Geometry (all content)
Area of Koch snowflake (1 of 2)
Starting to figure out the area of a Koch Snowflake (which has an infinite perimeter). This is an advanced video. Created by Sal Khan.
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- this is a little confusing can you give me a small summery?(3 votes)
- Sure. He starts with an equilateral triangle, and pastes smaller triangles to the sides. This makes new sides. Then he adds even smaller triangles. For some crazy reason, he's never going to stop. :)
He wants to know how big his snowflake is. It gets a little bigger each time. Two ways it gets bigger are
Area: Adds up the area of of all the triangles.
Perimeter: This is a little trickier. When he pastes new triangles, they cover some of the old perimeter. He would have to subtract the edges that are now inside, and add the new edges. Here, Sal keeps track of the number of triangles but does not calculate the perimeter.
So the process is fairly simple: paste some triangles, add up the perimeter and area. Paste some more, add up the perimeter and area.
The interesting part is that the perimeter keeps on getting bigger... if you give me a big number and say, "the perimeter can never be longer than this", I can tell you exactly how many triangles I need to add, to make the perimeter bigger than that.
....but the same is not true for area. The snowflake will never reach the edges of the Sal's blackboard. You can draw a box around the snowflake and the area will never be as large as that box.
So there area is limited. Limited to what number? Sal is trying to figure out that number.
Adding up infinite sums isn't easy. Is that the part that's confusing?
Hope that helps.(59 votes)
- Why is the area divided by 4?!(7 votes)
- The value needs to stay constant so your to divide and the multiply. I got it !!!(12 votes)
- If it goes on forever wouldn't tha area also be infinite(2 votes)
- No; for the area to be infinite, the snowflake would have to take up an infinite amount of space. It doesn't take up and infinite amount of space, so the area can't be infinite.
A more mathematical way to explain it is this: as the snowflake continues growing, the extra parts get smaller and smaller; they get small so fast that the area of the snowflake slows down, and the area never gets larger than a certain amount.(8 votes)
- At9:40or so, I get REALLY puzzled. I was having a hard time following before that, but when Sal started tossing fours all over the place he lost me. If one multiplies by 4 and then divides by 4, doesn't one end up right back where one started? If so, what's the point?(2 votes)
- This is a common algebraic move that Sal used to "clean up" his equation, and it works because you are absolutely right: multiplying by 4 and dividing by 4 leaves the value of the equation unchanged. What's happening is that by leaving the division alone for the moment (Sal leaves it out front as 1/4), he can use the multiplication by 4 where he knows he wants it, which is distributed out to the terms of the summation so he can raise the power of the 4's. Sal does mention what he's doing, even though he doesn't go into it in detail. If you're interested, try watching some videos on algebra, or try writing down the equation Sal had and see if you can follow along his reasoning.(6 votes)
- So the area is finite and the perimeter is infinate?(2 votes)
this clarifies on that point(4 votes)
- what are some applications of the koch snowflake fractal?(3 votes)
- It's more of a math theory- it wouldn't actually work in the real world. It is a nice way to show how fractals work without a SUPER complicated shape. Edit: Well, you could do it to a point in the real world, but it would not be infinite. In fact, you see fractals A LOT in nature.(1 vote)
- When he starts factoring at6:29, why doesn't the square (the 2 on the blue section) get factored out with the rest of the equation? Is it because it's not exactly s^2 like the original yellow equation?(2 votes)
sqrt(3)•s^2/4 + 3•sqrt(3)•(s/3)^2/4 + 12•sqrt(3)•(s/9)^2/4 + 48•sqrt(3)•(s/27)^2/4 ...
sqrt(3)•(s^2/4 + 3•(s/3)^2/4 + 12•(s/9)^2/4 + 48•(s/27)^2/4 ...)
sqrt(3)/4•(s^2 + 3•(s/3)^2 + 12•(s/9)^2 + 48•(s/27)^2 ...)
sqrt(3)/4•(s^2 + 3•s^2•(1/3)^2 + 12•s^2•(1/9)^2 + 48•s^2•(1/27)^2 ...)
sqrt(3)•s^2/4•(1 + 3•(1/3)^2 + 12•(1/9)^2 + 48•(1/27)^2 ...)(1 vote)
- It irked me that the third simplified equation used a blue addition sign rather than it being pink.
Isn't something to the first power pointless? Math hurts my brain sometimes.(2 votes)
- So, is this like the Koch curve?(1 vote)
- Yes, but instead of one side, he uses the three sides of a triangle.(1 vote)
- At approximately8:59, why are we allowed to divide the "outside bit" by four, and multiply the "inside bit" by four?(1 vote)
- You can multiply both the outside and the inside because any number is itself multiplied by 1, and 1 = 4 * (1/4)
You can hence multiply an expression by 4 * (1/4) while keeping its value. Using the property of multiplication m * (a * b) = (m * a) * b = a * (m * b) you can write that as 1/4 times the outside multiplied by 4 times the inside.
I hope this helps.
--Phi φ(1 vote)
We now know how to find the area of an equilateral triangle. What I want to do in this video is attempt to find the area of a-- and I know I'm mispronouncing it-- a "kawch," or "coach" snowflake. And the way you construct one is you start with an equilateral triangle. And then on each of the sides you split them into thirds and in the middle third you put another smaller equilateral triangle. And that's after one pass. And on the next pass, you do that for all of the sides here. So a little one here, here, here, here, here, here, here, here, here. I think you get the general idea. That's the next pass. And on the next pass you do it for all of these sides. And what's really neat about this-- and we showed this in a previous video-- is that you have a figure here that has an infinite perimeter. But we're about to see in this video it actually has a finite area, which is kind of interesting to think about. So let's start with a clean equilateral triangle right over here. We're going to assume that each of the sides have length s. So it's going to be a clean equilateral triangle. Each of the sides-- let me draw that a little bit neater. Each of the sides have length s. And so what I'm going to do is I'm going to keep track of two things. I'm going to keep track of the sides on this triangle as it turns into a snowflake . I'm going to keep track of the number of sides, and I'm going to keep track of the area after each pass of adding more smaller triangles. So this is going to be our count of the area. Actually, let me give myself a little bit more real estate, just because I have a feeling I might need to use it. So this is the sides. I'm going to write up here. And then this is our running count of the area down here. So right when we start, we have 3 sides. And our area, we already figured out in a previous video, is going to be-- if we assume each of the sides are length s-- is going to be square root of 3s squared over 4. Fair enough. That was just a simple equilateral triangle. Now we're going to take each of these sides, divide them into thirds, and then in that middle third, we're going to add another smaller equilateral triangle. So it'll look like that on that side right over there. I want you to think about what we're doing to each side right here. So before I did this, this was just one side. Then I split it into thirds. And that middle third, I essentially put two sides in there. I put an equilateral triangle. So one side is now turned into one, two, three, four sides. So every time we do a pass of making the snowflake more intricate, each side will turn into four sides. So you could imagine if we do this on all three sides, we have 4 times 3, which is now twelve sides. So if you multiply this times 4, this gets us to twelve sides now. We can count them out just to make sure where our logic is correct. 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 sides. And now what is the area, now? Well it's going to be the area of our original yellow equilateral triangle, plus the area of each of these smaller ones. And what's the area of each of these smaller ones? Well first of all, we have three of them. There are three of each of these smaller ones and then we use the formula for the area of an equilateral triangle again. So it's going to be square root of 3 times s squared. But now the length of each of the sides of each of these smaller equilateral triangles-- they aren't s anymore, they are s over 3. Remember, this length right over here is s over 3. So this is going to be s over 3, as well. Every pass, the sides of the equilateral triangle become 1/3 of the previous pass. So it's not going to be s squared anymore. It's going to be s over 3 squared. And then all of that over 4. And then let's do another pass. So we're going to add these triangles right over here. Going to add these right over there. And this is the last pass where I'll actually attempt to draw all of the triangles over there. So how many sides am I going to have, first of all, after I do another pass? Well, the previous pass, I had 12 sides. Each of those 12 sides are now going to turn to 4 new sides when I add these little orange bumps there. So I'm going to multiply it times 4 again. I'm going to multiply it times 4. So now I'm going to have 48 sides. And how many new triangles? So what's the area? Well it's going to be the yellow area plus the blue area plus the orange area. So how many new orange triangles do I have.? Well I'm adding a new orange triangle to each of the sides for the previous pass. In the previous pass I had 12 sides. So now I'm going to add 12 orange triangles. And actually let me write that. I'll just write 12 orange triangles. But it's really I just multiplied it times 4. And then I'm going to have times the square root of 3. And now this isn't going to be s over 3 any more. These are now going to be s over 9. These have 1/3 of the dimensions of these blue triangle. So this is going to be s over 9. s over 9 squared over 4. And so I think you might start to see the pattern building if we do another pass after this one. Move to the right a little bit. What will that look like? Let me do this in a different color that I haven't used yet. Let me see I haven't used this pink, yet. So now I'm going to have the previous number of sides, that's my number of new triangles, 48 times the square root of 3 times s over-- now these are going to be even 1/3 of this-- s over 27 to the second power. All of that over 4. And I'm going to keep adding an infinite number of terms of this to get the area of a true koch snowflake. So I'm just going to keep doing this over, and over again. So the trick really is finding this infinite sum and seeing if we get a finite number over here. So the first thing I want to do, just to simplify, well, let me just rewrite it a little bit differently over here. So the first thing that might be obvious is that we can factor out a square root of 3s squared over 4. So let me factor that out. So if we factor out a square root of 3s squared over 4 from all of the terms. Then this term right over here will become a 1. This term right over here is going to become a 3. Let's see, we factored out a square root of 3. We factor out a four. And we factored out the s squared. We factor out only the s squared. So now it's going to have plus 3 times 1/3 squared. That's all we have left here. We have the 1/3 squared. And then we have this 3. And I'm not simplifying this on purpose, so that we see a pattern emerge. And then this next term, right over here, plus-- so this 12 is still going to be there. But I'm going to write that as 3 times 4. Let's see, we're factoring out the square root of 3, we're factoring out the 4, we're factoring out the s squared. And so we're going to be left with 3 squared. That's what this is down here, squared. So this is 1/3 squared. And then that, squared. So that's what we're left with, with that orange term. And then we go to the pink term. This pink term. 48 is just 3 times 4 times 4. Three times 4. I'll write 4 squared here, because each time we're going to multiply it times 4 again. So the next one's going to be 4 to the third, because we're really-- each side turns into four sides. That's where that came from. 4 squared, we're factoring out the square root of 3, we're factoring out the four, we're factoring out the s squared. And all we're left is 1 over 3 to the third power, squared. So times 1 over 3 to the third power, squared. And we're just going to keep going like that forever. So on each step, we're multiplying by 4, and we're also multiplying-- I guess we say, the power of this 4 is incrementing. So there's really 4 to the 0-th power here. We have a 1 here you can imagine, implicitly. The 4 to the first power, 4 squared, then it'll go 4 to the third. This power is also incrementing-- 3 to the first, 3 to the second, 3 to the third. But we see that this power is always one more than that. And it'll be much easier to calculate this infinite-- what's going to turn into an infinite-- geometric series, if those are actually the same power. So what I want to do is I want to increase the power of 4 in all of those. But I can't just willy-nilly multiply everything by 4. If I'm going to multiply everything by 4, I also need to divide everything by 4. So what I'm going to do in this step right over here is I'm going to multiply and divide everything by 4. So if we divide by 4, I can do that on the outside. So let me multiply 1/4 times this right over here. And so I'm dividing by 4 out here. And then I'm going to multiply this by 4. And so I'm not going to be changing the value of the actual thing. This is going to be 4 plus 3 times 4 plus 3 times 4 to the third. And so what was cool about this is now that the power of 4 and the power of this 3 down here are going to be the same power. But it still seems a little weird because we're taking this 1 over 3 squared and then we're squaring it. And here we just have to realize-- so this is always going to be squared. And this is the thing that's incrementing. But in general, if I have 1 over 3 to the n, and I'm squaring it, this is equal to 1 over 3 to the 2n power. So I'm just multiplying it by 2. If I'm raising something to the exponent, then raising that to an exponent, that's just raising it to the n times 2 exponent. And this is the exact same thing as 1 over 3 squared, raised to the nth power. So we can actually switch these two exponents in a very legitimate way. And then let me rewrite everything. Because I don't want to do too much on this one step right over here. So this part right over here gives us square root of 3s squared over 16. And then that's going to be times-- I'll open and close the parentheses. So then we have 4 plus. Then in blue, I'll write 3 times 4 to the first power. And then I can rewrite this as times 1/3. We could view this is 1/3 squared. We could view this as 1 over 3 to the first power, squared, or we could view this as 1 over 3 squared to the first power. And I'm going to write it that way. So times 1/9 to the first power. And then plus 3 times 4 squared. And then this we can write as, times 1/9 to the second power. And then this one we could write, plus 3 times 4 to the third times-- we could write this is 1 over 27 to the second power. But we could also write this based on what we saw over here. We could write this as 1 over 3 squared to the third power, based on this right over here. Let me make this clear. 1 over 3 to the third, to the second power. This is the same thing as 1 over 3 squared to the third power. That's what we showed right over here. So this is the equivalent to 1/9 to the third power. And now we start to see the pattern is starting to clean up a little bit. And let me just do one more step. And then we'll finish this in the next video. So this is equal to square root of 3s squared over 16 times 4 plus 3 times, this is 4/9, plus the next term is three times 4/9 squared. And then we have plus 3 times 4 over 9 to the third power. And then we're just going to keep going on, and on, and on, and on taking the 3 times 4/9 to the successively larger, and larger powers. So this is what we have to find the sum of to find our area. And we're going to do that in the next video. And we're going to use some of the tools we've used to find the sums of infinite geometric series. But we're going to re-do it in this video just so that you don't have to remember that formula or that proof.