What is the arc length of #f(t)=(t^2-4t,5-t) # over #t in [3,4] #?

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Answer 1 · 2018-03-16T08:10:57

#L=1/2(2sqrt17-sqrt5)+1/4ln((4+sqrt17)/(2+sqrt5))# units.

#f(t)=(t^2−4t,5−t)#

#f'(t)=(2t−4,−1)#

Arc length is given by:

#L=int_3^4sqrt((2t-4)^2+1)dt#

Apply the substitution #2t-4=tantheta#:

#L=1/2intsec^3thetad theta#

This is a known integral:

#L=1/4[secthetatantheta+ln|sectheta+tantheta|]_3^4#

Reverse the substitution:

#L=1/4[(2t-4)sqrt((2t-4)^2+1)+ln|(2t-4)+sqrt((2t-4)^2+1)|]_3^4#

Insert the limits of integration:

#L=1/2(2sqrt17-sqrt5)+1/4ln((4+sqrt17)/(2+sqrt5))#