Find arc length given #x=t\sint#, #y=t\cost# and #0\let\le1#?

I got up to #\int_0^1\sqrt(1+t^2)dt# using the parametric arc length formula #\color(indianred)(\int_a^b\sqrt((\dotx)^2+(\doty)^2)dt#, but a solution I found online changes the bounds from #\int_0^1# to #\color(palevioletred)(\int_0^(\tan^-1(1))#...

I assume the 0 is because #\tan^-1(0)=0#, but why arctangent? Is it because #t=\tan\theta# from trigonometric substitution?

1 Answer
May 28, 2018

#L=1/2(sqrt2+ln(1+sqrt2))# units.

Explanation:

#x=tsint#
#x'=sint+tcost#

#y=tcost#
#y'=cost-tsint#

Arc length is given by:

#L=int_0^1sqrt((sint+tcost)^2+(cost-tsint)^2)dt#

Expand and simplify:

#L=int_0^1sqrt(1+t^2)dt#

Apply the substitution #t=tantheta#:

#L=int_0^(tan^(-1)(1))sec^3thetad theta#

This is a known integral. If you do not have it memorized look it up in a table of integrals or apply integration by parts:

#L=1/2[secthetatantheta+ln|sectheta+tantheta|]_0^(tan^(-1)(1))#

Insert the limits of integration:

#L=1/2(sqrt2+ln(1+sqrt2))#