Expressing velocity in polar coordiantes

Question

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.

Answer 1

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.

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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

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

Answer 2

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.