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.
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.
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.
WyzAnt.com
Comments
First off, thanks for the help! I'm taking calc 2 online and I think there is a big gap in my curriculum... where did sqrt(1+t^2) come from? (not in my text), why did you choose t= tanx? How did you know to change the limits of integration to (0, pi/4)?
- John R. from Apex, NC 11/30/2012(cost - tsint)^2 + (sint + tcost)^2 = 1 + t^2, since -/+2tsintcost terms cancelled, and cos^2t + sin^2t = 1
Pick t = tanx because 1+tan^2x = sec^2x. In this way, you can get rid of radical.
Using symmetry, integral [-1, 1] sqrt(1+t^2) dt = 2integral [0, 1] sqrt(1+t^2) dt. When t = 0, tanx = 0, x = 0, the lower limit; and when t = 1, tanx = 1, x = pi/4, the upper limit.
- Robert J. 11/30/2012