# What is the arc length of r(t)=(t^2,2t,4-t) on tin [0,5]?

May 24, 2018

$L = 5 \sqrt{105} + \frac{5}{2} \ln \left(2 \sqrt{5} + \sqrt{21}\right)$

#### Explanation:

r(t)=(t^2,2t,4−t)

r'(t)=(2t,2,−1)

Arc length is given by:

$L = {\int}_{0}^{5} \sqrt{4 {t}^{2} + 4 + 1} \mathrm{dt}$

Simplify:

$L = {\int}_{0}^{5} \sqrt{4 {t}^{2} + 5} \mathrm{dt}$

Apply the substitution $2 t = \sqrt{5} \tan \theta$:

$L = \frac{5}{2} \int {\sec}^{3} \theta d \theta$

This is a known integral:

$L = \frac{5}{2} \left[\sec \theta \tan \theta + \ln | \sec \theta + \tan \theta |\right]$

Reverse the substitution:

$L = {\left[t \sqrt{4 {t}^{2} + 5} + \frac{5}{2} \ln | 2 t + \sqrt{4 {t}^{2} + 5} |\right]}_{0}^{5}$

Hence

$L = 5 \sqrt{105} + \frac{5}{2} \ln \left(2 \sqrt{5} + \sqrt{21}\right)$