What is the arc length of the polar curve #f(theta) = cos(3theta-pi/2) +thetacsc(-theta) # over #theta in [pi/12, pi/8] #?

1 Answer
Feb 10, 2016

Answer:

#L = int_(alpha1)^(alpha2) sqrt(f'^" 2"(alpha) + f^2(alpha)) " "dalpha #
#L ~~ .2037 #

Explanation:

The Arc Length, L is given:
from parametric length equation we have the familiar:
#s = int_a^b sqrt(((dx)/dt)^2 + ((dy)/dt)^2) dt # it is just the aggregate distance formula that you may have seen many time. If you think the polar coordinate as a parametric representation in #r, alpha# then:
#r = f(alpha)#
#x= rcosalpha=f(alpha) cosalpha#
# y=rsinalpha=f(alpha) sinalpha #
#[(dx)/(dalpha)]^2 = [f'(alpha)cosalpha - f(alpha) sin(alpha)]^2 #
#[(dy)/(dalpha)]^2 = [f'(alpha)sinalpha + f(alpha) cos(alpha)]^2 #
Add and Simplify:
#[(dx)/(dalpha)]^2+[(dy)/(dalpha)]^2= f'^" 2"(alpha) + f^2(alpha) #
#L = int_(alpha1)^(alpha2) sqrt(f'^" 2"(alpha) + f^2(alpha)) " "dalpha #
Now evaluate:
# f'^" 2"(alpha) =[ d/(dalpha)(cos(3alpha-pi/2) + alphacsc(-alpha))]^2 #
# f^2(alpha) =[ cos(3alpha-pi/2) + alphacsc(-alpha)]^2 #

Integrand, #I =sqrt( f'^" 2"(alpha) + f^2(alpha)); alpha = t#
#I =sqrt(3cos3t + ((t cott - 1)csct)^2 + (sin3t - tcsct)^2 #
#L = int_(pi/12)^(pi/8)(I) dt #
At this point you ca use an Online Integration Calculator and get
#L ~~ .2037 #