Question

A particle moves along the curve y = x² + 2x. At what point(s) on the curve are the x and y coordinates of the particle changing at the same rate?

Answer 1

\[\text { Here, } \]

\[y = x^2 + 2x\]

\[ \Rightarrow \frac{dy}{dt} = \left( 2x + 2 \right)\frac{dx}{dt}\]

\[ \Rightarrow 2x + 2 = 1 \left[ \because \frac{dy}{dt} = \frac{dx}{dt} \right]\]

\[ \Rightarrow 2x = - 1\]

\[ \Rightarrow x = \frac{- 1}{2}\]

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

\[y = \frac{- 3}{4}\]

\[\text { Hence, the coordinates of the point are } \left( \frac{- 1}{2}, \frac{- 3}{4} \right) .\]