| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2006 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differential equations |
| Type | Related rates |
| Difficulty | Standard +0.3 This is a standard related rates problem requiring chain rule applications and separation of variables. Parts (a) and (b) are straightforward 'show that' exercises using S = 6x² and V = x³. Part (c) involves routine separation and integration of V^(-1/3), which is typical C4 material. The problem requires multiple steps but follows a predictable pattern with no novel insight needed, making it slightly easier than average. |
| Spec | 1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates1.08k Separable differential equations: dy/dx = f(x)g(y) |
7.
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At time $t$ seconds the length of the side of a cube is $x \mathrm {~cm}$, the surface area of the cube is $S \mathrm {~cm} ^ { 2 }$, and the volume of the cube is $V \mathrm {~cm} ^ { 3 }$.
The surface area of the cube is increasing at a constant rate of $8 \mathrm {~cm} ^ { 2 } \mathrm {~s} ^ { - 1 }$.\\
Show that
\begin{enumerate}[label=(\alph*)]
\item $\frac { \mathrm { d } x } { \mathrm {~d} t } = \frac { k } { x }$, where $k$ is a constant to be found,
\item $\frac { \mathrm { d } V } { \mathrm {~d} t } = 2 V ^ { \frac { 1 } { 3 } }$.
Given that $V = 8$ when $t = 0$,
\item solve the differential equation in part (b), and find the value of $t$ when $V = 16 \sqrt { } 2$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 2006 Q7 [15]}}