# Basic Time-Derivative Problem

• Cosmophile
In summary, the author is trying to find velocity in polar coordinates. They first need to keep two cartesian components and express the resulting vector in polar coordinates. Then, they need to remember that sin(theta) is not dependent on time when \theta is present. They also note that the time derivative of sin(theta) is zero when \theta is not dependent on time.f

#### Cosmophile

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!)

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

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!

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:

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

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

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.

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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Well done.

For the record: 1 typo ##(cos \theta \hat\imath - sin \theta \hat\imath) = \hat{ \theta}## should be ##(cos \theta \;\hat\jmath - sin \theta\; \hat\imath) = \hat{ \theta}## but you processed it correctly.

That probably occurred when I went back to change the tex for all of my unit vectors. I saw how you had done it in a way to get rid of the dots for the i's and j's, and probably typed it in wrong. Anyway, thank you for the help! This text is pretty rigorous, especially for someone who has taught himself the Calculus that I know, but I'm really enjoying it!