Question-214199

Question Number 214199 by mr W last updated on 01/Dec/24

Commented by mr W last updated on 01/Dec/24

[see Q214176] find the radius of the maximal circle inscribed between the curves f(x) and g(x) as shown.

$$\left[{see}\:{Q}\mathrm{214176}\right] \\ $$$${find}\:{the}\:{radius}\:{of}\:{the}\:{maximal} \\ $$$${circle}\:{inscribed}\:{between}\:{the}\:{curves} \\ $$$${f}\left({x}\right)\:{and}\:{g}\left({x}\right)\:{as}\:{shown}. \\ $$

Answered by issac last updated on 01/Dec/24

Osculate circle radious=(1/κ) κ=curvature of f(x) at x=α κ=((∣f^((2)) (α)∣)/( (√((1+(f^((1)) (α))^2 )^3 ))))

$$\mathrm{Osculate}\:\mathrm{circle}\:\mathrm{radious}=\frac{\mathrm{1}}{\kappa} \\ $$$$\kappa=\mathrm{curvature}\:\mathrm{of}\:{f}\left({x}\right)\:\mathrm{at}\:{x}=\alpha \\ $$$$\kappa=\frac{\mid{f}^{\left(\mathrm{2}\right)} \left(\alpha\right)\mid}{\:\sqrt{\left(\mathrm{1}+\left({f}^{\left(\mathrm{1}\right)} \left(\alpha\right)\right)^{\mathrm{2}} \right)^{\mathrm{3}} }}\: \\ $$

Commented by mr W last updated on 01/Dec/24

$${this}\:{has}\:{nothing}\:{to}\:{do}\:{with}\:{the} \\ $$$${current}\:{question},\:{therefore}\:{useless}! \\ $$

Answered by mr W last updated on 01/Dec/24

due to symmetry the center of circle lies on the y−axis, say at (0, h). the equation of circle is then x^2 +(y−h)^2 =R^2 we only need to consider x>0. say the circle touches g(x) at (s, t) and touches f(x) at (p, q). at touching point circle and curve have same tangent line. x^2 +(y−h)^2 =R^2 2x+2(y−h)(dy/dx)=0 y=ln x (dy/dx)=(1/x) s+(t−h)(1/s)=0 ⇒t−h=−s^2 s^2 +s^4 =R^2 ⇒s^2 =(((√(1+4R^2 ))−1)/2) t=h−s^2 =h−(((√(1+4R^2 ))−1)/2) t=ln s=(1/2)ln ((((√(1+4R^2 ))−1)/2)) ⇒h−(((√(1+4R^2 ))−1)/2)=(1/2)ln ((((√(1+4R^2 ))−1)/2)) ⇒h=(1/2)[(√(1+4R^2 ))−1+ln ((((√(1+4R^2 ))−1)/2))] y=(3/x^3 )+(1/(15x)) (dy/dx)=−(9/x^4 )−(1/(15x^2 )) 2p−2(q−h)((9/p^4 )+(1/(15p^2 )))=0 ⇒q−h=((15p^5 )/(p^2 +135)) p^2 +(((15p^5 )/(p^2 +135)))^2 =R^2 ⇒R=(√(p^2 +(((15p^5 )/(p^2 +135)))^2 )) ...(i) q=h+((15p^5 )/(p^2 +135)) q=(3/p^3 )+(1/(15p)) ⇒h+((15p^5 )/(p^2 +135))=(3/p^3 )+(1/(15p)) ⇒h=(3/p^3 )+(1/(15p))−((15p^5 )/(p^2 +135)) (1/2)[(√(1+4R^2 ))−1+ln ((((√(1+4R^2 ))−1)/2))] =(3/p^3 )+(1/(15p))−((15p^5 )/(p^2 +135)) ...(ii) we can solve (i) and (ii) and get p≈1.312274765822 R≈1.379982232083

$${due}\:{to}\:{symmetry}\:{the}\:{center}\:{of}\:{circle} \\ $$$${lies}\:{on}\:{the}\:{y}−{axis},\:{say}\:{at}\:\left(\mathrm{0},\:{h}\right). \\ $$$${the}\:{equation}\:{of}\:{circle}\:{is}\:{then} \\ $$$${x}^{\mathrm{2}} +\left({y}−{h}\right)^{\mathrm{2}} ={R}^{\mathrm{2}} \\ $$$$ \\ $$$${we}\:{only}\:{need}\:{to}\:{consider}\:{x}>\mathrm{0}. \\ $$$$ \\ $$$${say}\:{the}\:{circle}\:{touches}\:{g}\left({x}\right)\:{at}\:\left({s},\:{t}\right) \\ $$$${and}\:{touches}\:{f}\left({x}\right)\:{at}\:\left({p},\:{q}\right). \\ $$$${at}\:{touching}\:{point}\:{circle}\:{and}\:{curve} \\ $$$${have}\:{same}\:{tangent}\:{line}. \\ $$$$ \\ $$$${x}^{\mathrm{2}} +\left({y}−{h}\right)^{\mathrm{2}} ={R}^{\mathrm{2}} \\ $$$$\mathrm{2}{x}+\mathrm{2}\left({y}−{h}\right)\frac{{dy}}{{dx}}=\mathrm{0} \\ $$$${y}=\mathrm{ln}\:{x} \\ $$$$\frac{{dy}}{{dx}}=\frac{\mathrm{1}}{{x}} \\ $$$${s}+\left({t}−{h}\right)\frac{\mathrm{1}}{{s}}=\mathrm{0}\: \\ $$$$\Rightarrow{t}−{h}=−{s}^{\mathrm{2}} \\ $$$${s}^{\mathrm{2}} +{s}^{\mathrm{4}} ={R}^{\mathrm{2}} \\ $$$$\Rightarrow{s}^{\mathrm{2}} =\frac{\sqrt{\mathrm{1}+\mathrm{4}{R}^{\mathrm{2}} }−\mathrm{1}}{\mathrm{2}} \\ $$$${t}={h}−{s}^{\mathrm{2}} ={h}−\frac{\sqrt{\mathrm{1}+\mathrm{4}{R}^{\mathrm{2}} }−\mathrm{1}}{\mathrm{2}} \\ $$$${t}=\mathrm{ln}\:{s}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\left(\frac{\sqrt{\mathrm{1}+\mathrm{4}{R}^{\mathrm{2}} }−\mathrm{1}}{\mathrm{2}}\right) \\ $$$$\Rightarrow{h}−\frac{\sqrt{\mathrm{1}+\mathrm{4}{R}^{\mathrm{2}} }−\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\left(\frac{\sqrt{\mathrm{1}+\mathrm{4}{R}^{\mathrm{2}} }−\mathrm{1}}{\mathrm{2}}\right) \\ $$$$\Rightarrow{h}=\frac{\mathrm{1}}{\mathrm{2}}\left[\sqrt{\mathrm{1}+\mathrm{4}{R}^{\mathrm{2}} }−\mathrm{1}+\mathrm{ln}\:\left(\frac{\sqrt{\mathrm{1}+\mathrm{4}{R}^{\mathrm{2}} }−\mathrm{1}}{\mathrm{2}}\right)\right] \\ $$$$ \\ $$$${y}=\frac{\mathrm{3}}{{x}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{15}{x}} \\ $$$$\frac{{dy}}{{dx}}=−\frac{\mathrm{9}}{{x}^{\mathrm{4}} }−\frac{\mathrm{1}}{\mathrm{15}{x}^{\mathrm{2}} } \\ $$$$\mathrm{2}{p}−\mathrm{2}\left({q}−{h}\right)\left(\frac{\mathrm{9}}{{p}^{\mathrm{4}} }+\frac{\mathrm{1}}{\mathrm{15}{p}^{\mathrm{2}} }\right)=\mathrm{0} \\ $$$$\Rightarrow{q}−{h}=\frac{\mathrm{15}{p}^{\mathrm{5}} }{{p}^{\mathrm{2}} +\mathrm{135}} \\ $$$${p}^{\mathrm{2}} +\left(\frac{\mathrm{15}{p}^{\mathrm{5}} }{{p}^{\mathrm{2}} +\mathrm{135}}\right)^{\mathrm{2}} ={R}^{\mathrm{2}} \\ $$$$\Rightarrow{R}=\sqrt{{p}^{\mathrm{2}} +\left(\frac{\mathrm{15}{p}^{\mathrm{5}} }{{p}^{\mathrm{2}} +\mathrm{135}}\right)^{\mathrm{2}} }\:\:\:…\left({i}\right) \\ $$$${q}={h}+\frac{\mathrm{15}{p}^{\mathrm{5}} }{{p}^{\mathrm{2}} +\mathrm{135}} \\ $$$${q}=\frac{\mathrm{3}}{{p}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{15}{p}} \\ $$$$\Rightarrow{h}+\frac{\mathrm{15}{p}^{\mathrm{5}} }{{p}^{\mathrm{2}} +\mathrm{135}}=\frac{\mathrm{3}}{{p}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{15}{p}} \\ $$$$\Rightarrow{h}=\frac{\mathrm{3}}{{p}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{15}{p}}−\frac{\mathrm{15}{p}^{\mathrm{5}} }{{p}^{\mathrm{2}} +\mathrm{135}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left[\sqrt{\mathrm{1}+\mathrm{4}{R}^{\mathrm{2}} }−\mathrm{1}+\mathrm{ln}\:\left(\frac{\sqrt{\mathrm{1}+\mathrm{4}{R}^{\mathrm{2}} }−\mathrm{1}}{\mathrm{2}}\right)\right] \\ $$$$\:\:\:\:\:\:\:\:\:=\frac{\mathrm{3}}{{p}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{15}{p}}−\frac{\mathrm{15}{p}^{\mathrm{5}} }{{p}^{\mathrm{2}} +\mathrm{135}}\:\:\:…\left({ii}\right) \\ $$$${we}\:{can}\:{solve}\:\left({i}\right)\:{and}\:\left({ii}\right)\:{and}\:{get} \\ $$$${p}\approx\mathrm{1}.\mathrm{312274765822} \\ $$$${R}\approx\mathrm{1}.\mathrm{379982232083} \\ $$

Commented by mr W last updated on 01/Dec/24