Post a New Question

calculus

posted by on .

The slope of the curve x^3y^2 + 2x - 5y + 2 = 0 at the point (1,1) is

  • calculus - ,

    using implicit derivatives :

    x^3(2y)dy/dx + y^2 (3x^2) + 2 - 5 dy/dx = 0
    dy/dx(2x^3 y - 5) = -3x^2 y^2 - 2
    dy/dx = (-3x^2y^2 - 2)/(2x^3y - 5)
    at the point (1,1)
    dy/dx = (-3 - 2)/(2-5) = 5/3

    now you have the slope of the tangent, and a point on that tangent.
    Use the method you commonly use to find the equation of that straight line

Answer This Question

First Name:
School Subject:
Answer:

Related Questions

More Related Questions

Post a New Question