# How do you find the length of the curve x=t/(1+t), y=ln(1+t), where 0<=t<=2 ?

Mar 30, 2018

$s \left(K\right) = {\int}_{a}^{b} {\left({\dot{x}}^{2} \left(t\right) + {\dot{y}}^{2} \left(t\right)\right)}^{\frac{1}{2}} \mathrm{dt}$
where:
$K$ - parameterized curve $K \left(x \left(t\right) , y \left(t\right)\right)$
$t$ - parameter
$s$ - lenght
$\left[a , b\right]$ - interval of the parameter

#### Explanation:

$a = 0 , b = 2$
$\dot{x} = \frac{\left(1 + t\right) - t \cdot 1}{1 + t} ^ 2 = \frac{1}{1 + t} ^ 2$
$\dot{y} = \frac{1}{1 + t}$
$s \left(K\right) = {\int}_{0}^{2} \left({\left({\left(\frac{1}{1 + t} ^ 2\right)}^{2} + {\left(\frac{1}{1 + t}\right)}^{2}\right)}^{\frac{1}{2}}\right) \mathrm{dt} =$
$= {\int}_{0}^{2} \left({\left(\frac{1}{1 + t} ^ 4 + \frac{1}{1 + t} ^ 2\right)}^{\frac{1}{2}}\right) \mathrm{dt} =$
$= {\int}_{0}^{2} {\left(\frac{1 + {\left(1 + t\right)}^{2}}{1 + t} ^ 4\right)}^{\frac{1}{2}} \mathrm{dt} =$
$= {\int}_{0}^{2} {\left(\frac{2 + 2 t + {t}^{2}}{1 + t} ^ 4\right)}^{\frac{1}{2}} \mathrm{dt}$