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(adsbygoogle = window.adsbygoogle || []).push({});

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

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# Homework Help: C1 parameterisations

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