Gradient vector for polar coordinates

In summary, the gradients for Cartesian coordinates can be calculated using the chain rule. However, the directional derivative must be converted from polar to Cartesian coordinates.
  • #1
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Homework Statement


Find the gradient vector of:
[tex]g(r, \theta) = e^{-r} sin \theta[/tex]


Homework Equations





The Attempt at a Solution


I know how to get gradients for Cartesian - partially derive the equation of the surface wrt each variable. But I have no idea how to do it for non-Cartesian coordinate systems.

I tried using the chain rule to get [tex]\frac{{\partial g}{\partial x}}[/tex] and [tex]\frac{{\partial g}{\partial y}}[/tex] so I could plug them into [tex]\nabla g[/tex]. But then I'm ending up with tans and sines and cosines all over the place.
 
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  • #2
What coordinate system do you have in mind? Different coordinate systems are set up differently with different basis vectors, for cylindrical/spherical coordinates refer to http://en.wikipedia.org/wiki/Del_in_cylindrical_and_spherical_coordinates

For clearer understanding I think it's best to work through the basis vectors, and then use the definition of gradient to work out the required expressions, which you can then finally (whew!) apply to the eqn you want.
 
  • #3
I want to convert del g into cartesians so I can find a directional derivative, considering the direction vector is in cartesians. I know how to calculate the directional derivative; I just don't know how to convert del g into cartesians from polars.
 
  • #4
Why don't you just convert g into cartesians and take the grad there? It is a bit of a pain in the neck, but...
 
  • #5
Dick said:
Why don't you just convert g into cartesians and take the grad there? It is a bit of a pain in the neck, but...

because I don't know how, but I'll give it a shot.


r = x/cos (theta) = y/sin(theta)
theta = arccos (r/x) = arcsin (r/y)

partial g/partial x = (partial g/partial r)(partial r/partial x) + (partial g/partial theta)(partial theta/partial x)

= (-e^(-r))(1/cos(theta)) + (e^(-r) cos(theta))(-r ln x/sqrt(1-(r/x)^2))


partial g/partial y = (-e^(-r))(1/sin(theta)) + (e^(-r) cos(theta))(r ln y/sqrt(1-(r/y)^2))
 
  • #6
I think you're making it even more complicated for yourself.

What's r? sqrt of r^2, yes? Now what's r^2? And then what's sin (theta)?

Draw it out if you don't see what I mean!
 
  • #7
ahh. of course.

r^2 = x^2 + y^2, and sin theta = y/r, where r is the square root of r^2. I indeed was making it too complicated for myself. thanks everyone.
 

1. What is the gradient vector for polar coordinates?

The gradient vector for polar coordinates is a vector that represents the direction and magnitude of the steepest increase in a function at a given point. It is defined as the partial derivative of the function with respect to each coordinate in the polar coordinate system.

2. How is the gradient vector calculated for a function in polar coordinates?

The gradient vector in polar coordinates is calculated by taking the partial derivatives of the function with respect to the radial and angular coordinates, and then converting these derivatives to Cartesian coordinates using the chain rule.

3. What is the physical interpretation of the gradient vector in polar coordinates?

In polar coordinates, the gradient vector represents the direction and rate of change of a physical quantity, such as temperature or pressure, at a given point. It can also be thought of as the direction in which an object will move if it is subjected to a force or gradient field.

4. Can the gradient vector change direction in polar coordinates?

Yes, the gradient vector can change direction in polar coordinates. This is because the direction of the gradient vector is always perpendicular to the level curves of the function, and the shape of the level curves can vary in polar coordinates.

5. How is the gradient vector used in polar coordinates?

The gradient vector in polar coordinates is used in many applications, such as optimization problems, fluid dynamics, and electromagnetism. It can also be used to determine the direction and rate of change of a physical quantity, as well as to calculate the directional derivative of a function in a specific direction.

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