| Exam Board | Edexcel |
|---|---|
| Module | P4 (Pure Mathematics 4) |
| Year | 2022 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Connected Rates of Change |
| Type | Tablet or dissolving object |
| Difficulty | Standard +0.3 This is a standard connected rates of change problem requiring the chain rule with given relationships. Part (a) involves differentiating the area formula πx² and substituting a given rate and value. Part (b) requires finding dV/dt using the product/chain rule with V = 3πx³. Both parts follow routine procedures for this topic with straightforward algebra and no conceptual surprises, making it slightly easier than average. |
| Spec | 1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{dA}{dt}=-0.5\) | B1 | Seen or implied from working |
| \(A=\pi x^2\Rightarrow\frac{dA}{dx}=2\pi x\) | B1 | Seen or implied; must be in terms of \(x\) (allow recovery if in terms of \(r\), later using \(r=7\)) |
| \(\frac{dx}{dt}=\frac{dA}{dt}\div\frac{dA}{dx}=\frac{"-0.5"}{2\pi x"}\) \(\left(=\frac{-1}{4\pi x}\right)\) | M1 | Attempts to use appropriate chain rule with their \(\frac{dA}{dt}\) and \(\frac{dA}{dx}\) |
| \(\frac{dx}{dt}=-0.011368...\) | A1cso | awrt \(-0.0114\) or \(-\frac{1}{28\pi}\) cso (must have negative sign) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(V=\pi x^2(3x)=3\pi x^3\) | B1 | \(V=\pi x^2(3x)\) or \(V=3\pi x^3\) |
| \(\frac{dV}{dx}=9\pi x^2\) | B1ft | Or ft from their equation for \(V\) in one variable |
| \(\frac{dV}{dt}=\frac{dV}{dx}\times\frac{dx}{dt}=9\pi x^2\times"-\frac{1}{4\pi x}"\ (=-2.25x)\) | M1 | Their \(\frac{dV}{dx}\times\) their \(\frac{dx}{dt}\); note \(\frac{dx}{dt}\) must be in terms of \(x\) or with \(x=4\) substituted first |
| \(\left(\frac{dV}{dt}=\right)-9\Rightarrow\) Rate of decrease \(= 9\ (\text{mm}^3\text{s}^{-1})\) | A1 | With or without negative sign; may be scored following \(\frac{dA}{dt}=0.5\) in part (a) |
# Question 3:
## Part (a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{dA}{dt}=-0.5$ | B1 | Seen or implied from working |
| $A=\pi x^2\Rightarrow\frac{dA}{dx}=2\pi x$ | B1 | Seen or implied; must be in terms of $x$ (allow recovery if in terms of $r$, later using $r=7$) |
| $\frac{dx}{dt}=\frac{dA}{dt}\div\frac{dA}{dx}=\frac{"-0.5"}{2\pi x"}$ $\left(=\frac{-1}{4\pi x}\right)$ | M1 | Attempts to use appropriate chain rule with their $\frac{dA}{dt}$ and $\frac{dA}{dx}$ |
| $\frac{dx}{dt}=-0.011368...$ | A1cso | awrt $-0.0114$ or $-\frac{1}{28\pi}$ cso (must have negative sign) |
## Part (b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $V=\pi x^2(3x)=3\pi x^3$ | B1 | $V=\pi x^2(3x)$ or $V=3\pi x^3$ |
| $\frac{dV}{dx}=9\pi x^2$ | B1ft | Or ft from their equation for $V$ in one variable |
| $\frac{dV}{dt}=\frac{dV}{dx}\times\frac{dx}{dt}=9\pi x^2\times"-\frac{1}{4\pi x}"\ (=-2.25x)$ | M1 | Their $\frac{dV}{dx}\times$ their $\frac{dx}{dt}$; note $\frac{dx}{dt}$ must be in terms of $x$ or with $x=4$ substituted first |
| $\left(\frac{dV}{dt}=\right)-9\Rightarrow$ Rate of decrease $= 9\ (\text{mm}^3\text{s}^{-1})$ | A1 | With or without negative sign; may be scored following $\frac{dA}{dt}=0.5$ in part (a) |
---3.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{2dffe245-b18a-4486-af8e-bad598ceb6e8-08_401_652_246_708}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
A tablet is dissolving in water.\\
The tablet is modelled as a cylinder, shown in Figure 1.\\
At $t$ seconds after the tablet is dropped into the water, the radius of the tablet is $x \mathrm {~mm}$ and the length of the tablet is $3 x \mathrm {~mm}$.
The cross-sectional area of the tablet is decreasing at a constant rate of $0.5 \mathrm {~mm} ^ { 2 } \mathrm {~s} ^ { - 1 }$
\begin{enumerate}[label=(\alph*)]
\item Find $\frac { \mathrm { d } x } { \mathrm {~d} t }$ when $x = 7$
\item Find, according to the model, the rate of decrease of the volume of the tablet when $x = 4$
\end{enumerate}
\hfill \mbox{\textit{Edexcel P4 2022 Q3 [8]}}