Basic Time-Derivative Problem

I'm attempting to find velocity in polar coordinates (Kleppner/Kolenkow text). What I've got is:

$$\vec{v} = \frac {d}{dt} (x \hat{i} + y \hat{j})$$
$$= \frac {d}{dt} (rcos \theta) + \frac {d}{dt} (rsin \theta)$$
$$= r \frac {d}{dt} (cos \theta) + cos \theta \frac {d}{dt} (r) + r \frac {d}{dt} (sin \theta) + sin \theta \frac {d}{dt} (r)$$

From here on, I get stuck. How do I take the time derivative of $$sin \theta$$ or other trig functions whose subject is not time? So far, I've got this:

$$-r \dot {\theta} sin \theta + \dot{r}cos \theta + r \dot{\theta} cos \theta + \dot{r} sin \theta$$

That being said, I also have a thought that the time-derivative of, say, $$sin \theta$$ would be zero, because sin(theta) is not dependent upon time. Though, I may be wrong. (Which is entirely why I'm asking here!)

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(Also, there is the obvious grouping to be done on the 5th line of LaTeX)

Stephen Tashi
I'm attempting to find velocity in polar coordinates (Kleppner/Kolenkow text). What I've got is:

$$\vec{v} = \frac {d}{dt} (x \hat{i} + y \hat{j})$$
$$= \frac {d}{dt} (rcos \theta) + \frac {d}{dt} (rsin \theta)$$
What happened to $i$ and $j$ ? You need to keep two cartesian components in this calculation. Then express the resulting cartesian vector in polar coordinates.

That being said, I also have a thought that the time-derivative of, say, $$sin \theta$$ would be zero, because sin(theta) is not dependent upon time.

It does depend on time when $\theta$ does.

What happened to $i$ and $j$ ? You need to keep two cartesian components in this calculation. Then express the resulting cartesian vector in polar coordinates.

It does depend on time when $\theta$ does.
Alright! I'll work with it and post when I think I've got something. Thanks!

BvU
Homework Helper
I very much liked the presentation here. It took me an awful long time to reproduce the expressions for the straight line motion example (page 4,5)

Alright! I've got it. All I needed was to remember to include my unit vectors (beginner's mistake, I hope), and an identity derived from the geometry of polar coordinates:

$$\vec {r} = (x \hat{i} )$$

BvU
Homework Helper
Nope, ##\vec r = x\hat\imath+y\hat\jmath## !

BvU
Homework Helper
To work out your ##\vec v## in the way you started, you need to convert ##\hat\imath## and ##\hat \jmath## to polar coordinates too. Do you have those expressions ?

(Hint: Think of a simple rotation to go from ##\hat\imath,\;\hat \jmath## to ##\hat r,\;\hat \theta## and the inverse rotation (= over minus same angle) to go the other way).

Very instructive to see how only two terms emerge from a rather long expression !

That post was actually accidentally posted prematurely; I'm working on the full one right now, haha.

BvU
Homework Helper
That post was actually accidentally posted prematurely; I'm working on the full one right now, haha.
Happens to me all the time: pressing that button too early. They do have an edit button on the left, but sometimes doing that only increases confusion and creates misunderstandings

Alright! I think I've got it. All I needed was to remember to include my unit vectors (beginner's mistake, I hope), and some identities derived from the geometry of polar coordinates:

$$\vec {r} = (x \hat\imath + y \hat\jmath )$$
$$\frac {d \vec {r}}{dt} = \frac {d}{dt} (x \hat\imath + y \hat\jmath)$$
$$x = r cos \theta \hat\imath$$ and $$y = r sin \theta \hat\jmath$$
Thus, $$\frac {d}{dt} [x \hat\imath + y \hat{j}] = \frac {d}{dt}(rcos \theta \hat{i} + r sin \theta \hat\jmath)$$
$$= r \frac {d}{dt} [cos \theta \hat\imath] + cos \theta \hat{i} \frac {d}{dt}(r) + r \frac {d}{dt} (sin \theta \hat\jmath) + sin \theta \hat{j} \frac {d}{dt}(r)$$
$$= r \frac {d}{dt} [cos \theta \hat\imath + sin \theta \hat\jmath] + (cos \theta \hat {i} + sin \theta \hat {j})\frac {d}{dt}[r]$$
$$= r( \frac {d}{dt}[cos \theta \hat\imath] + \frac {d}{dt}[sin \theta \hat\jmath]) + \frac {dr}{dt}(cos \theta \hat{i} + sin \theta \hat{j})$$
$$= r( \dot{ \theta} cos \theta \hat\jmath - \dot{ \theta} sin \theta \hat\imath) + \dot {r}(cos \theta \hat{i}+sin \theta \hat{j})$$
$$= r \dot{ \theta} (cos \theta \hat\jmath - sin \theta \hat\imath) + \dot {r}(cos \theta \hat{i}+sin \theta \hat\jmath)$$
From the geometry of polar coordinates:
$$(cos \theta \hat\imath - sin \theta \hat\imath) = \hat{ \theta}$$
$$(cos \theta \hat\imath+sin \theta \hat\jmath) = \hat{r}$$
So,
$$\frac {d \vec{r}}{dt} = \dot{r} \hat{r} + r \dot { \theta} \hat { \theta} = \vec {v}$$

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BvU