Find the radius of curvature of √x+√y = √a
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Given : curve √x+√y=1.
To find : Radius of curvature at the point (1/4,1/4)
Solution: √x+√y=1. => 1/2√x + (1/2√y )(dy/dx)= 0.
=> dy/dx = -√y /√x. d²y/dx² = (-√y(-1/2) /x√x - (1/2√x.√y)dy/dx. ...
=> d²y/dx² = √y /2x√x + 1/2x. R = ( 1 + (dy/dx)²)^(3/2) / (d²y/dx²) ...
radius of curvature of the curve = 1/√2.
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