# Expressing velocity in polar coordiantes

My question has to with an excerpt from my textbook:

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And I'm completely unaware as to how this was the derivative of vector $r$. I'm also unaware as to how this is derived from the chain rule.

I'm aware that finding the derivative of vectors is a different process from that of normal real-valued functions. In fact, the textbook addresses it here:

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Explaining the first identity to me would be helpful, but if it's too involved to explain here or against the taste of this forum to just ask people to explain things so lazily like this, a link involving some helpful explanation would be much appreciated, since my searching has not been helpful as the vector calculus identities are a bit hard for me to understand.

Here is my attempt differentiating $\frac{d}{dt}(c \vec A)$:

After this, I don't know what to do next. I could take the scalars $a, b$ and $c$ from the vectors but I don't know how to differentiate unit vectors.

• It is not really different. If it helps, think of the three components of the vector as separate functions. Apply the chain rule to each and then just observe that the three equations you get are the three components of the vector equation. Sep 25, 2017 at 18:15
• The first identity is just the product rule Sep 25, 2017 at 18:17
• I'm going to attach my attempt at using the product rule to better understand this here, and I'll edit it with my attempt if I run into any problems! :) Sep 25, 2017 at 18:18
• I've attached my attempt at differentiating. I don't see how the product rule applies here. Sep 25, 2017 at 18:23
• Note that the product rule is a special case of the chain rule. Most of us don't think of the product rule as being the chain rule, but evidently the author does. Sep 25, 2017 at 18:28

As others have said, the example that is tripping you up is the product rule:

$$\frac{d}{dt}(c \vec A) = \frac{dc}{dt}\vec A + c \frac{d\vec A}{dt}$$

where $c$ is a time-dependent scalar and $\vec A$ is a time-dependent vector.

The real question appears to relate to expressing the velocity of an object in different coordinate systems. $\frac{d\vec r}{dt}$ is a geometric object, and isn't attached to any particular coordinates. However, we could express it using whatever coordinates we want.

$\textbf{Cartesian Coordinates}$

In Cartesian coordinates,

$$\vec r = x \hat e_x + y \hat e_y$$

The unit vectors $\hat e_x$ and $\hat e_y$ are constant everywhere, so

$$\frac{d\hat e_x}{dt} = \frac{d\hat e_y}{dt} = 0$$

Therefore, $$\frac{d\vec r}{dt} = \frac{d x}{dt} \hat e_x + \frac{d y}{dt} \hat e_y$$

$\textbf{Polar Coordinates}$

On the other hand, in polar coordinates,

$$\vec r = R \hat e_R$$

It's crucial to note that unlike the Cartesian unit vectors, the unit vector $\hat e_R$ is not constant. In general, the object will be moving around, and the radial unit vector points in different directions at different points. Therefore,

$$\frac{d\vec r}{dt} = \frac{dR}{dt} \hat e_R + R \frac{d \hat e_R}{dt}$$

If you do a bit of thinking, you can figure out that $\hat e_R$ is actually a function of $\theta$, and that

$$\frac{d \hat e_R}{dt} = \frac{d\theta}{dt} \hat e_\theta$$

This follows from the fact that you can express $$\hat e_R = \cos(\theta) \hat e_x + \sin(\theta) \hat e_y$$ $$\hat e_\theta = \cos(\theta) \hat e_y - \sin(\theta) \hat e_x$$ Try differentiating $\hat e_R$ with respect to time, and you'll see this explicitly.

So, putting it all together, expressed in polar coordinates, we have that

$$\frac{d\vec r}{dt} = \frac{dR}{dt} \hat e_R + R \frac{d\theta}{dt} \hat e_\theta$$

• Thanks so much for this. And, to be clear, $R$ here, is it a vector, or a time-dependent scalar? Sep 26, 2017 at 13:50
• $R$ is the distance between the object and the coordinate origin, and is therefore a scalar. However, it's important to recognize that in most coordinate systems, the components of a vector are not scalars. This is where the idea that scalars are simply things without a direction attached breaks down. Sep 26, 2017 at 14:23
• Which is why the derivative of $R$ is not merely zero? It can be considered a time-dependent scalar? Sep 26, 2017 at 14:42
• Whether a quantity is time-dependent and whether it is a scalar, a vector, or something else are two completely unrelated questions. You seem to think that they are related, when they are emphatically not. Sep 26, 2017 at 14:44
• Well I'm asking because the product rule would not be used if $R$ was not some function, so I'm trying to work out what that means since we are in fact using it. I know you still could use the product rule if it was a scalar, but it will differentiate to $0$ and you'll be left with the scalar times the derivative of the function. I'm a bit confused with your explanation about how in some coordinate systems the components of a vector are not scalars, but I'm just hoping I'm right in saying $R$ here is some function with time that outputs the distance from the origin to an object. Sep 26, 2017 at 14:52

Well, basically, derivatives of vector work in exactly the same way. The reason is just that it's only a convention: we call

$\frac{d\vec{A}}{dt} := \left(\frac{dA_x}{dt}, \frac{dA_y}{dt} , \frac{dA_z}{dt}\right)$

Which means, we state that "the derivative of a vector is the vector of the derivatives of its components".

Once this is accepted (check it carefully), let's go to your problem. We should actually separate each coordinate and do it one by one:

$\frac{d}{dt}(c\cdot \vec{A})\longrightarrow \frac{d}{dt}(c\cdot A_x)$

Then it is obviously the derivative of a product:

$\frac{dc}{dt}\cdot A_x + c\cdot \frac{dA_x}{dt}$

And the same for the rest of the components.

What we say is, well... since they are not mixing with each other, we can rejoin them:

$\frac{dc}{dt}\cdot A_x + c\cdot \frac{dA_x}{dt}$

$\frac{dc}{dt}\cdot A_y + c\cdot \frac{dA_y}{dt}$

$\frac{dc}{dt}\cdot A_z + c\cdot \frac{dA_z}{dt}$

$\frac{dc}{dt}\cdot (A_x, A_y, A_z) + c\cdot \left(\frac{dA_x}{dt}, \frac{dA_y}{dt} , \frac{dA_z}{dt}\right)$

and so

$\frac{d}{dt} (c\cdot \vec{A})= \frac{dc}{dt} \vec{A}+ c\frac{d\vec{A}}{dt}$

So there is it.