What is the arc length of the curve given by #r(t)= (e^-t,e^t,1)# on # t in [1, 2]#?

Question

1442 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-05-29T16:06:08

#L=sum_(n=0)^oo((1/2),(n))(e^(2(1-4n))-e^(1-4n))/(1-4n)# units.

#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 #t in [1,2]#, #e^(-4t)<1#. Take the series expansion of the square root:

#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)#