The radius of curvature of curve y2 = x3 + 8 at (-2, 0)
⇒ the formula for radius of curvature is given by R = [1+ (y')2]3/2 ÷ (y")2
⇒ differentiate y2 = x3 + 8
\(2\frac{dy}{dx}=3(x^{3-1})+0\)
\(2\frac{dy}{dx}=3x^2\)
\(\frac{dy}{dx}=\frac{3}{2}x^2\) ⇒ \(y'=\frac{3}{2}x^2\)
⇒ differentiate y' = \(\frac{1}{2}x^2\)
\(y''=\frac{3}{2}(2^{x^2-1})\)
y'' = 3x
y' = 3/2 x2 at (-2, 0) = y' = 3/2 (-2)2 = 3/2 (4) = 6
y" = 3x at (-2, 0) = y" = 3(-2) = -6
radius R = [1 + (6)2]3/2 ÷ (-6)2
= [1 + 63] ÷ 36
= 37 ÷ 36 = 1.028 units
