The unit tangent/normal vectors to motion+their derivatives

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Homework Statement


http://img124.imageshack.us/img124/4485/clipboard01hb6.jpg



Homework Equations


magnitude of n, u=1, u is in the direction of the velocity u=V/V



The Attempt at a Solution


The 1st part is easy, I wrote:
http://img129.imageshack.us/img129/1244/clipboard01mv3.jpg

But I can't do the second part, I read about 10 different sources about these unit vectors, but now I'm even more confused. Especially about so called "radial" and "transverse" unit vectors, they are not "tangential" and "normal" unit vectors?

Anyway, for this part I tried analytical approach:
http://img129.imageshack.us/img129/7342/clipboard01tu6.jpg
But not sure what to do next. Or another approach:
http://img129.imageshack.us/img129/287/clipboard02ve7.jpg

Any help will be appreciated, I spent about 2 days on this problem alone :cry:
 
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Use the chain rule to express \math d \hat u / dt in terms of \math d \hat u / ds.
 
Your alternate approach using theta will work. Let

\rho\mbox{d\theta}=ds

Integrate with respect time. V(s) = ds/dt and

\rho\frac{d\theta}{dt}=\rho\omega

and

\omega{ds}= \mbox{magnitude of V}
 
ok, but I don't know what to do next :(

\frac{d \hat u}{ds}=\frac{d \hat u}{ds}\frac{ds}{dt}

Added: will try your method now, chrisk. thanks
 
You know \frac{d\hat u}{ds} from part (a), and you should know \frac{ds}{dt}.
 
I still can't complete it :(
http://img89.imageshack.us/img89/2286/clipboard01da6.jpg

The last step, why du/d(theta)=n?
 
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Express the unit vectors n and u in terms of unit vectors x and y using cosine and sine. The unit vectors x and y do not change direction. Now, take du/d(theta) and this will show how this derivative is related to n.
 
What is the relation between ds/dt and v?
 
D H said:
What is the relation between ds/dt and v?
Are they equal? I don't know, this is too hard :( please just tell me how to do it, so I know for future.
 
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