# What is the arc length of f(t)=(sqrt(t-1),2-8t)  over t in [1,3]?

Mar 12, 2018

$L = \frac{3}{2} \sqrt{114} + \frac{1}{32} \ln \left(16 \sqrt{2} + 3 \sqrt{57}\right)$ units.

#### Explanation:

f(t)=(sqrt(t−1),2−8t)

f'(t)=(1/(2sqrt(t−1)),−8)

Arc length is given by:

L=int_1^3sqrt(1/(4(t−1))+64)dt

Apply the substitution $t - 1 = {u}^{2}$:

$L = {\int}_{0}^{\sqrt{2}} \sqrt{\frac{1}{4 {u}^{2}} + 64} \left(2 u \mathrm{du}\right)$

Simplify:

$L = {\int}_{0}^{\sqrt{2}} \sqrt{1 + 256 {u}^{2}} \mathrm{du}$

Apply the substitution $16 u = \tan \theta$:

$L = \frac{1}{16} \int {\sec}^{3} \theta d \theta$

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:

$L = \frac{1}{32} \left[\sec \theta \tan \theta + \ln | \sec \theta + \tan \theta |\right]$

Reverse the last substitution:

$L = {\left[\frac{1}{2} u \sqrt{1 + 256 {u}^{2}} + \frac{1}{32} \ln | 16 u + \sqrt{1 + 256 {u}^{2}} |\right]}_{0}^{\sqrt{2}}$

Insert the limits of integration:

$L = \frac{3}{2} \sqrt{114} + \frac{1}{32} \ln \left(16 \sqrt{2} + 3 \sqrt{57}\right)$