Arturo O. answered • 04/02/18

Tutor

5.0
(66)
Experienced Physics Teacher for Physics Tutoring

I think your confusion is regarding the magnitude of the polar unit vector

**R**. It is exactly 1 regardless of θ, as shown below.|

**R**| = |cosθ**i**+ sinθ**j**| = √(cos^{2}θ + sin^{2}θ) = √1 = 1Arturo 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 identitycos

^{2}θ + sin^{2}θ = 1for 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.
Report

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.
Report

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

"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

Report

04/03/18

Arturo O.

Very good, Callista

Report

04/03/18

Callista H.

iandjare always 1 too though so how canR=cos(Θ)i+sin(Θ)jthen equal 1 for all Θ?1=cos(45)*1+sin(45)*1

04/02/18