Find a piecewise smooth parametric curve to the astroid (a hypocycloid with four cusps)

  • #1
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Homework Statement


Find a piecewise smooth parametric curve to the astroid. The astroid, given by $\phi(\theta) = (cos^3(\theta),sin^3(\theta))$, is not smooth, as we see singular points at 0, pi/2, 3pi/2, and 2pi. However is there a piecewise smooth curve?

Homework Equations


$\phi(\theta) = (cos^3(\theta),sin^3(\theta))$

The Attempt at a Solution


I have tried to use the cartesian equation x^(2/3) + y^(2/3) = 1 but that didn't help. I tried to change the periodicity of the cos and sine functions but obviously that was pointless. I thought at one time maybe this is not possible, but I see examples of line integrals over the astroid so it must be piecewise smooth. Can I get a hint?
 

Answers and Replies

  • #2
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How can phi be a vector?

Just by looking at the plot of the curve, which pieces seem useful?
 
  • #3
LCKurtz
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Homework Statement


Find a piecewise smooth parametric curve to the astroid. The astroid, given by $\phi(\theta) = (cos^3(\theta),sin^3(\theta))$, is not smooth, as we see singular points at 0, pi/2, 3pi/2, and 2pi. However is there a piecewise smooth curve?

Homework Equations


##\phi(\theta) = (cos^3(\theta),sin^3(\theta))##
That is poorly written, even after fixing the tex as I did. Using phi and theta as your variables suggests spherical coordinates somehow. Let's rewrite it as ##\vec R(t) = \langle \cos^3(t),\sin^3(t) \rangle##. Now I am confused. That equation gives a graph that is piecewise smooth already, doesn't it? Or is your problem really to find the points of the vector function where it violates the definition of "smooth", which is not the same thing as the graph having sharp corners?
 
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