# What is the arclength of (t-3t^2,t^2-t) on t in [1,2]?

May 19, 2018

$L = \frac{1}{10} \left(9 \sqrt{130} - 4 \sqrt{26}\right) + \frac{1}{10 \sqrt{10}} \ln \left(\frac{18 + 5 \sqrt{13}}{8 + \sqrt{65}}\right)$ units.

#### Explanation:

$f \left(t\right) = \left(t - 3 {t}^{2} , {t}^{2} - t\right)$

$f ' \left(t\right) = \left(1 - 6 t , 2 t - 1\right)$

Arclength is given by:

$L = {\int}_{1}^{2} \sqrt{{\left(1 - 6 t\right)}^{2} + {\left(2 t - 1\right)}^{2}} \mathrm{dt}$

Expand the squares and combine terms:

$L = {\int}_{1}^{2} \sqrt{40 {t}^{2} - 16 t + 2} \mathrm{dt}$

Complete the square in the square root:

$L = \sqrt{\frac{2}{5}} {\int}_{1}^{2} \sqrt{{\left(10 t - 2\right)}^{2} + 1} \mathrm{dt}$

Apply the substitution $10 t - 2 = \tan \theta$:

$L = \frac{1}{5 \sqrt{10}} \int {\sec}^{3} \theta d \theta$

This is a known integral:

$L = \frac{1}{10 \sqrt{10}} \left[\sec \theta \tan \theta + \ln | \sec \theta + \tan \theta |\right]$

Reverse the substitution:

$L = \frac{1}{10 \sqrt{10}} {\left[\left(10 t - 2\right) \sqrt{{\left(10 t - 2\right)}^{2} + 1} + \ln | \left(10 t - 2\right) + \sqrt{{\left(10 t - 2\right)}^{2} + 1} |\right]}_{1}^{2}$

Hence

$L = \frac{1}{10 \sqrt{10}} \left\{90 \sqrt{13} - 8 \sqrt{65} + \ln \left(\frac{18 + 5 \sqrt{13}}{8 + \sqrt{65}}\right)\right\}$

Simplify:

$L = \frac{1}{10} \left(9 \sqrt{130} - 4 \sqrt{26}\right) + \frac{1}{10 \sqrt{10}} \ln \left(\frac{18 + 5 \sqrt{13}}{8 + \sqrt{65}}\right)$