arc length and parametric curves

Question

I'm totally lost here:

x= tcost

y= tsint

-1<or= t <or=1

find the arc length...   I understand the point is to find an anti-derivative using the sqrt of each term's derivative squared and integrate from -1 to 1.  Accomplishing this seems impossible using the parametric arc length formula without a CAS.  I'm supposed to do this by hand...  please help.  

Answer 1

Robert J. did great work.

For John let me explain more detail processes.

x = t cost                         dx/dt = cost - tsint

y = t sint                          dy/dt = sint + tcost

(dβ/dt)² = (dx/dt)² + (dy/dt)² ; in here β is a arc length and dβ is a small segment of this

dβ/dt = √((dx/dt)2 + (dy/dt)2 )  

         = √( ( cost^2 -2tcost sint + t² sint^2 ) + (sint^2 +2tsint cost +t²cost^2) )

          = √(  (cost^2 + sint^2) + t²(sint^2 +cost^2)  )

           = √( 1 + t² )

β = ∫(from -1 to 1) √( 1 + t² ) dt

to solve above equation, let t = tanx , ( this is an easier way to solve this type's of problem. If you solve many problems you will know it by experiences )

t = tanx      dt = secx^2 dx  (and when t = 1, x = inverse tan 1, therefore x = pi/4)

so, when t varies from -1 to 1 it is better to calculate 2 times of integration of x from 0 to pi/4.

therefore, β = 2∫(from 0 to pi/4) √(1+tant^2) secx^2 dx = 2∫(from 0 to pi/4) √( secx^3 ) dx

after this step, you have to do the partial integration. (this process is also hard).

The result of Robert J. is correct.

Good luck.

 

Answer 2

Take derivative with respect to t,

x' = cost - tsint

y' = sint + tcost

x'^2 + y'^2 = 1 + t^2

 

arc length

= integral [-1, 1] sqrt(1+t^2) dt

Let t = tanx

dt = sec^2x dx

integral [-1, 1] sqrt(1+t^2) dt

= 2integral [0, pi/4] sec^3x dx

= sqrt(2) + ln[1+sqrt(2)], after integration by parts.