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Question Number 189608 by mr W last updated on 19/Mar/23

Commented by mr W last updated on 19/Mar/23

$$abinhastheshapeofafrustum$$$$asshown.anantclimbsontheouter$$$$surfaceofthebinfrompoint$$$$AtopointBfollowingapathwith$$$$constantslope.findthetotalway$$$$lengthoftheant.$$

Answered by aleks041103 last updated on 19/Mar/23

$$letsusecillindricalcoords$$$$r\left(z\right)=R-\frac{R-r}{h}z$$$$slope=\frac{vertical}{horizontal}=\frac{dz}{r\left(z\right)d\theta }=a=const.$$$$l={\int }_0^{\pi }\sqrt{{dz}^2+{\left(rd\theta \right)}^2}={\int }_0^{\pi }\sqrt{1+{\left(\frac{rd\theta }{dz}\right)}^2}dz=$$$$=\sqrt{1+\frac{1}{a^2}}{\int }_0^{\pi }dz=h\sqrt{1+\frac{1}{a^2}}$$$$\frac{dz}{R-\frac{R-r}{h}z}=ad\theta$$$$\Rightarrow {\int }_0^h\frac{dz}{R-\frac{R-r}{h}z}=a{\int }_0^{\pi }d\theta$$$$-\frac{h}{R-r}{\int }_0^h\frac{d(R-\frac{R-r}{h}z)}{R-\frac{R-r}{h}z}=a\pi$$$$-\frac{h}{R-r}ln\left(\frac{r}{R}\right)=a\pi \Rightarrow \frac{1}{a}=\frac{\pi (R-r)}{hln\left(R/r\right)}$$$$\Rightarrow l=\sqrt{h^2+{\left(\frac{\pi (R-r)}{ln\left(R/r\right)}\right)}^2}$$

Commented by mr W last updated on 19/Mar/23

$$verynicesolutionsir!thanks!$$

Answered by mr W last updated on 19/Mar/23

Commented by mr W last updated on 19/Mar/23

$$slope=\frac{dz}{\rho d\theta }=k=constant$$$$\frac{z}{R-\rho }=\frac{h}{R-r}\Rightarrow z=\frac{(R-\rho )h}{R-r}$$$$dz=-\frac{hd\rho }{R-r}=k\rho d\theta$$$$\frac{hd\rho }{\rho }=-k(R-r)d\theta$$$$h{\int }_R^r\frac{d\rho }{\rho }=-k(R-r){\int }_0^{\pi }d\theta$$$$h\ln \frac{r}{R}=-k(R-r)\pi$$$$\Rightarrow k=\frac{h}{\pi (R-r)}\ln \frac{R}{r}$$$$lengthofpath$$$$L=h\sqrt{1+\frac{1}{k^2}}=h\sqrt{1+{\left[\frac{\pi (R-r)}{h\ln \frac{R}{r}}\right]}^2}$$

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