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Question Number 158749 by ajfour last updated on 08/Nov/21

Commented by ajfour last updated on 08/Nov/21

$F i n d e x t r e m e v a l u e s o f R .$
	Answered by mr W last updated on 08/Nov/21

	Commented by mr W last updated on 08/Nov/21
	$let c=a+b let μ=(b/a), λ=(R/a) BD=a+c sin θ DC=(√((R−b)^2 −(a+c sin θ)^2 )) CE=(√((R−a)^2 −a^2 ))=(√(R^2 −2aR)) DE=BF=c cos θ c cos θ=(√((R−b)^2 −(a+c sin θ)^2 ))+(√(R^2 −2aR)) c cos θ−(√(R^2 −2aR))=(√((R−b)^2 −(a+c sin θ)^2 )) c^2 cos^2 θ−2c cos θ(√(R^2 −2aR))=2(a−b)R+b^2 −(a+c sin θ)^2 a(1+sin θ)−((a−b)/(a+b))R=cos θ(√(R^2 −2aR)) (1+sin θ)−((1−μ)/(1+μ))λ=cos θ(√(λ^2 −2λ)) (((1−μ)/(1+μ)))^2 λ^2 +(1+sin θ)^2 −2(((1−μ)/(1+μ)))(1+sin θ)λ=cos^2 θλ^2 −2cos^2 θλ [(((1−μ)/(1+μ)))^2 −cos^2 θ]λ^2 −2[(1+sin θ)(((1−μ)/(1+μ)))−cos^2 θ]λ+(1+sin θ)^2 =0 [(((1−μ)/(1+μ)))^2 −cos^2 θ]λ^2 −2(1+sin θ)(sin θ−((2μ)/(1+μ)))λ+(1+sin θ)^2 =0 λ=(((1+sin θ)(sin θ−((2μ)/(1+μ)))−(1+sin θ)(√((sin θ−((2μ)/(1+μ)))^2 −(((1−μ)/(1+μ)))^2 +cos^2 θ)))/((((1−μ)/(1+μ)))^2 −cos^2 θ)) λ=(((1+sin θ){sin θ−((2μ)/(1+μ))−(√((4μ(1−sin θ))/(1+μ)))})/((((1−μ)/(1+μ)))^2 −cos^2 θ)) or λ=(((1+sin θ)[ξ+(√(2ξ(1−sin θ)))−sin θ])/(ξ(2−ξ)−sin^2 θ)) with ξ=((2μ)/(1+μ))=((2b)/(a+b))$
	$l e t c = a + b$ $l e t μ = \frac{b}{a}, λ = \frac{R}{a}$ $B D = a + c \sin θ$ $D C = \sqrt{{(R - b)}^{2} - {(a + c \sin θ)}^{2}}$ $C E = \sqrt{{(R - a)}^{2} - a^{2}} = \sqrt{R^{2} - 2 a R}$ $D E = B F = c \cos θ$ $c \cos θ = \sqrt{{(R - b)}^{2} - {(a + c \sin θ)}^{2}} + \sqrt{R^{2} - 2 a R}$ $c \cos θ - \sqrt{R^{2} - 2 a R} = \sqrt{{(R - b)}^{2} - {(a + c \sin θ)}^{2}}$ $c^{2} \cos^{2} θ - 2 c \cos θ \sqrt{R^{2} - 2 a R} = 2 (a - b) R + b^{2} - {(a + c \sin θ)}^{2}$ $a (1 + \sin θ) - \frac{a - b}{a + b} R = \cos θ \sqrt{R^{2} - 2 a R}$ $(1 + \sin θ) - \frac{1 - μ}{1 + μ} λ = \cos θ \sqrt{λ^{2} - 2 λ}$ ${(\frac{1 - μ}{1 + μ})}^{2} λ^{2} + {(1 + \sin θ)}^{2} - 2 (\frac{1 - μ}{1 + μ}) (1 + \sin θ) λ = \cos^{2} θ λ^{2} - {2 \cos}^{2} θ λ$ $[{(\frac{1 - μ}{1 + μ})}^{2} - \cos^{2} θ] λ^{2} - 2 [(1 + \sin θ) (\frac{1 - μ}{1 + μ}) - \cos^{2} θ] λ + {(1 + \sin θ)}^{2} = 0$ $[{(\frac{1 - μ}{1 + μ})}^{2} - \cos^{2} θ] λ^{2} - 2 (1 + \sin θ) (\sin θ - \frac{2 μ}{1 + μ}) λ + {(1 + \sin θ)}^{2} = 0$ $λ = \frac{(1 + \sin θ) (\sin θ - \frac{2 μ}{1 + μ}) - (1 + \sin θ) \sqrt{{(\sin θ - \frac{2 μ}{1 + μ})}^{2} - {(\frac{1 - μ}{1 + μ})}^{2} + \cos^{2} θ}}{{(\frac{1 - μ}{1 + μ})}^{2} - \cos^{2} θ}$ $λ = \frac{(1 + \sin θ) {\sin θ - \frac{2 μ}{1 + μ} - \sqrt{\frac{4 μ (1 - \sin θ)}{1 + μ}}}}{{(\frac{1 - μ}{1 + μ})}^{2} - \cos^{2} θ}$ $o r$ $λ = \frac{(1 + \sin θ) [ξ + \sqrt{2 ξ (1 - \sin θ)} - \sin θ]}{ξ (2 - ξ) - \sin^{2} θ}$ $w i t h ξ = \frac{2 μ}{1 + μ} = \frac{2 b}{a + b}$
	Commented by ajfour last updated on 08/Nov/21

	$T h a n k y o u S i r$ $S e e m s s a t i s f a c t o r y!$
	Commented by mr W last updated on 08/Nov/21

	Commented by mr W last updated on 08/Nov/21

	Commented by mr W last updated on 08/Nov/21

	Commented by mr W last updated on 08/Nov/21

	Commented by mr W last updated on 08/Nov/21

	$u s u a l l y R_{m i n} e x i s t s .$ $b u t R_{m a x} {d o e s n}^{'} t e x i s t .$
	Commented by mr W last updated on 09/Nov/21

	Commented by ajfour last updated on 09/Nov/21

	$T h e s i t u a t i o n i s : a ⩾ b . A i s$ $i n c o n t a c t w i t h g r o u n d a n d$ $B . B i n c o n t a c t w i t h w a l l$ $a n d A; (j u s t t h e 1^{s t} q u a d r a n t) .$
	Commented by mr W last updated on 09/Nov/21

	$i f b o t h c i r c l e s a a n d b a r e o n l y i n$ $q u a d r a n t I a n d a > b, t h e n$ $- \sin^{- 1} \frac{a - b}{a + b} ⩽ θ ⩽ \cos^{- 1} \frac{a - b}{a + b}$
	Commented by mr W last updated on 09/Nov/21

	Commented by Tawa11 last updated on 16/Nov/21

	$Great sir$
	Answered by ajfour last updated on 08/Nov/21
	$(R−b)cos φ+(a+b)cos ψ =R−a (R−b)sin φ=(a+b)sin ψ φ+ψ−θ=(π/2) sin φcos ψ+cos φsin ψ=cos θ {(((a+b)/(R−b)))cos ψ+cos φ}sin ψ =cos θ ⇒(R−a)sin ψ=(R−b)cos θ {cos ψ+(((R−b)/(a+b)))cos φ}sin φ =cos θ ⇒ (R−a)sin φ=(a+b)cos θ cos φcos ψ−sin φsin ψ=sin θ sin φsin ψ=f(R,θ) =(((R−b)(a+b)cos^2 θ)/((R−a)^2 )) (sin φsin ψ+sin θ)^2 =(1−sin^2 φ)(1−sin^2 ψ) (f+sin θ)^2 =(1−f_1 ^( 2) )(1−f_2 ^( 2) ) ⇒ h(θ,R)=0$
	$(R - b) \cos ϕ + (a + b) \cos ψ$ $= R - a$ $(R - b) \sin ϕ = (a + b) \sin ψ$ $ϕ + ψ - θ = \frac{π}{2}$ $\sin ϕ \cos ψ + \cos ϕ \sin ψ = \cos θ$ ${(\frac{a + b}{R - b}) \cos ψ + \cos ϕ} \sin ψ$ $= \cos θ$ $\Rightarrow (R - a) \sin ψ = (R - b) \cos θ$ ${\cos ψ + (\frac{R - b}{a + b}) \cos ϕ} \sin ϕ$ $= \cos θ$ $\Rightarrow (R - a) \sin ϕ = (a + b) \cos θ$ $\cos ϕ \cos ψ - \sin ϕ \sin ψ = \sin θ$ $\sin ϕ \sin ψ = f (R, θ)$ $= \frac{(R - b) (a + b) \cos^{2} θ}{{(R - a)}^{2}}$ ${(\sin ϕ \sin ψ + \sin θ)}^{2}$ $= (1 - \sin^{2} ϕ) (1 - \sin^{2} ψ)$ ${(f + \sin θ)}^{2} = (1 - f_{1}^{2}) (1 - f_{2}^{2})$ $\Rightarrow h (θ, R) = 0$
	Commented by ajfour last updated on 08/Nov/21
	$(R−a)sin ψ=(R−b)cos θ ⇒ sin ψ=(((z−c)/(z−1)))cos θ (R−a)sin φ=(a+b)cos θ ⇒ sin φ=(((z−1)/(1+c)))cos θ (sin φsin ψ+sin θ)^2 =(1−sin^2 φ)(1−sin^2 ψ) ⇒ sin^2 θ+2sin θ(sin φ+sin ψ) +sin^2 φ+sin^2 ψ=1 ⇒ (sin φ+sin θ)^2 +(sin ψ+sin θ)^2 =1+sin^2 θ (((z−1)/(1+c))cos θ+sin θ)^2 +(((z−c)/(z−1))cos θ+sin θ)^2 =1+sin^2 θ let tan θ=1+m {((z−1+(1+mc)+(c+m))/(1+c))}^2 {(((2+m)(z−1)+(1−c))/(z−1))}^2 =1+2(1+m)^2 z−1=(R/a)−1=x {x+(1+c)(1+m)}^2 {(2+m)x+(1−c)}^2 =(1+c)^2 {2(m+1)^2 +1}x^2 (1+c)(1+m)=A (((1−c)/(2+m)))=B (((1+c)/(2+m)))^2 {2(m+1)^2 +1}=C (x+A)^2 (x+B)^2 =Cx^2 differentiating ((dA/dm))(x+A)(x+B)^2 +((dB/dm))(x+B)(x+A)^2 =(x^2 /2)(dC/dm) ⇒ (dA/(x+A))+(dB/(x+B))=(dC/(2C)) ..............................................$
	$(R - a) \sin ψ = (R - b) \cos θ$ $\Rightarrow \sin ψ = (\frac{z - c}{z - 1}) \cos θ$ $(R - a) \sin ϕ = (a + b) \cos θ$ $\Rightarrow \sin ϕ = (\frac{z - 1}{1 + c}) \cos θ$ ${(\sin ϕ \sin ψ + \sin θ)}^{2}$ $= (1 - \sin^{2} ϕ) (1 - \sin^{2} ψ)$ $\Rightarrow \sin^{2} θ + 2 \sin θ (\sin ϕ + \sin ψ)$ $+ \sin^{2} ϕ + \sin^{2} ψ = 1$ $\Rightarrow$ ${(\sin ϕ + \sin θ)}^{2} + {(\sin ψ + \sin θ)}^{2}$ $= 1 + \sin^{2} θ$ ${(\frac{z - 1}{1 + c} \cos θ + \sin θ)}^{2} + {(\frac{z - c}{z - 1} \cos θ + \sin θ)}^{2}$ $= 1 + \sin^{2} θ$ $l e t \tan θ = 1 + m$ ${\frac{z - 1 + (1 + m c) + (c + m)}{1 + c}}^{2} {\frac{(2 + m) (z - 1) + (1 - c)}{z - 1}}^{2}$ $= 1 + 2 {(1 + m)}^{2}$ $z - 1 = \frac{R}{a} - 1 = x$ ${x + (1 + c) (1 + m)}^{2} {(2 + m) x + (1 - c)}^{2}$ $= {(1 + c)}^{2} {2 {(m + 1)}^{2} + 1} x^{2}$ $(1 + c) (1 + m) = A$ $(\frac{1 - c}{2 + m}) = B$ ${(\frac{1 + c}{2 + m})}^{2} {2 {(m + 1)}^{2} + 1} = C$ ${(x + A)}^{2} {(x + B)}^{2} = {C x}^{2}$ $d i f f e r e n t i a t i n g$ $(\frac{d A}{d m}) (x + A) {(x + B)}^{2} + (\frac{d B}{d m}) (x + B) {(x + A)}^{2}$ $= \frac{x^{2}}{2} \frac{d C}{d m}$ $\Rightarrow \frac{d A}{x + A} + \frac{d B}{x + B} = \frac{d C}{2 C}$ $. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .$
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