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September 19, 2014

September 19, 2014

Posted by **Jessica Chaney** on Thursday, March 1, 2012 at 1:13pm.

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 3:55pmx^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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