X and y components of polar unit vectors.

[jhosamelly]

Homework Statement

What are the x- and y-components of the polar unit vectors \hat{r} and \hat{\theta} when
a. \theta = 180°
b. \theta = 45°
c. \theta = 215°

Homework Equations

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.

\hat{r_{x}} = r cos \theta
\hat{r_{x}} = 1 cos 180°
\hat{r_{x}} = -1

\hat{r_{y}} = r sin \theta
\hat{r_{y}} = 1 sin 180°
\hat{r_{y}} = 0

in terms of theta... i don't have any idea how... please help

 

[tjackson3]

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.

 

[jhosamelly]

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.

What about theta?? who could I find its x and y component?

 

[jhosamelly]

Can someone help me how to find \hat{\theta}? I don't know how. thanks

 

[SammyS]

Can someone help me how to find \hat{\theta}? I don't know how. thanks

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

The unit vector \hat{\theta} lies in the xy-plane and is 90° counter-clockwise from \hat{r}\,.

 

[jhosamelly]

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

The unit vector \hat{\theta} lies in the xy-plane and is 90° counter-clockwise from \hat{r}\,.

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.

 

[SammyS]

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.

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

Use the angle addition identity to simplify cos(θ+90°) .

 

[jhosamelly]

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

Use the angle addition identity to simplify cos(θ+90°) .

so for a


(\hat{\theta})_x=cos(180+90)
(\hat{\theta})_x=cos(270)
(\hat{\theta})_x= 0

then

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

am i correct?

 

[SammyS]

Yes. That's correct.