Question #149109
For the velocity potential function, φ = x^2 - y^2 . Find the velocity components at the point (4, 5).
1
Expert's answer
2020-12-14T09:35:50-0500

In velocity we can write 


u=φx,v=φyu=\dfrac{\partial\varphi}{\partial x}, v=\frac{\partial\varphi}{\partial y}


Given φ=x2y2\varphi=x^2-y^2

u=φx=2x,v=φy=2yu=\dfrac{\partial\varphi}{\partial x}=2x, v=\dfrac{\partial\varphi}{\partial y}=-2y

Calculate the velocity at the point (4, 5)


u=8,v=10u=8, v=-10

velocity=u2+v2=(8)2+(10)2=241|velocity|=\sqrt{u^2+v^2 }=\sqrt{(8)^2+(-10)^2}=2\sqrt{41}


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