Question Number 189608 by mr W last updated on 19/Mar/23
Commented by mr W last updated on 19/Mar/23
$${a}\:{bin}\:{has}\:{the}\:{shape}\:{of}\:{a}\:{frustum} \\ $$$${as}\:{shown}.\:{an}\:{ant}\:{climbs}\:{on}\:{the}\:{outer} \\ $$$${surface}\:{of}\:{the}\:{bin}\:{from}\:{point} \\ $$$${A}\:{to}\:{point}\:{B}\:{following}\:{a}\:{path}\:{with} \\ $$$${constant}\:{slope}.\:{find}\:{the}\:{total}\:{way} \\ $$$${length}\:{of}\:{the}\:{ant}. \\ $$
Answered by aleks041103 last updated on 19/Mar/23
$${lets}\:{use}\:{cillindrical}\:{coords} \\ $$$${r}\left({z}\right)={R}−\frac{{R}−{r}}{{h}}{z} \\ $$$${slope}\:=\:\frac{{vertical}}{{horizontal}}=\frac{{dz}}{{r}\left({z}\right){d}\theta}={a}={const}. \\ $$$${l}=\int_{\mathrm{0}} ^{\pi} \sqrt{{dz}^{\mathrm{2}} +\left({rd}\theta\right)^{\mathrm{2}} }=\int_{\mathrm{0}} ^{\pi} \sqrt{\mathrm{1}+\left(\frac{{rd}\theta}{{dz}}\right)^{\mathrm{2}} }{dz}= \\ $$$$=\sqrt{\mathrm{1}+\frac{\mathrm{1}}{{a}^{\mathrm{2}} }}\int_{\mathrm{0}} ^{\pi} {dz}={h}\sqrt{\mathrm{1}+\frac{\mathrm{1}}{{a}^{\mathrm{2}} }} \\ $$$$\frac{{dz}}{{R}−\frac{{R}−{r}}{{h}}{z}}={ad}\theta \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{\:{h}} \frac{{dz}}{{R}−\frac{{R}−{r}}{{h}}{z}}={a}\int_{\mathrm{0}} ^{\:\pi} {d}\theta \\ $$$$−\frac{{h}}{{R}−{r}}\int_{\mathrm{0}} ^{\:{h}} \frac{{d}\left({R}−\frac{{R}−{r}}{{h}}{z}\right)}{{R}−\frac{{R}−{r}}{{h}}{z}}={a}\pi \\ $$$$−\frac{{h}}{{R}−{r}}{ln}\left(\frac{{r}}{{R}}\right)={a}\pi\Rightarrow\frac{\mathrm{1}}{{a}}=\frac{\pi\left({R}−{r}\right)}{{h}\:{ln}\left({R}/{r}\right)} \\ $$$$\Rightarrow{l}=\sqrt{{h}^{\mathrm{2}} +\left(\frac{\pi\left({R}−{r}\right)}{{ln}\left({R}/{r}\right)}\right)^{\mathrm{2}} } \\ $$
Commented by mr W last updated on 19/Mar/23
$${very}\:{nice}\:{solution}\:{sir}!\:{thanks}! \\ $$
Answered by mr W last updated on 19/Mar/23
Commented by mr W last updated on 19/Mar/23
![slope =(dz/(ρdθ))=k=constant (z/(R−ρ))=(h/(R−r)) ⇒z=(((R−ρ)h)/(R−r)) dz=−((hdρ)/(R−r))=kρdθ ((hdρ)/ρ)=−k(R−r)dθ h∫_R ^r (dρ/ρ)=−k(R−r)∫_0 ^π dθ hln (r/R)=−k(R−r)π ⇒k=(h/(π(R−r)))ln (R/r) length of path L=h(√(1+(1/k^2 )))=h(√(1+[((π(R−r))/(h ln (R/r)))]^2 ))](https://www.tinkutara.com/question/Q189623.png)
$${slope}\:=\frac{{dz}}{\rho{d}\theta}={k}={constant} \\ $$$$\frac{{z}}{{R}−\rho}=\frac{{h}}{{R}−{r}}\:\Rightarrow{z}=\frac{\left({R}−\rho\right){h}}{{R}−{r}} \\ $$$${dz}=−\frac{{hd}\rho}{{R}−{r}}={k}\rho{d}\theta \\ $$$$\frac{{hd}\rho}{\rho}=−{k}\left({R}−{r}\right){d}\theta \\ $$$${h}\int_{{R}} ^{{r}} \frac{{d}\rho}{\rho}=−{k}\left({R}−{r}\right)\int_{\mathrm{0}} ^{\pi} {d}\theta \\ $$$${h}\mathrm{ln}\:\frac{{r}}{{R}}=−{k}\left({R}−{r}\right)\pi \\ $$$$\Rightarrow{k}=\frac{{h}}{\pi\left({R}−{r}\right)}\mathrm{ln}\:\frac{{R}}{{r}} \\ $$$${length}\:{of}\:{path} \\ $$$${L}={h}\sqrt{\mathrm{1}+\frac{\mathrm{1}}{{k}^{\mathrm{2}} }}={h}\sqrt{\mathrm{1}+\left[\frac{\pi\left({R}−{r}\right)}{{h}\:\mathrm{ln}\:\frac{{R}}{{r}}}\right]^{\mathrm{2}} } \\ $$