What is the unit of $\frac{d\theta}{dt}$?

Let's say I have a triangle with an angle $\theta$ and the opposite and hypotenuse sides $y$ and $h$, respectively, so that

$\sin \theta$ = $\frac{y}{h}$

Taking the derivative of both sides with respect to $t$:

$\cos \theta$ $\frac{d\theta}{dt}$ = $\dfrac{\frac{dy}{dt}h-\frac{dh}{dt}y}{h^2}$

The units of the right side, where $m$ is distance (meters) and $s$ is time (seconds):

$\dfrac{\dfrac{m}{s}m-\dfrac{m}{s}m}{m^2}$ = $\dfrac{1}{s}$

Since $\cos \theta$ is a ratio, it has no units, therefore:

$\dfrac{d\theta}{dt}$ = $\dfrac{1}{s}$

However, by common sense, $\dfrac{d\theta}{dt}$ should be in $\dfrac{^o}{s}$ (if using degrees).

Why is it different, and if I am making a mistake, how could it be corrected?


Because the default unit for angle measure in calculus is the radian, which is dimensionless. Remember the definition of a radian: it is the ratio between two lengths (an arc length to the radius). Hence the units of length cancel out from either side.

So in calculus, you shouldn't be using degree measure unless you're absolutely certain it won't affect the final answer.

  • $\begingroup$ Edited to take out the part about it not making a difference in this case. It will make a numerical difference. If your rate is given as (say) a steady increase in angle of $30$ degrees per second, and you're asked for the rate of change of $\sin \theta$ at the instant when the angle is $60$ degrees, then the correct answer is not simply $\frac{1}{2}\cdot (30) = 15$ (and what units?), but rather $\frac{1}{2} \cdot \frac{\pi}{6} = \frac{\pi}{12}$, and the unit is $s^{-1}$. $\endgroup$ – Deepak Oct 3 '14 at 1:11

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