Question Number 65022 by ajfour last updated on 24/Jul/19
Commented by MJS last updated on 24/Jul/19
$$\mathrm{I}\:\mathrm{used}\:\mathrm{a}\:\mathrm{calculator}.\:\mathrm{I}'\mathrm{ve}\:\mathrm{got}\:\mathrm{an}\:\mathrm{old}\:\mathrm{TI}−\mathrm{89} \\ $$$$\mathrm{you}\:\mathrm{can}\:\mathrm{use}\:\mathrm{any}\:\mathrm{calculator}\:\mathrm{which}\:\mathrm{is}\:\mathrm{able}\:\mathrm{to} \\ $$$$\mathrm{approximately}\:\mathrm{solve}\:\mathrm{equations}\:\mathrm{and}\:\mathrm{integrals} \\ $$$$\mathrm{select}\:\mathrm{a}\:\mathrm{value}\:\mathrm{for}\:{r} \\ $$$$\mathrm{calculate}\:{P},\:{Q}\:\mathrm{and}\:\mathrm{the}\:\mathrm{integrals} \\ $$$$\mathrm{it}\:\mathrm{takes}\:\mathrm{some}\:\mathrm{time}\:\mathrm{though}… \\ $$
Commented by ajfour last updated on 07/Aug/19
$${Area}\:{of}\:{region}\:{A}={Area}\:{of}\:{region}\:{B}. \\ $$$${Find}\:{radius}\:{of}\:{circle}.\:\:\:\:\: \\ $$
Commented by ajfour last updated on 24/Jul/19
$${You}\:{are}\:{invincible}\:{Sir},\:{but}\:{how} \\ $$$${could}\:{you}\:{even}\:{approximately} \\ $$$${arrive}\:{at}\:{a}\:{value}\:{of}\:\boldsymbol{{r}};\:{do}\:{explain} \\ $$$${Sir}. \\ $$
Commented by MJS last updated on 24/Jul/19
$$\mathrm{I}\:\mathrm{can}\:\mathrm{only}\:\mathrm{solve}\:\mathrm{approximately}\:\mathrm{and}\:\mathrm{get} \\ $$$${r}\approx.\mathrm{9269083316} \\ $$
Answered by ajfour last updated on 07/Aug/19
![(x−r)^2 +(y−r)^2 =r^2 x^2 +y^2 −2r(x+y)+r^2 =0 but also y=x^2 ⇒ x^2 +x^4 −2r(x+x^2 )+r^2 =0 x^4 +(1−2r)x^2 −2rx+r^2 =0 a=1−2r , b=−2r , c=r^2 p^6 +2ap^4 +(a^2 −4c)p^2 −b^2 =0 ⇒ p^6 +2(1−2r)p^4 +(1−4r)p^2 −4r^2 =0 let p^2 =z−(2/3)(1−2r) z^3 +(4/3)(1−2r)^2 z−(8/(27))(1−2r)^3 +2(1−2r)[−(4/3)(1−2r)z+(4/9)(1−2r)^2 ] +(1−4r)[z−(2/3)(1−2r)]−4r^2 =0 ⇒ z^3 −z[(4/3)(1−2r)^2 −(1−4r)] +(1−2r)^3 [(8/9)−(8/(27))] −(2/3)(1−4r)(1−2r)−4r^2 =0 ....](https://www.tinkutara.com/question/Q65991.png)
$$\left({x}−{r}\right)^{\mathrm{2}} +\left({y}−{r}\right)^{\mathrm{2}} ={r}^{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{2}{r}\left({x}+{y}\right)+{r}^{\mathrm{2}} =\mathrm{0} \\ $$$${but}\:{also}\:\:{y}={x}^{\mathrm{2}} \\ $$$$\Rightarrow\:{x}^{\mathrm{2}} +{x}^{\mathrm{4}} −\mathrm{2}{r}\left({x}+{x}^{\mathrm{2}} \right)+{r}^{\mathrm{2}} =\mathrm{0} \\ $$$${x}^{\mathrm{4}} +\left(\mathrm{1}−\mathrm{2}{r}\right){x}^{\mathrm{2}} −\mathrm{2}{rx}+{r}^{\mathrm{2}} =\mathrm{0} \\ $$$${a}=\mathrm{1}−\mathrm{2}{r}\:,\:{b}=−\mathrm{2}{r}\:,\:{c}={r}^{\mathrm{2}} \\ $$$${p}^{\mathrm{6}} +\mathrm{2}{ap}^{\mathrm{4}} +\left({a}^{\mathrm{2}} −\mathrm{4}{c}\right){p}^{\mathrm{2}} −{b}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow\:{p}^{\mathrm{6}} +\mathrm{2}\left(\mathrm{1}−\mathrm{2}{r}\right){p}^{\mathrm{4}} +\left(\mathrm{1}−\mathrm{4}{r}\right){p}^{\mathrm{2}} −\mathrm{4}{r}^{\mathrm{2}} =\mathrm{0} \\ $$$$\:{let}\:{p}^{\mathrm{2}} ={z}−\frac{\mathrm{2}}{\mathrm{3}}\left(\mathrm{1}−\mathrm{2}{r}\right) \\ $$$${z}^{\mathrm{3}} +\frac{\mathrm{4}}{\mathrm{3}}\left(\mathrm{1}−\mathrm{2}{r}\right)^{\mathrm{2}} {z}−\frac{\mathrm{8}}{\mathrm{27}}\left(\mathrm{1}−\mathrm{2}{r}\right)^{\mathrm{3}} \\ $$$$+\mathrm{2}\left(\mathrm{1}−\mathrm{2}{r}\right)\left[−\frac{\mathrm{4}}{\mathrm{3}}\left(\mathrm{1}−\mathrm{2}{r}\right){z}+\frac{\mathrm{4}}{\mathrm{9}}\left(\mathrm{1}−\mathrm{2}{r}\right)^{\mathrm{2}} \right] \\ $$$$+\left(\mathrm{1}−\mathrm{4}{r}\right)\left[{z}−\frac{\mathrm{2}}{\mathrm{3}}\left(\mathrm{1}−\mathrm{2}{r}\right)\right]−\mathrm{4}{r}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow\:{z}^{\mathrm{3}} −{z}\left[\frac{\mathrm{4}}{\mathrm{3}}\left(\mathrm{1}−\mathrm{2}{r}\right)^{\mathrm{2}} −\left(\mathrm{1}−\mathrm{4}{r}\right)\right] \\ $$$$\:\:\:+\left(\mathrm{1}−\mathrm{2}{r}\right)^{\mathrm{3}} \left[\frac{\mathrm{8}}{\mathrm{9}}−\frac{\mathrm{8}}{\mathrm{27}}\right] \\ $$$$\:\:\:\:−\frac{\mathrm{2}}{\mathrm{3}}\left(\mathrm{1}−\mathrm{4}{r}\right)\left(\mathrm{1}−\mathrm{2}{r}\right)−\mathrm{4}{r}^{\mathrm{2}} =\mathrm{0} \\ $$$$…. \\ $$
Commented by ajfour last updated on 07/Aug/19
$${hopeless}..{to}\:{carry} \\ $$