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

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  • #1
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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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  • #2
D H
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What are the cartesian coordinates of those polar coordinate unit vectors? What happens when you differentiate with respect to time?
 
  • #3
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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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  • #4
D H
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Is ##\hat x + \hat y## a unit vector? (No.)

And that is not ##\hat r##.
 
  • #5
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As DH is requesting, do you know how to express [itex]\hat{r}[/itex] and [itex]\hat{\theta}[/itex] in terms of [itex]\hat{x}[/itex], [itex]\hat{y}[/itex], sinθ, and cosθ?
 
  • #6
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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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  • #7
D H
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You've been given a very good hint. Look in your textbook and your class notes. This homework question was not asked out of thin air. If this is a homework problem from your textbook, that information is right there in your book. It's in the same chapter as the question is in.
 

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