| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2019 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Connected Rates of Change |
| Type | Chain rule with three variables |
| Difficulty | Standard +0.3 Part (i) is straightforward algebra showing S = 28x² and V = 8x³, then eliminating x. Part (ii) applies the chain rule dV/dt = (dV/dS)(dS/dt) with given values—a standard connected rates question requiring implicit differentiation but following a predictable template with no novel insight needed. |
| Spec | 1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(S = 28x^2\), \(V = 8x^3\) | B1B1 | SOI |
| \(7V^{\frac{2}{3}} = 7 \times 4x^2 = S\) | B1 | AG, WWW |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\left(\frac{dS}{dV}\right) = \frac{14V^{-\frac{1}{3}}}{3} = \frac{14}{30}\) SOI when \(V = 1000\) | *M1, A1 | Attempt to differentiate; for M mark \(\left(\frac{dS}{dV}\right)\) to be of form \(kV^{-\frac{1}{3}}\) |
| \(\left(\frac{dV}{dt} = \frac{dS}{dt} \times \frac{dV}{dS}\right)\) OE used with \(\frac{dS}{dt} = 2\) and \(\frac{1}{their \frac{14}{30}}\) | DM1 | |
| \(\frac{30}{7}\) or \(4.29\) | A1 | OE |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(V = \frac{S^{\frac{3}{2}}}{7\sqrt{7}} \rightarrow \left(\frac{dV}{dS}\right) = \frac{3}{2} \times S^{\frac{1}{2}} \times \frac{1}{7\sqrt{7}} = \frac{30}{14}\) SOI when \(S = 700\) | *M1, A1 | Attempt to differentiate; for M mark \(\left(\frac{dV}{dS}\right)\) to be of form \(kS^{\frac{1}{2}}\) |
| \(\left(\frac{dV}{dt} = \frac{dS}{dt} \times \frac{dV}{dS}\right)\) OE used with \(\frac{dS}{dt} = 2\) and \(\frac{1}{their \frac{14}{30}}\) | DM1 | |
| \(\frac{30}{7}\) or \(4.29\) | A1 | OE |
| Attempt to find either \(\frac{dV}{dx}\) or \(\left(\frac{dS}{dx}\text{ and }\frac{dV}{dS}\right)\) together with either \(\frac{dx}{dt}\) or \(x\) | *M1 | |
| \(\frac{dV}{dx} = 24x^2\) or \(\left(\frac{dS}{dx} = 56x\text{ and }\frac{dV}{dS} = \frac{3x}{7}\right)\), \(\frac{dx}{dt} = \frac{1}{140}\) or \(x = 5\) (A1) | A1 | |
| Correct method for \(\frac{dV}{dt}\) | DM1 | |
| \(\frac{30}{7}\) or \(4.29\) | A1 | OE |
## Question 5(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $S = 28x^2$, $V = 8x^3$ | B1B1 | SOI |
| $7V^{\frac{2}{3}} = 7 \times 4x^2 = S$ | B1 | AG, WWW |
## Question 5(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\left(\frac{dS}{dV}\right) = \frac{14V^{-\frac{1}{3}}}{3} = \frac{14}{30}$ SOI when $V = 1000$ | *M1, A1 | Attempt to differentiate; for M mark $\left(\frac{dS}{dV}\right)$ to be of form $kV^{-\frac{1}{3}}$ |
| $\left(\frac{dV}{dt} = \frac{dS}{dt} \times \frac{dV}{dS}\right)$ OE used with $\frac{dS}{dt} = 2$ and $\frac{1}{their \frac{14}{30}}$ | DM1 | |
| $\frac{30}{7}$ or $4.29$ | A1 | OE |
**Alternative method:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $V = \frac{S^{\frac{3}{2}}}{7\sqrt{7}} \rightarrow \left(\frac{dV}{dS}\right) = \frac{3}{2} \times S^{\frac{1}{2}} \times \frac{1}{7\sqrt{7}} = \frac{30}{14}$ SOI when $S = 700$ | *M1, A1 | Attempt to differentiate; for M mark $\left(\frac{dV}{dS}\right)$ to be of form $kS^{\frac{1}{2}}$ |
| $\left(\frac{dV}{dt} = \frac{dS}{dt} \times \frac{dV}{dS}\right)$ OE used with $\frac{dS}{dt} = 2$ and $\frac{1}{their \frac{14}{30}}$ | DM1 | |
| $\frac{30}{7}$ or $4.29$ | A1 | OE |
| Attempt to find either $\frac{dV}{dx}$ or $\left(\frac{dS}{dx}\text{ and }\frac{dV}{dS}\right)$ together with either $\frac{dx}{dt}$ or $x$ | *M1 | |
| $\frac{dV}{dx} = 24x^2$ or $\left(\frac{dS}{dx} = 56x\text{ and }\frac{dV}{dS} = \frac{3x}{7}\right)$, $\frac{dx}{dt} = \frac{1}{140}$ or $x = 5$ (A1) | A1 | |
| Correct method for $\frac{dV}{dt}$ | DM1 | |
| $\frac{30}{7}$ or $4.29$ | A1 | OE |
---5\\
\includegraphics[max width=\textwidth, alt={}, center]{17e813c6-890f-4198-b20a-557b133e8c34-06_462_878_258_635}
The dimensions of a cuboid are $x \mathrm {~cm} , 2 x \mathrm {~cm}$ and $4 x \mathrm {~cm}$, as shown in the diagram.\\
(i) Show that the surface area $S \mathrm {~cm} ^ { 2 }$ and the volume $V \mathrm {~cm} ^ { 3 }$ are connected by the relation
$$S = 7 V ^ { \frac { 2 } { 3 } }$$
(ii) When the volume of the cuboid is $1000 \mathrm {~cm} ^ { 3 }$ the surface area is increasing at $2 \mathrm {~cm} ^ { 2 } \mathrm {~s} ^ { - 1 }$. Find the rate of increase of the volume at this instant.\\
\hfill \mbox{\textit{CAIE P1 2019 Q5 [7]}}