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## Homework Statement

## 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.

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- Thread starter Panphobia
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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.

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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 e

(a) x (e

and

(b) x (e

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^Yea that was pretty much what I was going to say.

Last edited:

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√x

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, e

I understand your confusion as e

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(e

e

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.

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- #7

ehild

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

ehild

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Chestermiller

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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

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So the angle between the radius and the x axis is 2t?

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Chestermiller

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Yes.So the angle between the radius and the x axis is 2t?

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- #12

Chestermiller

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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.

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Chestermiller

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Yes. Exactlly.

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- #16

Chestermiller

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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.

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- #18

Chestermiller

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You're going to kick yourself.

d(kt)/dt = k(dt/dt)=k

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Brain fart :s I am kicking myself

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ehild

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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

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This was a mispost

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- #23

ehild

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ehild

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v = (-10sin(2t), 10cos(2t))

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ehild

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ehild

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