As we know, two functions f ( x , y ) f(x,y) f ( x , y ) g ( x , y ) g(x,y) g ( x , y )
J = ∂ ( f , g ) ∂ ( x , y ) = ∣ f x ′ f y ′ g x ′ g y ′ ∣ J=\frac{\partial\left(f,g\right)}{\partial\left(x,y\right)}=
\left|\begin{array}{cc}
f'_x&f'_y\\[0.3cm]
g'_x&g'_y
\end{array}\right| J = ∂ ( x , y ) ∂ ( f , g ) = ∣ ∣ f x ′ g x ′ f y ′ g y ′ ∣ ∣
In our case,
f ( x , y ) = ln x − ln y ⟶ { f x ′ = 1 x f y ′ = − 1 y g ( x , y ) = x 2 + 3 y 2 2 x y ≡ x 2 y + 3 y 2 x ⟶ { g x ′ = 1 2 y − 3 y 2 x 2 g y ′ = − x 2 y 2 + 3 2 x f(x,y)=\ln x-\ln y\longrightarrow\left\{
\begin{array}{l}
f'_x=\displaystyle\frac{1}{x}\\[0.3cm]
f'_y=-\displaystyle\frac{1}{y}
\end{array}\right.\\[0.3cm]
g(x,y)=\frac{x^2+3y^2}{2xy}\equiv\frac{x}{2y}+\frac{3y}{2x}\longrightarrow\left\{
\begin{array}{l}
g'_x=\displaystyle\frac{1}{2y}-\displaystyle\frac{3y}{2x^2}\\[0.3cm]
g'_y=-\displaystyle\frac{x}{2y^2}+\displaystyle\frac{3}{2x}
\end{array}\right.\\[0.3cm] f ( x , y ) = ln x − ln y ⟶ ⎩ ⎨ ⎧ f x ′ = x 1 f y ′ = − y 1 g ( x , y ) = 2 x y x 2 + 3 y 2 ≡ 2 y x + 2 x 3 y ⟶ ⎩ ⎨ ⎧ g x ′ = 2 y 1 − 2 x 2 3 y g y ′ = − 2 y 2 x + 2 x 3
Then,
J = ∣ f x ′ f y ′ g x ′ g y ′ ∣ = ∣ 1 x − 1 y 1 2 y − 3 y 2 x 2 − x 2 y 2 + 3 2 x ∣ = = 1 x ⋅ ( − x 2 y 2 + 3 2 x ) − ( − 1 y ) ⋅ ( 1 2 y − 3 y 2 x 2 ) = = − 1 2 y 2 + 3 2 x 2 + 1 2 x 2 − 3 2 x 2 ≡ 0 ⟶ Conclusion, J ( f , g ) ≡ 0 , ∀ ( x , y ) ∈ D = { ( x , y ) ∣ x > 0 , y > 0 } J=\left|\begin{array}{cc}
f'_x&f'_y\\[0.3cm]
g'_x&g'_y
\end{array}\right|=\left|\begin{array}{cc}
\displaystyle\frac{1}{x} &-\displaystyle\frac{1}{y}\\[0.5cm]
\displaystyle\frac{1}{2y}-\displaystyle\frac{3y}{2x^2}&
-\displaystyle\frac{x}{2y^2}+\displaystyle\frac{3}{2x}
\end{array}\right|=\\[0.3cm]
=\frac{1}{x}\cdot\left(-\frac{x}{2y^2}+\frac{3}{2x}\right)-
\left(-\frac{1}{y}\right)\cdot\left(\frac{1}{2y}-\frac{3y}{2x^2}\right)=\\[0.3cm]
=-\frac{1}{2y^2}+\frac{3}{2x^2}+\frac{1}{2x^2}-\frac{3}{2x^2}\equiv0\longrightarrow\\[0.3cm]
\boxed{\text{Conclusion,}\,\,\,J\left(f,g\right)\equiv0,\,\,\forall (x,y)\in D=\left\{(x,y)\left|x>0,y>0\right.\right\}} J = ∣ ∣ f x ′ g x ′ f y ′ g y ′ ∣ ∣ = ∣ ∣ x 1 2 y 1 − 2 x 2 3 y − y 1 − 2 y 2 x + 2 x 3 ∣ ∣ = = x 1 ⋅ ( − 2 y 2 x + 2 x 3 ) − ( − y 1 ) ⋅ ( 2 y 1 − 2 x 2 3 y ) = = − 2 y 2 1 + 2 x 2 3 + 2 x 2 1 − 2 x 2 3 ≡ 0 ⟶ Conclusion, J ( f , g ) ≡ 0 , ∀ ( x , y ) ∈ D = { ( x , y ) ∣ x > 0 , y > 0 }
To find the functional relationship between f ( x , y ) f(x,y) f ( x , y ) g ( x , y ) g(x,y) g ( x , y )
z = f ( x , y ) = ln x − ln y ≡ ln ( x / y ) u = g ( x , y ) = x 2 + 3 y 2 2 x y = 1 2 ⋅ ( x y + 3 ⋅ y x ) u = 1 2 ⋅ ( x y + 3 ⋅ ( x y ) − 1 ) z=f(x,y)=\ln x-\ln y\equiv\ln\left(x/y\right)\\[0.3cm]
u=g(x,y)=\frac{x^2+3y^2}{2xy}=\frac{1}{2}\cdot\left(\frac{x}{y}+3\cdot\frac{y}{x}\right)\\[0.3cm]
u=\frac{1}{2}\cdot\left(\frac{x}{y}+3\cdot\left(\frac{x}{y}\right)^{-1}\right) z = f ( x , y ) = ln x − ln y ≡ ln ( x / y ) u = g ( x , y ) = 2 x y x 2 + 3 y 2 = 2 1 ⋅ ( y x + 3 ⋅ x y ) u = 2 1 ⋅ ( y x + 3 ⋅ ( y x ) − 1 )
Our task is reduced to finding a function φ \varphi φ z z z u u u
As we know
e z = e ln ( x / y ) = x y e − z = e − ln ( x / y ) = e ln ( y / x ) = y x φ ( z ) = 1 2 ⋅ ( e z + 3 e − z ) e^z=e^{\ln\left(x/y\right)}=\frac{x}{y}\\[0.3cm]
e^{-z}=e^{-\ln\left(x/y\right)}=e^{\ln\left(y/x\right)}=\frac{y}{x}\\[0.3cm]
\varphi(z)=\frac{1}{2}\cdot\left(e^z+3e^{-z}\right) e z = e l n ( x / y ) = y x e − z = e − l n ( x / y ) = e l n ( y / x ) = x y φ ( z ) = 2 1 ⋅ ( e z + 3 e − z )
Or in the original notation
φ ( f ( x , y ) ) = 1 2 ⋅ ( e f ( x , y ) + 3 e − f ( x , y ) ) ≡ g ( x , y ) \boxed{\varphi\left(f(x,y)\right)=\frac{1}{2}\cdot\left(e^{f(x,y)}+3e^{-f(x,y)}\right)\equiv g(x,y)} φ ( f ( x , y ) ) = 2 1 ⋅ ( e f ( x , y ) + 3 e − f ( x , y ) ) ≡ g ( x , y )
#340153 on Dec 2023
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