Pre-Calculus

Polar. Graph. Contest.

Here’s what I hung up last week:

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Here’s a closeup of some of them…

 

 

These are polar graphs that students designed using Desmos. Then I printed them out on photopaper and hung them up.

This was something I wanted to do after introducing polar graphing. Why? Because one day during the polar unit, I started playing around with desmos and accidentally created:

pic1

… from something so simple …

pic2

(Now to be fair, desmos isn’t great with creating great complicated polar graphs… and it’s better to write them parametrically to get a bit more accuracy… so this is a bit of a lie of a graph in that it isn’t totally accurate… but it’s oh so pretty.) [1]

So after our unit on polar graphing, I took 10 minutes at the start of class to introduce this idea of a Polar. Graph. Contest. First I threw this image up:

Polar.Graph.Contest..png

I then pulled up desmos and asked my kids to shout out some polar function. I graphed it. Then I put in a slider or two. So for example, if they said r=\cos(\theta), I might have added the slider r=\cos(a\theta)+b. And then I started changing the sliders. Then I might have altered the function a bit more, like r=\cos(a\theta)+b\theta and we saw what happened. Then I gave everyone 7 minutes to just come up with something pretty.

It was magical.

Kids just started playing. They dug into old functions they had learned about. They got excited by what they were seeing. They gasped and turned their screens to show their friends. Some who were getting boring graphs saw the cool graphs their classmates were getting and were inspired to mix things up since they knew they could make neat things. #mathjoy in the house.

My heart was singing.

Then I showed kids a google doc which had all the info for the contest — and the link to the google form to submit their entries. There were initially two contests. Students needed to create the coolest polar graph with one equation. And students needed to create the coolest polar art using multiple equations. However, some students were animating the sliders and coming up with fun animations (like this or this… watch both for a while). So I added an optional third animated polar graph category.

I haven’t yet told my kids who the winners are. I want to just let them appreciate the work of their classmates for now.

After creating the bulletin board, I’ve seen kids look at the artwork. Kids from my class, but also kids from other grades. And what I’ve found fascinating is that so far, very few kids pick the same polar art pieces as their favorites. I expected everyone to love the same ones I do. But it just isn’t the case. I think when I announce the winners, I’ll have the class go to the board, have everyone point out a few that they like, and then I’ll make my grand pronouncement.

Student Feedback:

I asked my kids, when submitting their artwork, “This is something new I came up with this year. I want to know if you enjoyed doing it or not. No judgments if you didn’t. Y’all tend to be honest when I ask for feedback, and I appreciate it! I genuinely want to know. I also am a bit curious if you had any mathematical thought as you were playing on Desmos? You don’t have to say what thoughts you had (if any) — just if you had any.”

Every student responded positively. Some responses included:

  • I think this was so awesome! I love art and this felt like art to me. It is so fun when art and math intersect, I loved it!!!!!
  • Messing around with the graphs was actually more entertaining than I thought it would be. I spent a lot more time on this than I thought I would, and I feel like I’ll probably spend more time on this trying to find a really cool design I like (and possibly gaining a better understanding why the graphs look the way they do…).
  • I had so much more fun doing this than I thought I would, honestly. Once I finished my multi-equation graph, I looked at the clock on my computer and realized I had been working on it for nearly 20 minutes; it had seemed like maybe 5.
  • I really enjoyed doing this assignment. I felt that I learned a lot about polar through it. I didn’t think too much about math while making my graphs, however I thought about math a lot in order to observe and think about patterns I found in my graphs.

Now I want to be frank: there isn’t much “learning” that happens when kids are doing this assignment. This isn’t a way to teach polar. But it is a way to get kids to appreciate the power of polar when they are done working with polar, and what sorts of different kinds of graphs compared to the boring ol’ rectangular coordinate system. I just wanted kids to play, like I played, and get excited, like I got excited. It’s a slightly different way to appreciate the power of math, and I am good for that. Especially since it only took 10 minutes of classtime!

 

As an aside, I love that when I tweeted this out, a tweep said he was going to be doing this in his class after his kids learn about circles. Um, hell yeah!

Circle.PNG

 

[1] So there are two ways to graph polar in desmos. First is the straight up polar way, and the second is the parametric way. It turns out that the polar way is solid for most things, but it loses refinement at times. Let me show an example. If we graphed r=\cos(57\theta), we should get a flower with 57 petals.

And happily, if we graphed it in both polar and parametric, we get the same looking graph:

 

However if we zoom in a bunch, we can see that the red graph (the polar equation) is interesting and stunning, but just isn’t correct. While the zoomed in blue graph (the parametric equation) is more boring, but is technically correct.

 

It turns out desmos samples more points using parametrics than polar.

As a result, a few of the polar artworks my kids made aren’t “true.” Their pieces are a desmos quirk, like the red graph is above. But what a lovely desmos quirk.

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A simple vector problem with a rich set of approaches

In precalculus, we do a little bit with vectors. And last year and this year, I gave a basic problem to my kids. (I think I found it in some standard textbook.) What I love about this problem is that there are so many ways kids approached it. All essentially the same, but all different enough. Because my kids weren’t all doing it the same way, it has shown me that we are teaching them well. And also, it reminds me that a super basic problem can be a super rich problem.

vector problem

Approach 1: Since we were in vector land, a few kids solved it like this. They found the vector from P to Q, and then set the magnitude equal to 5 and did the algebra.

approach1

Approach 2: Similarly, some students just used the distance formula that they had memorized.

approach2

I like that this kid expanded out (-3-x)^2 and then later eventually factored. Because that was slightly different than how the student in approach 1 solved it (leaving (x+3)^2 as is, and then taking the square root of both sides.

Approach 3: One student found the equation of a circle of radius 5 around the point (-3,1). Then they realized that they were looking for the solution to a system of equations for the circle and the line y=4. So they substituted y=4 into the equation of the circle and solved!

approach3

Approach 4: Most students took a geometric/visual approach. They drew the point (-3,1) and the line y=4. Then they drew these two triangles (seeing that the vertical distance from the point to the line was 3 and the diagonal distance from the point to the line was 5, since we want a distance of 5 away from (-3,1). Then they used 3-4-5 right triangles to get the horizontal distance.

approach4.PNG

All of these were lovely. I enjoyed seeing them all together and drawing some connections among them. Most kids were awed when they saw Approach 3. And since so many students didn’t take a straight up algebraic approach, they were like “ooooh” when they saw Approach 1/2. I supposed what I most like about this is that it really highlights how circles, distance, and vectors are all essentially tied together. I mean kids should know that circles and distance are fundamentally related (but of course they don’t always remember that). But this problem connects those two concepts with something new: vectors. And that the magnitude of vectors simply being an equation involving a circle, secretly.

***

Actually, while I’m writing this, I might as well share this other problem that had a couple of approaches. This was the basic question:

approach5

Approach 1: Most students took this approach. They drew the vector, and then drew a smaller vector with unit length, and then used similarity to find this new vector (with a scale factor of \frac{1}{\sqrt{5}}.

approach6.PNG

Related to this were students who simply saw the scalar that was multiplied by the vector <5,12> to be a “scale factor” that stretches/shrinks the vector by a particular factor. But why this works is because of this similar triangles argument.

Approach 2: A bunch of students used trig. They first found “the angle” and then realized that angle put on a unit circle would work! The fact that so many students saw the problem this way made me happy. I then asked what if we wanted the vector to have length 2 or 3 (instead of unit length), and they were able to answer it. We also talked about one huge deficit of this approach: you lose exactitude since they approximated the angle with their calculator. Even if they didn’t round, they wouldn’t “see” the square root of 5 pop out, when they would with the similarity argument.

approach7.PNG

Approach 3: Okay, so strangely this year, none of my students used this approach. But it is related to the similar triangles approach, and in years past, I’ve had students come up with it. So I showed it to them so they could see another approach. It’s an algebraic approach to find the scale factor.

approach8.PNG

Fin!

A nice proof for the Law of Cosines

We’ve been working on the Law of Cosines in my precalculus classes. And I am having them prove it by scaffolding up from specific triangles to more general triangles. And then with the most general triangles, students consider acute, obtuse, and right triangles.

Kids tend to struggle a bit on the first triangle, but as soon as they realize they need to draw an altitude, they see all that opens up with right triangles and are good. After that, for the rest of the concrete ones, they tend to breeze through. The place where they first stumble again is when they get to the fifth triangle, the one with the angle \beta. They get to L^2=(5\sin\beta)^2+(4-5\cos\beta)^2 but then don’t go any further. But since I know I want them to get to the law of cosines, I tell them to expand and look for something nice. Sometimes I’ll give them the answer (L^2=41-40\cos\beta) and then say: work your work until it looks like this, with one trig function in it. From that point on, kids are in the zone.

For years, I used to teach this by giving kids waaaay too much information.

triangle.png

And I kinda told them what to do… Meh. I was jumping way ahead to get to the formula. We weren’t savoring the thinking to get to the formula. Now we are.

That being said, I ran across something quite beautiful. A stunning proof of the Law of Cosines (at least for acute triangles) on the site trigonography.

stunning.png

I love it because it looks like a proof for the Pythagorean theorem.  Which is nice because  the Law of Cosines is essentially a more generalized version of the Pythagoren theorem.

The area of the bottom square (the green one) is clearly the area of the top two squares (the red and blue ones) minus two green areas. Ummm…. c^2=a^2+b^2-2ab\cos(C) anyone? You see the c^2 and the a^2 and b^2, but what you also see is that you’re taking out some area (the green bits). [1]

When introducing it to my class, I showed them this image:

triangle.png

and said it was a proof of c^2=a^2+b^2-2ab\cos(C).

And I just said to observe. To make statements based on what they see visually…Anything and everything. And if students could, see if they could make connections to the equation (but without writing anything down). After a short while of observations, I opened this geogebra applet and played around. I showed them what happened when we made angle C a right angle.

applet.png

They saw the green rectangles disappear, and how this would be a proof of the Pythagorean theorem if the blue areas and the pink areas were equal to each other. And then I squiggled and smushed the triangle about and eventually kids conjectured that the blue rectangle areas were equal, the pink rectangle areas were equal, and the green rectangle areas were equal. I told them that was true, and they were going to prove that. But before doing that, I asked them: if this was true, do you see a connection between this diagram and the law of cosines?

And kids eventually got there. They saw this argument, essentially…

daigram.png

…and they realized that the green rectangles were probably the thing that was being subtracted out in the law of cosines!

At this point, I gave my kids a blank paper copy of the diagram, and groups work on proving that the blue rectangle areas were equal, the pink rectangle areas were equal, and the green rectangle areas were equal. They had seen all these right triangles before,  when they were looking at the diagram and making observations, so this went pretty quickly for most of them.

I love this proof of the law of cosines! Of course when I went online, I saw so many other beautiful proofs (look here, and the links at the bottom, for some). Troll the internet and be amazed! They are so elegant! This “scaling up” one might be my favorite. And here’s one that David Butler sent me (that is on the site I linked to above). And I remember proving the Pythagorean Theorem in geometry using the crossed chord theorem, and now the same argument here can be used for the law of cosines.

 

[1] To be clear, this diagram only works for acute triangles. I haven’t yet modified the argument to work for obtuse triangles.

Some very cool things about the Law of Sines

So yeah, I’m teaching the law of sines and cosines, and I’m finding some awesome things. What’s totally ridiculous is that after I introduced some of it to my class, I looked back at my stuff from last year and apparently I had done the same thing. Like… I don’t remember it at all.

In any case, when I teach the Law of Sines, I tend to have kids derive it by finding the area of a triangle in three different ways.

Pic1.png

We set these three different ways to get the area of a triangle equal to each other, and divide by \frac{1}{2}abc to get the Law of Sines:

\frac{\frac{1}{2}ba\sin(C)}{\frac{1}{2}abc}=\frac{\frac{1}{2}cb\sin(A)}{\frac{1}{2}abc}=\frac{\frac{1}{2}ac\sin(B)}{\frac{1}{2}abc}

\frac{\sin(C)}{c}=\frac{\sin(A)}{a}=\frac{\sin(B)}{b}

Now, to be clear, there are some subtleties that have to be addressed here. Like for example, this argument clearly works for acute triangles, but what about an obtuse triangle or right triangle, like:

 

It turns out with just a tiiiiny little bit of extra work, we can show that the Law of Sines holds. (Here’s a fun little applet you can play with for this… one important thing that can help you for the obtuse triangle proof is that \sin(\theta)=\sin(180-\theta).)

So… yeah. That’s a pretty traditional way to teach the Law of Sines. But did you know that the ratio that pops up with the Law of Sines has a geometric interpretation?

Like, look at this triangle. And look at the ratio of (side length)/(the sine of the angle opposite the side).

pic4.png

That 10.36 has a geometric meaning. Ready for it? READY? I don’t think you are, but I’m going to show it to you anyway…

pic5.png

Dang! HOLY MOTHERFATHER! Yuppers. That triangle has one circle that can perfectly circumscribe it. And twice the radius of that circumscribed circle is that ratio!!! Don’t believe me? Ok, I know you do, but play with this applet I made to see it happen! Maybe try to create a right triangle and see if that reminds you of something you learned in geometry?

Now this year I told my kids that \frac{a}{\sin(A)}=\frac{b}{\sin(B)}=\frac{c}{\sin(C)}=2R. And I sent ’em up to the whiteboards and asked them to prove it. I gave some hints. Like, for example, the inscribed angle theorem. But eventually kids got it!

pic6.png

This is actually another proof of the Law of Sines! (To be clear, you will also have to make an argument for an obtuse triangle, which requires a tiny bit of modification. You have to see a central angle and one of the angles of the triangle are the same because they both subtend the same arc. And a right triangle.)

So I had a follow up after this… I asked kids to prove that the area of any triangle is: \frac{abc}{4R}, where R is the radius of the circumscribed circle. I asked them to prove it algebraically, and they did:

\text{Area}=\frac{1}{2}ab\sin(C). But we know that \frac{c}{\sin(C}=2R. So let’s manipulate the rea equation to get an R in it.

\text{Area}=\frac{1}{2}abc\frac{\sin(C)}{c}.

Now we have \text{Area}=\frac{1}{2}abc\frac{1}{2R}=\frac{abc}{4R}.

I asked kids to show this algebraically. They did it in various ways (all correct), similar to the argument above. However I had a student present me with a stunning geometric argument that proved this area formula. I honestly don’t know if I would have been able to come up with it. It was so stunning I had to take a photo of it. I leave this as an exercise for the reader. MWAHAHAHA.

(All of this Law of Sine stuff was inspired by this webpage.)

A beautiful combinatorics argument

Today a teacher in my department was struggling to understand algebraically and conceptually why:

binom

He was on the way to making a neat Pascal’s Triangle argument. Look at that 70. That’s \binom{8}{4}:

pascal

He started working backwards, and saw that 70=35+35.

But each of those 35s came from 15 and 20. So 70=(15+20)+(20+15).

And then going backwards more, we see the 15 comes from 5 and 10, and the 20 comes from 10 and 10. So 70=((5+10)+(10+10))+((10+10)+(10+5)).

And then going backwards once more, we see the 5 comes from 1 and 4, and the 10 comes from 4 and 6. So 70=(((1+4)+(4+6))+((4+6)+(4+6)))+(((6+4)+(6+4))+((6+4)+(4+1)))

In other words, 70=1*1+4*4+6*6+4*4+1*1.

By the time we make our way down from the 1-4-6-4-1 row to 70, we see that:

The first one in the 1-4-6-4-1 row just is added once when making the 70.
The first four in the 1-4-6-4-1 row is added four times when making the 70.
The six in the 1-4-6-4-1 row is added six times when making the 70.
The second four in the 1-4-6-4-1 row is added four times when making the 70.
The second one in the 1-4-6-4-1 is added just once when making the 70.

In other words: 70=1^2+4^2+6^2+4^2+1^2.

The other teacher and I realized we could generalize this. But we were left unsatisfied. It was a Pascal’s Triangle argument, but I wanted to see the answer with an understanding of combinations. I wanted something even more conceptual. So my friend and I started thinking, and he had an awesome insight. And I want to record it here so I don’t lose it! It made me so happy — little mathematical endorphins exploding in my head!

Let’s assume we have a set of 2n letters, where n letters are A and n letters are B.

Blergity blerg, let’s just keep things concrete, and have 8 letters, where 4 are As and 4 are Bs. (We can generalize later, but I want to just see this happen!) Given 4As and 4Bs, there are \binom{8}{4} ways to arrange them to make different 8 letter words [1]. Great! That was the easy part.

Now we are going to construct a whole bunch of different sets of 8 letter words, in a particular way (using AAAABBBB), so that when we add up all those sets, we’re going to get all possible 8 letter arrangements of AAAABBBB.

How are we going to do this? We are going to create special of 4 letter words and concatenate them together to make 8 letter words. 

Set 1: We are going to create a 4 letter word with 0As (and thus by default, 4Bs) and a 4 letter word with 4As (and thus by default, 0Bs).

How many ways can we create 4 letter words with 0As? \binom{4}{0}. To be clear, this is just 1. The word is {BBBB}.

How many ways can we create 4 letter words with 4As? \binom{4}{4}. To be clear, this is just 1. The word is {AAAA}.

And when we concatenate them, we are going to have \binom{4}{0}\cdot\binom{4}{4} eight letter words. But we know \binom{4}{0}=\binom{4}{4}. So this is simply \binom{4}{0}\cdot\binom{4}{0}. And this is just 1, because the only eight letter word possible is {BBBBAAAA}.

This is a degenerate case, so it’s hard to really see what’s going on here. So let’s move on.

Set 2: We are going to create a 4 letter word with 1A (and thus by default, 3Bs) and a 4 letter word with 3As (and thus by default, 1Bs).

How many ways can we create 4 letter words with 1As? \binom{4}{1}. To be clear, this is just 4. The words are {ABBB, BABB, BBAB, BBBA}.

How many ways can we create 4 letter words with 3As? \binom{4}{3}. To be clear, this is just 4. The words are {AAAB, AABA, ABAA, BAAA}.

And when we concatenate them, we are going to have \binom{4}{1}\cdot\binom{4}{3} eight letter words. But we know \binom{4}{1}=\binom{4}{3}. So this is simply \binom{4}{1}\cdot\binom{4}{1}. And this is 16 eight letter words. (Each of the first four letter words can be paired with each of the second four letter words… so this is merely 4*4. Just to be clear, I’ll list the first few eight letter words out: ABBBAAAB, ABBBAABA, ABBBABAA, ABBBBAAA, BABBAAAB, BABBAABA, …

Set 3: We are going to create a 4 letter word with 2As (and thus by default, 2Bs) and a 4 letter word with 2As (and thus by default, 2Bs).

How many ways can we create 4 letter words with 2As? \binom{4}{2}. To be clear, this is just 6. The words are {AABB, ABAB, ABBA, BAAB, BABA, BBAA}.

How many ways can we create 4 letter words with 2As? \binom{4}{2}. To be clear, this is just 6. The words are {AABB, ABAB, ABBA, BAAB, BABA, BBAA}.

And when we concatenate them, we are going to have \binom{4}{2}\cdot\binom{4}{2} eight letter words. And this is 36. (Each of the first four letter words can be paired with each of the second four letter words… so 6*6 eight letter words.)

Set 4: We are going to create a 4 letter word with 3As (and thus by default, 1B) and a 4 letter word with 1As (and thus by default, 3Bs). By the same logic as above, we are going to end up with \binom{4}{3}\cdot\binom{4}{3} eight letter words. This is just 4*4 eight letter words.

Set 5: We are going to create a 4 letter word with 4As (and thus by default, 0Bs) and a 4 letter word with 0As (and thus by default, 1B). By the same logic as above, we are going to end up with \binom{4}{4}\cdot\binom{4}{4} eight letter words. This is just 1*1 eight letter words.

 

Now look at all the different eight letter words created by this process, from Set 1, Set 2, Set 3, Set 4, and Set 5. We have captured every single possible eight letter word with four As and four Bs. Let’s check a few random words:

AABABBAB… okay this is in Set 4.
BAABAABB… okay this is in Set 3.
BBBABAAA… okay this is in Set 2.

Cool! I only have to look at the first four letters to decide which set it is going to be in!

But look at what we’ve done. We’ve shown that we can get all eight letter words in these five sets… so the number of eight letter words is:

\binom{4}{0}\cdot\binom{4}{0}+\binom{4}{1}\cdot\binom{4}{1}+\binom{4}{2}\cdot\binom{4}{2}+\binom{4}{3}\cdot\binom{4}{3}+\binom{4}{4}\cdot\binom{4}{4}

If we simply write the squares out…

\binom{4}{0}^2+\binom{4}{1}^2+\binom{4}{2}^2+\binom{4}{3}^2+\binom{4}{4}^2

But we saw at the very start that the number of eight letter words is simply \binom{8}{4}

So the two are equal.

All the hard work is done, so I leave it as an exercise to the reader to generalize.

P.S. I take no credit for this amazingly wonderful letter rearrangement solution. I just bore witness as my friend figured it out, and I got giddier and giddier. I love it because it’s abstract, but still understandable to me. But it’s close to my threshhold of abstraction!

 

[1] If you don’t quite see this, imagine 8 blank slots.

___ ___ ___ ___ ___ ___ ___ ___

You choose four of them to put the As into. There are \binom{8}{4} ways to choose four of these slots. Put As into those four. By default the rest of the slots must be filled with Bs — they are forced! So there are \binom{8}{4} ways to create eight letter words with four As and four Bs.

Double Angle Formulae

I posted this on my Adv. Precalculus google classroom site. I don’t know if I’ll get any responses, but I loved the problem, so I thought I’d share it here.

***

I mentioned in class that I had stumbled across a beautiful different proof for the double angle formulae for sine and cosine, and I would post it to the classroom. But instead of *giving* you the proof, I thought I’d share it as an (optional) challenge. Can you use this diagram to derive the formulae? You are going to have to remember a tiiiiny bit of geometry! I already included one bit (the 2*theta) using the “inscribed angle theorem.”

If you do solve it, please share it with me! If you attempt it but get stuck, feel free to show me and I can nudge you along!

nice

***

Below this fold, I’m posting an image of my solutions! But I say to get maximal enjoyment, you don’t look further, take out a piece of paper, and take a stab at this!

(more…)

What felt like Forever… It was maybe 20 minutes?

On Friday in one of my Advanced Precalculus classes, kids were working on figuring out the double angle formulas for sine and cosine. They got \sin(2\theta)=2\sin(\theta)\cos(\theta) and \cos(2\theta)=\cos^2(\theta)-\sin^2(\theta).

And then… they got stuck.

You see, I showed them two alternative forms for the double angle formula for cosine (\cos(2\theta)=1-2\sin^2(\theta) and \cos(2\theta)=2\cos^2(\theta)-1). I showed them these forms. And I said: figure out where they came from.

All groups in a few minutes were on yellow cups (“our progress is slowing down, but we’re not totally stuck yet”). I didn’t want to give anything away, but I didn’t have any group have a solid insight that I could have them share with others. I let things remain a bit more, no luck, so then I said: “this looks related to something we’ve seen before… a trig identity… maybe that will be helpful. Bring in something you know to open up the problem for you.” Eventually kids realized they needed to bring in some outside information (namely: \sin^2(\theta)+\cos^2(\theta)=1).

I was sure that was going to be enough. Totally certain. But after another 5 minutes of watching them struggle, I wasn’t so sure. I didn’t want to give anything more away, but I had to because we had to move forward. But what more could I give without giving the whole show away? Since many groups were trying some crazy stuff, I said: “this is a simple one or two step thing…” Why? I just wanted them to take fresh eyes and see what they could do thinking simply. They kept on saying I was trying to trick them, but I told them it wasn’t a trick!

And then, in the span of the next five minutes, all my groups got it.

But what was more interesting was that we had three different ways to do it. As kids moved on to the next set of questions (and I breathed a sigh of relief that they figured this out), I reflected on how awesome it was that they persevered and then came up with different approaches. So while they worked, I put up the three different approaches.

20180209_094429

And with a few minutes to go at the end of class, I had everyone put everything away and I just pointed out the embarrassment of riches they came up with. And it was great to hear the audible reactions when kids who had one way saw the other ways and say things like “ooooh, I never would have thought of that!” or “that’s so clever!”

I had (have?) so many mixed feelings when I saw how difficult this question was for my kids. And I was hyperconscious about how much time we had to spend on this. But the ending made me feel like it was time well-spent.