What is the arclength of #f(t) = (-(t+3)^2,3t-4)# on #t in [0,1]#?

Question

1502 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-06-15T07:09:05

#L=2sqrt73-9/2sqrt5+9/4ln((8+sqrt73)/(6+3sqrt5))# units.

#f(t)=(−(t+3)2,3t−4)#

#f'(t)=(−2(t+3),3)#

Arclength is given by:

#L=int_0^1sqrt(4(t+3)^2+9)dt#

Apply the substitution #2(t+3)=3tantheta#:

#L=9/2intsec^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=9/4[secthetatantheta+ln|sectheta+tantheta|]#

Reverse the substitution:

#L=[1/2(t+3)sqrt(4(t+3)^2+9)+9/4ln|2(t+3)+sqrt(4(t+3)^2+9)|]_0^1#

Insert the limits of integration:

#L=2sqrt73-9/2sqrt5+9/4ln((8+sqrt73)/(6+3sqrt5))#