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)
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.
1 answer