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

Explanation

`y=x^{2}+2 x `


`Rightarrow frac{d y}{d t}=(2 x+2) frac{d x}{d t} `


`Rightarrow 2 x+2=1[because frac{d y}{d t}=frac{d x}{d t}] `


`Rightarrow 2 x=-1 `


`Rightarrow x=frac{-1}{2}`



Substituting `{x}=frac{-1}{2}` in `{y}=x^{2}+2 x`, we get `y=frac{-3}{4}`


Hence, the coordinates of the point are `(frac{-1}{2}, frac{-3}{4})`.