| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2015 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Related rates with cones, hemispheres, and bowls (variable depth) |
| Difficulty | Standard +0.3 This is a standard related rates problem requiring basic geometry (30-60-90 triangle), volume calculation, and implicit differentiation. The geometry is given away by the 30° angles, and the chain rule application dV/dt = dV/dh × dh/dt is routine for A-level mechanics. Slightly easier than average due to the straightforward setup and standard technique. |
| Spec | 1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\tan 60 = \frac{x}{h} \rightarrow x = h\tan 60\) | B1 | Any correct unsimplified length |
| \(A = h \times x\) | M1 | Correct method for area |
| \(V = 40\sqrt{(3h^2)}\) | A1 [3] | ag |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{dV}{dh} = 80\sqrt{(3h)}\) | B1 | B1 |
| If \(h=5\), \(\frac{dh}{dt} = \frac{1}{2\sqrt{(3)}}\) or \(0.289\) | M1A1 [3] | M1 (must be \(\div\), not \(\times\)) |
## Question 3:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\tan 60 = \frac{x}{h} \rightarrow x = h\tan 60$ | **B1** | Any correct unsimplified length |
| $A = h \times x$ | **M1** | Correct method for area |
| $V = 40\sqrt{(3h^2)}$ | **A1** [3] | ag |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{dV}{dh} = 80\sqrt{(3h)}$ | **B1** | **B1** |
| If $h=5$, $\frac{dh}{dt} = \frac{1}{2\sqrt{(3)}}$ or $0.289$ | **M1A1** [3] | M1 (must be $\div$, not $\times$) |
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\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{9cdb00a6-1e86-4185-bb73-ed3ecab981ba-2_369_863_799_296}
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\caption{Fig. 1}
\end{center}
\end{figure}
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{9cdb00a6-1e86-4185-bb73-ed3ecab981ba-2_225_652_943_1192}
\captionsetup{labelformat=empty}
\caption{Fig. 2}
\end{center}
\end{figure}
Fig. 1 shows an open tank in the shape of a triangular prism. The vertical ends $A B E$ and $D C F$ are identical isosceles triangles. Angle $A B E =$ angle $B A E = 30 ^ { \circ }$. The length of $A D$ is 40 cm . The tank is fixed in position with the open top $A B C D$ horizontal. Water is poured into the tank at a constant rate of $200 \mathrm {~cm} ^ { 3 } \mathrm {~s} ^ { - 1 }$. The depth of water, $t$ seconds after filling starts, is $h \mathrm {~cm}$ (see Fig. 2).\\
(i) Show that, when the depth of water in the tank is $h \mathrm {~cm}$, the volume, $V \mathrm {~cm} ^ { 3 }$, of water in the tank is given by $V = ( 40 \sqrt { } 3 ) h ^ { 2 }$.\\
(ii) Find the rate at which $h$ is increasing when $h = 5$.
\hfill \mbox{\textit{CAIE P1 2015 Q3 [6]}}