
Let us now transform this expression into polar coördinates by means
of the substitutions x = \rho cos \theta, y = \rho sin \theta.

Then dx = -\rho sin \theta d\theta + cos \theta d\rho,

dy = \rho cos \theta d\theta + sin \theta d\rho,

and we have

[dx^2 + dy^2]^1/2 = [(-\rho sin \theta d\theta + cos \theta d\rho)^2 + (\rho cos \theta d\theta + sin \theta d\rho)^2]^1/2
= [\rho^2 d\theta^2 + d\rho^2]^1/2.

If the equation of the curve is

\rho = f(\theta),

then d\rho = f'(\theta) d\theta = d\rho /d\theta d\theta.

Substituting this in the above differential expression, we get

[\rho^2 + (d\rho / d\theta)^2]^1/2 d\theta.

If then \alpha and \beta are the limits of the independent variable \theta corresponding
to the limits in (A) and (B), p. 374,
we get the formula for the length of the arc,

(A) s = $$ [\rho^2 + (d\rho / d\theta)^2]^1/2 d\theta,

where \rho and d\rho / d\theta in terms of \theta must be substituted
from the equation of the given curve.

In case it is more convenient to use \rho as the independent variable,
and the equation is in the form

\theta = \phi(\rho),

then d\theta = \phi'(\rho) d\rho = d\theta /d\rho d\rho.

Substituting this in [\rho^2 d\theta^2 + d\rho^2]^1/2
gives [\rho^2 (d\theta /d\rho)^2 + 1]^1/2 d\rho.

Hence if \rho_1 and \rho_2 are the corresponding limits of the independent
variable \rho, we get the formula for the length of the arc,

(B) s = $$ [\rho^2 (d\theta /d\rho)^2 + 1]^1/2 d\rho,

where d\theta /d\rho in terms of \rho must be substituted from the equation of the
given curve. 
