What is the arc length of #f(t)=(sqrt(t-1),2-8t) # over #t in [1,3]#?

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Answer 1 · 2018-03-12T08:58:49

#L=3/2sqrt114+1/32ln(16sqrt2+3sqrt57)# units.

#f(t)=(sqrt(t−1),2−8t)#

#f'(t)=(1/(2sqrt(t−1)),−8)#

Arc length is given by:

#L=int_1^3sqrt(1/(4(t−1))+64)dt#

Apply the substitution #t-1=u^2#:

#L=int_0^sqrt2sqrt(1/(4u^2)+64)(2udu)#

Simplify:

#L=int_0^sqrt2sqrt(1+256u^2)du#

Apply the substitution #16u=tantheta#:

#L=1/16intsec^3thetad theta#

This is a known integral. If you do not have it memorized apply integration by parts or look it up in a table of integrals:

#L=1/32[secthetatantheta+ln|sectheta+tantheta|]#

Reverse the last substitution:

#L=[1/2usqrt(1+256u^2)+1/32ln|16u+sqrt(1+256u^2)|]_0^sqrt2#

Insert the limits of integration:

#L=3/2sqrt114+1/32ln(16sqrt2+3sqrt57)#