A curve $C$ has parametric equations \[x = \frac{\lambda}{1 + \lambda^3} \quad \text{and} \quad y

Question

A curve $C$ has parametric equations $$x = \frac{\lambda}{1 + \lambda^3} \quad \text{and} \quad y = \frac{\lambda^2}{1 + \lambda^3}, \quad \text{for}\hspace{0.5em} \lambda > 0.$$ The point $P$ is a variable point on $C$. With reference to the origin $O$, $OP$ forms the diagonal of the rectangle $OQPR$, where vertices $Q$ and $R$ lie on the $x$- and $y$-axis respectively. Using differentiation, find the value of $\lambda$ which maximises the area of rectangle $OQPR$. You need to show that your answer gives a maximum.[5] When the area of rectangle $OQPR$ is at its maximum, the rate of change of the $x$-coordinate of the point $P$ is $1$ unit per second. Find $\frac{\mathrm{d}y}{\mathrm{d}x}$ and hence determine the rate of change of the $y$-coordinate of the point $P$ at this instant.[4]

Accepted Answer

(a) $\lambda=1$ (b) $\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{2\lambda-\lambda^4}{1-2\lambda^3}$, $\dfrac{\mathrm dy}{\mathrm dt}=-1\text{ unit per second}$

2024 RI Promo Q4

2024 RI Promo Q4

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