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 of the particle change at the same rate,we have $\frac{dy}{dt}

Vedclass · Accepted Answer

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

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(C) Given the equation of the curve is $y = x^2 + 2x$.Since the $x$ and $y$ coordinates of the particle change at the same rate,we have $\frac{dy}{dt} = \frac{dx}{dt}$.Differentiating the equation of the curve 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 can divide both sides by $\frac{dx}{dt}$:$1 = 2x + 2$$2x = -1$$x = -\frac{1}{2}$.Now,substitute $x = -\frac{1}{2}$ into the curve equation to find $y$:$y = \left(-\frac{1}{2}ight)^2 + 2\left(-\frac{1}{2}ight)$$y = \frac{1}{4} - 1 = -\frac{3}{4}$.Therefore,the required point is $\left(-\frac{1}{2}, -\frac{3}{4}ight)$.

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

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