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

Arc length $l = \sqrt{2} \cdot \left(\ln 4 - \ln 3\right)$

$l = 0.406844$

#### Explanation:

$l = \int \sqrt{{\left(\frac{\mathrm{dy}}{\mathrm{dt}}\right)}^{2} + {\left(\frac{\mathrm{dx}}{\mathrm{dt}}\right)}^{2}} \mathrm{dt}$

Solve for $\frac{\mathrm{dx}}{\mathrm{dt}}$ given $x = \ln t$

$\frac{\mathrm{dx}}{\mathrm{dt}} = \left(\frac{1}{t}\right)$

Solve for $\frac{\mathrm{dy}}{\mathrm{dt}}$ given $y = 5 - \ln t$

$\frac{\mathrm{dy}}{\mathrm{dt}} = - 1 \cdot \frac{1}{t}$

Solve for the length $l$

$l = {\int}_{a}^{b} \sqrt{{\left(\frac{\mathrm{dy}}{\mathrm{dt}}\right)}^{2} + {\left(\frac{\mathrm{dx}}{\mathrm{dt}}\right)}^{2}} \mathrm{dt}$

$a = 3$ and $b = 4$

$l = {\int}_{3}^{4} \sqrt{{\left(\frac{\mathrm{dy}}{\mathrm{dt}}\right)}^{2} + {\left(\frac{\mathrm{dx}}{\mathrm{dt}}\right)}^{2}} \mathrm{dt}$

$l = {\int}_{3}^{4} \sqrt{{\left(- \frac{1}{t}\right)}^{2} + {\left(\frac{1}{t}\right)}^{2}} \mathrm{dt}$

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

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

$l = \sqrt{2} \cdot \left(\ln 4 - \ln 3\right)$

$l = 0.406844$

God bless....I hope the explanation is useful.