Answer on Question #76041 – Math – Differential Equations Question
Find the surface which is orthogonal to the one parameter system
z = c x y ( x 2 + y 2 ) z = c x y (x ^ {2} + y ^ {2}) z = c x y ( x 2 + y 2 )
and which passes through the hyperbola
x 2 − y 2 = a , z = 0 x ^ {2} - y ^ {2} = a, z = 0 x 2 − y 2 = a , z = 0 Solution
f ( x , y , z ) = z x y ( x 2 + y 2 ) = c f (x, y, z) = \frac {z}{x y (x ^ {2} + y ^ {2})} = c f ( x , y , z ) = x y ( x 2 + y 2 ) z = c ∂ f ∂ x = − z y 3 x 2 + y 2 x 2 ( x 2 + y 2 ) 2 \frac {\partial f}{\partial x} = - \frac {z}{y} \frac {3 x ^ {2} + y ^ {2}}{x ^ {2} (x ^ {2} + y ^ {2}) ^ {2}} ∂ x ∂ f = − y z x 2 ( x 2 + y 2 ) 2 3 x 2 + y 2 ∂ f ∂ y = − z x 3 y 2 + x 2 y 2 ( x 2 + y 2 ) 2 \frac {\partial f}{\partial y} = - \frac {z}{x} \frac {3 y ^ {2} + x ^ {2}}{y ^ {2} (x ^ {2} + y ^ {2}) ^ {2}} ∂ y ∂ f = − x z y 2 ( x 2 + y 2 ) 2 3 y 2 + x 2 ∂ f ∂ z = 1 x y ( x 2 + y 2 ) \frac {\partial f}{\partial z} = \frac {1}{x y (x ^ {2} + y ^ {2})} ∂ z ∂ f = x y ( x 2 + y 2 ) 1
Let z ( x , y ) z(x, y) z ( x , y )
( f x , f y , f z ) ⋅ ( z x , z y , − 1 ) = 0 \left(f _ {x}, f _ {y}, f _ {z}\right) \cdot \left(z _ {x}, z _ {y}, - 1\right) = 0 ( f x , f y , f z ) ⋅ ( z x , z y , − 1 ) = 0
So, we have differential equation:
− z y 3 x 2 + y 2 x 2 ( x 2 + y 2 ) 2 z x − z x 3 y 2 + x 2 y 2 ( x 2 + y 2 ) 2 z y − 1 x y ( x 2 + y 2 ) = 0 - \frac {z}{y} \frac {3 x ^ {2} + y ^ {2}}{x ^ {2} (x ^ {2} + y ^ {2}) ^ {2}} z _ {x} - \frac {z}{x} \frac {3 y ^ {2} + x ^ {2}}{y ^ {2} (x ^ {2} + y ^ {2}) ^ {2}} z _ {y} - \frac {1}{x y (x ^ {2} + y ^ {2})} = 0 − y z x 2 ( x 2 + y 2 ) 2 3 x 2 + y 2 z x − x z y 2 ( x 2 + y 2 ) 2 3 y 2 + x 2 z y − x y ( x 2 + y 2 ) 1 = 0
Divide by − z x y ( x 2 + y 2 ) -\frac{z}{xy(x^2 + y^2)} − x y ( x 2 + y 2 ) z
3 x 2 + y 2 x ( x 2 + y 2 ) z x + 3 y 2 + x 2 y ( x 2 + y 2 ) z y + 1 z = 0 \frac {3 x ^ {2} + y ^ {2}}{x (x ^ {2} + y ^ {2})} z _ {x} + \frac {3 y ^ {2} + x ^ {2}}{y (x ^ {2} + y ^ {2})} z _ {y} + \frac {1}{z} = 0 x ( x 2 + y 2 ) 3 x 2 + y 2 z x + y ( x 2 + y 2 ) 3 y 2 + x 2 z y + z 1 = 0
The auxiliary equations:
x ( x 2 + y 2 ) d x 3 x 2 + y 2 = y ( x 2 + y 2 ) d y 3 y 2 + x 2 = − z d z \frac {x (x ^ {2} + y ^ {2}) d x}{3 x ^ {2} + y ^ {2}} = \frac {y (x ^ {2} + y ^ {2}) d y}{3 y ^ {2} + x ^ {2}} = - z d z 3 x 2 + y 2 x ( x 2 + y 2 ) d x = 3 y 2 + x 2 y ( x 2 + y 2 ) d y = − z d z
Adding the first and the second equations and equating to the third one:
( x 2 + y 2 ) ( x d x + y d y ) 4 ( x 2 + y 2 ) = ( x d x + y d y ) 4 = − z d z \frac{(x^2 + y^2)(xdx + ydy)}{4(x^2 + y^2)} = \frac{(xdx + ydy)}{4} = -zdz 4 ( x 2 + y 2 ) ( x 2 + y 2 ) ( x d x + y d y ) = 4 ( x d x + y d y ) = − z d z x 2 + y 2 + 4 z 2 = c 1 x^2 + y^2 + 4z^2 = c_1 x 2 + y 2 + 4 z 2 = c 1
Subtracting the second equation from the first one:
( x 2 + y 2 ) ( x d x − y d y ) 2 ( x 2 − y 2 ) = ( c 1 − 4 z 2 ) ( x d x − y d y ) 2 ( x 2 − y 2 ) = − z d z \frac{(x^2 + y^2)(xdx - ydy)}{2(x^2 - y^2)} = \frac{(c_1 - 4z^2)(xdx - ydy)}{2(x^2 - y^2)} = -zdz 2 ( x 2 − y 2 ) ( x 2 + y 2 ) ( x d x − y d y ) = 2 ( x 2 − y 2 ) ( c 1 − 4 z 2 ) ( x d x − y d y ) = − z d z x 2 − y 2 = c 2 c 1 − 4 z 2 x^2 - y^2 = c_2 \sqrt{c_1 - 4z^2} x 2 − y 2 = c 2 c 1 − 4 z 2 c 2 = x 2 − y 2 x 2 + y 2 c_2 = \frac{x^2 - y^2}{\sqrt{x^2 + y^2}} c 2 = x 2 + y 2 x 2 − y 2
Using the given conditions x 2 − y 2 = a , z = 0 x^2 - y^2 = a, z = 0 x 2 − y 2 = a , z = 0
a 2 = c 2 2 c 1 a^2 = c_2^2 c_1 a 2 = c 2 2 c 1
Then:
a 2 = ( x 2 + y 2 + 4 z 2 ) ⋅ ( x 2 − y 2 ) 2 x 2 + y 2 a^2 = (x^2 + y^2 + 4z^2) \cdot \frac{(x^2 - y^2)^2}{x^2 + y^2} a 2 = ( x 2 + y 2 + 4 z 2 ) ⋅ x 2 + y 2 ( x 2 − y 2 ) 2 ( x 2 + y 2 ) a 2 = ( x 2 + y 2 + 4 z 2 ) ( x 2 − y 2 ) 2 (x^2 + y^2)a^2 = (x^2 + y^2 + 4z^2)(x^2 - y^2)^2 ( x 2 + y 2 ) a 2 = ( x 2 + y 2 + 4 z 2 ) ( x 2 − y 2 ) 2 z 2 = ( x 2 + y 2 ) a 2 4 ( x 2 − y 2 ) 2 − x 2 + y 2 4 z^2 = \frac{(x^2 + y^2)a^2}{4(x^2 - y^2)^2} - \frac{x^2 + y^2}{4} z 2 = 4 ( x 2 − y 2 ) 2 ( x 2 + y 2 ) a 2 − 4 x 2 + y 2
Answer:
z = ( x 2 + y 2 ) a 2 4 ( x 2 − y 2 ) 2 − x 2 + y 2 4 z = \sqrt{\frac{(x^2 + y^2)a^2}{4(x^2 - y^2)^2} - \frac{x^2 + y^2}{4}} z = 4 ( x 2 − y 2 ) 2 ( x 2 + y 2 ) a 2 − 4 x 2 + y 2
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#340153 on Dec 2023