Question-215126

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

Commented by Ghisom last updated on 29/Dec/24

L must be horizontal for r → max or am I missing something?

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

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

$${an}\:{uniform}\:{rod}\:{with}\:{length}\:{L} \\ $$$${rests}\:{inclined}\:{in}\:{a}\:{slippery}\:{parabolic} \\ $$$${cup}\:{as}\:{shown}. \\ $$$${find}\:{the}\:{radius}\:{of}\:{the}\:{largest}\:{circle} \\ $$$${which}\:{can}\:{be}\:{placed}\:{under}\:{the}\:{rod}. \\ $$

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

$${the}\:{question}\:{requests}\:“{inclined}'' \\ $$$${position}\:{of}\:{the}\:{rod}. \\ $$

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

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

let′s look at the general case y=(x^2 /k) with k=1 or not say A(−p, (p^2 /k)) B(q, (q^2 /k)) if p=q=(L/2), the rod rests horizontally. for inclined positions of rod, p≠q. tan φ=−(dy/dx)=((2p)/k) tan ϕ=(dy/dx)=((2q)/k) α=(π/2)−φ−θ β=(π/2)−ϕ+θ ((sin α)/(sin φ))=((DC)/(AC))=((DC)/(BC))=((sin β)/(sin ϕ)) ((cos (φ+θ))/(sin φ))=((cos (ϕ−θ))/(sin ϕ)) ((1−tan φ tan θ)/(tan φ))=((1+tan ϕ tan θ)/(tan ϕ)) ⇒tan θ=(1/2)((1/(tan φ))−(1/(tan ϕ)))=(k/4)((1/p)−(1/q))=((k(q−p))/(4pq)) on the other side tan θ=((q^2 −p^2 )/(k(q+p)))=((q−p)/k) ((q−p)/k)=((k(q−p))/(4pq)) since q≠p, ⇒pq=(k^2 /4) ...(i) say m=tan θ ⇒q−p=mk ...(ii) q, −p are roots of z^2 −mkz−(k^2 /4)=0 ⇒q=(k/2)((√(1+m^2 ))+m) ⇒p=(k/2)((√(1+m^2 ))−m) AB=(√((p+q)^2 +((q^2 /k)−(p^2 /k))^2 ))=L (p+q)(√(1+(((q−p)/k))^2 ))=L (p+q)(√(1+m^2 ))=L k(1+m^2 )=L ⇒1+m^2 =(L/k) ⇒m=(√((L/k)−1)) >0 ⇒L>k ⇒p=(k/2)((√(L/k))−(√((L/k)−1))) ⇒q=(k/2)((√(L/k))+(√((L/k)−1))) we can see tan φ tan ϕ=((4pq)/k^2 )=(4/k^2 )×(k^2 /4)=1, that means the tangent lines (and also the normals) at the ends of the rod are perpendicular to each other. eqn. of rod: y=(p^2 /k)+(x+p)m mx−y+(p^2 /k)+pm=0 ⇒mx−y+(k/4)=0 say the circle with radius r tangents the parabola at S(s, (s^2 /k)) tan δ=(dy/dx)=((2s)/k) G=center of circle x_G =s−r sin δ=s−((2sr)/( (√(4s^2 +k^2 )))) y_G =(s^2 /k)+r cos δ=(s^2 /k)+((kr)/( (√(4s^2 +k^2 )))) r=((m(s−((2sr)/( (√(4s^2 +k^2 )))))−(s^2 /k)−((kr)/( (√(4s^2 +k^2 ))))+(k/4))/( (√(1+m^2 )))) ((√(1+m^2 ))+((2ms+k)/( (√(4s^2 +k^2 )))))r=ms−(s^2 /k)+(k/4) r=((ms−(s^2 /k)+(k/4))/( (√(1+m^2 ))+((2ms+k)/( (√(4s^2 +k^2 )))))) (r/k)=((((ms)/k)−(s^2 /k^2 )+(1/4))/( (√(1+m^2 ))+((((2ms)/k)+1)/( (√(((4s^2 )/k^2 )+1)))))) let ξ=(s/k) ⇒(r/k)=((mξ−ξ^2 +(1/4))/( (√(1+m^2 ))+((2mξ+1)/( (√(4ξ^2 +1))))))=f(ξ) for maximum r, f′(ξ)=0 alternative way: for maximum circle the tangent at S(s, (s^2 /k)) must be parallel to the rod. tan δ=((2s)/k)=m ⇒s=((km)/2) 2r_(max) =((((km^2 )/2)−((k^2 m^2 )/(4k))+(k/4))/( (√(1+m^2 ))))=((k(√(1+m^2 )))/4)=((√(Lk))/4) ⇒r_(max) =((√(Lk))/8)

$${let}'{s}\:{look}\:{at}\:{the}\:{general}\:{case}\: \\ $$$${y}=\frac{{x}^{\mathrm{2}} }{{k}}\:{with}\:{k}=\mathrm{1}\:{or}\:{not} \\ $$$${say}\:{A}\left(−{p},\:\frac{{p}^{\mathrm{2}} }{{k}}\right) \\ $$$${B}\left({q},\:\frac{{q}^{\mathrm{2}} }{{k}}\right) \\ $$$${if}\:{p}={q}=\frac{{L}}{\mathrm{2}},\:{the}\:{rod}\:{rests}\:{horizontally}. \\ $$$${for}\:{inclined}\:{positions}\:{of}\:{rod},\:{p}\neq{q}. \\ $$$$\mathrm{tan}\:\phi=−\frac{{dy}}{{dx}}=\frac{\mathrm{2}{p}}{{k}} \\ $$$$\mathrm{tan}\:\varphi=\frac{{dy}}{{dx}}=\frac{\mathrm{2}{q}}{{k}} \\ $$$$\alpha=\frac{\pi}{\mathrm{2}}−\phi−\theta \\ $$$$\beta=\frac{\pi}{\mathrm{2}}−\varphi+\theta \\ $$$$\frac{\mathrm{sin}\:\alpha}{\mathrm{sin}\:\phi}=\frac{{DC}}{{AC}}=\frac{{DC}}{{BC}}=\frac{\mathrm{sin}\:\beta}{\mathrm{sin}\:\varphi} \\ $$$$\frac{\mathrm{cos}\:\left(\phi+\theta\right)}{\mathrm{sin}\:\phi}=\frac{\mathrm{cos}\:\left(\varphi−\theta\right)}{\mathrm{sin}\:\varphi} \\ $$$$\frac{\mathrm{1}−\mathrm{tan}\:\phi\:\mathrm{tan}\:\theta}{\mathrm{tan}\:\phi}=\frac{\mathrm{1}+\mathrm{tan}\:\varphi\:\mathrm{tan}\:\theta}{\mathrm{tan}\:\varphi} \\ $$$$\Rightarrow\mathrm{tan}\:\theta=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{tan}\:\phi}−\frac{\mathrm{1}}{\mathrm{tan}\:\varphi}\right)=\frac{{k}}{\mathrm{4}}\left(\frac{\mathrm{1}}{{p}}−\frac{\mathrm{1}}{{q}}\right)=\frac{{k}\left({q}−{p}\right)}{\mathrm{4}{pq}} \\ $$$${on}\:{the}\:{other}\:{side} \\ $$$$\mathrm{tan}\:\theta=\frac{{q}^{\mathrm{2}} −{p}^{\mathrm{2}} }{{k}\left({q}+{p}\right)}=\frac{{q}−{p}}{{k}} \\ $$$$\frac{{q}−{p}}{{k}}=\frac{{k}\left({q}−{p}\right)}{\mathrm{4}{pq}} \\ $$$${since}\:{q}\neq{p}, \\ $$$$\Rightarrow{pq}=\frac{{k}^{\mathrm{2}} }{\mathrm{4}}\:\:\:…\left({i}\right) \\ $$$${say}\:{m}=\mathrm{tan}\:\theta \\ $$$$\Rightarrow{q}−{p}={mk}\:\:\:…\left({ii}\right) \\ $$$${q},\:−{p}\:{are}\:{roots}\:{of} \\ $$$${z}^{\mathrm{2}} −{mkz}−\frac{{k}^{\mathrm{2}} }{\mathrm{4}}=\mathrm{0} \\ $$$$\Rightarrow{q}=\frac{{k}}{\mathrm{2}}\left(\sqrt{\mathrm{1}+{m}^{\mathrm{2}} }+{m}\right) \\ $$$$\Rightarrow{p}=\frac{{k}}{\mathrm{2}}\left(\sqrt{\mathrm{1}+{m}^{\mathrm{2}} }−{m}\right) \\ $$$${AB}=\sqrt{\left({p}+{q}\right)^{\mathrm{2}} +\left(\frac{{q}^{\mathrm{2}} }{{k}}−\frac{{p}^{\mathrm{2}} }{{k}}\right)^{\mathrm{2}} }={L} \\ $$$$\left({p}+{q}\right)\sqrt{\mathrm{1}+\left(\frac{{q}−{p}}{{k}}\right)^{\mathrm{2}} }={L} \\ $$$$\left({p}+{q}\right)\sqrt{\mathrm{1}+{m}^{\mathrm{2}} }={L} \\ $$$${k}\left(\mathrm{1}+{m}^{\mathrm{2}} \right)={L} \\ $$$$\Rightarrow\mathrm{1}+{m}^{\mathrm{2}} =\frac{{L}}{{k}} \\ $$$$\Rightarrow{m}=\sqrt{\frac{{L}}{{k}}−\mathrm{1}}\:>\mathrm{0}\:\Rightarrow{L}>{k} \\ $$$$\Rightarrow{p}=\frac{{k}}{\mathrm{2}}\left(\sqrt{\frac{{L}}{{k}}}−\sqrt{\frac{{L}}{{k}}−\mathrm{1}}\right) \\ $$$$\Rightarrow{q}=\frac{{k}}{\mathrm{2}}\left(\sqrt{\frac{{L}}{{k}}}+\sqrt{\frac{{L}}{{k}}−\mathrm{1}}\right) \\ $$$${we}\:{can}\:{see} \\ $$$$\mathrm{tan}\:\phi\:\mathrm{tan}\:\varphi=\frac{\mathrm{4}{pq}}{{k}^{\mathrm{2}} }=\frac{\mathrm{4}}{{k}^{\mathrm{2}} }×\frac{{k}^{\mathrm{2}} }{\mathrm{4}}=\mathrm{1}, \\ $$$${that}\:{means}\:{the}\:{tangent}\:{lines} \\ $$$$\left({and}\:{also}\:{the}\:{normals}\right)\:{at}\:{the}\:{ends}\: \\ $$$${of}\:{the}\:{rod}\:{are}\:{perpendicular}\:{to}\:{each} \\ $$$${other}. \\ $$$$ \\ $$$${eqn}.\:{of}\:{rod}: \\ $$$${y}=\frac{{p}^{\mathrm{2}} }{{k}}+\left({x}+{p}\right){m} \\ $$$${mx}−{y}+\frac{{p}^{\mathrm{2}} }{{k}}+{pm}=\mathrm{0} \\ $$$$\Rightarrow{mx}−{y}+\frac{{k}}{\mathrm{4}}=\mathrm{0} \\ $$$$ \\ $$$${say}\:{the}\:{circle}\:{with}\:{radius}\:{r}\: \\ $$$${tangents}\:{the}\:{parabola}\:{at}\:{S}\left({s},\:\frac{{s}^{\mathrm{2}} }{{k}}\right) \\ $$$$\mathrm{tan}\:\delta=\frac{{dy}}{{dx}}=\frac{\mathrm{2}{s}}{{k}} \\ $$$${G}={center}\:{of}\:{circle} \\ $$$${x}_{{G}} ={s}−{r}\:\mathrm{sin}\:\delta={s}−\frac{\mathrm{2}{sr}}{\:\sqrt{\mathrm{4}{s}^{\mathrm{2}} +{k}^{\mathrm{2}} }} \\ $$$${y}_{{G}} =\frac{{s}^{\mathrm{2}} }{{k}}+{r}\:\mathrm{cos}\:\delta=\frac{{s}^{\mathrm{2}} }{{k}}+\frac{{kr}}{\:\sqrt{\mathrm{4}{s}^{\mathrm{2}} +{k}^{\mathrm{2}} }} \\ $$$${r}=\frac{{m}\left({s}−\frac{\mathrm{2}{sr}}{\:\sqrt{\mathrm{4}{s}^{\mathrm{2}} +{k}^{\mathrm{2}} }}\right)−\frac{{s}^{\mathrm{2}} }{{k}}−\frac{{kr}}{\:\sqrt{\mathrm{4}{s}^{\mathrm{2}} +{k}^{\mathrm{2}} }}+\frac{{k}}{\mathrm{4}}}{\:\sqrt{\mathrm{1}+{m}^{\mathrm{2}} }} \\ $$$$\left(\sqrt{\mathrm{1}+{m}^{\mathrm{2}} }+\frac{\mathrm{2}{ms}+{k}}{\:\sqrt{\mathrm{4}{s}^{\mathrm{2}} +{k}^{\mathrm{2}} }}\right){r}={ms}−\frac{{s}^{\mathrm{2}} }{{k}}+\frac{{k}}{\mathrm{4}} \\ $$$${r}=\frac{{ms}−\frac{{s}^{\mathrm{2}} }{{k}}+\frac{{k}}{\mathrm{4}}}{\:\sqrt{\mathrm{1}+{m}^{\mathrm{2}} }+\frac{\mathrm{2}{ms}+{k}}{\:\sqrt{\mathrm{4}{s}^{\mathrm{2}} +{k}^{\mathrm{2}} }}} \\ $$$$\frac{{r}}{{k}}=\frac{\frac{{ms}}{{k}}−\frac{{s}^{\mathrm{2}} }{{k}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{4}}}{\:\sqrt{\mathrm{1}+{m}^{\mathrm{2}} }+\frac{\frac{\mathrm{2}{ms}}{{k}}+\mathrm{1}}{\:\sqrt{\frac{\mathrm{4}{s}^{\mathrm{2}} }{{k}^{\mathrm{2}} }+\mathrm{1}}}} \\ $$$${let}\:\xi=\frac{{s}}{{k}} \\ $$$$\Rightarrow\frac{{r}}{{k}}=\frac{{m}\xi−\xi^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{4}}}{\:\sqrt{\mathrm{1}+{m}^{\mathrm{2}} }+\frac{\mathrm{2}{m}\xi+\mathrm{1}}{\:\sqrt{\mathrm{4}\xi^{\mathrm{2}} +\mathrm{1}}}}={f}\left(\xi\right) \\ $$$${for}\:{maximum}\:{r},\:{f}'\left(\xi\right)=\mathrm{0} \\ $$$$ \\ $$$$\underline{{alternative}\:{way}:} \\ $$$${for}\:{maximum}\:{circle}\:{the}\:{tangent}\:{at} \\ $$$${S}\left({s},\:\frac{{s}^{\mathrm{2}} }{{k}}\right)\:{must}\:{be}\:{parallel}\:{to}\:{the}\:{rod}. \\ $$$$\mathrm{tan}\:\delta=\frac{\mathrm{2}{s}}{{k}}={m} \\ $$$$\Rightarrow{s}=\frac{{km}}{\mathrm{2}} \\ $$$$\mathrm{2}{r}_{{max}} =\frac{\frac{{km}^{\mathrm{2}} }{\mathrm{2}}−\frac{{k}^{\mathrm{2}} {m}^{\mathrm{2}} }{\mathrm{4}{k}}+\frac{{k}}{\mathrm{4}}}{\:\sqrt{\mathrm{1}+{m}^{\mathrm{2}} }}=\frac{{k}\sqrt{\mathrm{1}+{m}^{\mathrm{2}} }}{\mathrm{4}}=\frac{\sqrt{{Lk}}}{\mathrm{4}} \\ $$$$\Rightarrow{r}_{{max}} =\frac{\sqrt{{Lk}}}{\mathrm{8}} \\ $$

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