Application of Derivatives

We are given that a point is moving along the curve described by the equation $y=x^3-3 x^2+2 x-1$ and that the $y$-coordinate of the point is increasing at a rate of 6 units per second. We need to determine the rate of chang of the $x$-coordinate of the point when the point is at $(2,-1)$.Steps to Find the Rate of Change of the $x$-coordinate:Differentiating the Curve Equation:Given $y=x^3-3 x^2+2 x-1$, differentiate $y$ with respect to $t$ to express $\;\frac{dy}{dt}$ :$\;\frac{dy}{dt}= \;\frac{d}{dt} (x^3-3 x^2+2 x-1) = (3 x^2-6 x+2) \;\frac{dx}{dt}$Using Given Information:We know $\;\frac{dy}{dt}=6$. Substitute this value into the differentiated equation:$(3 x^2-6 x+2) \;\frac{dx}{dt}=6$Solving for $\;\frac{dx}{dt}$ :Rearrange the equation to solve for $\;\frac{dx}{dt}$ :$\;\frac{dx}{dt}= \;\frac{6}{3 x^2-6 x+2}$Substitute the Point $(2,-1)$ :Substitute $x=2$ into the equation to find $\;\frac{dx}{dt}$ at this specific point:$\;\frac{dx}{dt}= \;\frac{6}{3(2)^2-6(2)+2}= \;\frac{6}{3 \times 4-6 \times 2+2}$Simplify the Expression:$\;\frac{dx}{dt}= \;\frac{6}{12-12+2}= \;\frac{6}{2}=3$Therefore, the rate of change of the $x$-coordinate of the point when at $(2,-1)$ is 3 units per second.