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

\[\text { Here,} \]

\[y = \frac{2}{3} x^3 + 1\]

\[ \Rightarrow \frac{dy}{dt} = 2 x^2 \frac{dx}{dt}\]

\[ \Rightarrow 2\frac{dx}{dt} = 2 x^2 \frac{dx}{dt} \left[ \because \frac{dy}{dt} = 2\frac{dx}{dt} \right]\]

\[ \Rightarrow x = \pm 1\]

\[\text { Substituting the value of }x=1 \text { and } x=-1\text { in y } = \frac{2}{3} x^3 + 1, \text { we get }\]

\[ \Rightarrow y = \frac{5}{3} \text { and } y = \frac{1}{3}\]

\[\text { So, the points are }\left( 1, \frac{5}{3} \right)\text { and }\left( - 1, \frac{1}{3} \right).\]