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Question Number 190615 by mr W last updated on 07/Apr/23

Commented by mr W last updated on 08/Apr/23

1. find ratio (a_1 /a)=? such that the right  circular cone is cut into two parts   with equal volume as shown.  2. find in which ratio the lateral   surface of the right circular cone is  cut.

1.findratioa1a=?suchthattherightcircularconeiscutintotwopartswithequalvolumeasshown.2.findinwhichratiothelateralsurfaceoftherightcircularconeiscut.

Answered by mr W last updated on 08/Apr/23

Commented by mr W last updated on 11/Apr/23

Part I  volume of cone V=((πr^2 h)/3)  a=AC=(√(r^2 +h^2 ))  ((DC)/(sin α))=((BD)/(sin θ))=((2r)/(sin (θ+α)))=((2r(√(r^2 +h^2 )))/(r sin α+h cos α))  DC=((2r sin α(√(r^2 +h^2 )))/(r sin α+h cos α))  BD=((2rh)/(r sin α+h cos α))  a_1 =AC−DC=(√(r^2 +h^2 ))(1−((2r sin α)/(r sin α+h cos α)))  λ=(a_1 /a)=1−((2r sin α)/(r sin α+h cos α))=1−(2/(1+(h/(r tan α))))  the cut section is an ellipse with semi  axes p and q.  2p=BD=((2rh)/(r sin α+h cos α))  ⇒p=((rh)/(r sin α+h cos α))  BE=(r/(cos α))  FE=a−BE=((rh)/(r sin α+h cos α))−(r/(cos α))  OE=r tan α  AE=h−r tan α  ((EG)/r)=((AE)/h)=1−((r tan α)/h)  EG=r−((r^2  tan α)/h)  (((FE)/p))^2 +(((EG)/q))^2 =1  (((((rh)/(r sin α+h cos α))−(r/(cos α)))/((rh)/(r sin α+h cos α))))^2 +(((r−((r^2  tan α)/h))/q))^2 =1  tan^2  α+(((h−r tan α)/q))^2 =(h^2 /r^2 )  q=((h−r tan α)/( (√((h^2 /r^2 )−tan^2  α))))=r(√((h cos α−r sin α)/(h cos α+r sin α)))  h_1 =AE sin β=AE cos α=h cos α−r sin α  V_1 =((πpqh_1 )/3)=(π/3)×((rh)/(r sin α+h cos α))×r(√((h cos α−r sin α)/(h cos α+r sin α)))×(h cos α−r sin α)    V_1 =((πr^2 h)/3)×(((h cos α−r sin α)/(h cos α+r sin α)))^(3/2)   V_1 =V×(((h cos α−r sin α)/(h cos α+r sin α)))^(3/2) =(V/2)  (((h cos α−r sin α)/(h cos α+r sin α)))^(3/2) =(1/2)  ⇒((h cos α−r sin α)/(h cos α+r sin α))=(1/( (4)^(1/3) ))  ⇒tan α=((((4)^(1/3) −1)h)/(((4)^(1/3) +1)r))  λ=(a_1 /a)=1−(2/(1+(((4)^(1/3) +1)/( (4)^(1/3) −1))))=(1/( (4)^(1/3) ))≈0.63 ✓  generally for (V_1 /V)=(1/n):         (a_1 /a)=(1/( (n^2 )^(1/3) ))

PartIvolumeofconeV=πr2h3a=AC=r2+h2DCsinα=BDsinθ=2rsin(θ+α)=2rr2+h2rsinα+hcosαDC=2rsinαr2+h2rsinα+hcosαBD=2rhrsinα+hcosαa1=ACDC=r2+h2(12rsinαrsinα+hcosα)λ=a1a=12rsinαrsinα+hcosα=121+hrtanαthecutsectionisanellipsewithsemiaxespandq.2p=BD=2rhrsinα+hcosαp=rhrsinα+hcosαBE=rcosαFE=aBE=rhrsinα+hcosαrcosαOE=rtanαAE=hrtanαEGr=AEh=1rtanαhEG=rr2tanαh(FEp)2+(EGq)2=1(rhrsinα+hcosαrcosαrhrsinα+hcosα)2+(rr2tanαhq)2=1tan2α+(hrtanαq)2=h2r2q=hrtanαh2r2tan2α=rhcosαrsinαhcosα+rsinαh1=AEsinβ=AEcosα=hcosαrsinαV1=πpqh13=π3×rhrsinα+hcosα×rhcosαrsinαhcosα+rsinα×(hcosαrsinα)V1=πr2h3×(hcosαrsinαhcosα+rsinα)32V1=V×(hcosαrsinαhcosα+rsinα)32=V2(hcosαrsinαhcosα+rsinα)32=12hcosαrsinαhcosα+rsinα=143tanα=(431)h(43+1)rλ=a1a=121+43+1431=1430.63generallyforV1V=1n:a1a=1n23

Commented by mr W last updated on 10/Apr/23

Commented by mr W last updated on 09/Apr/23

Commented by mr W last updated on 11/Apr/23

Part II  we have got tan α=((h((4)^(1/3) −1))/(r((4)^(1/3) +1))) or  ((r tan α)/h)=(((4)^(1/3) −1)/( (4)^(1/3) +1))=(1/m)  rγ=aφ  ⇒γ=((aφ)/r)=(√(1+((h/r))^2 )) φ=kφ  with k=(√(1+((h/r))^2 ))  ϕ=((πr)/a)=(π/k)  (ρ/a)=(r_1 /r) ⇒r_1 =(ρ/k)  (ρ/a)=1−((r(1−(ρ/a) cos γ)tan α)/h)  ⇒(ρ/a)=((m−1)/(m−cos kφ))  S_1 =2∫_0 ^ϕ ((ρ^2 dφ)/2)       =a^2 ∫_0 ^ϕ (((m−1)^2 dφ)/((m−cos kφ)^2 ))  S=ϕa^2 =((πa^2 )/k)  (S_1 /S)=((k(m−1)^2 )/π)∫_0 ^ϕ (dφ/((m−cos kφ)^2 ))    =(((m−1)^2 )/π)∫_0 ^π (dγ/((m−cos γ)^2 ))    =(((m−1)^2 )/π)×((mπ)/((m−1)^(3/2) (m+1)^(3/2) ))    =(m/(m+1))(√((m−1)/(m+1)))     =((1+(4)^(1/3) )/4) ≈0.6469  generally for (V_1 /V)=(1/n):          (S_1 /S)=((1+(n^2 )^(1/3) )/(2n))

PartIIwehavegottanα=h(431)r(43+1)orrtanαh=43143+1=1mrγ=aϕγ=aϕr=1+(hr)2ϕ=kϕwithk=1+(hr)2φ=πra=πkρa=r1rr1=ρkρa=1r(1ρacosγ)tanαhρa=m1mcoskϕS1=20φρ2dϕ2=a20φ(m1)2dϕ(mcoskϕ)2S=φa2=πa2kS1S=k(m1)2π0φdϕ(mcoskϕ)2=(m1)2π0πdγ(mcosγ)2=(m1)2π×mπ(m1)32(m+1)32=mm+1m1m+1=1+4340.6469generallyforV1V=1n:S1S=1+n232n

Commented by mr W last updated on 11/Apr/23

Commented by mr W last updated on 11/Apr/23

Commented by mr W last updated on 11/Apr/23

Commented by mr W last updated on 11/Apr/23

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