How do we prove that two parametric equations are drawing the same thing?

For example, if I have \begin {align} x(t) &= r\sin t\cos t\\ y(t) &= r\sin^2 t\\ \end {align} and \begin {align} x(t) &= \frac r 2 \cos t\\ y(t) &= \frac r 2 (\sin t + 1) \end {align} How do we show that the two parametric equations draw the same line?

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What if the parametric equations are all rational functions? In this case, it's often possible to implicitize -- i.e. convert to equations of the form $f(x,y)=0$ and $g(x,y)=0$. Then, if the two curves are the same point set, I would guess that something can be said about $f$ and $g$? maybe one is a multiple of the other?? Need comments from someone who knows more about algebraic geometry than I do. – bubba Apr 28 '13 at 7:30
Maybe I should turn my comment above into a separate question, so as not to divert this. So, that's what I did. – bubba Apr 28 '13 at 8:15

You should find bijection $t_2=f(t_1)$, so that $x_1(t_1)=x_2(f(t_1))$ and $y_1(t_1)=y_2(f(t_1))$.

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To show that $\{(x,y) \;|\; \exists{t}:x=x_1(t),y=y_1(t)\}$ and $\{(x,y) \;|\; \exists{t}:x=x_2(t),y=y_2(t)\}$ are equal sets (as subsets of $\mathbb{R}^2$), you need a bijection $f$ such that $x_1(t)=x_2(f(t))$ and $y_1(t)=y_2(f(t))$. What one typically means by "traces the same curve" is slightly stronger, in that $f$ should be continuous as well, and perhaps even monotonically increasing, if the sense (e.g., clockwise vs. counterclockwise) of the curve is considered. In your case, the two curves are the same in the stronger sense: the second is simply being traversed twice as fast, and is out of phase with the first at $t=0$.

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Some efforts to establish that two parametrizations represent the same curve are trickier than others. In your first set, you will need the "double-angle" trig identities, or their relatives (I am also going to write '$r$' as '$R$' , since it is intended to be a constant, rather than radius in polar coordinates):

$$\sin 2 \theta \ = \ 2 \sin\theta \cos\theta \ \rightarrow \ x(t) \ = \ R \ \sin t \cos t \ = \ R \cdot \frac{1}{2} \sin2t \ ,$$

$$\sin^2 \theta \ = \ \frac{1}{2} (1 - \cos 2\theta) \ \rightarrow \ y(t) \ = \ R \ \sin^2 t \ = \ R \cdot \frac{1}{2} (1- \cos 2t)$$

$$R \ \sin 2t \ = \ 2x \ , \ 2y \ = \ R - R \ \cos 2t \ \Rightarrow \ R \ \cos 2t \ = \ R - 2y$$

$$R^2 \sin^2 2t \ + \ R^2 \cos^2 2t \ = \ R^2 \ = \ (2x)^2 \ + \ ( R - 2y)^2 \ = \ 4x^2 + \ R^2 \ - \ 4Ry \ + 4y^2$$

$$\Rightarrow \ 0 \ = \ 4x^2 \ - \ 4Ry \ + \ 4y^2 \ \Rightarrow \ x^2 \ + \ y^2 \ - \ Ry \ = \ 0$$

$$\Rightarrow \ x^2 \ + \ (y^2 \ - \ Ry \ + \frac{R^2}{4}) \ = \ \frac{R^2}{4} \ \Rightarrow \ x^2 \ + \ (y - \frac{R}{2})^2 \ = \ (\frac{R}{2})^2 \ ,$$

that is, a circle centered at $(0, \frac{R}{2})$ with radius $\frac{R}{2}$ . Since the trig functions both have a period of $\pi$, the parametrization starts on the circle with $t = 0$ at $(0,0)$, advances counter-clockwise to $(\frac{R}{2}, \frac{R}{2})$ at $t = \frac{\pi}{4}$, and continues around to return to the origin at $t = \pi$ .

With the second parametrization, we follow a similar process (so I won't show all the steps):

$$x(t) \ = \ \frac{R}{2} \cos t \ , \ y(t) \ = \ \frac{R}{2} ( \sin t \ + \ 1)$$

$$R \ \cos t \ = \ 2x \ , \ 2y \ = R \ \sin t \ + \ R \ \Rightarrow \ R \ \sin t \ = \ R - 2y$$

$$R^2 \sin^2 t \ + \ R^2 \cos^2 t \ = \ R^2 \ = \ (2x)^2 \ + \ ( R - 2y)^2 \ \Rightarrow \ x^2 \ + \ (y - \frac{R}{2})^2 \ = \ (\frac{R}{2})^2 \ .$$

So we have the same circle again. However, this parametrization has a period of $2 \pi$ , starts at $(\frac{R}{2}, \frac{R}{2})$ for $t = 0$ , advances counter-clockwise to $(0,R)$ at $t = \frac{\pi}{2}$ , and completes its cycle at $(\frac{R}{2}, \frac{R}{2})$ for $t = 2 \pi$ . (As mjqxxx has said, the circular trajectory is followed in the same direction, but at different "speeds" and "starting points".)

You may well imagine that an unlimited number of other parametrizations can be chosen, differing not only in "rate of travel", "direction of travel", and "initial point", but also using "speed functions" which are not constant (or even reverse themselves!). Because this is possible, there are many ways to "disguise" a parametric curve; consequently, there is no general method for checking whether two descriptions actually produce the same figure.

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Consider position vectors of the two points and check linear independence by evaluating the zero Jacobian matrix determinant for functional dependence.

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\begin{align} \hline x(s) &= r\sin s\cos s\\ y(s) &= r\sin^2 s\\ \hline x(t) &= \frac r2 \cos t\\ y(t) &= \frac r2 (\sin t + 1)\\ \hline \end {align}

Let's rewrite the equations to make them more comparable.

\begin{align} \hline x(s) &= \dfrac r2 \sin 2s\\ y(s) &= \dfrac r2(1 - \cos 2s)\\ \hline x(t) &= \dfrac r2 \cos t\\ y(t) &= \dfrac r2 (1 + \sin t)\\ \hline \end {align}

It seems we need to find a relationship between $s$ and $t$ such that

$$\cos t = \sin 2s \quad \text{and} \quad \sin t = -\cos 2s$$

When we let $2s = \dfrac{\pi}{6} = 30^\circ$, we get

$$\cos t=\dfrac 12 \quad \text{and} \quad \sin t=-\dfrac{\sqrt 3}{2}$$

Which corresponds to the angle $t = -\dfrac{\pi}{3} = -60^\circ$. So it seems that $$t = 2s - \dfrac{\pi}{2} = 2s - 90^\circ.$$ We need to check that.

\begin{align} x(t)&= \dfrac r2 \cos t\\ &= \dfrac r2 \cos( 2s - \dfrac{\pi}{2})\\ &= \dfrac r2 ( \cos 2s \, \cos \dfrac{\pi}{2} + \sin 2s \, \sin \dfrac{\pi}{2} ) \\ &= \dfrac r2 \sin 2s \\ &= r \sin s \cos s \\ &= x(s)\\ y(t)&= \dfrac r2 (1 + \sin t)\\ &= \dfrac r2 (1 + \sin( 2s - \dfrac{\pi}{2}))\\ &= \dfrac r2 (1 + \sin 2s \, \cos \dfrac{\pi}{2} - \cos 2s \, \sin \dfrac{\pi}{2} ) \\ &= \dfrac r2(1-\cos 2s)\\ &= r \sin^2 s\\ &= y(s) \end{align}

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