| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2013 |
| Session | November |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differential equations |
| Type | Conical geometry differential equations |
| Difficulty | Standard +0.3 This is a standard differential equations question requiring volume-rate relationships, separation of variables, and straightforward substitution. The geometry setup is given, the differential equation form is provided, and the solution method is routine for P3 level. Slightly easier than average due to the scaffolded structure and standard techniques. |
| Spec | 1.07t Construct differential equations: in context1.08k Separable differential equations: dy/dx = f(x)g(y)3.02f Non-uniform acceleration: using differentiation and integration |
| Answer | Marks | Guidance |
|---|---|---|
| (i) State or imply \(V = \pi h^{3}\) | B1 | |
| State or imply \(\frac{dV}{dt} = -k\sqrt{h}\) | B1 | |
| Use \(\frac{dV}{dt} = \frac{dV}{dh} \cdot \frac{dh}{dt}\) or equivalent | M1 | |
| Obtain the given equation | A1 | [4] |
| Answer | Marks | Guidance |
|---|---|---|
| (ii) Separate variables and integrate at least one side | M1 | |
| Obtain terms \(\frac{2}{5}h^{5}\) and \(-At\), or equivalent | A1 | |
| Use \(t = 0, h = H\) in a solution containing terms of the form \(ah^{\frac{5}{}}\) and \(bt + c\) | M1 | |
| Use \(t = 60, h = 0\) in a solution containing terms of the form \(ah^{\frac{5}{}}\) and \(bt + c\) | M1 | |
| Obtain a correct solution in any form, e.g. \(\frac{2}{5}h^{2} = \frac{1}{150}H^{2}t + \frac{2}{5}H^{2}\) | A1 | |
| (iii) Obtain final answer \(t = 60\left[1 - \left(\frac{h}{H}\right)^{\frac{5}{2}}\right]\), or equivalent | A1 | [6] |
| (iv) Substitute \(h = \frac{1}{2}H\) and obtain answer \(t = 49.4\) | B1 | [1] |
**(i)** State or imply $V = \pi h^{3}$ | B1 |
State or imply $\frac{dV}{dt} = -k\sqrt{h}$ | B1 |
Use $\frac{dV}{dt} = \frac{dV}{dh} \cdot \frac{dh}{dt}$ or equivalent | M1 |
Obtain the given equation | A1 | [4]
[The M1 is only available if $\frac{dV}{dh}$ is in terms of $h$ and has been obtained by a correct method.]
[Allow B1 for $\frac{dV}{dt} = k\sqrt{h}$ but withhold the final A1 until the polarity of the constant $\frac{k}{3\pi}$ has been justified.]
**(ii)** Separate variables and integrate at least one side | M1 |
Obtain terms $\frac{2}{5}h^{5}$ and $-At$, or equivalent | A1 |
Use $t = 0, h = H$ in a solution containing terms of the form $ah^{\frac{5}{}}$ and $bt + c$ | M1 |
Use $t = 60, h = 0$ in a solution containing terms of the form $ah^{\frac{5}{}}$ and $bt + c$ | M1 |
Obtain a correct solution in any form, e.g. $\frac{2}{5}h^{2} = \frac{1}{150}H^{2}t + \frac{2}{5}H^{2}$ | A1 |
**(iii)** Obtain final answer $t = 60\left[1 - \left(\frac{h}{H}\right)^{\frac{5}{2}}\right]$, or equivalent | A1 | [6]
**(iv)** Substitute $h = \frac{1}{2}H$ and obtain answer $t = 49.4$ | B1 | [1]\includegraphics{figure_10}
A tank containing water is in the form of a cone with vertex $C$. The axis is vertical and the semi-vertical angle is $60°$, as shown in the diagram. At time $t = 0$, the tank is full and the depth of water is $H$. At this instant, a tap at $C$ is opened and water begins to flow out. The volume of water in the tank decreases at a rate proportional to $\sqrt{h}$, where $h$ is the depth of water at time $t$. The tank becomes empty when $t = 60$.
\begin{enumerate}[label=(\roman*)]
\item Show that $h$ and $t$ satisfy a differential equation of the form
$$\frac{dh}{dt} = -Ah^{-\frac{1}{2}},$$
where $A$ is a positive constant. [4]
\item Solve the differential equation given in part (i) and obtain an expression for $t$ in terms of $h$ and $H$. [6]
\item Find the time at which the depth reaches $\frac{1}{2}H$. [1]
\end{enumerate}
[The volume $V$ of a cone of vertical height $h$ and base radius $r$ is given by $V = \frac{1}{3}\pi r^2 h$.]
\hfill \mbox{\textit{CAIE P3 2013 Q10 [11]}}