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Polar Coordinate Unit Vectors

  • Thread starter Panphobia
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  • #1
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Homework Statement



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The Attempt at a Solution


I already know how to do a), but what I am wondering is what the question means by expressing position in the terms of those unit vectors.
 

Answers and Replies

  • #2
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In the initial equation of the vector r, you can see that there are given unit vectors, i, and unit vector j . These are simply denoting a direction in Cartesian coordinates. Because a unit vector is equal to one, it's only there to designate movement along a direction, as multiplying by one doesn't change a given value.

But in polar coordinates, you're multiplying by a unit vector r, to give you the magnitude, and then multiplying by unit vector of angle theta, to find the angle off of the positive x-axis. Unlike Cartesian coordinates, this unit vector r is always changing, making a full circle as you revolve around. Because a vector has both direction, and magnitude, as r revolves, its direction revolves, and the vector is no longer considered constant.

In any case, the polar unit vectors er and eθ are there to take the place of the i, j, and k unit vectors in Cartesian coordinates. Such that,

(a) x (er) denotes the magnitude of r, where (a) is some value

and

(b) x (eθ) denotes the magnitude of θ, where (b) is some value
 
  • #3
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^Yea that was pretty much what I was going to say.
 
Last edited:
  • #4
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Rawr!! I still don't understand completely what you're trying to get across. I get Er that it is a unit vector in the direction of r, but what is eθ in the direction of? since theta is just an angle, and an angle has no direction per se.
 
  • #5
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Where you have i, j , and k, that remain constant along the axes, polar coordinates are just another way of addressing the same problem of denoting a position in space. Suppose, in just using the xy-plane, you have a circle of radius one, centered at the origin. Using Cartesian coordinates, you'll constantly have to change the values of x and y, to satisfy the equation,

√x2 + y2 = 1

But in Polar coordinates, the radius r, remains 1, where it is simply the angle θ that changes as you revolve around. As you said, er is the unit vector in the direction of r, (in this case, r is just 1). And eθ is there TO literally denote the direction of the vector. Rather than breaking down the vector into x, y, and z components, each with its own direction, you simply designate the direction with an angle θ.

I understand your confusion as eθ doesn't have a fixed position, but could give any direction, based on the value. eθ is just there to designate that you're moving in relation to θ, and that it's your coefficient in front that denotes the magnitude of that direction.

For example, if you have a point at (1i,1j), you have a value of 1 along the x-axis, and a value of 1 along the y-axis. Converting this to polar coordinates, you have r=1 and an angle of (pi/4) which turns to (1(er),pi/4(eθ)).

eθ is just there, denoting that you're changing the angle with respect to the positive x-axis.

I hope this makes better sense. If not, please ask any questions, or address what you're having trouble with so i may attempt to help.
 
  • #6
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Actually I think the position would just be 5 [itex]\hat{r}[/itex] since at any point its radial position would be 5. In my dynamics class we also used the equation [itex]\vec{r}[/itex]=r [itex]\hat{e}[/itex][itex]_{r}[/itex]. [itex]\hat{e}[/itex][itex]_{θ}[/itex] was only used with velocity and acceleration equations because those would change with θ.
 
  • #7
ehild
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[itex]\hat e_\theta[/itex] is a unit vector perpendicular to [itex]\hat e_r[/itex], enclosing positive 90° angle with it.

Maiq is right, the position vector is [itex]5 \hat e_r[/itex]

ehild
 
  • #8
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Actually I think the position would just be 5 [itex]\hat{r}[/itex] since at any point its radial position would be 5. In my dynamics class we also used the equation [itex]\vec{r}[/itex]=r [itex]\hat{e}[/itex][itex]_{r}[/itex]. [itex]\hat{e}[/itex][itex]_{θ}[/itex] was only used with velocity and acceleration equations because those would change with θ.
Think of polar coordinates as radial lines out from the origin (the lines of constant θ) and concentric circles surrounding the origin (the curves of constant r). The unit vector [itex]\hat{e}_{θ}[/itex] is tangent to these circles at each point.

Chet
 
  • #9
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So the angle between the radius and the x axis is 2t?
 
  • #10
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So the angle between the radius and the x axis is 2t?
Yes.
 
  • #11
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So to get the velocity and acceleration vectors would I just take the derivative and second derivative of the position equation?
 
  • #12
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So to get the velocity and acceleration vectors would I just take the derivative and second derivative of the position equation?
Yes. To do this in cartesian coordinates is no problem, but, if you are going to do it in polar coordinates, you have to remember to include terms involving the derivatives of the unit vectors.
 
  • #13
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so another question is asking me to express velocity in terms of Er and Eθ, so is this what you are saying, the derivative of the unit vectors too?
 
  • #14
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so another question is asking me to express velocity in terms of Er and Eθ, so is this what you are saying, the derivative of the unit vectors too?
Yes. Exactlly.
 
  • #15
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When you say derivative of a unit vector do you mean the derivative of the unit vector with respect to time? If so would it be something like dEr/dt? and dEθ/dt?
 
  • #16
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When you say derivative of a unit vector do you mean the derivative of the unit vector with respect to time? If so would it be something like dEr/dt? and dEθ/dt?
Not exactly. Each or these two unit vectors changes direction with θ (i.e., are functions of θ). So, their time derivatives are their spatial derivatives with respect to θ times the derivative of θ with respect to time:

dEr/dt = dEr/dθ x dθ/dt

dEθ/dt = dEθ/dθ x dθ/dt

What you need to do is figure out how dEr/dθ and dEθ/dθ can each be expressed as a function of Er and Eθ.
 
  • #17
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I know how to get velocity in polar coordinates if it was cosθ and sinθ, but it is 2t so how do you take the derivative of 2t with respect to t?
 
  • #18
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I know how to get velocity in polar coordinates if it was cosθ and sinθ, but it is 2t so how do you take the derivative of 2t with respect to t?
You're going to kick yourself.

d(kt)/dt = k(dt/dt)=k
 
  • #19
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Brain fart :s I am kicking myself
 
  • #20
ehild
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so another question is asking me to express velocity in terms of Er and Eθ, so is this what you are saying, the derivative of the unit vectors too?
Before you take the derivatives of the position vector, you need to know the derivatives of the unit vectors. It is easiest to understand, if you recall that [itex]\hat e_r =\cos(\theta) \hat i + \sin (\theta ) \hat j[/itex] and [itex]\hat e_{theta} =-\sin(\theta) \hat i + \cos (\theta) \hat j[/itex]. (You see, they are orthogonal unit vectors.)

What are their derivatives with respect to time when θ=2t?

ehild
 
  • #21
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This was a mispost
 
  • #22
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The question asks me to prove that the position and velocity vectors are orthogonal, so I just dot producted them got 0, and that is the proof. But then the question asks "Using this knowledge, express v in terms of Er and Eθ." The only thing I don't fully understand is how I am supposed to use that knowledge.
 
  • #23
ehild
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Show the velocity you got in Cartesian coordinates. How is it related to [itex]\hat e_{theta}?[/itex]

ehild
 
  • #24
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v = (-10sin(2t), 10cos(2t))
 
  • #25
ehild
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You can write it in the form v=10(-sin(θ), cos(θ)), θ=2t. Check my post #20. What is the vector (-sin(θ), cos(θ)) equal to?

ehild
 

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