| Exam Board | Edexcel |
|---|---|
| Module | Paper 2 (Paper 2) |
| Year | 2024 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric differentiation |
| Type | Find tangent equation at parameter |
| Difficulty | Moderate -0.3 This is a straightforward parametric differentiation question requiring standard techniques: finding the parameter value at P, computing dy/dx using the chain rule, and forming the tangent equation. Part (b) requires finding maximum y by setting dy/dt = 0. All steps are routine A-level procedures with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x=4,\ y=2 \Rightarrow t=-1\) | B1 | 2.2a |
| \(\frac{dy}{dx} = \frac{dy}{dt} \times \frac{dt}{dx} = -3t^2 \times \frac{1}{2(t+3)}\) | M1 | 1.1b – Attempt parametric differentiation; both parameters must be differentiated |
| \(\frac{dy}{dx} = -3(-1)^2 \times \frac{1}{2(-1+3)} = -\frac{3}{4}\) | M1 | 1.1b – Substitutes numerical value of \(t\) (not 4) into \(\frac{dy}{dx}\) |
| \(y-2 = -\frac{3}{4}(x-4)\) or \(y = -\frac{3}{4}x + c \Rightarrow c=5\) | ddM1 | 2.1 – Correct straight line method using their \(\frac{dy}{dx}\) with \(x=4, y=2\) correctly placed |
| \(3x + 4y = 20\) | A1* | 1.1b – Correct equation (printed answer, needs intermediate step shown) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Maximum height is \(9\text{m}\) | B1 | 3.4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dy}{dx} = \frac{dy}{dt} \times \frac{dt}{dx} = -3t^2 \times \frac{1}{2(t+3)}\) | B1 | Either a correct expression for \(\frac{dy}{dx}\) in terms of \(x\) and/or \(y\) following a correct \(\frac{dy}{dx}\) in terms of \(t\), or for \(t = -1\) seen anywhere |
| \(-3t^2 \times \frac{1}{2(t+3)} = -3(1-y)^{\frac{2}{3}} \times \frac{1}{2\sqrt{x}} = -3(1-2)^{\frac{2}{3}} \times \frac{1}{2\sqrt{4}} = -\frac{3}{4}\) | M1 | Attempts to use \(\frac{dy}{dx} = \frac{dy}{dt} \times \frac{dt}{dx}\) with differentiated equations; must attempt to differentiate both parameters and divide; using \(\frac{dy}{dx} = \frac{y}{x}\) scores M0 |
| Expressing \(\frac{dy}{dx}\) in terms of \(x\) and/or \(y\), using \(x=4\), \(y=2\) | M1 | Attempts to express their \(\frac{dy}{dx}\) (in terms of \(t\)) in terms of \(x\) and/or \(y\) and uses \(x=4\) and \(y=2\) correctly |
| \(y - 2 = -\frac{3}{4}(x-4) \Rightarrow 3x + 4y = 20\) | ddM1 A1* | Applies correct straight line method with their \(\frac{dy}{dx}\) from parametric differentiation, with \(x=4\), \(y=2\). Correct equation with no errors; needs at least one intermediate step. Depends on both previous M marks |
## Question 10:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x=4,\ y=2 \Rightarrow t=-1$ | B1 | 2.2a |
| $\frac{dy}{dx} = \frac{dy}{dt} \times \frac{dt}{dx} = -3t^2 \times \frac{1}{2(t+3)}$ | M1 | 1.1b – Attempt parametric differentiation; both parameters must be differentiated |
| $\frac{dy}{dx} = -3(-1)^2 \times \frac{1}{2(-1+3)} = -\frac{3}{4}$ | M1 | 1.1b – Substitutes numerical value of $t$ (not 4) into $\frac{dy}{dx}$ |
| $y-2 = -\frac{3}{4}(x-4)$ or $y = -\frac{3}{4}x + c \Rightarrow c=5$ | ddM1 | 2.1 – Correct straight line method using their $\frac{dy}{dx}$ with $x=4, y=2$ correctly placed |
| $3x + 4y = 20$ | A1* | 1.1b – Correct equation (printed answer, needs intermediate step shown) |
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### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Maximum height is $9\text{m}$ | B1 | 3.4 |
# Question 10 (Parametric Differentiation Alternative):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dy}{dx} = \frac{dy}{dt} \times \frac{dt}{dx} = -3t^2 \times \frac{1}{2(t+3)}$ | B1 | Either a correct expression for $\frac{dy}{dx}$ in terms of $x$ and/or $y$ following a correct $\frac{dy}{dx}$ in terms of $t$, **or** for $t = -1$ seen anywhere |
| $-3t^2 \times \frac{1}{2(t+3)} = -3(1-y)^{\frac{2}{3}} \times \frac{1}{2\sqrt{x}} = -3(1-2)^{\frac{2}{3}} \times \frac{1}{2\sqrt{4}} = -\frac{3}{4}$ | M1 | Attempts to use $\frac{dy}{dx} = \frac{dy}{dt} \times \frac{dt}{dx}$ with differentiated equations; must attempt to differentiate both parameters and divide; using $\frac{dy}{dx} = \frac{y}{x}$ scores M0 |
| Expressing $\frac{dy}{dx}$ in terms of $x$ and/or $y$, using $x=4$, $y=2$ | M1 | Attempts to express their $\frac{dy}{dx}$ (in terms of $t$) in terms of $x$ and/or $y$ **and** uses $x=4$ and $y=2$ correctly |
| $y - 2 = -\frac{3}{4}(x-4) \Rightarrow 3x + 4y = 20$ | ddM1 A1* | Applies correct straight line method with their $\frac{dy}{dx}$ from parametric differentiation, with $x=4$, $y=2$. Correct equation with no errors; needs at least one intermediate step. Depends on both previous M marks |
---10.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{2ce10759-9ce6-47a1-b55d-d22082f88f55-26_707_992_246_539}
\captionsetup{labelformat=empty}
\caption{Figure 4}
\end{center}
\end{figure}
Figure 4 shows a sketch of the curve $C$ with parametric equations
$$x = ( t + 3 ) ^ { 2 } \quad y = 1 - t ^ { 3 } \quad - 2 \leqslant t \leqslant 1$$
The point $P$ with coordinates $( 4,2 )$ lies on $C$.
\begin{enumerate}[label=(\alph*)]
\item Using parametric differentiation, show that the tangent to $C$ at $P$ has equation
$$3 x + 4 y = 20$$
The curve $C$ is used to model the profile of a slide at a water park.\\
Units are in metres, with $y$ being the height of the slide above water level.
\item Find, according to the model, the greatest height of the slide above water level.
\end{enumerate}
\hfill \mbox{\textit{Edexcel Paper 2 2024 Q10 [6]}}