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sudhindra katre Grade:
        can anyone give me the proof of the formula of radius of curvature of a curve?
7 years ago

Answers : (1)

sushant singh
66 Points

hey Sudhindra....

To understand the radius of curvature proof. you need to understand it like wise as:

1.  The radius of curvature is the radius of the "osculating circle,"
    i.e., the circle that is tangent to the curve at that point.  
    Clearly the circle itself is its own osculating circle everywhere,
    and the radius is R, so that the radius of curvature is 1/K = R.
2.  Another "cheat" is to use the polar equation for the radius of
    curvature.  If the curve in polar coordinates is given by 
    r = r(@), then the radius of curvature is 

                  [r^2 + (dr/d@)^2]^(3/2)
      1/K =  --------------------------------.
              r^2 + 2(dr/d@)^2 - r*(d^2r/d@^2)

    This makes things trivial for a circle, because r(@) = R and 
    all the derivatives of r with respect to @ vanish, so we obtain

      1/K = (R^2)^(3/2) / R^2 = R^3/R^2 = R.

3.  Now let's do this in x-y coordinates, using the straightforward
    expression for the top half of the circle.  It's helpful to 
    define the variable u = R^2 - x^2.

       y = sqrt(R^2 - x^2) = sqrt(u)

    We will need the first and second derivatives of y with respect 
    to x.  

       du/dx = -2x

       dy/dx = [1/(2sqrt(u))]*(du/dx) = -2x/[2sqrt(u)] = -x/sqrt(u).

       d^2y/dx^2 = [sqrt(u)*(-1) - (-x)(dy/dx)]/u
                 = [-sqrt(u) - x^2/sqrt(u)]/u
    So we plug this into the formula for the curvature K:

                 d^2y/dx^2           [-sqrt(u) - x^2/sqrt(u)]/u
       K = ---------------------  =  --------------------------
            [1 + (dy/dx)^2]^(3/2)         [1 + x^2/u]^(3/2)

              -u^(-1/2) * [u + x^2] * u^-1     multiply terms in
           = -----------------------------     brackets by sqrt(u).
               u^(-3/2) * [u + x^2]^(3/2)
           = ----------------                  lots of cancellation!
              (u + x^2)^(1/2)

           = 1/(R^2 - x^2 + x^2)^(1/2)

           = 1/R

    and again we arrive at the expected result that the radius of
    curvature is 1/K = R for the top semicircle. You can argue the
    result for the bottom semicircle (y = -sqrt(R^2 - x^2)) either
    by symmetry or by grinding through the derivatives again.
6 years ago
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