A particle moves along the curve $y=x^2+2 x$. Then, the point on the curve such that $x$ and $y$ coordinates of the particle change with the same rate is

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Aicharya · Accepted Answer

Given equation of curve is $ y=x^2+2 x $ On differentiating both sides w.r.t. $t$, we get $ \frac{d y}{d t}=(2 x+2) \frac{d x}{d t} $

$ \begin{aligned} \Rightarrow & & 2 x & =-1 \\ \Rightarrow & & x & =-1 / 2, y=-3 / 4 \end{aligned} $ $\therefore$ Point on the curve is $\left(-\frac{1}{2},-\frac{3}{4}\right)$

A particle moves along the curve $y=x^2+2 x$. Then, the point on the curve such that $x$ and $y$ coordinates…

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