calculus
posted by Little on .
Consider the curve given by the equation y^3+3x^2y+13=0
a.find dy/dx
b. Write an equation for the line tangent to the curve at the point (2,1)
c. Find the minimum ycoordinate of any point on the curve.
the work for these would be appreciated i don't need the answers.

differentiate implicitly.
3y^2 dy/dx + 3x^2 dy/dx + 6xy + 0 = 0
dy/dx(3y^2 + 3x^2) = 6xy
dy/dx/ = 6xy/(3y^2 + 3x^2)
b) sub in x = 2, to get the slope, then use the grade 9 way to find the equation of a line given the slope and a point.
c) set dy/dx = 0
so 6xy = 0
x = 0 or y = 0
when x = 0, y^3 +1 = 13
y = (13)^(1/3) = appr.  2.35
when y = 0, 0 + 0 + 13 = 0 , not possible.
so we have one turning point at (0,2.35)
what about when x becomes large either + or 
investigate for x = 100
y^3 + 30000y = 13
y = 13/30000 which approaches zero.
the same thing is true if x = 100
so the minimum value of y is 2.35 when x = 0