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Proof for the Radius of Curvature

Date: 08/17/2003 at 18:02:05
From: Cam Coward
Subject: A proof for the radius of curvature

The equation of a circle of radius R centered at the origin is 
x^2+y^2 = R^2. Demonstrate that the radius of curvature is equal to R.

I've found the first and second derivative but I can't seem to 
simplify the equation. I've tried several routes to solving this 
equation. The most productive seems to have been by using the trig 
identities to manipulate the equation.

Date: 08/18/2003 at 19:10:49
From: Doctor Douglas
Subject: Re: A proof for the radius of curvature

Hi Cam,

Thanks for writing to the Math Forum. You can certainly do this with 
trigonometric identities, but it's also fruitful to exploit the 
circular symmetry of the problem.  

Here are three suggestions, two of which are "cheats" and one which
uses calculus, but in a clever way.

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.  

    If you insist on parametrizing the circle by x = R*cos(t) and 
    y = R*sin(t), you should find that in the expression for the 
    curvature, things cancel in a way similar to that in method #3,

- Doctor Douglas, The Math Forum 
Associated Topics:
College Calculus
High School Calculus

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