What is the arc length of the curve given by #r(t)= (9sqrt(2),e^(9t),e^(-9t))# on # t in [3,4]#?

Question

9507 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-12T08:44:55

A zeroth-order approximation gives #2sinh(36)-2sinh(27)# units.

#r(t)=(9sqrt2, e^(9t), e^(−9t))#

#r'(t)=(0, 9e^(9t), -9e^(−9t))#

Arc length is given by:

#L=int_3^4sqrt(0+81e^(18t)+81e^(-18t))dt#

Simplify:

#L=9int_3^4sqrt(e^(18t)+e^(-18t))dt#

Apply the identity #coshx=1/2(e^x+e^-x)#:

#L=9int_3^4sqrt(2cosh(18t))dt#

Apply the identity #cosh2x=2cosh^2x-1#:

#L=9sqrt2int_3^4sqrt(2cosh^2(9t)-1)dt#

Factor out the larger piece:

#L=18int_3^4cosh(9t)sqrt(1-1/2sech^2(9t))dt#

For #t in [3, 4]#, #1/2sech^2(9t)<1#. Take the series expansion of the square root:

#L=18int_3^4cosh(9t){sum_(n=0)^oo((1/2),(n))(-1/2sech^2(9t))^n}dt#

Isolate the #n=0# term:

#L=18int_3^4cosh(9t)dt+18sum_(n=1)^oo((1/2),(n))(-1/2)^nint_3^4sech^(2n-1)(9t)dt#

A zeroth-order approximation gives:

#L~~2[sinh(9t)]_ 3^4#

Hence:

#L~~2sinh(36)-2sinh(27)#