Consider the curve given by the equation $6xy = 2 + y^{3}$. (a) Show that $\frac{dy}{dx} = \frac{2y}{y^{2} - 2x}$. (b) Find the coordinates of a point on the curve at which the line tangent to the curve is horizontal, or explain why no such point exists. (c) Find the coordinates of a point on the curve at which the line tangent to the curve is vertical, or explain why no such point exists. (d) A particle is moving along the curve. At the instant when the particle is at the point $\left(\frac{1}{2}, -2\right)$, its horizontal position is increasing at a rate of $\frac{dx}{dt} = \frac{2}{3}$ unit per second. What is the value of $\frac{dy}{dt}$, the rate of change of the particle's vertical position, at that instant?
Consider the curve given by the equation $6xy = 2 + y^{3}$.
(a) Show that $\frac{dy}{dx} = \frac{2y}{y^{2} - 2x}$.
(b) Find the coordinates of a point on the curve at which the line tangent to the curve is horizontal, or explain why no such point exists.
(c) Find the coordinates of a point on the curve at which the line tangent to the curve is vertical, or explain why no such point exists.
(d) A particle is moving along the curve. At the instant when the particle is at the point $\left(\frac{1}{2}, -2\right)$, its horizontal position is increasing at a rate of $\frac{dx}{dt} = \frac{2}{3}$ unit per second. What is the value of $\frac{dy}{dt}$, the rate of change of the particle's vertical position, at that instant?