# A parabolic arc is not semicircular. But students think so

I'm teaching a Calc 2 class now (integration and applications) and I'm surprised that more than a handful of students seem to think the graph of $y=x^2$ on $-1\le x\le 1$ is part of a circle!

Here is a quiz question I posed:

Consider the part of the parabola $y=x^2$ that goes from the point $(-1,1)$ to $(1,1)$. Suppose I used the arc length formula to find the length of that curve and I got an answer of $2\sqrt{2}$. Explain to me how you can be sure that my answer is too small.

Here are two answers I got from students:

• "The answer is too small because the graph only extends from (-1,1) to (1,1), which means that the shape is half of a circle and not a whole circle."
• "The distance between these two points is 2 units, and this portion of the parabola resembles a circle. If the diameter is 2, then the circumference should be about 2pi, or 6.28. Half of this would be 3.14, and 2 root 2 is only about 2.83."

Questions:

1. Where does this come from? Is there evidence that students tend to confuse parabolas with circles, in general? Or is it only in particular cases, like this one where they were already asked to think geometrically?
2. Is there a short presentation I can give in class to, once and for all, convince students these are two different curves? Ideally, this could be just a 5-minute (max) discussion in class to say, "Look at how these are different. Remember this!" This could involve basic geometry or calculus or both.
3. Has anyone else noticed this? What was your audience? How did you first notice this phenomenon? What did you do to address it?

(Note: I realize there are likely other issues at play with this particular quiz question. For instance, Student #2 might think that the only way to show some value is too small is to find the correct value and point out that it's bigger. Perhaps that made them make a mistake they wouldn't usually make: conflating parabola with circle. But the fact remains, they conflated the two, and it's that phenomenon about which I'm curious.)

• To be fair, the second student said it "resembled" a circle. So they seem to know that it is not a circle. They also say circumference is "about 2pi", which could indicate that they know the arc length is not exactly $2\pi$. This does not make the argument correct (is the circle "close enough" to the parabola for this purpose?), but it gives some hope that student 2 doesn't have this mistaken belief. Student 1 is probably just used to all "geometric reasoning" problems being about circles and squares (a lot of "evaluate this geometrically" problems are about these two). – Steven Gubkin Feb 9 '15 at 5:20
• My calculus I students didn't recognize that y = sqrt(1-x^2) was a half circle. It made me change what I'm doing in pre-calc. It comes from not being able to reason with math. I don't have an answer for it, I just keep trying to get them to think. – Sue VanHattum Feb 9 '15 at 5:51
• I agree with @StevenGubkin regarding the second student. It seems clear to me that the student knows the parabolic arc is not a semicircle, but is using the superficial resemblance to argue that the its length should be approximately $\pi$, and therefore the answer of $2\sqrt{2}$ is too small. I don't think this is a very good answer, to be sure, but I think it is flawed for reasons that are different from the first student's. – mweiss Feb 9 '15 at 17:26
• My initial reaction on seeing your question is that this could be a nice "teaching moment". Can the students think of ways to prove (or to disprove) this is an arc of a circle? I can think of at least one method that should be within their reach. Suppose these points lie on a circle. Let $(a,b)$ be the center. Then $(a+1)^2+(b-1)^2=a^2+b^2=(a-1)^2+(b-1)^2.$ Equating the left-most and right-most expressions gives $a=0.$ Putting $a=0$ and equating the first two expressions gives $b=1.$ Now pick another point on the parabola, say let $x=1/2,$ and see how far $(0,1)$ is from this other point. – Dave L Renfro Feb 9 '15 at 20:06
• Personally, I think the second answer is pretty good, even if it doesn't allow you "to be sure". It demonstrates creativity, geometric thinking, a willingness to compare, the courage to venture an argument when you're not sure what formula to plug-and-chug, and imprecision ("resembles", "about") where it is appropriate. And, it is essentially correct. I would personally be quite happy if a student offered such an explanation, unless the student gave some sign that he/she believed it actually was a semicircle. – Frank Thorne Mar 4 '15 at 2:15

1. This is pretty natural, I think. People understand things in terms of things that they already know, and while Calc 1/2/3 students should theoretically have a reasonably well developed 'catalog' of curves, it typically isn't the case in practice. They do all know, understand, and love circles though. Second, any infinitesimal portion of a parabola actually is a portion of a circle, ie: can be well approximated by a portion of a circle, and this is 'maximally true' (lets not be too rigorous at 5am) at and around the vertex. Third, all of the reference points which are easy to check ($(-1, 1), (0,0)$, and $(1,1)$) line up correctly.
2. Calc 2 students might, in general, be deficient in their ability to classify different curves, but I'll bet that very few of them will have trouble differentiating $x^2$ and finding that its slopes at $-1$ and $1$ are $-2$ and $2$ respectively. They also should have little trouble finding that the slopes at the 'sides' of the circle of radius $1$ centered at $(0,1)$ are infinite/undefined. Talking about this contradiction is likely to be sufficient, but be sure to be clear about drawing out 'natural language' statements about each curve, since most students at this point aren't familiar with mathematical argument.

...This discussion with your students might be a decent jumping-off point to start talking about curves approximating other curves and getting into Taylor series if that's part of your course.

• Indeed, we will get to Taylor Series later in the semester, but this can serve as a good preview. And I think you're right, that their "curve catalog" is just not well indexed and annotated, and their brains jumped to "circle" because it "looked close". If anything, working through the derivative argument you outlined will show them how to think something through and, in the future, they might jump to erroneous conclusions less willingly. – Brendan W. Sullivan Feb 9 '15 at 20:21

This might be a good opportunity to inject some history into your course. Aristotle thought that cannonball trajectories were line segments, up at an angle then straight down. Look in Chapter 5 of "The Day the Universe Changed" by James Burke. Leonardo realized that the trajectory was curved, but he did not seem to further quantify. In about 1537 Tartaglia attempted to quantify: the cannonball angles up in a straight line, then follows the arc of a circle along the top, then follows a line segment down. There are many other references. One is "Exploring the Limits of Preclassical Mechanics" by Peter Damerow.

And then of course people used think that planets followed circular orbits, presumably because "circles are perfect."

• Good idea. I'm always looking for opportunities to mention math history, and this seems like a good time to do so! As I noted in another comment (matheducators.stackexchange.com/questions/7368/…), we can have a class discussion about Newton, physics, and integration here, too! – Brendan W. Sullivan Feb 9 '15 at 20:19
• Though to be precise for many centuries most astronomers had models that consisted of paths constructed from circles. Ptolemaic orbits (and Copernican ones) aren't circles at all, but beautiful spirograph patterns. – Francis Davey Feb 9 '15 at 21:27
• And those were necessary because retrograde motion was a known thing, and couldn't be explained with a model where Earth is in the center and the planets revolve around it in pure circles. – RemcoGerlich Feb 10 '15 at 14:09

What is most concerning here is that this is coming from calculus students. This sort of misconception should have been removed during their algebra and calculus courses. To have this happen in a second calculus course is of great concern. Calculus students seem to make most of their mistakes on their algebra anyway, so if you can, pay attention to these students and how they handle the algebra required to implement the things they're learning in your calculus class.

1. Now, is this common? I would expect this to be common at an algebra-level class. I would expect this to not be common once someone has gone through a trig (tangent lines should make this a giveaway, and they had to at least touch on the different conic sections, right?) or multiple algebra courses (calculating slope enough times would raise a flag that they're different).

2. Is there something you could do? Well, you could try putting a real semicircle up and showing they're two different shapes might do it!! Alternatively, you could get out a flashlight and start making shapes on the blackboard showing how the parabola turns into an ellipse that turns into a circle as you rotate the flashlight (i.e. teach them about conic sections.)

• I don't know if there's any reason to believe the students will have seen conic sections even in passing. I certainly never did. – Jack M Mar 3 '15 at 13:47

People don't always speak precisely; lacking good words to precisely express their meaning, they may be falling back on more familiar words and ideas that they believe are 'good enough' to use, and may not have expected you to read their words so literally.

e.g. the second student does even take care to talk about it being a resemblance than meant literally. And his intuition is that the difference is small enough that looking at the arclength of the circle makes for a decent sanity check on the value. (and didn't realize you were asking something rigorously provable, or at maybe they did, but weren't able to come up with a better idea after getting this one in their head)

As for your second question, there are many ways to get a contradiction from the assumption that that curve is half a circle:

• If that curve were half a circle, we should be able construct a full cirle by putting two of such curves together. However we get this:

• If that curve were half a circle, the point $\left (\tfrac{1}{\sqrt{2}}, 1-\tfrac{1}{\sqrt{2}} \right)$ would have to be on it, as its distance to the center of the circle at $(0,1)$ is 1. However, $\left (\tfrac{1}{\sqrt{2}} \right )^2 ≠ 1-\tfrac{1}{\sqrt{2}}$

• I like your second suggestion better. It appeals to the fundamental definition of a circle: all the points are the same distance from the center. Perhaps that's the important concept the students are neglecting and, if anything, it could be helpful to just remind them of it! – Brendan W. Sullivan Feb 9 '15 at 20:17
• As @DaveL.Renfro pointed out in a slightly later comment (matheducators.stackexchange.com/questions/7368/…), we should be careful not to indulge the student's magical thinking: the centre of the 'circle' is 'obviously' at $(0, 1)$, but how do we know that (other than the kind of pictorial reasoning that led to the problem in the first place)? Better to prove it geometrically or, as Dave L. Renfro does, algebraically. – LSpice Apr 2 '15 at 17:17
1. depending on your students' backgrounds this could come from plain ignorance, oversimplification at a younger age, poor teaching, etc. I don't think that this necessarily matters when moving forward but it is a good thing to know and may allow you to predict future gaps your students have. I would just simply ask them and if they don't remember or don't know then move on, but they may be able to give you some interesting insight.

2. I think a very effective way to demonstrate that projectile motion is not semi-circular is to show them a "time lapse" or "stop motion" video of a projectile and have them trace the path on the board. an easy way to do this is to film with an iPhone or something similar a slow-mo video from the side of you throwing a basketball through the air. Slow the video down and project it onto the black/whiteboard in your classroom. Have the students watch the video a few times, having them focus on the path of the ball. Then, have a student come up and trace the path of the ball on the board as it moves through the air with a marker/chalk. This should make it pretty clear that the path is not circular. Well then what is it?? It is a parabola! Which can lead right into a discussion on parabolas. Alternatively, challenge the students to throw a ball of paper or a pencil in the air so that its path is a semi-circle and they will very quickly find out it is impossible. It sounds like there is a very fundamental misunderstanding, so showing them sophisticated proofs etc. might confuse even more.

3. I have definitely noticed this, as well as many other misunderstandings! the joys of teaching!
• I like your suggestion in #2, although I don't want to take up too much class time, since this mistake doesn't seem to be totally widespread. But, since we are studying applications of integration to physics currently, it would be great to take $a(t)=g$ and "discover" that position $s(t)$ is quadratic! – Brendan W. Sullivan Feb 9 '15 at 20:16
• Another physical activity is to use a stream of water from a hose. – DavidButlerUofA Feb 9 '15 at 21:39

Based on the "evidence" given, my best guess is that the matter is not about confusing parabolas with circles. Rather, it is the result of a more general phenomenon that is indeed very common:

If you cannot solve the problem, solve what you can solve!

Consider that the problem explicitly state "the part of the parabola", so why one should replace that with a circle. The reason is that he or she does not know the arc length for a parabola by heart, but knows that for a circle.

Second, I strongly doubt that students confuse "big" part of a parabola with a circle. For years I have challenged cal 1 students by asking them to guess the shape of a hanging chain. They always come with parabola, never with circle (of course, the answer is neither of them).

Third, it is likely that they don't know the formula neither for a parabola nor for a circle. Second and third together means if you still want to go with that five minutes lecture, it is better to focus on the difference in formulas not in shapes. That said, it seems that even the first student you mentioned knows the general shape of a parabola since he or she has replaced it with "partly" a similar looking figure, not for example, with two segments joining $(-1,1)$ to $(0,0)$ and $(0,0)$ to $(1,1)$.

• I think that it is dangerous to confuse "students don't always make this mistake" with "students don't have this confusion." In the context of hanging chains, students are lining up their library of curves, and consciously observing their differences as they look for a best fit. In the context of @brendansullivan07's question, their focus is (unfortunately) on the numerical answer, not the geometry behind it, so their brains are (evidently!) perfectly willing to make the substitution of one curved shape for another. – LSpice Apr 2 '15 at 17:21

I think your second student is closer than you think but maybe didn't couch it correctly. A parabola that passes through the two endpoints and the midpoint of a half circle is intuitively always smaller than the half circle. It's also intuitively larger than an abs function that passes through the same points. The lengths of both of those shapes are easy to find and the length of the parabolic arc should be between them. His problem is that he found the bigger one when he needed to find the smaller (the abs function has a length of 2 sqrt(2) and since the parabola is always bigger...) I would ask him to prove or disprove his implied assumption that the arc of the parabola is always smaller than the arc of the circle (ie there is no possbile transform that could make the parabola bigger than the circle) for some extra credit.

A simple proof by contradiction answers your question 2. Suppose that the arc of the parabola $y=x^2$ between $A\;(1,1)$ and $B\;(-1,1)$, which touches the $x$ axis at the origin $O$, is also an arc of a circle. This arc is symmetric about the $y$ axis, and so the centre must lie on the $y$ axis: say at $C\;(0,a)$. Equating the squared radii $CO^2$ and $CA^2$ (or $CB^2$), using Pythagoras, after one line of algebraic simplification, gives $a=1$. But that would make the tangent to the parabola at $A$ or $B$ (which is perpendicular to the radius) vertical, which we know is not the case.

This is a case where an answer that is too detailed (in a way that includes a mistake) is worse than an answer that is just detailed enough.

Here is a simple proof for the original poster's question. Notice that this proof does not describe the shape of the parabola, other than not being just two line segments. Also, notice that this proof (with a minor tweak at step 7) is equally valid for a circle:

1. The graph y = x2 goes through the (x, y) points (-1, 0), (0, 0), and (1, 1).
2. The distance between (-1, 0) and (0, 0) is √2.
3. The distance between ( 0, 0) and (0, 1) is also √2.
4. We have accounted for the entire alleged length of 2√2, using just two straight lines.
5. A straight line is the shortest distance between two points.
6. If the curve between (-1, 1) and (1, 1) goes through any points that are not on these two line segments, then the arclength must be greater than 2√2.
7. The graph y = x^2 goes through (1/2, 1/4).
8. The two-segment approximation goes through (1/2, 1/2).
9. Therefore the curve between (-1, 1) and (1, 1) goes through a point that is on neither of the two line segments.
10. The arclength must be greater than 2√2. Q.E.D.

For a very informal proof, steps 6 - 8 can be summarized as:

• A parabola is curved; it does not have any straight-line segments.
• Optionally: We know this because a straight-line has either a zero second derivative, or an infinite second derivative. Whereas a parabola has a non-zero, non-infinite second derivative.

In terms of the difference between a parabola and a circle, that last "optional" explanation gets to the heart of the matter. A circle has constant curvature -- it is everywhere bending (at the same rate) toward a single point. A parabola has a constant second derivative -- it is everywhere bending away from a single line. This allows the parabola to extend forever (or until it hits the ground, whichever happens first), whereas the circle wraps back around onto itself.

For a sufficiently short distance, a parabola can approximate a circle, and vice versa. But the approximation cannot be perfect, because they are two different shapes (that are not defined piece-wise).

• Yes, this is pretty much what I expected the students to say for a correct answer. But I don't think this addresses any of the questions in OP. – Brendan W. Sullivan Feb 9 '15 at 20:14