Question

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

Answer 1

`s(K)=int_a^b (dot x^2(t) + dot y^2(t))^(1/2)dt`
where:
`K` - parameterized curve `K(x(t),y(t))`
`t` - parameter
`s` - lenght
`[a,b]` - interval of the parameter

Explanation:

`a=0, b=2`
`dot x=((1+t)-t*1)/(1+t)^2=1/(1+t)^2`
`dot y=1/(1+t)`
`s(K)=int_0^2 (((1/(1+t)^2)^2+ (1/(1+t))^2)^(1/2)) dt=`
`=int_0^2 ((1/(1+t)^4+ 1/(1+t)^2)^(1/2)) dt=`
`=int_0^2 ((1+(1+t)^2)/(1+t)^4)^(1/2) dt=`
`=int_0^2 ((2+2t+t^2)/(1+t)^4)^(1/2) dt`