# What is the arc length of f(t)=(ln(1/t) ,5-lnt)  over t in [3,4] ?

Jun 28, 2018

$L = \sqrt{2} \ln \left(\frac{4}{3}\right)$ units.

#### Explanation:

$f \left(t\right) = \left(\ln \left(\frac{1}{t}\right) , 5 - \ln t\right) = \left(- \ln t , 5 - \ln t\right)$

$f ' \left(t\right) = \left(- \frac{1}{t} , - \frac{1}{t}\right)$

Arc length is given by:

$L = {\int}_{3}^{4} \sqrt{\frac{1}{t} ^ 2 + \frac{1}{t} ^ 2} \mathrm{dt}$

Simplify:

$L = \sqrt{2} {\int}_{3}^{4} \frac{1}{t} \mathrm{dt}$

Integrate directly:

$L = \sqrt{2} {\left[\ln t\right]}_{3}^{4}$

Insert the limits of integration:

$L = \sqrt{2} \ln \left(\frac{4}{3}\right)$