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!

Cartesian to Polar Vectors

  1. Mar 25, 2010 #1
    Hi I pretty much can't get past the first chapter in my physics book until I master the vector representation of polar coordinates.

    Every explanation I've read thus far has confused me, I keep thinking in Cartesian terms so I think it'd be great to convert a vector equation from cartesian to polar description and then differentiate (with somebodies help, which I need!).

    I have a cartesian based vector,

    [tex] r(t) = (2t^2)i + (3t - 2)j + (3t^2 - 1)k [/tex]

    I don't know how to go about converting it, could anybody give a helping hint?

    Is it even possible? Every book I see polar coordinates mentioned have a [tex] coswt [/tex] etc... already mentioned.

    Edit: Thinking about it, just a two dimensional vector would be easier on everyone!

    [tex] r(t) = (2t^2)i + (3t - 2)j [/tex]
     
    Last edited: Mar 25, 2010
  2. jcsd
  3. Mar 25, 2010 #2
    http://img716.imageshack.us/img716/4241/vec.jpg [Broken]

    Hopefully this information will help somewhat!
     
    Last edited by a moderator: May 4, 2017
  4. Mar 26, 2010 #3
    [itex] \overline{r} (t) = (2t^2)i + (3t - 2)j [/itex]

    [itex] convert \ to \ polar \ coordinates \ of \ the \ form \ \overline{r}(t) = \ r \hat{r} \ then \ differentiate![/itex]

    1: First we want to calculate |r| = r.

    [itex] r \ = \ | \overline{r} | \ = \ \sqrt{(2t^2)^2 + (3t - 2)^2} \ = \sqrt{4t^4 + 9 t^2- 12t + 4} [/itex]

    2: We know our unit vector will be;

    [itex] \hat{r} \ = \ cos \theta \ + \ \sin \theta [/itex]

    3. Calculate θ!

    [itex] \theta \ = \ \arctan( \frac{y}{x} ) \ = \ \arctan(\frac{3t - 2}{2t^2} ) [/itex]

    4. We see that [itex] \hat{r} [/itex] must be;

    [itex] \hat{r} \ = \ cos \theta \ + \ \sin \theta \ = \ \cos [\arctan(\frac{3t - 2}{2t^2} )] \ + \ \sin [\arctan(\frac{3t - 2}{2t^2} )][/itex]

    5. Put the position vector all together, (all are equivalent).

    [itex] \overline{r}(t) = \ r \hat{r} [/itex]

    [itex] \overline{r}(t) \ = \ [ \sqrt{4t^4 + 9 t^2- 12t + 4} ] \hat{r}[/itex]

    [itex] \overline{r}(t) \ = \ [ \sqrt{4t^4 + 9 t^2- 12t + 4} ] [\ cos \theta \ + \ \sin \theta ] [/itex]

    [itex] \overline{r}(t) \ = \ ( \sqrt{4t^4 + 9 t^2- 12t + 4} ) (cos \theta ) \ + \ ( \sqrt{4t^4 + 9 t^2- 12t + 4} ) (sin \theta ) [/itex]

    [itex] \overline{r}(t) \ = \ ( \sqrt{4t^4 + 9 t^2- 12t + 4} ) ( \cos [\arctan(\frac{3t - 2}{2t^2} )] ) \ + \ \sin ( \arctan(\frac{3t - 2}{2t^2} ) ] ) [/itex]

    [itex] \overline{r}(t) \ = ( \sqrt{4t^4 + 9 t^2- 12t + 4} ) ( \cos [\arctan(\frac{3t - 2}{2t^2} ) ] ) \ + ( \sqrt{4t^4 + 9 t^2- 12t + 4} ) ( \ \sin [\arctan(\frac{3t - 2}{2t^2} ) ] ) [/itex]

    Alright, so I've set up the position vector as best I can. I'm assuming there's nothing wrong so I'll now take the derivative to find the velocity vector.

    6. Using;

    [itex] \overline{r}(t) \ = \ [ \sqrt{4t^4 + 9 t^2- 12t + 4} ] [\ cos \theta \ + \ \sin \theta ] [/itex]

    I'll differentiate to obtain an equation of the form;

    [itex] \overline{v} (t) \ = \ \frac{d \overline{r} }{dt} \ = \ r \frac{d \hat{r}}{dt} \ + \ \frac{dr}{dt} \hat{r} [/itex]

    Here I go!

    [itex] \frac{d \overline{r} }{dt} \ = \ ( \sqrt{4t^4 + 9 t^2- 12t + 4} ) (- sin \theta \frac{d \theta} {dt} \ + \ cos \theta \frac{d \theta}{dt} ) \ + \ ( \frac{16t^3 \ + \ 18t \ - \ 12}{2 \sqrt{4t^4 + \ 9t^2 - 12t + 4} } ) (\ cos \theta \ + \ \sin \theta ) [/itex]

    [itex] \frac{d \overline{r} }{dt} \ = \ ( \sqrt{4t^4 + 9 t^2- 12t + 4} ) (- sin \theta \ + \ cos \theta ) \frac{d \theta}{dt} \ + \ ( \frac{16t^3 \ + \ 18t \ - \ 12}{2 \sqrt{4t^4 + \ 9t^2 - 12t + 4} } ) (\ cos \theta \ + \ \sin \theta ) [/itex]


    I didn't dare try to bring in the arctan and it's craziness into this one, (yet...).

    7. Look for patterns!

    Well,

    [itex] r \ = \ \sqrt{4t^4 + 9 t^2- 12t + 4} [/itex]

    [itex] \dot{r} \ = \ ( \frac{16t^3 \ + \ 18t \ - \ 12}{2 \sqrt{4t^4 + \ 9t^2 - 12t + 4} } ) [/itex]

    [itex] \hat{r} \ = \ cos \theta \ + \ \sin \theta [/itex]

    [itex] \frac{ d \hat{r}}{dt} \ = \ ( - \sin \theta \ + \ \cos \theta ) \frac{ d \theta}{dt}[/itex]

    We'll set;

    [itex] \hat{ \theta } \ = \ (- sin \theta \ + \ cos \theta )[/itex]

    &

    [itex] \dot{ \theta} \ = \ \frac{d \theta}{dt} [/itex]

    to get

    [itex] \frac{ d \hat{r}}{dt} \ = \ \hat{ \theta } \dot{ \theta}[/itex]

    8. Write the final, simplified equation for velocity.

    [itex] \overline{v} (t) \ = \ r \dot{ \theta} \hat{ \theta } \ + \ \dot{r} \hat{r} [/itex]

    Comparing the result of 8. with the result of 6. using 7. to be sure we've labelled everything correctly I'd say we have taken a deceptively simple equation in cartesian form, converted it to it's equivalent polar coordinate description & take the derivative of this to obtain the velocity.

    Q.E.D.

    [tex] \alpha : [/tex]Have I missed anything?

    [tex] \beta : [/tex] Have you ever actually had to do a calculation like this??
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook