Consider the curve defined by $2y^3 + 6x^2y - 12x^2 + 6y = 1$.
(a) Show that $\dfrac{dy}{dx} = \dfrac{4x - 2xy}{x^2 + y^2 + 1}$.
(b) Write an equation of each horizontal tangent line to the curve.
(c) The line through the origin with slope $-1$ is tangent to the curve at point $P$. Find the $x$- and $y$-coordinates of point $P$.

Consider the curve given by $y^2 = 2 + xy$.
(a) Show that $\dfrac{dy}{dx} = \dfrac{y}{2y - x}$.
(b) Find all points $(x, y)$ on the curve where the line tangent to the curve has slope $\dfrac{1}{2}$.
(c) Show that there are no points $(x, y)$ on the curve where the line tangent to the curve is horizontal.
(d) Let $x$ and $y$ be functions of time $t$ that are related by the equation $y^2 = 2 + xy$. At time $t = 5$, the value of $y$ is 3 and $\dfrac{dy}{dt} = 6$. Find the value of $\dfrac{dx}{dt}$ at time $t = 5$.

Consider the curve given by the equation $y^3 - xy = 2$. It can be shown that $\dfrac{dy}{dx} = \dfrac{y}{3y^2 - x}$.
(a) Write an equation for the line tangent to the curve at the point $(-1, 1)$.
(b) Find the coordinates of all points on the curve at which the line tangent to the curve at that point is vertical.
(c) Evaluate $\dfrac{d^2y}{dx^2}$ at the point on the curve where $x = -1$ and $y = 1$.

Consider the function $y = f(x)$ whose curve is given by the equation $2y^{2} - 6 = y\sin x$ for $y > 0$.
(a) Show that $\frac{dy}{dx} = \frac{y\cos x}{4y - \sin x}$.
(b) Write an equation for the line tangent to the curve at the point $(0, \sqrt{3})$.
(c) For $0 \leq x \leq \pi$ and $y > 0$, find the coordinates of the point where the line tangent to the curve is horizontal.
(d) Determine whether $f$ has a relative minimum, a relative maximum, or neither at the point found in part (c). Justify your answer.

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 defined by the equation $x^2 + 3y + 2y^2 = 48$. It can be shown that $\frac{dy}{dx} = \frac{-2x}{3 + 4y}$.
(a) There is a point on the curve near $(2, 4)$ with $x$-coordinate 3. Use the line tangent to the curve at $(2, 4)$ to approximate the $y$-coordinate of this point.
(b) Is the horizontal line $y = 1$ tangent to the curve? Give a reason for your answer.
(c) The curve intersects the positive $x$-axis at the point $(\sqrt{48}, 0)$. Is the line tangent to the curve at this point vertical? Give a reason for your answer.
(d) For time $t \geq 0$, a particle is moving along another curve defined by the equation $y^3 + 2xy = 24$. At the instant the particle is at the point $(4, 2)$, the $y$-coordinate of the particle's position is decreasing at a rate of 2 units per second. At that instant, what is the rate of change of the $x$-coordinate of the particle's position with respect to time?

Consider the curve $G$ defined by the equation $y ^ { 3 } - y ^ { 2 } - y + \frac { 1 } { 4 } x ^ { 2 } = 0$.
A. Show that $\frac { d y } { d x } = \frac { - x } { 2 \left( 3 y ^ { 2 } - 2 y - 1 \right) }$.
B. There is a point $P$ on the curve $G$ near $( 2 , - 1 )$ with $x$-coordinate 1.6. Use the line tangent to the curve at $( 2 , - 1 )$ to approximate the $y$-coordinate of point $P$.
C. For $x > 0$ and $y > 0$, there is a point $S$ on the curve $G$ at which the line tangent to the curve at that point is vertical. Find the $y$-coordinate of point $S$. Show the work that leads to your answer.
D. A particle moves along the curve $H$ defined by the equation $2 x y + \ln y = 8$. At the instant when the particle is at the point $( 4,1 ) , \frac { d x } { d t } = 3$. Find $\frac { d y } { d t }$ at that instant. Show the work that leads to your answer.