How do you find the length of the curve x=t/(1+t)x=t1+t, y=ln(1+t)y=ln(1+t), where 0<=t<=20≤t≤2 ?

1 Answer
AltairSafir
Mar 30, 2018

s(K)=int_a^b (dot x^2(t) + dot y^2(t))^(1/2)dts(K)=∫ba(.x2(t)+.y2(t))12dt
where:
KK - parameterized curve K(x(t),y(t))K(x(t),y(t))
tt - parameter
ss - lenght
[a,b][a,b] - interval of the parameter

Explanation:

a=0, b=2a=0,b=2
dot x=((1+t)-t*1)/(1+t)^2=1/(1+t)^2.x=(1+t)−t⋅1(1+t)2=1(1+t)2
dot y=1/(1+t).y=11+t
s(K)=int_0^2 (((1/(1+t)^2)^2+ (1/(1+t))^2)^(1/2)) dt=s(K)=∫20⎛⎜ ⎜⎝⎛⎝(1(1+t)2)2+(11+t)2⎞⎠12⎞⎟ ⎟⎠dt=
=int_0^2 ((1/(1+t)^4+ 1/(1+t)^2)^(1/2)) dt==∫20⎛⎜⎝(1(1+t)4+1(1+t)2)12⎞⎟⎠dt=
=int_0^2 ((1+(1+t)^2)/(1+t)^4)^(1/2) dt==∫20(1+(1+t)2(1+t)4)12dt=
=int_0^2 ((2+2t+t^2)/(1+t)^4)^(1/2) dt=∫20(2+2t+t2(1+t)4)12dt

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