# What is the arc length of the curve given by r(t)= (1,t,t^2) on  t in [0, 1]?

Jun 20, 2018

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

#### Explanation:

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

$r ' \left(t\right) = \left(0 , 1 , 2 t\right)$

Arc length is given by:

$L = {\int}_{0}^{1} \sqrt{{0}^{2} + {1}^{2} + {\left(2 t\right)}^{2}} \mathrm{dt}$

Simplify:

$L = {\int}_{0}^{1} \sqrt{1 + 4 {t}^{2}} \mathrm{dt}$

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:

$L = \frac{1}{4} {\left[2 t \sqrt{1 + 4 {t}^{2}} + \ln | 2 t + \sqrt{1 + 4 {t}^{2}} |\right]}_{0}^{1}$

Insert the limits of integration:

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