# Vector Calculus

1. Mar 21, 2009

### MrB3nn

1. The problem statement, all variables and givenknown data
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. Relevant equations
1 $$\nabla$$x($$\nabla$$$$\Psi$$)=0
2 $$\nabla$$.($$\nabla$$xv)=0
3 $$\nabla$$x($$\nabla$$xv)=$$\nabla$$($$\nabla$$.v)-$$\nabla$$$$\^{}2$$v
4 $$\nabla$$.($$\Psi$$v)=$$\Psi$$$$\nabla$$.v+v.$$\nabla$$$$\Psi$$

3. 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. I hope someone can give me that.

2. Mar 21, 2009

### MrB3nn

More relevant equations (the site was crashing when I tried to put them all into one post):
5 $$\nabla$$x($$\Psi$$v)=$$\Psi$$$$\nabla$$xv+($$\nabla$$$$\Psi$$)xv
6 $$\nabla$$.(v.w)=w.($$\nabla$$xv)-v.($$\nabla$$xw)
7 $$\nabla$$x(vxw)=v($$\nabla$$.w)-w($$\nabla$$.v)+(w.$$\nabla$$)v-(v.$$\nabla$$)w

3. Mar 21, 2009

### gabbagabbahey

Why not start by computing the divergence and curl of f(r)r....

4. Mar 21, 2009

### MrB3nn

Ok, I did that using the correct formulae.
Div:
3f(r)+x$$\partial$$f(r)/$$\partial$$x+y$$\partial$$f(r)/$$\partial$$y+z$$\partial$$f(r)/$$\partial$$z
Curl:
(z$$\partial$$f(r)/$$\partial$$y-y$$\partial$$f(r)/$$\partial$$z)i+(x$$\partial$$f(r)/$$\partial$$z-z$$\partial$$f(r)/$$\partial$$x)j+(y$$\partial$$f(r)/$$\partial$$x-x$$\partial$$f(r)/$$\partial$$y)k

I really can't see how that helps since I don't know f(r), and f(r) could be positive or negative. I know that the divergence of a vector field is the (flux of an infinitessimal box/unit volume) placed at a point in the field but I don't know how that helps. I don't know what curl is exactly so I don't know how that helps me visualise the field.

5. Mar 21, 2009

### gabbagabbahey

Yuck!:yuck:...since f is a function only of r, use spherical coordinates to find the div and curl instead......remember, $$\frac{\partial f(r)}{\partial{r}}=f'(r)$$

6. Mar 21, 2009

### MrB3nn

When I switch to spherical polars, r becomes r=r$$\widehat{r}$$ and grad changes appropriately.
For the div I get:
Div = f(r) + rf'(r)
and for the curl I get zero.
I still cant see what this means. The div could be positive or negative depending on the actual function.

7. Mar 21, 2009

### gabbagabbahey

The curl is zero, which tells you that the field lines don't 'rotate'....You are correct that Div (f(r)r) can be pos/neg or zero, so why not sketch and label all 3 cases?

8. Mar 21, 2009

### MrB3nn

Ok I see. Thanks very much for your consistent help.