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Polar coordinates: e_r and e_theta

  • Thread starter sara_87
  • Start date
763
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1. Homework Statement [/b]

Let e_r=(cos[tex]\theta[/tex],sin[tex]\theta[/tex]) and e_theta=(-sin[tex]\theta[/tex],cos[tex]\theta[/tex]).
Let P(r,[tex]\theta[/tex]) be a point with e_r and e_theta at that point.
What can you say about the three quantities (e_r, e_theta and the point P) as r and [tex]\theta[/tex] vary?

2. Homework Equations

r: distance from origin
[tex]\theta[/tex]: angle

3. The Attempt at a Solution

As r moves around, the e_r and e_theta change place. As r increases or decreases e_r and e_theta don't change place but as theta changes, they do change place since e_theta is always orthogonal to point P.

I feel like I'm not putting enough information. Is there something i didn't mention?
Thank you.
 

djeitnstine

Gold Member
614
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Yes there is some information missing. You have to note that as r changes, [tex]\theta[/tex] changes with the rate that r changes, because [tex]e_{\theta}[/tex] is the derivative of [tex]e_{r}[/tex]
 
763
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What do you mean, I dont understand why.
 

djeitnstine

Gold Member
614
0
Ok, starting with a (position) vector function, (I will use the cartesian coordinate system to make things appear obvious) lets say [tex]v=t^{2}i+2tj[/tex] The derivative is [tex]v'=2ti+2j[/tex]. Then how does [tex]v[/tex] vary with [tex]v'[/tex]? Since [tex]v'[/tex] is the derivative of [tex]v[/tex], then [tex]v'[/tex] must vary with [tex]v[/tex]'s rate of change. Looking at your position functions, [tex]-\sin{(\theta)}[/tex] is the derivative of [tex]\cos{(\theta)}[/tex]. Likewise for [tex]sin{(\theta)}[/tex] and [tex]cos{(\theta)}[/tex]
 
763
0
Oh right, i see.
Thank you
 

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