| Exam Board | Edexcel |
|---|---|
| Module | P4 (Pure Mathematics 4) |
| Year | 2022 |
| Session | January |
| Marks | 6 |
| 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 requiring students to differentiate both sides with respect to x using the product rule, solve for dy/dx, substitute the point (2,3) to find the gradient, then write the tangent equation. While it involves multiple steps and careful algebraic manipulation, it follows a standard procedure taught in P4/Further Pure with no novel insight required, making it slightly easier than average. |
| Spec | 1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| E.g. \(xy^2 \rightarrow (\ldots+)2xy\frac{dy}{dx}\) or \(x^2y \rightarrow (\ldots+)x^2\frac{dy}{dx}\) | M1 | Attempt at implicit differentiation in some form. Allow \(y'\) for \(\frac{dy}{dx}\). For \(xy^2\) look for \(kxy\frac{dy}{dx}\), condone \(ky\frac{dy}{dx}\). For \(x^2y\) look for \(kx^2\frac{dy}{dx}\), condone \(kx\frac{dy}{dx}\) |
| \(xy^2 \rightarrow y^2 + kxy\frac{dy}{dx}\) or \(x^2y \rightarrow kxy + x^2\frac{dy}{dx}\) | dM1 | Attempting product rule on either \(xy^2\) or \(x^2y\). Award for \(xy^2 \rightarrow y^2 + kxy\frac{dy}{dx}\) or \(x^2y \rightarrow kxy + x^2\frac{dy}{dx}\) |
| \(y^2 + 2xy\frac{dy}{dx} = 2xy + x^2\frac{dy}{dx}\) | A1 | Ignore any spurious \("\frac{dy}{dx}="\) in front of equation |
| \((x^2-2xy)\frac{dy}{dx} = y^2 - 2xy \Rightarrow \frac{dy}{dx} = \frac{y^2-2xy}{x^2-2xy} \Rightarrow \frac{dy}{dx}\bigg | _P = \frac{3^2-12}{2^2-12} = \ldots\) or \(3^2 + 2\times2\times3\frac{dy}{dx}\bigg | _P = 2\times2\times3 + 2^2\frac{dy}{dx}\bigg |
| Tangent is \(y - 3 = "\frac{3}{8}"(x-2)\) | dM1 | Uses their non-zero \(\frac{dy}{dx}\) at \((2,3)\) to find equation of tangent. Look for \(y-3="m"(x-2)\) or \(y="m"x+c\) where \(m\) is gradient attempt at \(P\), followed by attempt to find \(c\) using \((2,3)\) with 2 and 3 positioned correctly. Depends on previous M |
| \(\Rightarrow 3x - 8y + 18 = 0\) | A1 | \(3x-8y+18=0\) or any non-zero integer multiple with all terms on one side |
## Question 1:
| Answer/Working | Mark | Guidance |
|---|---|---|
| E.g. $xy^2 \rightarrow (\ldots+)2xy\frac{dy}{dx}$ or $x^2y \rightarrow (\ldots+)x^2\frac{dy}{dx}$ | M1 | Attempt at implicit differentiation in some form. Allow $y'$ for $\frac{dy}{dx}$. For $xy^2$ look for $kxy\frac{dy}{dx}$, condone $ky\frac{dy}{dx}$. For $x^2y$ look for $kx^2\frac{dy}{dx}$, condone $kx\frac{dy}{dx}$ |
| $xy^2 \rightarrow y^2 + kxy\frac{dy}{dx}$ or $x^2y \rightarrow kxy + x^2\frac{dy}{dx}$ | dM1 | Attempting product rule on either $xy^2$ or $x^2y$. Award for $xy^2 \rightarrow y^2 + kxy\frac{dy}{dx}$ **or** $x^2y \rightarrow kxy + x^2\frac{dy}{dx}$ |
| $y^2 + 2xy\frac{dy}{dx} = 2xy + x^2\frac{dy}{dx}$ | A1 | Ignore any spurious $"\frac{dy}{dx}="$ in front of equation |
| $(x^2-2xy)\frac{dy}{dx} = y^2 - 2xy \Rightarrow \frac{dy}{dx} = \frac{y^2-2xy}{x^2-2xy} \Rightarrow \frac{dy}{dx}\bigg|_P = \frac{3^2-12}{2^2-12} = \ldots$ or $3^2 + 2\times2\times3\frac{dy}{dx}\bigg|_P = 2\times2\times3 + 2^2\frac{dy}{dx}\bigg|_P \Rightarrow \frac{dy}{dx}\bigg|_P = \frac{9-12}{4-12} = \ldots$ | M1 | Attempts to find $\frac{dy}{dx}$ at $P$. Must substitute **both** $x$ and $y$ values and rearrange. Allow if at least two terms in $\frac{dy}{dx}$ in their differentiation. Must be some correct substitution |
| Tangent is $y - 3 = "\frac{3}{8}"(x-2)$ | dM1 | Uses their **non-zero** $\frac{dy}{dx}$ at $(2,3)$ to find equation of tangent. Look for $y-3="m"(x-2)$ or $y="m"x+c$ where $m$ is gradient attempt at $P$, followed by attempt to find $c$ using $(2,3)$ with 2 and 3 **positioned correctly**. Depends on previous M |
| $\Rightarrow 3x - 8y + 18 = 0$ | A1 | $3x-8y+18=0$ or any non-zero integer multiple with all terms on one side |
---\begin{enumerate}
\item The curve $C$ has equation
\end{enumerate}
$$x y ^ { 2 } = x ^ { 2 } y + 6 \quad x \neq 0 \quad y \neq 0$$
Find an equation for the tangent to $C$ at the point $P ( 2,3 )$, giving your answer in the form $a x + b y + c = 0$ where $a , b$ and $c$ are integers.\\
(6)
\hfill \mbox{\textit{Edexcel P4 2022 Q1 [6]}}