# What is the arclength of f(t) = (-(t+3)^2,3t-4) on t in [0,1]?

Jun 15, 2018

$L = 2 \sqrt{73} - \frac{9}{2} \sqrt{5} + \frac{9}{4} \ln \left(\frac{8 + \sqrt{73}}{6 + 3 \sqrt{5}}\right)$ units.

#### Explanation:

f(t)=(−(t+3)2,3t−4)

f'(t)=(−2(t+3),3)

Arclength is given by:

$L = {\int}_{0}^{1} \sqrt{4 {\left(t + 3\right)}^{2} + 9} \mathrm{dt}$

Apply the substitution $2 \left(t + 3\right) = 3 \tan \theta$:

$L = \frac{9}{2} \int {\sec}^{3} \theta d \theta$

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:

$L = \frac{9}{4} \left[\sec \theta \tan \theta + \ln | \sec \theta + \tan \theta |\right]$

Reverse the substitution:

$L = {\left[\frac{1}{2} \left(t + 3\right) \sqrt{4 {\left(t + 3\right)}^{2} + 9} + \frac{9}{4} \ln | 2 \left(t + 3\right) + \sqrt{4 {\left(t + 3\right)}^{2} + 9} |\right]}_{0}^{1}$

Insert the limits of integration:

$L = 2 \sqrt{73} - \frac{9}{2} \sqrt{5} + \frac{9}{4} \ln \left(\frac{8 + \sqrt{73}}{6 + 3 \sqrt{5}}\right)$