What is the arc length of the curve given by x = t^2-t and y= t^2 -1, for 1<t<5?

1 Answer
Jun 18, 2018

L=1/8(19sqrt181-3sqrt5)+1/(8sqrt2)ln((19+sqrt262)/(3+sqrt10)) units.

Explanation:

x=t^2-t
x'=2t-1

y=t^2-1
y'=2t

Arc length is given y:

L=int_1^5sqrt((2t-1)^2+(2t)^2)dt

Expand the squares:

L=int_1^5sqrt(8t^2-4t+1)dt

Complete the square:

L=1/sqrt2int_1^5sqrt((4t-1)^2+1)dt

Apply the substitution 4t-1=tantheta:

L=1/(4sqrt2)intsec^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/(8sqrt2)[secthetatantheta+ln|sectheta+tantheta|]

Reverse the substitution:

L=1/(8sqrt2)[(4t-1)sqrt((4t-1)^2+1)+ln|(4t-1)+sqrt((4t-1)^2+1)|]_1^5

Insert the limits of integration:

L=1/(8sqrt2)(19sqrt262-3sqrt10+ln((19+sqrt262)/(3+sqrt10)))

Hence

L=1/8(19sqrt181-3sqrt5)+1/(8sqrt2)ln((19+sqrt262)/(3+sqrt10))