| Exam Board | Edexcel |
|---|---|
| Module | P4 (Pure Mathematics 4) |
| Year | 2021 |
| Session | June |
| Marks | 7 |
| 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 the given volume formula and application of the chain rule. Part (a) involves substituting h=20 and dividing by the flow rate (routine calculation). Part (b) requires implicit differentiation to find dh/dt when dV/dt and h are known—a textbook application of related rates with no conceptual surprises. Slightly easier than average due to the formula being provided and straightforward algebraic manipulation. |
| Spec | 1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates1.08a Fundamental theorem of calculus: integration as reverse of differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{\frac{1}{3}\times 20^2\times 24}{160} = 20\) seconds | M1 A1 | M1: Attempts to find volume using formula at \(h=20\) and divides by 160. Condone slips. A1: 20 seconds — correct units required |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{dV}{dh} = h^2 + \frac{8}{3}h\) | M1 | For attempt to differentiate \(V\). Multiply out then differentiate, or product rule to get \(\frac{dV}{dh} = \alpha h^2 + \beta h\) |
| \(\frac{dV}{dt} = \frac{dV}{dh}\times\frac{dh}{dt} \Rightarrow 160 = \left(h^2 + \frac{8}{3}h\right)\times\frac{dh}{dt}\) | M1 A1 | M1: Use of chain rule with \(\frac{dV}{dt}=160\) and their \(\frac{dV}{dh}\). A1: Correct expression involving \(\frac{dh}{dt}\) and \(h\) |
| \(h=5 \Rightarrow 160 = \left(25 + \frac{40}{3}\right)\times\frac{dh}{dt} \Rightarrow \frac{dh}{dt} = \frac{96}{23}\) (4.2) cm s\(^{-1}\) | dM1 A1 | dM1: Substitutes \(h=5\) and proceeds to value for \(\frac{dh}{dt}\). A1: \(\frac{96}{23}\) or awrt 4.2 cm s\(^{-1}\). No requirement for units |
# Question 3:
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{\frac{1}{3}\times 20^2\times 24}{160} = 20$ seconds | M1 A1 | M1: Attempts to find volume using formula at $h=20$ and divides by 160. Condone slips. A1: 20 seconds — correct units required |
## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{dV}{dh} = h^2 + \frac{8}{3}h$ | M1 | For attempt to differentiate $V$. Multiply out then differentiate, or product rule to get $\frac{dV}{dh} = \alpha h^2 + \beta h$ |
| $\frac{dV}{dt} = \frac{dV}{dh}\times\frac{dh}{dt} \Rightarrow 160 = \left(h^2 + \frac{8}{3}h\right)\times\frac{dh}{dt}$ | M1 A1 | M1: Use of chain rule with $\frac{dV}{dt}=160$ and their $\frac{dV}{dh}$. A1: Correct expression involving $\frac{dh}{dt}$ and $h$ |
| $h=5 \Rightarrow 160 = \left(25 + \frac{40}{3}\right)\times\frac{dh}{dt} \Rightarrow \frac{dh}{dt} = \frac{96}{23}$ (4.2) cm s$^{-1}$ | dM1 A1 | dM1: Substitutes $h=5$ and proceeds to value for $\frac{dh}{dt}$. A1: $\frac{96}{23}$ or awrt 4.2 cm s$^{-1}$. No requirement for units |
---3.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{960fe82f-c180-422c-b409-a5cdc5fae924-08_524_878_255_532}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
A bowl with circular cross section and height 20 cm is shown in Figure 2.\\
The bowl is initially empty and water starts flowing into the bowl.\\
When the depth of water is $h \mathrm {~cm}$, the volume of water in the bowl, $V \mathrm {~cm} ^ { 3 }$, is modelled by the equation
$$V = \frac { 1 } { 3 } h ^ { 2 } ( h + 4 ) \quad 0 \leqslant h \leqslant 20$$
Given that the water flows into the bowl at a constant rate of $160 \mathrm {~cm} ^ { 3 } \mathrm {~s} ^ { - 1 }$, find, according to the model,
\begin{enumerate}[label=(\alph*)]
\item the time taken to fill the bowl,
\item the rate of change of the depth of the water, in $\mathrm { cm } \mathrm { s } ^ { - 1 }$, when $h = 5$
\end{enumerate}
\hfill \mbox{\textit{Edexcel P4 2021 Q3 [7]}}