Question

A particle moves along the curve y = (2/3)x³ + 1. Find the points on the curve at which the y-coordinate is changing twice as fast as the x-coordinate.

Answer 1

$$\begin{gathered} \mathrm{Here}, \\ y=\frac{2}{3}{x}^3+1 \\ \Rightarrow \frac{dy}{dt}=2x^2\frac{dx}{dt} \\ \Rightarrow 2\frac{dx}{dt}=2x^2\frac{dx}{dt}\left[\because \frac{dy}{dt}=2\frac{dx}{dt}\right] \\ \Rightarrow x=\pm 1 \\ \text{Substituting the value of}\mathit{\text{x}}\text{=1 and}\mathit{\text{x}}\text{=-1 in}y=\frac{2}{3}{x}^3+1,\text{we get} \\ \Rightarrow y=\frac{5}{3}\mathrm{and}y=\frac{1}{3} \\ \text{So, the points are}\left(1,\frac{5}{3}\right)\text{and}\left(-1,\frac{1}{3}\right)\text{.} \end{gathered}$$