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Question Number 170005 by ajfour last updated on 13/May/22

Commented by ajfour last updated on 14/May/22

A moves in circle with speed u, while B down the vertical diameter with speed ucos θ, where θ is shown, what to mean. Given initial positions, if they collide in P(0,a) , find b/a.

$${A}\:{moves}\:{in}\:{circle}\:{with}\: \\ $$$${speed}\:{u},\:{while}\:{B}\:{down}\:{the} \\ $$$${vertical}\:{diameter}\:{with}\:{speed} \\ $$$${u}\mathrm{cos}\:\theta,\:{where}\:\theta\:{is}\:{shown}, \\ $$$${what}\:{to}\:{mean}.\:{Given}\:{initial} \\ $$$${positions},\:{if}\:{they}\:{collide}\:{in} \\ $$$${P}\left(\mathrm{0},{a}\right)\:\:,\:{find}\:{b}/{a}. \\ $$

Commented by mr W last updated on 14/May/22

((b−a)/(u cos θ))=((πa)/(2u)) (((b−a)(√(b^2 +a^2 )))/b)=((πa)/2) let λ=(b/a) (λ−1)(√(1+λ^2 ))=((πλ)/2) ⇒λ≈2.4547

$$\frac{{b}−{a}}{{u}\:\mathrm{cos}\:\theta}=\frac{\pi{a}}{\mathrm{2}{u}} \\ $$$$\frac{\left({b}−{a}\right)\sqrt{{b}^{\mathrm{2}} +{a}^{\mathrm{2}} }}{{b}}=\frac{\pi{a}}{\mathrm{2}} \\ $$$${let}\:\lambda=\frac{{b}}{{a}} \\ $$$$\left(\lambda−\mathrm{1}\right)\sqrt{\mathrm{1}+\lambda^{\mathrm{2}} }=\frac{\pi\lambda}{\mathrm{2}} \\ $$$$\Rightarrow\lambda\approx\mathrm{2}.\mathrm{4547} \\ $$

Commented by ajfour last updated on 14/May/22

θ is variable, along instantaeous BA; sir.

$$\theta\:{is}\:{variable},\:{along}\: \\ $$$${instantaeous}\:{BA};\:{sir}. \\ $$

Answered by ajfour last updated on 14/May/22

let at time t, A(p,q) ∀ p^2 +q^2 =a^2 B(0,y) (ds/dt)=u tan θ=(p/(y−q)) ucos θ=−(dy/dt)=−((udy)/ds) (1/( (√(1+(((acos φ)/(y−asin φ)))^2 ))))=−(dy/(adφ)) ⇒ a^2 ((dφ/dy))^2 =1+(((acos φ)/(y−asin φ)))^2 .......

$${let}\:{at}\:{time}\:{t}, \\ $$$${A}\left({p},{q}\right)\:\:\:\:\forall\:\:{p}^{\mathrm{2}} +{q}^{\mathrm{2}} ={a}^{\mathrm{2}} \\ $$$${B}\left(\mathrm{0},{y}\right) \\ $$$$\frac{{ds}}{{dt}}={u} \\ $$$$\mathrm{tan}\:\theta=\frac{{p}}{{y}−{q}} \\ $$$${u}\mathrm{cos}\:\theta=−\frac{{dy}}{{dt}}=−\frac{{udy}}{{ds}} \\ $$$$\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\left(\frac{{a}\mathrm{cos}\:\phi}{{y}−{a}\mathrm{sin}\:\phi}\right)^{\mathrm{2}} }}=−\frac{{dy}}{{ad}\phi} \\ $$$$\Rightarrow\:\:{a}^{\mathrm{2}} \left(\frac{{d}\phi}{{dy}}\right)^{\mathrm{2}} =\mathrm{1}+\left(\frac{{a}\mathrm{cos}\:\phi}{{y}−{a}\mathrm{sin}\:\phi}\right)^{\mathrm{2}} \\ $$$$....... \\ $$$$ \\ $$

Answered by mr W last updated on 14/May/22

Commented by mr W last updated on 14/May/22

at time t: P at ϕ=((ut)/a)=ωt with ω=(u/a) Q(0,h) tan θ=((a cos ϕ)/(h−a sin ϕ))=((a cos ωt)/(h−a sin ωt)) (dh/dt)=−u cos θ=− (u/( (√(1+(((a cos ωt)/(h−a sin ωt)))^2 )))) let ξ=(h/a) (dξ/dt)+ (ω/( (√(1+(((cos ωt)/(ξ−sin ωt)))^2 ))))=0 or (dξ/dϕ)+ (1/( (√(1+(((cos ϕ)/(ξ−sin ϕ)))^2 ))))=0 ξ(ϕ=0)=(b/a)=λ ξ(ϕ=(π/2))=1 ....

$${at}\:{time}\:{t}: \\ $$$${P}\:{at}\:\varphi=\frac{{ut}}{{a}}=\omega{t}\:{with}\:\omega=\frac{{u}}{{a}} \\ $$$${Q}\left(\mathrm{0},{h}\right) \\ $$$$\mathrm{tan}\:\theta=\frac{{a}\:\mathrm{cos}\:\varphi}{{h}−{a}\:\mathrm{sin}\:\varphi}=\frac{{a}\:\mathrm{cos}\:\omega{t}}{{h}−{a}\:\mathrm{sin}\:\omega{t}} \\ $$$$\frac{{dh}}{{dt}}=−{u}\:\mathrm{cos}\:\theta=−\:\frac{{u}}{\:\sqrt{\mathrm{1}+\left(\frac{{a}\:\mathrm{cos}\:\omega{t}}{{h}−{a}\:\mathrm{sin}\:\omega{t}}\right)^{\mathrm{2}} }} \\ $$$${let}\:\xi=\frac{{h}}{{a}} \\ $$$$\frac{{d}\xi}{{dt}}+\:\frac{\omega}{\:\sqrt{\mathrm{1}+\left(\frac{\mathrm{cos}\:\omega{t}}{\xi−\mathrm{sin}\:\omega{t}}\right)^{\mathrm{2}} }}=\mathrm{0} \\ $$$${or} \\ $$$$\frac{{d}\xi}{{d}\varphi}+\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\left(\frac{\mathrm{cos}\:\varphi}{\xi−\mathrm{sin}\:\varphi}\right)^{\mathrm{2}} }}=\mathrm{0} \\ $$$$\xi\left(\varphi=\mathrm{0}\right)=\frac{{b}}{{a}}=\lambda \\ $$$$\xi\left(\varphi=\frac{\pi}{\mathrm{2}}\right)=\mathrm{1} \\ $$$$.... \\ $$

Commented by ajfour last updated on 14/May/22

thanks sir, difficult d.e. !

$${thanks}\:{sir},\:{difficult}\:{d}.{e}.\:! \\ $$

Commented by mr W last updated on 14/May/22

yes! i don′t know how to solve.

$${yes}!\:{i}\:{don}'{t}\:{know}\:{how}\:{to}\:{solve}. \\ $$

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