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Question-175938

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Question Number 175938 by ajfour last updated on 09/Sep/22
Answered by mahdipoor last updated on 09/Sep/22
point of B1: (a,b)+a(sinθ,cosθ) θ=wt=(v/a)t point of B2: (ut,0) ⇒ D=∣∣B_1 B_2 ∣∣^2 = (a(1+sinθ)−ut)^2 +(b+a.cosθ)^2 when θ_m =α+(π/2)=((2π)/3) ⇒ t_m =((aθ)/v)=((2aπ)/(3v)) (dθ/dt)(t_m )=(v/a) (dD/dt)(t_m )= 2(a+a.sinθ_m −ut_m ).(a.cosθ_m .(dθ/dt)(t_m )−u)+ 2(b+a.cosθ_m )(−a.sinθ_m .(dθ/dt)(t_m ))=0 ⇒ 2a(1+((√3)/2)−((2uπ)/(3v)))(−(v/2)−u)+ 2(b−(a/2))(−((v(√3))/2))=0 ⇒b=(a/2)−((a(1+((√3)/2)−((2uπ)/(3v)))((v/2)+u))/((v(√3))/2))= a((((2uπ)/(3v))(2u+v)−(v+u(2+(√3))))/(v(√3)))
$$ \\ $$$${point}\:{of}\:{B}\mathrm{1}: \\ $$$$\left({a},{b}\right)+{a}\left({sin}\theta,{cos}\theta\right)\:\:\:\:\:\:\:\:\:\theta={wt}=\frac{{v}}{{a}}{t} \\ $$$${point}\:{of}\:{B}\mathrm{2}: \\ $$$$\left({ut},\mathrm{0}\right) \\ $$$$\Rightarrow \\ $$$${D}=\mid\mid{B}_{\mathrm{1}} {B}_{\mathrm{2}} \mid\mid^{\mathrm{2}} = \\ $$$$\left({a}\left(\mathrm{1}+{sin}\theta\right)−{ut}\right)^{\mathrm{2}} +\left({b}+{a}.{cos}\theta\right)^{\mathrm{2}} \\ $$$${when}\:\theta_{{m}} =\alpha+\frac{\pi}{\mathrm{2}}=\frac{\mathrm{2}\pi}{\mathrm{3}}\:\:\Rightarrow\:{t}_{{m}} =\frac{{a}\theta}{{v}}=\frac{\mathrm{2}{a}\pi}{\mathrm{3}{v}} \\ $$$$\frac{{d}\theta}{{dt}}\left({t}_{{m}} \right)=\frac{{v}}{{a}} \\ $$$$\frac{{dD}}{{dt}}\left({t}_{{m}} \right)= \\ $$$$\mathrm{2}\left({a}+{a}.{sin}\theta_{{m}} −{ut}_{{m}} \right).\left({a}.{cos}\theta_{{m}} .\frac{{d}\theta}{{dt}}\left({t}_{{m}} \right)−{u}\right)+ \\ $$$$\mathrm{2}\left({b}+{a}.{cos}\theta_{{m}} \right)\left(−{a}.{sin}\theta_{{m}} .\frac{{d}\theta}{{dt}}\left({t}_{{m}} \right)\right)=\mathrm{0} \\ $$$$\Rightarrow \\ $$$$\mathrm{2}{a}\left(\mathrm{1}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}−\frac{\mathrm{2}{u}\pi}{\mathrm{3}{v}}\right)\left(−\frac{{v}}{\mathrm{2}}−{u}\right)+ \\ $$$$\mathrm{2}\left({b}−\frac{{a}}{\mathrm{2}}\right)\left(−\frac{{v}\sqrt{\mathrm{3}}}{\mathrm{2}}\right)=\mathrm{0} \\ $$$$\Rightarrow{b}=\frac{{a}}{\mathrm{2}}−\frac{{a}\left(\mathrm{1}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}−\frac{\mathrm{2}{u}\pi}{\mathrm{3}{v}}\right)\left(\frac{{v}}{\mathrm{2}}+{u}\right)}{\frac{{v}\sqrt{\mathrm{3}}}{\mathrm{2}}}= \\ $$$${a}\frac{\frac{\mathrm{2}{u}\pi}{\mathrm{3}{v}}\left(\mathrm{2}{u}+{v}\right)−\left({v}+{u}\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)\right)}{{v}\sqrt{\mathrm{3}}} \\ $$
Commented by ajfour last updated on 10/Sep/22
Thanks both sirs. Excellent sol^n s.
$${Thanks}\:{both}\:{sirs}.\:{Excellent}\:{sol}^{{n}} {s}. \\ $$
Commented by Tawa11 last updated on 15/Sep/22
Great sir
$$\mathrm{Great}\:\mathrm{sir} \\ $$
Answered by mr W last updated on 09/Sep/22
θ=((vt)/a) ⇒t=((aθ)/v) x_A =a+a sin θ y_A =b+a cos θ x_B =ut=((uaθ)/v) y_B =0 Φ=d^2 =(x_A −x_B )^2 +(y_A −y_B )^2 Φ=(a+a sin θ−((uaθ)/v))^2 +(b+a cos θ)^2 (dΦ/dθ)=2(a+a sin θ−((uaθ)/v))(a cos θ−((ua)/v))−2(b+a cos θ)(a sin θ)=0 let μ=(b/a), λ=(u/v) (1+sin θ−λθ)(cos θ−λ)−(μ+cos θ) sin θ=0 ⇒μ=(((1+sin θ−λθ)(cos θ−λ))/(sin θ))−cos θ with θ=(π/2)+(π/6)=((2π)/3) ⇒μ=((2(1+((√3)/2)−((2λπ)/3))(−(1/2)−λ))/( (√3)))+(1/2) ⇒μ=(1/2)+(((4λπ−6−3(√3))(1+2λ))/( 6(√3)))
$$\theta=\frac{{vt}}{{a}}\:\Rightarrow{t}=\frac{{a}\theta}{{v}} \\ $$$${x}_{{A}} ={a}+{a}\:\mathrm{sin}\:\theta \\ $$$${y}_{{A}} ={b}+{a}\:\mathrm{cos}\:\theta \\ $$$${x}_{{B}} ={ut}=\frac{{ua}\theta}{{v}} \\ $$$${y}_{{B}} =\mathrm{0} \\ $$$$\Phi={d}^{\mathrm{2}} =\left({x}_{{A}} −{x}_{{B}} \right)^{\mathrm{2}} +\left({y}_{{A}} −{y}_{{B}} \right)^{\mathrm{2}} \\ $$$$\Phi=\left({a}+{a}\:\mathrm{sin}\:\theta−\frac{{ua}\theta}{{v}}\right)^{\mathrm{2}} +\left({b}+{a}\:\mathrm{cos}\:\theta\right)^{\mathrm{2}} \\ $$$$\frac{{d}\Phi}{{d}\theta}=\mathrm{2}\left({a}+{a}\:\mathrm{sin}\:\theta−\frac{{ua}\theta}{{v}}\right)\left({a}\:\mathrm{cos}\:\theta−\frac{{ua}}{{v}}\right)−\mathrm{2}\left({b}+{a}\:\mathrm{cos}\:\theta\right)\left({a}\:\mathrm{sin}\:\theta\right)=\mathrm{0} \\ $$$${let}\:\mu=\frac{{b}}{{a}},\:\lambda=\frac{{u}}{{v}} \\ $$$$\left(\mathrm{1}+\mathrm{sin}\:\theta−\lambda\theta\right)\left(\mathrm{cos}\:\theta−\lambda\right)−\left(\mu+\mathrm{cos}\:\theta\right)\:\mathrm{sin}\:\theta=\mathrm{0} \\ $$$$\Rightarrow\mu=\frac{\left(\mathrm{1}+\mathrm{sin}\:\theta−\lambda\theta\right)\left(\mathrm{cos}\:\theta−\lambda\right)}{\mathrm{sin}\:\theta}−\mathrm{cos}\:\theta \\ $$$${with}\:\theta=\frac{\pi}{\mathrm{2}}+\frac{\pi}{\mathrm{6}}=\frac{\mathrm{2}\pi}{\mathrm{3}} \\ $$$$\Rightarrow\mu=\frac{\mathrm{2}\left(\mathrm{1}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}−\frac{\mathrm{2}\lambda\pi}{\mathrm{3}}\right)\left(−\frac{\mathrm{1}}{\mathrm{2}}−\lambda\right)}{\:\sqrt{\mathrm{3}}}+\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow\mu=\frac{\mathrm{1}}{\mathrm{2}}+\frac{\left(\mathrm{4}\lambda\pi−\mathrm{6}−\mathrm{3}\sqrt{\mathrm{3}}\right)\left(\mathrm{1}+\mathrm{2}\lambda\right)}{\:\mathrm{6}\sqrt{\mathrm{3}}} \\ $$
Commented by Tawa11 last updated on 15/Sep/22
Great sir.
$$\mathrm{Great}\:\mathrm{sir}. \\ $$

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