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Question Number 61303 by mr W last updated on 31/May/19

Commented by mr W last updated on 31/May/19

$A s m a l l s p h e r e (s a y t h e m o o n) m o v e s$ $a r o u n d a b i g s p h e r e (s a y t h e e a r t h) i n$ $a c i r c u l a r t r a c k . A p o i n t s o u r c e (s a y$ $t h e s u n) i s a t a d i s t a n c e e f r o m t h e$ $c e n t e r o f t h e b i g s p h e r e .$ $F i n d t h e a r e a o f s h a d o w o f t h e s m a l l$ $s p h e r e o n t h e s u r f a c e o f t h e b i g$ $s p h e r e i n t e r m s o f α . F i n d t h e$ $m a x i m u m s h a d o w a r e a .$ $W e o n l y n e e d t o t r e a t t h e c a s e t h a t$ $t h e s h a d o w i s c o m p l e t e l y o n t h e s u r f a c e$ $o f t h e b i g s p h e r e .$
Commented by ajfour last updated on 02/Jun/19

$P l a n O n l y (d i f f i c u l t t o o b t a i n$ $a n a n s w e r t h i s w a y)$ $__________________________$ $F (e, 0, 0)$ $T (d \cos α + r \sin ϕ \cos ψ, r \sin ϕ \sin ψ, d \sin α + r \cos ϕ)$ $l e t ∠ P O G = δ$ $P (R \cos δ \cos θ, R \cos δ \sin θ, R \sin δ)$ $__________________________$ ${F T}^{2} = {C F}^{2} + {C T}^{2}$ ${(e - d \cos α - r \sin ϕ \cos ψ)}^{2} + r^{2} \sin^{2} ϕ \sin^{2} ψ$ $+ {(d \sin α + r \cos ϕ)}^{2} + r^{2} =$ ${(e - d \cos α)}^{2} + d^{2} \sin^{2} α . . . (i)$ $__________________________$ $A s F T P i s s t r a i g h t :$ $\frac{R \cos δ \cos θ - e}{d \cos α + r \sin ϕ \cos ψ - e} = \frac{R \cos δ \sin θ}{r \sin ϕ \sin ψ}$ $= \frac{R \sin δ}{d \sin α + r \cos ϕ} . . (i i), (i i i)$ $f r o m (i), (i i), (i i i)$ $w e f i n d θ = f (δ) e l i m i n a t i n g$ $ϕ a n d ψ f r o m t h e m .$ $A = 2 \int_{δ_{m i n}}^{δ_{m a x}} (R \cos δ) θ (R d δ)$ $f o r o b t a i n i n g δ_{m i n}, δ_{m a x} w e t a k e$ $θ = 0, ψ = 0$ $\Rightarrow \frac{R \cos δ - e}{d \cos α + r \sin ϕ - e} = \frac{R \sin δ}{d \sin α + r \cos ϕ}$ $f o r ϕ_{m i n} a n d ϕ_{m a x}$ ${(e - d \cos α - r \sin ϕ)}^{2} + {(d \sin α + r \cos ϕ)}^{2}$ $+ r^{2} = {(e - d \cos α)}^{2} + d^{2} \sin^{2} α$ $\Rightarrow (e - d \cos α) \sin ϕ - d \sin α \cos ϕ = r$ $\sin (ϕ + \tan^{- 1} \frac{d \sin α}{e - d \cos α}) = \frac{r}{\sqrt{{(e - d \cos α)}^{2} + d^{2} \sin^{2} α}}$ $\Rightarrow ϕ_{1} = \sin^{- 1} (\frac{r}{\sqrt{{(e - d \cos α)}^{2} + d^{2} \sin^{2} α}})$ $- \tan^{- 1} \frac{d \sin α}{e - d \cos α}$ $& ϕ_{2} = π - \sin^{- 1} (\frac{r}{\sqrt{{(e - d \cos α)}^{2} + d^{2} \sin^{2} α}})$ $- \tan^{- 1} \frac{d \sin α}{e - d \cos α}$ $W e t h e n o b t a i n δ_{m i n} a n d δ_{m a x}$ $f r o m u p p e r e q .$ $. . . . .$
Commented by mr W last updated on 01/Jun/19

$t h a n k s f o r t r y i n g s i r! i h a v e n o t g o t$ $a s o l u t i o n . b u t y o u r m e t h o d w o u l d b e$ $t h a t w h a t i h a d a l s o u s e d . b u t o n t h e$ $o t h e r s i d e i t h i n k t h e r e s h o u l d b e a n$ $o t h e r m a y b e e a s i e r w a y . i^{'} l l t r y i t$ $l a t e r .$
	Answered by ajfour last updated on 02/Jun/19

	Answered by mr W last updated on 02/Jun/19

	Commented by mr W last updated on 02/Jun/19

	$t h e l i g h t r a y s f o r m a r i g h t c i r c u l a r$ $c o n e . o u r t a s k i s t o d e t e r m i n e$ $t h e i n t e r s e c t i o n f r o m$ $t h i s c o n e a n d t h e b i g s p h e r e .$ $t o s i m p l i f y t h e c a l c u l t i o n i s e l e c t t h e$ $s o u r c e S a s o r i g i n a n d S M a s x - a x i s .$ $t h e s h a p e o f c o n e i s d e s c r i b e d b y t h e$ $a n g l e β .$ $t h e p o s i t i o n o f t h e b i g s p h e r e i s$ $d e s c r i b e d b y h a n d k .$ $S O = e, O M = d$ $S M = \sqrt{{(e - d \cos α)}^{2} + {(d \sin α)}^{2}}$ $= \sqrt{e^{2} + d^{2} - 2 e d \cos α}$ $\sin β = \frac{r}{\sqrt{e^{2} + d^{2} - 2 e d \cos α}}$ $\Rightarrow β = \sin^{- 1} \frac{r}{\sqrt{e^{2} + d^{2} - 2 e d \cos α}}$ $\tan β = \frac{r}{\sqrt{e^{2} + d^{2} - r^{2} - 2 e d \cos α}} = m (s a y)$ $\frac{k}{d \sin α} = \frac{e}{S M}$ $\Rightarrow k = \frac{e d \sin α}{\sqrt{e^{2} + d^{2} - 2 e d \cos α}}$ $h = \sqrt{e^{2} - k^{2}} = \sqrt{e^{2} - \frac{e^{2} d^{2} \sin^{2} α}{e^{2} + d^{2} - 2 e d \cos α}}$ $\Rightarrow h = \frac{e (e - d \cos α)}{\sqrt{e^{2} + d^{2} - 2 e d \cos α}}$ $c o n d i t i o n s u c h t h a t t h e s h a d o w i s$ $c o m p l e t e l y o n t h e b i g s p h e r e i s$ $\sin β = \frac{R - \frac{k}{\cos β}}{h - k \tan β} = \frac{R \cos β - k}{h \cos β - k \sin β}$
	Commented by mr W last updated on 02/Jun/19

	Commented by mr W last updated on 03/Jun/19
	$eqn. of cone: x=((√(y^2 +z^2 ))/(tan β))=((√(y^2 +z^2 ))/m) or y^2 =m^2 x^2 −z^2 eqn. of sphere: (x−h)^2 +y^2 +(z+k)^2 =R^2 intersection of cone and sphere: (x−h)^2 +m^2 x^2 −z^2 +(z+k)^2 =R^2 ⇒(1+m^2 )x^2 −2hx+2kz+(h^2 +k^2 −R^2 )=0 ...(i) (this is a parabola in xz−plane) at θ: let ρ=R−Δh x=h−ρ cos θ z=−k+ρ sin θ put this into (i): (1+m^2 )[h−ρ cos θ]^2 −2h[h−ρ cos θ]+2k[−k+ρ sin θ]+(h^2 +k^2 −R^2 )=0 (1+m^2 )cos^2 θ ρ^2 −2(m^2 h cos θ−k sin θ)ρ+(m^2 h^2 −k^2 −R^2 )=0 ⇒ρ=(((m^2 h cos θ−k sin θ)+(√(m^4 h^2 cos^2 θ+k^2 sin^2 θ−2m^2 hk sin θ cos θ−(1+m^2 )cos^2 θ(m^2 h^2 −k^2 −R^2 ))))/((1+m^2 )cos^2 θ)) (two intersections, we take + for the left one) ⇒ρ=R cos ϕ=(((m^2 h cos θ−k sin θ)+(√(k^2 +[(1+m^2 )R^2 −m^2 (h^2 −k^2 )]cos^2 θ−m^2 hk sin 2θ)))/((1+m^2 )cos^2 θ)) ⇒ϕ=cos^(−1) {(((m^2 h cos θ−k sin θ)+(√(k^2 +[(1+m^2 )R^2 −m^2 (h^2 −k^2 )]cos^2 θ−m^2 hk sin 2θ)))/((1+m^2 )R cos^2 θ))} for ρ=(((m^2 h cos θ−k sin θ)+(√(k^2 +[(1+m^2 )R^2 −m^2 (h^2 −k^2 )]cos^2 θ−m^2 hk sin 2θ)))/((1+m^2 )cos^2 θ))=R we get θ_1 and θ_2 . Area of shadow=A A=∫_θ_1 ^θ_2 2RϕRdθ ⇒A=2R^2 ∫_θ_1 ^θ_2 cos^(−1) {(((m^2 h cos θ−k sin θ)+(√(k^2 +[(1+m^2 )R^2 −m^2 (h^2 −k^2 )]cos^2 θ−m^2 hk sin 2θ)))/((1+m^2 )R cos^2 θ))}dθ with μ=(k/R),λ=(h/R) ⇒A=2R^2 ∫_θ_1 ^θ_2 cos^(−1) {((m^2 λ cos θ−μ sin θ+(√(μ^2 +[1+m^2 (1+μ^2 −λ^2 )]cos^2 θ−m^2 λμ sin 2θ)))/((1+m^2 )cos^2 θ))}dθ$
	$e q n . o f c o n e :$ $x = \frac{\sqrt{y^{2} + z^{2}}}{\tan β} = \frac{\sqrt{y^{2} + z^{2}}}{m}$ $o r y^{2} = m^{2} x^{2} - z^{2}$ $e q n . o f s p h e r e :$ ${(x - h)}^{2} + y^{2} + {(z + k)}^{2} = R^{2}$ $i n t e r s e c t i o n o f c o n e a n d s p h e r e :$ ${(x - h)}^{2} + m^{2} x^{2} - z^{2} + {(z + k)}^{2} = R^{2}$ $\Rightarrow (1 + m^{2}) x^{2} - 2 h x + 2 k z + (h^{2} + k^{2} - R^{2}) = 0 . . . (i)$ $(t h i s i s a p a r a b o l a i n x z - p l a n e)$ $a t θ :$ $l e t ρ = R - Δ h$ $x = h - ρ \cos θ$ $z = - k + ρ \sin θ$ $p u t t h i s i n t o (i) :$ $(1 + m^{2}) {[h - ρ \cos θ]}^{2} - 2 h [h - ρ \cos θ] + 2 k [- k + ρ \sin θ] + (h^{2} + k^{2} - R^{2}) = 0$ $(1 + m^{2}) \cos^{2} θ ρ^{2} - 2 (m^{2} h \cos θ - k \sin θ) ρ + (m^{2} h^{2} - k^{2} - R^{2}) = 0$ $\Rightarrow ρ = \frac{(m^{2} h \cos θ - k \sin θ) + \sqrt{m^{4} h^{2} \cos^{2} θ + k^{2} \sin^{2} θ - 2 m^{2} h k \sin θ \cos θ - (1 + m^{2}) \cos^{2} θ (m^{2} h^{2} - k^{2} - R^{2})}}{(1 + m^{2}) \cos^{2} θ}$ $(t w o i n t e r s e c t i o n s, w e t a k e + f o r t h e l e f t o n e)$ $\Rightarrow ρ = R \cos φ = \frac{(m^{2} h \cos θ - k \sin θ) + \sqrt{k^{2} + [(1 + m^{2}) R^{2} - m^{2} (h^{2} - k^{2})] \cos^{2} θ - m^{2} h k \sin 2 θ}}{(1 + m^{2}) \cos^{2} θ}$ $\Rightarrow φ = \cos^{- 1} {\frac{(m^{2} h \cos θ - k \sin θ) + \sqrt{k^{2} + [(1 + m^{2}) R^{2} - m^{2} (h^{2} - k^{2})] \cos^{2} θ - m^{2} h k \sin 2 θ}}{(1 + m^{2}) R \cos^{2} θ}}$ $f o r ρ = \frac{(m^{2} h \cos θ - k \sin θ) + \sqrt{k^{2} + [(1 + m^{2}) R^{2} - m^{2} (h^{2} - k^{2})] \cos^{2} θ - m^{2} h k \sin 2 θ}}{(1 + m^{2}) \cos^{2} θ} = R$ $w e g e t θ_{1} a n d θ_{2} .$ $A r e a o f s h a d o w = A$ $A = \int_{θ_{1}}^{θ_{2}} 2 R φ R d θ$ $\Rightarrow A = 2 R^{2} \int_{θ_{1}}^{θ_{2}} \cos^{- 1} {\frac{(m^{2} h \cos θ - k \sin θ) + \sqrt{k^{2} + [(1 + m^{2}) R^{2} - m^{2} (h^{2} - k^{2})] \cos^{2} θ - m^{2} h k \sin 2 θ}}{(1 + m^{2}) R \cos^{2} θ}} d θ$ $w i t h μ = \frac{k}{R}, λ = \frac{h}{R}$ $\Rightarrow A = 2 R^{2} \int_{θ_{1}}^{θ_{2}} \cos^{- 1} {\frac{m^{2} λ \cos θ - μ \sin θ + \sqrt{μ^{2} + [1 + m^{2} (1 + μ^{2} - λ^{2})] \cos^{2} θ - m^{2} λ μ \sin 2 θ}}{(1 + m^{2}) \cos^{2} θ}} d θ$
	Commented by mr W last updated on 02/Jun/19

	Commented by mr W last updated on 02/Jun/19

	Commented by mr W last updated on 02/Jun/19

	Commented by mr W last updated on 02/Jun/19

	Commented by ajfour last updated on 02/Jun/19

	$G r e a t S i r, b u t {y o u}^{'} l l h a v e t o$ $g u i d e m e t o t h e i d e a u s e d, i d o n t$ $t h i n k i^{'} l l f o l l o w i t o t h e r w i s e . .$
	Commented by mr W last updated on 02/Jun/19

	$t h e b o u n d a r y o f t h e s h a d o w o n t h e$ $s u r f a c e o f b i g s p h e r e i s t h e i n t e r s e c t i o n$ $(a c u r v e i n t h e s p a c e) f r o m t h e c o n e$ $a n d t h e b i g s p h e r e . t h e s h a p e o f t h e$ $c o n e i s g i v e n b y t h e s i z e o f t h e s m a l l$ $s p h e r e . w i t h t h e c o o r d i n a t e s y s t e m$ $i s e l e c t t h e e q n . o f t h e c o n e a n d t h e$ $b i g s p h e r e c a n b e d e s c r i b e d i n s i m p l e$ $f o r m . t h e e q n . o f t h e i n t e r s e c t i o n$ $l i n e i n t h e x z - p l a n e c a n b e o b t a i n e d$ $q u i t e e a s i l y, a n d {i t}^{'} s a p a r a b o l a . w i t h$ $t h i s e q n . w e c a n d e t e r m i n e t h e$ $p a r a m e t e r s o f t h e s h a d o w a r e a o n$ $t h e s p h e r e s u r f a c e . t h e p a r a m e t e r s$ $a r e : θ a n d ρ (i . e . φ) .$
	Commented by ajfour last updated on 06/Jun/19

	$I a m v e r y t h a n k f u l b u t i l a c k$ $t h e r e q u i r e d p a t i e n c e (t h e s e d a y s)$ $S i r, i h a v e s a v e d y o u r s o l u t i o n,$ $a n d s h a l l a n a l y s e i t s o o n .$
	Commented by mr W last updated on 06/Jun/19

	$t h a n k y o u s i r!$ $i {w a s n}^{'} t a b l e t o p r o v i d e a b e t t e r a n d$ $c o m p l e t e s o l u t i o n . b u t i^{'} l l t r y t o$ $c o n t i n u e t o t h i n k a b o u t t h e q u e s t i o n .$ $i h o p e o n e d a y y o u m a y g i v e a p e r f e c t$ $s o l u t i o n s i r .$
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