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Consider on of those rising balloon related rates Calc problems. Based on the actual problem, you'd label a triangle with a few sides and one of the angles as $\theta$. You set up some trig relationship, and then take the derivative. $$tan(\theta)=\frac{y}{100}$$ $$ sec^2(\theta)\frac{d\theta}{dt}=\frac{1}{100}\frac{dy}{dt}$$

Right there, now that we have a $\frac{d\theta}{dt}$, there is an implied unit, since this refers to the rate of change of the angle SIZE. Is the unit degrees or radians? We never actually refer to a specific size, so it stays ambiguous.

Sub in some givens, and get something like $$\frac{d\theta}{dt}=7 \frac{rad}{min}$$

Yet, nowhere in the problem have I actually referred to the actual size of $\theta$. All we ever dealt with was the trig relationship of the sides. So, can't it be measured in either radians or degrees?

So, my question is why can't the solution just as be $$\frac{d\theta}{dt}=7 \frac{degrees}{min}??$$ Who ever stated it's measure was in radians? Is this other link related? Something to do with implied radians when you take the derivative?

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JackOfAll
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3 Answers3

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You are correct that when you write $\tan \theta = \frac y{100}$ there is nothing with a scale and you can measure the angle in degrees or radians. When you write $\frac {d\tan \theta}{d\theta}=\sec^2 \theta$ you are measuring the change in angle in some unit. Radians are the natural one and the one we prove this identity (as well as all the others) in. You could write $\frac {d\tan \theta^\circ}{d\theta^\circ}=\frac {\pi}{180}\sec^2 \theta^\circ$ if you want, where the $\frac \pi{180}$ comes from applying the chain rule.

Ross Millikan
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  • Can you explain how the chain rule would apply when doing the derivative of tan in degrees? Does units come up when you take the derivative of tan(x) using the limit definition of the derivative? I have seen it derived this way based on $tan(x) = \frac{sin(x)}{cos(x)}$ but there is no mention of units. http://www.wyzant.com/resources/lessons/math/calculus/derivative_proofs/tanx – JackOfAll Feb 27 '15 at 23:32
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    With $x$ in radians, $d\tan x/dx = \sec^2 x$, which can be proven by basic trig sum and difference rules and the special limit that $\lim_{x\to0} \sin x / x = 1$ for $x$ in radians. However if $x^\circ$ is in degrees we really have $\tan (x^\circ) = \tan(\pi/180 * x)$, so $d\tan x ^\circ /dx = d\tan(\pi/180 * x)/dx = \pi/180 * \sec^2(x^\circ)$ (by the chain rule). – David Etler Feb 27 '15 at 23:39
  • ie: Why do you have to convert the degree version into radians via the pi/180 in the first place? Who says radians is the default? Why doesn't it work both ways? You don't take the radian version of tan(x) and convert it to degree, by introducing 180/pi, do you? Why not just take the derivative in degree units? – JackOfAll Feb 28 '15 at 17:41
  • @JackOfAll, it comes from the fact that $\lim_{x\to 0}\frac{\sin x}{x}=1$ when and only when $x$ is in radians (which itself comes from squeeze theorem). When we set up $d\sin x/dx$ in radians we get $\cos x$ because of this property as well as trig sum-and-difference formulas. In degrees, this is not the case (the special $\sin x /x=1$ limit doesn't hold) so our derivatives are a bit more complicated. In fact, $\lim_{x\to 0}\sin x^{\circ}/x = \pi/180$. Because of this, our trig derivatives are only other trig functions multiplied by $\pm1$ when $x$ is in radians. – David Etler Mar 06 '15 at 16:26
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This is in radians because $\frac{d\tan\theta}{d\theta} $ only equals $(\sec^2 \theta)\frac{d\theta}{dt}$ if $\theta$ is in radians. Otherwise it is equal to $(\sec^2 \theta)\frac{d\theta}{dt} * \frac{\pi}{180}$ by the chain rule (and essentially that factor is a conversion from degrees to radians). At the end of the day it comes down to our derivatives of trig functions being other trig functions times one when $\theta$ is in radians, which is given mainly from the derivitive rules for sine and cosine.

David Etler
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In general, with mathematics, assume use radians unless otherwise stated.

The main reason for this is related to what David and Ross said and how the trig functions relate to the Taylor Series (you should learn about this in the future).

John Nicholson
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