| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9794/2 (Pre-U Mathematics Paper 2) |
| Year | 2013 |
| Session | June |
| Marks | 11 |
| Topic | Differential equations |
| Type | Tank/container - constant cross-section (cuboid/cylinder) |
| Difficulty | Standard +0.3 This is a standard differential equations question requiring separation of variables and application of boundary conditions. The rectangular cross-section simplifies the relationship between volume and depth (V ∝ h), making part (i) straightforward. Parts (ii) and (iii) involve routine integration and algebraic manipulation with given conditions. While it requires multiple steps, each technique is standard A-level material with no novel insight needed. |
| Spec | 1.08k Separable differential equations: dy/dx = f(x)g(y)1.08l Interpret differential equation solutions: in context |
(i) $\frac{dV}{dt} \propto \sqrt{h}$; Since the tank is a prism $V \propto h$, so $\frac{dV}{dt} = a\sqrt{V}$ where $a$ is a constant — M1, A1 **[2]**
(ii) Separating variables: $\int \frac{1}{\sqrt{V}}\,dv = \int a\,dt$ — M1; $2\sqrt{V} = at\ (+c)$ — M1, A1; Use $t = 0$, $V = V_0$ to obtain $c = 2\sqrt{V_0}$ — B1; and $t = 1$, $V = \frac{1}{2}V_0$ in an equation involving $a$ and $c$ (or using definite integrals) to find $a$ in terms of $V_0$ only — M1; $a = 2\sqrt{V_0}\left(\frac{1}{\sqrt{2}} - 1\right)$ — A1; convincingly substitute and rearrange to get $V = V_0\left(\left(\frac{1}{\sqrt{2}}-1\right)t + 1\right)^2$ — A1 **[7]**
(iii) $V = 0$ implies $t = \frac{-1}{\frac{1}{\sqrt{2}}-1} = 2 + \sqrt{2} = 3.41\ldots$ — M1; 3.41 hours is 3 hours 24 mins and 51 seconds — A1 **[2]**
Condone verification only if $5.16 \times 10^{-6} V_0$ seen.
**Total: [11]**10 A tank with vertical sides and rectangular cross-section is initially full of water. The water is leaking out of a hole in the base of the tank at a rate which is proportional to the square root of the depth of the water. $V \mathrm {~m} ^ { 3 }$ is the volume of water in the tank at time $t$ hours.\\
(i) Show that $\frac { \mathrm { d } V } { \mathrm {~d} t } = a \sqrt { V }$, where $a$ is a constant.\\
(ii) Given that the tank is half full after one hour, show that $V = V _ { 0 } \left( \left( \frac { 1 } { \sqrt { 2 } } - 1 \right) t + 1 \right) ^ { 2 }$, where $V _ { 0 } \mathrm {~m} ^ { 3 }$ is the initial volume of water in the tank.\\
(iii) Hence show that the tank will be empty after approximately 3 hours and 25 minutes.
\hfill \mbox{\textit{Pre-U Pre-U 9794/2 2013 Q10 [11]}}