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

Explanation

`y=frac{2}{3} x^{3}+1 `


`Rightarrow frac{d y}{d t}=2 x^{2} frac{d x}{d t} `


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


`Rightarrow x=pm 1`



Substituting the value of `x=1` and `x=-1` in `y=frac{2}{3} x^{3}+1`, we get `Rightarrow y=frac{5}{3}` and `y=frac{1}{3}`


So, the points are `(1, frac{5}{3})` and `(-1, frac{1}{3})`.