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
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
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))