Question-80977

Question Number 80977 by ajfour last updated on 08/Feb/20

Commented by ajfour last updated on 08/Feb/20

$${Regions}\:{A}\:{and}\:{B}\:{have}\:{equal} \\ $$$${area},\:{find}\:{radius}\:{of}\:{circle}, \\ $$$${centered}\:{at}\:{origin}. \\ $$

Commented by ajfour last updated on 08/Feb/20

Also find radius ratio of incircle in region A to that of incircle in region B. (say a/b=? ).

$${Also}\:{find}\:{radius}\:{ratio}\:{of} \\ $$$${incircle}\:{in}\:{region}\:{A}\:{to}\:{that}\:{of} \\ $$$${incircle}\:{in}\:{region}\:{B}.\: \\ $$$$\left({say}\:\:{a}/{b}=?\:\right). \\ $$

Commented by mr W last updated on 08/Feb/20

$${i}\:{got}\:{R}\approx\mathrm{2}.\mathrm{35223} \\ $$

Commented by ajfour last updated on 08/Feb/20

Commented by ajfour last updated on 08/Feb/20

let F(h,h^2 ) S=∫_0 ^( h) [hx−x^2 ]dx=(h^3 /6) R=h(√(1+h^2 )) ((R^2 (tan^(−1) h))/2)−((2h^3 )/6)=(R^2 /2)((π/2)−tan^(−1) h) h^2 (1+h^2 ){tan^(−1) h−(π/4)}=(h^3 /3) ⇒ tan^(−1) h=(π/4)+(h/(3(1+h^2 ))) R=h(√(1+h^2 )) ..........

$${let}\:{F}\left({h},{h}^{\mathrm{2}} \right) \\ $$$${S}=\int_{\mathrm{0}} ^{\:{h}} \left[{hx}−{x}^{\mathrm{2}} \right]{dx}=\frac{{h}^{\mathrm{3}} }{\mathrm{6}} \\ $$$${R}={h}\sqrt{\mathrm{1}+{h}^{\mathrm{2}} } \\ $$$$\frac{\mathrm{R}^{\mathrm{2}} \left(\mathrm{tan}^{−\mathrm{1}} {h}\right)}{\mathrm{2}}−\frac{\mathrm{2}{h}^{\mathrm{3}} }{\mathrm{6}}=\frac{{R}^{\mathrm{2}} }{\mathrm{2}}\left(\frac{\pi}{\mathrm{2}}−\mathrm{tan}^{−\mathrm{1}} {h}\right) \\ $$$${h}^{\mathrm{2}} \left(\mathrm{1}+{h}^{\mathrm{2}} \right)\left\{\mathrm{tan}^{−\mathrm{1}} {h}−\frac{\pi}{\mathrm{4}}\right\}=\frac{{h}^{\mathrm{3}} }{\mathrm{3}} \\ $$$$\Rightarrow\:\mathrm{tan}^{−\mathrm{1}} {h}=\frac{\pi}{\mathrm{4}}+\frac{{h}}{\mathrm{3}\left(\mathrm{1}+{h}^{\mathrm{2}} \right)} \\ $$$${R}={h}\sqrt{\mathrm{1}+{h}^{\mathrm{2}} }\: \\ $$$$………. \\ $$

Commented by ajfour last updated on 08/Feb/20

thats right sir, at least i got the same, R≈2.35223015

$${thats}\:{right}\:{sir},\:{at}\:{least}\:{i}\:{got}\:{the} \\ $$$${same},\:\:{R}\approx\mathrm{2}.\mathrm{35223015} \\ $$

Answered by mr W last updated on 08/Feb/20

Commented by mr W last updated on 08/Feb/20

F(h,h^2 ) R=OF=(√(h^2 +h^4 ))=h(√(1+h^2 )) tan α=(h^2 /h)=h β=(π/2)−α Area B=((2h^3 )/3)+(R^2 /4)(2β−sin 2β)=((πR^2 )/8) ((16h^3 )/(3R^2 ))+2π−4α−2sin (π−2α)=π ((16h)/(3(1+h^2 )))+π−4α−2sin 2α=0 ((16h)/(3(1+h^2 )))+π−4α−((4h)/(1+h^2 ))=0 ⇒(h/(3(1+h^2 )))+(π/4)−tan^(−1) h=0 ⇒h≈1.380139 ⇒R≈2.352230 T(t,t^2 ) (θ is marked as α at B in diagram!) tan θ=2t ⇒t=((tan θ)/2) x_B =t−b sin θ=b ⇒b=(t/(1+sin θ))=((tan θ)/(2(1+sin θ))) y_B =t^2 +b cos θ x_B ^2 +y_B ^2 =(R−b)^2 (t^2 +b cos θ)^2 =R(R−2b) (((tan^2 θ)/4)+((sin θ)/(2(1+sin θ))))^2 =R(R−((tan θ)/(1+sin θ))) ⇒θ≈66.999803° ⇒b≈0.613336 S(s,s^2 ) (ϕ is marked as β at A in diagram!) tan ϕ=2s ⇒s=((tan ϕ)/2) x_A =s+a sin ϕ y_A =s^2 −a cos ϕ=a ⇒a=(s^2 /(1+cos ϕ))=((tan^2 ϕ)/(4(1+cos ϕ))) x_A ^2 +y_A ^2 =(R−a)^2 (s+a sin ϕ)^2 +a^2 =(R−a)^2 (s+a sin ϕ)^2 =R(R−2a) tan^2 ϕ(1+(1/(cos ϕ)))^2 =16R(R−((1−cos ϕ)/(2cos^2 ϕ))) ⇒ϕ≈62.917105° ⇒a≈0.656990 (a/b)≈1.071174

$${F}\left({h},{h}^{\mathrm{2}} \right) \\ $$$${R}={OF}=\sqrt{{h}^{\mathrm{2}} +{h}^{\mathrm{4}} }={h}\sqrt{\mathrm{1}+{h}^{\mathrm{2}} } \\ $$$$\mathrm{tan}\:\alpha=\frac{{h}^{\mathrm{2}} }{{h}}={h} \\ $$$$\beta=\frac{\pi}{\mathrm{2}}−\alpha \\ $$$${Area}\:{B}=\frac{\mathrm{2}{h}^{\mathrm{3}} }{\mathrm{3}}+\frac{{R}^{\mathrm{2}} }{\mathrm{4}}\left(\mathrm{2}\beta−\mathrm{sin}\:\mathrm{2}\beta\right)=\frac{\pi{R}^{\mathrm{2}} }{\mathrm{8}} \\ $$$$\frac{\mathrm{16}{h}^{\mathrm{3}} }{\mathrm{3}{R}^{\mathrm{2}} }+\mathrm{2}\pi−\mathrm{4}\alpha−\mathrm{2sin}\:\left(\pi−\mathrm{2}\alpha\right)=\pi \\ $$$$\frac{\mathrm{16}{h}}{\mathrm{3}\left(\mathrm{1}+{h}^{\mathrm{2}} \right)}+\pi−\mathrm{4}\alpha−\mathrm{2sin}\:\mathrm{2}\alpha=\mathrm{0} \\ $$$$\frac{\mathrm{16}{h}}{\mathrm{3}\left(\mathrm{1}+{h}^{\mathrm{2}} \right)}+\pi−\mathrm{4}\alpha−\frac{\mathrm{4}{h}}{\mathrm{1}+{h}^{\mathrm{2}} }=\mathrm{0} \\ $$$$\Rightarrow\frac{{h}}{\mathrm{3}\left(\mathrm{1}+{h}^{\mathrm{2}} \right)}+\frac{\pi}{\mathrm{4}}−\mathrm{tan}^{−\mathrm{1}} {h}=\mathrm{0} \\ $$$$\Rightarrow{h}\approx\mathrm{1}.\mathrm{380139} \\ $$$$\Rightarrow{R}\approx\mathrm{2}.\mathrm{352230} \\ $$$$ \\ $$$${T}\left({t},{t}^{\mathrm{2}} \right) \\ $$$$\left(\theta\:{is}\:{marked}\:{as}\:\alpha\:{at}\:{B}\:{in}\:{diagram}!\right) \\ $$$$\mathrm{tan}\:\theta=\mathrm{2}{t}\:\Rightarrow{t}=\frac{\mathrm{tan}\:\theta}{\mathrm{2}} \\ $$$${x}_{{B}} ={t}−{b}\:\mathrm{sin}\:\theta={b}\:\Rightarrow{b}=\frac{{t}}{\mathrm{1}+\mathrm{sin}\:\theta}=\frac{\mathrm{tan}\:\theta}{\mathrm{2}\left(\mathrm{1}+\mathrm{sin}\:\theta\right)} \\ $$$${y}_{{B}} ={t}^{\mathrm{2}} +{b}\:\mathrm{cos}\:\theta \\ $$$${x}_{{B}} ^{\mathrm{2}} +{y}_{{B}} ^{\mathrm{2}} =\left({R}−{b}\right)^{\mathrm{2}} \\ $$$$\left({t}^{\mathrm{2}} +{b}\:\mathrm{cos}\:\theta\right)^{\mathrm{2}} ={R}\left({R}−\mathrm{2}{b}\right) \\ $$$$\left(\frac{\mathrm{tan}^{\mathrm{2}} \:\theta}{\mathrm{4}}+\frac{\mathrm{sin}\:\theta}{\mathrm{2}\left(\mathrm{1}+\mathrm{sin}\:\theta\right)}\right)^{\mathrm{2}} ={R}\left({R}−\frac{\mathrm{tan}\:\theta}{\mathrm{1}+\mathrm{sin}\:\theta}\right) \\ $$$$\Rightarrow\theta\approx\mathrm{66}.\mathrm{999803}° \\ $$$$\Rightarrow{b}\approx\mathrm{0}.\mathrm{613336} \\ $$$$ \\ $$$${S}\left({s},{s}^{\mathrm{2}} \right) \\ $$$$\left(\varphi\:{is}\:{marked}\:{as}\:\beta\:{at}\:{A}\:{in}\:{diagram}!\right) \\ $$$$\mathrm{tan}\:\varphi=\mathrm{2}{s}\:\Rightarrow{s}=\frac{\mathrm{tan}\:\varphi}{\mathrm{2}} \\ $$$${x}_{{A}} ={s}+{a}\:\mathrm{sin}\:\varphi \\ $$$${y}_{{A}} ={s}^{\mathrm{2}} −{a}\:\mathrm{cos}\:\varphi={a}\:\Rightarrow{a}=\frac{{s}^{\mathrm{2}} }{\mathrm{1}+\mathrm{cos}\:\varphi}=\frac{\mathrm{tan}^{\mathrm{2}} \:\varphi}{\mathrm{4}\left(\mathrm{1}+\mathrm{cos}\:\varphi\right)} \\ $$$${x}_{{A}} ^{\mathrm{2}} +{y}_{{A}} ^{\mathrm{2}} =\left({R}−{a}\right)^{\mathrm{2}} \\ $$$$\left({s}+{a}\:\mathrm{sin}\:\varphi\right)^{\mathrm{2}} +{a}^{\mathrm{2}} =\left({R}−{a}\right)^{\mathrm{2}} \\ $$$$\left({s}+{a}\:\mathrm{sin}\:\varphi\right)^{\mathrm{2}} ={R}\left({R}−\mathrm{2}{a}\right) \\ $$$$\mathrm{tan}^{\mathrm{2}} \:\varphi\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{cos}\:\varphi}\right)^{\mathrm{2}} =\mathrm{16}{R}\left({R}−\frac{\mathrm{1}−\mathrm{cos}\:\varphi}{\mathrm{2cos}^{\mathrm{2}} \:\varphi}\right) \\ $$$$\Rightarrow\varphi\approx\mathrm{62}.\mathrm{917105}° \\ $$$$\Rightarrow{a}\approx\mathrm{0}.\mathrm{656990} \\ $$$$ \\ $$$$\frac{{a}}{{b}}\approx\mathrm{1}.\mathrm{071174} \\ $$

Commented by mr W last updated on 08/Feb/20