Posted by mark on Monday, September 22, 2008 at 7:26pm.
Consider a plane curve which is described in polar coordinates (r, @) by r = g(@) for @ for all [a,b] (@ representing theta).
Starting from the known expression for the length of a plane curve in Cartesian coordinates and the equation x = rcos@, y = rsin@, obtain an integral expression in terms of the function g and the numbers a and b for the length of the given curve.
How would I do this question. Thanks.

calculus  Count Iblis, Monday, September 22, 2008 at 9:14pm
If ds is a length element, then:
ds^2 = dx^2 + dy^2
In polar coordinates:
x = r cos(theta)
y = r sin(theta)
Using Leibnitz rule:
dx = cos(theta)dr sin(theta) r dtheta
dy = sin(theta)dr +cos(theta) r dtheta
ds^2 = dx^2 + dy^2 =
[cos^2(theta) + sin^2(theta)] dr^2 +
[cos^2(theta) + sin^2(theta)] r^2 dtheta^2 +
2 [sin(theta)cos(theta)  cos(theta)sin(theta)] r dr dtheta.
The contents of first two square brackets are equal to one, the last is zero. So, we have:
ds^2 = dr^2 + r^2 dtheta^2
You could have imediately written down this equation. At any point along the curve, dr is a length element in the radial direction and r dtheta is the length element in the tangential direction, which is orthogonal to the radial direction. So, by Pythagoras' theorem, ds^2 is the sum of the suares of the two length elements.
It then follows that:
ds = sqrt[dr^2 + r^2 dtheta^2] =
sqrt[(dr/dtheta)^2 + r^2] dtheta
Integrating both sides over theta from theta = a to theta = b then gives the curve length.
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