Question

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?

Answer 1

`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))`