An Insight into Polar Co-ordinate Velocity

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SUMMARY

The discussion focuses on deriving the velocity and acceleration vectors of a particle in polar coordinates. The position vector is expressed as r = xi + yj, where x = rCos(θ) and y = rSin(θ). The correct differentiation approach involves treating both r and θ as functions of time, applying the product and chain rules to derive the velocity vector v = ṙe_r + rθ̇e_θ. The basis vectors are defined as e_r = ⟨cos(θ), sin(θ)⟩ and e_θ = ⟨-sin(θ), cos(θ)⟩.

PREREQUISITES
  • Understanding of polar coordinates and their representation
  • Familiarity with vector calculus and differentiation techniques
  • Knowledge of the product and chain rules in calculus
  • Basic concepts of velocity and acceleration in physics
NEXT STEPS
  • Study the application of the product rule in vector differentiation
  • Learn about the chain rule in the context of polar coordinates
  • Explore the derivation of acceleration vectors in polar coordinates
  • Investigate the use of basis vectors in vector calculus
USEFUL FOR

Students studying physics or mathematics, particularly those focusing on mechanics and vector calculus, as well as educators teaching polar coordinate systems.

mooneyes
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Homework Statement



Derive the equations for the velocity and acceleration vectors of a particle in polar coordinates.

2. The attempt at a solution

r = xi + yj
where x = rCos[tex]\Theta[/tex], y = rSin[tex]\Theta[/tex]
r = r(Cos[tex]\Theta[/tex]i + Sin[tex]\Theta[/tex]j)
v = [tex]\frac{d}{dt}[/tex]r
v = [tex]\frac{d}{dt}[/tex][r(Cos[tex]\Theta[/tex]i + Sin[tex]\Theta[/tex]j)]

And I know the answer is:
v = r[dot] + r(theta[dot])
But I just don't know how to actually differentiate the expression for r with respect to time, do I treat r and theta as constants, or how does it work?

Thanks.
 
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mooneyes said:

Homework Statement



Derive the equations for the velocity and acceleration vectors of a particle in polar coordinates.

2. The attempt at a solution

r = xi + yj
where x = rCos[tex]\Theta[/tex], y = rSin[tex]\Theta[/tex]
r = r(Cos[tex]\Theta[/tex]i + Sin[tex]\Theta[/tex]j)
v = [tex]\frac{d}{dt}[/tex]r
v = [tex]\frac{d}{dt}[/tex][r(Cos[tex]\Theta[/tex]i + Sin[tex]\Theta[/tex]j)]

And I know the answer is:
v = r[dot] + r(theta[dot])
But I just don't know how to actually differentiate the expression for r with respect to time, do I treat r and theta as constants, or how does it work?

Thanks.

Both r and θ are functions of t. So differentiate both your rcos(θ) and your rsin(θ) in your position vector using the product and chain rules. When you are done you should be able to express your answer in terms of the basis vectors

[tex]e_r = \langle \cos\theta,\sin\theta\rangle,\ e_\theta=\langle -\sin\theta,\cos\theta\rangle[/tex]
 

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