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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