## Sunday, September 22, 2013

### Teaching Question: Students Overusing Proportional Reasoning

I have heard, from a colleague who works with prospective elementary teachers, that many of them are not good at proportional thinking. My students (in pre-calc) seem to be fine at it, but ... they're using it even when it doesn't apply. My question for you is how to help them see why proportional reasoning is not always a sensible choice.

Problem #54. Determining a Distance: A woman standing on a hill sees a flagpole she knows is 60 feet tall [yeah, right]. The angle of depression to the bottom of the pole is 14 degrees, and the angle of elevation to the top of the pole is 18 degrees. Find her distance x from the pole.

One student wanted to average the two angles at 16 degrees each. Another said the observer could stand on a stool to be a little higher, so the angles would be 16 degrees each. Their answers were very close. There were other good (but wrong) methods that all came down to assuming this relationship was linear in a way that it's not. Since their answers were very close, it was hard to help them see what was wrong with their reasoning.

Can anyone help me here?

1. This example you've picked is a toughie, for two reasons. As you point out, 18 and 14 degrees are both darn close to 16 degrees, so if you make that simplifying assumption 1) the problem is much easier and 2) the answer is pretty close to what it should be. But, you have to realize you are making a simplifying assumption that will throw your answer off a little bit, which I take it was not what your students were doing. They thought the result should lead to the exact, correct answer.

The other thing that makes this tough is that the diagram looks for all the world like relationships ought to be linear -- look at all those line segments! The catch is that when the angles change by d degrees, the amount of flagpole intercepted by the angle's rays doesn't change by the same length.

I realize you know all this already. Just thinking it through.

I suppose for this example, I'd try an extreme case. If we drew the angle that was only the topmost 2 degrees, it would intercept more of the flagpole than 2 degrees of that angle near the altitude x. This would show that you can't just redistribute parts of the angle and expect it to correspond to the same distance along the flagpole.

2. I like the thinking you're doing. My hope is to come up with some exercises that are mixed - some problems can use proportional thinking, and others can't. But before they can do those, they need to really see why this method won't work. I tried to draw angles that were extreme, but my diagram got too messy. I might be able to make a handout that explains this one situation well, but I want them to get more experience with evaluating when they can do this sort of thing and when they can't.

3. If you move the position around so that the sum of the angles is the same (32 = 18+14), you end up moving in a circle (in fact, the circumcircle of the triangle defined by the person and the two ends of the flagpole) with the flagpole as a chord (inscribed angle is constant for the same subtended arc).

The distance to the flagpole is always computed perpendicular to the chord, so it can change pretty dramatically. It can get arbitrarily small, and is maximized when the person is on a diameter (and the two angles both 16 degrees). [So that person found at least the upper-bound, I guess.]

4. This sounds perfect! Hao, I need to draw some pictures to see it, but I think this will help!

I still want to make a collection of problems that students might mistake as solvable with proportions mixed with problems that aren't.

5. That reminds me of this essay by James Tanton regarding an distance problem. The MAA test-writer made a similar relationship linear in the "correct" answer to a 2004 AMC8 question.

6. @Denise: Thanks for the link! In addition to the linearity error for the "correct" answer, an additional issue is that that the graph also implies a different amount of time necessary to run two identical length segments of the rectangle. (Or the time axis isn't linear or the runner is not running at uniform speed.)

7. Thanks, Denise! I hadn't looked at it right away. Hao's comment came when I had a moment, so I finally got to check it out. I never would have noticed the connection. Good catch! I do notice that Tanton did not give the correct graph.

Hao, could you explain this? "The graph also implies a different amount of time necessary to run two identical length segments of the rectangle."

8. I think Hao means that the two long sides of the block should take the same amount of time to run, but the graph shows the first long side taking more time than the second.

As for the "correct" graph, I suppose there really isn't one unless we had more information. After all, the problem didn't say she ran at a steady speed, did it? So perhaps she adjusted her speed to make the linearity fit and then ran double-speed once home was in sight, in which case graph D really would be correct. Personally, I like the science fiction explanation for graph B.

9. Sue, I believe the correct formula is:
d = h/(tan(a)+tan(b))
Are the students able to derive that?

The students appear to be using the formula:
d = h/(2*tan((a+b)/2))
Do you know how they arrived at that?
Do the students think the two formulae are equivalent?

I'm guessing the students are both using:
tan((a+b)/2) = (h/2)/d

The first student is effectively forming a triangle using the bisector of the angle subtended by the flagpole and assuming that it bisects the flagpole. But even if it did, that is not a right triangle, because the angle bisector is not perpendicular to the flagpole. The second student is forming a right triangle by moving the observation point vertically onto the perpendicular bisector of the flagpole and assuming that the subtended angle is unchanged. But the subtended angle is always smaller the farther you move vertically from the perpendicular bisector, because the flagpole center is farther away and the flagpole is viewed at an increasingly oblique angle.

10. Trying to understand how the students might be thinking about this problem, I presume they know that:

from which:

So they know that they can determine the distance if they have a right triangle where the length of the adjacent side is the distance and they know the angle and the length of the opposite side. In this problem they have two right triangles where the length of the adjacent side is the distance and they have the two angles, but they don't have the length of the two opposite sides, only their sum.

The two triangles do not appear very different, because the angles are roughly the same. The students can see that if the triangles were congruent, they could find the length of the opposite side by simply diving the sum of the lengths in half. They also know that if the triangles are congruent, the corresponding angles of the two triangles would also have to be equal. Assuming the total subtended angle is the same, the angles of the congruent triangles could be found by dividing the total angle in half. That gives them both the angle and the length of the opposite side, which is what they need to determine the distance.

The students are attempting to employ the problem-solving technique of transforming the given problem into a simpler problem which they are able to solve. Unfortunately, if the distance is the same, then the total subtended angle is not the same. So the problem they're solving does not give the same result as the original problem.

The students may think that the subtended angle wouldn't change because they are keeping the same distance from the flagpole, and if the distance is the same then the apparent size should not change. But although the perpendicular distance doesn't change, the distance to the center of the flagpole does change. The distance is smaller, so the object should appear larger, meaning the subtended angle would be greater. Also, the relative orientation of the flagpole changes, becoming more direct (perpendicular), so that the apparent size increases for that reason as well.

Taking a more direct approach, the students know that if they could find a way to determine the lengths of the opposite sides, then they could solve the problem. They probably recognize that the opposite side lengths are related to the angle in some way. They likely can see that as the angle increases, the length of the opposite side also increases. It is natural for them to consider whether the sides might proportional to the angles, since that is the simplest possible relation. The relation isn't that simple, but fortunately this error is easily detected because the two triangles will give different results. That should tell the students that the side lengths can't be proportional to the angles.

The students may also be able to see that the side length increases at at accelerating rate when the angle increases at a constant rate, which should tell them the relationship is not proportional. But that observation alone isn't enough for them to determine what the relationship is.

The students shouldn't have to guess at the relation between the angles and the opposite side, since they know that:

from which:

This formula should tell them that when the adjacent side is fixed, the opposite side is proportional, not to the angle, but to the tan() of the angle. And that is what they need to find the solution to the problem.

Were the students unable to discover this? Did they not realize that they have a formula which shows the relation of the opposite side to the angle? Or did they not realize that the formula indicated that the opposite side was proportional to the tan() of the angle? Or was there something else that prevented them from seeing this?

11. Zeno, you have made so many great observations here. I don't know why my students didn't see the things you see, but I suspect that most of them treat math very mechanically, and don't check whether a procedure really makes sense.

When you wrote "the two triangles will give different results", which two triangles were you referring to?

12. Sue,

The two triangles I was referring to are the right triangles formed by the observing position, the point on the flagpole where the horizontal line thru the observing position intersects it, and either the top (for the upper triangle) or bottom (for the lower triangle) of the flagpole. These two triangles share a side formed by the horizontal line from the observing position to the flagpole, whose length (x) is the distance to the flagpole. The sides of these two triangles opposite the observing point are the parts of the flagpole above (for the upper triangle) and below (for the lower triangle) the horizontal line. The sum of the lengths of the two opposite sides is the height of the flagpole (60 feet). The angles of the two triangles at the observing point are 18 and 14 degrees. If the sides of the two triangles opposite the observing point had the same ratio as the two angles, they would be found to have lengths of 33.75 and 26.25 feet, respectively. Using the fact that tan(angle) = opposite/adjacent, so that adjacent = opposite/tan(angle), the value of x can be found using either triangle. For the upper triangle, x = 33.75/tan(18) = 103.87 feet. For the lower triangle, x = 26.25/tan(14) = 105.28 feet. The two results are different, which shows that the opposite sides are not proportional to the angles.

As I mentioned, the fact that tan(angle) = opposite/adjacent tells you that opposite = adjacent*tan(angle). And that tells you that for a fixed adjacent side length, the length of the opposite side is proportional to tan(angle), not the angle itself. But also, it gives you a way to check your answer, because the sum of the opposite sides of the two triangles is x*tan(18)+x*tan(14) = 60. So you can substitute your answer for x and verify that the equation holds (or find that it doesn't). Of course, if you know algebra you can solve the equation to find that x = 60/(tan(14)+tan(18)).

One of the frustrating things about many of the errors at the "Math Mistakes" website is that not only are the answers wrong, but they are obviously wrong, and the students should be able to see for themselves that the answer is wrong with an easy check. If you're trying to find the solution to an equation, you should always check that the answer you get actually satisfies the equation! 