| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2023 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Connected Rates of Change |
| Type | Container filling: find volume or depth |
| Difficulty | Standard +0.3 This is a straightforward connected rates of change problem requiring chain rule application (dV/dt = dV/dh × dh/dt). Part (a) involves differentiating the given formula and substituting known values. Part (b) reverses the process to find h then V. While it requires careful algebraic manipulation, it follows a standard template with no novel insight needed, making it slightly easier than average. |
| Spec | 1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\dfrac{dV}{dh} = \dfrac{4}{3} \times 3(25+h)^2\) [= 4900 when \(h=10\)] | B1 | Correct expression for \(\dfrac{dV}{dh}\) |
| \(\dfrac{dV}{dh} \times \dfrac{dh}{dt} = \dfrac{dV}{dt} \Rightarrow \textit{their } 4(25+10)^2 \times \dfrac{dh}{dt} = 500 \Rightarrow \dfrac{dh}{dt} = \left[\dfrac{500}{4900}\right]\) | M1 | Use chain rule correctly to find a numerical expression for \(\dfrac{dh}{dt}\). Accept e.g. \(\dfrac{500}{2500+2000+400}\) |
| \(\dfrac{dh}{dt} = 0.102\left[\text{cms}^{-1}\right]\) | A1 | AWRT OE e.g. \(\dfrac{5}{49}\) ISW |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\dfrac{dV}{dt} = \dfrac{dV}{dh} \times \dfrac{dh}{dt} \Rightarrow 500 = \textit{their } 4(25+h)^2 \times 0.075\) | \*M1 | SOI Use chain rule correctly to form equation in \(h\) |
| \(\left[(25+h)^2 = \dfrac{5000}{3}\right] \Rightarrow h = [15.8248\ldots]\) | DM1 | Solve quadratic to find \(h\). Exact value of \(h\) is \(\sqrt{\dfrac{5000}{3}}-25\) or \(\dfrac{50\sqrt{6}}{3}-25\). \(h+25=40.82\ldots\) |
| \(V = 69900 \text{ cm}^3\) | A1 | AWRT ISW Look for \(698(88.5)\) |
## Question 9(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\dfrac{dV}{dh} = \dfrac{4}{3} \times 3(25+h)^2$ [= 4900 when $h=10$] | B1 | Correct expression for $\dfrac{dV}{dh}$ |
| $\dfrac{dV}{dh} \times \dfrac{dh}{dt} = \dfrac{dV}{dt} \Rightarrow \textit{their } 4(25+10)^2 \times \dfrac{dh}{dt} = 500 \Rightarrow \dfrac{dh}{dt} = \left[\dfrac{500}{4900}\right]$ | M1 | Use chain rule correctly to find a numerical expression for $\dfrac{dh}{dt}$. Accept e.g. $\dfrac{500}{2500+2000+400}$ |
| $\dfrac{dh}{dt} = 0.102\left[\text{cms}^{-1}\right]$ | A1 | AWRT OE e.g. $\dfrac{5}{49}$ ISW |
---
## Question 9(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\dfrac{dV}{dt} = \dfrac{dV}{dh} \times \dfrac{dh}{dt} \Rightarrow 500 = \textit{their } 4(25+h)^2 \times 0.075$ | \*M1 | SOI Use chain rule correctly to form equation in $h$ |
| $\left[(25+h)^2 = \dfrac{5000}{3}\right] \Rightarrow h = [15.8248\ldots]$ | DM1 | Solve quadratic to find $h$. Exact value of $h$ is $\sqrt{\dfrac{5000}{3}}-25$ or $\dfrac{50\sqrt{6}}{3}-25$. $h+25=40.82\ldots$ |
| $V = 69900 \text{ cm}^3$ | A1 | AWRT ISW Look for $698(88.5)$ |9 Water is poured into a tank at a constant rate of $500 \mathrm {~cm} ^ { 3 }$ per second. The depth of water in the tank, $t$ seconds after filling starts, is $h \mathrm {~cm}$. When the depth of water in the tank is $h \mathrm {~cm}$, the volume, $V \mathrm {~cm} ^ { 3 }$, of water in the tank is given by the formula $V = \frac { 4 } { 3 } ( 25 + h ) ^ { 3 } - \frac { 62500 } { 3 }$.
\begin{enumerate}[label=(\alph*)]
\item Find the rate at which $h$ is increasing at the instant when $h = 10 \mathrm {~cm}$.
\item At another instant, the rate at which $h$ is increasing is 0.075 cm per second.
Find the value of $V$ at this instant.
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2023 Q9 [6]}}