Question-214020

Question Number 214020 by ajfour last updated on 24/Nov/24

Commented by ajfour last updated on 24/Nov/24

Red area = Blue area. Find r in terms of k, φ.

$${Red}\:{area}\:=\:{Blue}\:{area}.\:{Find}\:{r}\:{in} \\ $$$${terms}\:{of}\:{k},\:\phi. \\ $$

Answered by dionigi last updated on 24/Nov/24

kp^2 = (1+ cos ϕ) R p^2 = (((1+ cos ϕ) R)/k) A_1 = 2 (k p^3 − ((k p^3 )/3)) = ((4 k p^3 )/3) A_2 = π R^2 − ϕ R^2 + sin ϕ cos ϕ R^2 A_2 = (π− ϕ + sin ϕ cos ϕ) R^2 A_1 = A_2 A_1 ^2 = A_2 ^2 ((16 k^2 p^6 )/9) = (π− ϕ + sin ϕ cos ϕ)^2 R^4 ((16 (1+ cos ϕ)^3 R^3 )/(9 k)) = (π− ϕ + sin ϕ cos ϕ)^2 R^4 R = ((16 (1+ cos ϕ)^3 )/(9 k (π− ϕ + sin ϕ cos ϕ)^2 ))

$$ \\ $$$${kp}^{\mathrm{2}} \:=\:\left(\mathrm{1}+\:\mathrm{cos}\:\varphi\right)\:{R} \\ $$$${p}^{\mathrm{2}} \:=\:\frac{\left(\mathrm{1}+\:\mathrm{cos}\:\varphi\right)\:{R}}{{k}} \\ $$$${A}_{\mathrm{1}} \:=\:\mathrm{2}\:\left({k}\:{p}^{\mathrm{3}} \:−\:\frac{{k}\:{p}^{\mathrm{3}} }{\mathrm{3}}\right)\:=\:\frac{\mathrm{4}\:{k}\:{p}^{\mathrm{3}} }{\mathrm{3}} \\ $$$${A}_{\mathrm{2}} \:=\:\pi\:{R}^{\mathrm{2}} \:−\:\varphi\:{R}^{\mathrm{2}} \:+\:\mathrm{sin}\:\varphi\:\mathrm{cos}\:\varphi\:{R}^{\mathrm{2}} \\ $$$${A}_{\mathrm{2}} \:=\:\left(\pi−\:\varphi\:+\:\mathrm{sin}\:\varphi\:\mathrm{cos}\:\varphi\right)\:{R}^{\mathrm{2}} \\ $$$${A}_{\mathrm{1}} \:=\:{A}_{\mathrm{2}} \\ $$$${A}_{\mathrm{1}} ^{\mathrm{2}} \:=\:{A}_{\mathrm{2}} ^{\mathrm{2}} \\ $$$$\frac{\mathrm{16}\:{k}^{\mathrm{2}} \:{p}^{\mathrm{6}} }{\mathrm{9}}\:=\:\left(\pi−\:\varphi\:+\:\mathrm{sin}\:\varphi\:\mathrm{cos}\:\varphi\right)^{\mathrm{2}} \:{R}^{\mathrm{4}} \\ $$$$\frac{\mathrm{16}\:\left(\mathrm{1}+\:\mathrm{cos}\:\varphi\right)^{\mathrm{3}} \:{R}^{\mathrm{3}} }{\mathrm{9}\:{k}}\:=\:\left(\pi−\:\varphi\:+\:\mathrm{sin}\:\varphi\:\mathrm{cos}\:\varphi\right)^{\mathrm{2}} \:{R}^{\mathrm{4}} \\ $$$${R}\:=\:\frac{\mathrm{16}\:\left(\mathrm{1}+\:\mathrm{cos}\:\varphi\right)^{\mathrm{3}} \:}{\mathrm{9}\:{k}\:\left(\pi−\:\varphi\:+\:\mathrm{sin}\:\varphi\:\mathrm{cos}\:\varphi\right)^{\mathrm{2}} } \\ $$

Commented by ajfour last updated on 24/Nov/24

$${Thank}\:{you}. \\ $$

Answered by mr W last updated on 24/Nov/24

r(1+cos ϕ)=kp^2 ⇒p=(√((r(1+cos ϕ))/k)) ((r^2 (π−ϕ))/2)+((r^2 sin ϕ cos ϕ)/2)=((2pr(1+cos ϕ))/3) ((r^2 (π−ϕ))/2)+((r^2 sin ϕ cos ϕ)/2)=((2r(1+cos ϕ)(√(r(1+cos ϕ))))/(3(√k))) ⇒r=((16(1+cos ϕ)^3 )/( 9k(π−ϕ+sin ϕ cos ϕ)^2 ))

$${r}\left(\mathrm{1}+\mathrm{cos}\:\varphi\right)={kp}^{\mathrm{2}} \\ $$$$\Rightarrow{p}=\sqrt{\frac{{r}\left(\mathrm{1}+\mathrm{cos}\:\varphi\right)}{{k}}} \\ $$$$\frac{{r}^{\mathrm{2}} \left(\pi−\varphi\right)}{\mathrm{2}}+\frac{{r}^{\mathrm{2}} \:\mathrm{sin}\:\varphi\:\mathrm{cos}\:\varphi}{\mathrm{2}}=\frac{\mathrm{2}{pr}\left(\mathrm{1}+\mathrm{cos}\:\varphi\right)}{\mathrm{3}} \\ $$$$\frac{{r}^{\mathrm{2}} \left(\pi−\varphi\right)}{\mathrm{2}}+\frac{{r}^{\mathrm{2}} \:\mathrm{sin}\:\varphi\:\mathrm{cos}\:\varphi}{\mathrm{2}}=\frac{\mathrm{2}{r}\left(\mathrm{1}+\mathrm{cos}\:\varphi\right)\sqrt{{r}\left(\mathrm{1}+\mathrm{cos}\:\varphi\right)}}{\mathrm{3}\sqrt{{k}}} \\ $$$$\Rightarrow{r}=\frac{\mathrm{16}\left(\mathrm{1}+\mathrm{cos}\:\varphi\right)^{\mathrm{3}} }{\:\mathrm{9}{k}\left(\pi−\varphi+\mathrm{sin}\:\varphi\:\mathrm{cos}\:\varphi\right)^{\mathrm{2}} } \\ $$

Commented by mr W last updated on 24/Nov/24

Commented by mr W last updated on 24/Nov/24

Commented by ajfour last updated on 24/Nov/24

Thanks Sir ⇒ for ϕ=(π/2) , k=1 r=((16)/(9π^2 ))=((4/(3π)))^2 (I am reminded of c.m. height of semi- disc)

$${Thanks}\:{Sir} \\ $$$$\Rightarrow\:\:{for}\:\varphi=\frac{\pi}{\mathrm{2}}\:,\:{k}=\mathrm{1} \\ $$$${r}=\frac{\mathrm{16}}{\mathrm{9}\pi^{\mathrm{2}} }=\left(\frac{\mathrm{4}}{\mathrm{3}\pi}\right)^{\mathrm{2}} \:\:\:\left({I}\:{am}\:{reminded}\:{of}\:\right. \\ $$$$\left.{c}.{m}.\:{height}\:{of}\:{semi}-\:{disc}\right) \\ $$$$ \\ $$

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