Why is dθ/dt Always in Radians?

[Seung Tai Kang]

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?

 

[Charles Link]

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.

 

[anorlunda]

I have to solve for dθ/dt and whenever I do it the answer comes naturally in radians.

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.

 

[kuruman]

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.

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.)

 

[CollinsArg]

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.

Please, could you extend this? What do you mean by it, if I calculate in degrees won't it work? In which situation?

 

[anorlunda]

Please, could you extend this? What do you mean by it, if I calculate in degrees won't it work? In which situation?

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

In radians $\sin{0.017}\approx{0.017}$, but in degrees $sin{1}\neq{1}$.

 

[_PJ_]

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.

 

[Charles Link]

Please, could you extend this? What do you mean by it, if I calculate in degrees won't it work? In which situation?

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.

 

[BP Leonard]

We often see warnings such as: "such-and-such a formula is true only if theta is 'measured in radians'." For example: "s = r theta" or "d[sin(theta)]/d(theta) = cos(theta)." Unfortunately, the warning is not correctly (or completely) stated. What is meant is: "theta is the (dimensionless) NUMERICAL VALUE of the angle when the angle is expressed in radians," in other words: theta = (angle of interest)/rad.

In order to avoid confusion, let me use theta to represent an angle (dimension A); and theta* to represents the numerical value of theta when theta is expressed in radians: theta* = theta/rad. The above (correct) formulas are: s = r theta* = r theta/rad and d[sin(theta*)]/d(theta*) = cos(theta*) or d[sin(theta/rad)]/d(theta) = cos(theta/rad). Then it doesn't matter what units are used for (the angle) theta.

We need to distinguish between "geometric" trigonometric functions (ratios of lengths of sides in a right-angled triangle) for which the argument is the angle theta (dimension A) and "mathematical" trigonometric functions: solutions of certain differential equations for which the argument is the independent variable, a (dimensionless) number (dimension: 1). Let me use an initial capital for the former. Then:

Sin(theta) = opp/hyp = sin(theta*) = sin(theta/rad)

and so on for the other trigonometric functions. The length ratios and mathematical functions are themselves dimensionless.

In terms of geometric functions of the (dimensional) angle theta, we have, for example:

d[Sin(theta)]/d(theta) = Cos(theta)/rad

easily proved geometrically, using properties of similar triangles (and s = r theta/rad).

Note the dimensional consistency: dim(LHS) = 1/A; dim(RHS) = 1/A. And, since d[Cos(theta)]/d(theta) = –Sin(theta)/rad, if we define y(theta) = Sin(theta) we find that y is the solution of:

d^2y/d(theta)^2 + y/rad^2 = 0

which, with appropriate "initial" conditions, yields: y(theta) = sin(theta/rad).

 

[anorlunda]

We often see warnings such as: "such-and-such a formula is true only if theta is 'measured in radians'."

You may be interested in this PF Insights article in which the author, Derek Bolton, offers arguments for assigning a unit for angle. Also in the ensuing discussion of that article where may others argue against that.

 

[Mark44]

Sin(theta) = opp/hyp = sin(theta*) = sin(theta/rad)

The notation you used here and elsewhere in your post is confusing -- it looks like you're dividing theta by "rad".

d[Sin(theta)]/d(theta) = Cos(theta)/rad

Note the dimensional consistency: dim(LHS) = 1/A; dim(RHS) = 1/A.

1/A? What does that mean?

 

[anorlunda]

The OP question has been well answered.

Thread closed.