Question-190615

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.

$$\mathrm{1}.\:{find}\:{ratio}\:\frac{{a}_{\mathrm{1}} }{{a}}=?\:{such}\:{that}\:{the}\:{right} \\ $$$${circular}\:{cone}\:{is}\:{cut}\:{into}\:{two}\:{parts}\: \\ $$$${with}\:{equal}\:{volume}\:{as}\:{shown}. \\ $$$$\mathrm{2}.\:{find}\:{in}\:{which}\:{ratio}\:{the}\:{lateral}\: \\ $$$${surface}\:{of}\:{the}\:{right}\:{circular}\:{cone}\:{is} \\ $$$${cut}. \\ $$

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) ))

$$\boldsymbol{{Part}}\:\boldsymbol{{I}} \\ $$$${volume}\:{of}\:{cone}\:{V}=\frac{\pi{r}^{\mathrm{2}} {h}}{\mathrm{3}} \\ $$$${a}={AC}=\sqrt{{r}^{\mathrm{2}} +{h}^{\mathrm{2}} } \\ $$$$\frac{{DC}}{\mathrm{sin}\:\alpha}=\frac{{BD}}{\mathrm{sin}\:\theta}=\frac{\mathrm{2}{r}}{\mathrm{sin}\:\left(\theta+\alpha\right)}=\frac{\mathrm{2}{r}\sqrt{{r}^{\mathrm{2}} +{h}^{\mathrm{2}} }}{{r}\:\mathrm{sin}\:\alpha+{h}\:\mathrm{cos}\:\alpha} \\ $$$${DC}=\frac{\mathrm{2}{r}\:\mathrm{sin}\:\alpha\sqrt{{r}^{\mathrm{2}} +{h}^{\mathrm{2}} }}{{r}\:\mathrm{sin}\:\alpha+{h}\:\mathrm{cos}\:\alpha} \\ $$$${BD}=\frac{\mathrm{2}{rh}}{{r}\:\mathrm{sin}\:\alpha+{h}\:\mathrm{cos}\:\alpha} \\ $$$${a}_{\mathrm{1}} ={AC}−{DC}=\sqrt{{r}^{\mathrm{2}} +{h}^{\mathrm{2}} }\left(\mathrm{1}−\frac{\mathrm{2}{r}\:\mathrm{sin}\:\alpha}{{r}\:\mathrm{sin}\:\alpha+{h}\:\mathrm{cos}\:\alpha}\right) \\ $$$$\lambda=\frac{{a}_{\mathrm{1}} }{{a}}=\mathrm{1}−\frac{\mathrm{2}{r}\:\mathrm{sin}\:\alpha}{{r}\:\mathrm{sin}\:\alpha+{h}\:\mathrm{cos}\:\alpha}=\mathrm{1}−\frac{\mathrm{2}}{\mathrm{1}+\frac{{h}}{{r}\:\mathrm{tan}\:\alpha}} \\ $$$${the}\:{cut}\:{section}\:{is}\:{an}\:{ellipse}\:{with}\:{semi} \\ $$$${axes}\:{p}\:{and}\:{q}. \\ $$$$\mathrm{2}{p}={BD}=\frac{\mathrm{2}{rh}}{{r}\:\mathrm{sin}\:\alpha+{h}\:\mathrm{cos}\:\alpha} \\ $$$$\Rightarrow{p}=\frac{{rh}}{{r}\:\mathrm{sin}\:\alpha+{h}\:\mathrm{cos}\:\alpha} \\ $$$${BE}=\frac{{r}}{\mathrm{cos}\:\alpha} \\ $$$${FE}={a}−{BE}=\frac{{rh}}{{r}\:\mathrm{sin}\:\alpha+{h}\:\mathrm{cos}\:\alpha}−\frac{{r}}{\mathrm{cos}\:\alpha} \\ $$$${OE}={r}\:\mathrm{tan}\:\alpha \\ $$$${AE}={h}−{r}\:\mathrm{tan}\:\alpha \\ $$$$\frac{{EG}}{{r}}=\frac{{AE}}{{h}}=\mathrm{1}−\frac{{r}\:\mathrm{tan}\:\alpha}{{h}} \\ $$$${EG}={r}−\frac{{r}^{\mathrm{2}} \:\mathrm{tan}\:\alpha}{{h}} \\ $$$$\left(\frac{{FE}}{{p}}\right)^{\mathrm{2}} +\left(\frac{{EG}}{{q}}\right)^{\mathrm{2}} =\mathrm{1} \\ $$$$\left(\frac{\frac{{rh}}{{r}\:\mathrm{sin}\:\alpha+{h}\:\mathrm{cos}\:\alpha}−\frac{{r}}{\mathrm{cos}\:\alpha}}{\frac{{rh}}{{r}\:\mathrm{sin}\:\alpha+{h}\:\mathrm{cos}\:\alpha}}\right)^{\mathrm{2}} +\left(\frac{{r}−\frac{{r}^{\mathrm{2}} \:\mathrm{tan}\:\alpha}{{h}}}{{q}}\right)^{\mathrm{2}} =\mathrm{1} \\ $$$$\mathrm{tan}^{\mathrm{2}} \:\alpha+\left(\frac{{h}−{r}\:\mathrm{tan}\:\alpha}{{q}}\right)^{\mathrm{2}} =\frac{{h}^{\mathrm{2}} }{{r}^{\mathrm{2}} } \\ $$$${q}=\frac{{h}−{r}\:\mathrm{tan}\:\alpha}{\:\sqrt{\frac{{h}^{\mathrm{2}} }{{r}^{\mathrm{2}} }−\mathrm{tan}^{\mathrm{2}} \:\alpha}}={r}\sqrt{\frac{{h}\:\mathrm{cos}\:\alpha−{r}\:\mathrm{sin}\:\alpha}{{h}\:\mathrm{cos}\:\alpha+{r}\:\mathrm{sin}\:\alpha}} \\ $$$${h}_{\mathrm{1}} ={AE}\:\mathrm{sin}\:\beta={AE}\:\mathrm{cos}\:\alpha={h}\:\mathrm{cos}\:\alpha−{r}\:\mathrm{sin}\:\alpha \\ $$$${V}_{\mathrm{1}} =\frac{\pi{pqh}_{\mathrm{1}} }{\mathrm{3}}=\frac{\pi}{\mathrm{3}}×\frac{{rh}}{{r}\:\mathrm{sin}\:\alpha+{h}\:\mathrm{cos}\:\alpha}×{r}\sqrt{\frac{{h}\:\mathrm{cos}\:\alpha−{r}\:\mathrm{sin}\:\alpha}{{h}\:\mathrm{cos}\:\alpha+{r}\:\mathrm{sin}\:\alpha}}×\left({h}\:\mathrm{cos}\:\alpha−{r}\:\mathrm{sin}\:\alpha\right) \\ $$$$ \\ $$$${V}_{\mathrm{1}} =\frac{\pi{r}^{\mathrm{2}} {h}}{\mathrm{3}}×\left(\frac{{h}\:\mathrm{cos}\:\alpha−{r}\:\mathrm{sin}\:\alpha}{{h}\:\mathrm{cos}\:\alpha+{r}\:\mathrm{sin}\:\alpha}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \\ $$$${V}_{\mathrm{1}} ={V}×\left(\frac{{h}\:\mathrm{cos}\:\alpha−{r}\:\mathrm{sin}\:\alpha}{{h}\:\mathrm{cos}\:\alpha+{r}\:\mathrm{sin}\:\alpha}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} =\frac{{V}}{\mathrm{2}} \\ $$$$\left(\frac{{h}\:\mathrm{cos}\:\alpha−{r}\:\mathrm{sin}\:\alpha}{{h}\:\mathrm{cos}\:\alpha+{r}\:\mathrm{sin}\:\alpha}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} =\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow\frac{{h}\:\mathrm{cos}\:\alpha−{r}\:\mathrm{sin}\:\alpha}{{h}\:\mathrm{cos}\:\alpha+{r}\:\mathrm{sin}\:\alpha}=\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\mathrm{4}}} \\ $$$$\Rightarrow\mathrm{tan}\:\alpha=\frac{\left(\sqrt[{\mathrm{3}}]{\mathrm{4}}−\mathrm{1}\right){h}}{\left(\sqrt[{\mathrm{3}}]{\mathrm{4}}+\mathrm{1}\right){r}} \\ $$$$\lambda=\frac{{a}_{\mathrm{1}} }{{a}}=\mathrm{1}−\frac{\mathrm{2}}{\mathrm{1}+\frac{\sqrt[{\mathrm{3}}]{\mathrm{4}}+\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\mathrm{4}}−\mathrm{1}}}=\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\mathrm{4}}}\approx\mathrm{0}.\mathrm{63}\:\checkmark \\ $$$${generally}\:{for}\:\frac{{V}_{\mathrm{1}} }{{V}}=\frac{\mathrm{1}}{{n}}:\: \\ $$$$\:\:\:\:\:\:\frac{{a}_{\mathrm{1}} }{{a}}=\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{{n}^{\mathrm{2}} }} \\ $$

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))

$$\boldsymbol{{Part}}\:\boldsymbol{{II}} \\ $$$${we}\:{have}\:{got}\:\mathrm{tan}\:\alpha=\frac{{h}\left(\sqrt[{\mathrm{3}}]{\mathrm{4}}−\mathrm{1}\right)}{{r}\left(\sqrt[{\mathrm{3}}]{\mathrm{4}}+\mathrm{1}\right)}\:{or} \\ $$$$\frac{{r}\:\mathrm{tan}\:\alpha}{{h}}=\frac{\sqrt[{\mathrm{3}}]{\mathrm{4}}−\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\mathrm{4}}+\mathrm{1}}=\frac{\mathrm{1}}{{m}} \\ $$$${r}\gamma={a}\phi \\ $$$$\Rightarrow\gamma=\frac{{a}\phi}{{r}}=\sqrt{\mathrm{1}+\left(\frac{{h}}{{r}}\right)^{\mathrm{2}} }\:\phi={k}\phi \\ $$$${with}\:{k}=\sqrt{\mathrm{1}+\left(\frac{{h}}{{r}}\right)^{\mathrm{2}} } \\ $$$$\varphi=\frac{\pi{r}}{{a}}=\frac{\pi}{{k}} \\ $$$$\frac{\rho}{{a}}=\frac{{r}_{\mathrm{1}} }{{r}}\:\Rightarrow{r}_{\mathrm{1}} =\frac{\rho}{{k}} \\ $$$$\frac{\rho}{{a}}=\mathrm{1}−\frac{{r}\left(\mathrm{1}−\frac{\rho}{{a}}\:\mathrm{cos}\:\gamma\right)\mathrm{tan}\:\alpha}{{h}} \\ $$$$\Rightarrow\frac{\rho}{{a}}=\frac{{m}−\mathrm{1}}{{m}−\mathrm{cos}\:{k}\phi} \\ $$$${S}_{\mathrm{1}} =\mathrm{2}\int_{\mathrm{0}} ^{\varphi} \frac{\rho^{\mathrm{2}} {d}\phi}{\mathrm{2}} \\ $$$$\:\:\:\:\:={a}^{\mathrm{2}} \int_{\mathrm{0}} ^{\varphi} \frac{\left({m}−\mathrm{1}\right)^{\mathrm{2}} {d}\phi}{\left({m}−\mathrm{cos}\:{k}\phi\right)^{\mathrm{2}} } \\ $$$${S}=\varphi{a}^{\mathrm{2}} =\frac{\pi{a}^{\mathrm{2}} }{{k}} \\ $$$$\frac{{S}_{\mathrm{1}} }{{S}}=\frac{{k}\left({m}−\mathrm{1}\right)^{\mathrm{2}} }{\pi}\int_{\mathrm{0}} ^{\varphi} \frac{{d}\phi}{\left({m}−\mathrm{cos}\:{k}\phi\right)^{\mathrm{2}} } \\ $$$$\:\:=\frac{\left({m}−\mathrm{1}\right)^{\mathrm{2}} }{\pi}\int_{\mathrm{0}} ^{\pi} \frac{{d}\gamma}{\left({m}−\mathrm{cos}\:\gamma\right)^{\mathrm{2}} } \\ $$$$\:\:=\frac{\left({m}−\mathrm{1}\right)^{\mathrm{2}} }{\pi}×\frac{{m}\pi}{\left({m}−\mathrm{1}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \left({m}+\mathrm{1}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} } \\ $$$$\:\:=\frac{{m}}{{m}+\mathrm{1}}\sqrt{\frac{{m}−\mathrm{1}}{{m}+\mathrm{1}}}\: \\ $$$$\:\:=\frac{\mathrm{1}+\sqrt[{\mathrm{3}}]{\mathrm{4}}}{\mathrm{4}}\:\approx\mathrm{0}.\mathrm{6469} \\ $$$${generally}\:{for}\:\frac{{V}_{\mathrm{1}} }{{V}}=\frac{\mathrm{1}}{{n}}: \\ $$$$\:\:\:\:\:\:\:\:\frac{{S}_{\mathrm{1}} }{{S}}=\frac{\mathrm{1}+\sqrt[{\mathrm{3}}]{{n}^{\mathrm{2}} }}{\mathrm{2}{n}}\: \\ $$

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