Question Number 11433 by FilupS last updated on 26/Mar/17

for r=(1/θ), show that the arc length between  θ=3π^(−1)   and θ=nπ^(−1)    (where  n>3)  is aproxiately  equal to the length of the line y=3π^(−1)   between the same bounds. Or show otherwise.

Commented byFilupS last updated on 26/Mar/17

Commented by@ANTARES_VY last updated on 26/Mar/17

what  is the  name  of  the  program..

Commented byFilupS last updated on 26/Mar/17

desmos

Commented bymrW1 last updated on 26/Mar/17

y=rsin θ=((sin θ)/θ)  lim_(θ→0) y=lim_(θ→0) ((sin θ)/θ)=1  that means for small θ the curve  r=(1/θ) ≈ the line y=1  is this maybe the reason for your  assumption? but it is true only for small  values of θ.

Commented byFilupS last updated on 26/Mar/17

This makes sense! Thanks

Answered by mrW1 last updated on 26/Mar/17

curve r=(1/θ):  (dr/dθ)=−(1/θ^2 )  (√(r^2 +((dr/dθ))^2 ))=(√((1/θ^2 )+(1/θ^4 )))=((√(1+θ^2 ))/θ^2 )  L=∫_θ_1  ^θ_2  (√(r^2 +((dr/dθ))^2 ))dθ=∫_θ_1  ^θ_2  ((√(1+θ^2 ))/θ^2 )dθ  [−((√(1+θ^2 ))/θ)+ln (θ+(√(1+θ^2 )))]_θ_1  ^θ_2    =[((√(1+θ_1 ^2 ))/θ_1 )−((√(1+θ_2 ^2 ))/θ_2 )+ln (((ϑ_2 +(√(1+θ_2 ^2 )))/(θ_1 +(√(1+θ_1 ^2 )))))]  with θ_1 =3π^(−1)  and θ_2 =nπ^(−1)       line y=3π^(−1) :  ⇒x=y×cot θ=3π^(−1) cot θ  x_1 =3π^(−1) cot θ_1   x_2 =3π^(−1) cot θ_2   L_1 =∣∫_x_1  ^x_2  (√(1+(y′)^2 ))dx∣=∣∫_x_1  ^x_2  dx∣=x_1 −x_2   =3π^(−1) (cot θ_1 −cot θ_2 )  =3π^(−1) [cot (3π^(−1) )−cot (nπ^(−1) )]    L≠L_1

Commented bymrW1 last updated on 26/Mar/17

Commented byajfour last updated on 26/Mar/17

length of curve = ∫(√((rdθ)^2 +(dr)^2 ))  = ∫(√(r^2 +((dr/dθ))^2 )) dθ

Commented bymrW1 last updated on 26/Mar/17

I=∫((√(1+x^2 ))/x^2 )dx  u=(√(1+x^2 ))  u′=(x/(√(1+x^2 )))  v=−(1/x)  v^′ =(1/x^2 )  I=∫uv′dx=uv−∫vu′dx  =−((√(1+x^2 ))/x)+∫((xdx)/(x(√(1+x^2 ))))  =−((√(1+x^2 ))/x)+∫(dx/(√(1+x^2 )))  =−((√(1+x^2 ))/x)+ln (x+(√(1+x^2 )))+C

Commented bymrW1 last updated on 26/Mar/17

you are right!

Commented byajfour last updated on 26/Mar/17

∫(√(r^(2 ) +((dr/dθ))^2 )) =∫(√((1/θ^2 )+(1/θ^4 ))) dθ  =∫((√(1+θ^2 ))/θ^2 )dθ

Commented byajfour last updated on 26/Mar/17

and if φ^2 =1+θ^2   φdφ =θdθ  ∫((√(1+θ^2 ))/θ^2 ) dθ = ∫(φ^2 /((φ^2 −1)^(3/2) )) d∅  otherwise if θ=cot φ  ∫((√(1+θ^2 ))/θ^2 ) dθ = −∫ ((cosec  ^3 φ)/(cot ^2 φ)) dφ  =∫ ((sec ^2 φ)/(sin φ)) dφ    unable to integrate it..