1
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ 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 '15 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 '15 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 '15 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 '15 at 11:13
  • 2
    $\begingroup$ 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 $r\pi\,\cfrac{\mathrm d\theta}{180^\circ}$... $\endgroup$ – Rahul Nov 5 '15 at 11:21
-1
$\begingroup$

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 $= r dA$ ...

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.