Trig without Trig

Over the summer I made a Donors Choose page in the hope of getting some clinometers, so that we can go out into the world and use them to calculate the heights of objects like trees and buildings. And we did!
I created the Clinometer Lab as an introduction to Trigonometry. As such, prior to the lab, they had not seen any trig at all. So I started off with this video as the homework from the night before.

In class, we talked about how, when the angle is the same, the ratio of the height and shadow is the same. And how, long ago, mathematicians made huge lists of all of these ratios.
So if they told me any angle, I could tell them the ratio, and then they could just set up the proportion and solve.

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So we left the building and went to the park, armed with clinometers, measuring tape, calculators, and INBs, so do some calculations. I had the students get the angles from their eyes to the height of a tree, from two different spots, and the distances, so they could set up a proportion and discover the height of the tree. (When they finished, they did a nearby building. If they finished that too, the extension problem was to switch it up: given the height of a famous building, the Metlife Tower, that they could see from the park, and determine how far away they were.)

It went really well, and I think they got the idea. The fact that they could choose which tree to measure and were given free range of the park, and could choose where to stand to look at the top, but always had to come back to me to get their ratio, worked seamlessly. I could check in on them easily and keep them on the right path. (Especially with my co-teacher there to keep them all wrangled, or the AP who observed/helped chaperone the classes without my co-teacher.)

The next class, I revealed to them that I was not using that crazy chart to give them their ratio, but rather they can just get it themselves from the calculator, which basically internalized the list. Then they got it, that is was just ratios and proportions, not some crazy function thing.

Trig as a topic did not go great in my class, but that is the fault of my follow-up lesson, where I tried to squeeze in too much and didn’t do any practice, not that fault of this lesson, which still sticks in their minds. Later in the year we were doing a project about utilizing unused space, so we picked empty lots but couldn’t get in. So in order to figure out how big they were, we used clinometers, and trig. Now that’s a real world application.

The Lab

Clinometer Lab Instructions

Clinometer Lab Sheet

Recursive Combinations with Replacement

So I was in my classroom last night with my boyfriend, waiting for his phone to charge before we went to dinner. Since we had some time, we played some of the math games I have in my room. (He’s a math PhD student, so he was all for it.) We played Set, of course, and then played a bit of 24. We idly wondered if it were possible to get 24 with any combination of 4 digits. So I looked at the box, and saw it came with 192 possible configurations. Well, if we determined how many possibilities there were (maybe there were 192), that might give us an idea of the feasibility.

So we tried to calculate how many configurations there were. Shouldn’t be too hard, right? Well, it kinda is, especially when you’re not already familiar with combinations with replacement. So we started using what we did know of combinations, but were stuck because we could use the same number multiple times, which made it trickier. Otherwise it would just be 9 C 4.

So, unsure how to solve, we tried to make a simpler case. What if we only had 2 digits to choose from, not 9? There’s there’s 5 possibilities. (1111, 1112, 1122, 1222, 2222.) And with 3 digits, there’s 15 (1111, 1112, 1113, 1122, 1123, 1133, 1222, 1223, 1233, 1333, 2223, 2233, 2333, 3333). We got a lot of fruitful thinking out of this, finding patterns, but didn’t really get closer to the answer. (Four digits had 35, btw. But we didn’t want to list all the ones for 5 digits and beyond.)

At this point it was time to go to dinner, so we put the whiteboard aside. But that couldn’t stop us thinking and talking about it, which we did as we walked to the restaurant and waited for out table, when we finally had a breakthrough.

Instead of trying to figure out the pattern with fewer digits but the same number of slots, let’s try to iterate up with the same number of digits, but using increasing number of slots. Let me explain, using 4 possible digits.

If we only have 1 number slot on the card, there are only 4 possibilities. (1, 2, 3, 4.) When we increase to 2 slots, we could start by putting a 1 in front of each of those possibilities. (11, 12, 13, 14). But, because order doesn’t matter, we can’t also put 2 in front of everything, because 21 is the same as 22. So we don’t use the one, and get 22, 23, 24. Same logic for 3 gives us 33, 34, and then finally 44.

This gives us a total of 10 possibilities. (4 + 3 + 2 + 1.) Now let’s think about 3 slots. In the same way, we can add a 1 in front of everything we’ve done so far. So for 3 digit possibilities there are 10 that start 1. Since we have to eliminate the four that two-digit configurations that have 1, there are 6 remaining, so that’s how many will start with 2. (3 + 2 + 1). Then three will start with 3. (2 + 1) And 1 will start with 4.

The process here is to add up all of what we had before, chopping off the start, to get the total number of new possibilities. So now, with 3 slots, we have 20 possibilities. (10 + 6 + 3 + 1.) To get for 4 slots, we use the same process: 20 start with 1, 10 start with 2 (6 + 3 + 1), 4 start with 3 (3 + 1), and 1 starts with 4, for a total of 35. Which is what we found before.

(If there were 5 slots, it would be 35 + 15 + 5 + 1, or 56.)

I don’t know of this recursive method of solving for combinations with replacements has been done before. I’m sure it is, but I haven’t found it in a very short google search. If someone knows of it, please let me know. But I wanted to share what I did. You can tell I love math, and so does my boyfriend, because we got completely distracted from a board game by solving a problem. He told me I’d make a good mathematician, because of how I tackled the problem. That may be true.

No Right Answer

A bit ago I got yelled at by a commenter on Kate’s blog who claimed that being always right is why we like math. The problem with that point of view is that, while yes, you can always be right while doing computation, math isn’t just computation. So the other day I was talking with a friend of mine, and that prompted me to post the following tweets:

My friend Phil (@albrecht_letao) responded to the question, and he came up with an answer of $20/hr. When I worked it out with my friend, we came up with $14.25. Does that mean one of us is wrong, since we got different numbers?

No, of course not. What happened is we approached the problems in different ways. Phil only calculated the monetary value: with his amount, my friend would earn the same amount of money she does now. He figured this was an important way to look at it, for paying bills and whatnot. Our calculation came from thinking about how her time is being compensated. Since those 16 hours are being wasted (she has to work them for free; actually, she pays to lose that time), we calculated her “real” hourly rate and used that.

There can be more answers than even these two, depending on what you think is important. But it’s a clear example of a problem, solved using math, with no one right answer. That’s what math is about. I tweeted it thinking maybe it could be a problem worth considering in class, to show that essential idea to students.

What do you think?

P.S. The right answer, of course, came from @calcdave:

 

The Math of Nail Clipping?

To demonstrate how I’m such a nerd (or such a math teacher, or both):

I was just clipping my nails, and started thinking about the math involved. Often when I clip I’ll only do 1 or 2 clips per nail, and they can come out really jagged, pointy, and sharp. But this time I did about five clips, closely following the curve of the nail and it came out much smoother.

Which makes sense, because I’m basically approximating the shape of my nail (a curve) with the nail clipper (a tangent line), and so the more tangent lines I used, the closer the approximation is.

Now the question just is if I can turn that into a WCYDWT, or if it’s too gross for that….