A particle moves along the curve y=x^2+2x. Th | vedclass

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(C) Given the equation of the curve is $y=x^2+2x$.Since the $x$ and $y$ coordinates change at the same rate,we have $\frac{dy}{dt} = \frac{dx}{dt}$.Di

Vedclass · Accepted Answer

$\left(-\frac{1}{2}, -\frac{3}{4}\right)$

Vedclass · Answer

(C) Given the equation of the curve is $y=x^2+2x$.Since the $x$ and $y$ coordinates change at the same rate,we have $\frac{dy}{dt} = \frac{dx}{dt}$.Differentiating the equation $y=x^2+2x$ with respect to $t$,we get:$\frac{dy}{dt} = (2x+2) \frac{dx}{dt}$.Substituting $\frac{dy}{dt} = \frac{dx}{dt}$ into the equation,we get:$\frac{dx}{dt} = (2x+2) \frac{dx}{dt}$.Assuming $\frac{dx}{dt} 
eq 0$,we divide both sides by $\frac{dx}{dt}$:$1 = 2x+2$.$2x = -1 \implies x = -\frac{1}{2}$.Now,find the corresponding $y$ coordinate by substituting $x = -\frac{1}{2}$ into the curve equation:$y = \left(-\frac{1}{2}ight)^2 + 2\left(-\frac{1}{2}ight) = \frac{1}{4} - 1 = -\frac{3}{4}$.Therefore,the point on the curve is $\left(-\frac{1}{2}, -\frac{3}{4}ight)$.

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

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