Need Help With Gradient (Spherical Coordinates)

AI Thread Summary
The discussion focuses on finding the gradient of the function f(r) = rcos(θ) in spherical coordinates. The gradient is calculated using the formula involving partial derivatives with respect to r, θ, and φ. The user correctly identifies the partial derivatives as ∂f/∂r = cos(θ), ∂f/∂θ = -rsin(θ), and ∂f/∂φ = 0. The resulting gradient is expressed as cos(θ) ȧr - sin(θ) ȧθ. The solution is confirmed as correct by another participant in the discussion.
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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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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! :)
 
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