Question and Answers Forum

All Questions      Topic List

Differentiation Questions

Previous in All Question      Next in All Question      

Previous in Differentiation      Next in Differentiation      

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.

$$\mathrm{thanks}\:\mathrm{alot}. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com