What is the arclength of #(sint/(t+cos2t),cost/(2t))# on #t in [pi/12,pi/3]#?

Question

1652 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-10-14T04:15:56

Use #L = int_(pi/12)^(pi/3)sqrt((dx/dt)^2+ (dy/dt)^2)dt~~ 2.34#

Given:
#x = sin(t)/(t + cos(2t))#
#y = cos(t)/2t#

#dx/dt = (sin(t) (2 sin(2 t)-1)+cos(t) (t+cos(2 t)))/(t+cos(2 t))^2#

#dy/dt = -(t sin(t)+cos(t))/(2 t^2)#

#L = int_(alpha)^(beta)sqrt((dx/dt)^2+ (dy/dt)^2)dt#

#L = int_(pi/12)^(pi/3)sqrt(((sin(t) (2 sin(2 t)-1)+cos(t) (t+cos(2 t)))/(t+cos(2 t))^2)^2 + (-(t sin(t)+cos(t))/(2 t^2))^2)dt#

#L ~~ 2.34#