Question-81485

Question Number 81485 by ajfour last updated on 13/Feb/20

Commented by ajfour last updated on 13/Feb/20

$${If}\:{area}\:{of}\:{region}\:{A}\:{is}\:{equal}\:{to} \\ $$$${that}\:{of}\:{B},\:{find}\:{eq}.\:{of}\:{parabola}. \\ $$

Commented by jagoll last updated on 13/Feb/20

$$\mathrm{area}\:\mathrm{B}\:=\:\frac{\mathrm{1}}{\mathrm{4}}\pi \\ $$

Commented by ajfour last updated on 13/Feb/20

$${how}\:{can}\:{you}\:{be}\:{so}\:{quick},\:{MjS} \\ $$$${Sir}\:?,\:{thanks}\:{for}\:{answer}. \\ $$$${Can}\:{we}\:{have}\:{an}\:{equation}\:{in} \\ $$$${a}\:{yielding}\:{this}\:{answer},\:{Sir}? \\ $$

Commented by MJS last updated on 13/Feb/20

approximation leads to y=ax^2 with a≈.427395

$$\mathrm{approximation}\:\mathrm{leads}\:\mathrm{to} \\ $$$${y}={ax}^{\mathrm{2}} \:\mathrm{with}\:{a}\approx.\mathrm{427395} \\ $$

Commented by ajfour last updated on 13/Feb/20

$${and}\:\left({b}−\mathrm{1}\right)^{\mathrm{2}} +{ab}^{\mathrm{2}} =\mathrm{1} \\ $$

Commented by jagoll last updated on 13/Feb/20

area B = ∫^b _0 ((√(1−(1−x^2 )))−ax^2 ) dx = (π/4)

$$\mathrm{area}\:\mathrm{B}\:=\:\underset{\mathrm{0}} {\int}^{\mathrm{b}} \left(\sqrt{\mathrm{1}−\left(\mathrm{1}−\mathrm{x}^{\mathrm{2}} \right)}−\mathrm{ax}^{\mathrm{2}} \right)\:\mathrm{dx}\:=\:\frac{\pi}{\mathrm{4}} \\ $$

Commented by jagoll last updated on 13/Feb/20

$$\mathrm{typo} \\ $$

Answered by mr W last updated on 13/Feb/20

circle: y=(√(1−(x−1)^2 )) parabola: y=Ax^2 intersection at (h,Ah^2 ) Ah^2 =(√(1−(h−1)^2 ))=(√(h(2−h))) ⇒A=(√((2−h)/h^3 )) ∫_0 ^( h) ((√(1−(x−1)^2 ))−Ax^2 )dx=(π/4) ∫_(−1) ^( h−1) (√(1−u^2 ))du−((Ah^3 )/3)=(π/4) (1/2)[sin^(−1) u+u(√(1−u^2 ))]_(−1) ^(h−1) −((Ah^3 )/3)=(π/4) (1/2)[sin^(−1) (h−1)+(h−1)(√(h(2−h)))]−((h(√(h(2−h))))/3)=0 ⇒sin^(−1) (h−1)=(((3−h)(√(h(2−h))))/3) ⇒h≈1.446798 ⇒A≈0.427396

$${circle}:\:{y}=\sqrt{\mathrm{1}−\left({x}−\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${parabola}:\:{y}={Ax}^{\mathrm{2}} \\ $$$${intersection}\:{at}\:\left({h},{Ah}^{\mathrm{2}} \right) \\ $$$${Ah}^{\mathrm{2}} =\sqrt{\mathrm{1}−\left({h}−\mathrm{1}\right)^{\mathrm{2}} }=\sqrt{{h}\left(\mathrm{2}−{h}\right)} \\ $$$$\Rightarrow{A}=\sqrt{\frac{\mathrm{2}−{h}}{{h}^{\mathrm{3}} }} \\ $$$$\int_{\mathrm{0}} ^{\:{h}} \left(\sqrt{\mathrm{1}−\left({x}−\mathrm{1}\right)^{\mathrm{2}} }−{Ax}^{\mathrm{2}} \right){dx}=\frac{\pi}{\mathrm{4}} \\ $$$$\int_{−\mathrm{1}} ^{\:{h}−\mathrm{1}} \sqrt{\mathrm{1}−{u}^{\mathrm{2}} }{du}−\frac{{Ah}^{\mathrm{3}} }{\mathrm{3}}=\frac{\pi}{\mathrm{4}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{sin}^{−\mathrm{1}} {u}+{u}\sqrt{\mathrm{1}−{u}^{\mathrm{2}} }\right]_{−\mathrm{1}} ^{{h}−\mathrm{1}} −\frac{{Ah}^{\mathrm{3}} }{\mathrm{3}}=\frac{\pi}{\mathrm{4}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{sin}^{−\mathrm{1}} \left({h}−\mathrm{1}\right)+\left({h}−\mathrm{1}\right)\sqrt{{h}\left(\mathrm{2}−{h}\right)}\right]−\frac{{h}\sqrt{{h}\left(\mathrm{2}−{h}\right)}}{\mathrm{3}}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{sin}^{−\mathrm{1}} \left({h}−\mathrm{1}\right)=\frac{\left(\mathrm{3}−{h}\right)\sqrt{{h}\left(\mathrm{2}−{h}\right)}}{\mathrm{3}} \\ $$$$\Rightarrow{h}\approx\mathrm{1}.\mathrm{446798} \\ $$$$\Rightarrow{A}\approx\mathrm{0}.\mathrm{427396} \\ $$

Commented by mr W last updated on 13/Feb/20