| Exam Board | Edexcel |
|---|---|
| Module | P4 (Pure Mathematics 4) |
| Year | 2023 |
| Session | October |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Related rates with spheres, circles, and cubes |
| Difficulty | Standard +0.3 This is a standard related rates problem requiring chain rule application with familiar geometric formulas (S = 6x², V = x³). Part (a) involves straightforward differentiation and algebraic manipulation, while part (b) requires combining results using dV/dt = dV/dx × dx/dt. The 'show that' format guides students to the answer, and the techniques are routine for P4 level, making this slightly easier than average. |
| Spec | 1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| States or uses \(S=6x^2\) or \(\frac{dS}{dt}=4\) | B1 | Ignore any units. \(S=6x^2\) may be implied by \(\frac{dS}{dx}=12x\) |
| States or uses \(S=6x^2\) and \(\frac{dS}{dt}=4\) | B1 | Both required |
| Attempts \(\frac{dS}{dt}=\frac{dS}{dx}\times\frac{dx}{dt} \Rightarrow 4=12x\times\frac{dx}{dt}\Rightarrow\frac{dx}{dt}=...\) | M1 | Attempt chain rule with \(\frac{dS}{dx}=kx\) and \(\frac{dS}{dt}\) a constant |
| \(\Rightarrow\frac{dx}{dt}=\frac{1}{3x}\) o.e. | A1 | Allow e.g. \(\frac{1/3}{x}\), \(\frac{4}{12x}\). The "\(\frac{dx}{dt}=\)" must appear |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| States or uses \(S=6x^2\) or \(\frac{dS}{dt}=4\) | B1 | |
| States or uses \(S=6x^2\) and \(\frac{dS}{dt}=4\) | B1 | |
| \(\frac{dS}{dt}=4\Rightarrow S=4t+c\Rightarrow S=6x^2\Rightarrow 6x^2=4t+c\Rightarrow 12x\frac{dx}{dt}=4\Rightarrow\frac{dx}{dt}=...\) | M1 | Integrates \(\frac{dS}{dt}=4\) to get \(S=4t+c\), substitutes \(S=6x^2\), differentiates |
| \(\Rightarrow\frac{dx}{dt}=\frac{1}{3x}\) o.e. | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| States or uses \(\frac{dV}{dx}=3x^2\) | B1 | Allow this mark anywhere in the question |
| Attempts \(\frac{dV}{dt}=\frac{dV}{dx}\times\frac{dx}{dt}\Rightarrow\frac{dV}{dt}=3x^2\times\frac{1}{3x}=x\) | M1 | Chain rule attempt |
| \(\Rightarrow\frac{dV}{dt}=V^{\frac{1}{3}}\) | A1 | May be implied by e.g. \(\frac{dV}{dt}=x,\ x=V^p\Rightarrow p=\frac{1}{3}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(V=x^3,\ S=6x^2\Rightarrow S=6V^{\frac{2}{3}}\Rightarrow\frac{dS}{dV}=4V^{-\frac{1}{3}}\) | B1 | |
| Attempts \(\frac{dV}{dt}=\frac{dV}{dS}\times\frac{dS}{dt}=\frac{1}{4V^{-\frac{1}{3}}}\times 4\) | M1 | |
| \(\Rightarrow\frac{dV}{dt}=V^{\frac{1}{3}}\) | A1 |
# Question 2:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| States or uses $S=6x^2$ **or** $\frac{dS}{dt}=4$ | B1 | Ignore any units. $S=6x^2$ may be implied by $\frac{dS}{dx}=12x$ |
| States or uses $S=6x^2$ **and** $\frac{dS}{dt}=4$ | B1 | Both required |
| Attempts $\frac{dS}{dt}=\frac{dS}{dx}\times\frac{dx}{dt} \Rightarrow 4=12x\times\frac{dx}{dt}\Rightarrow\frac{dx}{dt}=...$ | M1 | Attempt chain rule with $\frac{dS}{dx}=kx$ and $\frac{dS}{dt}$ a constant |
| $\Rightarrow\frac{dx}{dt}=\frac{1}{3x}$ o.e. | A1 | Allow e.g. $\frac{1/3}{x}$, $\frac{4}{12x}$. The "$\frac{dx}{dt}=$" must appear |
**Alternative (a):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| States or uses $S=6x^2$ **or** $\frac{dS}{dt}=4$ | B1 | |
| States or uses $S=6x^2$ **and** $\frac{dS}{dt}=4$ | B1 | |
| $\frac{dS}{dt}=4\Rightarrow S=4t+c\Rightarrow S=6x^2\Rightarrow 6x^2=4t+c\Rightarrow 12x\frac{dx}{dt}=4\Rightarrow\frac{dx}{dt}=...$ | M1 | Integrates $\frac{dS}{dt}=4$ to get $S=4t+c$, substitutes $S=6x^2$, differentiates |
| $\Rightarrow\frac{dx}{dt}=\frac{1}{3x}$ o.e. | A1 | |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| States or uses $\frac{dV}{dx}=3x^2$ | B1 | Allow this mark anywhere in the question |
| Attempts $\frac{dV}{dt}=\frac{dV}{dx}\times\frac{dx}{dt}\Rightarrow\frac{dV}{dt}=3x^2\times\frac{1}{3x}=x$ | M1 | Chain rule attempt |
| $\Rightarrow\frac{dV}{dt}=V^{\frac{1}{3}}$ | A1 | May be implied by e.g. $\frac{dV}{dt}=x,\ x=V^p\Rightarrow p=\frac{1}{3}$ |
**Alternative (b):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $V=x^3,\ S=6x^2\Rightarrow S=6V^{\frac{2}{3}}\Rightarrow\frac{dS}{dV}=4V^{-\frac{1}{3}}$ | B1 | |
| Attempts $\frac{dV}{dt}=\frac{dV}{dS}\times\frac{dS}{dt}=\frac{1}{4V^{-\frac{1}{3}}}\times 4$ | M1 | |
| $\Rightarrow\frac{dV}{dt}=V^{\frac{1}{3}}$ | A1 | |2.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{7f5fc83d-ab7c-4edb-a2c6-7a58f1357d5a-04_271_223_246_922}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Figure 1 shows a cube which is increasing in size.\\
At time $t$ seconds,
\begin{itemize}
\item the length of each edge of the cube is $x \mathrm {~cm}$
\item the surface area of the cube is $S \mathrm {~cm} ^ { 2 }$
\item the volume of the cube is $V \mathrm {~cm} ^ { 3 }$
\end{itemize}
Given that the surface area of the cube is increasing at a constant rate of $4 \mathrm {~cm} ^ { 2 } \mathrm {~s} ^ { - 1 }$
\begin{enumerate}[label=(\alph*)]
\item show that $\frac { \mathrm { d } x } { \mathrm {~d} t } = \frac { k } { x }$ where $k$ is a constant to be found,
\item show that $\frac { \mathrm { d } V } { \mathrm {~d} t } = V ^ { p }$ where $p$ is a constant to be found.
\end{enumerate}
\hfill \mbox{\textit{Edexcel P4 2023 Q2 [7]}}