Need Help With Gradient (Spherical Coordinates)

Click For Summary
SUMMARY

The gradient of the function f(r) = rcos(θ) in spherical coordinates is calculated using the formula ∇f = ∂f/∂r ȧr + (1/r) ∂f/∂θ ȧθ + (1/(rsinθ)) ∂f/∂φ ȧφ. The partial derivatives are determined as follows: ∂f/∂r = cosθ, ∂f/∂θ = -rsinθ, and ∂f/∂φ = 0. Consequently, the gradient is expressed as cosθ ȧr - sinθ ȧθ. This confirms the correct application of spherical coordinates in gradient calculations.

PREREQUISITES
  • Understanding of spherical coordinates
  • Knowledge of partial derivatives
  • Familiarity with vector calculus
  • Ability to apply gradient formulas
NEXT STEPS
  • Study the derivation of the gradient in spherical coordinates
  • Learn about vector fields and their properties
  • Explore applications of gradients in physics and engineering
  • Practice calculating gradients for various functions in different coordinate systems
USEFUL FOR

Students studying multivariable calculus, physicists working with vector fields, and engineers applying spherical coordinates in their analyses.

Dopplershift
Messages
58
Reaction score
9

Homework Statement


Find te gradient of the following function f(r) = rcos(##\theta##) in spherical coordinates.

Homework Equations


\begin{equation}
\nabla f = \frac{\partial f}{\partial r} \hat{r} + (\frac{1}{r}) \frac{\partial f}{\partial \theta} \hat{\theta} + \frac{1}{rsin\theta} \frac{\partial f}{\partial \phi} \hat{\phi}
\end{equation}

The Attempt at a Solution


I know that z = rcos##\theta## But I don't know where to go from there since I don't see any ##\hat{r}##, ##\hat{\theta}##, ##\hat{\phi}##
 
Physics news on Phys.org
Do you know how to calculate ## \frac{\partial f}{\partial r} ##,## \frac{\partial f}{\partial \theta} ## and ## \frac{\partial f}{\partial \phi} ##?
 
ShayanJ said:
Do you know how to calculate ## \frac{\partial f}{\partial r} ##,## \frac{\partial f}{\partial \theta} ## and ## \frac{\partial f}{\partial \phi} ##?

if you're asking if I know how to take partial derivatives, then yes. The issue lies in I don't know where to begin since there is ##\hat{r}##, ##\hat{\theta}##, ##\hat{\phi}## in the equation.
 
Its just like gradient in the Cartesian coordinates. You calculate ## \frac{\partial f}{\partial x} ##,etc. and assume they are components of a vector field. So you have ## \vec \nabla f=(\frac{\partial f}{\partial x},\frac{\partial f}{\partial y},\frac{\partial f}{\partial z})=\frac{\partial f}{\partial x}\hat x+\frac{\partial f}{\partial y}\hat y+\frac{\partial f}{\partial z}\hat z ##.
What is so different about spherical coordinates?
 
So would it be:

\begin{equation}
\begin{split}
f(r) = rcos(\theta) \\
\frac{\partial f}{\partial r} = cos\theta \\
\frac{\partial f}{\partial \theta} = - rsin \theta \\
\frac{\partial f}{\partial \phi} = 0
\end{split}
\end{equation}
So the gradient is
\begin{equation}
cos\theta \ \hat{r} + -sin\theta \ \hat{\theta}
\end{equation}

Is this correct?
 
ShayanJ said:
Yes, that's correct.

Thank you, very much! :)
 

Similar threads

  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 9 ·
Replies
9
Views
4K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 19 ·
Replies
19
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
2
Views
2K
Replies
2
Views
3K
Replies
13
Views
3K
Replies
29
Views
2K
  • · Replies 3 ·
Replies
3
Views
4K