Find arc length of #r=2\cos\theta# in the range #0\le\theta\le\pi#?

#L=\intds=\int_(\theta_1)^(\theta_2)rd\theta#

#s=r\theta# so #ds=rd\theta#, thus
#2\int_0^\pi\cos\theta=2(\sin\theta)|_0^\pi=2(\sin\pi-\sin0)=...#

Considering that this is for a semicircle perimeter of a full cycle.. #...4(\sin\theta)|_0^(\pi/2)=4(\sin(\pi/2)-\sin0)=4(1-0)=4#

It is easy to see that the curve is a circle of radius 1. It's length is obviously #2pi#

A more analytic solution would go as follows

#ds^2 = dr^2+r^2d theta^2#

So, for #r = 2 cos theta#, we have

#dr = -2 sin theta d theta#

and hence

#ds^2 = (-2 sin theta d theta)^2+(2 cos theta)^2 d theta^2 = 4d theta^2 implies#

#ds = 2 d theta#

Thus, the arc length is

#L = int_{theta = 0}^{theta=pi}ds = 2pi#

We seek the arc length of #r=2cos theta# for #theta in [0,pi]#

We calculate polar arc length using the formula:

# l = int_alpha^beta \ sqrt(r^2 + ((dr)/(d theta))^2 ) \ d theta #

Then, given that #r=2cos theta# then differentiating wrt #theta# we get:

# (dr)/(d theta) = -2sin theta #

So then:

# l = int_(0)^(pi) \ sqrt((2cos theta)^2 + (-2sin theta)^2 ) \ d theta #

# \ = int_(0)^(pi) \ sqrt(4(cos^2 theta + sin^2 theta) ) \ d theta #

# \ = int_(0)^(pi) \ sqrt(4 ) \ d theta \ \ \ \ \ \ \ \ \ (because cos^2 theta + sin^2 theta -= 1)#

# \ = int_(0)^(pi) \ 2 \ d theta #

# \ = 2[ \ theta \ ]_(0)^(pi) #

# \ = 2(pi - 0) #

# \ = 2pi#

Notes:

The observant reader will note that that #r=2cos theta# represents a circle of radius #1# centred on #(1,0)#, so we can calculate the arc length as the perimeter of a unit circle, thus>

# P = (2pi)(1) = 2pi#, as above