Menu Close

Question-210694

Tinku Tara 0 Comments
Pin




Question Number 210694 by ajfour last updated on 16/Aug/24
Answered by mr W last updated on 17/Aug/24
Commented by mr W last updated on 17/Aug/24
R=(√(a^2 +b^2 )) ((sin α)/b)=((sin ((π/2)−θ+α))/R)=((cos θ cos α+sin θ sin α)/R) (R/b)=((cos θ )/(tan α))+sin θ ⇒(1/(tan α))=((R/b)−sin θ)(1/(cos θ)) ((sin α)/a)=((sin (θ+α))/R)=((sin θ cos α+cos θ sin α)/R) (R/a)=((sin θ)/(tan α))+cos θ ⇒(1/(tan α))=((R/a)−cos θ)(1/(sin θ)) ((R/b)−sin θ)(1/(cos θ))=((R/a)−cos θ)(1/(sin θ)) (R/a) cos θ−(R/b) sin θ=cos 2θ ⇒((cos θ)/(cos φ))−((sin θ)/(sin φ))=cos 2θ L_(min) =(√(a^2 +R^2 −2aR cos θ))+(√(b^2 +R^2 −2bR sin θ)) (L_(min) /R)=(√(1+cos^2 φ−2 cos φ cos θ))+(√(1+sin^2 φ−2 sin φ sin θ))
$${R}=\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} } \\ $$$$\frac{\mathrm{sin}\:\alpha}{{b}}=\frac{\mathrm{sin}\:\left(\frac{\pi}{\mathrm{2}}−\theta+\alpha\right)}{{R}}=\frac{\mathrm{cos}\:\theta\:\mathrm{cos}\:\alpha+\mathrm{sin}\:\theta\:\mathrm{sin}\:\alpha}{{R}} \\ $$$$\frac{{R}}{{b}}=\frac{\mathrm{cos}\:\theta\:}{\mathrm{tan}\:\alpha}+\mathrm{sin}\:\theta \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{tan}\:\alpha}=\left(\frac{{R}}{{b}}−\mathrm{sin}\:\theta\right)\frac{\mathrm{1}}{\mathrm{cos}\:\theta} \\ $$$$\frac{\mathrm{sin}\:\alpha}{{a}}=\frac{\mathrm{sin}\:\left(\theta+\alpha\right)}{{R}}=\frac{\mathrm{sin}\:\theta\:\mathrm{cos}\:\alpha+\mathrm{cos}\:\theta\:\mathrm{sin}\:\alpha}{{R}} \\ $$$$\frac{{R}}{{a}}=\frac{\mathrm{sin}\:\theta}{\mathrm{tan}\:\alpha}+\mathrm{cos}\:\theta \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{tan}\:\alpha}=\left(\frac{{R}}{{a}}−\mathrm{cos}\:\theta\right)\frac{\mathrm{1}}{\mathrm{sin}\:\theta} \\ $$$$\left(\frac{{R}}{{b}}−\mathrm{sin}\:\theta\right)\frac{\mathrm{1}}{\mathrm{cos}\:\theta}=\left(\frac{{R}}{{a}}−\mathrm{cos}\:\theta\right)\frac{\mathrm{1}}{\mathrm{sin}\:\theta} \\ $$$$\frac{{R}}{{a}}\:\mathrm{cos}\:\theta−\frac{{R}}{{b}}\:\mathrm{sin}\:\theta=\mathrm{cos}\:\mathrm{2}\theta \\ $$$$\Rightarrow\frac{\mathrm{cos}\:\theta}{\mathrm{cos}\:\phi}−\frac{\mathrm{sin}\:\theta}{\mathrm{sin}\:\phi}=\mathrm{cos}\:\mathrm{2}\theta \\ $$$$ \\ $$$${L}_{{min}} =\sqrt{{a}^{\mathrm{2}} +{R}^{\mathrm{2}} −\mathrm{2}{aR}\:\mathrm{cos}\:\theta}+\sqrt{{b}^{\mathrm{2}} +{R}^{\mathrm{2}} −\mathrm{2}{bR}\:\mathrm{sin}\:\theta} \\ $$$$\frac{{L}_{{min}} }{{R}}=\sqrt{\mathrm{1}+\mathrm{cos}^{\mathrm{2}} \:\phi−\mathrm{2}\:\mathrm{cos}\:\phi\:\mathrm{cos}\:\theta}+\sqrt{\mathrm{1}+\mathrm{sin}^{\mathrm{2}} \:\phi−\mathrm{2}\:\mathrm{sin}\:\phi\:\mathrm{sin}\:\theta} \\ $$
Commented by mr W last updated on 17/Aug/24
an exact solution seems not to be possible.
$${an}\:{exact}\:{solution}\:{seems}\:{not}\:{to}\:{be} \\ $$$${possible}. \\ $$
Commented by ajfour last updated on 17/Aug/24
((sin α)/b)=((sin ((π/2)+θ−α))/R)=((cos θ cos α−sin θ sin α)/R) (R/a)=((sin θ)/(tan α))−cos θ (((sin θ)/m)−(R/a))^2 =1−sin^2 θ L_(min) =(√(a^2 +R^2 −2aR cos θ))+(√(b^2 +R^2 −2bR sin θ)) =(√(a^2 +R^2 −2R(((asin θ)/m)−R))) +(√(b^2 +R^2 −2bRsin θ)) =(√(a^2 +3R^2 −((2aR)/m)sin θ))+(√(b^2 +R^2 −2bRsin θ)) .......
$$\frac{\mathrm{sin}\:\alpha}{{b}}=\frac{\mathrm{sin}\:\left(\frac{\pi}{\mathrm{2}}+\theta−\alpha\right)}{{R}}=\frac{\mathrm{cos}\:\theta\:\mathrm{cos}\:\alpha−\mathrm{sin}\:\theta\:\mathrm{sin}\:\alpha}{{R}} \\ $$$$\frac{{R}}{{a}}=\frac{\mathrm{sin}\:\theta}{\mathrm{tan}\:\alpha}−\mathrm{cos}\:\theta \\ $$$$\left(\frac{\mathrm{sin}\:\theta}{{m}}−\frac{{R}}{{a}}\right)^{\mathrm{2}} =\mathrm{1}−\mathrm{sin}\:^{\mathrm{2}} \theta \\ $$$${L}_{{min}} =\sqrt{{a}^{\mathrm{2}} +{R}^{\mathrm{2}} −\mathrm{2}{aR}\:\mathrm{cos}\:\theta}+\sqrt{{b}^{\mathrm{2}} +{R}^{\mathrm{2}} −\mathrm{2}{bR}\:\mathrm{sin}\:\theta} \\ $$$$\:\:=\sqrt{{a}^{\mathrm{2}} +{R}^{\mathrm{2}} −\mathrm{2}{R}\left(\frac{{a}\mathrm{sin}\:\theta}{{m}}−{R}\right)} \\ $$$$\:\:\:\:\:\:\:\:+\sqrt{{b}^{\mathrm{2}} +{R}^{\mathrm{2}} −\mathrm{2}{bR}\mathrm{sin}\:\theta} \\ $$$$=\sqrt{{a}^{\mathrm{2}} +\mathrm{3}{R}^{\mathrm{2}} −\frac{\mathrm{2}{aR}}{{m}}\mathrm{sin}\:\theta}+\sqrt{{b}^{\mathrm{2}} +{R}^{\mathrm{2}} −\mathrm{2}{bR}\mathrm{sin}\:\theta} \\ $$$$……. \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *