Callista H.

asked • 04/02/18

Cartesian Unit Vector to Polar Unit Vector

I'm having trouble understanding the following equation conceptually. 
Where....
R= Polar unit vector
I, J= Cartesian unit vectors
 
R=cos(Θ)I+sin(Θ)J
 
Please explain how this equation can be true when R=√(I2+J2) is also true. I understand the conversion in the second equation, but don't understand it in the first. It seems to me that both of these being true would contradict each other. 
 
For example, if you assume that I, and J are both equal to 1 and that theta =45 degrees:
  • Using the first equation, I get: (cos(45)*1)+(sin(45)*1)=1.376
  • Using the second, I get: √(1^2+1^2)=1.41
 
Please help me figure out how to properly use these two equations, as I'm sure I am misunderstanding something simple. Please give a numerical answer to supplement your reasoning!
 
 

1 Expert Answer

By:

Arturo O. answered • 04/02/18

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Callista H.

and j are always 1 too though so how can R=cos(Θ)i +sin(Θ)then equal 1 for all Θ?

1=cos(45)*1+sin(45)*1 
is not true...
 
 
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04/02/18

Arturo O.

I just showed you in my answer that |R| = 1 for all θ.  Do you know how to find the magnitude of a vector?  Are you familiar with the trigonometric identity
 
cos2θ + sin2θ = 1
 
for all θ?  Apply these two, and you will see that |R| = 1 for all θ.  The equation in your Comment is not the correct equation for the magnitude of the vector R.  I gave you the correct equation in my Answer.
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04/02/18

Arturo O.

Your equation for the magnitude of R is not correct.  I gave you the correct equation in my Answer.  Evaluate the magnitude of R using the correct equation and you will see that it gives 1 for all θ. 
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04/02/18

Callista H.

I think I understand now. The source of my confusion was that I was not adding them like vectors should be added. According to the equation provided by my MIT OpenCoursework textbook:

"The unit vectors (r,θ) at the point P also are related to the Cartesian unit vectors (i, j)
by the transformations
r = cosθi + sinθj , (3.2.6)
θ = −sinθi + cosθj . (3.2.7) "

Which is true. My problem is that I was using it as though you could add vectors like any other number. (see above 1=cos(45)*1+sin(45)*1 ). Using that equation correctly, I can derive the equation you provided when I add those vectors via tail-to-tip method.

  /|
r/ |sinθj=sinθ*1=sinθ                    **r2=cos2θ+sin2θ
/_|                                               **r=√(cos2θ+sin2θ)
cosθi=cosθ*1=cosθ

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04/03/18

Arturo O.

Very good, Callista
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04/03/18

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