| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2017 |
| Session | March |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Volumes of Revolution |
| Type | Multi-part: volume and related rates |
| Difficulty | Standard +0.3 Part (i) is straightforward substitution into the pyramid volume formula with given side length. Part (ii) is a standard related rates problem requiring differentiation of V with respect to time. The geometry part appears unrelated and involves basic arc/sector calculations. All components are routine A-level techniques with no novel problem-solving required, making this slightly easier than average. |
| Spec | 1.02z Models in context: use functions in modelling1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(V = \frac{1}{12}h^3\) oe | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{dV}{dh} = \frac{1}{4}h^2\) or \(\frac{dh}{dV} = 4(12v)^{-2/3}\) | M1A1 | Attempt differentiation. Allow incorrect notation for M. For A mark accept *their* letter for volume, but otherwise correct notation. Allow \(V'\) |
| \(\frac{dh}{dt} = \frac{dh}{dV} \times \frac{dV}{dt} = \frac{4}{h^2} \times 20\) soi | DM1 | Use chain rule correctly with \(\frac{d(V)}{dt} = 20\). Any equivalent formulation. Accept non-explicit chain rule |
| \(\left(\frac{dh}{dt}\right) = \frac{4}{10^2} \times 20 = 0.8\) or equivalent fraction | A1 |
## Question 3(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $V = \frac{1}{12}h^3$ oe | B1 | |
**Total: 1 mark**
## Question 3(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{dV}{dh} = \frac{1}{4}h^2$ or $\frac{dh}{dV} = 4(12v)^{-2/3}$ | M1A1 | Attempt differentiation. Allow incorrect notation for M. For A mark accept *their* letter for volume, but otherwise correct notation. Allow $V'$ |
| $\frac{dh}{dt} = \frac{dh}{dV} \times \frac{dV}{dt} = \frac{4}{h^2} \times 20$ soi | DM1 | Use chain rule correctly with $\frac{d(V)}{dt} = 20$. Any equivalent formulation. Accept non-explicit chain rule |
| $\left(\frac{dh}{dt}\right) = \frac{4}{10^2} \times 20 = 0.8$ or equivalent fraction | A1 | |
**Total: 4 marks**
---3\\
\includegraphics[max width=\textwidth, alt={}, center]{f759ce41-708e-4fe7-80b9-adc2be2972ac-04_489_465_258_840}
The diagram shows a water container in the form of an inverted pyramid, which is such that when the height of the water level is $h \mathrm {~cm}$ the surface of the water is a square of side $\frac { 1 } { 2 } h \mathrm {~cm}$.\\
(i) Express the volume of water in the container in terms of $h$.\\[0pt]
[The volume of a pyramid having a base area $A$ and vertical height $h$ is $\frac { 1 } { 3 } A h$.]\\
Water is steadily dripping into the container at a constant rate of $20 \mathrm {~cm} ^ { 3 }$ per minute.\\
(ii) Find the rate, in cm per minute, at which the water level is rising when the height of the water level is 10 cm .\\
\includegraphics[max width=\textwidth, alt={}, center]{f759ce41-708e-4fe7-80b9-adc2be2972ac-06_403_773_258_685}
In the diagram, $A B = A C = 8 \mathrm {~cm}$ and angle $C A B = \frac { 2 } { 7 } \pi$ radians. The circular $\operatorname { arc } B C$ has centre $A$, the circular arc $C D$ has centre $B$ and $A B D$ is a straight line.\\
\hfill \mbox{\textit{CAIE P1 2017 Q3 [5]}}