What is the arclength of #(t-3,t^2)# on #t in [1,2]#?

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Answer 1 · 2018-03-08T03:55:55

The arclength is #1/2(2sqrt17-sqrt5)+1/4ln((4+sqrt17)/(2+sqrt5))# units.

#f(t)=(t-3,t^2)#

#f'(t)=(1,2t)#

Arclength is given by:

#L=int_1^2sqrt(1+4t^2)dt#

Apply the substitution #2t=tantheta#:

#L=1/2intsec^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=1/4[secthetatantheta+ln|sectheta+tantheta|]#

Reverse the substitution:

#L=1/4[2tsqrt(1+4t^2)+ln|2t+sqrt(1+4t^2)|]_1^2#

Insert the limits of integration:

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