Question Number 189693 by mr W last updated on 20/Mar/23
Commented by mr W last updated on 20/Mar/23
$${additional}\:{question}\:{to}\:{Q}\mathrm{189608}: \\ $$$${what}\:{is}\:{the}\:{shortest}\:{way}\:{length}\:{for}\: \\ $$$${the}\:{ant}\:{from}\:{A}\:{to}\:{B}? \\ $$
Answered by aleks041103 last updated on 20/Mar/23
$${we}\:{can}\:{unravel}\:{the}\:{truncated}\:{cone} \\ $$$${l}=\sqrt{{h}^{\mathrm{2}} +\left({R}−{r}\right)^{\mathrm{2}} } \\ $$$$\pi{r}=\theta{q} \\ $$$$\pi{R}=\theta\left({q}+{l}\right)=\theta{p} \\ $$$$\Rightarrow\pi\left({R}−{r}\right)=\theta{l}\Rightarrow\theta=\frac{{R}−{r}}{\:\sqrt{{h}^{\mathrm{2}} +\left({R}−{r}\right)^{\mathrm{2}} }}\pi \\ $$$$\Rightarrow\theta=\pi{sin}\left(\alpha\right) \\ $$$${q}=\frac{\pi{r}}{\theta}=\frac{{r}}{{sin}\left(\alpha\right)} \\ $$$${p}=\frac{{r}}{{sin}\left(\alpha\right)}+{l}=\frac{{r}}{{sin}\left(\alpha\right)}+\frac{{R}−{r}}{{sin}\left(\alpha\right)}=\frac{{R}}{{sin}\left(\alpha\right)} \\ $$$$\Rightarrow\begin{cases}{\theta=\pi{sin}\left(\alpha\right)}\\{{p}={R}/{sin}\left(\alpha\right)}\\{{q}={r}/{sin}\left(\alpha\right)}\end{cases} \\ $$$${Now},\:{to}\:{find}\:{the}\:{shortest}\:{path}\:{we}\: \\ $$$${can}\:{find}\:{the}\:{shortest}\:{path}\:{between}\:{A}\:{and}\:{B} \\ $$$${on}\:{the}\:{unravelled}\:{surface}. \\ $$
Commented by aleks041103 last updated on 20/Mar/23
Commented by mr W last updated on 20/Mar/23
$${thanks}\:{sir}! \\ $$$${can}\:{it}\:{be}\:{that}\:{in}\:{some}\:{cases}\:{the}\: \\ $$$${path}\:{II}\:{is}\:{the}\:{shortest}\:{path}? \\ $$
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=AA′=(√((R−r)^2 +h^2 ))=(R−r)(√(1+((h/(R−r)))^2 )) OA′=((rs)/(R−r))=r(√(1+((h/(R−r)))^2 )) case 1: φ=((πr)/(r(√(1+((h/(R−r)))^2 ))))≤cos^(−1) ((r(√(1+((h/(R−r)))^2 )))/( (√((R−r)^2 +h^2 ))+r(√(1+((h/(R−r)))^2 )))) ⇒(r/( R)) ≥cos (π/( (√(1+((h/(R−r)))^2 )))) AB=(√([R^2 +r^2 −2Rr cos (π/( (√(1+((h/(R−r)))^2 ))))][1+((h/(R−r)))^2 ]))](https://www.tinkutara.com/question/Q189960.png)
$${s}={AA}'=\sqrt{\left({R}−{r}\right)^{\mathrm{2}} +{h}^{\mathrm{2}} }=\left({R}−{r}\right)\sqrt{\mathrm{1}+\left(\frac{{h}}{{R}−{r}}\right)^{\mathrm{2}} } \\ $$$${OA}'=\frac{{rs}}{{R}−{r}}={r}\sqrt{\mathrm{1}+\left(\frac{{h}}{{R}−{r}}\right)^{\mathrm{2}} } \\ $$$$\boldsymbol{{case}}\:\mathrm{1}: \\ $$$$\phi=\frac{\pi{r}}{{r}\sqrt{\mathrm{1}+\left(\frac{{h}}{{R}−{r}}\right)^{\mathrm{2}} }}\leqslant\mathrm{cos}^{−\mathrm{1}} \frac{{r}\sqrt{\mathrm{1}+\left(\frac{{h}}{{R}−{r}}\right)^{\mathrm{2}} }}{\:\sqrt{\left({R}−{r}\right)^{\mathrm{2}} +{h}^{\mathrm{2}} }+{r}\sqrt{\mathrm{1}+\left(\frac{{h}}{{R}−{r}}\right)^{\mathrm{2}} }} \\ $$$$\Rightarrow\frac{{r}}{\:{R}}\:\geqslant\mathrm{cos}\:\frac{\pi}{\:\sqrt{\mathrm{1}+\left(\frac{{h}}{{R}−{r}}\right)^{\mathrm{2}} }} \\ $$$${AB}=\sqrt{\left[{R}^{\mathrm{2}} +{r}^{\mathrm{2}} −\mathrm{2}{Rr}\:\mathrm{cos}\:\frac{\pi}{\:\sqrt{\mathrm{1}+\left(\frac{{h}}{{R}−{r}}\right)^{\mathrm{2}} }}\right]\left[\mathrm{1}+\left(\frac{{h}}{{R}−{r}}\right)^{\mathrm{2}} \right]} \\ $$
Commented by mr W last updated on 25/Mar/23
Commented by mr W last updated on 26/Mar/23
$$\boldsymbol{{case}}\:\mathrm{2} \\ $$$${let}\:{m}=\sqrt{\mathrm{1}+\left(\frac{{h}}{{R}−{r}}\right)^{\mathrm{2}} } \\ $$$$\phi=\frac{{r}\theta}{{rm}}=\frac{\theta}{{m}} \\ $$$${AC}={m}\sqrt{{R}^{\mathrm{2}} +{r}^{\mathrm{2}} −\mathrm{2}{Rr}\:\mathrm{cos}\:\frac{\theta}{{m}}} \\ $$$${CB}=\mathrm{2}{r}\:\mathrm{cos}\:\frac{\theta}{\mathrm{2}} \\ $$$${L}={m}\sqrt{{R}^{\mathrm{2}} +{r}^{\mathrm{2}} −\mathrm{2}{Rr}\:\mathrm{cos}\:\frac{\theta}{{m}}}+\mathrm{2}{r}\:\mathrm{cos}\:\frac{\theta}{\mathrm{2}} \\ $$