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

##### 1 Answer
Jun 16, 2018

$L = \frac{1}{8} \left(3 \sqrt{5} + 5 \sqrt{13}\right) + \frac{1}{8 \sqrt{2}} \ln \left(\frac{\sqrt{10} + 3}{\sqrt{26} - 5}\right)$ units.

#### Explanation:

$f \left(t\right) = \left({t}^{2} - t , {t}^{2} - 1\right)$

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

Arclength is given by:

$L = {\int}_{-} {1}^{1} \sqrt{{\left(2 t - 1\right)}^{2} + 4 {t}^{2}} \mathrm{dt}$

Expand the square:

$L = {\int}_{-} {1}^{1} \sqrt{8 {t}^{2} - 4 t + 1} \mathrm{dt}$

Complete the square:

$L = \frac{1}{\sqrt{2}} {\int}_{-} {1}^{1} \sqrt{{\left(4 t - 1\right)}^{2} + 1} \mathrm{dt}$

Apply the substitution $4 t - 1 = \tan \theta$:

$L = \frac{1}{4 \sqrt{2}} \int {\sec}^{3} \theta d \theta$

This is a known integral. If you do not have it memorized apply integration by parts or look it up in a table of integrals:

$L = \frac{1}{8 \sqrt{2}} \left[\sec \theta \tan \theta + \ln | \sec \theta + \tan \theta |\right]$

Reverse the substitution:

$L = \frac{1}{8 \sqrt{2}} {\left[\left(4 t - 1\right) \sqrt{{\left(4 t - 1\right)}^{2} + 1} + \ln | 4 t - 1 + \sqrt{{\left(4 t - 1\right)}^{2} + 1} |\right]}_{-} {1}^{1}$

Insert the limits of integration:

$L = \frac{1}{8 \sqrt{2}} \left(3 \sqrt{10} + 5 \sqrt{26} + \ln \left(\frac{3 + \sqrt{10}}{- 5 + \sqrt{26}}\right)\right)$

Hence:

$L = \frac{1}{8} \left(3 \sqrt{5} + 5 \sqrt{13}\right) + \frac{1}{8 \sqrt{2}} \ln \left(\frac{\sqrt{10} + 3}{\sqrt{26} - 5}\right)$