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.
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.