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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

$$1.findratio\frac{a_1}{a}=?suchthattheright$$$$circularconeiscutintotwoparts$$$$withequalvolumeasshown.$$$$2.findinwhichratiothelateral$$$$surfaceoftherightcircularconeis$$$$cut.$$

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

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

$$\mathit{P}\mathit{a}\mathit{r}\mathit{t}\mathit{I}$$$$volumeofconeV=\frac{\pi r^{2}h}{3}$$$$a=AC=\sqrt{r^2+h^2}$$$$\frac{DC}{\sin \alpha }=\frac{BD}{\sin \theta }=\frac{2r}{\sin (\theta +\alpha )}=\frac{2r\sqrt{r^2+h^2}}{r\sin \alpha +h\cos \alpha }$$$$DC=\frac{2r\sin \alpha \sqrt{r^2+h^2}}{r\sin \alpha +h\cos \alpha }$$$$BD=\frac{2rh}{r\sin \alpha +h\cos \alpha }$$$$a_1=AC-DC=\sqrt{r^2+h^2}(1-\frac{2r\sin \alpha }{r\sin \alpha +h\cos \alpha })$$$$\lambda =\frac{a_1}{a}=1-\frac{2r\sin \alpha }{r\sin \alpha +h\cos \alpha }=1-\frac{2}{1+\frac{h}{r\tan \alpha }}$$$$thecutsectionisanellipsewithsemi$$$$axespandq.$$$$2p=BD=\frac{2rh}{r\sin \alpha +h\cos \alpha }$$$$\Rightarrow p=\frac{rh}{r\sin \alpha +h\cos \alpha }$$$$BE=\frac{r}{\cos \alpha }$$$$FE=a-BE=\frac{rh}{r\sin \alpha +h\cos \alpha }-\frac{r}{\cos \alpha }$$$$OE=r\tan \alpha$$$$AE=h-r\tan \alpha$$$$\frac{EG}{r}=\frac{AE}{h}=1-\frac{r\tan \alpha }{h}$$$$EG=r-\frac{r^2\tan \alpha }{h}$$$${\left(\frac{FE}{p}\right)}^2+{\left(\frac{EG}{q}\right)}^2=1$$$${\left(\frac{\frac{rh}{r\sin \alpha +h\cos \alpha }-\frac{r}{\cos \alpha }}{\frac{rh}{r\sin \alpha +h\cos \alpha }}\right)}^2+{\left(\frac{r-\frac{r^2\tan \alpha }{h}}{q}\right)}^2=1$$$${\tan }^2\alpha +{\left(\frac{h-r\tan \alpha }{q}\right)}^2=\frac{h^2}{r^2}$$$$q=\frac{h-r\tan \alpha }{\sqrt{\frac{h^2}{r^2}-{\tan }^2\alpha }}=r\sqrt{\frac{h\cos \alpha -r\sin \alpha }{h\cos \alpha +r\sin \alpha }}$$$$h_1=AE\sin \beta =AE\cos \alpha =h\cos \alpha -r\sin \alpha$$$$V_1=\frac{\pi {pqh}_1}{3}=\frac{\pi }{3}\times \frac{rh}{r\sin \alpha +h\cos \alpha }\times r\sqrt{\frac{h\cos \alpha -r\sin \alpha }{h\cos \alpha +r\sin \alpha }}\times (h\cos \alpha -r\sin \alpha )$$$$$$$$V_1=\frac{\pi r^{2}h}{3}\times {\left(\frac{h\cos \alpha -r\sin \alpha }{h\cos \alpha +r\sin \alpha }\right)}^{\frac{3}{2}}$$$$V_1=V\times {\left(\frac{h\cos \alpha -r\sin \alpha }{h\cos \alpha +r\sin \alpha }\right)}^{\frac{3}{2}}=\frac{V}{2}$$$${\left(\frac{h\cos \alpha -r\sin \alpha }{h\cos \alpha +r\sin \alpha }\right)}^{\frac{3}{2}}=\frac{1}{2}$$$$\Rightarrow \frac{h\cos \alpha -r\sin \alpha }{h\cos \alpha +r\sin \alpha }=\frac{1}{\sqrt[3]{4}}$$$$\Rightarrow \tan \alpha =\frac{(\sqrt[3]{4}-1)h}{(\sqrt[3]{4}+1)r}$$$$\lambda =\frac{a_1}{a}=1-\frac{2}{1+\frac{\sqrt[3]{4}+1}{\sqrt[3]{4}-1}}=\frac{1}{\sqrt[3]{4}}\approx 0.63✓$$$$generallyfor\frac{V_1}{V}=\frac{1}{n}:$$$$\frac{a_1}{a}=\frac{1}{\sqrt[3]{n^2}}$$

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

$$\mathit{P}\mathit{a}\mathit{r}\mathit{t}\mathit{I}\mathit{I}$$$$wehavegot\tan \alpha =\frac{h(\sqrt[3]{4}-1)}{r(\sqrt[3]{4}+1)}or$$$$\frac{r\tan \alpha }{h}=\frac{\sqrt[3]{4}-1}{\sqrt[3]{4}+1}=\frac{1}{m}$$$$r\gamma =a\varphi$$$$\Rightarrow \gamma =\frac{a\varphi }{r}=\sqrt{1+{\left(\frac{h}{r}\right)}^2}\varphi =k\varphi$$$$withk=\sqrt{1+{\left(\frac{h}{r}\right)}^2}$$$$\phi =\frac{\pi r}{a}=\frac{\pi }{k}$$$$\frac{\rho }{a}=\frac{r_1}{r}\Rightarrow r_1=\frac{\rho }{k}$$$$\frac{\rho }{a}=1-\frac{r(1-\frac{\rho }{a}\cos \gamma )\tan \alpha }{h}$$$$\Rightarrow \frac{\rho }{a}=\frac{m-1}{m-\cos k\varphi }$$$$S_1=2{\int }_0^{\phi }\frac{{\rho }^{2}d\varphi }{2}$$$$=a^2{\int }_0^{\phi }\frac{{(m-1)}^{2}d\varphi }{{(m-\cos k\varphi )}^2}$$$$S=\phi a^2=\frac{\pi a^2}{k}$$$$\frac{S_1}{S}=\frac{k{(m-1)}^2}{\pi }{\int }_0^{\phi }\frac{d\varphi }{{(m-\cos k\varphi )}^2}$$$$=\frac{{(m-1)}^2}{\pi }{\int }_0^{\pi }\frac{d\gamma }{{(m-\cos \gamma )}^2}$$$$=\frac{{(m-1)}^2}{\pi }\times \frac{m\pi }{{(m-1)}^{\frac{3}{2}}{(m+1)}^{\frac{3}{2}}}$$$$=\frac{m}{m+1}\sqrt{\frac{m-1}{m+1}}$$$$=\frac{1+\sqrt[3]{4}}{4}\approx 0.6469$$$$generallyfor\frac{V_1}{V}=\frac{1}{n}:$$$$\frac{S_1}{S}=\frac{1+\sqrt[3]{n^2}}{2n}$$

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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