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

Mar 8, 2018

The arclength is $\frac{1}{2} \left(2 \sqrt{17} - \sqrt{5}\right) + \frac{1}{4} \ln \left(\frac{4 + \sqrt{17}}{2 + \sqrt{5}}\right)$ units.

#### Explanation:

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

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

Arclength is given by:

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

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

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

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:

$L = \frac{1}{4} \left[\sec \theta \tan \theta + \ln | \sec \theta + \tan \theta |\right]$

Reverse the substitution:

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

Insert the limits of integration:

$L = \frac{1}{2} \left(2 \sqrt{17} - \sqrt{5}\right) + \frac{1}{4} \ln \left(\frac{4 + \sqrt{17}}{2 + \sqrt{5}}\right)$