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
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))#
