Question

Find the curvature of the curve y=x^3 at the point (1,1).

Answer 1

Answer:

$\frac{3\sqrt{10} }{50}$

Step-by-step explanation:

The curvature of the curve y = f(x) is given by

k(x) = $\frac{|f"(x)|}{(1+[f'(x)]^2 )^{\frac{3}{2} } }$

~~~~~~~~~~~~~~~~

y = x³

y' = 3x²

y" = 6x

k(x) = $\frac{|6x|}{(1+[3x^2]^2)^{\frac{3}{2} } }$

The curvature of the curve y = x³ at the point (1, 1) is k(1)

k(1) = $\frac{6}{(1+[3]^2 )^{\frac{3}{2} } }$ = $\frac{6}{10\sqrt{10} }$ = $\frac{3\sqrt{10} }{50}$