CAIE P1 2022 November — Question 4 5 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2022
SessionNovember
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicConnected Rates of Change
TypeContainer filling: find depth rate
DifficultyStandard +0.3 This is a straightforward connected rates of change problem requiring differentiation of the given volume formula, substitution of dV/dt = 3.6, and solving for dx/dt at x = 4. The main steps are mechanical: differentiate V with respect to x, apply the chain rule dV/dt = (dV/dx)(dx/dt), and convert units. While it requires careful algebra and unit conversion, it follows a standard template with no conceptual surprises, making it slightly easier than average.
Spec1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates
4 A large industrial water tank is such that, when the depth of the water in the tank is \(x\) metres, the volume \(V \mathrm {~m} ^ { 3 }\) of water in the tank is given by \(V = 243 - \frac { 1 } { 3 } ( 9 - x ) ^ { 3 }\). Water is being pumped into the tank at a constant rate of \(3.6 \mathrm {~m} ^ { 3 }\) per hour. Find the rate of increase of the depth of the water when the depth is 4 m , giving your answer in cm per minute.
Question 4:
AnswerMarks Guidance
\(\left\frac{dv}{dx}\right = (9-x)^2\)
Substitute \(x = 4\) into *their* differentiated \(v\)\*M1 Expect 25.
\(\frac{dx}{dt} = \frac{1}{\text{their derivative}} \times 3.6\) (accept \(\frac{dt}{dx} = \frac{\text{their derivative}}{3.6}\))M1 Correct use of the chain rule, ignore incorrect conversions at this point. Expect 0.144.
\(= \frac{1}{\text{their numerical derivative}} \times 3.6 \times \frac{100}{60}\)DM1 Correct use of the conversion factors.
\(= \frac{1}{25} \times 3.6 \times \frac{100}{60} = 0.24\)A1 
## Question 4:

$\left|\frac{dv}{dx}\right| = (9-x)^2$ | **B1** | Allow unsimplified forms. Allow any or no notation.

Substitute $x = 4$ into *their* differentiated $v$ | **\*M1** | Expect 25.

$\frac{dx}{dt} = \frac{1}{\text{their derivative}} \times 3.6$ (accept $\frac{dt}{dx} = \frac{\text{their derivative}}{3.6}$) | **M1** | Correct use of the chain rule, ignore incorrect conversions at this point. Expect 0.144.

$= \frac{1}{\text{their numerical derivative}} \times 3.6 \times \frac{100}{60}$ | **DM1** | Correct use of the conversion factors.

$= \frac{1}{25} \times 3.6 \times \frac{100}{60} = 0.24$ | **A1** |

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4 A large industrial water tank is such that, when the depth of the water in the tank is $x$ metres, the volume $V \mathrm {~m} ^ { 3 }$ of water in the tank is given by $V = 243 - \frac { 1 } { 3 } ( 9 - x ) ^ { 3 }$. Water is being pumped into the tank at a constant rate of $3.6 \mathrm {~m} ^ { 3 }$ per hour.

Find the rate of increase of the depth of the water when the depth is 4 m , giving your answer in cm per minute.\\

\hfill \mbox{\textit{CAIE P1 2022 Q4 [5]}}
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