Finding Nabla Operator for f(r) with r = |R|

  • Thread starter Thread starter MCooltA
  • Start date Start date
  • Tags Tags
    Nabla Operator
Click For Summary
To find the Nabla operator for f(r) with r = |R|, where R is a vector defined as R = (x, y, z), the gradient of r is calculated as ∇(r) = (x/r, y/r, z/r). The correct approach to find ∇f(r) involves using the chain rule, resulting in ∇f(r) = (df/dr)∇r, where df/dr is the derivative of f with respect to r. It is emphasized that ∇(∇r) does not yield ∇f(r) since ∇r is a vector and ∇f must be applied to a scalar function. The discussion also raises uncertainty regarding the expression for df/dr and its calculation.
MCooltA
Messages
8
Reaction score
0

Homework Statement


I need to find \nablaf(r). I am given r = |R| where R is a vector, R =(x,y,z). I also have the function f(r) which is a differentiable function of r.

Homework Equations



So i know \nabla(g) = (\partialg/\partialx, \partialg/\partialy, \partialg/\partialz)

The Attempt at a Solution



So I've got;

r=√(x^2 + y^2 + z^2)

\nabla(r) = (2x,2y,2z)

Do i need to apply \nabla operator again to the above \nabla(r)?

Does \nabla(\nabla(r)) equal \nablaf(r)?
 
Physics news on Phys.org
No, of course not. For one thing, \nabla \nabla r makes no sense. \nabla r is a vector function and "\nabla f has to be applied to a scalar function. Also, there is NO "f" in that. You would be saying that nabla f(r) is the same for all functions! You need the chain rule:
\nabla f(r)= \frac{df}{dr}\nabla r
df/dr will be a scalar function aand \nabla f a vector function so this is scalar multiplication.

Or, equivalently, replace r in the function with (x^2+ y^2+ z^2)^{1/2} and find the gradient.

What is f?
 
Thanks for your reply. So far I've got;

\nabla(r) = (1/r) (x,y,z).

And then for df/dr = \partialf_{x}/\partialx + \partialf_{y}/\partialy + \partialf_{z}/\partialz

Im unsure about what df/dr is.
 
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?

Similar threads

  • · Replies 26 ·
Replies
26
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 13 ·
Replies
13
Views
4K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
2
Views
2K
  • · Replies 10 ·
Replies
10
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 23 ·
Replies
23
Views
3K