| Exam Board | Edexcel |
|---|---|
| Module | Paper 2 (Paper 2) |
| Session | Specimen |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Connected Rates of Change |
| Type | Container filling: find depth rate |
| Difficulty | Standard +0.3 This is a standard connected rates of change question requiring differentiation of the given volume formula and application of the chain rule. The volume formula is provided (no derivation needed), and students simply need to find dV/dh, substitute given values, and use dV/dt = (dV/dh)(dh/dt). While it requires careful algebraic manipulation, it follows a well-practiced procedure with no novel 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 |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dV}{dh} = 50\pi h - \pi h^2\) | M1 | Differentiates \(V\) with respect to \(h\) to give \(\pm\alpha h \pm \beta h^2\), \(\alpha \neq 0\), \(\beta \neq 0\) |
| \(\frac{dV}{dh} = 50\pi h - \pi h^2\) | A1 | Correct expression |
| \(\left(\frac{dV}{dh} \times \frac{dh}{dt} = \frac{dV}{dt}\right) \Rightarrow (50\pi h - \pi h^2)\frac{dh}{dt} = 160\pi\) | M1 | Attempts complete method applying \(\left(\text{their } \frac{dV}{dh}\right) \times \frac{dh}{dt} = 160\pi\) |
| When \(h=10\): \(\frac{dh}{dt} = \frac{dV}{dt} \div \frac{dV}{dh} \Rightarrow \frac{160\pi}{50\pi(10)-\pi(10)^2} \left\{=\frac{160\pi}{400\pi}\right\}\) | dM1 | Depends on previous M mark. Substitutes \(h=10\) into model for \(\frac{dh}{dt}\) in form \(\frac{160\pi}{\text{their } \frac{dV}{dh}}\) |
| \(\frac{dh}{dt} = 0.4\) (cms⁻¹) | A1 | Obtains correct answer 0.4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dh}{dt} = \frac{300\pi}{50\pi(20) - \pi(20)^2}\) | M1 | Realises rate of \(160\pi\) cm³s⁻¹ for \(0 \leq h \leq 12\) has no effect when rate is increased to \(300\pi\) cm³s⁻¹ for \(12 < h \leq 24\), substitutes \(h=20\) into model for \(\frac{dh}{dt}\) in form \(\frac{300\pi}{\text{their } \frac{dV}{dh}}\) |
| \(\frac{dh}{dt} = 0.5\) (cms⁻¹) | A1 | Obtains correct answer 0.5 |
# Question 8:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dV}{dh} = 50\pi h - \pi h^2$ | M1 | Differentiates $V$ with respect to $h$ to give $\pm\alpha h \pm \beta h^2$, $\alpha \neq 0$, $\beta \neq 0$ |
| $\frac{dV}{dh} = 50\pi h - \pi h^2$ | A1 | Correct expression |
| $\left(\frac{dV}{dh} \times \frac{dh}{dt} = \frac{dV}{dt}\right) \Rightarrow (50\pi h - \pi h^2)\frac{dh}{dt} = 160\pi$ | M1 | Attempts complete method applying $\left(\text{their } \frac{dV}{dh}\right) \times \frac{dh}{dt} = 160\pi$ |
| When $h=10$: $\frac{dh}{dt} = \frac{dV}{dt} \div \frac{dV}{dh} \Rightarrow \frac{160\pi}{50\pi(10)-\pi(10)^2} \left\{=\frac{160\pi}{400\pi}\right\}$ | dM1 | Depends on previous M mark. Substitutes $h=10$ into model for $\frac{dh}{dt}$ in form $\frac{160\pi}{\text{their } \frac{dV}{dh}}$ |
| $\frac{dh}{dt} = 0.4$ (cms⁻¹) | A1 | Obtains correct answer 0.4 |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dh}{dt} = \frac{300\pi}{50\pi(20) - \pi(20)^2}$ | M1 | Realises rate of $160\pi$ cm³s⁻¹ for $0 \leq h \leq 12$ has no effect when rate is increased to $300\pi$ cm³s⁻¹ for $12 < h \leq 24$, substitutes $h=20$ into model for $\frac{dh}{dt}$ in form $\frac{300\pi}{\text{their } \frac{dV}{dh}}$ |
| $\frac{dh}{dt} = 0.5$ (cms⁻¹) | A1 | Obtains correct answer 0.5 |
---8.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{659a0479-c8c6-418b-b8a9-67ad68474023-18_367_709_280_676}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
A bowl is modelled as a hemispherical shell as shown in Figure 3.\\
Initially the bowl is empty and water begins to flow into the bowl.
When the depth of the water is $h \mathrm {~cm}$, the volume of water, $V \mathrm {~cm} ^ { 3 }$, according to the model is given by
$$V = \frac { 1 } { 3 } \pi h ^ { 2 } ( 75 - h ) , \quad 0 \leqslant h \leqslant 24$$
The flow of water into the bowl is at a constant rate of $160 \pi \mathrm {~cm} ^ { 3 } \mathrm {~s} ^ { - 1 }$ for $0 \leqslant h \leqslant 12$
\begin{enumerate}[label=(\alph*)]
\item Find the rate of change of the depth of the water, in $\mathrm { cm } \mathrm { s } ^ { - 1 }$, when $h = 10$
Given that the flow of water into the bowl is increased to a constant rate of $300 \pi \mathrm {~cm} ^ { 3 } \mathrm {~s} ^ { - 1 }$ for $12 < h \leqslant 24$
\item find the rate of change of the depth of the water, in $\mathrm { cms } ^ { - 1 }$, when $h = 20$
\end{enumerate}
\hfill \mbox{\textit{Edexcel Paper 2 Q8 [7]}}