What is the arc length of the curve given by #r(t)= (e^-t,e^t,1)# on # t in [1, 2]#?
1 Answer
May 29, 2018
Explanation:
#r(t)=(e^-t,e^t,1)#
#r'(t)=(-e^-t,e^t,0)#
Arc length is given by:
#L=int_1^2sqrt(e^(-2t)+e^(2t)+0)dt#
Rearrange:
#L=int_1^2e^tsqrt(1+e^(-4t))dt#
For
#L=int_1^2e^t{sum_(n=0)^oo((1/2),(n))e^(-4nt)}dt#
Simplify:
#L=sum_(n=0)^oo((1/2),(n))int_1^2e^((1-4n)t)dt#
Integrate directly:
#L=sum_(n=0)^oo((1/2),(n))[e^((1-4n)t)]_1^2/(1-4n)#
Hence
#L=sum_(n=0)^oo((1/2),(n))(e^(2(1-4n))-e^(1-4n))/(1-4n)#