1. Sep 11, 2009

### Tocqueville

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!!!

2. Sep 12, 2009

### Tocqueville

I am still waiting online.......

3. Sep 12, 2009

### HallsofIvy

Staff Emeritus

What does "order of contact" of two curves mean? Does it have something to do with how many derivatives of the difference between the functions at that point are 0?

4. Sep 12, 2009

### Tocqueville

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 concept is introduced in the course "differential geometry",but Courant's textbook contains lots of material on higher geometry and numerical analysis.

Ha,I really felt good when I saw your reply.Hope you could take me out from the hopeless and lonely pit (That's how I feel when I spend 2 days on a single problem and still cannot find a way out).Thank you again ,HallsofIvy.

Last edited: Sep 12, 2009
5. Sep 12, 2009

### Tocqueville

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:" little T ,the problem is difficult and definitely beyond your ability,so take it easy,forget it ,and keep on studying math.Maybe after 4 or 5 years you will suddenly consider it as just a piece of cake.

6. Sep 12, 2009

### HallsofIvy

Staff Emeritus

Okay "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. "
And, since you are not talking about and endpoint here, P'= [(3/2)(1+ f'(x)^2)^(1/2)(2f'(x)f"(x)^2- (1+ f'(x)^2)^(3/2)f'''(x)/(f''(x))^2= 0.

What about the osculating circle? What must its equation be and what are its derivatives?

7. Sep 12, 2009

### Tocqueville

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" sculating circle have the same derivatives up to 2nd order with the primitive curve at x0.
OK,since a,b,r are 3 unknow variables,we get 3 equations,it's easy to find a,b and r.

The problem asks us to prove g'''(x0)=f'''(x0),I am stuck exactly here.

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)

8. Sep 12, 2009

### Tocqueville

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 the value of "the reciprocal of curvature",so up to this step,there is no mistakes.

when x0 is critical point
from P'(x0)=0 we get f'''(x0)={3f'(x0)*[f(x0)]^2}/[1+f'(x0)]
differentiate the function in both side up to 3rd order :(x-a)^2+(g(x)-b)^2=r^2
we get g'''(x0)={2g'(x0)g''(x0)+[g''(x0)]^2}*g''(x0)/[1+f'(x0)^2]
g(x0)=f(x0) g'(x0)=f'(x0) g''(x0)=f''(x0),we cannot get g'''(x0)=f'''(x0)

It's just the critical point ,what to do with endpoint???

This is how the problem annoys me.
My method does not work at all!!!
I need a little fresh air.

9. Sep 13, 2009