CAIE P1 2018 June — Question 2 4 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2018
SessionJune
Marks4
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TopicConnected Rates of Change
TypeCurve motion: find dy/dt
DifficultyModerate -0.5 This is a straightforward connected rates of change problem requiring differentiation of a simple function and substitution of given values. The calculation involves basic chain rule application (dy/dt = dy/dx × dx/dt) with no conceptual challenges, making it slightly easier than average for A-level.
Spec1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates
2 A point is moving along the curve \(y = 2 x + \frac { 5 } { x }\) in such a way that the \(x\)-coordinate is increasing at a constant rate of 0.02 units per second. Find the rate of change of the \(y\)-coordinate when \(x = 1\).
Question 2:
AnswerMarks Guidance
\(y = 2x + \dfrac{5}{x} \rightarrow \dfrac{dy}{dx} = 2 - \dfrac{5}{x^2} = -3\) (may be implied) when \(x = 1\)M1 A1 Reasonable attempt at differentiation. CAO \((-3)\)
\(\dfrac{dy}{dt} = \dfrac{dy}{dx} \times \dfrac{dx}{dt} \rightarrow -0.06\)M1 A1 Ignore notation, but needs to multiply \(\dfrac{dy}{dx}\) by \(0.02\) 
## Question 2:

$y = 2x + \dfrac{5}{x} \rightarrow \dfrac{dy}{dx} = 2 - \dfrac{5}{x^2} = -3$ (may be implied) when $x = 1$ | **M1 A1** | Reasonable attempt at differentiation. CAO $(-3)$

$\dfrac{dy}{dt} = \dfrac{dy}{dx} \times \dfrac{dx}{dt} \rightarrow -0.06$ | **M1 A1** | Ignore notation, but needs to multiply $\dfrac{dy}{dx}$ by $0.02$
2 A point is moving along the curve $y = 2 x + \frac { 5 } { x }$ in such a way that the $x$-coordinate is increasing at a constant rate of 0.02 units per second. Find the rate of change of the $y$-coordinate when $x = 1$.\\

\hfill \mbox{\textit{CAIE P1 2018 Q2 [4]}}
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