| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2012 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Related rates with spheres, circles, and cubes |
| Difficulty | Moderate -0.3 This is a standard related rates problem from C4 requiring chain rule application. Part (a) is trivial differentiation of V=x³, part (b) uses dV/dt = (dV/dx)(dx/dt) with given values, and part (c) extends this to surface area. All steps are routine textbook exercises with no problem-solving insight required, making it slightly easier than average. |
| Spec | 1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates |
| Answer | Marks | Guidance |
|---|---|---|
| \(V = x^3 \Rightarrow \frac{dV}{dx} = 3x^2\) | B1 (1) | cso |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dx}{dt} = \frac{dx}{dV} \times \frac{dV}{dt} = \frac{0.048}{3x^2}\) | M1 | |
| At \(x = 8\): \(\frac{dx}{dt} = \frac{0.048}{3(8^2)} = 0.00025\) (cm s\(^{-1}\)) | A1 (2) | \(2.5 \times 10^{-4}\) |
| Answer | Marks |
|---|---|
| \(S = 6x^2 \Rightarrow \frac{dS}{dx} = 12x\) | B1 |
| \(\frac{dS}{dt} = \frac{dS}{dx} \times \frac{dx}{dt} = 12x\left(\frac{0.048}{3x^2}\right)\) | M1 |
| At \(x = 8\): \(\frac{dS}{dt} = 0.024\) (cm\(^2\) s\(^{-1}\)) | A1 (3) |
## Question 2:
### Part (a)
| $V = x^3 \Rightarrow \frac{dV}{dx} = 3x^2$ | B1 (1) | cso |
### Part (b)
| $\frac{dx}{dt} = \frac{dx}{dV} \times \frac{dV}{dt} = \frac{0.048}{3x^2}$ | M1 | |
| At $x = 8$: $\frac{dx}{dt} = \frac{0.048}{3(8^2)} = 0.00025$ (cm s$^{-1}$) | A1 (2) | $2.5 \times 10^{-4}$ |
### Part (c)
| $S = 6x^2 \Rightarrow \frac{dS}{dx} = 12x$ | B1 | |
| $\frac{dS}{dt} = \frac{dS}{dx} \times \frac{dx}{dt} = 12x\left(\frac{0.048}{3x^2}\right)$ | M1 | |
| At $x = 8$: $\frac{dS}{dt} = 0.024$ (cm$^2$ s$^{-1}$) | A1 (3) | |
---2.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{12fbfe89-60fe-4890-9a22-2b1988d05d33-03_424_465_228_721}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Figure 1 shows a metal cube which is expanding uniformly as it is heated. At time $t$ seconds, the length of each edge of the cube is $x \mathrm {~cm}$, and the volume of the cube is $V \mathrm {~cm} ^ { 3 }$.
\begin{enumerate}[label=(\alph*)]
\item Show that $\frac { \mathrm { d } V } { \mathrm {~d} x } = 3 x ^ { 2 }$
Given that the volume, $V \mathrm {~cm} ^ { 3 }$, increases at a constant rate of $0.048 \mathrm {~cm} ^ { 3 } \mathrm {~s} ^ { - 1 }$,
\item find $\frac { \mathrm { d } x } { \mathrm {~d} t }$, when $x = 8$
\item find the rate of increase of the total surface area of the cube, in $\mathrm { cm } ^ { 2 } \mathrm {~s} ^ { - 1 }$, when $x = 8$
\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 2012 Q2 [6]}}