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

Here,


`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}`

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