Proof for the Radius of CurvatureDate: 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) u^0 = ---------------- 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, above. - Doctor Douglas, The Math Forum http://mathforum.org/dr.math/ |
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