Asked by jump
The slope of the curve x^3y^2 + 2x - 5y + 2 = 0 at the point (1,1) is
Answers
Answered by
Reiny
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
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