Deriving a forumla for the gradient in cylindrical coordinates

In summary, the gradient of a scalar function in cartesian coordinates is given by \frac{\partial T}{\partial x}\hat{x}+\frac{\partial T}{\partial y}\hat{y}+\frac{\partial T}{\partial z}\hat{z} and the gradient of a scalar function in cylindrical coordinates is given by \frac{\partial T}{\partial s}\hat{s}+\frac{\partial T}{\partial ϕ}\hat{ϕ}+\frac{\partial T}{\partial z}\hat{
  • #1
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Problem:

Starting from the gradient of a scalar function T(x,y,z) in cartesian coordinates find the formula for the gradient of T(s,ϕ,z) in cylindrical coordinates.

Solution (so far):

I know that the gradient is given by [itex] \nabla T = \frac{\partial T}{\partial x}\hat{x}+\frac{\partial T}{\partial y}\hat{y}+\frac{\partial T}{\partial z}\hat{z}[/itex]. We must use the chain rule, so we have [itex] \frac{\partial T}{\partial x}(\frac{\partial z}{\partial x})=\frac{\partial T}{\partial s}(\frac{\partial s}{\partial x})+\frac{\partial T}{\partial ϕ}(\frac{\partial ϕ}{\partial x})+\frac{\partial T}{\partial z}(\frac{\partial z}{\partial x}) [/itex], [itex] \frac{\partial T}{\partial y}=\frac{\partial T}{\partial s}(\frac{\partial s}{\partial y})+\frac{\partial T}{\partial ϕ}(\frac{\partial ϕ}{\partial y})+\frac{\partial T}{\partial z}(\frac{\partial z}{\partial y}) [/itex], and [itex] \frac{\partial T}{\partial x}=\frac{\partial T}{\partial s}(\frac{\partial s}{\partial z})+\frac{\partial T}{\partial ϕ}(\frac{\partial ϕ}{\partial z})+\frac{\partial T}{\partial z}(\frac{\partial z}{\partial z}) [/itex].

So then I went and calculated ##\hat{x}## and ##\hat{y}## in terms of ##\hat{s}## and ##\hat{ϕ}##. I got ##\hat{x}=cosϕ\hat{s}-sinϕ\hat{ϕ}## and ##\hat{y}=sinϕ\hat{s}+cosϕ\hat{ϕ}##. Note that ##\hat{x} \cdot \hat{x} =1##, ##\hat{y} \cdot \hat{y}=1##, and ##\hat{x} \cdot \hat{y} =0## - as they should.

Since I need partials of s and ϕ with respect to x and y, I found the inversion formulas ##s=\sqrt{x^2+y^2}## and ##ϕ=sin^{1}(\frac{y}{\sqrt{x^2+y^2}})##. (I used x=s cosϕ and y=s sinϕ)

Obviously ##\frac{\partial ϕ}{\partial z}=0## and ##\frac{\partial s}{\partial z} =0##, but
I'm concerned with how complicated ##\frac{\partial ϕ}{\partial x}## and ##\frac{\partial ϕ}{\partial y}## are going to turn out.

Am I on the right track? Thanks in advance.
 
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  • #2
Actually, after differentiating and putting it in terms of ϕ, I got ## \frac{\partial ϕ}{\partial x}=-\frac{1}{s}sinϕ## and ##\frac{\partial ϕ}{\partial y}=\frac{1}{s}cosϕ## - That's not too bad. I'd still like to know if this is the correct approach, though.
 
  • #3
Any input is appreciated!
 
  • #4
Yup, keep going.
 
  • #5
It worked! Praise the Lord!
 

What are cylindrical coordinates and how are they different from Cartesian coordinates?

Cylindrical coordinates are a type of three-dimensional coordinate system commonly used in mathematics and physics. They consist of a radius, an angle, and a height, and are often denoted as (r, θ, z). They differ from Cartesian coordinates in that they use a polar angle (θ) instead of the traditional x, y, and z axes.

Why is it important to derive a formula for the gradient in cylindrical coordinates?

The gradient is an important mathematical tool used to measure the rate and direction of change of a function. By deriving a formula for the gradient in cylindrical coordinates, we can solve problems involving curved surfaces and cylindrical objects, which is essential in many fields such as engineering and physics.

What is the process for deriving a formula for the gradient in cylindrical coordinates?

The formula for the gradient in cylindrical coordinates can be derived using vector calculus techniques. It involves converting the cylindrical coordinates into Cartesian coordinates, finding the partial derivatives of the function with respect to each coordinate, and then using vector operations to calculate the gradient.

Can the formula for the gradient in cylindrical coordinates be applied to any function?

Yes, the formula for the gradient in cylindrical coordinates can be applied to any scalar field, which is a function that assigns a scalar value to each point in space. It is a general formula that can be used to find the gradient of any function in cylindrical coordinates.

Are there any limitations to using the formula for the gradient in cylindrical coordinates?

The formula for the gradient in cylindrical coordinates is limited to only working with scalar fields. It cannot be used with vector fields, which are functions that assign a vector value to each point in space. Additionally, the formula may become more complex and difficult to use for higher-dimensional problems.

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