Piecewise C1 Parameterisation of Curve Defined in Polar Coordinates

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for the Curve defined in polar coordinates as r=2(1-cos(t)), t in [0,2pi] show it is not a C1 curve but is a piecewise C1 curve


My notes say for a parameterisation (x(t),y(t)) to be C1 both x(t) and y(t) must be C1 and (x'(t),y'(t)) not = (0,0)


using parameterisation

(rcos(t),rsin(t)) = (2cos(t)-2cos(t)^2 , 2sin(t)-2cos(t)sin(t))

hence

(x'(t),y'(t)) = (-2sin(t)+4cos(t)sin(t) , 2cos(t)+2sin(t)^2-2cos(t)^2)

which is equivalent to (0,0) at t=0,2pi


so I am wondering, if the parameterisaton is only not C1 at its endpoints how do you break it up into piecewise C1 curves as surely no matter what parameterisation you use this will occur at the endpoints...?

is it valid to say C is 0 at t=0,2pi and (x(t),y(t)) (as defined above) for t in (0,2pi), and if so, how do you write down an integral for the length of C if it is broken up like this...?


cheers for any help...

bart
 
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Not necessarily. What happens if you "shift" \theta, say by subtracting \pi/2 or \pi/4? You will still have a point at which (x', y')= (0,0) but it will be a different point. That way, you can get a "piecewise" C1 parameterization by using the different parameterizations in areas that do NOT include the point where (x', y')= (0, 0).
 
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im sorry. i think i understand what you mean by "shifting" t, but in which equation (i understand you are trying to make me think more about this)...

at pi/2,

(x(t),y(t)) = (-2,2)

at pi/4

(x(t),y(t)) not = (0,0)

so i don't understand why these points would be of significance.

and, if we shifted the parameter t=t'=t-pi/2 say, then wouldn't the interval for t change anyway from t in [0,2pi] to t in [-pi/2,3pi/2] otherwise wouldn't the curve be different (though its a closed curve so maybe it wouldnt. I am thinking as i type...sorry)

so are you saying...well despite all this thinking and wasteful typing i guess I am trying to say i don't understand what you are suggesting...(sorry bout you having to read the above drivel)
 
either way. "shifting or not" at the point (0,0), isn't (x'(t),y'(t)) = (0,0) that's just the nature of the curve at that point...im starting to wonder how anything you do could stop that happening...
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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