Parametrization of a Cycloid Example Solution

Question

In this example solution, for $\overrightarrow{OC}$, why is it $12\sqrt{2}\lt\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\gt$ ? Isn't the formula for the general equation supposed to be $\lt a\theta, a \gt$ where $a$ is the radius? How do we figure out what $\theta$ is? More precisely, why do they multiply the speed by the components $\lt\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\gt$? Also for $\overrightarrow{CP}$, shouldn't the $x$ component have the sin argument while the $y$ component should have the cosine argument?

Answer 1

The unit vector in the direction of $(1,1)$ is $$ \left \langle \frac{1}{\sqrt{2}} , \frac{1}{\sqrt{2}} \right \rangle $$ Assuming the disc's center starts at the origin, since it is moving in that direction at a rate of $12 \sqrt 2$ units per second, its position is $$ 12 \sqrt 2 t\left \langle \frac{1}{\sqrt{2}} , \frac{1}{\sqrt{2}} \right \rangle = \langle 12 t , 12 t \rangle $$ after $t$ seconds.

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The "standard" parameterization of the origin-centered circle of radius $r$ is given by $$ \begin{cases} x(t) = r \cos t \\ y(t) = r \sin t \\ t \in [0,2\pi) \end{cases}$$

This makes a revolution every $2\pi$ units of time. To make it do $f$ revolutions in that timeframe, replace $t$ with $ft$: $$ \begin{cases} x(t) = r \cos ft \\ y(t) = r \sin ft \\ t \in [0,2\pi) \end{cases}$$ (This is entirely natural if you've seen this parameterization animated, e.g. here or here.) To simplify matters, it is often convenient to let $t \in [0,1)$ instead, so we need to multiply $ft$ by $2\pi$ to account for that: $$ \begin{cases} x(t) = r \cos 2\pi ft \\ y(t) = r \sin 2\pi ft \\ t \in [0,1) \end{cases}$$ Hence, if you want to do $3$ revolutions per second, and $t$ measures in seconds, then $f=3$. Here, $r=2$ as given, so we get $$ \begin{cases} x(t) = 2 \cos 6\pi t \\ y(t) = 2 \sin 6\pi t \\ t \in [0,1) \end{cases}$$ which the linked text expresses in a vector form in the obvious manner.

You can, technically, use $\cos \theta$ for $x$ and $\sin \theta$ for $y$, if you please, but it tweaks some things, makes some aspects of mathematics and calculus a little unpleasant, and is not the usual parameterization.

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Isn't the formula for the general equation supposed to be $\lt a\theta, a \gt$ where $a$ is the radius? How do we figure out what $\theta$ is?

No clue what you're talking about here.