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X and y components of polar unit vectors.

  1. Jan 31, 2012 #1
    1. The problem statement, all variables and given/known data
    What are the x- and y-components of the polar unit vectors [itex]\hat{r}[/itex] and [itex]\hat{\theta}[/itex] when
    a. [itex]\theta[/itex] = 180°
    b. [itex]\theta[/itex] = 45°
    c. [itex]\theta[/itex] = 215°

    2. Relevant equations



    3. The attempt at a solution
    Please check if I'm correct, i'll just show my answer for a since the process is the same for a, b and c

    for a.

    [itex]\hat{r_{x}}[/itex] = r cos [itex]\theta[/itex]
    [itex]\hat{r_{x}}[/itex] = 1 cos 180°
    [itex]\hat{r_{x}}[/itex] = -1

    [itex]\hat{r_{y}}[/itex] = r sin [itex]\theta[/itex]
    [itex]\hat{r_{y}}[/itex] = 1 sin 180°
    [itex]\hat{r_{y}}[/itex] = 0

    in terms of theta... i don't have any idea how... please help
     
    Last edited: Jan 31, 2012
  2. jcsd
  3. Jan 31, 2012 #2
    That's exactly right. If you imagine a unit circle, at 180 degrees, you're on the opposite side of the circle from 0 degrees. x = -1, y = 0.
     
  4. Jan 31, 2012 #3
    What about theta?? who could I find its x and y component?
     
  5. Jan 31, 2012 #4
    Can someone help me how to find [itex]\hat{\theta}[/itex]? I don't know how. thanks
     
  6. Jan 31, 2012 #5

    SammyS

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    Do you not have a definition of the unit vector [itex]\hat{\theta}\,?[/itex]

    The unit vector [itex]\hat{\theta}[/itex] lies in the xy-plane and is 90° counter-clockwise from [itex]\hat{r}\,.[/itex]
     
  7. Jan 31, 2012 #6
    I didn't really get what you said. sorry. can you show me an example on how to get x and y component for a then i'll do it for b and c. thanks. much appreciated.
     
  8. Jan 31, 2012 #7

    SammyS

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    Well, if [itex](\hat{r})_x=\cos(\theta)\,,\text{ then }(\hat{\theta})_x=\cos(\theta+90^\circ)\,.[/itex] ... etc.

    Use the angle addition identity to simplify cos(θ+90°) .
     
  9. Jan 31, 2012 #8
    so for a


    [itex](\hat{\theta})_x=cos(180+90)[/itex]
    [itex](\hat{\theta})_x=cos(270)[/itex]
    [itex](\hat{\theta})_x= 0[/itex]

    then

    [itex](\hat{\theta})_y=sin (180+90)[/itex]
    [itex](\hat{\theta})_y=sin (270)[/itex]
    [itex](\hat{\theta})_y= -1 [/itex]

    am i correct?
     
  10. Jan 31, 2012 #9

    SammyS

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    Yes. That's correct.
     
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