| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2021 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Related rates with explicitly given non-geometric algebraic relationships |
| Difficulty | Standard +0.3 This is a standard related rates problem requiring differentiation of the given volume formula with respect to time, then substituting known values. Part (a) involves straightforward chain rule application and algebraic manipulation. Part (b) requires working backwards from dr/dt to find V. While it involves multiple steps, the techniques are routine for A-level students who have learned related rates, making it 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 | Mark | Guidance |
| \(\left[\frac{dV}{dr} =\right] \frac{9}{2}\left(r - \frac{1}{2}\right)^2\) | B1 | OE. Accept unsimplified |
| \(\frac{dr}{dt} = \frac{dr}{dV} \times \frac{dV}{dt} = \frac{1.5}{\text{their } \frac{dV}{dr}} \left[= \frac{1.5}{\frac{9}{2}\left(5.5 - \frac{1}{2}\right)^2} = \frac{1.5}{112.5}\right]\) | M1 | Correct use of chain rule with 1.5, *their* differentiated expression for \(\frac{dV}{dr}\) and using \(r = 5.5\) |
| \(0.0133\) or \(\frac{3}{225}\) or \(\frac{1}{75}\) [metres per second] | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{dV}{dr}\) or their \(\frac{dV}{dr} = \frac{1.5}{0.1}\) or \(15\) OR \(0.1 = \frac{1.5}{\text{their } \frac{dV}{dr}} \left[= \frac{2\times1.5}{9\left(r-\frac{1}{2}\right)^2} \text{ OE}\right]\) | B1 FT | Correct statement involving \(\frac{dV}{dr}\) or *their* \(\frac{dV}{dr}\), 1.5 and 0.1 |
| \(\left[\frac{9}{2}\left(r-\frac{1}{2}\right)^2 = 15 \Rightarrow\right] r = \frac{1}{2} + \sqrt{\frac{10}{3}}\) | B1 | OE e.g. AWRT 2.3. Can be implied by correct volume |
| [Volume \(=\)] \(8.13\) AWRT | B1 | OE e.g. \(\frac{-3+5\sqrt{30}}{3}\). CAO |
## Question 9(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\left[\frac{dV}{dr} =\right] \frac{9}{2}\left(r - \frac{1}{2}\right)^2$ | B1 | OE. Accept unsimplified |
| $\frac{dr}{dt} = \frac{dr}{dV} \times \frac{dV}{dt} = \frac{1.5}{\text{their } \frac{dV}{dr}} \left[= \frac{1.5}{\frac{9}{2}\left(5.5 - \frac{1}{2}\right)^2} = \frac{1.5}{112.5}\right]$ | M1 | Correct use of chain rule with 1.5, *their* differentiated expression for $\frac{dV}{dr}$ and using $r = 5.5$ |
| $0.0133$ or $\frac{3}{225}$ or $\frac{1}{75}$ [metres per second] | A1 | |
---
## Question 9(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{dV}{dr}$ or their $\frac{dV}{dr} = \frac{1.5}{0.1}$ or $15$ OR $0.1 = \frac{1.5}{\text{their } \frac{dV}{dr}} \left[= \frac{2\times1.5}{9\left(r-\frac{1}{2}\right)^2} \text{ OE}\right]$ | B1 FT | Correct statement involving $\frac{dV}{dr}$ or *their* $\frac{dV}{dr}$, 1.5 and 0.1 |
| $\left[\frac{9}{2}\left(r-\frac{1}{2}\right)^2 = 15 \Rightarrow\right] r = \frac{1}{2} + \sqrt{\frac{10}{3}}$ | B1 | OE e.g. AWRT 2.3. Can be implied by correct volume |
| [Volume $=$] $8.13$ AWRT | B1 | OE e.g. $\frac{-3+5\sqrt{30}}{3}$. CAO |
---9 The volume $V \mathrm {~m} ^ { 3 }$ of a large circular mound of iron ore of radius $r \mathrm {~m}$ is modelled by the equation $V = \frac { 3 } { 2 } \left( r - \frac { 1 } { 2 } \right) ^ { 3 } - 1$ for $r \geqslant 2$. Iron ore is added to the mound at a constant rate of $1.5 \mathrm {~m} ^ { 3 }$ per second.\\[0pt]
\begin{enumerate}[label=(\alph*)]
\item Find the rate at which the radius of the mound is increasing at the instant when the radius is 5.5 m . [3]
\item Find the volume of the mound at the instant when the radius is increasing at 0.1 m per second.
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2021 Q9 [6]}}