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The task is to express the length of an arc of a circle trapped between two radii named $r$ if the angle between them is infinitesimally small, named $d\theta$.

The solution to this problem is supposed to be:

$$l=r \cdot {d\theta}$$

but I do not understand why this would be the case.

I've provided a simple Paint sketch for more details: enter image description here

Since $d\theta$ is infinitesimally small, line $\overline{BE}$ is infinitesimally small as well, so we note that with $\overline {BE} = dx$. Since $dx$ is so small, its projection $\overline {BD}$ and circular chord $\overline{BC}$ can be approximated as: $$\overline{BD} \approx \overline{BC} \approx dx$$ The similar can be concluded for the radii: $$\overline{AB} = \overline{AC} = r$$ $$r \approx \overline{AD} \approx \overline {AE}$$

The task was to find the length of an arc limited with the chord $\overline {BC}$, that is, trapped between $\overline{AB}$ and $\overline{AC}$. The solution should be $l=r \cdot {d\theta}$, but how?

Formula for deducing arc length is: $$l=2r\pi \frac{d\theta}{360°} = r\pi \frac{d\theta}{180°}$$

The only way for this to become $l=r\cdot d\theta$ is if $\pi$ and $180°$ would somehow cancel each other out. $\pi rad$ indeed has the value of $180°$, but this $\pi$ has the meaning of length, that is $\pi \approx 3.14$, and not radians.

If there I did not specify something enough, please let me know, so I can explain myself better. Thank you in advance.

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    $\begingroup$ The solution $r \cos d\theta$ is wrong. It should be $r d\theta$, from the exact solution $2\sin(\frac12 d\theta)$. $\endgroup$
    – TonyK
    Nov 5, 2015 at 11:07
  • $\begingroup$ @TonyK with all due respect, I highly doubt this. The solution comes from a second edition of a validated textbook and is used in every example following this one. I am aware that the solution is only an approximation, but if you have a proof to support your statement, I would highly appreciate if you could share it. $\endgroup$
    – 0lt
    Nov 5, 2015 at 11:11
  • $\begingroup$ With all due respect, all you have to do is let $d\theta$ tend to zero to see that your esteemed textbook has screwed up. $\endgroup$
    – TonyK
    Nov 5, 2015 at 11:13
  • $\begingroup$ @postmortes I edited the question. The solution is supposed to be $r \cdot d\theta$ , NOT $r \cdot \cos{d\theta}$ $\endgroup$
    – 0lt
    Nov 5, 2015 at 11:13
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    $\begingroup$ This is a problem of units. If $\mathrm d\theta$ is in radians then the length is $r\,\mathrm d\theta$. But if $\mathrm d\theta$ is in degrees, the length is $r\pi\,\cfrac{\mathrm d\theta}{180^\circ}$... $\endgroup$
    – user856
    Nov 5, 2015 at 11:21

1 Answer 1

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Consider a circle has radius $r$ units. length of circumference is $2\pi r$ which produces total angle $2 \pi$ radians. unit radian angle produce arc length$= \frac{2\pi r}{2\pi} = r$ units "$\theta$ radian (no matter how small or big it is )" angle produces arc length $= \theta r$.

ANOTHER WAY

Two end points of the arc is very close to each other forming an angle $dA$ in radians. For small angle $\tan dA = dA$. you can also draw a tangent at that point.

For right-angle triangle: $$\tan dA = dA = \text{length of arc / radius} $$

so length of arc $= r dA$ ...

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