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I'm studying about involute of a plane curve from here and there's a small point that is really bothering me and I can not understand it.

Assuming curve is parameterized by arc length, the involute is defined as $$\gamma(t) = \beta(t) - t \beta'(t) $$

Why is there a negative sign? Like it says in the link I added, the bob's position is at distance $t$ in direction $-\beta'(t)$. I can't see how this negative sign has to be there for every curve in general. If the tangent line of $\beta$ at point $t$ is $\beta(t) + \lambda \beta'(t)$, then starting from $\beta(t)$ I can move along the line is either direction depending on $\lambda$. Why must it be negative in the case of the involute?

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  • $\begingroup$ $s = t$ in your linked page. Why not here as well? Cheers! $\endgroup$ – Robert Lewis May 9 at 0:47
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    $\begingroup$ Edited it, thanks! $\endgroup$ – 9Sp May 9 at 0:56
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The model for the involute is this: Take a circle, with scotch tape wrapped around it. Start to peel the scotch tape off and follow the point $P$ at the end of the tape. As you go counterclockwise around the circle, the tape is tangent, but the line segment from the point of contact to $P$ goes in the direction opposite to the (counterclockwise) tangent vector to the circle.

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    $\begingroup$ Thank you! so if I understood correctly, the tangent vector is pointing in the direction of increasing $t$, but in order to actually draw the involute, we start from from point of contact going to $P$ in the opposide direction $\endgroup$ – 9Sp May 9 at 1:10
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    $\begingroup$ Right. You want to think of unwrapping a string from the curve :) $\endgroup$ – Ted Shifrin May 9 at 1:10
  • $\begingroup$ +1 For explaining the model with scotch tape. $\endgroup$ – Ernie060 May 9 at 13:41

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