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

  1. Apr 11, 2012 #1
    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
     
  2. jcsd
  3. Apr 11, 2012 #2

    Fredrik

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

    I have no idea what you're doing. You haven't even defined K. You need to include a lot more information.
     
  4. Apr 11, 2012 #3
    Re: Proof: K(x)= l f"(x) l / (1+(f'(x)^2)^3/2) for y=f(x)

    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
     
    Last edited: Apr 11, 2012
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