| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2011 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Related rates with cones, hemispheres, and bowls (variable depth) |
| Difficulty | Standard +0.3 This is a standard related rates problem requiring differentiation of a given formula and application of the chain rule (dV/dt = dV/dh × dh/dt). Part (a) is routine differentiation, part (b) is straightforward substitution. The formula is provided, making this slightly easier than average for C4 related rates questions. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(\frac{dV}{dh} = \frac{1}{2}\pi h - \pi h^2\) or equivalent | M1 A1 | |
| At \(h = 0.1\): \(\frac{dV}{dh} = \frac{1}{2}\pi(0.1) - \pi(0.1)^2 = 0.04\pi\) | M1 A1 | |
| \(\frac{\pi}{25}\) | (4) | |
| (b) \(\frac{dh}{dt} = \frac{dV}{dt} \div \frac{dV}{dh} = \frac{\pi}{800} \times \frac{1}{\frac{1}{2}\pi h - \pi h^2}\) or \(\frac{\pi}{800} \div\) their (a) | M1 | |
| At \(h = 0.1\): \(\frac{dh}{dt} = \frac{\pi}{800} \times \frac{25}{\pi} = \frac{1}{32}\) | A1 | awrt 0.031 |
**(a)** $\frac{dV}{dh} = \frac{1}{2}\pi h - \pi h^2$ or equivalent | M1 A1 |
At $h = 0.1$: $\frac{dV}{dh} = \frac{1}{2}\pi(0.1) - \pi(0.1)^2 = 0.04\pi$ | M1 A1 |
| $\frac{\pi}{25}$ | (4) |
**(b)** $\frac{dh}{dt} = \frac{dV}{dt} \div \frac{dV}{dh} = \frac{\pi}{800} \times \frac{1}{\frac{1}{2}\pi h - \pi h^2}$ or $\frac{\pi}{800} \div$ their (a) | M1 |
At $h = 0.1$: $\frac{dh}{dt} = \frac{\pi}{800} \times \frac{25}{\pi} = \frac{1}{32}$ | A1 | awrt 0.031 | (2) [6] |
---3.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{9d513d77-b8f9-4223-832f-f566c5f50457-04_391_741_274_605}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
A hollow hemispherical bowl is shown in Figure 1. Water is flowing into the bowl. When the depth of the water is $h \mathrm {~m}$, the volume $V \mathrm {~m} ^ { 3 }$ is given by
$$V = \frac { 1 } { 12 } \pi \cdot h ^ { 2 } ( 3 - 4 h ) , \quad 0 \leqslant h \leqslant 0.25$$
\begin{enumerate}[label=(\alph*)]
\item Find, in terms of $\pi , \frac { \mathrm { d } V } { \mathrm {~d} h }$ when $h = 0.1$
Water flows into the bowl at a rate of $\frac { \pi } { 800 } \mathrm {~m} ^ { 3 } \mathrm {~s} ^ { - 1 }$.
\item Find the rate of change of $h$, in $\mathrm { m } \mathrm { s } ^ { - 1 }$, when $h = 0.1$
\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 2011 Q3 [6]}}