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for-r-1-show-that-the-arc-length-between-3pi-1-and-npi-1-where-n-gt-3-is-aproxiately-equal-to-the-length-of-the-line-y-3pi-1-between-the-same-bounds-Or-show-otherwise-




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.
forr=1θ,showthatthearclengthbetweenθ=3π1andθ=nπ1(wheren>3)isaproxiatelyequaltothelengthoftheliney=3π1betweenthesamebounds.Orshowotherwise.
Commented by FilupS last updated on 26/Mar/17
Commented by @ANTARES_VY last updated on 26/Mar/17
what  is the  name  of  the  program..
whatisthenameoftheprogram..
Commented by FilupS last updated on 26/Mar/17
desmos
desmos
Commented by mrW1 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.
y=rsinθ=sinθθlimθ0y=limθ0sinθθ=1thatmeansforsmallθthecurver=1θtheliney=1isthismaybethereasonforyourassumption?butitistrueonlyforsmallvaluesofθ.
Commented by FilupS last updated on 26/Mar/17
This makes sense! Thanks
Thismakessense!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
curver=1θ:drdθ=1θ2r2+(drdθ)2=1θ2+1θ4=1+θ2θ2L=θ1θ2r2+(drdθ)2dθ=θ1θ21+θ2θ2dθ[1+θ2θ+ln(θ+1+θ2)]θ1θ2=[1+θ12θ11+θ22θ2+ln(ϑ2+1+θ22θ1+1+θ12)]withθ1=3π1andθ2=nπ1liney=3π1:x=y×cotθ=3π1cotθx1=3π1cotθ1x2=3π1cotθ2L1=∣x1x21+(y)2dx∣=∣x1x2dx∣=x1x2=3π1(cotθ1cotθ2)=3π1[cot(3π1)cot(nπ1)]LL1
Commented by mrW1 last updated on 26/Mar/17
Commented by ajfour last updated on 26/Mar/17
length of curve = ∫(√((rdθ)^2 +(dr)^2 ))  = ∫(√(r^2 +((dr/dθ))^2 )) dθ
lengthofcurve=(rdθ)2+(dr)2=r2+(drdθ)2dθ
Commented by mrW1 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
I=1+x2x2dxu=1+x2u=x1+x2v=1xv=1x2I=uvdx=uvvudx=1+x2x+xdxx1+x2=1+x2x+dx1+x2=1+x2x+ln(x+1+x2)+C
Commented by mrW1 last updated on 26/Mar/17
you are right!
youareright!
Commented by ajfour last updated on 26/Mar/17
∫(√(r^(2 ) +((dr/dθ))^2 )) =∫(√((1/θ^2 )+(1/θ^4 ))) dθ  =∫((√(1+θ^2 ))/θ^2 )dθ
r2+(drdθ)2=1θ2+1θ4dθ=1+θ2θ2dθ
Commented by ajfour 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..
andifϕ2=1+θ2ϕdϕ=θdθ1+θ2θ2dθ=ϕ2(ϕ21)3/2dotherwiseifθ=cotϕ1+θ2θ2dθ=cosec3ϕcot2ϕdϕ=sec2ϕsinϕdϕunabletointegrateit..

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