| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2024 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differential equations |
| Type | Tank/container - constant cross-section (cuboid/cylinder) |
| Difficulty | Moderate -0.3 This is a standard differential equations question requiring students to form and solve a separable DE from a rates problem. Part (a) is straightforward bookwork showing the DE formation, and part (b) involves routine separation of variables and integration. While it requires multiple steps, the techniques are entirely standard for P3 level with no novel insight needed, making it slightly easier than average. |
| Spec | 1.07t Construct differential equations: in context1.08k Separable differential equations: dy/dx = f(x)g(y) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| State that \(\frac{dV}{dt} = 50000 - 600h\) | B1 | May be seen as \(\frac{dV}{dt} = 50000\) and \(\frac{dV}{dt} = [-]600h\). When put together (may be in the chain rule) B1 can be awarded. |
| Use \(V = 40000h\) to obtain \(\frac{dV}{dh} = 40000\) and use this and their \(\frac{dV}{dt}\) in the correct chain rule to obtain \(\frac{dh}{dt}\), OR use \(V = 40000h\) to obtain \(\frac{dV}{dt} = 40000\frac{dh}{dt}\) and equate to their \(\frac{dV}{dt}\) | M1 | \(\frac{dh}{dt} = \frac{dV}{dt} \div \frac{dV}{dh}\), e.g. \(\frac{50000-600h}{40000} = \frac{dh}{dt}\), e.g. \(40000\frac{dh}{dt} = 50000 - 600h\) |
| Obtain \(200\frac{dh}{dt} = 250 - 3h\) from full and correct working | A1 | AG; \(\frac{dh}{dt} = \frac{50000-600h}{40000}\) OE, leading to given answer with no other working or incorrect working seen. SC B1 (1/3) or (2/3) for various partial approaches |
| Total | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Separate variables correctly and integrate one side correctly | M1 | E.g. \(\int \frac{1}{250-3h}\,dh = \int \frac{1}{200}\,dt\). Integral signs may be omitted, 200 may be on opposite side. |
| Obtain \(-\frac{1}{3}\ln\ | 250-3h\ | = \frac{t}{200}\ (+C)\) |
| Use \(t=0\), \(h=50\) in an expression containing \(\ln(250-3h)\) or \(\ln\ | 250-3h\ | \) to find the constant of integration |
| Obtain \(C = -\frac{1}{3}\ln 100\) | A1 | OE, e.g. \(\frac{1}{3}\ln\frac{100}{250-3h} = \frac{t}{200}\), or \(-\frac{200}{3}\ln\!\left(\frac{10}{100}\right)\). With or without modulus signs on the log terms. |
| \(t = 150\) | A1 | |
| Total | 5 |
## Question 10(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| State that $\frac{dV}{dt} = 50000 - 600h$ | B1 | May be seen as $\frac{dV}{dt} = 50000$ and $\frac{dV}{dt} = [-]600h$. When put together (may be in the chain rule) B1 can be awarded. |
| Use $V = 40000h$ to obtain $\frac{dV}{dh} = 40000$ and use this and their $\frac{dV}{dt}$ in the correct chain rule to obtain $\frac{dh}{dt}$, OR use $V = 40000h$ to obtain $\frac{dV}{dt} = 40000\frac{dh}{dt}$ and equate to their $\frac{dV}{dt}$ | M1 | $\frac{dh}{dt} = \frac{dV}{dt} \div \frac{dV}{dh}$, e.g. $\frac{50000-600h}{40000} = \frac{dh}{dt}$, e.g. $40000\frac{dh}{dt} = 50000 - 600h$ |
| Obtain $200\frac{dh}{dt} = 250 - 3h$ from full and correct working | A1 | AG; $\frac{dh}{dt} = \frac{50000-600h}{40000}$ OE, leading to given answer with no other working or incorrect working seen. **SC B1** (1/3) or (2/3) for various partial approaches |
| **Total** | **3** | |
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## Question 10(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| Separate variables correctly and integrate one side correctly | M1 | E.g. $\int \frac{1}{250-3h}\,dh = \int \frac{1}{200}\,dt$. Integral signs may be omitted, 200 may be on opposite side. |
| Obtain $-\frac{1}{3}\ln\|250-3h\| = \frac{t}{200}\ (+C)$ | A1 | OE. Condone missing "$+C$" and lack of modulus signs. |
| Use $t=0$, $h=50$ in an expression containing $\ln(250-3h)$ or $\ln\|250-3h\|$ to find the constant of integration | M1 | Or equivalent use of limits 50 and 80. |
| Obtain $C = -\frac{1}{3}\ln 100$ | A1 | OE, e.g. $\frac{1}{3}\ln\frac{100}{250-3h} = \frac{t}{200}$, or $-\frac{200}{3}\ln\!\left(\frac{10}{100}\right)$. With or without modulus signs on the log terms. |
| $t = 150$ | A1 | |
| **Total** | **5** | |
---10 A water tank is in the shape of a cuboid with base area $40000 \mathrm {~cm} ^ { 2 }$. At time $t$ minutes the depth of water in the tank is $h \mathrm {~cm}$. Water is pumped into the tank at a rate of $50000 \mathrm {~cm} ^ { 3 }$ per minute. Water is leaking out of the tank through a hole in the bottom at a rate of $600 \mathrm {~cm} ^ { 3 }$ per minute.
\begin{enumerate}[label=(\alph*)]
\item Show that $200 \frac { \mathrm {~d} h } { \mathrm {~d} t } = 250 - 3 h$.\\
\begin{center}
\end{center}
\includegraphics[max width=\textwidth, alt={}, center]{6280ab81-0bdb-47b4-8651-bff1261a0adf-17_2723_33_99_22}
\item It is given that when $t = 0 , h = 50$.
Find the time taken for the depth of water in the tank to reach 80 cm . Give your answer correct to 2 significant figures.
\end{enumerate}
\hfill \mbox{\textit{CAIE P3 2024 Q10 [8]}}