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

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(B) Given the equation of the curve is $y = x^2 + 2x$. We are given that the $x$ and $y$ coordinates change at the same rate,which means $\frac{dx}{dt

Vedclass · Accepted Answer

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

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(B) Given the equation of the curve is $y = x^2 + 2x$. We are given that the $x$ and $y$ coordinates change at the same rate,which means $\frac{dx}{dt} = \frac{dy}{dt}$. This implies $\frac{dy/dt}{dx/dt} = 1$. Differentiating the equation $y = x^2 + 2x$ with respect to $t$,we get: $\frac{dy}{dt} = (2x + 2) \frac{dx}{dt}$. Dividing both sides by $\frac{dx}{dt}$,we get: $\frac{dy/dt}{dx/dt} = 2x + 2$. Since $\frac{dy/dt}{dx/dt} = 1$,we have: $1 = 2x + 2$. $2x = 1 - 2 = -1$. $x = -\frac{1}{2}$. Now,substitute $x = -\frac{1}{2}$ into the original equation to find $y$: $y = (-\frac{1}{2})^2 + 2(-\frac{1}{2}) = \frac{1}{4} - 1 = -\frac{3}{4}$. Thus,the required point is $\left( -\frac{1}{2}, -\frac{3}{4} \right)$.

$A$ particle moves along the curve $y = x^2 + 2x$. At what point on the curve do the $x$ and $y$ coordinates of the particle change at the same rate?

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