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 . f i n d r a t i o \frac{a_{1}}{a} = ? s u c h t h a t t h e r i g h t

c i r c u l a r c o n e i s c u t i n t o t w o p a r t s

w i t h e q u a l v o l u m e a s s h o w n .

2 . f i n d i n w h i c h r a t i o t h e l a t e r a l

s u r f a c e o f t h e r i g h t c i r c u l a r c o n e i s

c u t .

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

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

P a r t I

v o l u m e o f c o n e V = \frac{π r^{2} h}{3}

a = A C = \sqrt{r^{2} + h^{2}}

\frac{D C}{\sin α} = \frac{B D}{\sin θ} = \frac{2 r}{\sin (θ + α)} = \frac{2 r \sqrt{r^{2} + h^{2}}}{r \sin α + h \cos α}

D C = \frac{2 r \sin α \sqrt{r^{2} + h^{2}}}{r \sin α + h \cos α}

B D = \frac{2 r h}{r \sin α + h \cos α}

a_{1} = A C - D C = \sqrt{r^{2} + h^{2}} (1 - \frac{2 r \sin α}{r \sin α + h \cos α})

λ = \frac{a_{1}}{a} = 1 - \frac{2 r \sin α}{r \sin α + h \cos α} = 1 - \frac{2}{1 + \frac{h}{r \tan α}}

t h e c u t s e c t i o n i s a n e l l i p s e w i t h s e m i

a x e s p a n d q .

2 p = B D = \frac{2 r h}{r \sin α + h \cos α}

\Rightarrow p = \frac{r h}{r \sin α + h \cos α}

B E = \frac{r}{\cos α}

F E = a - B E = \frac{r h}{r \sin α + h \cos α} - \frac{r}{\cos α}

O E = r \tan α

A E = h - r \tan α

\frac{E G}{r} = \frac{A E}{h} = 1 - \frac{r \tan α}{h}

E G = r - \frac{r^{2} \tan α}{h}

{(\frac{F E}{p})}^{2} + {(\frac{E G}{q})}^{2} = 1

{(\frac{\frac{r h}{r \sin α + h \cos α} - \frac{r}{\cos α}}{\frac{r h}{r \sin α + h \cos α}})}^{2} + {(\frac{r - \frac{r^{2} \tan α}{h}}{q})}^{2} = 1

\tan^{2} α + {(\frac{h - r \tan α}{q})}^{2} = \frac{h^{2}}{r^{2}}

q = \frac{h - r \tan α}{\sqrt{\frac{h^{2}}{r^{2}} - \tan^{2} α}} = r \sqrt{\frac{h \cos α - r \sin α}{h \cos α + r \sin α}}

h_{1} = A E \sin β = A E \cos α = h \cos α - r \sin α

V_{1} = \frac{π {p q h}_{1}}{3} = \frac{π}{3} \times \frac{r h}{r \sin α + h \cos α} \times r \sqrt{\frac{h \cos α - r \sin α}{h \cos α + r \sin α}} \times (h \cos α - r \sin α)

V_{1} = \frac{π r^{2} h}{3} \times {(\frac{h \cos α - r \sin α}{h \cos α + r \sin α})}^{\frac{3}{2}}

V_{1} = V \times {(\frac{h \cos α - r \sin α}{h \cos α + r \sin α})}^{\frac{3}{2}} = \frac{V}{2}

{(\frac{h \cos α - r \sin α}{h \cos α + r \sin α})}^{\frac{3}{2}} = \frac{1}{2}

\Rightarrow \frac{h \cos α - r \sin α}{h \cos α + r \sin α} = \frac{1}{\sqrt[3]{4}}

\Rightarrow \tan α = \frac{(\sqrt[3]{4} - 1) h}{(\sqrt[3]{4} + 1) r}

λ = \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 ✓

g e n e r a l l y f o r \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

P a r t I I

w e h a v e g o t \tan α = \frac{h (\sqrt[3]{4} - 1)}{r (\sqrt[3]{4} + 1)} o r

\frac{r \tan α}{h} = \frac{\sqrt[3]{4} - 1}{\sqrt[3]{4} + 1} = \frac{1}{m}

r γ = a ϕ

\Rightarrow γ = \frac{a ϕ}{r} = \sqrt{1 + {(\frac{h}{r})}^{2}} ϕ = k ϕ

w i t h k = \sqrt{1 + {(\frac{h}{r})}^{2}}

φ = \frac{π r}{a} = \frac{π}{k}

\frac{ρ}{a} = \frac{r_{1}}{r} \Rightarrow r_{1} = \frac{ρ}{k}

\frac{ρ}{a} = 1 - \frac{r (1 - \frac{ρ}{a} \cos γ) \tan α}{h}

\Rightarrow \frac{ρ}{a} = \frac{m - 1}{m - \cos k ϕ}

S_{1} = 2 \int_{0}^{φ} \frac{ρ^{2} d ϕ}{2}

= a^{2} \int_{0}^{φ} \frac{{(m - 1)}^{2} d ϕ}{{(m - \cos k ϕ)}^{2}}

S = φ a^{2} = \frac{π a^{2}}{k}

\frac{S_{1}}{S} = \frac{k {(m - 1)}^{2}}{π} \int_{0}^{φ} \frac{d ϕ}{{(m - \cos k ϕ)}^{2}}

= \frac{{(m - 1)}^{2}}{π} \int_{0}^{π} \frac{d γ}{{(m - \cos γ)}^{2}}

= \frac{{(m - 1)}^{2}}{π} \times \frac{m π}{{(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

g e n e r a l l y f o r \frac{V_{1}}{V} = \frac{1}{n} :

\frac{S_{1}}{S} = \frac{1 + \sqrt[3]{n^{2}}}{2 n}

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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