Posted by Jessica Chaney on Thursday, March 1, 2012 at 1:13pm.
Please help this is due tomorrow and I don't know how to Ive missed a lot of school sick
Consider the curve given by the equation x^3+3xy^2+y^3=1
a.Find dy/dx
b. Write an equation for the tangent line to the curve when x = 0.
c. Write an equation for the normal line to the curve when x = 0.
d. Find all the points on the curve where the curve has a horizontal or vertical tangent.

Calculus  Steve, Thursday, March 1, 2012 at 3:55pm
x^3+3xy^2+y^3=1
3x^2 + 3y^2 + 6xyy' + 3y^2 y' = 0
y' = (x^2+y^2)/(2xy + y^2)
when x=0, y=1
y'(0) = (0+1)/(0+1) = 1
tangent line at (0,1) is y=x+1
slope of normal = 1
line is y=x+1
only place where tangent could be horizontal (y'=0) would be at (0,0) but that point is not on the graph.
Vertical tangents where 2xy+y^2 = 0
at y=0 x=1
y' = 1/0  vertical tangent
y(2x+y) = 0
y = 2x
y' = (x^2 + 4x^2)/(2x(2x) + 4x^2) = 5x^2/0  vertical tangent
when y=2x,
x^3+3xy^2+y^3=1
x^3 + 3x(4x^2)  8x^3 = 1
x^3 + 12x^3  8x^3 = 1
5x^3 = 1
x = 0.5848
so, vertical tangent at (.5848,1.1696)
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