What is the arclength of #r=3/4theta # on #theta in [-pi,pi]#?

1 Answer
May 19, 2018

Answer:

#L=3/4pisqrt(pi^2+1)+3/4ln(pi+sqrt(pi^2+1))# units.

Explanation:

#r=3/4theta#
#r^2=9/16theta^2#

#r'=3/4#
#(r')^2=9/16#

Arclength is given by:

#L=int_-pi^pisqrt(9/16theta^2+9/16)d theta#

Simplify:

#L=3/4int_-pi^pisqrt(theta^2+1)d theta#

From symmetry:

#L=3/2int_0^pisqrt(theta^2+1)d theta#

Apply the substitution #theta=tanphi#:

#L=3/2intsec^3phidphi#

This is a known integral:

#L=3/4[secphitanphi+ln|secphi+tanphi|]#

Reverse the substitution:

#L=3/4[thetasqrt(theta^2+1)+ln|theta+sqrt(theta^2+1)|]_0^pi#

Insert the limits of integration:

#L=3/4pisqrt(pi^2+1)+3/4ln(pi+sqrt(pi^2+1))#