Question Number 183344 by mr W last updated on 25/Dec/22
Commented by mr W last updated on 25/Dec/22
$${the}\:{circle}\:{rolls}\:{along}\:{the}\:{parabola}. \\ $$$${find}\:{the}\:{locus}\:{of}\:{the}\:{fix}\:{point}\:{P}\:{on} \\ $$$${the}\:{circle}. \\ $$
Answered by TheSupreme last updated on 25/Dec/22
![let′s H the intersection between parablla and circle s(H−O)=∫_0 ^x (√(1+((4x^2 )/a^2 )))dx arcsin(((2x)/a))=u ((2x)/a)=sinh(u) x=(a/2)sinh(u) dx=(a/2)cosh(u)du s(H−0)=∫_0 ^x (a/2)cosh(u)(√(1+sinh^2 (u)))du= =∫_0 ^x (a/2)cosh^2 (u)du=(a/4)(arcsinh(((2x)/a))+sinh(arcsinh(((2x)/a))cosh(arcsinh(((2x)/a))) =(a/4)(((2x)/a)+((2x)/a)(√(1+((4x^2 )/a^2 ))))=(x/2)[1+(√(1+((4x^2 )/a^2 )))] θ=(s/R)=(x/(2R))[1+(√(1+((4x^2 )/a^2 )))] x_p =x_c −Rcos(θ) y_p =y_c +Rsin(θ) .... x_c =x_H +Rcos(∅) y_c =y_h +Rsin(∅) tan(−∅)=y′=−2(x/a) tan(∅)=(a/(2x)) cos(∅)=(1/( (√(1+(a^2 /(4x^2 ))))))=((2x)/( (√(4x^2 +a^2 )))) sin(∅)=((a/(2x))/( (√(1+(a^2 /(4x^2 ))))))=(a/( (√(4x^2 +a^2 )))) x_p =x+((2x)/( (√(4x^2 +a^2 ))))−Rcos((x/(2R))[1+(√(1+((4x^2 )/a^2 )))]) y_p =−(x^2 /a)+(a/( (√(4x^2 +a^2 ))))+Rsin((x/(2R))[1+(√(1+((4x^2 )/a^2 )))])](https://www.tinkutara.com/question/Q183365.png)
$${let}'{s}\:{H}\:{the}\:{intersection}\:{between}\:{parablla}\:{and}\:{circle} \\ $$$${s}\left({H}−{O}\right)=\int_{\mathrm{0}} ^{{x}} \sqrt{\mathrm{1}+\frac{\mathrm{4}{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }}{dx} \\ $$$${arcsin}\left(\frac{\mathrm{2}{x}}{{a}}\right)={u} \\ $$$$\frac{\mathrm{2}{x}}{{a}}={sinh}\left({u}\right) \\ $$$${x}=\frac{{a}}{\mathrm{2}}{sinh}\left({u}\right) \\ $$$${dx}=\frac{{a}}{\mathrm{2}}{cosh}\left({u}\right){du} \\ $$$${s}\left({H}−\mathrm{0}\right)=\int_{\mathrm{0}} ^{{x}} \frac{{a}}{\mathrm{2}}{cosh}\left({u}\right)\sqrt{\mathrm{1}+{sinh}^{\mathrm{2}} \left({u}\right)}{du}= \\ $$$$=\int_{\mathrm{0}} ^{{x}} \frac{{a}}{\mathrm{2}}{cosh}^{\mathrm{2}} \left({u}\right){du}=\frac{{a}}{\mathrm{4}}\left({arcsinh}\left(\frac{\mathrm{2}{x}}{{a}}\right)+{sinh}\left({arcsinh}\left(\frac{\mathrm{2}{x}}{{a}}\right){cosh}\left({arcsinh}\left(\frac{\mathrm{2}{x}}{{a}}\right)\right)\right.\right. \\ $$$$=\frac{{a}}{\mathrm{4}}\left(\frac{\mathrm{2}{x}}{{a}}+\frac{\mathrm{2}{x}}{{a}}\sqrt{\left.\mathrm{1}+\frac{\mathrm{4}{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }\right)}=\frac{{x}}{\mathrm{2}}\left[\mathrm{1}+\sqrt{\mathrm{1}+\frac{\mathrm{4}{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }}\right]\right. \\ $$$$ \\ $$$$\theta=\frac{{s}}{{R}}=\frac{{x}}{\mathrm{2}{R}}\left[\mathrm{1}+\sqrt{\mathrm{1}+\frac{\mathrm{4}{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }}\right]\: \\ $$$$ \\ $$$${x}_{{p}} ={x}_{{c}} −{Rcos}\left(\theta\right) \\ $$$${y}_{{p}} ={y}_{{c}} +{Rsin}\left(\theta\right) \\ $$$$…. \\ $$$${x}_{{c}} ={x}_{{H}} +{Rcos}\left(\emptyset\right) \\ $$$${y}_{{c}} ={y}_{{h}} +{Rsin}\left(\emptyset\right) \\ $$$${tan}\left(−\emptyset\right)={y}'=−\mathrm{2}\frac{{x}}{{a}} \\ $$$${tan}\left(\emptyset\right)=\frac{{a}}{\mathrm{2}{x}} \\ $$$${cos}\left(\emptyset\right)=\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\frac{{a}^{\mathrm{2}} }{\mathrm{4}{x}^{\mathrm{2}} }}}=\frac{\mathrm{2}{x}}{\:\sqrt{\mathrm{4}{x}^{\mathrm{2}} +{a}^{\mathrm{2}} }} \\ $$$${sin}\left(\emptyset\right)=\frac{\frac{{a}}{\mathrm{2}{x}}}{\:\sqrt{\mathrm{1}+\frac{{a}^{\mathrm{2}} }{\mathrm{4}{x}^{\mathrm{2}} }}}=\frac{{a}}{\:\sqrt{\mathrm{4}{x}^{\mathrm{2}} +{a}^{\mathrm{2}} }} \\ $$$$ \\ $$$${x}_{{p}} ={x}+\frac{\mathrm{2}{x}}{\:\sqrt{\mathrm{4}{x}^{\mathrm{2}} +{a}^{\mathrm{2}} }}−{Rcos}\left(\frac{{x}}{\mathrm{2}{R}}\left[\mathrm{1}+\sqrt{\mathrm{1}+\frac{\mathrm{4}{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }}\right]\right) \\ $$$${y}_{{p}} =−\frac{{x}^{\mathrm{2}} }{{a}}+\frac{{a}}{\:\sqrt{\mathrm{4}{x}^{\mathrm{2}} +{a}^{\mathrm{2}} }}+{Rsin}\left(\frac{{x}}{\mathrm{2}{R}}\left[\mathrm{1}+\sqrt{\mathrm{1}+\frac{\mathrm{4}{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }}\right]\right) \\ $$$$ \\ $$
Commented by mr W last updated on 25/Dec/22
$${thanks}\:{sir}! \\ $$$${we}\:{got}\:{different}\:{result}\:{for}\:{the}\:{length} \\ $$$${of}\:{curve}. \\ $$
Answered by mr W last updated on 25/Dec/22
Commented by mr W last updated on 25/Dec/22
![y=−(x^2 /a) tan θ=−(dy/dx)=((2x)/a) say point Q(q,−(q^2 /a)) tan θ=((2q)/a)=t ⇒q=((at)/2) s=∫_0 ^q (√(1+y′^2 ))dx=∫_0 ^q (√(1+((4x^2 )/a^2 ))) dx s=(a/2)∫_0 ^t (√(1+ξ^2 )) dξ s=a[t(√(1+t^2 ))+ln (t+(√(1+t^2 )))] s=rϕ ⇒ϕ=[t(√(1+t^2 ))+ln (t+(√(1+t^2 )))](a/r) x_C =q+r sin θ=((at)/2)+r sin θ y_C =−(q^2 /a)+r cos θ=−((at^2 )/4)+r cos θ x_P =x_C −ρ sin (θ+ϕ)=((at)/2)+r sin θ−ρ sin (θ+ϕ) y_P =y_C −ρ cos (θ+ϕ)=−((at^2 )/4)+r cos θ−ρ cos (θ+ϕ) let λ=(a/r), μ=(ρ/r) { (((x_P /r)=((λ tan θ)/2)+sin θ−μ sin (θ+ϕ))),(((y_P /r)=−((λ tan^2 θ)/4)+cos θ−μ cos (θ+ϕ))) :} with ϕ=λ[tan θ (√(1+tan^2 θ))+ln (tan θ+(√(1+tan^2 θ)))]](https://www.tinkutara.com/question/Q183368.png)
$${y}=−\frac{{x}^{\mathrm{2}} }{{a}} \\ $$$$\mathrm{tan}\:\theta=−\frac{{dy}}{{dx}}=\frac{\mathrm{2}{x}}{{a}} \\ $$$${say}\:{point}\:{Q}\left({q},−\frac{{q}^{\mathrm{2}} }{{a}}\right) \\ $$$$\mathrm{tan}\:\theta=\frac{\mathrm{2}{q}}{{a}}={t} \\ $$$$\Rightarrow{q}=\frac{{at}}{\mathrm{2}} \\ $$$${s}=\int_{\mathrm{0}} ^{{q}} \sqrt{\mathrm{1}+{y}'^{\mathrm{2}} }{dx}=\int_{\mathrm{0}} ^{{q}} \sqrt{\mathrm{1}+\frac{\mathrm{4}{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }}\:{dx} \\ $$$${s}=\frac{{a}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{t}} \sqrt{\mathrm{1}+\xi^{\mathrm{2}} }\:{d}\xi \\ $$$${s}={a}\left[{t}\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }+\mathrm{ln}\:\left({t}+\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }\right)\right] \\ $$$${s}={r}\varphi \\ $$$$\Rightarrow\varphi=\left[{t}\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }+\mathrm{ln}\:\left({t}+\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }\right)\right]\frac{{a}}{{r}} \\ $$$${x}_{{C}} ={q}+{r}\:\mathrm{sin}\:\theta=\frac{{at}}{\mathrm{2}}+{r}\:\mathrm{sin}\:\theta \\ $$$${y}_{{C}} =−\frac{{q}^{\mathrm{2}} }{{a}}+{r}\:\mathrm{cos}\:\theta=−\frac{{at}^{\mathrm{2}} }{\mathrm{4}}+{r}\:\mathrm{cos}\:\theta \\ $$$${x}_{{P}} ={x}_{{C}} −\rho\:\mathrm{sin}\:\left(\theta+\varphi\right)=\frac{{at}}{\mathrm{2}}+{r}\:\mathrm{sin}\:\theta−\rho\:\mathrm{sin}\:\left(\theta+\varphi\right) \\ $$$${y}_{{P}} ={y}_{{C}} −\rho\:\mathrm{cos}\:\left(\theta+\varphi\right)=−\frac{{at}^{\mathrm{2}} }{\mathrm{4}}+{r}\:\mathrm{cos}\:\theta−\rho\:\mathrm{cos}\:\left(\theta+\varphi\right) \\ $$$${let}\:\lambda=\frac{{a}}{{r}},\:\mu=\frac{\rho}{{r}} \\ $$$$\begin{cases}{\frac{{x}_{{P}} }{{r}}=\frac{\lambda\:\mathrm{tan}\:\theta}{\mathrm{2}}+\mathrm{sin}\:\theta−\mu\:\mathrm{sin}\:\left(\theta+\varphi\right)}\\{\frac{{y}_{{P}} }{{r}}=−\frac{\lambda\:\mathrm{tan}^{\mathrm{2}} \:\theta}{\mathrm{4}}+\mathrm{cos}\:\theta−\mu\:\mathrm{cos}\:\left(\theta+\varphi\right)}\end{cases} \\ $$$${with} \\ $$$$\varphi=\lambda\left[\mathrm{tan}\:\theta\:\sqrt{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\theta}+\mathrm{ln}\:\left(\mathrm{tan}\:\theta+\sqrt{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\theta}\right)\right] \\ $$
Commented by mr W last updated on 25/Dec/22
Commented by mr W last updated on 25/Dec/22
Commented by mr W last updated on 25/Dec/22
