What is the arclength of (t^2-lnt,lnt)(t2−lnt,lnt) on t in [1,2]t∈[1,2]?
1 Answer
Explanation:
f(t)=(t^2-lnt,lnt)f(t)=(t2−lnt,lnt)
f'(t)=(2t-1/t, 1/t)
Arclength is given by:
L=int_1^2sqrt((2t-1/t)^2+1/t^2)dt
Simplify:
L=int_1^2sqrt((2t^2-1)^2+1)/tdt
Apply the substitution
L=1/2int_1^7sqrt(u^2+1)/(u+1)du
Rearrange:
L=1/2int_1^7(u^2+1)/(sqrt(u^2+1)(u+1))du
color(white)(L)=1/2int_1^7(u^2-1+2)/(sqrt(u^2+1)(u+1))du
Factorize:
L=1/2int_1^7((u-1)(u+1)+2)/(sqrt(u^2+1)(u+1))du
Integration is distributive:
L=1/2int_1^7(u-1)/sqrt(u^2+1)du+int_1^7 1/(sqrt(u^2+1)(u+1))du
color(white)(L)=1/2int_1^7(u/sqrt(u^2+1)-1/sqrt(u^2+1))du+int_1^7 1/(sqrt(u^2+1)(u+1))du
Apply the substitution
L=1/2[sqrt(u^2+1)-sinh^(-1)u]_1^7+2int(sec2theta)/(tan2theta+1)d theta
Apply the double-angle Trigonometric identities:
L=1/2(sqrt50-sqrt2-sinh^(-1)7+sinh^(-1)1)+2intsec^2theta/(2tantheta+1-tan^2theta)d theta
Apply the substitution
L=2sqrt2+1/2(sinh^(-1)1-sinh^(-1)7)+2int_(sqrt2-1)^((5sqrt2-1)/7)1/(2v+1-v^2)dv
Factorize the denominator:
L=2sqrt2+1/2(sinh^(-1)1-sinh^(-1)7)+2int_(sqrt2-1)^((5sqrt2-1)/7)1/((sqrt2-1+v)(sqrt2+1-v))dv
Apply partial fraction decomposition:
L=2sqrt2+1/2(sinh^(-1)1-sinh^(-1)7)+1/sqrt2int_(sqrt2-1)^((5sqrt2-1)/7)(1/(sqrt2-1+v)+1/(sqrt2+1-v))dv
Integrate directly:
L=2sqrt2+1/2(sinh^(-1)1-sinh^(-1)7)+1/sqrt2[ln|sqrt2-1+v|-ln|sqrt2+1-v|]_(sqrt2-1)^((5sqrt2-1)/7)
Simplify:
L=2sqrt2+1/2(sinh^(-1)1-sinh^(-1)7)+1/sqrt2[ln|(sqrt2-1+v)/(sqrt2+1-v)|]_(sqrt2-1)^((5sqrt2-1)/7)
Insert the limits of integration:
L=2sqrt2+1/2(sinh^(-1)1-sinh^(-1)7)+1/sqrt2ln|(12sqrt2-8)/(2sqrt2+8)*2/(2sqrt2-2)|
Simplify:
L=2sqrt2+1/2(sinh^(-1)1-sinh^(-1)7)+1/sqrt2ln2
