Consider the curve given by y^2 = 4xy + 1. Show work for all parts.
A. Find dy/dx.
B. Find all points on the curve where the tangent line to the curve has slope 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 = 4xy + 1. At time t=6, y=2 and dy/dt=7. Find dx/dt at time t=6.
y^2 = 4xy+1
2yy' = 4y + 4xy'
y' = 2y/(y-2x)
2y/(y-2x) = 2
2y = 2y-4x
x=0
so the points (0,±1) have tangent slope of 2
y'=0 when y=0, but there is no such point on the curve. y=0 is the horizontal asymptote
when y=2, x = 3/8, and dy/dx = 16/5
dy/dx = (dy/dt) / (dx/dt)
16/5 = 7 / (dx/dt)
dx/dt = 35/16