1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Basic Time-Derivative Problem

  1. Jan 6, 2015 #1
    I'm attempting to find velocity in polar coordinates (Kleppner/Kolenkow text). What I've got is:

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

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

    [tex] -r \dot {\theta} sin \theta + \dot{r}cos \theta + r \dot{\theta} cos \theta + \dot{r} sin \theta [/tex]

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

    Thanks in advance!
     
  2. jcsd
  3. Jan 6, 2015 #2
    (Also, there is the obvious grouping to be done on the 5th line of LaTeX)
     
  4. Jan 6, 2015 #3

    Stephen Tashi

    User Avatar
    Science Advisor

    What happened to [itex] i [/itex] and [itex] j [/itex] ? 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 [itex]\theta [/itex] does.
     
  5. Jan 6, 2015 #4
    Alright! I'll work with it and post when I think I've got something. Thanks!
     
  6. Jan 7, 2015 #5

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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)
     
  7. Jan 7, 2015 #6
    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:

    [tex] \vec {r} = (x \hat{i} )[/tex]
     
  8. Jan 7, 2015 #7

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Nope, ##\vec r = x\hat\imath+y\hat\jmath## !
     
  9. Jan 7, 2015 #8

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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 !
     
  10. Jan 7, 2015 #9
    That post was actually accidentally posted prematurely; I'm working on the full one right now, haha.
     
  11. Jan 7, 2015 #10

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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
     
  12. Jan 7, 2015 #11
    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:

    [tex] \vec {r} = (x \hat\imath + y \hat\jmath ) [/tex]
    [tex] \frac {d \vec {r}}{dt} = \frac {d}{dt} (x \hat\imath + y \hat\jmath) [/tex]
    [tex] x = r cos \theta \hat\imath [/tex] and [tex] y = r sin \theta \hat\jmath [/tex]
    Thus, [tex] \frac {d}{dt} [x \hat\imath + y \hat{j}] = \frac {d}{dt}(rcos \theta \hat{i} + r sin \theta \hat\jmath) [/tex]
    [tex] = 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) [/tex]
    [tex] = r \frac {d}{dt} [cos \theta \hat\imath + sin \theta \hat\jmath] + (cos \theta \hat {i} + sin \theta \hat {j})\frac {d}{dt}[r] [/tex]
    [tex] = 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}) [/tex]
    [tex] = r( \dot{ \theta} cos \theta \hat\jmath - \dot{ \theta} sin \theta \hat\imath) + \dot {r}(cos \theta \hat{i}+sin \theta \hat{j}) [/tex]
    [tex] = r \dot{ \theta} (cos \theta \hat\jmath - sin \theta \hat\imath) + \dot {r}(cos \theta \hat{i}+sin \theta \hat\jmath) [/tex]
    From the geometry of polar coordinates:
    [tex] (cos \theta \hat\imath - sin \theta \hat\imath) = \hat{ \theta} [/tex]
    [tex] (cos \theta \hat\imath+sin \theta \hat\jmath) = \hat{r} [/tex]
    So,
    [tex] \frac {d \vec{r}}{dt} = \dot{r} \hat{r} + r \dot { \theta} \hat { \theta} = \vec {v} [/tex]
     
    Last edited: Jan 7, 2015
  13. Jan 7, 2015 #12

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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.
     
  14. Jan 7, 2015 #13
    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!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted