| Exam Board | WJEC |
|---|---|
| Module | Unit 3 (Unit 3) |
| Session | Specimen |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differential equations |
| Type | Tank/container - constant cross-section (cuboid/cylinder) |
| Difficulty | Standard +0.3 This is a standard differential equations question requiring rate of change setup, separation of variables, and integration with partial fractions. While it involves multiple steps (9 marks total), each technique is routine for A-level: the DE setup is guided by a hint, the separation and integration follow textbook methods, and the final evaluation is straightforward substitution. Slightly above average difficulty due to the algebraic manipulation required, but no novel insight needed. |
| Spec | 1.08k Separable differential equations: dy/dx = f(x)g(y) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dV}{dt} = 4\frac{dh}{dt}\) | ||
| \(4\frac{dh}{dt} = 0.004 - 0.0008h\) | M1 | (3 terms, at least 2 correct) |
| \(\frac{dh}{dt} = 0.001 - 0.0002h\) | A1 | (Correct) |
| \(5000\frac{dh}{dt} = 5 - h\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(5000\int\frac{dh}{5-h} = \int dt\) | M1 | (Separation of variables) |
| \(-5000\ln(5-h) = t + C\) (1) | A1, A1 | (-1 if C omitted) |
| \(h = 0\) at \(t = 0\): \(\therefore -5000\ln(5) = C\) | m1 | |
| Substitute in (1): \(-5000\ln(5-h) = t - 5000\ln(5)\) | ||
| \(t = 5000\ln\left(\frac{5}{5-h}\right)\) | A1 | |
| \(\therefore \left(\frac{5}{5-h}\right) = e^{\frac{t}{5000}}\) | M1 | (Attempt to invert) |
| \(5 - h = 5e^{\frac{-t}{5000}}\) | ||
| \(h = 5 - 5e^{\frac{-t}{5000}}\) | A1 |
| Answer | Marks |
|---|---|
| \(h = 5 - 5e^{\frac{-3600}{5000}}\) | |
| \(= 2.57\) m | B1 |
### Part (a):
$\frac{dV}{dt} = 4\frac{dh}{dt}$ | |
$4\frac{dh}{dt} = 0.004 - 0.0008h$ | M1 | (3 terms, at least 2 correct)
$\frac{dh}{dt} = 0.001 - 0.0002h$ | A1 | (Correct)
$5000\frac{dh}{dt} = 5 - h$ | |
### Part (b):
$5000\int\frac{dh}{5-h} = \int dt$ | M1 | (Separation of variables)
$-5000\ln(5-h) = t + C$ (1) | A1, A1 | (-1 if C omitted)
$h = 0$ at $t = 0$: $\therefore -5000\ln(5) = C$ | m1 |
Substitute in (1): $-5000\ln(5-h) = t - 5000\ln(5)$ | |
$t = 5000\ln\left(\frac{5}{5-h}\right)$ | A1 |
$\therefore \left(\frac{5}{5-h}\right) = e^{\frac{t}{5000}}$ | M1 | (Attempt to invert)
$5 - h = 5e^{\frac{-t}{5000}}$ | |
$h = 5 - 5e^{\frac{-t}{5000}}$ | A1 |
### Part (c):
$h = 5 - 5e^{\frac{-3600}{5000}}$ | |
$= 2.57$ m | B1 |
**Total: [10]**\begin{enumerate}[label=(\alph*)]
\item A cylindrical water tank has base area 4 m$^2$. The depth of the water at time $t$ seconds is $h$ metres. Water is poured in at the rate 0.004 m$^3$ per second. Water leaks from a hole in the bottom at a rate of 0.0008$h$ m$^3$ per second. Show that
$$5000 \frac{dh}{dt} = 5 - h.$$ [2]
[Hint: the volume, $V$, of the cylindrical water tank is given by $V = 4h$.]
\item Given that the tank is empty initially, find $h$ in terms of $t$. [7]
\item Find the depth of the water in the tank when $t = 3600$ s, giving your answer correct to 2 decimal places. [1]
\end{enumerate}
\hfill \mbox{\textit{WJEC Unit 3 Q14 [10]}}