Visualizing Vector Calculus: Sketching Field Lines with Scalar Functions

MrB3nn
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Homework Statement


Let r be a position vector from the origin (r=xi+yj+zk), whose magnitude is r, and let f(r) be a scalar function of r. Sketch the field lines of f(r)r

2. Homework Equations
1 \nablax(\nabla\Psi)=0
2 \nabla.(\nablaxv)=0
3 \nablax(\nablaxv)=\nabla(\nabla.v)-\nabla^{}2v
4 \nabla.(\Psiv)=\Psi\nabla.v+v.\nabla\Psi
5 \nablax(\Psiv)=\Psi\nablaxv+(\nabla\Psi)xv
6 \nabla.(v.w=w.(\nablaxv)-v.(\nablaxw)
7 \nablax(vxw=v(\nabla.w-w(\nabla.v+(w.\nabla)v-(v.\nabla)w


The Attempt at a Solution


I can't get started on this question. I have no idea how you can draw a sketch of the field lines when the scalar function is unknown. My intuition says you should be able to use some of those identities but I need a push in the right direction. Please, someone give me that.
 
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The field lines represent the direction of the vector field.

Hint: what is the direction of your vector field, and in what way does it depend on f(r)? (trick question)
 
oh didn't realize this had posted, the site was crashing when I was trying to post. Well I know the field is always radial from the origin due to it's cartesian components. I also know the magnitude of the field will increase as you move in any direction from the origin. But in my mind, I don't know what f(r) is. I am thinking that f(r) can be negative sometimes i.e. f(r)=5-er, or negative always i.e. f(r)=r. Therefore, in my mind, if I multiply this scalar field with the vector field, it the combined effect could change the whole thing. In some places the field may converge, in some it may diverge etc. What I mean is, the field lines may reverse in some places. Am I thinking of this the right way?
 
Yes, that is correct. You might have points where the vector field vanishes, diverges or reverses (i.e. points to the origin). But this is all legit :)

By the way, this forum supports latex, which is a lot easier to use than the equation you wrote. Might want to look into it!

Example:

\nabla \times (\nabla\Psi) = 0

(Press the quote button to see how you can write it)
 
Thanks a lot for your help dude.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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