Question

What is the arclength of `(t^2-lnt,lnt)` on `t in [1,2]`?

Answer 1

`L=2sqrt2+1/2(sinh^(-1)1-sinh^(-1)7)+1/sqrt2ln2` units.

Explanation:

`f(t)=(t^2-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 `2t^2-1=u`:

`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 `u=tan2theta`:

`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 `tantheta=v`:

`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`