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
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?