In velocity we can write
u = ∂ φ ∂ x , v = ∂ φ ∂ y u=\dfrac{\partial\varphi}{\partial x}, v=\frac{\partial\varphi}{\partial y} u = ∂ x ∂ φ , v = ∂ y ∂ φ
Given φ = 4 x ( 3 y − 4 ) \varphi=4x(3y-4) φ = 4 x ( 3 y − 4 )
u = ∂ φ ∂ x = 12 y − 16 , v = ∂ φ ∂ y = 12 x u=\dfrac{\partial\varphi}{\partial x}=12y-16, v=\dfrac{\partial\varphi}{\partial y}=12x u = ∂ x ∂ φ = 12 y − 16 , v = ∂ y ∂ φ = 12 x Calculate the velocity at the point (2, 3)
u = 20 , v = 24 u=20, v=24 u = 20 , v = 24
∣ v e l o c i t y ∣ = u 2 + v 2 = ( 20 ) 2 + ( 24 ) 2 = 4 61 |velocity|=\sqrt{u^2+v^2 }=\sqrt{(20)^2+(24)^2}=4\sqrt{61} ∣ v e l oc i t y ∣ = u 2 + v 2 = ( 20 ) 2 + ( 24 ) 2 = 4 61
u = ∂ ψ ∂ y , v = − ∂ ψ ∂ x u=\dfrac{\partial\psi}{\partial y}, v=-\dfrac{\partial\psi}{\partial x} u = ∂ y ∂ ψ , v = − ∂ x ∂ ψ
u = ∂ ψ ∂ y = 12 y − 16 u=\dfrac{\partial\psi}{\partial y}=12y-16 u = ∂ y ∂ ψ = 12 y − 16 Integrate with respect to y y y
ψ = 6 y 2 − 16 y + g ( x ) \psi=6y^2-16y+g(x) ψ = 6 y 2 − 16 y + g ( x )
− ∂ ψ ∂ x = − d g d x = 12 x -\frac{\partial\psi}{\partial x}=-\frac{dg}{d x}=12x − ∂ x ∂ ψ = − d x d g = 12 x Integrate with respect to x x x
g ( x ) = − 6 x 2 + C g(x)=-6x^2+C g ( x ) = − 6 x 2 + C Then
ψ ( x , y ) = − 6 x 2 + 6 y 2 − 16 y + C \psi(x,y)=-6x^2+6y^2-16y+C ψ ( x , y ) = − 6 x 2 + 6 y 2 − 16 y + C
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