How to check if a student really can perform the Chain Rule

In my experience, a problem like the following is the acid test for determining if a student really understands the Chain Rule:

Find $f'(x)$ if $f(x) = \left[6x^2 + \sin 5x \right]^3$

The correct answer (unsimplified):

$f'(x) = 3 \left[6x^2 + \sin 5x \right]^2 \left(12x + [\cos 5x] \cdot 5 \right)$

However, even students that are quite proficient with the Chain Rule can often provide the following incorrect answer:

$f'(x) = 3 \left[6x^2 + \sin 5x \right]^2 \left(12x + \cos 5x \right) \cdot 5$

Notice the slightly incorrect placement of the $5$ at the end of the derivative. Students can so easily get into the rhythm of just multiplying by the derivative of the inside that they can forget where the derivative of the inside should be placed.

Needless to say, a problem like this often appears on my exams as a way of separating the A students from the B students.

Teaching the Chain Rule inductively

I taught Calculus I every spring between 1996 and 2008. Perhaps the hardest topic to teach — at least for me — in the entire course was the Chain Rule. In the early years, I would show students the technique, but it seemed like my students accepted it on faith that their professor knew what he was talking about it. Also, it took them quite a while to become proficient with the Chain Rule… as opposed to the Product and Quotient Rules, which they typically mastered quite quickly (except for algebraic simplifications).

It took me several years before I found a way of teaching the Chain Rule so that the method really sunk into my students by the end of the class period. Here’s the way that I now teach the Chain Rule.

On the day that I introduce the Chain Rule, I teach inductively (as opposed to deductively). At this point, my students are familiar with how to differentiate $y = x^n$ for positive and negative integers $n$ , the trigonometric function, and $y = \sqrt{x}$ . They also know the Product and Quotient Rules.

I begin class by listing a whole bunch of functions that can be found by the Chain Rule if they knew the Chain Rule. However, since my students don’t know the Chain Rule yet, they have to find the derivatives some other way. For example:

Let $y = (3x - 5)^2$ . Then

$y = (3x - 5) \cdot (3x -5)$

$y' = 3 \cdot (3x -5) + (3x -5) \cdot 3$

$y' = 6(3x-5)$ .

Let $y = (x^3 + 4)^2$ . Then

$y = (x^3 + 4) \cdot (x^3 + 4)$

$y' = 3x^2 \cdot (x^3 + 4) + (x^3 + 4) \cdot 3x^2$

$y' = 6x^2 (x^3 + 4)$

Let $y = (\sqrt{x} + 5)^2$ . Then

$y = x + 10 \sqrt{x} + 25$

$y' = 1 + \displaystyle \frac{5}{\sqrt{x}}$

Let $y = \sin^2 x$ . Then

$y = \sin x \cdot \sin x$

$y' = \cos x \cdot \sin x + \sin x \cdot \cos x$

$y' = 2 \sin x \cos x$

Let $y = \sin 2x$. Then

$y = 2 \sin x \cos x$

$y' = 2 \cos x \cos x - 2 \sin x \sin x$

$y' = 2 (\cos^2 x - \sin^2 x)$

$y' = 2 \cos 2x$

The important thing is to list example after example after example, and have students compute the derivatives. All along, I keep muttering something like, “Boy, it would sure be nice if there was a short-cut that would save us from doing all this work.” Of course, there is a short-cut (the Chain Rule), but I don’t tell the students what it is. Instead, I make the students try to figure out the pattern for themselves. This is absolutely critical: I don’t spill the beans. I just wait and wait and wait until the students figure out the pattern for themselves… though I might give suggestive hints, like rewriting the $6$ in the first example as $\latex 3 \times 2$.

This can take 20-30 minutes, and perhaps over a dozen examples (like those above), as students are completely engaged and frustrated trying to figure out the short-cut. But my experience is that when it clicks, it really clicks. So this pedagogical technique requires a lot of patience on the part of the instructor to not “save time” by giving the answer but to allow the students the thrill of discovering the pattern for themselves.

Once the Chain Rule is discovered, then my experience is that students have been prepared for differentiating more complicated functions, like $y = \sqrt{4 + \sin 2x}$ and $y = \cos ( \sqrt{x} )$ . In other words, there’s a significant front-end investment of time as students discover the Chain Rule, but applying the Chain Rule generally moves along quite quickly once it’s been discovered.

Getting the right answer the wrong way

I just read “But My Physics Teacher Said… A Mathematical Approach to a Physical Problem,” which was a very interesting pedagogical article concerning the teaching of calculus. Here’s the central problem:

I included on their exam a question involving average velocity. I gave the students a quadratic function and asked them to calculate the average velocity over a given interval… One of my students… got the final numerical answer correct, but he hadn’t used the average velocity formula he had learned in our course. Instead… he had calculated the average of the velocities at the end points of the given interval. When I explained this to him, he stated that he didn’t understand the difference because he had learned the latter formula to calculate average velocity in his physics class.

It turns out that this alternative approach always work under the condition of constant acceleration (i.e., a quadratic function), and since constant acceleration is such an important special case in freshman physics, the formula was presented and the student remembered the formula. Of course, the student probably was not aware of the formula was only generally true under this specific circumstance.

After some pedagogical reflection, the author concluded

My student and I both learned from this experience. He gave me the opportunity to look at a familiar topic with the eye of a physicist, and I taught him the importance of context when using a formula. Specific adventures such as the one my student and I encountered will undoubtedly strengthen my approach to teaching this course and my students’ ability to think like mathematicians.

The full article can be found at http://digitaleditions.walsworthprintgroup.com/publication/?i=187509&p=19.

On derivatives

This note appeared in the October 2012 issue of the American Mathematical Monthly.

Day One of my Calculus I class: Part 6

In this series of posts, I’d like to describe what I tell my students on the very first day of Calculus I. On this first day, I try to set the table for the topics that will be discussed throughout the semester. I should emphasize that I don’t hold students immediately responsible for the content of this lecture. Instead, this introduction, which usually takes 30-45 minutes, depending on the questions I get, is meant to help my students see the forest for all of the trees. For example, when we start discussing somewhat dry topics like the definition of a continuous function and the Mean Value Theorem, I can always refer back to this initial lecture for why these concepts are ultimately important.

I’ve told students that the topics in Calculus I build upon each other (unlike the topics of Precalculus), but that there are going to be two themes that run throughout the course:

Approximating curved things by straight things, and
Passing to limits

I’ve then quickly used these themes to solve two completely different problems: (1) finding the speed of a falling object at impact and (2) finding the area under a parabola. I can usually cover these topics in less than 50 minutes, sometimes in 35 minutes. Again, because I’m not immediately holding my students responsible for the contents of this introduction, I feel freer to move a little quicker than I would otherwise in the hopes of showing the forest for all of the trees.

I then ask the obvious question: what do these two questions have to do with each other. One involves the distance-rate-time formula. The other involves the areas of rectangles. At first blush, these two questions seem completely unrelated. And at second blush. And at third blush.

I tell my class that these two apparently unrelated questions are indeed related by something called the Fundamental Theorem of Calculus. Somehow, the process of finding the area under a curve is intimately related to finding an instantaneous rate of change. I then make a bold, eye-catching statement: The Fundamental Theorem of Calculus is one of the greatest discoveries in the history of mankind, period. And, at the ripe old age of 17, 18, or 19 years old, my students are now privileged to understand this great accomplishment.

This ends my introduction to Calculus I. I’ll then begin the more mundane development of limits on the way to formally defining a derivative.

Day One of my Calculus I class: Part 3

I’ve just told students that the topics in Calculus I build upon each other (unlike the topics of Precalculus), but that there are going to be two themes that run throughout the course:

Approximating curved things by straight things, and
Passing to limits

We are now studying the following problem.

Problem #1. A building on campus is 144 feet tall. A professor takes a particularly annoying student to the top of the building, and throws him (or her) off to his (or her) certain demise. (Usually I pick a student that I know and like as the one to throw off the building. This became a badge of honor over the years.) The distance that the student travels (in feet) after $t$ seconds is $f(t) = 16t^2$ . How fast is the student going when he (or she) hits the concrete sidewalk?

At this point in the lecture, we have done some experimental numerical work with successfully smaller time intervals to find better and better approximations to the speed at impact.

With a time interval of length $3$ seconds, the approximation is $48$ ft/s.
With a time interval of length $1$ seconds, the approximation is $80$ ft/s.
With a time interval of length $0.5$ seconds, the approximation is $88$ ft/s.
With a time interval of length $0.1$ seconds, the approximation is $94.4$ ft/s.
With a time interval of length $0.01$ seconds, the approximation is $95.84$ ft/s.

By this point, students realize that we’re getting better and better approximations… however, we’re probably not going to get the correct answer by just plugging in numbers. And we certainly can’t just take a time interval of $0$ seconds since dividing by zero is a no-no.

Depending on my read of the class — on whether or not they’re ready for a little more abstraction — I’ll then ask the class, “How can we make these fractions without plugging in all of these numbers?” Usually students are at a loss at first. Perhaps someone will volunteer that we ought to introduce a variable… but, in my experience, even bright students at the start of calculus do not have this step of abstraction at the tips of their fingers. So I’ll lay out the fractions that we’ve studied so far, like

$\displaystyle \frac{f(3) - f(2.9)}{0.1} \qquad$ and $\qquad \displaystyle \frac{f(3)-f(2.99)}{0.01}$ ,

and ask, “How could we do this more systematically? Does anyone see a pattern in these fractions?” Hopefully someone will notice that the input of the second function call is 3 minus the denominator; if not, I’ll volunteer this observation to the class. So both of these fractions can be written as

$\displaystyle \frac{f(3) - f(3-h)}{h}$ ,

where $h$ is a small positive number. Let’s now simplify this fraction:

$\displaystyle \frac{f(3) - f(3-h)}{h} = \displaystyle \frac{16(3)^2 - 16(3-h)^2}{h}$

$= \displaystyle \frac{144 - 16(9-6h+h^2)}{h}$

$= \displaystyle \frac{144 - 144 + 96h - 16h^2}{h}$

$= \displaystyle \frac{96h - 16h^2}{h}$

$= \displaystyle 96 - 16h$ .

The last step is permitted because $h$ is assumed to be a nonzero number. I then check to see if the previous work matches this algebraic expression:

If $h=1$ , then $96 - 16h = 80$ , matching the previous answer.
If $h=0.1$ , then $96-16h = 94.8$ , matching the previous answer.

I then ask the class, what’s the ultimate goal with $h$ ? The answer: send $h$ to zero. So we conclude that the velocity at impact is $96 - 16(0) = 96$ ft/s, which is the final answer.

Reviewing, the curved thing was the changing speed of the falling object, which was approximated by the straight thing, the ordinary distance-rate-time formula. Finally, we passed to limits to find the real velocity at impact.

All of the above is eventually done more systematically later in the semester after the properties of derivatives have been more fully developed. However, I think that doing this calculation on the very first day of class gives my students a taste of what’s going to be happening in the days and weeks to come. Again, I emphasize that I probably cover this material in maybe 15-20 minutes, and that I don’t hold students immediately responsible for repeating such a calculation on their own. (I do hold them responsible for this, of course, after they know how to differentiate $f(t) = 16 t^2$ .

Day One of my Calculus I class: Part 2

I’ve just told students that the topics in Calculus I build upon each other (unlike the topics of Precalculus), but that there are going to be two themes that run throughout the course:

Approximating curved things by straight things, and
Passing to limits

I then transition to applying these two themes to two different problems. Here’s the first.

Problem #1. A building on campus is 144 feet tall. A professor takes a particularly annoying student to the top of the building, and throws him (or her) off to his (or her) certain demise. (Usually I pick a student that I know and like as the one to throw off the building. This became a badge of honor over the years.) The distance that the student travels (in feet) after $t$ seconds is $f(t) = 16t^2$ . How fast is the student going when he (or she) hits the concrete sidewalk?

And then I ask my students how to solve this. Usually, they can come up with the first few ideas.

1. When the student hits the sidewalk and meet his/her demise? So we must solve $16t^2 = 144$ , so that $t = \pm 3$ . (And I make sure that they remember that this quadratic equation has two roots.) The solution $t = -3$ is clearly extraneous, so the time elapsed until the student meets his/her demise is $3$ seconds.

2. How fast is the student going after $3$ seconds? Most students realize the inherent difficulty of this question because the student’s speed is increasing as he/she gets closer to the ground. Some students will volunteer the word “accelerate.”

At this point, I’ll volunteer that the changing speed is a curved thing. Back in pre-algebra, students were taught

$\hbox{rate} = \hbox{distance} / \hbox{time}$

under the assumption that the rate was constant. However, if the rate is changing, all bets are off.

Still, the question remains: how fast is the student moving after 3 seconds? How should we measure this? Usually, someone will suggest that we just divide $144$ feet by $3$ seconds, for a rate of $48$ ft/sec. I then point out that this is an example of approximating a curved thing by a straight thing. The straight thing is the usual distance-rate-time formula, while the curved thing is the changing speed of the student as he/she falls. So the answer of $48$ ft/sec is not the correct answer, but it’s an approximate answer.

This leads to the next question: is this estimate too high or too low? Unequivocally, students answer “too low” since the student travels the slowest at the start of the fall and the fastest at the end of the fall. So since this interval of 3 seconds includes the slower speeds at the start of the fall, the answer of $48$ ft/sec will underestimate the speed at impact.

Which then leads to the next obvious question: How can we get a better approximation? I leave the question open-ended like this and take suggestions from the class. This often takes a while, and I’ll get a lot of creative (but bad) ideas. And that’s OK… the next step is hardly the most intuitive thing that immediately jumps to mind. I think that the process of keeping the answer unknown until someone volunteers the correct next step is worth it.

Eventually (though it might take a couple of minutes), somebody will suggest using a shorter time interval, like the distance traveled between $t=2$ and $t=3$ . We see that $f(2) = 64$ and $f(3) = 144$ , and so the new approximation is $(144-64)/1 = 80$ ft/s. I store these two approximations ( $48$ ft/s with a time interval of $3$ seconds and $80$ ft/s with a time interval of $1$ second in a table on the side of the chalkboard. The values derived below are entered in the table as they’re found.

I then note that the previous approximation was $48$ ft/s, and then ask the class, “Do you think that $80$ ft/s will be a better or worse approximation than $48$ ft/s?” Invariably, they’ll say it’s a better approximation because the change in speed isn’t as great from $t=2$ to $t=3$ as from $t=0$ to $t=3$ . I’ll then ask if they think that $80$ ft/s is too high or too low. Again, they’ll answer too low for the same reason as before.

Then I ask the obvious next question, “How do we find a better approximation?” The class typically responds something to the effect of, “Take a smaller interval.” I ask for a suggestion, and I’ll usually get something like $t=2.5$ to $t=3$ . We see that $f(2.5)=100$ ft/s and $f(3)$ is still $144$ ft/s, so that the new approximation is $(144-100)/0.5 = 88$ ft/s. Students will volunteer that this should be better than the previous two approximations but still less than the correct answer.

Then I do it again: “How do we find a better approximation?” The class typically respond, “Take an even smaller interval.” I suggest $t=2.9$ to $t=3$ . We see that $f(2.9)=134.56$ ft/s (by this point, a calculator is certainly needed) and $f(3)$ is still $144$ ft/s, so that the new approximation is $(144-134.56)/0.1 = 94.4$ ft/s. If we do it again with $t =2.99$ , we see that $f(2.99) =143.0416$ ft/s, for an approximation of $(144-143.0416)/0.01 = 95.84$ ft/s.

I turn to the class and ask, “Have we found the right answer yet?” They’ll answer “No” in unison, but they’ll note that the approximations are probably pretty good right now. Astute students will notice that the approximations appear to be “leveling off” to some final value.

I’ll then tell the class that this is an example of passing to limits, the second theme of calculus. By making the time intervals smaller and smaller, we get better and better approximations to the true speed at impact. In the next post, I’ll describe how I informally introduce the concept of a limit with this example.

Day One of my Calculus I class: Part 1

I begin by noting the different topics that appear in Precalculus, which they should have taken in the recent past:

The definition of a function and an inverse function
Graphing polynomials and rational functions
Properties and applications of exponential and logarithmic functions
Trigonometry
Sequences and series

These different topics, when taught in Precalculus, really don’t talk to one another. With a couple of exceptions, it feels like five different units being squeezed into the same course. I’ll present a visual image of laying down an imaginary brick on the floor, and then laying down a second brick next to the first one, and so on. The above topics (with a couple of exceptions) really don’t build upon each other; they’re lateral to one another. In other words, these topics made the foundation necessary for the study of calculus. After all, the class was called Pre-Calculus.

Now that we’re in calculus, I tell my students, we’re going to have topics that build on this foundation, and the topics will build on each other. Continuing the building image, I’ll start laying imaginary bricks on the initial foundation, building vertically higher and higher, noting that the topics that we’ll see in Weeks 13 and 14 will ultimately be built upon the topics that we’ll talk about in Weeks 1 and 2. Unlike Precalculus, the topics in Calculus are explicitly interconnected, building up a body of thought from the foundation of Precalculus.

So the good news is that, unlike Precalculus, Calculus I will be an incrementally developed course from start to finish. The bad news, of course, is that Calculus I will be an incrementally developed course from start to finish. In Precalculus, if you didn’t particularly like one topic (say, logarithms), that really would not affect your success later on with a future topic (say, trigonometry). However, in Calculus, the whole course is put together from start to finish.

The good news is that while there are many interconnected topics in calculus, there are going to be two themes that run throughout the course:

Approximating curved things by straight things, and
Passing to limits

And we’re going to be applying these two themes again and again throughout the semester. (I wish I could take credit for synthesizing the topics of calculus into these two themes, but I learned this idea from my own calculus professor back in the mid-1980s.)

For the remainder of this first lecture, I show how these two themes apply to two completely different problems:

Finding the speed of a falling object when it hits the ground.
Finding the area under the curve $y = x^2$ between $x = 0$ and $x = 1$ .

I’ll describe how I present these to new calculus students in the coming posts.

Area of a circle (Part 1)

Math majors are completely comfortable with the formula $A = \pi r^2$ for the area of a circle. However, they often tell me that they don’t remember a proof or justification for why this formula is true. And they certainly don’t remember a justification that would be appropriate for showing geometry students.

In this series of posts, I’ll discuss several ways that the area of a circle can be found using calculus. I’ll also discuss a straightforward classroom activity by which students can discover for themselves why $A = \pi r^2$ .In the first few weeks after a calculus class, after students are introduced to the concept of limits, the derivative is introduced for the first time… often as the slope of a tangent line to the curve. Here it is: if $y = f(x)$, then

$\displaystyle \frac{dy}{dx} = y' = f'(x) = \lim_{h \to 0} \displaystyle \frac{f(x+h) - f(x)}{h}$

From this definition, the first few rules of differentiation are derived in approximately the following order:

1. If $f(x) = c$ , a constant, then $\displaystyle \frac{d}{dx} (c) = 0$ .

2. If $f(x)$ and $g(x)$ are both differentiable, then $(f+g)'(x) = f'(x) + g'(x)$ .

3. If $f(x)$ is differentiable and $c$ is a constant, then $(cf)'(x) = c f'(x)$ .

4. If $f(x) = x^n$ , where $n$ is a nonnegative integer, then $f'(x) = n x^{n-1}$ . This may be proved by at least two different techniques:

The binomial expansion $(x+h)^n = x^n + n x^{n-1} h + \displaystyle {n \choose 2} x^{n-2} h^2 + \dots + h^n$
The Product Rule (derived later) and mathematical induction

5. If $f(x) = a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0$ is a polynomial, then $f'(x) = n a_n x^{n-1} + (n-1) a_{n-1} x^{n-2} + a_1$ . In other words, taking the derivative of a polynomial is easy.

After doing a few examples to help these concepts sink in, I’ll show the following two examples with about 3-4 minutes left in class.

Example 1. Let $A(r) = \pi r^2$ . Notice I’ve changed the variable from $x$ to $r$ , but that’s OK. Does this remind you of anything? (Students answer: the area of a circle.) What’s the derivative? Remember, $\pi$ is just a constant. So $A'(r) = \pi \cdot 2r = 2\pi r$ . Does this remind you of anything? (Students answer: Whoa… the circumference of a circle.)

Example 2. Now let’s try $V(r) = \displaystyle \frac{4}{3} \pi r^3$ . Does this remind you of anything? (Students answer: the volume of a sphere.) What’s the derivative? Again, $\displaystyle \frac{4}{3} \pi$ is just a constant. So $V'(r) = \displaystyle \frac{4}{3} \pi \cdot 3r^2 = 4\pi r^2$ . Does this remind you of anything? (Students answer: Whoa… the surface area of a sphere.)

Hmmm. That’s interesting. The derivative of the area of a circle is the circumference of the circle, and the derivative of the area of a sphere is the surface area of the sphere. I wonder why this works. Any ideas? (Students: stunned silence.)

This is what’s known on television as a cliff-hanger, and I’ll give you the answer at the start of class tomorrow. (Students groan, as they really want to know the answer immediately.)

In the spirit of a cliff-hanger, I offer the following thought bubble before presenting the answer.

By definition, if $A(r) = \pi r^2$ , then

$A'(r) = \displaystyle \lim_{h \to 0} \frac{ A(r+h) - A(r) }{h} = 2\pi r$

The numerator may be viewed as the area of the ring between concentric circles with radii $r$ and $r+h$ . In other words, imagine starting with a solid red disk of radius $r +h$ and then removing a solid white disk of radius $r$ . The picture would look something like this:

Notice that the ring has a thickness of $r+h -r = h$ . If this ring were to be “unpeeled” and flattened, it would approximately resemble a rectangle. The height of the rectangle would be $h$ , while the length of the rectangle would be the circumference of the circle. So

$A(r + h) - A(r) \approx 2 \pi r h$

and we can conclude that

$A'(r) = \displaystyle \lim_{h \to 0} \frac{ 2 \pi r h}{h} = 2\pi r$

By the same reasoning, the derivative of the volume of a sphere ought to be the surface area of the sphere.

Pedagogically, I find that the above discussion helps reinforce the definition of a derivative at a time when students are most willing to forget about the formal definition in favor of the various rules of differentiation.

In the above work, we started with the formula for the area of the circle and then confirmed that its derivative matched the expected result. However, the above logic can be used to derive the formula for the area of a circle from the formula $C(r) = 2\pi r$ for the circumference. We begin with the observation that $A'(r) = C(r)$ , as above. Therefore, by the Fundamental Theorem of Calculus,

$A(r) - A(0) = \displaystyle \int_0^r C(t) \, dt$

$A(r) - A(0) = \displaystyle \int_0^r 2\pi t \, dt$

$A(r) - A(0) = \displaystyle \left[ \pi t^2 \right]_0^r$

$A(r) - A(0) = \pi r^2$

Since the area of a circle with radius $0$ is $0$ , we conclude that $A(r) = \pi r^2$ .

Pedagogically, I don’t particularly recommend this approach, as I think students would find this explanation more confusing than the first approach. However, I can see that this could be useful for reinforcing the statement of the Fundamental Theorem of Calculus.

By the way, the above reasoning works for a square or cube also, but with a little twist. For a square of side length $s$ , the area is $A(s) = s^2$ and the perimeter is $P(s) = 4s$ , which isn’t the derivative of $A(s)$ . The reason this didn’t work is because the side length $s$ of a square corresponds to the diameter of a circle, not the radius of a circle.

But, if we let $x$ denote half the side length of a square, then the above logic works out since

$A(x) = s^2 = (2x)^2 = 4x^2$

and

$P(x) = 4s = 4(2x) = 8x$

Written in terms of the half-sidelength $x$ , we see that $A'(x) = P(x)$ .

Why do we teach students about radians?

Throughout grades K-10, students are slowly introduced to the concept of angles. They are told that there are $90$ degrees in a right angle, $180$ degrees in a straight angle, and a circle has $60$ degrees. They are introduced to $30-60-90$ and $45-45-90$ right triangles. Fans of snowboarding even know the multiples of $180$ degrees up to $1440$ or even $1620$ degrees.

Then, in Precalculus, we make students get comfortable with $\pi$ , $\displaystyle \frac{\pi}{2}$ , $\displaystyle \frac{\pi}{3}$ , $\displaystyle \frac{\pi}{4}$ , $\displaystyle \frac{\pi}{6}$ , and multiples thereof.

We tell students that radians and degrees are just two ways of measuring angles, just like inches and centimeters are two ways of measuring the length of a line segment.

Still, students are extremely comfortable with measuring angles in degrees. They can easily visualize an angle of $75^o$ , but to visualize an angle of $2$ radians, they inevitably need to convert to degrees first. In his book Surely You’re Joking, Mr. Feynman!, Nobel-Prize laureate Richard P. Feynman described himself as a boy:

I was never any good in sports. I was always terrified if a tennis ball would come over the fence and land near me, because I never could get it over the fence – it usually went about a radian off of where it was supposed to go.

Naturally, students wonder why we make them get comfortable with measuring angles with radians.

The short answer, appropriate for Precalculus students: Certain formulas are a little easier to write with radians as opposed to degrees, which in turn make certain formulas in calculus a lot easier.

The longer answer, which Precalculus students would not appreciate, is that radian measure is needed to make the derivatives of $\sin x$ and $\cos x$ look palatable.

Source: http://mathworld.wolfram.com/CircularSector.html

1. In Precalculus, the length of a circle arc with central angle $\theta$ in a circle with radius $r$ is

$s = r\theta$

Also, the area of a circular sector with central angle $\theta$ in a circle with radius $r$ is

$A = \displaystyle \frac{1}{2} r^2 \theta$

In both of these formulas, the angle $\theta$ must be measured in radians.

Students may complain that it’d be easy to make a formula of $\theta$ is measured in degrees, and they’d be right:

$s = \displaystyle \frac{180 r \theta}{\pi}$ and $A = \displaystyle \frac{180}{\pi} r^2 \theta$

However, getting rid of the $180/\pi$ makes the following computations from calculus a lot easier.

2a. Early in calculus, the limit

$\displaystyle \lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$

is derived using the Sandwich Theorem (or Pinching Theorem or Squeeze Theorem). I won’t reinvent the wheel by writing out the proof, but it can be found here. The first step of the proof uses the formula for the above formula for the area of a circular sector.

2b. Using the trigonometric identity $\cos 2x = 1 - 2 \sin^2 x$ , we replace $x$ by $\theta/2$ to find

$\displaystyle \lim_{\theta \to 0} \frac{1 - \cos \theta}{\theta} = \displaystyle \lim_{\theta \to 0} \frac{2\sin^2 \displaystyle \left( \frac{\theta}{2} \right)}{ \theta}$

$\displaystyle \lim_{\theta \to 0} \frac{1 - \cos \theta}{\theta} = \displaystyle \lim_{\theta \to 0} \sin \left( \frac{\theta}{2} \right) \cdot \frac{\sin \displaystyle \left( \frac{\theta}{2} \right)}{ \displaystyle \frac{\theta}{2}}$

$\displaystyle \lim_{\theta \to 0} \frac{1 - \cos \theta}{\theta} =0 \cdot 1$

$\displaystyle \lim_{\theta \to 0} \frac{1 - \cos \theta}{\theta} =0$

3. Both of the above limits — as well as the formulas for $\sin(\alpha + \beta)$ and $\cos(\alpha + \beta)$ — are needed to prove that $\displaystyle \frac{d}{dx} \sin x = \cos x$ and $\displaystyle \frac{d}{dx} \cos x = -\sin x$ . Again, I won’t reinvent the wheel, but the proofs can be found here.

So, to make a long story short, radians are used to make the derivatives $y = \sin x$ and $y = \cos x$ easier to remember. It is logically possible to differentiate these functions using degrees instead of radians — see http://www.math.ubc.ca/~feldman/m100/sinUnits.pdf. However, possible is not the same thing as preferable, as calculus is a whole lot easier without these extra factors of $\pi/180$ floating around.