Question-219685

Question Number 219685 by alcohol last updated on 30/Apr/25

Answered by SdC355 last updated on 01/May/25

κ=(2/( (√((1+4x^2 )^3 )))) (curvature κ=((∣∣y^((2)) (t)∣∣)/( (√((1+(y^((1)) (t))^2 )^3 )))) ) r=(1/κ)=((√((1+4x^2 )^3 ))/2) (Radius r =(1/κ) )

$$\kappa=\frac{\mathrm{2}}{\:\sqrt{\left(\mathrm{1}+\mathrm{4}{x}^{\mathrm{2}} \right)^{\mathrm{3}} }}\:\:\:\left(\mathrm{curvature}\:\kappa=\frac{\mid\mid{y}^{\left(\mathrm{2}\right)} \left({t}\right)\mid\mid}{\:\sqrt{\left(\mathrm{1}+\left({y}^{\left(\mathrm{1}\right)} \left({t}\right)\right)^{\mathrm{2}} \right)^{\mathrm{3}} }}\:\right) \\ $$$${r}=\frac{\mathrm{1}}{\kappa}=\frac{\sqrt{\left(\mathrm{1}+\mathrm{4}{x}^{\mathrm{2}} \right)^{\mathrm{3}} }}{\mathrm{2}}\:\:\left(\mathrm{Radius}\:{r}\:=\frac{\mathrm{1}}{\kappa}\:\right) \\ $$

Commented by Ghisom last updated on 01/May/25

$$\mathrm{what}\:\mathrm{are}\:\mathrm{you}\:\mathrm{doing}??? \\ $$

Commented by SdC355 last updated on 01/May/25

$$\mathrm{isn}'\mathrm{t}\:\mathrm{that}\:\mathrm{Circle}\:\mathrm{osculating}\:\mathrm{circle}? \\ $$

Commented by Ghisom last updated on 01/May/25

no. the circle tangents both parabolas, the upper one from outside. the osculating circle must be inside

$$\mathrm{no}.\:\mathrm{the}\:\mathrm{circle}\:\mathrm{tangents}\:\mathrm{both}\:\mathrm{parabolas},\:\mathrm{the} \\ $$$$\mathrm{upper}\:\mathrm{one}\:\mathrm{from}\:\mathrm{outside}.\:\mathrm{the}\:\mathrm{osculating} \\ $$$$\mathrm{circle}\:\mathrm{must}\:\mathrm{be}\:\mathrm{inside} \\ $$

Answered by Ghisom last updated on 01/May/25

perpendicular lines P= ((p),(p^2 ) ) ⇒ l_p : y=−(1/(2p))x+p^2 +(1/2) Q= ((q),((q^2 −1/2)) ) ⇒ l_q : y=−(1/(2q))x+q^2 center of circle C= (((2p^3 +p)),(0) ) = (((2q^3 )),(0) ) radius r=p^2 (√(4p^2 +1))=((∣2q^2 −1∣(√(4q^2 +1)))/2) we can solve this exactly: p=((√(2((√(6((√3)−1)))+2(√3)−3))/6) ⇒ r=((√(3(4(5(√3)+6)(√(2((√3)−1)))−26(√3)+63))/(54)) but I guess C≈ (((.484378532869)),(0) ) r≈.178131348774 is more useable

$$\mathrm{perpendicular}\:\mathrm{lines} \\ $$$${P}=\begin{pmatrix}{{p}}\\{{p}^{\mathrm{2}} }\end{pmatrix}\:\Rightarrow\:{l}_{{p}} :\:{y}=−\frac{\mathrm{1}}{\mathrm{2}{p}}{x}+{p}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${Q}=\begin{pmatrix}{{q}}\\{{q}^{\mathrm{2}} −\mathrm{1}/\mathrm{2}}\end{pmatrix}\:\Rightarrow\:{l}_{{q}} :\:{y}=−\frac{\mathrm{1}}{\mathrm{2}{q}}{x}+{q}^{\mathrm{2}} \\ $$$$\mathrm{center}\:\mathrm{of}\:\mathrm{circle} \\ $$$${C}=\begin{pmatrix}{\mathrm{2}{p}^{\mathrm{3}} +{p}}\\{\mathrm{0}}\end{pmatrix}\:=\begin{pmatrix}{\mathrm{2}{q}^{\mathrm{3}} }\\{\mathrm{0}}\end{pmatrix} \\ $$$$\mathrm{radius} \\ $$$${r}={p}^{\mathrm{2}} \sqrt{\mathrm{4}{p}^{\mathrm{2}} +\mathrm{1}}=\frac{\mid\mathrm{2}{q}^{\mathrm{2}} −\mathrm{1}\mid\sqrt{\mathrm{4}{q}^{\mathrm{2}} +\mathrm{1}}}{\mathrm{2}} \\ $$$$\mathrm{we}\:\mathrm{can}\:\mathrm{solve}\:\mathrm{this}\:\mathrm{exactly}: \\ $$$${p}=\frac{\sqrt{\mathrm{2}\left(\sqrt{\mathrm{6}\left(\sqrt{\mathrm{3}}−\mathrm{1}\right)}+\mathrm{2}\sqrt{\mathrm{3}}−\mathrm{3}\right.}}{\mathrm{6}} \\ $$$$\Rightarrow\:{r}=\frac{\sqrt{\mathrm{3}\left(\mathrm{4}\left(\mathrm{5}\sqrt{\mathrm{3}}+\mathrm{6}\right)\sqrt{\mathrm{2}\left(\sqrt{\mathrm{3}}−\mathrm{1}\right)}−\mathrm{26}\sqrt{\mathrm{3}}+\mathrm{63}\right.}}{\mathrm{54}} \\ $$$$\mathrm{but}\:\mathrm{I}\:\mathrm{guess} \\ $$$${C}\approx\begin{pmatrix}{.\mathrm{484378532869}}\\{\mathrm{0}}\end{pmatrix} \\ $$$${r}\approx.\mathrm{178131348774} \\ $$$$\mathrm{is}\:\mathrm{more}\:\mathrm{useable} \\ $$

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