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Question-42448




Question Number 42448 by Tawa1 last updated on 25/Aug/18
Answered by tanmay.chaudhury50@gmail.com last updated on 25/Aug/18
y=(√(3x^2 .y))  y=3x^2   (dy/dx)=6x
$${y}=\sqrt{\mathrm{3}{x}^{\mathrm{2}} .{y}} \\ $$$${y}=\mathrm{3}{x}^{\mathrm{2}} \\ $$$$\frac{{dy}}{{dx}}=\mathrm{6}{x} \\ $$
Commented by Tawa1 last updated on 25/Aug/18
God bless you sir.  But is it possible to integrate the final answer 6x and get the  original question back ?
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$$$\mathrm{But}\:\mathrm{is}\:\mathrm{it}\:\mathrm{possible}\:\mathrm{to}\:\mathrm{integrate}\:\mathrm{the}\:\mathrm{final}\:\mathrm{answer}\:\mathrm{6x}\:\mathrm{and}\:\mathrm{get}\:\mathrm{the} \\ $$$$\mathrm{original}\:\mathrm{question}\:\mathrm{back}\:? \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 25/Aug/18
the value of y→3x^2 ....  i think it is approximation...  as y=(√(3x^2 (√(3x^2 (√(3x^2 (√(3x^2 ....→∞))))))))  y=(√(3x^2 ×y))  y^2 =3x^2 ×y      y=3x^2   supose you take a glass of water from ocean  does the quantity of water changes...apperently  no change of volume  but truly speaking there is a change of volume  v=v_0 −△v  where △v<<<<<v  so we consider v≈v_0   ∫6x dx=3x^2   +c
$${the}\:{value}\:{of}\:{y}\rightarrow\mathrm{3}{x}^{\mathrm{2}} …. \\ $$$${i}\:{think}\:{it}\:{is}\:{approximation}… \\ $$$${as}\:{y}=\sqrt{\mathrm{3}{x}^{\mathrm{2}} \sqrt{\mathrm{3}{x}^{\mathrm{2}} \sqrt{\mathrm{3}{x}^{\mathrm{2}} \sqrt{\mathrm{3}{x}^{\mathrm{2}} ….\rightarrow\infty}}}} \\ $$$${y}=\sqrt{\mathrm{3}{x}^{\mathrm{2}} ×{y}} \\ $$$${y}^{\mathrm{2}} =\mathrm{3}{x}^{\mathrm{2}} ×{y}\:\:\:\:\:\:{y}=\mathrm{3}{x}^{\mathrm{2}} \\ $$$${supose}\:{you}\:{take}\:{a}\:{glass}\:{of}\:{water}\:{from}\:{ocean} \\ $$$${does}\:{the}\:{quantity}\:{of}\:{water}\:{changes}…{apperently} \\ $$$${no}\:{change}\:{of}\:{volume} \\ $$$${but}\:{truly}\:{speaking}\:{there}\:{is}\:{a}\:{change}\:{of}\:{volume} \\ $$$${v}={v}_{\mathrm{0}} −\bigtriangleup{v} \\ $$$${where}\:\bigtriangleup{v}<<<<<{v} \\ $$$${so}\:{we}\:{consider}\:{v}\approx{v}_{\mathrm{0}} \\ $$$$\int\mathrm{6}{x}\:{dx}=\mathrm{3}{x}^{\mathrm{2}} \:\:+{c} \\ $$
Commented by Tawa1 last updated on 25/Aug/18
God bless you sir
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$
Answered by ajfour last updated on 25/Aug/18
y=(√(3x^2 y))  y=3x^2     ⇒   (dy/dx)= 6x .
$${y}=\sqrt{\mathrm{3}{x}^{\mathrm{2}} {y}} \\ $$$${y}=\mathrm{3}{x}^{\mathrm{2}} \:\:\:\:\Rightarrow\:\:\:\frac{{dy}}{{dx}}=\:\mathrm{6}{x}\:. \\ $$
Commented by Tawa1 last updated on 25/Aug/18
God bless you sir
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$
Answered by byaw last updated on 25/Aug/18
squaring both sides  y^2 =3x^2 (√(3x^2 (√(3x^2 (√(3x^2 (√(...))))))))  y^2 =3x^2 y  2y dy/dx=6xy+3x^2 dy/dx  (2y−3x^2 )(dy/dx)=6xy  (dy/dx)=((6xy)/(2y−3x^2 ))
$$\mathrm{squaring}\:\mathrm{both}\:\mathrm{sides} \\ $$$${y}^{\mathrm{2}} =\mathrm{3}{x}^{\mathrm{2}} \sqrt{\mathrm{3}{x}^{\mathrm{2}} \sqrt{\mathrm{3}{x}^{\mathrm{2}} \sqrt{\mathrm{3}{x}^{\mathrm{2}} \sqrt{…}}}} \\ $$$${y}^{\mathrm{2}} =\mathrm{3}{x}^{\mathrm{2}} {y} \\ $$$$\mathrm{2}{y}\:{dy}/{dx}=\mathrm{6}{xy}+\mathrm{3}{x}^{\mathrm{2}} {dy}/{dx} \\ $$$$\left(\mathrm{2}{y}−\mathrm{3}{x}^{\mathrm{2}} \right)\frac{{dy}}{{dx}}=\mathrm{6}{xy} \\ $$$$\frac{{dy}}{{dx}}=\frac{\mathrm{6}{xy}}{\mathrm{2}{y}−\mathrm{3}{x}^{\mathrm{2}} } \\ $$
Commented by Tawa1 last updated on 25/Aug/18
God bless you sir
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

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