EXAMPLE 3 Show that the curvature of a circle of radius k is 1/k. SOLUTION We can take the circle to have center the origin, and then a parametrization is r(t) = k cos(t) i + k sin(t) j. Therefore r'(t) = X and [r'(t) | = k SO T(t) = r'(t) |r'(t) 1 X and T'(t) = X This gives IT'(t) | = 1, so using this equation, we have K(t) = 1 IT'(t) 1 Ir' (t) 1 K