Prove that the arc length of the polar curve #r = a(1-cos theta)# is #8a#?
1 Answer
The polar arc length of a curve is given by:
# L = int_alpha^beta \ sqrt(r^2 + ((dr)/(d theta))^2) \ d theta #
We have:
# r = a(1-cos theta) #
# \ \ = a-acos theta #
Thus:
# (dr)/(d theta) = asin theta #
So, the arc length is:
# L = int_0^(2pi) \ sqrt( (a-acos theta)^2 + (asin theta)^2 ) \ d theta #
# \ \ \ = int_0^(2pi) \ sqrt( (a^2-2a^2cos theta + a^2 cos^2 theta) + (asin theta)^2 ) \ d theta #
# \ \ \ = int_0^(2pi) \ asqrt( 1-2cos theta + cos^2 theta + sin^2 theta ) \ d theta #
# \ \ \ = a \ int_0^(2pi) sqrt( 2-2cos theta ) \ d theta #
# \ \ \ = a \ int_0^(2pi) sqrt( 2)sqrt(1-cos theta ) \ d theta #
Using the trig identity:
# cos2x -= 1-2sin^2x => 2sin^2x-=1-cos2x#
We then have:
# L = a \ int_0^(2pi) sqrt( 2)sqrt(2sin^2 (theta/2) ) \ d theta #
# \ \ \ = a \ int_0^(2pi) 2sin (theta/2) \ d theta #
# \ \ \ = 2a \ [ -2cos (theta/2) ]_0^(2pi)#
# \ \ \ = -4a \ (cos pi - cos 0)#
# \ \ \ = -4a \ (-1-1)#
# \ \ \ = (-4a)(-2) #
# \ \ \ = 8a # QED