Finding the Path of a Piecewise Smooth Parametrized Curve

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Suppose $\phi: [0,5] \to \mathbb{R}^2$ is the piecewise smooth parametrized curve given by \begin{align*} \phi(t) &= \begin{cases} \left(-1 + \cos\left(\frac{5\pi}{4} + \frac{\pi t}{2}\right), 1 + \sin\left(\frac{5\pi}{4} + \frac{\pi t}{2}\right)\right) & \text{if $0\leq t < 1$} \\ \left((\sqrt{2}-1) + \cos\left(\frac{5\pi}{4} - \frac{\pi t}{2}\right), (\sqrt{2}-1)\sin\left(\frac{5\pi}{4} - \frac{\pi t}{2}\right)\right) & \text{if $1\leq t < 4$} \\ \left(-1 + \cos\left(\frac{-7\pi}{4} + \frac{\pi t}{2}\right), -1 + \sin\left(\frac{-7\pi}{4} + \frac{\pi t}{2}\right)\right) & \text{if $4\leq t \leq 5$} \end{cases} \end{align*}

What would the sketch of the path $C_{\phi}$ look like in $\mathbb{R}^2$? I tried to plug in the numbers corresponding to $t$, but I can't seem to figure out what the shape of the curve looks like geometrically. Furthermore, can we find another parametrization of this curve that is not equivalent to $\phi$ or $-\phi$?

I would very much appreciate any help with this question. Thanks!

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The first and third part are circular arcs. The second part is an elliptical arc. Parametrized curve

Python code

import numpy as np
from matplotlib import pyplot as plt

fig, ax = plt.subplots()

t = np.linspace(0, 1, 100)
x = -1 + np.cos((5/4 + t/2)*np.pi)
y = 1 + np.sin((5/4 + t/2)*np.pi)
ax.plot(x, y)

t = np.linspace(1, 4, 100)
x = np.sqrt(2)-1 + np.cos((5/4 - t/2)*np.pi)
y = (np.sqrt(2)-1) *  np.sin((5/4 - t/2)*np.pi)
ax.plot(x, y)

t = np.linspace(4, 5, 100)
x = -1 + np.cos((-7/4 + t/2)*np.pi)
y = -1 + np.sin((-7/4 + t/2)*np.pi)
ax.plot(x, y)
ax.axis('equal')

plt.show()
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The curves are a quarter circle, a 270 degree ellipse, and another quarter circle, respectively. Any curve plotting software would tell you this, if you couldn’t see it from the equations. Mathematica, for example, or this package.

For alternative parameterizations, there are a few things you could do.

  1. Shift the domains
  2. Scale the domains
  3. Replace the trig functions by rational functions using the Weierstrass substitution.