# C1 parameterisations

1. May 13, 2007

### bartieshaw

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 im 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

2. May 13, 2007

### HallsofIvy

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).

Last edited by a moderator: May 13, 2007
3. May 13, 2007

### bartieshaw

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 dont understand why these points would be of significance.

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

so are you saying....well despite all this thinking and wasteful typing i guess im trying to say i dont understand what you are suggesting.....(sorry bout you having to read the above drivel)

4. May 13, 2007

### bartieshaw

either way. "shifting or not" at the point (0,0), isnt (x'(t),y'(t)) = (0,0) thats just the nature of the curve at that point...im starting to wonder how anything you do could stop that happening...