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

Mar 19, 2018

$L = 2 \sqrt{2} + \frac{1}{2} \left({\sinh}^{- 1} 1 - {\sinh}^{- 1} 7\right) + \frac{1}{\sqrt{2}} \ln 2$ units.

#### Explanation:

$f \left(t\right) = \left({t}^{2} - \ln t , \ln t\right)$

$f ' \left(t\right) = \left(2 t - \frac{1}{t} , \frac{1}{t}\right)$

Arclength is given by:

$L = {\int}_{1}^{2} \sqrt{{\left(2 t - \frac{1}{t}\right)}^{2} + \frac{1}{t} ^ 2} \mathrm{dt}$

Simplify:

$L = {\int}_{1}^{2} \frac{\sqrt{{\left(2 {t}^{2} - 1\right)}^{2} + 1}}{t} \mathrm{dt}$

Apply the substitution $2 {t}^{2} - 1 = u$:

$L = \frac{1}{2} {\int}_{1}^{7} \frac{\sqrt{{u}^{2} + 1}}{u + 1} \mathrm{du}$

Rearrange:

$L = \frac{1}{2} {\int}_{1}^{7} \frac{{u}^{2} + 1}{\sqrt{{u}^{2} + 1} \left(u + 1\right)} \mathrm{du}$

$\textcolor{w h i t e}{L} = \frac{1}{2} {\int}_{1}^{7} \frac{{u}^{2} - 1 + 2}{\sqrt{{u}^{2} + 1} \left(u + 1\right)} \mathrm{du}$

Factorize:

$L = \frac{1}{2} {\int}_{1}^{7} \frac{\left(u - 1\right) \left(u + 1\right) + 2}{\sqrt{{u}^{2} + 1} \left(u + 1\right)} \mathrm{du}$

Integration is distributive:

$L = \frac{1}{2} {\int}_{1}^{7} \frac{u - 1}{\sqrt{{u}^{2} + 1}} \mathrm{du} + {\int}_{1}^{7} \frac{1}{\sqrt{{u}^{2} + 1} \left(u + 1\right)} \mathrm{du}$

$\textcolor{w h i t e}{L} = \frac{1}{2} {\int}_{1}^{7} \left(\frac{u}{\sqrt{{u}^{2} + 1}} - \frac{1}{\sqrt{{u}^{2} + 1}}\right) \mathrm{du} + {\int}_{1}^{7} \frac{1}{\sqrt{{u}^{2} + 1} \left(u + 1\right)} \mathrm{du}$

Apply the substitution $u = \tan 2 \theta$:

$L = \frac{1}{2} {\left[\sqrt{{u}^{2} + 1} - {\sinh}^{- 1} u\right]}_{1}^{7} + 2 \int \frac{\sec 2 \theta}{\tan 2 \theta + 1} d \theta$

Apply the double-angle Trigonometric identities:

$L = \frac{1}{2} \left(\sqrt{50} - \sqrt{2} - {\sinh}^{- 1} 7 + {\sinh}^{- 1} 1\right) + 2 \int {\sec}^{2} \frac{\theta}{2 \tan \theta + 1 - {\tan}^{2} \theta} d \theta$

Apply the substitution $\tan \theta = v$:

$L = 2 \sqrt{2} + \frac{1}{2} \left({\sinh}^{- 1} 1 - {\sinh}^{- 1} 7\right) + 2 {\int}_{\sqrt{2} - 1}^{\frac{5 \sqrt{2} - 1}{7}} \frac{1}{2 v + 1 - {v}^{2}} \mathrm{dv}$

Factorize the denominator:

$L = 2 \sqrt{2} + \frac{1}{2} \left({\sinh}^{- 1} 1 - {\sinh}^{- 1} 7\right) + 2 {\int}_{\sqrt{2} - 1}^{\frac{5 \sqrt{2} - 1}{7}} \frac{1}{\left(\sqrt{2} - 1 + v\right) \left(\sqrt{2} + 1 - v\right)} \mathrm{dv}$

Apply partial fraction decomposition:

$L = 2 \sqrt{2} + \frac{1}{2} \left({\sinh}^{- 1} 1 - {\sinh}^{- 1} 7\right) + \frac{1}{\sqrt{2}} {\int}_{\sqrt{2} - 1}^{\frac{5 \sqrt{2} - 1}{7}} \left(\frac{1}{\sqrt{2} - 1 + v} + \frac{1}{\sqrt{2} + 1 - v}\right) \mathrm{dv}$

Integrate directly:

$L = 2 \sqrt{2} + \frac{1}{2} \left({\sinh}^{- 1} 1 - {\sinh}^{- 1} 7\right) + \frac{1}{\sqrt{2}} {\left[\ln | \sqrt{2} - 1 + v | - \ln | \sqrt{2} + 1 - v |\right]}_{\sqrt{2} - 1}^{\frac{5 \sqrt{2} - 1}{7}}$

Simplify:

$L = 2 \sqrt{2} + \frac{1}{2} \left({\sinh}^{- 1} 1 - {\sinh}^{- 1} 7\right) + \frac{1}{\sqrt{2}} {\left[\ln | \frac{\sqrt{2} - 1 + v}{\sqrt{2} + 1 - v} |\right]}_{\sqrt{2} - 1}^{\frac{5 \sqrt{2} - 1}{7}}$

Insert the limits of integration:

$L = 2 \sqrt{2} + \frac{1}{2} \left({\sinh}^{- 1} 1 - {\sinh}^{- 1} 7\right) + \frac{1}{\sqrt{2}} \ln | \frac{12 \sqrt{2} - 8}{2 \sqrt{2} + 8} \cdot \frac{2}{2 \sqrt{2} - 2} |$

Simplify:

$L = 2 \sqrt{2} + \frac{1}{2} \left({\sinh}^{- 1} 1 - {\sinh}^{- 1} 7\right) + \frac{1}{\sqrt{2}} \ln 2$