2.
\begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{7f5fc83d-ab7c-4edb-a2c6-7a58f1357d5a-04_271_223_246_922}
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\caption{Figure 1}
\end{figure}
Figure 1 shows a cube which is increasing in size.
At time \(t\) seconds,
- the length of each edge of the cube is \(x \mathrm {~cm}\)
- the surface area of the cube is \(S \mathrm {~cm} ^ { 2 }\)
- the volume of the cube is \(V \mathrm {~cm} ^ { 3 }\)
Given that the surface area of the cube is increasing at a constant rate of \(4 \mathrm {~cm} ^ { 2 } \mathrm {~s} ^ { - 1 }\)
- show that \(\frac { \mathrm { d } x } { \mathrm {~d} t } = \frac { k } { x }\) where \(k\) is a constant to be found,
- show that \(\frac { \mathrm { d } V } { \mathrm {~d} t } = V ^ { p }\) where \(p\) is a constant to be found.