For a sphere, why is the derivative of the volume the equation for the surface area?

Yeah, so the title of the post says it all. I am teaching a standard calculus course, and I wanted my kids to see why this beautiful thing holds true.

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It’s not a coincidence. And in fact, a circle also has a nice property: the derivative of the area of a circle (A(r)=\pi r^2) gives you the circumference of the circle (C(r)=2\pi r). So yesterday I decided I wanted to come up with a short investigation that at least exposes my kids to this idea.

After working for around 90 minutes this morning, I ended up with a packet, and these things on my desk which I’m going to use for illustrations (blocks, dumdums, and tape):

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UPDATE: I was shopping yesterday and found these gems. YAAAS!

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I’ll post the packet I whipped up below. It goes through the standard argument, so in that way it’s nothing special. But in the past when I taught the course, I used to just kinda stand up at the board and give a 5 minute explanation. But I wasn’t sure who was really grokking it, and I was doing too much handwaving.

The big picture trajectory:

*At the start of class, but way before doing this activity, I’m going to have kids recall what a derivative is graphically (the slope of a tangent line), and then how we approximated it before we used limits (the slope between two points close to each other). And from that, I’m going to remind kids of the formal definition of the derivative:

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* I may also start class with this problem, suggested to me on twitter by Joey Kelly (@joeykelly89). It’s a classic problem that was featured on xkcd, but oh so unintuitive and surprising!:

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*Way later in class, I will transition to this activity. The first idea is to get kids to see the connection between the volume of a sphere and the surface area of a sphere. And then again for the area of a circle and the circumference of a circle.

*Then I try to get kids to understand what’s going on with the sphere first… followed by the circle.

*Then I show kids the “better explained” explanation. Why? Because at this point, kids are spending a lot of time thinking about the algebra, and I’m afraid they might have lost the bigger picture. The algebra focuses on one “shell” of the sphere, or one “ring” of the circle. But how does it all fit together? [@calcdave sent me this video, which I’d seen before but forgot about, which has the same argument… this is where the licorice wheels above come into play.]

*Finally, I problematize what they’ve learned. I have them mistakenly make a conjecture that the derivative of the volume of a cube is going to be the surface area of the cube, and the derivative of the area of a square is going to be the perimeter of the square. But quickly kids will see that isn’t quite true. So they have to tease out what’s happening.

My document/investigation [docx version to download/edit]:

My solutions:

I haven’t taught this yet. So it could be a complete disaster. I don’t have a sense of timing. I don’t know how much of this is me and how much will be them. I am just hoping tomorrow isn’t a disaster! Fingers crossed!

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More Things I Want To Highlight From Twitter

As I mentioned before, I often see neat stuff on Twitter and want to remember it, but I just “favorite it” and then forget it. Sometimes I remember to go back and look for it, but it’s arduous. So I decided if I get the time, I’ll blog about some of my favorite things which will help me remember them better.

At the very end, you’ll see a problem that nerdsniped me for a good while. It hit a sweet spot for me. 

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First off, though, something about me! Kara Newhouse contacted me and Joel Bezaire about how we use reading in our math classes, and then she wrote an article about it for KQED Mind/Shift. I didn’t quite know what it was, but I figured why not answer a few questions!

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Even though I hadn’t heard of Mind/Shift before, it apparently is something that gets around. Because my friend Julie told me someone sent it to her, and my sister saw it in the NAIS newsletter. And even Steve Strogatz tweeted it out (with some links to my blog). So that’s random and awesome. If you want more information about how I organized the book club, here is my post-mortem after the first year of doing it. Now I am not teaching the course I had the book club in, but what’s awesome is that I have a few kids who wanted to read math books with me… so informally I meet with them for 40 minutes each 7 school days, and over donuts, we talk. First we read Flatland, and now we’re reading The Man Who Knew Infinity. And I’m loving it! So don’t think you need to do it in a class. I’ve read books and discussed them with kids one-on-one!

Okay, now that’s over! Onto the other less self-aggrandizing things!

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I love Richard Feynman. I don’t think it would be too strong a statement to say that I would be a different person today if I hadn’t had been introduced to him when I was young, when my father gave me a copy of What Do You Care What Other People Think? And I saw this, and I immediately wanted to get it printed on a business card to hand to kids at the start of the year.

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Now that we’re talking about Feynman, James Propp wrote a powerful piece about “genius” which problematizes Feynman. I already knew Feynman was self-fashioning himself in the way he presented himself to the public-at-large (and his contemporaries). But this article goes further, in a reflection connecting to a powerful piece by Moon Duchin about the sexual politics of genius. He notes:

But Duchin makes me ask, for whom does Feynman’s advice work well? Who in our culture is forgiven for putting aside personal relationships in the name of single-minded pursuit of truth? Who is permitted to be a joker? And who in our culture is steered, from an early age, toward an excruciating attunement to what other people think?

I highly recommend reading James Propp’s piece.

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Steve Phelps (@giohio) shared this desmos applet which plots lines normal to a curve.  You can change the curve!

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It reminds me of my family of curves project (post 1, post 2, post 3). I wonder if I couldn’t have kids come up with a way to get any perpedicular line to a curve in calculus, and then have them play around with this applet to generate beautiful designs!

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Nanette Johnson tweeted out a powerful slide from a talk she was at given by @danluevanos. I don’t need the rest of the talk. I get it.

pic4.pngI often feel like a crappy teacher. Right now, I’m on day 3 of Spring Break, and since it started, I’ve been contemplating how crappy I feel about my teaching. I know I’m not a bad teacher, but … maybe I am? I don’t know. But yes, these two questions screamed at me. Because they are part of something I need to recommit myself to: focus on the positive and take the positive and multiply it. Because I always focus on questions #3 and #4, and rarely give myself time to think about #1 and #2.

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Mark Kaercher is using Desmos to do warmups.

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Here’s the first link in his tweet: https://www.desmos.com/calculator/xxxmahtp91
Here’s the second link in his tweet: https://teacher.desmos.com/activitybuilder/custom/5a770219c1b9e208ca83895c
I love this idea of doing warmups using Activity Builder. Must remember for next year!

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Patrick Honner wrote a great article in Quanta magazine: “How Math (And Vaccines) Keep You Safe From The Flu”. I’m just mad I didn’t send this article to my calculus kids after we did our point of inf(l)ection activity (adapted lightly from Bowman Dickson @bowmanimal).

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This tweet from Kara Imm just made me so happy. I always believe that formalism and stuff should come after something has been explored (whenever possibe). And second graders were absolutely doing that! Sixogon! Navada! So awesome!

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So Anna Blinstein asked about higher level mathematics and this happened. I learned that arithmetic with complex numbers is akin to arithmetic with polynomials mod x^2+1. WHAAA?!

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Of course I took out pencil and paper and had some fun with this. Blows my mind.

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Steve Strogatz tweeted out this interesting article by Maria A. Vitulli about Writing Women in Mathematics in Wikipedia. The abstract is here:

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When it comes to polynomial division, I’ve seen the connection to standard division (where x=10). But I think I need to exploit this more in my teaching, especially to make the polynomial remainder theorem seem obvious to kids (and not like magic, which they often feel, even when we’ve figured it out). Erick Lee tweeted a perfect reminder:

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Joe Cossette tweeted out a neat idea — a stop motion photography race between two figures. (Click here for original tweet so you can watch the video.) I wonder if this can be adapted to calculus when we talk about rectilinear motion. Regardless, I could see it be interesting to give a physical understanding to various functions. Especially when comparing them (like exponential versus quadratic). “Which will eventually win?” Plotting x^2 versus e^x is one thing, but seeing them in a race is another.

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David Butler read my recent post about The Law of Cosines and shared his own post which talks about how the Law of Cosines doesn’t actually need cosine in it. Worth seeing! Trig without trig!

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I leave you with a problem from Abram Jopp that nerdsniped me!

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More constraints: You can’t use domain restrictions. You can use compound inequalities, but desmos only allows simple ones (e.g. 2<x<3 or x<y<x+2) and not complicated ones. A good number of tweeps got obsessed. There were many different proposed solutions, but I am proud of mine. I don’t think anyone else’s was quite like it.

In the process of working on this problem, David Butler and Suzanne Von Oy reminded me of this beautiful relationship: min(A,B)=1/2(A+B)-1/2|A-B| and max(A,B)=1/2(A+B)+1/2(A-B). And so I wanted to illustrate that with this Desmos graph. I definitely want to remember this gem when I teach Advanced Algebra II at some point when we’re exploring the power of absolute value. It’s so awesome.

 

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.

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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.

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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:

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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.

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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…

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…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.

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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).

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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…

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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!

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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:

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

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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.

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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

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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.

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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.