Proof: Everywhere Tangent to Curve?

[bakra904]

Proof: Everywhere Tangent to Curve??

If the function v depends on x and y, v(x,y) and we know there exists some function psi(x,y) such that
vx = partial w.r.t (y) of psi
vy= -(partial w.r.t (x) of psi)

show that the curves psi(x,y) = constant, are everywhere tangent to v.

 

[DavidWhitbeck]

Usually you are supposed to show effort to get there, but I think this is a case where either you get it or you don't.

$$\nabla \psi$$ is normal to surfaces of constant $$\psi$$ and $$v\cdot \nabla \psi = 0$$. Fill in the rest.

 

[bakra904]

Usually you are supposed to show effort to get there, but I think this is a case where either you get it or you don't.

$$\nabla \psi$$ is normal to surfaces of constant $$\psi$$ and $$v\cdot \nabla \psi = 0$$. Fill in the rest.

Thanks a bunch! I'm a new poster and did not know about the effort rule...I had worked on it but did not post what I had worked on.

I was trying to use the fact that if v = $$\nabla \times$$ $$\psi$$,

then that would imply that $$\psi$$ is a stream function, which in cartesian co-ordinates would reduce to:

Vx = $$\frac{\partial\psi}{\partial y}$$ and Vy = - $$\frac{\partial\psi}{\partial x}$$

which is basically what the problem had to begin with. Then, since I know that $$\psi$$ (x,y) is a stream function, doesn't it have to be tangent to v by virtue of the fact that its a streamline?

 

[DavidWhitbeck]

Are you trying to curl a scalar field??

 

[bakra904]

oh right...i overlooked that part. thanks!

 

[bakra904]

so basically $$v. \nabla\psi = 0$$ which proves that $$v$$ and $$\nabla\psi$$ are perpendicular (since their dot product is 0) and so $$\psi$$ must be tangent to $$v$$