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Why is d(theta)/dt in radians?

  1. Oct 16, 2017 #1
    This is just a conceptual question I don't understand.
    I am learning Calculus and rate of changes. There are many times where I have to solve for dθ/dt and whenever I do it the answer comes naturally in radians.
    Could someone explain to me why this is so?
     
  2. jcsd
  3. Oct 16, 2017 #2

    Charles Link

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    The arc distance on a circle is ## s=r \theta ##, if ## \theta ## is measured in radians. In addition, ## \frac{ d \sin(\theta)}{d \theta}=\cos(\theta) ##, and ## \frac{ d \cos(\theta)}{d \theta}=- \sin(\theta) ##, but these derivatives only hold if ## \theta ## is measured in radians. Likewise ## \sin x=x-\frac{x^3}{3!} +... ## as a Taylor series expansion, but only if ## x ## is measured in radians.
     
  4. Oct 16, 2017 #3

    anorlunda

    Staff: Mentor

    Don't you mean radians/second?

    You can use any unit you choose for angle, degree, radians, cycles. We prefer radians because of the advantages that @Charles Link said.
     
  5. Oct 16, 2017 #4

    kuruman

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    If you do dimensional analysis, θ is measured in radians, therefore dθ/dt is measured in rad/s or just s-1. Note that the θ is a dimensionless quantity defined as follows: Given an arc s on a circle of radius r, the angle subtended by the arc is θ = s/r. Being the ratio of two lengths, the angle is a dimensional quantity. We usually tack the units "radians" to an angle measure (of which there are 2π in a full circle) to distinguish the number from "degrees" (of which there are 360 in a full circle.)
     
  6. Oct 17, 2017 #5
    Please, could you extend this? What do you mean by it, if I calculate in degrees wont it work? In which situation?
     
  7. Oct 17, 2017 #6

    anorlunda

    Staff: Mentor

    Even easier. For small angles, we can use the approximation ##\sin\theta=\theta##.

    In radians ##\sin{0.017}\approx{0.017}##, but in degrees ##sin{1}\neq{1}##.
     
  8. Oct 17, 2017 #7
    That's really an issue with the approximation and what the relationships of the trig identities really mean.

    A circle is still a circle if the radius is changed from 0.0192512 microns or 581236823196589 lightyears or 1 arbitraryunitlength
    A circle is still a circle if the angular distance is 1440 degrees or 2*Pi radians or 1 period.

    Being a circle, the relations hold exactly perfectly.

    You can see this exemplified by drawing two radii from an origin, spaced 1 degree apart. The lines are extremely close at the small radial length scale, they may even may appear parrallel, but extend these radii to much larger distances and the separation between them grows accordingly. HOWEVER no properties of the angle are at all altered in any way by extending the lengths of these radii. Whatever units you choose to calculate the angle between them, the result is still the same, even if approximations may be used in one system only at certain scales.

    What's happening is that one presumes the question is given such as "Calculate the derivative of (some equation with Theta) with respect to time" whilst no mention is provided of the units in which Theta is given, and are largely irrelevant unless expected to evaluate. The result, then is typically as a ratio. Such a ratio is exactly what radians are, though ratios to exactly 2*pi rather than ratios to a single cycle (normalised wavelength). The relational values of ratios will hold regardless of units provided consistency is maintained.
     
  9. Oct 17, 2017 #8

    Charles Link

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    You can use either degrees or radians when plugging into your calculator, but the derivative calculation ## f'(\theta)=\frac{ sin(\theta+\Delta \theta)-sin(\theta)}{\Delta \theta} =cos(\theta) ## makes the assumption that ## limit \, \Delta \theta \rightarrow 0 \, \frac{\sin(\Delta \theta)}{\Delta \theta}=1 ##. That is only the case if ## \theta ## is measured in radians.
     
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