Question Number 211508 by MATHEMATICSAM last updated on 11/Sep/24
$$\mathrm{If}\:\sqrt{\mathrm{1}\:−\:{x}^{\mathrm{2}} }\:+\:\sqrt{\mathrm{1}\:−\:{y}^{\mathrm{2}} }\:=\:{a}\left({x}\:−\:{y}\right)\:\mathrm{then} \\ $$$$\mathrm{prove}\:\mathrm{that}\:\frac{{dy}}{{dx}}\:=\:\sqrt{\frac{\mathrm{1}\:−\:{y}^{\mathrm{2}} }{\mathrm{1}\:−\:{x}^{\mathrm{2}} }\:}\:. \\ $$
Commented by Frix last updated on 11/Sep/24
$$\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{think}\:\mathrm{it}'\mathrm{s}\:\mathrm{true}. \\ $$$$\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }−{ax}={ay}+\sqrt{\mathrm{1}−{y}^{\mathrm{2}} } \\ $$$$\Rightarrow \\ $$$$\frac{{dy}}{{dx}}=−\frac{\left({x}+{a}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\right)\sqrt{\mathrm{1}−{y}^{\mathrm{2}} }}{\left({y}−{a}\sqrt{\mathrm{1}−{y}^{\mathrm{2}} }\right)\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }} \\ $$
Commented by MATHEMATICSAM last updated on 11/Sep/24
$$\mathrm{nope}\:\mathrm{sir}\:\mathrm{I}\:\mathrm{found}\:\mathrm{a}\:\mathrm{process}\:\mathrm{which}\:\mathrm{can}\:\mathrm{be} \\ $$$$\mathrm{done}\:\mathrm{by}\:\mathrm{putting}\:{x}\:=\:\mathrm{sin}\alpha\:\mathrm{and} \\ $$$${y}\:=\:\mathrm{sin}\beta\:\mathrm{resulting}\:\mathrm{the}\:\mathrm{prove}.\:\mathrm{But}\:\mathrm{I}\:\mathrm{solved}\:\mathrm{by}\:\mathrm{general}\:\mathrm{method} \\ $$$$\mathrm{and}\:\mathrm{got}\:\mathrm{your}\:\mathrm{result}.\:\mathrm{So}\:\mathrm{I}\:\mathrm{have}\:\mathrm{confusion}. \\ $$
Commented by Spillover last updated on 12/Sep/24
$${Qn}\:\mathrm{211019} \\ $$
Answered by Frix last updated on 11/Sep/24
![We can solve for y: y=(((a^2 −1)x−2a(√(1−x^2 )))/(a^2 +1)) ⇒ y′=((a^2 −1)/(a^2 +1))+((2ax)/((a^2 +1)(√(1−x^2 )))) Now testing: ((dy/dx))^2 =((1−y^2 )/(1−x^2 ))=((1−((((a^2 −1)x−2a(√(1−x^2 )))/(a^2 +1)))^2 )/(1−x^2 )) (y′)^2 =(((a^2 −1)/(a^2 +1))+((2ax)/((a^2 +1)(√(1−x^2 )))))^2 [I′m too lazy to write out the expansions here but they′re indeed the same!] ⇒ (dy/dx)=y′ ⇒ (dx/dy)=((√(1−y^2 ))/( (√(1−x^2 ))))=((a^2 −1)/(a^2 +1))+((2ax)/((a^2 +1)(√(1−x^2 )))) ...which leaves me puzzled...](https://www.tinkutara.com/question/Q211519.png)
$$\mathrm{We}\:\mathrm{can}\:\mathrm{solve}\:\mathrm{for}\:{y}: \\ $$$${y}=\frac{\left({a}^{\mathrm{2}} −\mathrm{1}\right){x}−\mathrm{2}{a}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{{a}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\Rightarrow \\ $$$${y}'=\frac{{a}^{\mathrm{2}} −\mathrm{1}}{{a}^{\mathrm{2}} +\mathrm{1}}+\frac{\mathrm{2}{ax}}{\left({a}^{\mathrm{2}} +\mathrm{1}\right)\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }} \\ $$$$\mathrm{Now}\:\mathrm{testing}: \\ $$$$\left(\frac{{dy}}{{dx}}\right)^{\mathrm{2}} =\frac{\mathrm{1}−{y}^{\mathrm{2}} }{\mathrm{1}−{x}^{\mathrm{2}} }=\frac{\mathrm{1}−\left(\frac{\left({a}^{\mathrm{2}} −\mathrm{1}\right){x}−\mathrm{2}{a}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{{a}^{\mathrm{2}} +\mathrm{1}}\right)^{\mathrm{2}} }{\mathrm{1}−{x}^{\mathrm{2}} } \\ $$$$\left({y}'\right)^{\mathrm{2}} =\left(\frac{{a}^{\mathrm{2}} −\mathrm{1}}{{a}^{\mathrm{2}} +\mathrm{1}}+\frac{\mathrm{2}{ax}}{\left({a}^{\mathrm{2}} +\mathrm{1}\right)\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\right)^{\mathrm{2}} \\ $$$$\left[\mathrm{I}'\mathrm{m}\:\mathrm{too}\:\mathrm{lazy}\:\mathrm{to}\:\mathrm{write}\:\mathrm{out}\:\mathrm{the}\:\mathrm{expansions}\right. \\ $$$$\left.\mathrm{here}\:\mathrm{but}\:\mathrm{they}'\mathrm{re}\:\mathrm{indeed}\:\mathrm{the}\:\mathrm{same}!\right] \\ $$$$\Rightarrow\:\frac{{dy}}{{dx}}={y}' \\ $$$$\Rightarrow \\ $$$$\frac{{dx}}{{dy}}=\frac{\sqrt{\mathrm{1}−{y}^{\mathrm{2}} }}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}=\frac{{a}^{\mathrm{2}} −\mathrm{1}}{{a}^{\mathrm{2}} +\mathrm{1}}+\frac{\mathrm{2}{ax}}{\left({a}^{\mathrm{2}} +\mathrm{1}\right)\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }} \\ $$$$…\mathrm{which}\:\mathrm{leaves}\:\mathrm{me}\:\mathrm{puzzled}… \\ $$