What is the arclength of #f(t) = (t^3-t+55,t^2-1)# on #t in [2,3]#?
1 Answer
Explanation:
#f(t)=(t^3-t+55,t^2-1)#
#f'(t)=(3t^2-1,2t)#
Arclength is given by:
#L=int_2^3sqrt((3t^2-1)^2+(2t)^2)dt#
Expand the squares:
#L=int_2^3sqrt(9t^4-2t^2+1)dt#
Complete the square in the square root:
#L=1/3int_2^3sqrt((9t^2-1)^2+8)dt#
Rearrange:
#L=1/3int_2^3(9t^2-1)sqrt(1+8/(9t^2-1)^2)dt#
For
#L=1/3int_2^3(9t^2-1){sum_(n=0)^oo((1/2),(n))(8/(9t^2-1)^2)^n}dt#
Isolate the
#L=1/3int_2^3(9t^2-1)dt+1/3sum_(n=1)^oo((1/2),(n))8^nint_2^3 1/(9t^2-1)^(2n-1)dt#
Apply partial fraction decomposition:
#L=1/3[3t^3-t]_ 2^3+2/3sum_(n=1)^oo((1/2),(n))2^nint_2^3(1/(3t-1)-1/(3t+1))^(2n-1)dt#
For ease of reading, apply the substitution
#L=56/3+2/9sum_(n=1)^oo((1/2),(n))2^nint_6^9(1/(u-1)-1/(u+1))^(2n-1)du#
The