CAIE P1 2023 March — Question 3 4 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2023
SessionMarch
Marks4
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Mark schemeDownload PDF ↗
TopicConnected Rates of Change
TypeEqual rate condition
DifficultyModerate -0.3 This is a straightforward connected rates of change problem requiring implicit differentiation to find dy/dt, setting dy/dt = dx/dt, and solving a simple quadratic equation. While it requires understanding of the chain rule and connected rates, the algebraic manipulation is routine and the problem type is standard for P1 level.
Spec1.07k Differentiate trig: sin(kx), cos(kx), tan(kx)1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates
3 A curve has equation \(y = \frac { 1 } { 60 } ( 3 x + 1 ) ^ { 2 }\) and a point is moving along the curve.
Find the \(x\)-coordinate of the point on the curve at which the \(x\) - and \(y\)-coordinates are increasing at the same rate.
Question 3:
AnswerMarks Guidance
\(\frac{dy}{dx} = \left\{\frac{1}{60}(3x+1)\times 2\right\}\times\{3\}\)B1 B1 May see \(\frac{1}{60}(18x+6)\)
\(\frac{1}{10}(3x+1) = 1\)M1 Equate *their* \(\frac{dy}{dx}\) to 1
\(x = 3\)A1
4 
## Question 3:

| $\frac{dy}{dx} = \left\{\frac{1}{60}(3x+1)\times 2\right\}\times\{3\}$ | B1 B1 | May see $\frac{1}{60}(18x+6)$ |
|---|---|---|
| $\frac{1}{10}(3x+1) = 1$ | M1 | Equate *their* $\frac{dy}{dx}$ to 1 |
| $x = 3$ | A1 | |
| | **4** | |

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3 A curve has equation $y = \frac { 1 } { 60 } ( 3 x + 1 ) ^ { 2 }$ and a point is moving along the curve.\\
Find the $x$-coordinate of the point on the curve at which the $x$ - and $y$-coordinates are increasing at the same rate.\\

\hfill \mbox{\textit{CAIE P1 2023 Q3 [4]}}
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