Question

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

Answer 1

`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))`