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In this question, it's said that the path of a cycloid can be given as this parametric equation:

$$\begin{align*}x &= r(t - \sin t)\\ y &= r(1 - \cos t)\end{align*}$$

and is shown here:

cycloid animation

With unit radius, we can say

$$p(t)=<t-\sin t, 1- \cos t>.$$

I don't understand why we have minus signs in those terms. Imagine the vector from the origin to the point $p$ on the circumference as (a vector from the origin to the bottom of the circle at $t$) + (a vector from the bottom of the circle to the circle's center) + (a vector from the center of the circle to $p$), the only negatives would appear in the third summand. In particular, they would have both negative components when the point is both below and left of the center -- which should only happen 1/4 of the time, not all the time.

So why is it minus all the time, not just 1/4 of the time?

Community
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Hatshepsut
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  • For all $t$ you have $-1 \le \cos t \le 1$ so $1-\cos t \ge 0$ and is never negative, meaning the curve is never below the $x$-axis. Meanwhile $t- \sin t$ is negative precisely when $t$ is negative, i.e. when the curve is to the left of the $y$-axis – Henry Feb 23 '19 at 23:43

2 Answers2

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Note that $-\sin t$ and $-\cos t$ are both negative for exactly the $\frac14$ of the time that you are expecting.

The minus sign in the expression for $y$ is necessary to keep the starting point on the "ground." The minus sign in the expression for $x$ could be changed, in which case the wheel would be rolling to the left instead of to the right.

Mark Fischler
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Okay so we have
Cycloid Image
So we know that, like you said, we can express the position vector of $P$, as the sum of three other vectors, that is, according to this image, $$\vec{OP}=\vec{OT}+\vec{TC}+\vec{CP}$$ And I'm not going to go through it all because you already know it, but what seems to be giving issues is that $$\vec{CP}=<-r\sin(\theta),-r\cos(\theta)>$$ and you are wondering why it is always $-r$ because it looks like the signs of the components of this vector should be changing every quadrant of the circle. Well in this case it is helpful to refer back to the unit circle.
Unit Circle
And we can see that when $\theta \le \frac{\pi}{2}$, both components of the vector should be negative; which lines of with both sine and cosine being positive and then multiplied by the negative coefficient $-r$. Then for $\frac{\pi}{2}<\theta\le \pi$ we know that the vertical component should be positive and the horizontal should be negative, which once again lines up with the fact that sine remains positive but cosine becomes negative.
And so on, you can continue to check these and they all checkout, hence the coefficient $-r$ should always be negative and you end up with a minus in the final position vector.

Will Fisher
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