| Exam Board | Edexcel |
|---|---|
| Module | C34 (Core Mathematics 3 & 4) |
| Year | 2015 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Related rates with explicitly given non-geometric algebraic relationships |
| Difficulty | Standard +0.3 This is a straightforward related rates problem requiring students to differentiate the volume formula with respect to time using the chain rule and substitute given values. While it involves multiple variables and the product rule, the setup is clear, the algebra is routine, and it's a standard application of implicit differentiation—slightly easier than average for C3/C4 material. |
| Spec | 1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(A = \pi x^2 \Rightarrow \frac{dA}{dx} = 2\pi x\) | B1 | Condone use of \(r\) throughout instead of \(x\) |
| \(\frac{dA}{dt} = \frac{dA}{dx}\times\frac{dx}{dt} \Rightarrow \frac{\pi}{20} = 2\pi x\cdot\frac{dx}{dt} \Rightarrow \frac{dx}{dt} = \frac{1}{40x} = \left(\frac{1}{80}\right)\) | M1A1 | |
| \(V = 6\pi x^3 \Rightarrow \frac{dV}{dx} = 18\pi x^2\) | B1 | |
| \(\frac{dV}{dt} = \frac{dV}{dx}\times\frac{dx}{dt} \Rightarrow \frac{dV}{dt} = 18\pi\cdot 2^2 \times \frac{1}{80} = \frac{9}{10}\pi\) | dM1A1 | Dependent M; correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(V=6\pi x^3\), \(A=\pi x^2 \Rightarrow V = 6\pi\left(\frac{A}{\pi}\right)^{\frac{3}{2}}\) | M1 | |
| \(\frac{dV}{dA} = 9\left(\frac{A}{\pi}\right)^{\frac{1}{2}}\) | B1 B1 | |
| \(\frac{dV}{dt} = \frac{dV}{dA}\times\frac{dA}{dt} \Rightarrow \frac{dV}{dt} = 9\left(\frac{4\pi}{\pi}\right)^{\frac{1}{2}}\times\frac{\pi}{20} = \frac{9}{10}\pi\) | dM1A1A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(A=\pi x^2 \Rightarrow \frac{dA}{dx}=2\pi x\) | B1 | |
| \(V=6\pi x^3 \Rightarrow \frac{dV}{dx}=18\pi x^2\) | B1 | |
| \(\frac{dV}{dt}=\frac{dV}{dx}\times\frac{dx}{dA}\times\frac{dA}{dt} = 18\pi x^2\times\frac{1}{2\pi x}\times\frac{\pi}{20}\) | M1A1 | |
| Put \(x=2\): \(\frac{9}{10}\pi\) | dM1A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dA}{dx} = 2\pi x\) | B1 | Correct statement |
| Uses chain rule \(\frac{dA}{dt} = \frac{dA}{dx} \times \frac{dx}{dt}\) or equivalent e.g. \(\frac{dx}{dt} = \frac{dA}{dt} \div \frac{dA}{dx}\) | M1 | Must use correct chain rule; with \(\frac{dA}{dt} = \frac{\pi}{20}\) and their \(\frac{dA}{dx}\) to calculate \(\frac{dx}{dt}\) |
| \(\frac{dx}{dt} = \frac{\pi/20}{2\pi x}\) | A1 | Correct expression; award in misread cases where correct CSA or TSA used correctly |
| \(\frac{dV}{dx} = 18\pi x^2\) | B1 | Correct statement |
| Uses \(\frac{dV}{dt} = \frac{dV}{dx} \times \frac{dx}{dt}\) with their \(\frac{dx}{dt}\) and \(\frac{dV}{dx}\) to calculate \(\frac{dV}{dt}\), with \(x=2\) substituted | dM1 | \(\frac{dV}{dx}\) should be in terms of one variable \(x\) (form \(kx^2\) not \(kxh\)) |
| \(\frac{9}{10}\pi\) or \(0.9\pi\) or \(\frac{18}{20}\pi\) or \(k=0.9\) etc | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Writes \(V\) in terms of \(A\): \(V = ..A^{\frac{3}{2}}\) | M1 | First M on epen; indicates way 2 |
| \(\frac{dV}{dA} = 9\left(\frac{A}{\pi}\right)^{\frac{1}{2}}\) | B2 | Both Bs on epen |
| Uses \(\frac{dV}{dt} = \frac{dV}{dA} \times \frac{dA}{dt}\) with \(\frac{dA}{dt} = \frac{\pi}{20}\) to find \(\frac{dV}{dt}\) | dM1 | Dependent on first M; second M on epen; \(\frac{dV}{dA}\) in terms of one variable \(A\) (form \(kA^{\frac{1}{2}}\)) |
| \(\frac{9}{10}\pi\) or \(0.9\pi\) or \(\frac{18}{20}\pi\) or \(k=0.9\) etc | A2 | Both As on epen |
# Question 10:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $A = \pi x^2 \Rightarrow \frac{dA}{dx} = 2\pi x$ | B1 | Condone use of $r$ throughout instead of $x$ |
| $\frac{dA}{dt} = \frac{dA}{dx}\times\frac{dx}{dt} \Rightarrow \frac{\pi}{20} = 2\pi x\cdot\frac{dx}{dt} \Rightarrow \frac{dx}{dt} = \frac{1}{40x} = \left(\frac{1}{80}\right)$ | M1A1 | |
| $V = 6\pi x^3 \Rightarrow \frac{dV}{dx} = 18\pi x^2$ | B1 | |
| $\frac{dV}{dt} = \frac{dV}{dx}\times\frac{dx}{dt} \Rightarrow \frac{dV}{dt} = 18\pi\cdot 2^2 \times \frac{1}{80} = \frac{9}{10}\pi$ | dM1A1 | Dependent M; correct answer |
## Way 2:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $V=6\pi x^3$, $A=\pi x^2 \Rightarrow V = 6\pi\left(\frac{A}{\pi}\right)^{\frac{3}{2}}$ | M1 | |
| $\frac{dV}{dA} = 9\left(\frac{A}{\pi}\right)^{\frac{1}{2}}$ | B1 B1 | |
| $\frac{dV}{dt} = \frac{dV}{dA}\times\frac{dA}{dt} \Rightarrow \frac{dV}{dt} = 9\left(\frac{4\pi}{\pi}\right)^{\frac{1}{2}}\times\frac{\pi}{20} = \frac{9}{10}\pi$ | dM1A1A1 | |
## Way 3:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $A=\pi x^2 \Rightarrow \frac{dA}{dx}=2\pi x$ | B1 | |
| $V=6\pi x^3 \Rightarrow \frac{dV}{dx}=18\pi x^2$ | B1 | |
| $\frac{dV}{dt}=\frac{dV}{dx}\times\frac{dx}{dA}\times\frac{dA}{dt} = 18\pi x^2\times\frac{1}{2\pi x}\times\frac{\pi}{20}$ | M1A1 | |
| Put $x=2$: $\frac{9}{10}\pi$ | dM1A1 | |
# Question 10:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dA}{dx} = 2\pi x$ | B1 | Correct statement |
| Uses chain rule $\frac{dA}{dt} = \frac{dA}{dx} \times \frac{dx}{dt}$ or equivalent e.g. $\frac{dx}{dt} = \frac{dA}{dt} \div \frac{dA}{dx}$ | M1 | Must use **correct** chain rule; with $\frac{dA}{dt} = \frac{\pi}{20}$ and their $\frac{dA}{dx}$ to calculate $\frac{dx}{dt}$ |
| $\frac{dx}{dt} = \frac{\pi/20}{2\pi x}$ | A1 | Correct expression; award in misread cases where correct CSA or TSA used correctly |
| $\frac{dV}{dx} = 18\pi x^2$ | B1 | Correct statement |
| Uses $\frac{dV}{dt} = \frac{dV}{dx} \times \frac{dx}{dt}$ with their $\frac{dx}{dt}$ and $\frac{dV}{dx}$ to calculate $\frac{dV}{dt}$, with $x=2$ substituted | dM1 | $\frac{dV}{dx}$ should be in terms of one variable $x$ (form $kx^2$ not $kxh$) |
| $\frac{9}{10}\pi$ or $0.9\pi$ or $\frac{18}{20}\pi$ or $k=0.9$ etc | A1 | |
**Way 2:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Writes $V$ in terms of $A$: $V = ..A^{\frac{3}{2}}$ | M1 | First M on epen; indicates way 2 |
| $\frac{dV}{dA} = 9\left(\frac{A}{\pi}\right)^{\frac{1}{2}}$ | B2 | Both Bs on epen |
| Uses $\frac{dV}{dt} = \frac{dV}{dA} \times \frac{dA}{dt}$ with $\frac{dA}{dt} = \frac{\pi}{20}$ to find $\frac{dV}{dt}$ | dM1 | Dependent on first M; second M on epen; $\frac{dV}{dA}$ in terms of one variable $A$ (form $kA^{\frac{1}{2}}$) |
| $\frac{9}{10}\pi$ or $0.9\pi$ or $\frac{18}{20}\pi$ or $k=0.9$ etc | A2 | Both As on epen |
---10.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{4c08fbab-283e-4c92-89a4-10f68f37e133-16_319_508_237_719}
\captionsetup{labelformat=empty}
\caption{Figure 4}
\end{center}
\end{figure}
Figure 4 shows a right circular cylindrical rod which is expanding as it is heated.\\
At time $t$ seconds the radius of the rod is $x \mathrm {~cm}$ and the length of the rod is $6 x \mathrm {~cm}$.\\
Given that the cross-sectional area of the rod is increasing at a constant rate of $\frac { \pi } { 20 } \mathrm {~cm} ^ { 2 } \mathrm {~s} ^ { - 1 }$, find the rate of increase of the volume of the rod when $x = 2$
Write your answer in the form $k \pi \mathrm {~cm} ^ { 3 } \mathrm {~s} ^ { - 1 }$ where $k$ is a rational number.
\hfill \mbox{\textit{Edexcel C34 2015 Q10 [6]}}