Show magnitude of velocity vector in polar coordinates

AI Thread Summary
The discussion focuses on converting the magnitude of the velocity vector from Cartesian to polar coordinates. The initial equation in Cartesian coordinates is given as |v|^2 = Vx^2 + Vy^2. Participants clarify that to express this in polar coordinates, one must apply the chain rule to differentiate x and y in terms of r and θ. The correct expressions for dx/dt and dy/dt are derived, leading to the conclusion that the magnitude in polar coordinates can be represented as |v|^2 = Vr^2 + Vθ^2. The thread emphasizes the importance of proper differentiation and conversion techniques in this context.
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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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