Proof: K(x)= l f (x) l / (1+(f'(x)^2)^3/2) for y=f(x)

[mrmt]

Proof: K(x)= l f"(x) l / (1+(f'(x)^2)^3/2) for y=f(x)

Prove: K(x)= l f"(x) l / (1+(f'(x)^2)^3/2)


r(x)= xi + f(x)j = <x , f(x)>

r'(x)= 1i + f'(x)j= <1, f'(x)>

T(x) = r'(x)/llr'(x)ll

= <1, f'(x)> / ((1^2+(f'(x))^2)^1/2)

This is where I start to get even more lost:

T'(x) = <0, f"(x)> / ((-1/2)(1+(f'(x))^2)^(-3/2))*(0+2f'(x)f"(x))

=<0, f"(x)> / [ -(f'(x)f"(x))/(1+(f'(x))^2)^(3/2) ]
?

If anyone could help enlighten me that would be great

 

[Fredrik]

I have no idea what you're doing. You haven't even defined K. You need to include a lot more information.

 

[mrmt]

ooops...sorry

The question is determining the curvature of a curve defined by a vector valued function

K = curvature = ll T'(x) ll / ll r'(x) ll = ( ll r'(x) * r"(x) ll ) / ll r'(x) ll ^3