In velocity we can write
u = ∂ ψ ∂ y , v = − ∂ ψ ∂ x u=\dfrac{\partial\psi}{\partial y}, v=-\dfrac{\partial\psi}{\partial x} u = ∂ y ∂ ψ , v = − ∂ x ∂ ψ Given ψ = 8 x y \psi=8xy ψ = 8 x y
u = ∂ ψ ∂ y = 8 x , v = − ∂ ψ ∂ x = − 8 y u=\dfrac{\partial\psi}{\partial y}=8x, v=-\dfrac{\partial\psi}{\partial x}=-8y u = ∂ y ∂ ψ = 8 x , v = − ∂ x ∂ ψ = − 8 y Calculate the velocity at the point p(4, 5)
u = 32 , v = − 40 u=32, v=-40 u = 32 , v = − 40 ∣ v e l o c i t y ∣ = u 2 + v 2 = ( 32 ) 2 + ( − 40 ) 2 = 8 41 |velocity|=\sqrt{u^2+v^2 }=\sqrt{(32)^2+(-40)^2}=8\sqrt{41} ∣ v e l oc i t y ∣ = u 2 + v 2 = ( 32 ) 2 + ( − 40 ) 2 = 8 41 u = ∂ φ ∂ x , v = ∂ φ ∂ y u=\dfrac{\partial\varphi}{\partial x}, v=\frac{\partial\varphi}{\partial y} u = ∂ x ∂ φ , v = ∂ y ∂ φ u = ∂ φ ∂ x = 8 x u=\dfrac{\partial\varphi}{\partial x}=8x u = ∂ x ∂ φ = 8 x Integrate with respect to x x x
φ = 4 x 2 + g ( y ) \varphi=4x^2+g(y) φ = 4 x 2 + g ( y ) ∂ φ ∂ y = d g d y = − 8 y \frac{\partial\varphi}{\partial y}=\frac{dg}{d y}=-8y ∂ y ∂ φ = d y d g = − 8 y Integrate with respect to y y y
g ( y ) = − 4 y 2 + C g(y)=-4y^2+C g ( y ) = − 4 y 2 + C Then
φ ( x , y ) = 4 x 2 − 4 y 2 + C \varphi(x,y)=4x^2-4y^2+C φ ( x , y ) = 4 x 2 − 4 y 2 + C
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