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Question Number 186762 by ajfour last updated on 09/Feb/23

Commented by ajfour last updated on 09/Feb/23

$W h a t m i n i m u m l e n g t h t o w a l k$ $t o s t a r t f r o m h o m e w i t h e m p t y$ $b u c k e t g o g e t w a t e r f r o m w e l l n$ $w a t e r t h e t r e e .$
Commented by mr W last updated on 10/Feb/23

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

$f o r a g i v e n p a t h l e n g t h A C + C B t h e$ $l o c u s o f p o i n t C i s a n e l l i p s e w i t h$ $f o c i i a t A a n d B . t o f i n d t h e s h o r t e s t$ $l e n g t h o f p a t h f r o m A t o B v i a w e l l,$ $w e j u s t n e e d t o f i n d t h e e l l i p s e w h i c h$ $t o u c h e s t h e c i r c l e (w e l l) .$
Commented by mr W last updated on 11/Feb/23

	Answered by ajfour last updated on 10/Feb/23
	$w={(a+r−rcos θ)^2 +r^2 sin^2 θ}^(1/2) +{(b+r−rcos φ)^2 +r^2 sin^2 φ}^(1/2) +r{(π/2)−(φ+θ)} =Σ(√((a+2rsin^2 (θ/2))^2 +4r^2 sin^2 (θ/2)cos^2 (θ/2))) +r{(π/2)−(φ+θ)} =Σ(√(a^2 +4r(a+r)sin^2 (θ/2)))+r{(π/2)−(φ+θ)} =Σ(√(a^2 +2r(a+r)(1−cos θ)))+r{(π/2)−(φ+θ)} (∂w/∂θ)=((2r(a+r)sin θ)/( (√N_θ )))−r=0 ⇒ (√N_θ )=2(a+r)sin θ (∂w/∂φ)=0 ⇒ (√N_φ )=2(a+r)sin φ w=2(a+r)(sin θ−sin φ)+r{(π/2)−(φ+θ)} further 4(a+r)^2 sin^2 θ=a^2 +2r(a+r)(1−cos θ) ⇒ 4(a+r)^2 =4(a+r)^2 z^2 −2r(a+r)z +a^2 +2r(a+r) ⇒ z^2 −((rz)/(2(a+r)))+(1/4)+(r^2 /(4(a+r)^2 ))=1 z=(r/(4(a+r)))±(√((r^2 /(16(a+r)^2 ))−(r^2 /(4(a+r)^2 ))+1−(1/4))) cos θ=(r/(4(a+r))){1±(√((3/4)[((a/r)+1)^2 −1]))} similarly cos φ. Hence w.$
	$w = {{(a + r - r \cos θ)}^{2} + r^{2} \sin^{2} θ}^{1 / 2}$ $+ {{(b + r - r \cos ϕ)}^{2} + r^{2} \sin^{2} ϕ}^{1 / 2}$ $+ r {\frac{π}{2} - (ϕ + θ)}$ $= Σ \sqrt{{(a + 2 r \sin^{2} \frac{θ}{2})}^{2} + 4 r^{2} \sin^{2} \frac{θ}{2} \cos^{2} \frac{θ}{2}}$ $+ r {\frac{π}{2} - (ϕ + θ)}$ $= Σ \sqrt{a^{2} + 4 r (a + r) \sin^{2} \frac{θ}{2}} + r {\frac{π}{2} - (ϕ + θ)}$ $= Σ \sqrt{a^{2} + 2 r (a + r) (1 - \cos θ)} + r {\frac{π}{2} - (ϕ + θ)}$ $\frac{\partial w}{\partial θ} = \frac{2 r (a + r) \sin θ}{\sqrt{N_{θ}}} - r = 0$ $\Rightarrow \sqrt{N_{θ}} = 2 (a + r) \sin θ$ $\frac{\partial w}{\partial ϕ} = 0 \Rightarrow \sqrt{N_{ϕ}} = 2 (a + r) \sin ϕ$ $w = 2 (a + r) (\sin θ - \sin ϕ) + r {\frac{π}{2} - (ϕ + θ)}$ $f u r t h e r$ $4 {(a + r)}^{2} \sin^{2} θ = a^{2} + 2 r (a + r) (1 - \cos θ)$ $\Rightarrow 4 {(a + r)}^{2} = 4 {(a + r)}^{2} z^{2} - 2 r (a + r) z$ $+ a^{2} + 2 r (a + r)$ $\Rightarrow z^{2} - \frac{r z}{2 (a + r)} + \frac{1}{4} + \frac{r^{2}}{4 {(a + r)}^{2}} = 1$ $z = \frac{r}{4 (a + r)} \pm \sqrt{\frac{r^{2}}{16 {(a + r)}^{2}} - \frac{r^{2}}{4 {(a + r)}^{2}} + 1 - \frac{1}{4}}$ $\cos θ = \frac{r}{4 (a + r)} {1 \pm \sqrt{\frac{3}{4} [{(\frac{a}{r} + 1)}^{2} - 1]}}$ $s i m i l a r l y \cos ϕ . H e n c e w .$
	Commented by mr W last updated on 10/Feb/23

	Commented by mr W last updated on 10/Feb/23

	$i t h i n k o n l y i n c a s e 1 o n e m a y n e e d$ $t o w a l k a r o u n d t h e w e l l . i n a n y o t h e r$ $c a s e o n e s h o u l d n o t w a l k a r o u n d t h e$ $w e l l t o t a k e t h e s h o r t e s t w a y .$
	Commented by ajfour last updated on 10/Feb/23

	$y e a h b r i l l i a n t s h o r t i n t e r p r e t a t i o n!$
	Answered by mr W last updated on 10/Feb/23

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

	$c a s e 2 : \frac{a b}{\sqrt{a^{2} + b^{2}}} ⩾ r$ $(a s s u m e b ⩾ a)$ $e q n . o f e l l i p s e :$ $\frac{x^{2}}{p^{2}} + \frac{y^{2}}{q^{2}} = 1$ $A B = c = \sqrt{a^{2} + b^{2}}$ $\frac{c}{2} = \sqrt{p^{2} - q^{2}}$ $\Rightarrow p^{2} - q^{2} = \frac{a^{2} + b^{2}}{4} . . . (i)$ $x_{A} = - x_{B} = \frac{c}{2}$ $x_{D} = \frac{c}{2} - \frac{a^{2}}{c}$ $y_{D} = \frac{a b}{c}$ $s a y C (p \cos θ, q \sin θ)$ $\tan φ = \frac{q \cos θ}{p \sin θ} = \frac{q}{p \tan θ}$ $\frac{c}{2} - \frac{a^{2}}{c} = p \cos θ + r \sin φ = p \cos θ + \frac{r q}{\sqrt{p^{2} \tan^{2} θ + q^{2}}}$ $\frac{b^{2} - a^{2}}{2 \sqrt{a^{2} + b^{2}}} = p \cos θ + \frac{r q}{\sqrt{p^{2} \tan^{2} θ + q^{2}}}$ $l e t m = \tan θ$ $\Rightarrow \frac{b^{2} - a^{2}}{2 \sqrt{a^{2} + b^{2}}} = \frac{p}{\sqrt{m^{2} + 1}} + \frac{r q}{\sqrt{p^{2} m^{2} + q^{2}}} . . . (i i)$ $\frac{a b}{c} = q \sin θ + r \cos φ = q \sin θ + \frac{r p \tan θ}{\sqrt{p^{2} \tan^{2} θ + q^{2}}}$ $\Rightarrow \frac{a b}{\sqrt{a^{2} + b^{2}}} = m (\frac{q}{\sqrt{m^{2} + 1}} + \frac{r p}{\sqrt{p^{2} m^{2} + q^{2}}}) . . . (i i i)$ $w e c a n s o l v e (i) t o (i i i) f o r p, q, m .$ $t h e l e n g t h o f w a y A C + C B = L = 2 p$ $\underset{―}{e x a m p l e :}$ $a = 4, b = 6, r = 1.5$ $\Rightarrow p \approx 3.8645, q \approx 1.9344$ $\Rightarrow s h o r t e s t w a y L = 2 p \approx 7.7290$
	Commented by mr W last updated on 11/Feb/23

	Answered by mr W last updated on 11/Feb/23

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

	$c a s e 1 : \frac{a b}{\sqrt{a^{2} + b^{2}}} < r$ $A C = \sqrt{a^{2} + r^{2} - 2 a r \cos θ}$ $\overset{⌢}{C D} = r (\frac{π}{2} - θ - φ)$ $D B = \sqrt{b^{2} + r^{2} - 2 b r \cos φ}$ $L = \sqrt{a^{2} + r^{2} - 2 a r \sin θ} + \sqrt{b^{2} + r^{2} - 2 b r \sin φ} + r (\frac{π}{2} - θ - φ)$ $\frac{\partial L}{\partial θ} = \frac{a r \sin θ}{\sqrt{a^{2} + r^{2} - 2 a r \cos θ}} - r = 0$ $\Rightarrow a^{2} \sin^{2} θ = a^{2} + r^{2} - 2 a r \cos θ$ $\Rightarrow a^{2} \cos^{2} θ - 2 a r \cos θ + r^{2} = 0$ $\Rightarrow {(a \cos θ - r)}^{2} = 0$ $\Rightarrow \cos θ = \frac{r}{a}$ $s i m i l a r l y$ $\Rightarrow \cos φ = \frac{r}{b}$ $t h a t m e a n s f o r s h o r t e s t w a y A C$ $a n d B D t a n g e n t t h e c i r c l e r e s p e c t i v e l y$ $a s s h o w n i n t h e d i a g r a m a b o v e .$ $L_{m i n} = \sqrt{a^{2} - r^{2}} + \sqrt{b^{2} - r^{2}} + r (\frac{π}{2} - \cos^{- 1} \frac{r}{a} - \cos^{- 1} \frac{r}{b})$
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