| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2008 |
| Session | June |
| Marks | 8 |
| 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 application of the chain rule to standard geometric formulas. Part (a) involves differentiating A = πx² with given dA/dt to find dx/dt, then part (b) uses this to find dV/dt for V = πx²(5x). The setup is clear, the calculus is routine C4 material, and no geometric insight is required—slightly easier than average due to its mechanical nature. |
| Spec | 1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dA}{dt} = 0.032\) | B1 | Seen or implied from working |
| \(A = \pi x^2 \Rightarrow \frac{dA}{dx} = 2\pi x\) | B1 | \(2\pi x\) by itself seen or implied |
| \(\frac{dx}{dt} = \frac{dA}{dt} \div \frac{dA}{dx} = (0.032)\frac{1}{2\pi x}\) | M1 | \(0.032 \div\) candidate's \(\frac{dA}{dx}\) |
| When \(x=2\): \(\frac{dx}{dt} = \frac{0.016}{2\pi} = 0.002546...\) cm s\(^{-1}\) | A1 cso | awrt 0.00255 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(V = \pi x^2(5x) = 5\pi x^3\) | B1 | \(V = \pi x^2(5x)\) or \(5\pi x^3\) |
| \(\frac{dV}{dx} = 15\pi x^2\) | B1\(\checkmark\) | Or ft from candidate's \(V\) in one variable |
| \(\frac{dV}{dt} = \frac{dV}{dx} \times \frac{dx}{dt} = 15\pi x^2 \cdot \left(\frac{0.016}{\pi x}\right)\) | M1\(\checkmark\) | Candidate's \(\frac{dV}{dx} \times \frac{dx}{dt}\) |
| When \(x=2\): \(\frac{dV}{dt} = 0.24(2) = 0.48\) cm\(^3\) s\(^{-1}\) | A1 cso | \(0.48\) or awrt \(0.48\) |
# Question 3:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dA}{dt} = 0.032$ | B1 | Seen or implied from working |
| $A = \pi x^2 \Rightarrow \frac{dA}{dx} = 2\pi x$ | B1 | $2\pi x$ by itself seen or implied |
| $\frac{dx}{dt} = \frac{dA}{dt} \div \frac{dA}{dx} = (0.032)\frac{1}{2\pi x}$ | M1 | $0.032 \div$ candidate's $\frac{dA}{dx}$ |
| When $x=2$: $\frac{dx}{dt} = \frac{0.016}{2\pi} = 0.002546...$ cm s$^{-1}$ | A1 cso | awrt 0.00255 |
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $V = \pi x^2(5x) = 5\pi x^3$ | B1 | $V = \pi x^2(5x)$ or $5\pi x^3$ |
| $\frac{dV}{dx} = 15\pi x^2$ | B1$\checkmark$ | Or ft from candidate's $V$ in one variable |
| $\frac{dV}{dt} = \frac{dV}{dx} \times \frac{dx}{dt} = 15\pi x^2 \cdot \left(\frac{0.016}{\pi x}\right)$ | M1$\checkmark$ | Candidate's $\frac{dV}{dx} \times \frac{dx}{dt}$ |
| When $x=2$: $\frac{dV}{dt} = 0.24(2) = 0.48$ cm$^3$ s$^{-1}$ | A1 cso | $0.48$ or awrt $0.48$ |
---3.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{fb1924cc-9fa3-4fde-ba4d-6fb095f7f70b-04_444_705_205_623}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
Figure 2 shows a right circular cylindrical metal rod which is expanding as it is heated. After $t$ seconds the radius of the rod is $x \mathrm {~cm}$ and the length of the rod is $5 x \mathrm {~cm}$. The cross-sectional area of the rod is increasing at the constant rate of $0.032 \mathrm {~cm} ^ { 2 } \mathrm {~s} ^ { - 1 }$.
\begin{enumerate}[label=(\alph*)]
\item Find $\frac { \mathrm { d } x } { \mathrm {~d} t }$ when the radius of the rod is 2 cm , giving your answer to 3 significant figures.
\item Find the rate of increase of the volume of the rod when $x = 2$.\\
\section*{LU}
\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 2008 Q3 [8]}}