Question Number 137768 by mr W last updated on 06/Apr/21
Commented by mr W last updated on 06/Apr/21
$${find}\:{the}\:{maximal}\:{area}\:{of}\:{the}\:{shadow} \\ $$$${in}\:{terms}\:{of}\:{r},\:{h},\:{H}\:{and}\:{d}. \\ $$
Answered by mr W last updated on 07/Apr/21
Commented by ajfour last updated on 08/Apr/21
![let C be origin. CP^(−) =rcos φ(cos θi+sin θk) +rsin φj S≡(−d,0,H−h) r^ =−di+(H−h)k +λ{(rcos φcos θ+d)i+rsin φj +(rcos φsin θ−H+h)k} let H−h+λ(rcos φsin θ−H+h) =−h ⇒ λ=(H/(H−h−rcos φsin θ)) r^ =(−d+((H[rcos φcos θ+d])/(H−h−rcos φsin θ)))i +((Hrsin φj)/(H−h−rcos φsin θ)) let sin φ=0 ⇒ p_1 , p_2 =−d±((H(rcos θ+d))/(H−h∓rsin θ)) 2a=((H(rcos θ+d))/(H−h−rsin θ))+((H(rcos θ+d))/(H−h+rsin θ)) b=max(((Hrsin φ)/(H−h−rcos φsin θ))) (∂b/∂φ)=((Hr{cos φ(H−h−rcos φsin θ)−sin φ(rsin φsin θ)})/((H−h−rcos φsin θ)^2 )) (∂b/∂φ)=0 ⇒ tan φ=((H−h−rcos φsin θ)/(rsin φsin θ)) rsin θ=(H−h)cos φ cos φ=((rsin θ)/(H−h)) b^2 =(((Hr)^2 {1−((r^2 sin^2 θ)/((H−h)^2 ))})/({H−h−((r^2 sin^2 θ)/(H−h))}^2 )) A^2 =(πab)^2 =((π/2))^2 [(((Hr)^2 {1−((r^2 sin^2 θ)/((H−h)^2 ))})/({H−h−((r^2 sin^2 θ)/(H−h))}^2 ))] ×H^2 {((rcos θ+d)/(H−h−rsin θ))+((rcos θ+d)/(H−h+rsin θ))}^2 for A_(max) , ((d(A^2 ))/dθ)=0 A^2 =((π^2 H^4 (H−h)^2 r^2 (rcos θ+d)^2 )/({(H−h)^2 −r^2 sin^2 θ}^3 )) lets say rcos θ=t A^2 =((π^2 H^4 (H−h)^2 r^2 (t+d)^2 )/({(H−h)^2 −r^2 +t^2 }^3 )) ((d(A^2 ))/dt)=0 ⇒ 2(t+d){(H−h)^2 −r^2 +t^2 }^3 =3(2t)(t+d)^2 {(H−h)^2 −r^2 +t^2 }^2 ⇒ {(H−h)^2 −r^2 +t^2 }=3t(t+d) 2t^2 +(3d)t−(H−h)^2 +r^2 =0 t=((−3d+(√(9d^2 +8(H−h)^2 −8r^2 )))/4) A_m ^2 =((π^2 H^4 (H−h)^2 r^2 )/(27t^3 (t+d)))](Q137922.png)
$${let}\:{C}\:{be}\:{origin}. \\ $$$$\overline {{CP}}={r}\mathrm{cos}\:\phi\left(\mathrm{cos}\:\theta{i}+\mathrm{sin}\:\theta{k}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:+{r}\mathrm{sin}\:\phi{j} \\ $$$${S}\equiv\left(−{d},\mathrm{0},{H}−{h}\right) \\ $$$$\bar {{r}}=−{di}+\left({H}−{h}\right){k} \\ $$$$\:+\lambda\left\{\left({r}\mathrm{cos}\:\phi\mathrm{cos}\:\theta+{d}\right){i}+{r}\mathrm{sin}\:\phi{j}\right. \\ $$$$\left.\:\:\:\:\:\:\:\:\:\:\:\:\:+\left({r}\mathrm{cos}\:\phi\mathrm{sin}\:\theta−{H}+{h}\right){k}\right\} \\ $$$${let}\:\:\:{H}−{h}+\lambda\left({r}\mathrm{cos}\:\phi\mathrm{sin}\:\theta−{H}+{h}\right) \\ $$$$\:\:\:\:\:=−{h} \\ $$$$\Rightarrow\:\lambda=\frac{{H}}{{H}−{h}−{r}\mathrm{cos}\:\phi\mathrm{sin}\:\theta} \\ $$$$\bar {{r}}=\left(−{d}+\frac{{H}\left[{r}\mathrm{cos}\:\phi\mathrm{cos}\:\theta+{d}\right]}{{H}−{h}−{r}\mathrm{cos}\:\phi\mathrm{sin}\:\theta}\right){i} \\ $$$$\:\:\:+\frac{{Hr}\mathrm{sin}\:\phi{j}}{{H}−{h}−{r}\mathrm{cos}\:\phi\mathrm{sin}\:\theta} \\ $$$${let}\:\:\mathrm{sin}\:\phi=\mathrm{0} \\ $$$$\Rightarrow\:\:{p}_{\mathrm{1}} ,\:{p}_{\mathrm{2}} =−{d}\pm\frac{{H}\left({r}\mathrm{cos}\:\theta+{d}\right)}{{H}−{h}\mp{r}\mathrm{sin}\:\theta} \\ $$$$\mathrm{2}{a}=\frac{{H}\left({r}\mathrm{cos}\:\theta+{d}\right)}{{H}−{h}−{r}\mathrm{sin}\:\theta}+\frac{{H}\left({r}\mathrm{cos}\:\theta+{d}\right)}{{H}−{h}+{r}\mathrm{sin}\:\theta} \\ $$$${b}={max}\left(\frac{{Hr}\mathrm{sin}\:\phi}{{H}−{h}−{r}\mathrm{cos}\:\phi\mathrm{sin}\:\theta}\right) \\ $$$$\frac{\partial{b}}{\partial\phi}=\frac{{Hr}\left\{\mathrm{cos}\:\phi\left({H}−{h}−{r}\mathrm{cos}\:\phi\mathrm{sin}\:\theta\right)−\mathrm{sin}\:\phi\left({r}\mathrm{sin}\:\phi\mathrm{sin}\:\theta\right)\right\}}{\left({H}−{h}−{r}\mathrm{cos}\:\phi\mathrm{sin}\:\theta\right)^{\mathrm{2}} } \\ $$$$\frac{\partial{b}}{\partial\phi}=\mathrm{0}\:\:\Rightarrow \\ $$$$\mathrm{tan}\:\phi=\frac{{H}−{h}−{r}\mathrm{cos}\:\phi\mathrm{sin}\:\theta}{{r}\mathrm{sin}\:\phi\mathrm{sin}\:\theta} \\ $$$${r}\mathrm{sin}\:\theta=\left({H}−{h}\right)\mathrm{cos}\:\phi \\ $$$$\mathrm{cos}\:\phi=\frac{{r}\mathrm{sin}\:\theta}{{H}−{h}}\:\:\:\:\:\:\:\: \\ $$$${b}^{\mathrm{2}} =\frac{\left({Hr}\right)^{\mathrm{2}} \left\{\mathrm{1}−\frac{{r}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \theta}{\left({H}−{h}\right)^{\mathrm{2}} }\right\}}{\left\{{H}−{h}−\frac{{r}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \theta}{{H}−{h}}\right\}^{\mathrm{2}} } \\ $$$${A}^{\mathrm{2}} =\left(\pi{ab}\right)^{\mathrm{2}} \\ $$$$=\left(\frac{\pi}{\mathrm{2}}\right)^{\mathrm{2}} \left[\frac{\left({Hr}\right)^{\mathrm{2}} \left\{\mathrm{1}−\frac{{r}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \theta}{\left({H}−{h}\right)^{\mathrm{2}} }\right\}}{\left\{{H}−{h}−\frac{{r}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \theta}{{H}−{h}}\right\}^{\mathrm{2}} }\right] \\ $$$$×{H}^{\mathrm{2}} \left\{\frac{{r}\mathrm{cos}\:\theta+{d}}{{H}−{h}−{r}\mathrm{sin}\:\theta}+\frac{{r}\mathrm{cos}\:\theta+{d}}{{H}−{h}+{r}\mathrm{sin}\:\theta}\right\}^{\mathrm{2}} \\ $$$${for}\:\:{A}_{{max}} \:\:\:,\:\:\frac{{d}\left({A}^{\mathrm{2}} \right)}{{d}\theta}=\mathrm{0} \\ $$$${A}^{\mathrm{2}} =\frac{\pi^{\mathrm{2}} {H}^{\mathrm{4}} \left({H}−{h}\right)^{\mathrm{2}} {r}^{\mathrm{2}} \left({r}\mathrm{cos}\:\theta+{d}\right)^{\mathrm{2}} }{\left\{\left({H}−{h}\right)^{\mathrm{2}} −{r}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \theta\right\}^{\mathrm{3}} } \\ $$$${lets}\:{say}\:\:{r}\mathrm{cos}\:\theta={t} \\ $$$${A}^{\mathrm{2}} =\frac{\pi^{\mathrm{2}} {H}^{\mathrm{4}} \left({H}−{h}\right)^{\mathrm{2}} {r}^{\mathrm{2}} \left({t}+{d}\right)^{\mathrm{2}} }{\left\{\left({H}−{h}\right)^{\mathrm{2}} −{r}^{\mathrm{2}} +{t}^{\mathrm{2}} \right\}^{\mathrm{3}} } \\ $$$$\frac{{d}\left({A}^{\mathrm{2}} \right)}{{dt}}=\mathrm{0}\:\Rightarrow \\ $$$$\mathrm{2}\left({t}+{d}\right)\left\{\left({H}−{h}\right)^{\mathrm{2}} −{r}^{\mathrm{2}} +{t}^{\mathrm{2}} \right\}^{\mathrm{3}} \\ $$$$=\mathrm{3}\left(\mathrm{2}{t}\right)\left({t}+{d}\right)^{\mathrm{2}} \left\{\left({H}−{h}\right)^{\mathrm{2}} −{r}^{\mathrm{2}} +{t}^{\mathrm{2}} \right\}^{\mathrm{2}} \\ $$$$\Rightarrow\:\left\{\left({H}−{h}\right)^{\mathrm{2}} −{r}^{\mathrm{2}} +{t}^{\mathrm{2}} \right\}=\mathrm{3}{t}\left({t}+{d}\right) \\ $$$$\mathrm{2}{t}^{\mathrm{2}} +\left(\mathrm{3}{d}\right){t}−\left({H}−{h}\right)^{\mathrm{2}} +{r}^{\mathrm{2}} =\mathrm{0} \\ $$$${t}=\frac{−\mathrm{3}{d}+\sqrt{\mathrm{9}{d}^{\mathrm{2}} +\mathrm{8}\left({H}−{h}\right)^{\mathrm{2}} −\mathrm{8}{r}^{\mathrm{2}} }}{\mathrm{4}} \\ $$$${A}_{{m}} ^{\mathrm{2}} =\frac{\pi^{\mathrm{2}} {H}^{\mathrm{4}} \left({H}−{h}\right)^{\mathrm{2}} {r}^{\mathrm{2}} }{\mathrm{27}{t}^{\mathrm{3}} \left({t}+{d}\right)} \\ $$
Commented by mr W last updated on 08/Apr/21
$${great}! \\ $$