Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 123086 by aurpeyz last updated on 22/Nov/20

y(√(1−x^2 ))+x(√(1−y^2 ))=0 find dy/dx

$${y}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }+{x}\sqrt{\mathrm{1}−{y}^{\mathrm{2}} }=\mathrm{0}\:{find}\:{dy}/{dx} \\ $$

Commented by liberty last updated on 23/Nov/20

y(√(1−x^2 )) = −x(√(1−y^2 )) y^2 (1−x^2 ) = x^2 (1−y^2 ) y^2 −x^2 y^2 =x^2 −x^2 y^2 y^2 = x^2 ⇒2y.y′ = 2x ⇔ y′ = (x/y).

$$\:\:\:\:\:\:\:{y}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\:=\:−{x}\sqrt{\mathrm{1}−{y}^{\mathrm{2}} }\: \\ $$$$\:\:\:\:\:\:\:{y}^{\mathrm{2}} \left(\mathrm{1}−{x}^{\mathrm{2}} \right)\:=\:{x}^{\mathrm{2}} \left(\mathrm{1}−{y}^{\mathrm{2}} \right) \\ $$$$\:\:\:\:\:\:\:{y}^{\mathrm{2}} −{x}^{\mathrm{2}} {y}^{\mathrm{2}} ={x}^{\mathrm{2}} −{x}^{\mathrm{2}} {y}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:{y}^{\mathrm{2}} \:=\:{x}^{\mathrm{2}} \:\Rightarrow\mathrm{2}{y}.{y}'\:=\:\mathrm{2}{x}\: \\ $$$$\:\:\:\:\:\:\Leftrightarrow\:{y}'\:=\:\frac{{x}}{{y}}.\: \\ $$

Answered by Dwaipayan Shikari last updated on 22/Nov/20

y(√(1−x^2 )) +x(√(1−y^2 )) =0 (dy/dx)(√(1−x^2 )) −((xy)/( (√(1−x^2 ))))+(√(1−y^2 )) −((xy)/( (√(1−y^2 )))).(dy/dx)=0 (dy/dx)((√(1−x^2 ))−((xy)/( (√(1−y^2 )))))=((xy)/( (√(1−x^2 ))))−(√(1−y^2 )) ⇒(dy/dx)=−((√(1−y^2 ))/( (√(1−x^2 ))))

$${y}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\:+{x}\sqrt{\mathrm{1}−{y}^{\mathrm{2}} }\:=\mathrm{0} \\ $$$$\frac{{dy}}{{dx}}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\:−\frac{{xy}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}+\sqrt{\mathrm{1}−{y}^{\mathrm{2}} }\:−\frac{{xy}}{\:\sqrt{\mathrm{1}−{y}^{\mathrm{2}} }}.\frac{{dy}}{{dx}}=\mathrm{0} \\ $$$$\frac{{dy}}{{dx}}\left(\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }−\frac{{xy}}{\:\sqrt{\mathrm{1}−{y}^{\mathrm{2}} }}\right)=\frac{{xy}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}−\sqrt{\mathrm{1}−{y}^{\mathrm{2}} } \\ $$$$\Rightarrow\frac{{dy}}{{dx}}=−\frac{\sqrt{\mathrm{1}−{y}^{\mathrm{2}} }}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }} \\ $$

Answered by mathmax by abdo last updated on 22/Nov/20

y(√(1−x^2 ))+x(√(1−y^2 ))=0 ⇒x(√(1−y^2 ))=−y(√(1−x^2 ))with (y/x)<0 ⇒x^2 (1−y^2 )=y^2 (1−x^2 ) ⇒x^2 −x^2 y^2 −y^2 +x^2 y^(2 ) =0 ⇒ ⇒y^2 =x^2 ⇒y =+^− x ⇒(dy/dx)=+^− 1

$$\mathrm{y}\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }+\mathrm{x}\sqrt{\mathrm{1}−\mathrm{y}^{\mathrm{2}} }=\mathrm{0}\:\Rightarrow\mathrm{x}\sqrt{\mathrm{1}−\mathrm{y}^{\mathrm{2}} }=−\mathrm{y}\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }\mathrm{with}\:\frac{\mathrm{y}}{\mathrm{x}}<\mathrm{0} \\ $$$$\Rightarrow\mathrm{x}^{\mathrm{2}} \left(\mathrm{1}−\mathrm{y}^{\mathrm{2}} \right)=\mathrm{y}^{\mathrm{2}} \left(\mathrm{1}−\mathrm{x}^{\mathrm{2}} \right)\:\Rightarrow\mathrm{x}^{\mathrm{2}} −\mathrm{x}^{\mathrm{2}} \mathrm{y}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} +\mathrm{x}^{\mathrm{2}} \mathrm{y}^{\mathrm{2}\:} =\mathrm{0}\:\Rightarrow \\ $$$$\Rightarrow\mathrm{y}^{\mathrm{2}} =\mathrm{x}^{\mathrm{2}} \:\Rightarrow\mathrm{y}\:=\overset{−} {+}\mathrm{x}\:\Rightarrow\frac{\mathrm{dy}}{\mathrm{dx}}=\overset{−} {+}\mathrm{1} \\ $$

Answered by aurpeyz last updated on 22/Nov/20

wow. so which is correct answer. are they same?

$${wow}.\:{so}\:{which}\:{is}\:{correct}\:{answer}.\:{are}\:{they}\:{same}? \\ $$

Commented by mathmax by abdo last updated on 23/Nov/20

its better to explicit y(x)...!

$$\mathrm{its}\:\mathrm{better}\:\mathrm{to}\:\mathrm{explicit}\:\mathrm{y}\left(\mathrm{x}\right)...! \\ $$

Answered by MJS_new last updated on 23/Nov/20

y(√(1−x^2 ))+x(√(1−y^2 ))=0 let x>0∧y<0 [it makes no difference to x<0∧y>0] x(√(1−y^2 ))=−y(√(1−x^2 )) x^2 (1−y^2 )=y^2 (1−x^2 ) x^2 =y^2 ⇒ y=−x ⇒ (dy/dx)=−1

$${y}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }+{x}\sqrt{\mathrm{1}−{y}^{\mathrm{2}} }=\mathrm{0} \\ $$$$\mathrm{let}\:{x}>\mathrm{0}\wedge{y}<\mathrm{0}\:\left[\mathrm{it}\:\mathrm{makes}\:\mathrm{no}\:\mathrm{difference}\:\mathrm{to}\:{x}<\mathrm{0}\wedge{y}>\mathrm{0}\right] \\ $$$${x}\sqrt{\mathrm{1}−{y}^{\mathrm{2}} }=−{y}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} } \\ $$$${x}^{\mathrm{2}} \left(\mathrm{1}−{y}^{\mathrm{2}} \right)={y}^{\mathrm{2}} \left(\mathrm{1}−{x}^{\mathrm{2}} \right) \\ $$$${x}^{\mathrm{2}} ={y}^{\mathrm{2}} \:\Rightarrow\:{y}=−{x} \\ $$$$\Rightarrow\:\frac{{dy}}{{dx}}=−\mathrm{1} \\ $$

Commented by benjo_mathlover last updated on 23/Nov/20

x>0 ? for x=2 ⇒(√(1−x^2 )) ∉R sir

$${x}>\mathrm{0}\:?\:{for}\:{x}=\mathrm{2}\:\Rightarrow\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\:\notin\mathbb{R}\: \\ $$$${sir} \\ $$

Commented by MJS_new last updated on 23/Nov/20

it′s obvious that −1≤x≤1 and −1≤y≤1 but if we want to solve for y we need to square and we are only allowed to square when both sides are ≥0 ⇒ x=−y

$$\mathrm{it}'\mathrm{s}\:\mathrm{obvious}\:\mathrm{that}\:−\mathrm{1}\leqslant{x}\leqslant\mathrm{1}\:\mathrm{and}\:−\mathrm{1}\leqslant{y}\leqslant\mathrm{1} \\ $$$$\mathrm{but}\:\mathrm{if}\:\mathrm{we}\:\mathrm{want}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{for}\:{y}\:\mathrm{we}\:\mathrm{need}\:\mathrm{to} \\ $$$$\mathrm{square}\:\mathrm{and}\:\mathrm{we}\:\mathrm{are}\:\mathrm{only}\:\mathrm{allowed}\:\mathrm{to}\:\mathrm{square} \\ $$$$\mathrm{when}\:\mathrm{both}\:\mathrm{sides}\:\mathrm{are}\:\geqslant\mathrm{0}\:\Rightarrow\:{x}=−{y} \\ $$

Commented by benjo_mathlover last updated on 23/Nov/20

oo yes. i agree sir

$${oo}\:{yes}.\:{i}\:{agree}\:{sir} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com