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

$$\begin{gathered} \mathrm{Here}, \\ y=x^2+2x \\ \Rightarrow \frac{dy}{dt}=\left(2x+2\right)\frac{dx}{dt} \\ \Rightarrow 2x+2=1\left[\because \frac{dy}{dt}=\frac{dx}{dt}\right] \\ \Rightarrow 2x=-1 \\ \Rightarrow x=\frac{-1}{2} \\ \text{Substituting}\mathit{\text{x}}\text{=}\frac{-1}{2}\text{in}\mathit{\text{y}}\text{=}{\mathit{\text{x}}}^2\text{+2}\mathit{\text{x}}\text{, we get} \\ y=\frac{-3}{4} \\ \mathrm{Hence},\mathrm{the}\mathrm{coordinates}\mathrm{of}\mathrm{the}\mathrm{point}\mathrm{are}\left(\frac{-1}{2},\frac{-3}{4}\right). \end{gathered}$$