How Do You Prove the Time Derivatives of Polar Unit Vectors?

[A2Airwaves]

Homework Statement

Prove: $$\frac{d\hat{r}}{dt} = \dot{\phi} \hat{\phi }$$ and $$\frac{d\hat{\phi}}{dt} = -\dot{\phi} \hat{r }$$

Homework Equations

The Attempt at a Solution

I solved this for an Analytical Mechanics assignment a month ago, and completely forgot how it goes..
$$\hat{r} ⊥ \hat{\phi}$$
An change from r1 to r2 will create a $Δ\phi$ that is in the $\hat{\phi}$ direction...
and because $\hat{r} ⊥ \hat{\phi}$, we can say the same happens for a change from $\phi1$ to $\phi2$ except in the $-\hat{r}$ direction. Assuming the change is infinitesimal, we can write $Δr$ or $Δ\phi$ as d/dt.

But then I'm confused because, why are we assuming a change from r1 to r2 is a rotation by $Δ\phi,$ and not a change of the length r..? Am I getting something completely wrong here?

 

[geoffrey159]

Go back to definition:
$\hat r = \cos(\phi(t)) \vec i + \sin(\phi(t)) \vec j$
$\hat \phi = -\sin(\phi(t)) \vec i + \cos(\phi(t)) \vec j$

And use time derivative

 

[geoffrey159]

Go back to definition:
$\hat r = \cos(\phi(t)) \vec i + \sin(\phi(t)) \vec j$
$\hat \phi = -\sin(\phi(t)) \vec i + \cos(\phi(t)) \vec j$

And use time derivative

 

[A2Airwaves]

Go back to definition:
$\hat r = \cos(\phi(t)) \vec i + \sin(\phi(t)) \vec j$
$\hat \phi = -\sin(\phi(t)) \vec i + \cos(\phi(t)) \vec j$

And use time derivative

Thank you so much!
My professor emphasized the geometric interpretation of the answers that I completely forgot about those definitions.
Worked like magic, problem solved.

By the way, is there a reason why you're writing $\vec{i}$ and not $\hat{i}$?
I'm used to $\hat{i}$ as a notation for unit vectors, but do you mean the same thing or are you referring to something else?

 

[geoffrey159]

Yes you're right, I meant $\hat i$ and $\hat j$.
For clear explanations and nice drawings, look up Kleppner and Kolenkow first chapter.