Need Help With Gradient (Spherical Coordinates)

In summary, the gradient of the function f(r) = rcos(##\theta##) in spherical coordinates is given by: \begin{equation}\vec \nabla f = cos\theta \ \hat{r} + -sin\theta \ \hat{\theta}\end{equation}This is calculated by taking the partial derivatives with respect to r, ##\theta##, and ##\phi##, assuming they are components of a vector field. The only difference from Cartesian coordinates is the inclusion of the unit vectors ##\hat{r}## and ##\hat{\theta}##.
  • #1
Dopplershift
59
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}##
 
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  • #2
Do you know how to calculate ## \frac{\partial f}{\partial r} ##,## \frac{\partial f}{\partial \theta} ## and ## \frac{\partial f}{\partial \phi} ##?
 
  • #3
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.
 
  • #4
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?
 
  • #5
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?
 
  • #6
Yes, that's correct.
 
  • #7
ShayanJ said:
Yes, that's correct.

Thank you, very much! :)
 

Related to Need Help With Gradient (Spherical Coordinates)

1. What are gradient and spherical coordinates?

The gradient is a mathematical operation that describes the rate of change of a function at a given point. Spherical coordinates are a coordinate system used to describe points in three-dimensional space using angles and a distance from the origin.

2. How do you calculate gradient in spherical coordinates?

In spherical coordinates, the gradient is calculated using the partial derivatives of the function with respect to the spherical coordinates. The formula is:

∇f = (∂f/∂r)er + (1/r)(∂f/∂θ)eθ + (1/rsinθ)(∂f/∂φ)eφ

where r, θ, and φ are the spherical coordinates, and er, eθ, and eφ are the unit vectors in the radial, polar, and azimuthal directions, respectively.

3. What is the significance of gradient in spherical coordinates?

The gradient in spherical coordinates is important in various fields of science, especially in physics and engineering. It helps in calculating the direction and magnitude of the steepest slope of a function at a given point, which is useful in understanding the behavior of physical systems and optimizing designs.

4. Are there any applications of gradient in spherical coordinates?

Yes, there are many applications of gradient in spherical coordinates. Some examples include calculating the electric and magnetic fields of charged particles, analyzing fluid flow in pipes and channels, and determining the temperature distribution in a spherical object.

5. Can you provide an example of calculating gradient in spherical coordinates?

Sure, let's say we have a function f(r, θ, φ) = r2sin(θ)cos(φ). To calculate its gradient at the point (2, π/4, π/6), we first find the partial derivatives:

∂f/∂r = 2r
∂f/∂θ = 2rsin(φ)
∂f/∂φ = -2rsin(θ)sin(φ)

Substituting these values into the formula, we get:

∇f = (4)er + (1/2)(2√2)eθ + (1/2)(-√2)eφ

Therefore, the gradient at (2, π/4, π/6) is ∇f = 4er + √2eθ - √2eφ.

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