| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2017 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Volumes of Revolution |
| Type | Multi-part: volume and related rates |
| Difficulty | Standard +0.3 This is a straightforward multi-part question testing standard volumes of revolution and related rates. Part (a)(i) requires routine integration with respect to y (3 marks), part (a)(ii) is standard area calculation (4 marks), and part (b) is a textbook related rates problem using the chain rule (4 marks). All techniques are standard P1/C4 material with no novel insight required, making it slightly easier than average. |
| Spec | 1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates1.08e Area between curve and x-axis: using definite integrals4.08d Volumes of revolution: about x and y axes |
| Answer | Marks | Guidance |
|---|---|---|
| 10(a)(i) | Attempt to integrate V =( π ) ∫ ( y+1 ) dy | M1 |
| Answer | Marks |
|---|---|
| | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| | A1 | AG. Must be from clear use of limits 0→ h somewhere. |
| Total: | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| 10(ii) | ∫ ( y+1 )1/2 dy ALT 6 ‒ ∫ ( x2 −1 ) dx | M1 |
| ⅔ ( y+1 )3/2 oe ALT 6 ‒ (⅓x3−x) CAO | *A1 | Result of integration must be shown |
| Answer | Marks | Guidance |
|---|---|---|
| 3 3 | DM1 | Calculation seen with limits 0→3 for y. For ALT, limits are 1→2 and |
| Answer | Marks | Guidance |
|---|---|---|
| 14/3 ALT 6 ‒ 4/3 = 14/3 | A1 | 16/3 from ⅔×8 gets DM1A0 provided work is correct up to applying |
| Answer | Marks | Guidance |
|---|---|---|
| Total: | 4 | |
| Question | Answer | Marks |
| Answer | Marks | Guidance |
|---|---|---|
| 10(b) | Clear attempt to differentiate wrt h | M1 |
| Answer | Marks |
|---|---|
| Derivative = 4π SOI | *A1 |
| Answer | Marks |
|---|---|
| their derivative | DM1 |
| Answer | Marks | Guidance |
|---|---|---|
| 4π 2π | A1 | |
| Total: | 4 | |
| Question | Answer | Marks |
Question 10:
--- 10(a)(i) ---
10(a)(i) | Attempt to integrate V =( π ) ∫ ( y+1 ) dy | M1 | Use of h in integral e.g. ∫ ( h+1 )=½h2 +h is M0. Use of ∫y2dx is M0
= ( π ) y2 + y
2
| A1
h2
=π +h
2
| A1 | AG. Must be from clear use of limits 0→ h somewhere.
Total: | 3
--- 10(ii) ---
10(ii) | ∫ ( y+1 )1/2 dy ALT 6 ‒ ∫ ( x2 −1 ) dx | M1 | Correct variable and attempt to integrate
⅔ ( y+1 )3/2 oe ALT 6 ‒ (⅓x3−x) CAO | *A1 | Result of integration must be shown
[ ] 8 1
⅔8−1 ALT 6−[ −1 − −1 ]
3 3 | DM1 | Calculation seen with limits 0→3 for y. For ALT, limits are 1→2 and
rectangle.
14/3 ALT 6 ‒ 4/3 = 14/3 | A1 | 16/3 from ⅔×8 gets DM1A0 provided work is correct up to applying
limits.
Total: | 4
Question | Answer | Marks | Guidance
--- 10(b) ---
10(b) | Clear attempt to differentiate wrt h | M1 | Expect dV =π ( h+1 ) . Allow h + 1. Allow h.
dh
Derivative = 4π SOI | *A1
2
. Can be in terms of h
their derivative | DM1
2 1
or or 0.159
4π 2π | A1
Total: | 4
Question | Answer | Marks | Guidance\begin{enumerate}[label=(\alph*)]
\item
\includegraphics{figure_1}
Fig. 1 shows part of the curve $y = x^2 - 1$ and the line $y = h$, where $h$ is a constant.
\begin{enumerate}[label=(\roman*)]
\item The shaded region is rotated through $360°$ about the $y$-axis. Show that the volume of revolution, $V$, is given by $V = \pi\left(\frac{1}{2}h^2 + h\right)$. [3]
\item Find, showing all necessary working, the area of the shaded region when $h = 3$. [4]
\end{enumerate}
\item
\includegraphics{figure_2}
Fig. 2 shows a cross-section of a bowl containing water. When the height of the water level is $h$ cm, the volume, $V$ cm$^3$, of water is given by $V = \pi\left(\frac{1}{4}h^2 + h\right)$. Water is poured into the bowl at a constant rate of 2 cm$^3$ s$^{-1}$. Find the rate, in cm s$^{-1}$, at which the height of the water level is increasing when the height of the water level is 3 cm. [4]
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2017 Q10 [11]}}