Moderate -0.3 This is a straightforward connected rates of change question requiring the chain rule (dh/dt = dV/dt รท dV/dh). Students must differentiate the given volume formula, substitute h=6 and the given rate, then solve. It's slightly easier than average because the setup is clear, the differentiation is routine, and it's a standard textbook-style problem with no conceptual surprises.
\includegraphics{figure_2}
A vase with a circular cross-section is shown in Figure 2. Water is flowing into the vase.
When the depth of the water is \(h\) cm, the volume of water \(V\) cm\(^3\) is given by
$$V = 4\pi h(h + 4), \quad 0 \leq h \leq 25$$
Water flows into the vase at a constant rate of \(80\pi\) cm\(^3\)s\(^{-1}\)
Find the rate of change of the depth of the water, in cm s\(^{-1}\), when \(h = 6\) \hfill [5]
\includegraphics{figure_2}
A vase with a circular cross-section is shown in Figure 2. Water is flowing into the vase.
When the depth of the water is $h$ cm, the volume of water $V$ cm$^3$ is given by
$$V = 4\pi h(h + 4), \quad 0 \leq h \leq 25$$
Water flows into the vase at a constant rate of $80\pi$ cm$^3$s$^{-1}$
Find the rate of change of the depth of the water, in cm s$^{-1}$, when $h = 6$ \hfill [5]
\hfill \mbox{\textit{Edexcel C4 2014 Q4 [5]}}