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Homework Help: 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??
     
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