Determine where r(t) defines a smooth curve for r(t)= -pi

Question

Determine where r(t) defines a smooth curve for
r(t)=<4cos^3(t), 4sin^3(t)> -pi<t<pi

arabpride · Answer

Is this trig??

anonymous · Answer

No, it's a parametrically defined vector curve

arabpride · Answer

~ oh ~ sorry cant help you there ~ i have no idea what that is X"> ... yet

kainui · Answer

Yep, looks smooth to me.

anonymous · Answer

It's definitely not smooth at t=0

kainui · Answer

Why do you say that?

anonymous · Answer

Both equations of r'(0)=<0, 0>

kainui · Answer

So what definition of smoothness are you using? It appears to be infinitely differentiable on the interval.

kainui · Answer

Maybe I should say, piecewise smooth?

anonymous · Answer

A given parameterization, r(t), is called smooth on an interval if its derivative is continuous and not equal to the zero vector on the interval

kainui · Answer

Sounds right to me. So it's smooth on the intervals -pi<t<0 and 0<t<pi. That's what you're looking for right? Seems like you've answered your own question.

anonymous · Answer

Look at he graph and decide http://www.wolframalpha.com/input/?i=parametric+%284cos+%28t%29**3%2C+4sin%28t%29**3%29++t%3D-pi+to+pi