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

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Answer 1 · 2018-05-19T16:36:42

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

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

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

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:

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

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