Please help me!!!A Problem about radius of curvature Please forgive my poor English! I have been thinking about a problem in Courant's <Introduction to Calculus and Analysis> for 2 days.After trying all possible methods, I am now exhausted ,almost give up and lose hope.really need someone's help! Book 1,Page 480,Section5.6 problem 4: Show that the order of contact of a curve and its osculating circle is at least three at points where the radius of curvature is a maximum or minimum. Here is what I have figured out: Curve K:y=f(x) its osculating circle:(x-a)^2+(g(x)-b)^2=r^2 find 1st derivative:(x-a)+[g(x)-b]g'(x)=0 find 2nd derivative:1+[g'(x)]^2+[g(x)-b]g''(x)=0 find 3rd derivative:g'''(x) is a function of g(x),g'(x)and g''(X) the radius of curvature P(x)= (1+[f'(x)]^2)^(3/2)/f''(X) at point x0,the radius of curvature is a maximum or minimum,hence x0 must be critical point or end point. If x0 is a critical point ,then P'(X0)=0,but what I find is that f'''(x0) does not equal g'''(x)!!! If x0 is end point ,I have no idea what to do. Could you do me a favour to give me some hints?I really appreciate your help!!!