| Exam Board | Edexcel |
|---|---|
| Module | P4 (Pure Mathematics 4) |
| Year | 2024 |
| Session | January |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Related rates with explicitly given non-geometric algebraic relationships |
| Difficulty | Moderate -0.3 This is a straightforward related rates problem requiring Pythagoras' theorem, differentiation of a given formula, and substitution. Part (a) is trivial (l = √(r² + 25)), and part (b) follows a standard template: differentiate S with respect to t using the chain rule, substitute given values. While it requires multiple steps, it's a textbook application with no novel insight needed, making it slightly easier than average. |
| Spec | 1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(l = \sqrt{25 + r^2}\) | B1 | Correct expression. Accept \(l = \sqrt{5^2 + r^2}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(S = \pi r^2 + \pi r\sqrt{25+r^2} \Rightarrow \frac{dS}{dr} = 2\pi r + \pi\sqrt{25+r^2} + \pi r \cdot \frac{1}{2}(25+r^2)^{-\frac{1}{2}} \cdot 2r\) | M1 | Differentiates surface area using product and chain rules on term of form \(Kr\sqrt{\pm L \pm r^2}\) |
| Correct differentiation of curved surface area term | A1 | Need not be simplified |
| \(\frac{dS}{dt} = \frac{dS}{dr} \times \frac{dr}{dt} = \ldots \times 3\) | M1 | Applies chain rule correctly with \(\frac{dr}{dt} = 3\) and their expression for \(\frac{dS}{dr}\) |
| \(= 81.5 \text{ (cm}^2/\text{min)}\) | A1 | Accept \(\left(9 + \frac{177\sqrt{109}}{109}\right)\pi\) or equivalent; awrt 81.5 |
## Question 4:
### Part (a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $l = \sqrt{25 + r^2}$ | B1 | Correct expression. Accept $l = \sqrt{5^2 + r^2}$ |
### Part (b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $S = \pi r^2 + \pi r\sqrt{25+r^2} \Rightarrow \frac{dS}{dr} = 2\pi r + \pi\sqrt{25+r^2} + \pi r \cdot \frac{1}{2}(25+r^2)^{-\frac{1}{2}} \cdot 2r$ | M1 | Differentiates surface area using product and chain rules on term of form $Kr\sqrt{\pm L \pm r^2}$ |
| Correct differentiation of curved surface area term | A1 | Need not be simplified |
| $\frac{dS}{dt} = \frac{dS}{dr} \times \frac{dr}{dt} = \ldots \times 3$ | M1 | Applies chain rule correctly with $\frac{dr}{dt} = 3$ and their expression for $\frac{dS}{dr}$ |
| $= 81.5 \text{ (cm}^2/\text{min)}$ | A1 | Accept $\left(9 + \frac{177\sqrt{109}}{109}\right)\pi$ or equivalent; awrt 81.5 |
---4.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{6f577461-24b7-4615-b58b-e67597fd9675-12_595_588_248_740}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
A cone, shown in Figure 2, has
\begin{itemize}
\item fixed height 5 cm
\item base radius $r \mathrm {~cm}$
\item slant height $l \mathrm {~cm}$
\begin{enumerate}[label=(\alph*)]
\item Find an expression for $l$ in terms of $r$
\end{itemize}
Given that the base radius is increasing at a constant rate of 3 cm per minute,
\item find the rate at which the total surface area of the cone is changing when the radius of the cone is 1.5 cm . Give your answer in $\mathrm { cm } ^ { 2 }$ per minute to one decimal place.\\[0pt]
[The total surface area, $S$, of a cone is given by the formula $S = \pi r ^ { 2 } + \pi r l$ ]
\end{enumerate}
\hfill \mbox{\textit{Edexcel P4 2024 Q4 [5]}}