Math Question and Answers


Question and Answers Forum

All Questions Topic List
Geometry Questions
Previous in All Question Next in All Question
Previous in Geometry Next in Geometry
Question Number 189693 by mr W last updated on 20/Mar/23

Commented by mr W last updated on 20/Mar/23

$a d d i t i o n a l q u e s t i o n t o Q 189608 :$ $w h a t i s t h e s h o r t e s t w a y l e n g t h f o r$ $t h e a n t f r o m A t o B ?$
	Answered by aleks041103 last updated on 20/Mar/23
	$we can unravel the truncated cone l=(√(h^2 +(R−r)^2 )) πr=θq πR=θ(q+l)=θp ⇒π(R−r)=θl⇒θ=((R−r)/( (√(h^2 +(R−r)^2 ))))π ⇒θ=πsin(α) q=((πr)/θ)=(r/(sin(α))) p=(r/(sin(α)))+l=(r/(sin(α)))+((R−r)/(sin(α)))=(R/(sin(α))) ⇒ { ((θ=πsin(α))),((p=R/sin(α))),((q=r/sin(α))) :} Now, to find the shortest path we can find the shortest path between A and B on the unravelled surface.$
	$w e c a n u n r a v e l t h e t r u n c a t e d c o n e$ $l = \sqrt{h^{2} + {(R - r)}^{2}}$ $π r = θ q$ $π R = θ (q + l) = θ p$ $\Rightarrow π (R - r) = θ l \Rightarrow θ = \frac{R - r}{\sqrt{h^{2} + {(R - r)}^{2}}} π$ $\Rightarrow θ = π s i n (α)$ $q = \frac{π r}{θ} = \frac{r}{s i n (α)}$ $p = \frac{r}{s i n (α)} + l = \frac{r}{s i n (α)} + \frac{R - r}{s i n (α)} = \frac{R}{s i n (α)}$ $\Rightarrow {\begin{cases} θ = π s i n (α) \\ p = R / s i n (α) \\ q = r / s i n (α) \end{cases}$ $N o w, t o f i n d t h e s h o r t e s t p a t h w e$ $c a n f i n d t h e s h o r t e s t p a t h b e t w e e n A a n d B$ $o n t h e u n r a v e l l e d s u r f a c e .$
	Commented by aleks041103 last updated on 20/Mar/23

	Commented by mr W last updated on 20/Mar/23

	$t h a n k s s i r!$ $c a n i t b e t h a t i n s o m e c a s e s t h e$ $p a t h I I i s t h e s h o r t e s t p a t h ?$
	Commented by mr W last updated on 20/Mar/23

	Answered by mr W last updated on 25/Mar/23

	Commented by mr W last updated on 26/Mar/23

	$s = {A A}^{'} = \sqrt{{(R - r)}^{2} + h^{2}} = (R - r) \sqrt{1 + {(\frac{h}{R - r})}^{2}}$ ${O A}^{'} = \frac{r s}{R - r} = r \sqrt{1 + {(\frac{h}{R - r})}^{2}}$ $c a s e 1 :$ $ϕ = \frac{π r}{r \sqrt{1 + {(\frac{h}{R - r})}^{2}}} ⩽ \cos^{- 1} \frac{r \sqrt{1 + {(\frac{h}{R - r})}^{2}}}{\sqrt{{(R - r)}^{2} + h^{2}} + r \sqrt{1 + {(\frac{h}{R - r})}^{2}}}$ $\Rightarrow \frac{r}{R} ⩾ \cos \frac{π}{\sqrt{1 + {(\frac{h}{R - r})}^{2}}}$ $A B = \sqrt{[R^{2} + r^{2} - 2 R r \cos \frac{π}{\sqrt{1 + {(\frac{h}{R - r})}^{2}}}] [1 + {(\frac{h}{R - r})}^{2}]}$
	Commented by mr W last updated on 25/Mar/23

	Commented by mr W last updated on 26/Mar/23

	$c a s e 2$ $l e t m = \sqrt{1 + {(\frac{h}{R - r})}^{2}}$ $ϕ = \frac{r θ}{r m} = \frac{θ}{m}$ $A C = m \sqrt{R^{2} + r^{2} - 2 R r \cos \frac{θ}{m}}$ $C B = 2 r \cos \frac{θ}{2}$ $L = m \sqrt{R^{2} + r^{2} - 2 R r \cos \frac{θ}{m}} + 2 r \cos \frac{θ}{2}$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum