QUESTION (1i)
z = 2 x + a + 2 y + p ⟶ p = ∂ z ∂ x = 2 2 2 x + a ≡ 1 2 x + a q = ∂ z ∂ y = 2 2 2 y + b ≡ 1 2 y + b z=\sqrt{2x+a}+\sqrt{2y+p}\longrightarrow\\[0.3cm]
p=\frac{\partial z}{\partial x}=\frac{2}{2\sqrt{2x+a}}\equiv\frac{1}{\sqrt{2x+a}}\\[0.3cm]
q=\frac{\partial z}{\partial y}=\frac{2}{2\sqrt{2y+b}}\equiv\frac{1}{\sqrt{2y+b}}\\[0.3cm] z = 2 x + a + 2 y + p ⟶ p = ∂ x ∂ z = 2 2 x + a 2 ≡ 2 x + a 1 q = ∂ y ∂ z = 2 2 y + b 2 ≡ 2 y + b 1 Then,
1 p + 1 q = 1 1 2 x + a + 1 1 2 y + b ⟶ 1 p + 1 q = 2 x + a + 2 y + b ≡ z \frac{1}{p}+\frac{1}{q}=\frac{1}{\displaystyle\frac{1}{\sqrt{2x+a}}}+\frac{1}{\displaystyle\frac{1}{\sqrt{2y+b}}}\longrightarrow\\[0.3cm]
\frac{1}{p}+\frac{1}{q}=\sqrt{2x+a}+\sqrt{2y+b}\equiv z p 1 + q 1 = 2 x + a 1 1 + 2 y + b 1 1 ⟶ p 1 + q 1 = 2 x + a + 2 y + b ≡ z
Conclusion,
z = 2 x + a + 2 y + p → 1 p + 1 q = z \boxed{z=\sqrt{2x+a}+\sqrt{2y+p}\to\frac{1}{p}+\frac{1}{q}=z} z = 2 x + a + 2 y + p → p 1 + q 1 = z
QUESTION (1ii)
z 2 + μ = 2 ( 1 + λ − 1 ) ( x + λ y ) ⟶ 2 z ⋅ ∂ z ∂ x = 2 ( 1 + λ − 1 ) ⟶ p = ∂ z ∂ x = 1 + λ − 1 z = λ + 1 z λ 2 z ⋅ ∂ z ∂ y = 2 ( 1 + λ − 1 ) λ ⟶ q = ∂ z ∂ y = λ + 1 z z^2+\mu=2\left(1+\lambda^{-1}\right)(x+\lambda y)\longrightarrow\\[0.3cm]
2z\cdot\frac{\partial z}{\partial x}=2\left(1+\lambda^{-1}\right)\longrightarrow\\[0.3cm]
p=\frac{\partial z}{\partial x}=\frac{1+\lambda^{-1}}{z}=\frac{\lambda+1}{z\lambda}\\[0.3cm]
2z\cdot\frac{\partial z}{\partial y}=2\left(1+\lambda^{-1}\right)\lambda\longrightarrow\\[0.3cm]
q=\frac{\partial z}{\partial y}=\frac{\lambda+1}{z}\\[0.3cm] z 2 + μ = 2 ( 1 + λ − 1 ) ( x + λ y ) ⟶ 2 z ⋅ ∂ x ∂ z = 2 ( 1 + λ − 1 ) ⟶ p = ∂ x ∂ z = z 1 + λ − 1 = z λ λ + 1 2 z ⋅ ∂ y ∂ z = 2 ( 1 + λ − 1 ) λ ⟶ q = ∂ y ∂ z = z λ + 1
Then,
1 p + 1 q = 1 λ + 1 z λ + 1 λ + 1 z = z λ λ + 1 + z λ + 1 1 p + 1 q = z ( λ + 1 ) λ + 1 ≡ z \frac{1}{p}+\frac{1}{q}=\frac{1}{\displaystyle\frac{\lambda+1}{z\lambda}}+\frac{1}{\displaystyle\frac{\lambda+1}{z}}=\frac{z\lambda}{\lambda+1}+\frac{z}{\lambda+1}\\[0.3cm]
\frac{1}{p}+\frac{1}{q}=\frac{z\left(\lambda+1\right)}{\lambda+1}\equiv z p 1 + q 1 = z λ λ + 1 1 + z λ + 1 1 = λ + 1 z λ + λ + 1 z p 1 + q 1 = λ + 1 z ( λ + 1 ) ≡ z
Conclusion,
z 2 + μ = 2 ( 1 + λ − 1 ) ( x + λ y ) ⟶ 1 p + 1 q = z \boxed{z^2+\mu=2\left(1+\lambda^{-1}\right)(x+\lambda y)\longrightarrow\frac{1}{p}+\frac{1}{q}=z} z 2 + μ = 2 ( 1 + λ − 1 ) ( x + λ y ) ⟶ p 1 + q 1 = z
QUESTION 2
Suppose that b = b ( a ; λ ; μ ) b=b(a;\lambda;\mu) b = b ( a ; λ ; μ )
∂ ∂ a ∣ 2 x + a + 2 y + b − z = 0 ⟶ 1 2 2 x + a + 1 2 2 y + b ⋅ ∂ b ∂ a = 0 ⟶ 2 y + b = − 2 x + a ⋅ ∂ b ∂ a \frac{\partial }{\partial a}\left|\sqrt{2x+a}+\sqrt{2y+b}-z=0\right.\longrightarrow\\[0.3cm]
\frac{1}{2\sqrt{2x+a}}+\frac{1}{2\sqrt{2y+b}}\cdot\frac{\partial b}{\partial a}=0\longrightarrow\\[0.3cm]
\boxed{\sqrt{2y+b}=-\sqrt{2x+a}\cdot\frac{\partial b}{\partial a}} ∂ a ∂ ∣ ∣ 2 x + a + 2 y + b − z = 0 ⟶ 2 2 x + a 1 + 2 2 y + b 1 ⋅ ∂ a ∂ b = 0 ⟶ 2 y + b = − 2 x + a ⋅ ∂ a ∂ b
Suppose that
b ( a ; λ ; μ ) = − a λ + C ⟶ ∂ b ∂ a = − 1 λ b(a;\lambda;\mu)=-\frac{a}{\lambda}+C\longrightarrow\frac{\partial b}{\partial a}=-\frac{1}{\lambda} b ( a ; λ ; μ ) = − λ a + C ⟶ ∂ a ∂ b = − λ 1
Then,
2 y + b = − 2 x + a ⋅ ( − 1 λ ) ⟶ 2 y + b = 1 λ ⋅ 2 x + a [ 2 y + b ] 2 = [ 1 λ ⋅ 2 x + a ] 2 ⟶ 2 y = 2 x + a λ 2 − b \sqrt{2y+b}=-\sqrt{2x+a}\cdot\left(-\frac{1}{\lambda}\right)\longrightarrow\\[0.3cm]
\boxed{\sqrt{2y+b}=\frac{1}{\lambda}\cdot\sqrt{2x+a}}\\[0.3cm]
\left[\sqrt{2y+b}\right]^2=\left[\frac{1}{\lambda}\cdot\sqrt{2x+a}\right]^2\longrightarrow\\[0.3cm]
\boxed{2y=\frac{2x+a}{\lambda^2}-b} 2 y + b = − 2 x + a ⋅ ( − λ 1 ) ⟶ 2 y + b = λ 1 ⋅ 2 x + a [ 2 y + b ] 2 = [ λ 1 ⋅ 2 x + a ] 2 ⟶ 2 y = λ 2 2 x + a − b
Substituting the found relationship in the first equation and express z 2 z^2 z 2
z = 2 x + a + 2 y + b ⟶ z = 2 x + a + 1 λ ⋅ 2 x + a ⟶ z 2 = ( 1 + 1 λ ) 2 ⋅ ( 2 x + a ) z=\sqrt{2x+a}+\sqrt{2y+b}\longrightarrow\\[0.3cm]
z=\sqrt{2x+a}+\frac{1}{\lambda}\cdot\sqrt{2x+a}\longrightarrow\\[0.3cm]
\boxed{z^2=\left(1+\frac{1}{\lambda}\right)^2\cdot(2x+a)} z = 2 x + a + 2 y + b ⟶ z = 2 x + a + λ 1 ⋅ 2 x + a ⟶ z 2 = ( 1 + λ 1 ) 2 ⋅ ( 2 x + a )
Substituting the found relationship in the second equation :
z 2 + μ = 2 ( 1 + 1 λ ) ( x + λ y ) ⟶ ( 1 + λ λ ) 2 ( 2 x + a ) + μ = 2 ⋅ 1 + λ λ ( x + λ y ) ∣ ÷ 1 ( 1 + λ ) 2 2 x + a λ 2 + μ ( 1 + λ ) 2 = 2 ( x + λ y ) λ ( 1 + λ ) ⟶ 2 x + a λ 2 + μ ( 1 + λ ) 2 = 2 y ( 1 + λ ) + 2 x λ ( 1 + λ ) ⟶ 2 x + a λ 2 + μ ( 1 + λ ) 2 = 2 x + a λ 2 − b ( 1 + λ ) + 2 x λ ( 1 + λ ) ⟶ 2 x + a λ 2 + μ ( 1 + λ ) 2 = ( 2 x + a ) − λ 2 b λ 2 ( 1 + λ ) + 2 x λ ( 1 + λ ) ⟶ 2 x + a λ 2 + μ ( 1 + λ ) 2 = ( 2 x + a ) λ 2 ( 1 + λ ) − λ 2 b λ 2 ( 1 + λ ) + 2 x λ ( 1 + λ ) 2 x + a λ 2 + μ ( 1 + λ ) 2 − 2 x λ ( 1 + λ ) − ( 2 x + a ) λ 2 ( 1 + λ ) = − b ( 1 + λ ) ( 1 + λ ) ( 2 x + a ) − 2 x λ − ( 2 x + a ) λ 2 ( 1 + λ ) + μ ( 1 + λ ) 2 = − b ( 1 + λ ) 2 x + a + 2 x λ + a λ − 2 x λ − 2 x − a λ 2 ( 1 + λ ) + μ ( 1 + λ ) 2 = − b ( 1 + λ ) a λ λ 2 ( 1 + λ ) + μ ( 1 + λ ) 2 = − b ( 1 + λ ) ∣ × − ( 1 + λ ) b = − a λ − μ ( 1 + λ ) z^2+\mu=2\left(1+\frac{1}{\lambda}\right)(x+\lambda y)\longrightarrow\\[0.3cm]
\left.\left(\frac{1+\lambda}{\lambda}\right)^2(2x+a)+\mu=2\cdot\frac{1+\lambda}{\lambda}(x+\lambda y)\right|\div\frac{1}{(1+\lambda)^2}\\[0.3cm]
\frac{2x+a}{\lambda^2}+\frac{\mu}{(1+\lambda)^2}=\frac{2(x+\lambda y)}{\lambda(1+\lambda)}\longrightarrow\\[0.3cm]
\frac{2x+a}{\lambda^2}+\frac{\mu}{(1+\lambda)^2}=\frac{2y}{(1+\lambda)}+\frac{2x}{\lambda(1+\lambda)}\longrightarrow\\[0.3cm]
\frac{2x+a}{\lambda^2}+\frac{\mu}{(1+\lambda)^2}=\frac{\displaystyle\frac{2x+a}{\lambda^2}-b}{(1+\lambda)}+\frac{2x}{\lambda(1+\lambda)}\longrightarrow\\[0.3cm]
\frac{2x+a}{\lambda^2}+\frac{\mu}{(1+\lambda)^2}=\frac{(2x+a)-\lambda^2b}{\lambda^2(1+\lambda)}+\frac{2x}{\lambda(1+\lambda)}\longrightarrow\\[0.3cm]
\frac{2x+a}{\lambda^2}+\frac{\mu}{(1+\lambda)^2}=\frac{(2x+a)}{\lambda^2(1+\lambda)}-\frac{\lambda^2b}{\lambda^2(1+\lambda)}+\frac{2x}{\lambda(1+\lambda)}\\[0.3cm]
\frac{2x+a}{\lambda^2}+\frac{\mu}{(1+\lambda)^2}-\frac{2x}{\lambda(1+\lambda)}-\frac{(2x+a)}{\lambda^2(1+\lambda)}=-\frac{b}{(1+\lambda)}\\[0.3cm]
\frac{(1+\lambda)(2x+a)-2x\lambda-(2x+a)}{\lambda^2(1+\lambda)}+\frac{\mu}{(1+\lambda)^2}=-\frac{b}{(1+\lambda)}\\[0.3cm]
\frac{2x+a+2x\lambda+a\lambda-2x\lambda-2x-a}{\lambda^2(1+\lambda)}+\frac{\mu}{(1+\lambda)^2}=-\frac{b}{(1+\lambda)}\\[0.3cm]
\left.\frac{a\lambda}{\lambda^2(1+\lambda)}+\frac{\mu}{(1+\lambda)^2}=-\frac{b}{(1+\lambda)}\right|\times-(1+\lambda)\\[0.5cm]
\boxed{b=-\frac{a}{\lambda}-\frac{\mu}{(1+\lambda)}} z 2 + μ = 2 ( 1 + λ 1 ) ( x + λ y ) ⟶ ( λ 1 + λ ) 2 ( 2 x + a ) + μ = 2 ⋅ λ 1 + λ ( x + λ y ) ∣ ∣ ÷ ( 1 + λ ) 2 1 λ 2 2 x + a + ( 1 + λ ) 2 μ = λ ( 1 + λ ) 2 ( x + λ y ) ⟶ λ 2 2 x + a + ( 1 + λ ) 2 μ = ( 1 + λ ) 2 y + λ ( 1 + λ ) 2 x ⟶ λ 2 2 x + a + ( 1 + λ ) 2 μ = ( 1 + λ ) λ 2 2 x + a − b + λ ( 1 + λ ) 2 x ⟶ λ 2 2 x + a + ( 1 + λ ) 2 μ = λ 2 ( 1 + λ ) ( 2 x + a ) − λ 2 b + λ ( 1 + λ ) 2 x ⟶ λ 2 2 x + a + ( 1 + λ ) 2 μ = λ 2 ( 1 + λ ) ( 2 x + a ) − λ 2 ( 1 + λ ) λ 2 b + λ ( 1 + λ ) 2 x λ 2 2 x + a + ( 1 + λ ) 2 μ − λ ( 1 + λ ) 2 x − λ 2 ( 1 + λ ) ( 2 x + a ) = − ( 1 + λ ) b λ 2 ( 1 + λ ) ( 1 + λ ) ( 2 x + a ) − 2 x λ − ( 2 x + a ) + ( 1 + λ ) 2 μ = − ( 1 + λ ) b λ 2 ( 1 + λ ) 2 x + a + 2 x λ + aλ − 2 x λ − 2 x − a + ( 1 + λ ) 2 μ = − ( 1 + λ ) b λ 2 ( 1 + λ ) aλ + ( 1 + λ ) 2 μ = − ( 1 + λ ) b ∣ ∣ × − ( 1 + λ ) b = − λ a − ( 1 + λ ) μ
Conclusion,
The complete integral (1ii) is the envelope of one parameter sub-system obtained by taking
b = − a λ − μ ( 1 + λ ) b=-\frac{a}{\lambda}-\frac{\mu}{(1+\lambda)} b = − λ a − ( 1 + λ ) μ
#339914 on Dec 2023
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