Find the equation of the line tangent to the curve y3yx2x23y

Solution

To find the slope, take the derivative dy/dx and evaluate it at x=0. When x = 0, 0 + 0 + y^3 = 1, so y =1. x^3 + 3xy^2 + y^3 = 1 (d/dx)(x^3 + 3xy^2 + y^3) = (d/dx)1 3x^2 + 3x2y(dy/dx) + 3y^2 + 3y^2(dy/dx) = 0 (3x^2 + 3y^2) + (dy/dx)(6xy + 3y^2) = 0 dy/dx = (6xy+3y^2)/(-3x^2 + 3y^2) = -(2xy + y^2)/(x^2 + y^2) when x = 0, y = 1, and dy/dx = -1/1 = -1 Thus your slope is -1. Your tangent line equation is thus y = -x + b. To find the intercept, you know that the tangent contains the point x = 0, y = 1. Thus 1 = -0 + b, so b = 1 Thus the tangent line equation is y = -x + 1.