| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9794/1 (Pre-U Mathematics Paper 1) |
| Year | 2017 |
| Session | June |
| Marks | 8 |
| Topic | Connected Rates of Change |
| Type | Chain rule with three variables |
| Difficulty | Standard +0.3 This is a straightforward connected rates of change problem requiring the chain rule with two variables (not three as stated). Students must set up P = k/V, find k = 400, differentiate to get dP/dt = -k/V² × dV/dt, and substitute values. It's a standard textbook application with clear setup and direct calculation, slightly easier than average due to its routine nature. |
| Spec | 1.02r Proportional relationships: and their graphs1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates |
State $P = \dfrac{\pm k}{V}$ **B1**
Find $k$ by substituting $P = 5$ and $V = 80$ **M1**
$P = \dfrac{400}{V}$ (may be implied by correct working or $k = 400$) **A1**
Differentiate a correct expression for $P$: $\dfrac{\text{d}P}{\text{d}V} = \dfrac{-400}{V^2}$ **M1**
State $\dfrac{\text{d}V}{\text{d}t} = 10$ or implied by use in $\dfrac{\text{d}P}{\text{d}t} = \dfrac{\text{d}P}{\text{d}V} \times \dfrac{\text{d}V}{\text{d}t}$ **B1**
Use $\dfrac{\text{d}P}{\text{d}t} = \dfrac{\text{d}P}{\text{d}V} \times \dfrac{\text{d}V}{\text{d}t}$ to obtain an expression in $V$ **M1**
Substitute $V = 80$ into correct $\dfrac{\text{d}P}{\text{d}V} = \dfrac{-400}{V^2}$ **M1**
Obtain $0.625$ (pascals) **A1**
**Total: 8 marks**12 Boyle's Law states that when a gas is kept at a constant temperature, its pressure $P$ pascals is inversely proportional to its volume $V \mathrm {~m} ^ { 3 }$.
When the volume of a certain gas is $80 \mathrm {~m} ^ { 3 }$, its pressure is 5 pascals and the rate at which the volume is increasing is $10 \mathrm {~m} ^ { 3 } \mathrm {~s} ^ { - 1 }$. Find the rate at which the pressure is decreasing at this volume.
\hfill \mbox{\textit{Pre-U Pre-U 9794/1 2017 Q12 [8]}}