For arc length for which x is in terms of t and y also in terms of t, the formula is
#s=int sqrt(((dx)/dt)^2+((dy)/dt)^2) dt#
From the given #f(t)=(sin 2t-t csc t, t^2-1)# from #t=pi/12# to
#t=(5pi)/12#
this means #x=sin 2t-t csc t# and #y=t^2-1#
find #(dx)/dt# and #(dy)/dt" "#first
solve #dx/dt#
#x=sin 2t-t csc t#
#dx/dt=2 cos 2t-(-t*csc t *cot t+csc t)=2 cos 2t+t csc t*cot t-csc t#
solve #dy/dt#
#dy/dt=2t#
Solve for the length of arc #s#
#s=int_(pi/12)^((5pi)/12) sqrt(((dx)/dt)^2+((dy)/dt)^2) dt#
#s=int_(pi/12)^((5pi)/12) sqrt((2 cos 2t+t csc t*cot t-csc t)^2+(2t)^2) dt#
Using Simpson's Rule
#s=int_(pi/12)^((5pi)/12) sqrt((2 cos 2t+t csc t*cot t-csc t)^2+(2t)^2) dt#
#s=2.095350308" "#units
God bless....I hope the explanation is useful.
