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Question Number 215126 by mr W last updated on 29/Dec/24

Commented by Ghisom last updated on 29/Dec/24

$$L\mathrm{must}\mathrm{be}\mathrm{horizontal}\mathrm{for}r\to \max$$$$\mathrm{or}\mathrm{am}\mathrm{I}\mathrm{missing}\mathrm{something}?$$

Commented by mr W last updated on 29/Dec/24

$$anuniformrodwithlengthL$$$$restsinclinedinaslipperyparabolic$$$$cupasshown.$$$$findtheradiusofthelargestcircle$$$$whichcanbeplacedundertherod.$$

Commented by mr W last updated on 29/Dec/24

$$thequestionrequests‘‘{inclined}^{″}$$$$positionoftherod.$$

Answered by mr W last updated on 30/Dec/24

Commented by mr W last updated on 31/Dec/24

$${let}^{\prime }slookatthegeneralcase$$$$y=\frac{x^2}{k}withk=1ornot$$$$sayA(-p,\frac{p^2}{k})$$$$B(q,\frac{q^2}{k})$$$$ifp=q=\frac{L}{2},therodrestshorizontally.$$$$forinclinedpositionsofrod,p\ne q.$$$$\tan \varphi =-\frac{dy}{dx}=\frac{2p}{k}$$$$\tan \phi =\frac{dy}{dx}=\frac{2q}{k}$$$$\alpha =\frac{\pi }{2}-\varphi -\theta$$$$\beta =\frac{\pi }{2}-\phi +\theta$$$$\frac{\sin \alpha }{\sin \varphi }=\frac{DC}{AC}=\frac{DC}{BC}=\frac{\sin \beta }{\sin \phi }$$$$\frac{\cos (\varphi +\theta )}{\sin \varphi }=\frac{\cos (\phi -\theta )}{\sin \phi }$$$$\frac{1-\tan \varphi \tan \theta }{\tan \varphi }=\frac{1+\tan \phi \tan \theta }{\tan \phi }$$$$\Rightarrow \tan \theta =\frac{1}{2}(\frac{1}{\tan \varphi }-\frac{1}{\tan \phi })=\frac{k}{4}(\frac{1}{p}-\frac{1}{q})=\frac{k(q-p)}{4pq}$$$$ontheotherside$$$$\tan \theta =\frac{q^2-p^2}{k(q+p)}=\frac{q-p}{k}$$$$\frac{q-p}{k}=\frac{k(q-p)}{4pq}$$$$sinceq\ne p,$$$$\Rightarrow pq=\frac{k^2}{4}...\left(i\right)$$$$saym=\tan \theta$$$$\Rightarrow q-p=mk...\left(ii\right)$$$$q,-parerootsof$$$$z^2-mkz-\frac{k^2}{4}=0$$$$\Rightarrow q=\frac{k}{2}(\sqrt{1+m^2}+m)$$$$\Rightarrow p=\frac{k}{2}(\sqrt{1+m^2}-m)$$$$AB=\sqrt{{(p+q)}^2+{(\frac{q^2}{k}-\frac{p^2}{k})}^2}=L$$$$(p+q)\sqrt{1+{\left(\frac{q-p}{k}\right)}^2}=L$$$$(p+q)\sqrt{1+m^2}=L$$$$k(1+m^2)=L$$$$\Rightarrow 1+m^2=\frac{L}{k}$$$$\Rightarrow m=\sqrt{\frac{L}{k}-1}>0\Rightarrow L>k$$$$\Rightarrow p=\frac{k}{2}(\sqrt{\frac{L}{k}}-\sqrt{\frac{L}{k}-1})$$$$\Rightarrow q=\frac{k}{2}(\sqrt{\frac{L}{k}}+\sqrt{\frac{L}{k}-1})$$$$wecansee$$$$\tan \varphi \tan \phi =\frac{4pq}{k^2}=\frac{4}{k^2}\times \frac{k^2}{4}=1,$$$$thatmeansthetangentlines$$$$\left(andalsothenormals\right)attheends$$$$oftherodareperpendiculartoeach$$$$other.$$$$$$$$eqn.ofrod:$$$$y=\frac{p^2}{k}+(x+p)m$$$$mx-y+\frac{p^2}{k}+pm=0$$$$\Rightarrow mx-y+\frac{k}{4}=0$$$$$$$$saythecirclewithradiusr$$$$tangentstheparabolaatS(s,\frac{s^2}{k})$$$$\tan \delta =\frac{dy}{dx}=\frac{2s}{k}$$$$G=centerofcircle$$$$x_G=s-r\sin \delta =s-\frac{2sr}{\sqrt{4s^2+k^2}}$$$$y_G=\frac{s^2}{k}+r\cos \delta =\frac{s^2}{k}+\frac{kr}{\sqrt{4s^2+k^2}}$$$$r=\frac{m(s-\frac{2sr}{\sqrt{4s^2+k^2}})-\frac{s^2}{k}-\frac{kr}{\sqrt{4s^2+k^2}}+\frac{k}{4}}{\sqrt{1+m^2}}$$$$(\sqrt{1+m^2}+\frac{2ms+k}{\sqrt{4s^2+k^2}})r=ms-\frac{s^2}{k}+\frac{k}{4}$$$$r=\frac{ms-\frac{s^2}{k}+\frac{k}{4}}{\sqrt{1+m^2}+\frac{2ms+k}{\sqrt{4s^2+k^2}}}$$$$\frac{r}{k}=\frac{\frac{ms}{k}-\frac{s^2}{k^2}+\frac{1}{4}}{\sqrt{1+m^2}+\frac{\frac{2ms}{k}+1}{\sqrt{\frac{4s^2}{k^2}+1}}}$$$$let\xi =\frac{s}{k}$$$$\Rightarrow \frac{r}{k}=\frac{m\xi -{\xi }^2+\frac{1}{4}}{\sqrt{1+m^2}+\frac{2m\xi +1}{\sqrt{4{\xi }^2+1}}}=f\left(\xi \right)$$$$formaximumr,f^{\prime }\left(\xi \right)=0$$$$$$$$\underset{―}{alternativeway:}$$$$formaximumcirclethetangentat$$$$S(s,\frac{s^2}{k})mustbeparalleltotherod.$$$$\tan \delta =\frac{2s}{k}=m$$$$\Rightarrow s=\frac{km}{2}$$$$2r_{max}=\frac{\frac{{km}^2}{2}-\frac{k^2{m}^2}{4k}+\frac{k}{4}}{\sqrt{1+m^2}}=\frac{k\sqrt{1+m^2}}{4}=\frac{\sqrt{Lk}}{4}$$$$\Rightarrow r_{max}=\frac{\sqrt{Lk}}{8}$$

Commented by mr W last updated on 30/Dec/24

Commented by mr W last updated on 30/Dec/24

Commented by mr W last updated on 30/Dec/24

$$weseethesimilaritytotheproblem$$$$discussedinQ215094:$$

Commented by mr W last updated on 30/Dec/24

Commented by mr W last updated on 30/Dec/24

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