Show magnitude of velocity vector in polar coordinates

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SUMMARY

The discussion focuses on demonstrating the relationship between the magnitude of the velocity vector in polar coordinates and Cartesian coordinates. Specifically, it establishes that in polar coordinates, the squared magnitude of the velocity vector is expressed as |v|^2 = Vr^2 + Vθ^2. The participants clarify the conversion of derivatives from Cartesian to polar coordinates, emphasizing the use of the chain rule for differentiation. The correct expressions derived are dx/dt = (dr/dt) cosθ - r sinθ (dθ/dt) and dy/dt = (dr/dt) sinθ + r cosθ (dθ/dt).

PREREQUISITES
  • Understanding of polar and Cartesian coordinate systems
  • Knowledge of calculus, specifically differentiation and the chain rule
  • Familiarity with vector magnitude calculations
  • Basic physics concepts related to velocity vectors
NEXT STEPS
  • Study the derivation of polar coordinates from Cartesian coordinates
  • Learn about the chain rule in calculus and its applications in physics
  • Explore vector calculus, focusing on velocity and acceleration in different coordinate systems
  • Investigate the implications of converting between coordinate systems in physics problems
USEFUL FOR

Students studying physics or mathematics, particularly those focusing on vector calculus and coordinate transformations, as well as educators seeking to clarify these concepts for learners.

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Homework Statement


In Cartesian coordinates the magnitude of the velocity vector squared is
|v|^2=V*V= Vx^2 +Vy^2 =(dx/dt)^2+(dy/dt)^2
Show that in polar coordinated
|v|^2= Vr^2 +V@ ^2

Homework Equations





The Attempt at a Solution


Not really sure what the question is asking me to do, but i am guessing to convert (dx/dt)^2+(dy/dt)^2 into polar? or do i need to do it for all of it?

Well I got dy/dt=(dr/d@) sin@+rcos@ and dx/dt=(dr/d@) cos@-rsin@

Is this right? Were do i go after this?

Thanks
 
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It is not right.
If x = r cosθ, then
(dx/dt) = (dr/dt) cosθ - r sinθ (dθ/dt). That's the chain rule of differentiation.

Find a similar expression for (dy/dt), square each expression then add.
 
cool thanks for info ill try it out tomorrow
 

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