Re: Entropy - Actually a question about working in Polar Coordinates

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show that [itex]\frac{d\hat{r}}{dt}[/itex]=[itex]\hat{θ}[/itex][itex]\dot{θ}[/itex]

also, [itex]\frac{d\hat{θ}}{dt}[/itex]=-[itex]\dot{θ}[/itex]r


i've tried finding the relationship between r and theta via turning it into Cartesian coord.s, and I've tried the S=theta r but still no luck.

S=theta r

dS/dt=d(theta)/dt r which is similar to the RHS of the second equation I'm supposed to show. but i don't know how to turn dS/dt into dtheta hat /dt
 
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my appologies about the title


[itex]\hat{r}[/itex]=[itex]\hat{x}[/itex]+[itex]\hat{y}[/itex]

d[itex]\hat{r}[/itex]/dt = d([itex]\hat{x}[/itex]+[itex]\hat{y}[/itex])/dt=d[itex]\hat{x}[/itex]/dt+d[itex]\hat{y}[/itex]/dt

[itex]\hat{θ}[/itex]=?
 
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oops, sorry i misread aGAiN.. i have that habbit.

please clearify this for me

[itex]\hat{x}[/itex]= the x component of [itex]\hat{r}[/itex]? if it is, i can find x.. if it's not. then i am even more lost than i thought i was.
 
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