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if-x-3-y-3-3axy-find-dy-dx-in-terms-of-x-and-y-and-prove-that-dy-dx-cannot-be-equal-to-1-for-finite-values-of-x-and-y-except-x-y-please-help-

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Question Number 13226 by chux last updated on 16/May/17
if x^3 +y^3 =3axy,find dy/dx in terms of x and y and prove that dy/dx cannot be equal to -1 for finite values of x and y except x=y. please help
$$\mathrm{if}\:\mathrm{x}^{\mathrm{3}} +\mathrm{y}^{\mathrm{3}} =\mathrm{3axy},\mathrm{find}\:\mathrm{dy}/\mathrm{dx}\:\mathrm{in}\:\mathrm{terms} \\ $$$$\mathrm{of}\:\mathrm{x}\:\mathrm{and}\:\mathrm{y}\:\mathrm{and}\:\mathrm{prove}\:\mathrm{that}\:\mathrm{dy}/\mathrm{dx}\: \\ $$$$\mathrm{cannot}\:\mathrm{be}\:\mathrm{equal}\:\mathrm{to}\:-\mathrm{1}\:\mathrm{for}\:\mathrm{finite} \\ $$$$\mathrm{values}\:\mathrm{of}\:\mathrm{x}\:\mathrm{and}\:\mathrm{y}\:\mathrm{except}\:\mathrm{x}=\mathrm{y}. \\ $$$$ \\ $$$$ \\ $$$$\mathrm{please}\:\mathrm{help}\: \\ $$
Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 16/May/17
3x^2 +3y^2 (dy/dx)=3ay+3ax(dy/dx) ((dy/dx)=y^′ ) ⇒(3y^2 −3ax)y^′ =3(ay−x^2 )⇒ y^′ =((ay−x^2 )/(y^2 −ax)) .■ if :(y^′ =−1)⇒ay−x^2 =ax−y^2 ⇒ (y−x)(y+x)+a(y−x)=0⇒(y−x)(y+x+a)=0⇒ ⇒ { ((y−x=0⇒y=x)),((y+x+a=0⇒y=−(x+a) .■)) :}
$$\mathrm{3}{x}^{\mathrm{2}} +\mathrm{3}{y}^{\mathrm{2}} \frac{{dy}}{{dx}}=\mathrm{3}{ay}+\mathrm{3}{ax}\frac{{dy}}{{dx}}\:\:\:\:\:\:\left(\frac{{dy}}{{dx}}={y}^{'} \right) \\ $$$$\Rightarrow\left(\mathrm{3}{y}^{\mathrm{2}} −\mathrm{3}{ax}\right){y}^{'} =\mathrm{3}\left({ay}−{x}^{\mathrm{2}} \right)\Rightarrow \\ $$$${y}^{'} =\frac{{ay}−{x}^{\mathrm{2}} }{{y}^{\mathrm{2}} −{ax}}\:\:\:\:\:.\blacksquare \\ $$$${if}\::\left({y}^{'} =−\mathrm{1}\right)\Rightarrow{ay}−{x}^{\mathrm{2}} ={ax}−{y}^{\mathrm{2}} \Rightarrow \\ $$$$\left({y}−{x}\right)\left({y}+{x}\right)+{a}\left({y}−{x}\right)=\mathrm{0}\Rightarrow\left({y}−{x}\right)\left({y}+{x}+{a}\right)=\mathrm{0}\Rightarrow \\ $$$$\Rightarrow\begin{cases}{{y}−{x}=\mathrm{0}\Rightarrow{y}={x}}\\{{y}+{x}+{a}=\mathrm{0}\Rightarrow{y}=−\left({x}+{a}\right)\:\:\:.\blacksquare}\end{cases} \\ $$
Answered by ajfour last updated on 16/May/17
x^3 +y^3 =3axy ......(i) 3x^2 +3y^2 (dy/dx)=3ay+3ax(dy/dx) (dy/dx)=((ay−x^2 )/(y^2 −ax)) (dy/dx)+1=((ay−x^2 +y^2 −ax)/(y^2 −ax)) =(((y−x)(x+y+a))/(y^2 −ax)) ⇒ (dy/dx)+1=0 only when x=y , or (x+y+a)=0 (x+y)^3 =x^3 +y^3 +3xy(x+y) ...(ii) x^3 +y^3 = 3axy ...(i) (i)+(ii) : (x+y)^3 =3xy(x+y+a) ⇒ if (x+y+a)=0, x+y=0 so for (x+y+a)=0 condition is x+y=0 and even a=0 therefore if a≠0, x+y+a≠0 ⇒ (dy/dx)+1=0 or (dy/dx)=−1 only for x=y . (granted they are finite)
$${x}^{\mathrm{3}} +{y}^{\mathrm{3}} =\mathrm{3}{axy}\:\:\:\:\:\:\:\:\:\:\:\:\:\:……\left({i}\right) \\ $$$$\mathrm{3}{x}^{\mathrm{2}} +\mathrm{3}{y}^{\mathrm{2}} \frac{{dy}}{{dx}}=\mathrm{3}{ay}+\mathrm{3}{ax}\frac{{dy}}{{dx}} \\ $$$$\frac{{dy}}{{dx}}=\frac{{ay}−{x}^{\mathrm{2}} }{{y}^{\mathrm{2}} −{ax}} \\ $$$$\frac{{dy}}{{dx}}+\mathrm{1}=\frac{{ay}−{x}^{\mathrm{2}} +{y}^{\mathrm{2}} −{ax}}{{y}^{\mathrm{2}} −{ax}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\frac{\left({y}−{x}\right)\left({x}+{y}+{a}\right)}{{y}^{\mathrm{2}} −{ax}} \\ $$$$\Rightarrow\:\frac{{dy}}{{dx}}+\mathrm{1}=\mathrm{0}\:\:\:{only}\:{when}\: \\ $$$${x}={y}\:,\:\:{or}\:\:\left({x}+{y}+{a}\right)=\mathrm{0} \\ $$$$\left({x}+{y}\right)^{\mathrm{3}} ={x}^{\mathrm{3}} +{y}^{\mathrm{3}} +\mathrm{3}{xy}\left({x}+{y}\right)\:\:\:…\left({ii}\right) \\ $$$$\:\:{x}^{\mathrm{3}} +{y}^{\mathrm{3}} \:=\:\mathrm{3}{axy}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…\left({i}\right) \\ $$$$\left({i}\right)+\left({ii}\right)\:: \\ $$$$\left({x}+{y}\right)^{\mathrm{3}} =\mathrm{3}{xy}\left({x}+{y}+{a}\right) \\ $$$$\Rightarrow\:{if}\:\left({x}+{y}+{a}\right)=\mathrm{0},\:\:\:{x}+{y}=\mathrm{0} \\ $$$${so}\:{for}\:\left({x}+{y}+{a}\right)=\mathrm{0}\:\:{condition}\:{is} \\ $$$${x}+{y}=\mathrm{0}\:{and}\:{even}\:\:\boldsymbol{{a}}=\mathrm{0}\: \\ $$$${therefore}\:{if}\:{a}\neq\mathrm{0},\:\:{x}+{y}+{a}\neq\mathrm{0} \\ $$$$\Rightarrow\:\frac{{dy}}{{dx}}+\mathrm{1}=\mathrm{0}\:\:{or}\:\:\:\frac{{dy}}{{dx}}=−\mathrm{1}\:{only} \\ $$$${for}\:\:{x}={y}\:.\:\:\:\:\left({granted}\:{they}\:{are}\:{finite}\right) \\ $$
Commented by chux last updated on 17/May/17
thanks alot.
$$\mathrm{thanks}\:\mathrm{alot}. \\ $$

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