Finding level curves in relation to gradient vectors

[The1TL]

Homework Statement

Suppose f:R^2 - {0} → R is a differentiable function whose gradient is nowhere 0 and that satisfies -y(df/dx) + x(df/dy) = 0 everywhere.

a) find the level curves of f

b) Show that there is a differentiable function F defined on the set of positive real numbers so that f(x) = F(||x||)

Homework Equations

The Attempt at a Solution

a) I know that gradient vectors are orthogonal to level curves. So the fact that -y(df/dx) + x(df/dy) = 0 seems to show that the gradient vector is orthogonal to any vector of the form (x, -y). So would all vectors of this form be the level curves?

b)could I just show that F is a function that is the same as f but multiplies results that would be negative by -1?

 

[HallsofIvy]

Homework Statement

Suppose f:R^2 - {0} → R is a differentiable function whose gradient is nowhere 0 and that satisfies -y(df/dx) + x(df/dy) = 0 everywhere.

a) find the level curves of f

b) Show that there is a differentiable function F defined on the set of positive real numbers so that f(x) = F(||x||)

Homework Equations

The Attempt at a Solution

a) I know that gradient vectors are orthogonal to level curves. So the fact that -y(df/dx) + x(df/dy) = 0 seems to show that the gradient vector is orthogonal to any vector of the form (x, -y). So would all vectors of this form be the level curves?

Yes, that's exactly what it means. And that, in turn, means that dy/dx= -y/x.

b)could I just show that F is a function that is the same as f but multiplies results that would be negative by -1?

I think the (poor) notation is confusing you. f is a function on R² so f(x) makes no sense if x is a number. While x and y are numerical values in (a), the fact that f is defined on R² and the use of ||x||, rather than |x|, makes me think that, here, x is the vector in R² which woud be (x, y) in terms of (a).

 

[The1TL]

oh ok, so for b) is there some way that I could show that F exists based on the -y(df/dx) + x(df/dy) = 0 equation?
Or would it simply involve turning sqrt(x^2 + y^2) into an R^2 vector?

 

[The1TL]

actually, I've noticed that the function g(x) = sqrt(x^2 + y^2) has the property that -y(df/dx) + x(df/dy) = 0. Therefore I could let F be h(g(x)) where h is the identity function.