HallsofIvy,I have compute it again ,I am sure I am 100 percent correct.And my method
seems unfittable to the problem.
We solve the 3 variables linear equation,and get
a=x0-f(x0)*{1+[f'[x0]^2}/f''(x0)
b=f(x0)+{1+[f'[x0]}^2}/f''(x0)
r={1+[f'[x0]^(3/2)}/f''(x0)
as you can see,r equals to...
Thank you,HallsofIvy,you are so nice.
In order to find the the derivative of the osculating circle,we can only express it in implicit function:(x-a)^2+(g(x)-b)^2=r^2.
g(x0)=f(x0) g'(x0)=f'(x0) g''(x0)=f''(x0),this is how we define "osculating circle" :osculating circle have the same...
ohhhh,I kept thinking about the problem for another day.Today as I am waiting for replys online, I have read a differential geometry textbook about radius of curvature,but it helps little.
About 60 people saw my post, no one can give me a little hint?
Now I hope someone could say to me:"...
Hi,StarWrecker,I am 100 percent sure "p = 1/t = 1/2 * (1 + 1/(t^4))^(3/2) * |t^3|"
is correct.
And I am also sure that p(t) leads to the correct evolute for the curve.
Check your answer again and again,you will find out your errors on computation.
Hi,HallsofIvy,thank you for paying attention to my problem!
Two curves f(x) and g(x) have the nth order of contact at x0 if and only if the ith derivatives
of f(x0) and g(x0) is equal when i=(1,2,3...n),and the jth derivatives of f(x0) and g(x0) does not equal when j>n.
Usually the...
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...