| 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 problem requiring volume-rate relationships for a cone, separation of variables, and straightforward integration. While it involves multiple steps (relating volume to depth using geometry, forming and solving the DE, applying boundary conditions), each step follows a predictable pattern typical of A-level mechanics/pure maths. The geometric relationship r = h√3 from the 60° angle is straightforward, and the resulting DE is a standard separable form. Slightly above average difficulty due to the multi-step nature and need to connect geometry with calculus, but well within the scope of routine A-level pure maths questions. |
| Spec | 1.02z Models in context: use functions in modelling1.07t Construct differential equations: in context1.08k Separable differential equations: dy/dx = f(x)g(y) |
| 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 \(\frac{2}{5}h^{\frac{5}{2}}\) and \(-At\), or equivalent | A1 | |
| Use \(t=0, h=H\) in a solution containing terms of the form \(ah^{\frac{5}{2}}\) and \(bt+c\) | M1 | |
| Use \(t=60, h=0\) in a solution containing terms of the form \(ah^{\frac{5}{2}}\) and \(bt+c\) | M1 | |
| Obtain a correct solution in any form, e.g. \(\frac{2}{5}h^{\frac{5}{2}}=\frac{1}{150}H^{2}t+\frac{2}{5}H^{\frac{5}{2}}\) | 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 $\frac{2}{5}h^{\frac{5}{2}}$ and $-At$, or equivalent | A1 |
Use $t=0, h=H$ in a solution containing terms of the form $ah^{\frac{5}{2}}$ and $bt+c$ | M1 |
Use $t=60, h=0$ in a solution containing terms of the form $ah^{\frac{5}{2}}$ and $bt+c$ | M1 |
Obtain a correct solution in any form, e.g. $\frac{2}{5}h^{\frac{5}{2}}=\frac{1}{150}H^{2}t+\frac{2}{5}H^{\frac{5}{2}}$ | 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]10\\
\includegraphics[max width=\textwidth, alt={}, center]{dd7b2aee-4318-48e8-97c0-541e47f2e83a-4_335_875_262_635}
A tank containing water is in the form of a cone with vertex $C$. The axis is vertical and the semivertical angle is $60 ^ { \circ }$, 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 $t = 60$.\\
(i) Show that $h$ and $t$ satisfy a differential equation of the form
$$\frac { \mathrm { d } h } { \mathrm {~d} t } = - A h ^ { - \frac { 3 } { 2 } } ,$$
where $A$ is a positive constant.\\
(ii) Solve the differential equation given in part (i) and obtain an expression for $t$ in terms of $h$ and $H$.\\
(iii) Find the time at which the depth reaches $\frac { 1 } { 2 } H$.\\[0pt]
[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]}}