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September 16, 2014

September 16, 2014

Posted by **alexis** on Thursday, October 30, 2008 at 11:02pm.

r=5e^20

from 0 to 2pi

how do you find the length?

- calc -
**Count Iblis**, Thursday, October 30, 2008 at 11:20pmIf you call a length element on the curve ds, then we have:

(ds)^2 = (dr)^2 + r^2 (dtheta)^2 (1)

dr is the component of the length element in the radial direction and r d theta is the component in the direction orthogonal to that. So, by Pythagoras' therem you have to ad up the squares of both to find the squared length element.

From (1) it follws that:

ds = sqrt[(dr)^2 + r^2 (dtheta)^2] =

sqrt[r^2 + (dr/dtheta)^2] dtheta

If you integrate this from theta = 0 to 2 pi, you get the curve length:

s = Integral from zero to 2 pi of

sqrt[r^2 + (dr/dtheta)^2] dtheta

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