How to convert a Cartesian vector to polar coordinates and differentiate it?

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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,

r(t) = (2t^2)i + (3t - 2)j + (3t^2 - 1)k

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 coswt etc... already mentioned.

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

r(t) = (2t^2)i + (3t - 2)j

 

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http://img716.imageshack.us/img716/4241/vec.jpg

Hopefully this information will help somewhat!

 

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\overline{r} (t) = (2t^2)i + (3t - 2)j

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

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

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

2: We know our unit vector will be;

\hat{r} \ = \ cos \theta \ + \ \sin \theta

3. Calculate θ!

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

4. We see that \hat{r} must be;

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

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

\overline{r}(t) = \ r \hat{r}

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

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

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

\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} ) ] )

\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} ) ] )

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;

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

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

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

Here I go!

\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 )

\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 )


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

7. Look for patterns!

Well,

r \ = \ \sqrt{4t^4 + 9 t^2- 12t + 4}

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

\hat{r} \ = \ cos \theta \ + \ \sin \theta

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

We'll set;

\hat{ \theta } \ = \ (- sin \theta \ + \ cos \theta )

&

\dot{ \theta} \ = \ \frac{d \theta}{dt}

to get

\frac{ d \hat{r}}{dt} \ = \ \hat{ \theta } \dot{ \theta}

8. Write the final, simplified equation for velocity.

\overline{v} (t) \ = \ r \dot{ \theta} \hat{ \theta } \ + \ \dot{r} \hat{r}

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.

\alpha :Have I missed anything?

\beta : Have you ever actually had to do a calculation like this??