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How do the plane polar coordinates work?

OCW Scholar - Physics I: Classical Mechanics
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you know polar coordinate system right!!|dw:1349364467374:dw|
Yes.. a little... But NOT in vectors!!
no they are just same.|dw:1349364550790:dw|

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|dw:1349364634178:dw|
we know that we write (x, y) as our point. In polar coordinate we write (r, theta) <--- this means this is a vector space. you have been using it unknowingly.
Since it's related to sine, and cosine, it's not linear translation, right?
|dw:1349364748021:dw|
yep. now let's go back to unit vector for a while. in cartesian how do you represent a vector?
xi + yj? i and j are unit vectors
yes ... and in cartesian coordinate?
Huh?! Are they different??
Woops sorry ... polar coordinate??
Well... Isn't it what I got stuck?
\[ r \hat r + \theta \hat \theta\]
now we need to find the relation between \( \hat r , \hat \theta \) and \( \hat i, \hat j \)
for that we know
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how do we define unit vectors?
vector/length of vector?
yep ...
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can you write the relation between \( \hat r \) and i,j
unit vector of r = (xi + yj) / (x^2 + y^2) ? I know it's strange...
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Oh, should I express it in terms of theta?
yep
unit vector of r = (xi + yj) / (x^2 + y^2) \(= \frac{(rcos) \theta i +(rsin\theta)j}{r^2cos^2\theta+r^2 sin^2 \theta} \) \(= \frac{(rcos) \theta i +(rsin\theta)j}{r^2 } \) Hmm.. doesn't look good, sorry!!
|dw:1349366033743:dw|
can you check page no 14 again?
Yes....
you got your first relation right?
Eh.... I was about to say no, but then, yes :|
for second relation there are many tricks. this is my favourite. polar coordinate is orthogonal coordinate system. \[ \hat r \cdot \hat \theta = 0\] so change \( \theta = \theta + 90 \) you will get second relation.
Hmm, sorry!! OS is not loading properly here... And... please give me sometime to work it out first.. :(
sure till then i'll work out on the original method|dw:1349366641329:dw|
Sorry... May I know what that e is?? I meant he wrote \(\hat e_\theta\) or \(\hat e_r\) there.
|dw:1349366770653:dw|
|dw:1349367409678:dw|
So, they are the unit vectors?
yep!!
|dw:1349367720837:dw|
Right, your trick works :|
PQ as a vector? or...?
yep. PQ
|dw:1349367904332:dw|
i hope you are understanding.
PQ (vector) =\(r (cos(\theta +d\theta) -cos\theta )i + r (sin(\theta+d\theta) -sin\theta)j\) Sum-to-product formula?
yep .. .cos C - cos D .. and ...
\[PQ = r[cos(\theta+d\theta)-cos\theta] i + r[sin(\theta+d\theta)-sin\theta]j\] \[=-2rsin(\theta+\frac{d\theta}{2})sin\frac{d\theta}{2}i +2rcos(\theta+\frac{d\theta}{2})sin\frac{d\theta}{2}j\]
|dw:1349368740990:dw|
Isn't that |PQ| something horrible??
lol ... no. Looks like i forgot cos C - cos D
|dw:1349369192240:dw|
Enough hint already!! (shhh!!~)
|dw:1349369373117:dw|
|dw:1349369423357:dw|
|dw:1349369566965:dw|
|dw:1349369666573:dw|
**************************************************************************************************************************************************************
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|dw:1349369925726:dw|
similarly for theta |dw:1349370057135:dw|
*************************************************************************************** A more general approach.
|dw:1349370210626:dw|
Using this relation you can find relation between unit vectors on any coordinate system.
But from definition of derivative, you know |dw:1349370433162:dw|
And |dw:1349370596136:dw|