<s>[INST] 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?
 [/INST] $\begin{align}
\hline
x(s) &= r\sin s\cos s\\
y(s) &= r\sin^2 s\\
\hline
x(t) &= \frac r2 \cos  t\\
y(t) &= \frac r2 (\sin t + 1)\\
\hline
\end {align}$ 
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}
</s>