Question Number 186762 by ajfour last updated on 09/Feb/23
Commented by ajfour last updated on 09/Feb/23
$${What}\:{minimum}\:{length}\:{to}\:{walk}\: \\ $$$${to}\:{start}\:{from}\:{home}\:{with}\:{empty} \\ $$$${bucket}\:{go}\:{get}\:{water}\:{from}\:{well}\:{n} \\ $$$${water}\:{the}\:{tree}. \\ $$
Commented by mr W last updated on 10/Feb/23
Commented by mr W last updated on 11/Feb/23
$${for}\:{a}\:{given}\:{path}\:{length}\:{AC}+{CB}\:{the} \\ $$$${locus}\:{of}\:{point}\:{C}\:{is}\:{an}\:{ellipse}\:{with} \\ $$$${focii}\:{at}\:{A}\:{and}\:{B}.\:{to}\:{find}\:{the}\:{shortest} \\ $$$${length}\:{of}\:{path}\:{from}\:{A}\:{to}\:{B}\:{via}\:{well},\: \\ $$$${we}\:{just}\:{need}\:{to}\:{find}\:{the}\:{ellipse}\:{which} \\ $$$${touches}\:{the}\:{circle}\:\left({well}\right). \\ $$
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.](https://www.tinkutara.com/question/Q186789.png)
$${w}=\left\{\left({a}+{r}−{r}\mathrm{cos}\:\theta\right)^{\mathrm{2}} +{r}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \theta\right\}^{\mathrm{1}/\mathrm{2}} \\ $$$$+\left\{\left({b}+{r}−{r}\mathrm{cos}\:\phi\right)^{\mathrm{2}} +{r}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \phi\right\}^{\mathrm{1}/\mathrm{2}} \\ $$$$+{r}\left\{\frac{\pi}{\mathrm{2}}−\left(\phi+\theta\right)\right\} \\ $$$$=\Sigma\sqrt{\left({a}+\mathrm{2}{r}\mathrm{sin}\:^{\mathrm{2}} \:\frac{\theta}{\mathrm{2}}\right)^{\mathrm{2}} +\mathrm{4}{r}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \frac{\theta}{\mathrm{2}}\mathrm{cos}\:^{\mathrm{2}} \frac{\theta}{\mathrm{2}}} \\ $$$$\:\:\:\:\:\:+{r}\left\{\frac{\pi}{\mathrm{2}}−\left(\phi+\theta\right)\right\} \\ $$$$=\Sigma\sqrt{{a}^{\mathrm{2}} +\mathrm{4}{r}\left({a}+{r}\right)\mathrm{sin}\:^{\mathrm{2}} \frac{\theta}{\mathrm{2}}}+{r}\left\{\frac{\pi}{\mathrm{2}}−\left(\phi+\theta\right)\right\} \\ $$$$=\Sigma\sqrt{{a}^{\mathrm{2}} +\mathrm{2}{r}\left({a}+{r}\right)\left(\mathrm{1}−\mathrm{cos}\:\theta\right)}+{r}\left\{\frac{\pi}{\mathrm{2}}−\left(\phi+\theta\right)\right\} \\ $$$$\frac{\partial{w}}{\partial\theta}=\frac{\mathrm{2}{r}\left({a}+{r}\right)\mathrm{sin}\:\theta}{\:\sqrt{{N}_{\theta} }}−{r}=\mathrm{0} \\ $$$$\Rightarrow\:\:\sqrt{{N}_{\theta} }=\mathrm{2}\left({a}+{r}\right)\mathrm{sin}\:\theta \\ $$$$\:\frac{\partial{w}}{\partial\phi}=\mathrm{0}\:\:\Rightarrow\:\:\sqrt{{N}_{\phi} }=\mathrm{2}\left({a}+{r}\right)\mathrm{sin}\:\phi \\ $$$${w}=\mathrm{2}\left({a}+{r}\right)\left(\mathrm{sin}\:\theta−\mathrm{sin}\:\phi\right)+{r}\left\{\frac{\pi}{\mathrm{2}}−\left(\phi+\theta\right)\right\} \\ $$$$\:\:{further} \\ $$$$\mathrm{4}\left({a}+{r}\right)^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \theta={a}^{\mathrm{2}} +\mathrm{2}{r}\left({a}+{r}\right)\left(\mathrm{1}−\mathrm{cos}\:\theta\right) \\ $$$$\Rightarrow\:\mathrm{4}\left({a}+{r}\right)^{\mathrm{2}} =\mathrm{4}\left({a}+{r}\right)^{\mathrm{2}} {z}^{\mathrm{2}} −\mathrm{2}{r}\left({a}+{r}\right){z} \\ $$$$\:\:\:\:\:\:+{a}^{\mathrm{2}} +\mathrm{2}{r}\left({a}+{r}\right) \\ $$$$\Rightarrow\:\:{z}^{\mathrm{2}} −\frac{{rz}}{\mathrm{2}\left({a}+{r}\right)}+\frac{\mathrm{1}}{\mathrm{4}}+\frac{{r}^{\mathrm{2}} }{\mathrm{4}\left({a}+{r}\right)^{\mathrm{2}} }=\mathrm{1} \\ $$$${z}=\frac{{r}}{\mathrm{4}\left({a}+{r}\right)}\pm\sqrt{\frac{{r}^{\mathrm{2}} }{\mathrm{16}\left({a}+{r}\right)^{\mathrm{2}} }−\frac{{r}^{\mathrm{2}} }{\mathrm{4}\left({a}+{r}\right)^{\mathrm{2}} }+\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}} \\ $$$$\mathrm{cos}\:\theta=\frac{{r}}{\mathrm{4}\left({a}+{r}\right)}\left\{\mathrm{1}\pm\sqrt{\frac{\mathrm{3}}{\mathrm{4}}\left[\left(\frac{{a}}{{r}}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{1}\right]}\right\} \\ $$$${similarly}\:\mathrm{cos}\:\phi.\:{Hence}\:{w}. \\ $$$$ \\ $$
Commented by mr W last updated on 10/Feb/23
Commented by mr W last updated on 10/Feb/23
$${i}\:{think}\:{only}\:{in}\:{case}\:\mathrm{1}\:{one}\:{may}\:{need} \\ $$$${to}\:{walk}\:{around}\:{the}\:{well}.\:{in}\:{any}\:{other} \\ $$$${case}\:{one}\:{should}\:{not}\:{walk}\:{around}\:{the}\: \\ $$$${well}\:{to}\:{take}\:{the}\:{shortest}\:{way}. \\ $$
Commented by ajfour last updated on 10/Feb/23
$${yeah}\:{brilliant}\:{short}\:{interpretation}! \\ $$
Answered by mr W last updated on 10/Feb/23
Commented by mr W last updated on 11/Feb/23
$$\boldsymbol{{case}}\:\mathrm{2}:\:\frac{\boldsymbol{{ab}}}{\:\sqrt{\boldsymbol{{a}}^{\mathrm{2}} +\boldsymbol{{b}}^{\mathrm{2}} }}\geqslant\boldsymbol{{r}} \\ $$$$\left({assume}\:{b}\geqslant{a}\right) \\ $$$${eqn}.\:{of}\:{ellipse}: \\ $$$$\frac{{x}^{\mathrm{2}} }{{p}^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} }{{q}^{\mathrm{2}} }=\mathrm{1} \\ $$$${AB}={c}=\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} } \\ $$$$\frac{{c}}{\mathrm{2}}=\sqrt{{p}^{\mathrm{2}} −{q}^{\mathrm{2}} } \\ $$$$\Rightarrow{p}^{\mathrm{2}} −{q}^{\mathrm{2}} =\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{\mathrm{4}}\:\:\:…\left({i}\right) \\ $$$${x}_{{A}} =−{x}_{{B}} =\frac{{c}}{\mathrm{2}} \\ $$$${x}_{{D}} =\frac{{c}}{\mathrm{2}}−\frac{{a}^{\mathrm{2}} }{{c}} \\ $$$${y}_{{D}} =\frac{{ab}}{{c}} \\ $$$${say}\:{C}\left({p}\:\mathrm{cos}\:\theta,\:{q}\:\mathrm{sin}\:\theta\right) \\ $$$$\mathrm{tan}\:\varphi=\frac{{q}\:\mathrm{cos}\:\theta}{{p}\:\mathrm{sin}\:\theta}=\frac{{q}}{{p}\:\mathrm{tan}\:\theta} \\ $$$$\frac{{c}}{\mathrm{2}}−\frac{{a}^{\mathrm{2}} }{{c}}={p}\:\mathrm{cos}\:\theta+{r}\:\mathrm{sin}\:\varphi={p}\:\mathrm{cos}\:\theta+\frac{{rq}}{\:\sqrt{{p}^{\mathrm{2}} \:\mathrm{tan}^{\mathrm{2}} \:\theta+{q}^{\mathrm{2}} }} \\ $$$$\frac{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }{\mathrm{2}\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}={p}\:\mathrm{cos}\:\theta+\frac{{rq}}{\:\sqrt{{p}^{\mathrm{2}} \:\mathrm{tan}^{\mathrm{2}} \:\theta+{q}^{\mathrm{2}} }} \\ $$$${let}\:{m}=\mathrm{tan}\:\theta \\ $$$$\Rightarrow\frac{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }{\mathrm{2}\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}=\frac{{p}}{\:\sqrt{{m}^{\mathrm{2}} +\mathrm{1}}}+\frac{{rq}}{\:\sqrt{{p}^{\mathrm{2}} {m}^{\mathrm{2}} +{q}^{\mathrm{2}} }}\:\:\:…\left({ii}\right) \\ $$$$\frac{{ab}}{{c}}={q}\:\mathrm{sin}\:\theta+{r}\:\mathrm{cos}\:\varphi={q}\:\mathrm{sin}\:\theta+\frac{{rp}\:\mathrm{tan}\:\theta}{\:\sqrt{{p}^{\mathrm{2}} \:\mathrm{tan}^{\mathrm{2}} \:\theta+{q}^{\mathrm{2}} }} \\ $$$$\Rightarrow\frac{{ab}}{\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}={m}\left(\frac{{q}}{\:\sqrt{{m}^{\mathrm{2}} +\mathrm{1}}}+\frac{{rp}}{\:\sqrt{{p}^{\mathrm{2}} {m}^{\mathrm{2}} +{q}^{\mathrm{2}} }}\right)\:\:\:…\left({iii}\right) \\ $$$${we}\:{can}\:{solve}\:\left({i}\right)\:{to}\:\left({iii}\right)\:{for}\:{p},{q},{m}. \\ $$$${the}\:{length}\:{of}\:{way}\:{AC}+{CB}={L}=\mathrm{2}{p} \\ $$$$\underline{{example}:} \\ $$$${a}=\mathrm{4},\:{b}=\mathrm{6},\:{r}=\mathrm{1}.\mathrm{5} \\ $$$$\Rightarrow{p}\approx\mathrm{3}.\mathrm{8645},\:{q}\approx\mathrm{1}.\mathrm{9344} \\ $$$$\Rightarrow{shortest}\:{way}\:{L}=\mathrm{2}{p}\approx\mathrm{7}.\mathrm{7290} \\ $$
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
$$\boldsymbol{{case}}\:\mathrm{1}:\:\frac{\boldsymbol{{ab}}}{\:\sqrt{\boldsymbol{{a}}^{\mathrm{2}} +\boldsymbol{{b}}^{\mathrm{2}} }}<\boldsymbol{{r}} \\ $$$${AC}=\sqrt{{a}^{\mathrm{2}} +{r}^{\mathrm{2}} −\mathrm{2}{ar}\:\mathrm{cos}\:\theta} \\ $$$$\overset{\frown} {{CD}}={r}\left(\frac{\pi}{\mathrm{2}}−\theta−\varphi\right) \\ $$$${DB}=\sqrt{{b}^{\mathrm{2}} +{r}^{\mathrm{2}} −\mathrm{2}{br}\:\mathrm{cos}\:\varphi} \\ $$$${L}=\sqrt{{a}^{\mathrm{2}} +{r}^{\mathrm{2}} −\mathrm{2}{ar}\:\mathrm{sin}\:\theta}+\sqrt{{b}^{\mathrm{2}} +{r}^{\mathrm{2}} −\mathrm{2}{br}\:\mathrm{sin}\:\varphi}+{r}\left(\frac{\pi}{\mathrm{2}}−\theta−\varphi\right) \\ $$$$\frac{\partial{L}}{\partial\theta}=\frac{{ar}\:\mathrm{sin}\:\theta}{\:\sqrt{{a}^{\mathrm{2}} +{r}^{\mathrm{2}} −\mathrm{2}{ar}\:\mathrm{cos}\:\theta}}−{r}=\mathrm{0} \\ $$$$\Rightarrow{a}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \:\theta={a}^{\mathrm{2}} +{r}^{\mathrm{2}} −\mathrm{2}{ar}\:\mathrm{cos}\:\theta \\ $$$$\Rightarrow{a}^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \:\theta−\mathrm{2}{ar}\:\mathrm{cos}\:\theta+{r}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow\left({a}\:\mathrm{cos}\:\theta−{r}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow\mathrm{cos}\:\theta=\frac{{r}}{{a}} \\ $$$${similarly} \\ $$$$\Rightarrow\mathrm{cos}\:\varphi=\frac{{r}}{{b}} \\ $$$${that}\:{means}\:{for}\:{shortest}\:{way}\:{AC}\: \\ $$$${and}\:{BD}\:{tangent}\:{the}\:{circle}\:{respectively} \\ $$$${as}\:{shown}\:{in}\:{the}\:{diagram}\:{above}. \\ $$$${L}_{{min}} =\sqrt{{a}^{\mathrm{2}} −{r}^{\mathrm{2}} }+\sqrt{{b}^{\mathrm{2}} −{r}^{\mathrm{2}} }+{r}\left(\frac{\pi}{\mathrm{2}}−\mathrm{cos}^{−\mathrm{1}} \frac{{r}}{{a}}−\mathrm{cos}^{−\mathrm{1}} \frac{{r}}{{b}}\right) \\ $$