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If I change units of an angle for radians to degrees in the next expresion $$4\sin (\theta) \frac{d \theta}{d t}=\frac{dy}{dt}$$ the value of $$\frac{dy}{dt}$$ changes.

For example at a rate of change of $\frac{d\theta}{dt} = 30deg , \qquad $ and $\frac{d\theta}{dt}=\frac{\pi }{6}rad$ the rate of change is the same, but the final expresion is not.

So which is the correct unit? and mathematicaly why is the reason?.

I already know that the correct unit are radians, Im looking for a more formal and deeper explanation of why this units are the correct.

Thanks

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closed as unclear what you're asking by Ben Crowell, Brandon Enright, JamalS, Neuneck, ACuriousMind Nov 10 '14 at 13:36

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  • $\begingroup$ Related: physics.stackexchange.com/q/33542/2451 and links therein. $\endgroup$ – Qmechanic Nov 9 '14 at 18:23
  • $\begingroup$ Both are units of angle. Why should one be more correct than the other? (Note that the sine is defined in a way that the function takes radians input, though) $\endgroup$ – ACuriousMind Nov 10 '14 at 13:36
  • $\begingroup$ @ACuriousMind: One must be more correct, becouse if you use one or the other, the final expression of $\frac{dy}{dt}$ will change his value. $\endgroup$ – JuanMuñoz Nov 11 '14 at 17:30
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For dimensional consistency, firstly, I would expect the number '4' to be dimensionfull. Additionally, whether the angle is to be taken in radians or in degrees depends on where this equation came from. If I venture a guess, then I suppose at some point you differentiated a relation between $y$ and $\theta$. In that case, the relation you started of with will specify the units for $\theta$ after which the differentiation would have to carried out accordingly. Remember, $$ \dfrac{\mathrm{d}}{\mathrm{d}\theta}\cos{\theta}=-\sin\theta\\ \dfrac{\mathrm{d}}{\mathrm{d}\theta}\cos{\theta^{\circ}}=-\dfrac{\pi}{180}\sin\theta^{\circ}\\ \dfrac{\mathrm{d}}{\mathrm{d}\theta^{\circ}}\cos{\theta^{\circ}}=-\sin\theta^{\circ}\\ $$

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  • $\begingroup$ Where @surajshankar: did you get that only if $\theta$ is in radians then the derivative is $-\sin (\theta)$. I know that the demonstration if this rule, is a limit and never uses the hipotesis that $\theta$ is in radians $\endgroup$ – JuanMuñoz Nov 9 '14 at 20:54
  • $\begingroup$ Ok, small actually I have been sloppy with my notation. What is usually done is, when dealing with angles, they are invariably related to some length measures in the problem and hence radians is the most convenient unit to use and this usage follows through. $\endgroup$ – surajshankar Nov 10 '14 at 3:07
  • $\begingroup$ @JuanMuñoz I have made an edit in the last bit and changed the notation to make it clearer. Hope this helps. You can also look at math.stackexchange.com/questions/466299/…, math.stackexchange.com/questions/214912/… and math.stackexchange.com/questions/526414/… $\endgroup$ – surajshankar Nov 10 '14 at 3:12

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