Partial Derivatives in Cylindrical Coordinate Systems

Previously, we discussed the partial derivative of a function in Cartesian coordinates. Sometimes, it is more convenient to work with other coordinate systems, such as polar or cylindrical coordinates. Hence it is important to understand how partial derivatives work between different coordinate systems.

Table of Contents

• Table of Contents
• Polar/Cylindrical Coordinate Review
• Partial Derivatives in Cylindrical Coordinates
  • Partial Derivatives of r, θ, and z WRT x, y, and z
  • Partial Derivatives of r, θ, and z WRT time
  • Derivatives of Cylindrical Unit Vectors WRT time
  • Partial Derivatives of x, y, and z WRT r, θ, and z
  • Partial Derivatives of f
• Example Problem: Deriving Circular Motion
• Summary

Polar/Cylindrical Coordinate Review

Before we delve into the partial derivative in polar/cylindrical coordinates, let's review what polar/cylindrical coordinates are.

Cylindrical coordinates are a coordinate system that uses a distance from the origin, an angle from the -axis, and a height from the -plane to describe a point in space. Those three values are denoted as , respectively.

On the -plane, the cylindrical coordinates are equivalent to polar coordinates (which only use ). We will use 2D diagrams to illustrate the concepts, but remember that cylindrical coordinates are 3D.

For this text, we will use the following colors and notation to represent the different coordinates:

• , : The -coordinate and unit vector
• , : The -coordinate and unit vector
• , : The distance from the origin
• , : The angle from the -axis, and the tangent unit vector
• , : The height from the -plane, and the unit vector

Consider one point in the -plane with Cartesian coordinates :

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Polar basis vectors	Show Hide
Cartesian basis vectors	Show Hide

Using some trigonometry, we can convert the Cartesian coordinates to cylindrical coordinates :

And to convert back to Cartesian coordinates:

We can compute the basis vectors for cylindrical coordinates in a similar way:

Likewise, the unit vectors in Cartesian coordinates can be expressed in terms of cylindrical coordinates:

With this in mind, let's explore how to compute these partial derivatives in cylindrical coordinates.

Partial Derivatives in Cylindrical Coordinates

Partial Derivatives of r, θ, and z WRT x, y, and z

Consider the partial derivative of with respect to :

Next, consider taking the cosine of :

Therefore:

We can do the same for the partial derivative of with respect to :

For the partial derivative of with respect to , consider that the radius does not change no matter how high or low the point is. Therefore:

We can also compute the partial derivatives of and with respect to , , and . The results are:

Partial Derivatives of r, θ, and z WRT time

Consider a point that moves in space. The position of the point can be expressed as a function of time, , , and . We can compute the partial derivatives of , , and with respect to time.

First, consider the partial derivative of with respect to time:

Next, the partial derivative of with respect to time:

Derivatives of Cylindrical Unit Vectors WRT time

The unit vectors , , and can also be expressed as functions of time. Unlike Cartesian unit vectors, the cylindrical unit vectors are not constant, i.e., they change depending on where the point is in space.

The derivative of with respect to time is:

This result should make intuitive sense, as the derivative of is in the direction of . Consider a point moving in a circle. Take two snapshots of the point at very close times, and draw the vector at both points:

Notice that the change in , as this interval approaches zero, is almost perpendicular to . Additionally, its magnitude gets closer to the length of the arc traced by the point in that time interval, which is just .

Hence we can say that:

Then, the derivative of with respect to time is:

Similarly, the derivative of with respect to time is:

This should make sense as well, since the derivative of is in the direction of :

If you place the two vectors such that their tails are at the same point, the change in is almost perpendicular to :

(Notice that the red vectors have almost identical directions.)

Hence we can say that:

Then, the derivative of with respect to time is:

Partial Derivatives of x, y, and z WRT r, θ, and z

Just like how we computed the partial derivatives of , , and with respect to , , and , we can do the same in reverse.

First, consider the partial derivative of with respect to the cylindrical coordinates:

Next, the partial derivative of with respect to the cylindrical coordinates:

Finally, the partial derivative of with respect to the cylindrical coordinates:

Partial Derivatives of f

With that in mind, consider the partial derivative of a function . First, consider the partial derivative of with respect to the Cartesian coordinates. By using the chain rule, we can express these as:

Since we have already computed the partial derivatives of , , and with respect to the Cartesian coordinates, we can substitute them in:

We can also compute the partial derivatives of with respect to the cylindrical coordinates. The partial derivatives of with respect to the can be computed using the chain rule and Equations and :

Doing the same for the partial derivatives of with respect to :

And finally, the partial derivatives of with respect to :

Example Problem: Deriving Circular Motion

Consider a particle moving in a two-dimensional circle with radius . Let the angle be the angle from the -axis counterclockwise.

1. Derive the and components of the position vector .
2. The position can also be expressed in terms of the cylindrical unit vector in the form . Suppose the particle moves at a constant speed around the circle. Derive an expression for the velocity vector .

To help visualize the problem, plot a circle once again:

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Polar basis vectors	Show Hide
Cartesian basis vectors	Show Hide

1. This is simply the conversion from polar to Cartesian coordinates, as in Equations and :

2. The velocity vector is the derivative of the position vector with respect to time. Given that , we can express as:

   In mechanics, there's a special symbol for the derivative of with respect to time: , known as the angular velocity. Hence we can write:

   The geometric interpretation of this result is similar to the interpretation of , just scaled by .

Summary

That's a lot of partial derivatives, so we can summarize them into a tables:

Partial Derivative of	With Respect to
Partial Derivative of	With Respect to