A particle moves along the curve . At what point(s) on the curve, is the y-coordinate changing 8…

(c) $(4,11)$ and $(-4, -\frac{31}{3})$ Differentiating $6y = x^3 + 2$ with respect to $t$ gives $6\frac{dy}{dt} = 3x^2 \frac{dx}{dt}$. Substituting $\frac{dy}{dt} = 8\frac{dx}{dt}$ yields $6(8)\frac{dx}{dt} = 3x^2 \frac{dx}{dt} \Rightarrow 48 = 3x^2 \Rightarrow x^2 = 16 \Rightarrow x = \pm 4$. When $x = 4$, $6y = 4^3 + 2 = 66 \Rightarrow y = 11$. When $x = -4$, $6y = (-4)^3 + 2 = -62 \Rightarrow y = -\frac{31}{3}$. Therefore, the points are $(4, 11)$ and $(-4, -\frac{31}{3})$.