What is the arc length of f(t)=(t^2-4t,5-t) f(t)=(t24t,5t) over t in [3,4] t[3,4]?

1 Answer
Mar 16, 2018

L=1/2(2sqrt17-sqrt5)+1/4ln((4+sqrt17)/(2+sqrt5))L=12(2175)+14ln(4+172+5) units.

Explanation:

f(t)=(t^2−4t,5−t)f(t)=(t24t,5t)

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))