Proof: Everywhere Tangent to Curve?

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The discussion centers on proving that curves defined by the function psi(x,y) are everywhere tangent to the vector field v(x,y). It is established that the gradient of psi, denoted as ∇ψ, is normal to the surfaces of constant psi, leading to the conclusion that v·∇ψ = 0. This implies that the vector field v is perpendicular to the gradient of psi, confirming that the curves are indeed tangent to v. The relationship between v and psi is further clarified by recognizing that if v is derived from the curl of psi, it reinforces the concept of psi as a stream function. Ultimately, the proof hinges on the orthogonality of v and ∇ψ, establishing the tangential relationship.
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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.
 
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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.
 
DavidWhitbeck said:
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?
 
Are you trying to curl a scalar field??
 
oh right...i overlooked that part. thanks!
 
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
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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