, or substitutes $x=\frac{5}{2}$ into given equation; allow one slip |
| $15ky-100=0$ or $250-\frac{225}{4}ky+8y^3=875$ | A1 | Correct equation without fraction and with simplified coefficients |
| E.g. $16(\frac{5}{2})^3-9k(\frac{5}{2})^2(\frac{20}{3k})+8(\frac{20}{3k})^3=875\Rightarrow k^3=...\left(=\frac{64}{27}\right)\Rightarrow k=...$ | M1 | Complete method to find $k$; solves equations simultaneously to achieve value for $k$; may find $y$ first |
| $k=\frac{4}{3}$ | A1 | |\begin{enumerate}
\item In this question you must show all stages of your working.
\end{enumerate}
\section*{Solutions relying on calculator technology are not acceptable.}
A curve has equation
$$16 x ^ { 3 } - 9 k x ^ { 2 } y + 8 y ^ { 3 } = 875$$
where $k$ is a constant.\\
(a) Show that
$$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 6 k x y - 16 x ^ { 2 } } { 8 y ^ { 2 } - 3 k x ^ { 2 } }$$
Given that the curve has a turning point at $x = \frac { 5 } { 2 }$\\
(b) find the value of $k$
\hfill \mbox{\textit{Edexcel P4 2022 Q4 [8]}}