# Polar Coordinate Unit Vectors

so when you are doing the derivative of the position vector in polar form, do you have to use product rule to differentiate, because you need to differentiate the unit vectors too.

ehild
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so when you are doing the derivative of the position vector in polar form, do you have to use product rule to differentiate, because you need to differentiate the unit vectors too.
Yes.

ehild

so velocity = (dr/dt)*Er + 5*(dθ/dt)*Eθ?

ehild
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V= (dr/dt)*Er + r*dEr/dt. And dEr/dt=dθ/dt Eθ.

What is dr/dt equal to in the problem?

ehild

It is how fast it is moving radially outwards in the direction of Er. Right?

ehild
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Yes, but does it move radially outward now?

ehild

I am not grasping what you are trying to ask...

ehild
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The particle moves along a circle of radius R=5. You want its velocity vector in terms of Er and Eθ. Er is multiplied by dr/dt. r is the distance of the particle from the centre. Does that distance change while the particle moves along a circle? So what is the value of dr/dt?

ehild

it is 0 because dr/dt of a constant = 0

ehild
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dr/dt is not 5. It is zero, as the distance from the origin does not change. So how can you write the velocity vector? And what about the acceleration?

ehild

Yea after I wrote the wrong answer, I thought about it and changed it. But the velocity would just be this right? v = 5*(dθ/dt)*Eθ and the acceleration would be 0?

ehild
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Yes, the velocity vector is $$\vec v = 10 \hat e _{\theta}$$ in this case.

The acceleration is the time derivative of the velocity. It is not zero, as the velocity changes direction. You need the derivative of Eθ now. Go back to #20 and figure out how is it related to Er.

ehild

I only needed the velocity vector in polar coordinate form. Thank you so much for the help!

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