Find the equation of the tangent line to the curve x^3 + y^3 = 6xy at the point (3,3).

Question

Find the equation of the tangent line to the curve x³ + y³ = 6xy at the point (3,3).

Answer 1

good question! Try implicit differentiation..here goes:

If x^3 + y^3 = 6xy then d/dx(x^3 + y^3 = 6xy) - in other words just differentiate as you are both sides of equation and don't forget chain rule

3x^2 +3y^2*(dy/dx) = 6(y + x*(dy/dx))

now solve for dy/dx algrebraically

3x^2 +3y^2(dy/dx) = 6y + 6x(dy/dx)

3x^2 - 6y = (6x-3y^2)*dy/dx

Therefore dy/dx = (3x^2-6y)/(6x-3y^2) at (3,3), the m of line is -1

y = mx + b

3 = (-1)(3) + b

b = 6

y = -x +6 is the tangent line.