| Exam Board | Edexcel |
|---|---|
| Module | P4 (Pure Mathematics 4) |
| Year | 2022 |
| Session | January |
| Marks | 8 |
| 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 standard calculus techniques. Part (a) is simple geometry (area of equilateral triangle), part (b) uses chain rule with given formulas, and part (c) applies dV/dt = (dV/dA)(dA/dt). While it involves a 3D shape, no geometric insight is needed since formulas are provided. Slightly easier than average due to clear structure and routine differentiation. |
| Spec | 1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates6.02k Power: rate of doing work |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Area of one face \(= \frac{1}{2}x \times x \times \sin 60° = \frac{1}{2}x^2\frac{\sqrt{3}}{2}\) | M1 | Any correct method to establish a surd form for area of one face. Must evaluate trigonometric terms. Alternatives include \(\frac{1}{2} \times x \times \sqrt{x^2 - \left(\frac{1}{2}x\right)^2} = \frac{1}{2}x\sqrt{\frac{3}{4}x^2}\) |
| Surface area of icosahedron \(= 20 \times \frac{\sqrt{3}}{4}x^2 = 5\sqrt{3}x^2\) | A1* | Correct result achieved after showing side area of one face. If \(SA = 20\times\left(\frac{1}{2}\times x \times x\sin 60\right) = 5\sqrt{3}x^2\) scores M1 A0 for lack of working |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\frac{dA}{dx} = 10\sqrt{3}x\) and \(\frac{dV}{dx} = \frac{15}{12}(3+\sqrt{5})x^2\) | M1 | Attempts derivatives of both \(A\) and \(V\) wrt \(x\), achieving \(\frac{dA}{dx} = \alpha x\) and \(\frac{dV}{dx} = \beta x^2\). Note \(\frac{5}{12}(3+\sqrt{5})x^3\) sometimes expanded first |
| \(\frac{dV}{dA} = \frac{dV}{dx} \times \frac{dx}{dA} = \frac{dV}{dx} \div \frac{dA}{dx} = \ldots\) | M1 | Applies correct chain rule with their derivatives. Condone poor notation |
| \(= \frac{15(3+\sqrt{5})x^2}{12\times10\sqrt{3}x} = \frac{(3+\sqrt{5})x}{8\sqrt{3}}\) | A1* | Applies chain rule to achieve given result, with suitable intermediate step seen with \(\frac{dV}{dA} = \ldots\) appearing somewhere |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\frac{dA}{dt} = 0.025\) | B1 | Interprets the rate of change correctly to state or clearly imply \(\frac{dA}{dt} = 0.025\) |
| \(\frac{dV}{dt} = \frac{dV}{dA} \times \frac{dA}{dt} = \frac{(3+\sqrt{5})x}{8\sqrt{3}} \times 0.025 = \ldots\) | M1 | Applies the chain rule appropriately to find a value for \(\frac{dV}{dt}\) using \(x=2\) |
| When \(x=2\), \(\frac{dV}{dt} = \frac{2(3+\sqrt{5})}{8\sqrt{3}} \times 0.025 = \ldots\) | ||
| \(\approx 0.019 \text{ cm}^3\text{s}^{-1}\) | A1 | awrt \(0.019\), condone lack of units |
# Question 4:
## Part (a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Area of one face $= \frac{1}{2}x \times x \times \sin 60° = \frac{1}{2}x^2\frac{\sqrt{3}}{2}$ | M1 | Any correct method to establish a surd form for area of one face. Must evaluate trigonometric terms. Alternatives include $\frac{1}{2} \times x \times \sqrt{x^2 - \left(\frac{1}{2}x\right)^2} = \frac{1}{2}x\sqrt{\frac{3}{4}x^2}$ |
| Surface area of icosahedron $= 20 \times \frac{\sqrt{3}}{4}x^2 = 5\sqrt{3}x^2$ | A1* | Correct result achieved after showing side area of one face. If $SA = 20\times\left(\frac{1}{2}\times x \times x\sin 60\right) = 5\sqrt{3}x^2$ scores M1 A0 for lack of working |
## Part (b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\frac{dA}{dx} = 10\sqrt{3}x$ and $\frac{dV}{dx} = \frac{15}{12}(3+\sqrt{5})x^2$ | M1 | Attempts derivatives of both $A$ and $V$ wrt $x$, achieving $\frac{dA}{dx} = \alpha x$ and $\frac{dV}{dx} = \beta x^2$. Note $\frac{5}{12}(3+\sqrt{5})x^3$ sometimes expanded first |
| $\frac{dV}{dA} = \frac{dV}{dx} \times \frac{dx}{dA} = \frac{dV}{dx} \div \frac{dA}{dx} = \ldots$ | M1 | Applies correct chain rule with their derivatives. Condone poor notation |
| $= \frac{15(3+\sqrt{5})x^2}{12\times10\sqrt{3}x} = \frac{(3+\sqrt{5})x}{8\sqrt{3}}$ | A1* | Applies chain rule to achieve given result, with suitable intermediate step seen with $\frac{dV}{dA} = \ldots$ appearing somewhere |
## Part (c):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\frac{dA}{dt} = 0.025$ | B1 | Interprets the rate of change correctly to state or clearly imply $\frac{dA}{dt} = 0.025$ |
| $\frac{dV}{dt} = \frac{dV}{dA} \times \frac{dA}{dt} = \frac{(3+\sqrt{5})x}{8\sqrt{3}} \times 0.025 = \ldots$ | M1 | Applies the chain rule appropriately **to find a value for** $\frac{dV}{dt}$ using $x=2$ |
| When $x=2$, $\frac{dV}{dt} = \frac{2(3+\sqrt{5})}{8\sqrt{3}} \times 0.025 = \ldots$ | | |
| $\approx 0.019 \text{ cm}^3\text{s}^{-1}$ | A1 | awrt $0.019$, condone lack of units |
---4.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{fe07afad-9cfc-48c0-84f1-5717f81977d4-10_378_332_246_808}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
A regular icosahedron of side length $x \mathrm {~cm}$, shown in Figure 1, is expanding uniformly. The icosahedron consists of 20 congruent equilateral triangular faces of side length $x \mathrm {~cm}$.
\begin{enumerate}[label=(\alph*)]
\item Show that the surface area, $A \mathrm {~cm} ^ { 2 }$, of the icosahedron is given by
$$A = 5 \sqrt { 3 } x ^ { 2 }$$
Given that the volume, $V \mathrm {~cm} ^ { 3 }$, of the icosahedron is given by
$$V = \frac { 5 } { 12 } ( 3 + \sqrt { 5 } ) x ^ { 3 }$$
\item show that $\frac { \mathrm { d } V } { \mathrm {~d} A } = \frac { ( 3 + \sqrt { 5 } ) x } { 8 \sqrt { 3 } }$
The surface area of the icosahedron is increasing at a constant rate of $0.025 \mathrm {~cm} ^ { 2 } \mathrm {~s} ^ { - 1 }$
\item Find the rate of change of the volume of the icosahedron when $x = 2$, giving your answer to 2 significant figures.
\end{enumerate}
\hfill \mbox{\textit{Edexcel P4 2022 Q4 [8]}}