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So a classic thing to derive in calculus textbooks is something like a statement as follows

Is $\frac{d}{dx}\sin(u)$ the same as the derivative of $\frac{d}{dx}\sin(x)$ where $u$ is an angle measured in degrees and $x$ is measured in radians? and of course the answer is no because of the chain rule.

Except usually this is ambiguously worded as "Is the derivative of $\sin(u)$, where $u$ is measured in degrees, equal to the derivative of $\sin(x)$ where $x$ is the same angle but measured in in radians?"

Then the texts go on to say something like "No and this why we don't work in degrees and instead chose to work in radians, to avoid all the messy constants that come out of taking derivatives." Am I crazy by thinking this is an odd thing to say that will end up confusing students. If your independent variable was an angle measured in degrees, you are probably more interested in it's derivative with respect to degrees not radians, which would infact be equal at the corresponding degrees and radians of an angle. Is my understanding wrong here. Is what the books say fine? I think at minimum they should at least be clear that we are taking the derivative with respect to radians, no?

Note this is not a duplicate of

Derivative of the sine function when the argument is measured in degrees

Even though it is highly related.

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DO not use geometry to define sine :) –  Hagen von Eitzen Oct 14 '13 at 22:44
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@HagenvonEitzen, why not? What definition do you prefer? –  dfeuer Oct 14 '13 at 22:59
    
user1544793, why are you teaching degrees? –  dfeuer Oct 14 '13 at 23:32
    
I'm not teaching degrees, this is a standard result derived in most elementary calculus texts. I encourage my students to read the book. –  MHH Oct 14 '13 at 23:33
    
If the book words the problem incorrectly, you can reword it for them and discourage them from reading that bit... –  dfeuer Oct 14 '13 at 23:58
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3 Answers

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The essential problem is writing "$\sin u$, where $u$ is measured in degrees". This doesn't really mean anything. If you want degrees, you need to write them explicitly: $\sin u^\circ$. If you want the sine of an angle, you need to write that: $\sin\angle ABC$ or similar. At some point in geometry or trigonometry class, someone should be teaching that $\vphantom t^\circ=\frac{2\pi}{360}$.

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Very true, and I think this is where the confusion lies with most students. What they mean by $\sin(u)$ is $\sin(u)$ where $u=(\pi/180)x$ where $x$ corresponds to our usual definition of the domain of the sine function. You can still take the derivative with respect to $u$ though and I think its necessary to point out explicitly that this is not what they are doing. The question came up because one student thought this was the same as plugging sin(4) in radian mode into their calculator and plugging sin(4) in degree mode into their calculator and noting they were different. –  MHH Oct 14 '13 at 23:49
    
although hopefully they teach them degree$=2\pi/360$ ;) –  MHH Oct 14 '13 at 23:50
    
@user1544793, yes, of course. I briefly forgot that detail of ancient Babylonian geometry. –  dfeuer Oct 14 '13 at 23:52
    
@user1544793, what do you mean by that "What they mean" sentence? If any student writes $\sin 30$ to mean $\sin 30^\circ$, they should be dinged a point. It's best to catch bad habits early! Also, students should be strongly discouraged from using their calculators in "degree mode", and if calculators are used for the class they should be shown how to put them in radian mode. –  dfeuer Oct 14 '13 at 23:55
    
Oh no thats not what I was saying. We aren't teaching degrees, or using calculators, at all. The student was just making an analogy to the calculator, as to show that he thought the problem was not very insightful. But I pointed out that the analogy didn't quite make sense, although his general conclusion about the problem not being very insightful is probably a reasonable opinion. –  MHH Oct 15 '13 at 0:00
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  • If $x$ is measured in degrees then $\frac{d}{dx} \sin(x)= \cos(x)$.
  • If $x$ is measured in radians and $u$ in degrees then $\frac{d}{dx} \sin(u) = \frac{d}{dx} \sin(\frac{180^\circ}{\pi} x)= \frac{180^\circ}{\pi}\cos(u)$.

There nothing about the units of $x$ which will come into play when differentiating $f(x)$ with respect to $x$. If you are differentiating a function of some other variable with respect to $x$ then you must use that chain rule $\frac{d f(u)}{dx} = \frac{du}{dx} \frac{d f}{du}$.

In summary if you are differentiating a function of a variable with respect to the same variable you never have to worry about the units of the variable.

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Which is obviously what I said at the very beginning. I fully understand the concept, this was a pedagogical question about presentation. My point is $\frac{d}{du} sin(u) = cos(u)$ but students when reading "Is the derivative of sin(u), where u is measured in degrees, equal to the derivative of sin(x) where x is the same angle but measured in in radians?" often think the books explanation means $\frac{d}{du} sin(u) \ne cos(u)$. –  MHH Oct 14 '13 at 23:28
    
I see I misunderstood your question. On a first reading I thought you were asking if it was necessary to specify the units you were differentiation with respect to which is obviously a non-issue (the variables are either the same or not and if they aren't use chain rule). –  Spencer Oct 14 '13 at 23:43
    
This is just my opinion, but I think that students in a calculus course need to be expected to stand on their own a little more. If they take the wrong message away from what the book is saying its because the didn't read it carefully enough (like me with your question). –  Spencer Oct 14 '13 at 23:45
    
Also: I wouldn't every even bring up the distinction unless a student asked me about it. To me its not important enough to bring up in a lecture. –  Spencer Oct 14 '13 at 23:47
    
oh yes. This wasn't in lecture, it was in office hours. Although I think we should talk to students about their misunderstandings, especially if they are interested in the question. I want to give them a lot of positive reinforcement that they are reading the book. Most students these days don't do that, although potentially this is because the ratio of words to content in intro calc books is kind of getting out of control. –  MHH Oct 14 '13 at 23:54
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Yes, you're describing the same object $x$, but units matter. That is, the units you use to measure that object matter in describing what you mean. Like, 1 feet and 3.2808 meters are practically the same length, but are described through different units of measure.

And in Calculus, radians is a unit of measure that will give you the least amount of headaches (less tracking of constants, etc.) when calculating derivatives, integrals, etc. It's similar to why $\log_e=\ln$, in Calculus, is called the natural logarithm: $\frac{d}{dx}\ln x=\frac{1}{x}$, while $\frac{d}{dx}\log_a x=\frac{1}{x\ln a}$.

In fact, radians (as a measure of angles) is arguably a better measurement to define and use than degrees. Given a circle of radius $r$, if you're interested in the length of the arc of a circle, arc given by an angle $\theta$:

In degrees: the length of the arc of a circle is $r\left(\frac{\theta\pi}{180}\right)$

In radians: the length of the arc of a circle is simply $r\theta$.

Fundamentally, calculating $\frac{d}{dx}\sin x$ boils down to solving the limit $\lim_{x\to 0}\frac{\sin x}{x}$. And what does one mean by $\frac{\sin x}{x}$?

Geometrically, and seeing $x$ in radians, one can say $\frac{\sin x}{x}$ is simply the ratio of the length of a side of a right-triangle ($\sin x$) and the length of an arc of a circle ($x$). Describing $\frac{\sin x}{x}$ this way, we get a ratio of lengths.

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