## Posts Tagged ‘geometry’

### Visualizing Elementary Calculus: Introduction

March 25, 2011

Recently I’ve been trying to be more geometrical when discussing elementary calculus with high school students. I don’t want to write an entire introduction to calculus, but the next few posts will outline some ways I think the geometric view can be helpful.

This series
I – Introduction
II – Trigonometry

You know about $\Delta$, which means “the change in”. For example, if $w$ represents my weight, then $-\Delta w$ represents the weight of the poop I just took.

Let’s say $h$ is your height above sea level. $\Delta h$ is the change in that height, but what change? The change when you climb the stairs? When you jump out of a plane? When you step on a banana peel?

When we think about change, we usually think about two things changing together. You get higher when you climb another stair on the staircase. $h$ is changing, and so is $s$, the number of stairs climbed.

These two changes are related to each other. Say the stairs are 10 cm high. Then you gain 10 cm of height for each stair. We can write that as $\Delta h = 10 {\rm cm} \hspace{.5em} \Delta s$. We can also write it $\Delta h / \Delta s = 10 \hspace{.5em}{\rm cm}$. This says, “the height per stair is ten centimeters.”

This is the goal of calculus – to study the relationships between changing quantities. Let’s do a real example.

### The Area of a Square

Let’s say we have a square whose sides lengths are $x$. Its area is $x^2$. What is the relationship between changes in its area and changes in the length of a side? Draw the square, then expand the sides some. The amount the sides have expanded is $\Delta x$. The new area that’s been added is $\Delta (x^2)$.

We begin with the red square on the left, whose area is x^2. We add an extra amount Delta(x) to the sides, creating all the new green area.

From the picture we see

$\Delta(x^2) = 2x\Delta x + (\Delta x)^2$

This formula relates $\Delta (x^2)$, the change in the area, to $\Delta x$, the change in the length of a side.

### The Derivative of $x^2$

In the picture of the square, there is a little piece in the upper-right corner whose area is $(\Delta x)^2$. It is the smallest bit of area in the whole picture.

Look what happens when we make $\Delta x$ even smaller.

We shrink Delta(x) and observe what happens to the different areas being added on.

In the first picture, $\Delta x$ (no longer marked) is a quarter of $x$. $(\Delta x)^2$ is the dark green area, and it is one quarter as large as $x \Delta x$, the light green area. We see this because the dark patch fits inside the light one four times.

In the second picture, we shrink $\Delta x$ to one eighth of $x$. All the green areas shrink, but the dark patch shrinks on two sides while the light patches shrink on only one. As a result, the dark $(\Delta x)^2$ is now only one eighth the size of the light $x \Delta x$.

If we continued to shrink $\Delta x$, this ratio would continue to decrease. Eventually we could tile the dark patch a million times into the light one. So, as long as $\Delta x$ is very small, we can get a good estimate of the entire green area by ignoring the dark part $(\Delta x)^2$. Thus

$\Delta(x^2) \approx 2x\Delta x$

This approximation becomes better and better as $\Delta x$ shrinks, becoming perfect as $\Delta x$ becomes infinitesimally small.

When we want to indicate these infinitely small changes, we trade in the $\Delta$ for a ${\rm d}$ and write

$\textrm{d}(x^2) = 2x \textrm{d}x$

The terms $\textrm{d}(x^2)$ and $\textrm{d}x$ are called “differentials”. The equation expresses the relationship between two infinitely-small changes, one in $x$ and one in $x^2$.

Frequently, we divide by $\textrm{d}x$ on both sides to get

$\frac{\textrm{d}(x^2)}{\textrm{d}x} = 2x$

This is called “the derivative of $x^2$ with respect to $x$“.

#### Example 1: Estimating Squares

$20^2 = 400$. What is $21^2$?

Here $x$ = 20, and we’re looking at $x^2$. When $x$ goes from 20 to 21, it changes by 1, so $\textrm{d}x = 1$. Our formula tells us

$\textrm{d}(x^2) = 2x \textrm{d}x = 2*20*(1) = 40$

Hence, $x^2$ increases by about 40, from 400 to 440.

The real value is 441. We got the change in $x^2$ wrong by about 2%. That’s because $\textrm{d}x$ wasn’t infinitely small.

Let’s try again, this time estimating the square of 20.00458. Now $\textrm{d}x$ = .00458, so

$\textrm{d}(x^2) = 2 x \textrm{d}x = 2*20*.00458 = .1832$

The estimate is 400.1832. The real value is 400.183221. We did much better, under-estimating the change by only 0.01% this time. Also, it was not much harder to do this problem than the last, but squaring out 20.00458 by hand would be a pain. We saved some work.

#### Example 2: How Far Is the Horizon?

The beach is a good place to think about calculus. If you look out at the ocean, the horizon appears perfectly flat. Nonetheless, we know the Earth is really curved. In fact, we can deduce the curvature of the Earth by standing on the beach and enlisting the help of a friend in a boat.

It works like this: You stand on the beach with your head two meters above the water. Your friend sails away until the boat begins to disappear from sight. The reason the bottom of the boat is disappearing is that it is hidden behind the curvature of Earth.

When the bottom of the boat disappears, measure the distance to some part of the boat you can still see. What’s the relationship between your height, the distance to the boat, and the radius of Earth?

A picture will help. We’ll call your height $h$ and the distance to the horizon $z$.

You are the vertical stick on top, height h. The boat is the brown circle. It's at the horizon, a distance z away. The dotted line shows your line of sight. When the bottom of the boat begins disappearing, a right triangle forms.

Your height, the radius of Earth, and the distance to the horizon are related by the Pythagorean theorem to give

$R^2 + z^2 = (R+h)^2$

this is equivalent to

$z^2 = 2Rh + h^2$

As we have seen, if your height $h$ is small compared to the size of the Earth (and it is), the term $h^2$ drops away and the distance to the horizon is

$z = \sqrt{2Rh}$

You can see about $5 {\rm km}$ at the beach, making the radius of Earth about $6,000 {\rm km}$. (It’s actually $6378.1 {\rm km}$).

Next we want to know how much further you can see if you stand on your tiptoes. That would be a small change $\textrm{d}h$ to your height. It would let you see a small amount $\textrm{d}z$ further. How is $\textrm{d}h$ related to $\textrm{d}z$?

$\textrm{d}(x^2) = 2x\textrm{d}x$

So let $x^2 = h$, or $x = \sqrt{h}$, and we have

$\textrm{d}h = 2\sqrt{h}\hspace{.3em}\textrm{d}(\sqrt{h})$

But we also know

$\sqrt{h} = \frac{z}{\sqrt{2R}}$

so we can substitute that in to $\textrm{d}(\sqrt{h})$ and get

$\textrm{d}h = 2\sqrt{h}\hspace{.3em}\textrm{d}\left(\frac{z}{\sqrt{2R}}\right)$

or

$\frac{\textrm{d}z}{\textrm{d}h} = \sqrt{\frac{R}{2h}}$

This tells us how much further you can see if you get a little higher up. The interesting thing is it depends on $h$. The higher you go, the smaller $\textrm{d}z$. When you’re only two meters up, you get to see almost ten meters further out for every centimeter higher you go. However, if you’re 100m up on top a carousel, you get only 1 meter for each centimeter you rise.

It makes sense that the extra distance you see gets smaller and smaller the higher you go, and eventually shrinks down to zero. No matter how high you go, you can never see more than a quarter way around the globe.

(In reality, light bends due to refraction in the atmosphere, so you can sometimes see a bit further.)

### Circles

Suppose we have a circle with radius $r$. It has a certain area (you undoubtedly know the formula already, but play along). Suppose we increase $r$ by a small amount $\textrm{d}r$. What is the change $\textrm{d}A$ in the area?

The original circle is dark blue with area A and radius R. The radius increases an amount dR, increasing the area by the light blue ring with area dA.

$\textrm{d}A$ is the thin, light-blue ring. Imagine taking that ring and peeling it off the edge of the circle and laying it flat. We’d have a rectangle with width $\textrm{d}R$. Its length comes from the outside edge of the entire circle – the circumference. The circumference is $2 \pi R$, so

$\textrm{d}A = 2\pi R \textrm{d}R$

We saw earlier that $\textrm{d}(x^2) = 2x\textrm{d}x$, so let $x = R$ and we have

$\textrm{d}A = \pi \textrm{d}(R^2)$

Thus the quantities $A$ and $\pi R^2$ change in exactly the same way. Since they also start out the same (both zero when R is zero), we have

$A = \pi R^2$

### Next Post

We’ll look at trigonometry. Geometric arguments about the derivatives of trig functions are very simple ways of visualizing what’s going one, and are usually not introduced in a basic calculus course.

### Exercises

• Draw a cube with sides $x$ and show that $\textrm{d}(x^3) = 3x^2\textrm{d}x$. Thus the derivative of $x^3$ with respect to $x$ is $3x^2$.
• Draw a line with length $x$ and show that $\textrm{d}(x) = \textrm{d}x$, which is of course algebraically obvious. Thus the derivative of $x$ with respect to itself is 1.
• Draw a rectangle with width $w$ and length $c*w$ and show that $\textrm{d}(c*w^2) = 2cw\textrm{d}w = c\textrm{d}(w^2)$. Thus, whenever you have the differential of a variable multiplied by a constant, the constant can pop outside. Where was this property used implicitly in this post?
• Now that you know $\textrm{d}(x^3) = 3x^2\textrm{d}x$, let $x^3 = u$ and find the derivative of $u^{1/3}$ with respect to $u$. (Answer: $\frac{1}{3} u^{-2/3}$)
• What is $\textrm{d}(x^3)/\textrm{d}(x^2)$? Let $u = x^2$ and find the derivative of $u^{3/2}$ with respect to $u$. (Answer: $\frac{3}{2}u^{1/2}$).
• Examine $\textrm{d}(x^4)$ by letting $u = x^2$, so we’re looking at $\textrm{d}(u^2)$. Find the derivative of $x^4$ with respect to $x$. (Answer: $4x^3$)
• Draw an equilateral triangle with sides of length $s$. Increase the sides a small amount $\textrm{d}s$ and relate this to the change in area $\textrm{d}A$. Does this agree with our previous findings?
• Draw an ellipse with a fixed with semi-major axis $a$ and semi-minor axis $b$. Starting with a unit circle, argue by thinking about stretching that the area of the ellipse is $\pi ab$. Increase $a$ by a small amount $\textrm{d}a$ and increase $b$ proportionately. This adds a small area $\textrm{d}A$ to the ellipse. Show that this area is $\pi(a^2+b^2)/b\hspace{.3em}\textrm{d}a$. Does this let us find the circumference of the ellipse by the same thought process as we used for the circle? (Answer: no). Why not?
• Draw a sphere with radius $R$. Use the relationship between $\textrm{d}R$ and $\textrm{d}A$ to find the volume of a sphere, given its surface area is $4\pi R^2$. Check your answer against this post.

### My Brown Big Spiders

March 21, 2011

Professor: You have to learn to be able to play it blindfolded. The page, for God’s sake! The notes!

David: I’m sorry I was, uh, forgetting them, Professor.

Professor: Would it be asking too much to learn them first?

David: And-And then forget them?

Professor: Precisely.

from the movie Shine

If I want to find the volume and surface area of a sphere, I do it with calculus:

$V = \int_{r = 0}^R\int^{2\pi}_{\phi = 0}\int_{\theta = 0}^\pi r^2\sin\theta \textrm{d}\theta \textrm{d}\phi \textrm{d}r = \frac{4}{3}\pi R^3$

.

$S = \int_{\theta = 0}^\pi\int_{\phi = 0}^{2\pi} R^2 \sin\theta\textrm{d}\theta\textrm{d}\phi = 4\pi R^2$

This is correct, but I can’t use it with high school geometry students because they don’t know what an integral is, much less a Jacobian.

However, Archimedes came up with a beautiful way of discovering the volume and surface area of a sphere. He did it by relating the sphere to a known shape – a cylinder with a cone cut out of it.

He drew a picture like this:

On the left there’s a hemisphere with radius $R$. On the right, there’s a cylinder with radius and height both also $R$, so that the hemisphere would fit perfectly inside the cylinder. The cylinder has had a cone cut out from the top down tapering down to the center of the bottom. First, we’ll show that these two shapes have the same volume.

We imagine slicing the hemisphere horizontally at some certain height $h$. This would reveal a circle as seen in the picture. Call its radius $r$.

At the same height, we also slice the cylinder, leaving us with a disk. We’ll find the areas of this circle and disk.

The area of the circle is $\pi r^2$, which by the Pythagorean theorem is also $\pi (R^2 - h^2)$.

Looking at the cylinder, the outer edge of the disk has radius $R$ and the inner edge has radius $h$, so the area of the disk is also $\pi (R^2 - h^2)$.

Because every horizontal slice of the hemisphere has the same area as the corresponding horizontal slice of the drilled-out cylinder, they must have the same volume. The volume of the cylinder is its original volume minus the volume of the cone, or $\pi R^3 - 1/3 \pi R^3 = 2/3 \pi R^3$. Hence, the volume of a full sphere is

$V = 4/3 \pi R^3$

Next, we’ll show that the hemisphere has the same surface area as the outside of the cylinder (the cone is now unimportant).

Take a slice of the outside of the cylinder at height $h$ and of thickness $\textrm{d}h$. This forms a band around the cylinder whose area is

$\textrm{d}S = 2 \pi R \textrm{d}h$

Now slice the sphere at the same height with the same $\textrm{d}h$. This also forms a band. The band is a shorter distance around, but due to the slant of the edge of the circle, it’s also thicker. Let’s call the thickness of this band $\textrm{d}x$.

The area of the band around the hemisphere is the circumference at height $h$ multiplied by the thickness $\textrm{d}x$.

$\textrm{d}S = 2\pi\sqrt{R^2 - h^2}\textrm{d}x$

If we draw a tangent line on the sphere, it’s perpendicular to the radius. This gives us similar triangles.

So

$\frac{\textrm{d}x}{\textrm{d}h} = \frac{R}{\sqrt{R^2 - h^2}}$

Plugging back into the previous expression,

$\textrm{d}S = 2\pi\sqrt{R^2 - h^2}*\textrm{d}h * \frac{R}{\sqrt{R^2 - h^2}} = 2\pi R \textrm{d}h$

So the band around the outside of the cylinder and sphere have the same surface area, so the entire shapes have the same surface area. That makes the surface area of a sphere

$S = 4 \pi R^2$

This is a really lovely argument. The problem is pretty hard, but the solution is simple. (I’m not sure if this is quite how Archimedes did it. To be honest I never even met the guy. I learned the idea from this animation).

I was reviewing solid geometry with a high school junior the other day, so I showed her this argument (but only the volume part). I was proud of myself for offering this little example of how interesting mathematical ideas can be. At least, I was as we began.

“It’s all so complicated!” she moaned a few minutes later when I asked her to identify a certain quantity in our sketch.

Complicated? I had thought the argument was remarkably simple – just draw a sphere and a cylinder next to each other and you’re practically done. What could be simpler? Somehow my student was getting entangled in brambles I couldn’t even see.

I did not draw quite the same picture for her that I drew earlier in this post. I didn’t want to give it all away, so I drew something more like this and asked for $r$:

Finding $r$ is a simple application of something she knew well – the Pythagorean theorem. She didn’t see it, though, so I showed her this right triangle:

But then she didn’t see how long the new line I just drew was. It’s just $R$ because it’s a radius of the sphere, but although she knew that all radii of a sphere have the same length, she couldn’t easily identify the two lines as radii and call up the relevant information. So I showed her that step, too.

After a bit more prodding, she wrote down $r = \sqrt{R^2 + h^2}$, a mistake that comes from applying the Pythagorean theorem incorrectly. She knows better, and should have found $r^2 = R^2 - h^2$, but by this point she was already flustered from her earlier mistakes, confused about what we were trying to do, self-conscious, and generally unable to approach the problem equanimously.

When she realized she had applied the Pythagorean theorem wrong, her frustration mounted, and moments later, at my next question, I was shocked with, “It’s all so complicated!”

Why did this happen? Why did I so horribly misjudge the difficulty of the exercise?

The other day I read this comment on an essay on teaching

I used to teach English as a second language. It was a mind trip.

I remember one of my students saying something like “I saw a brown big spider”. I responded “No, it should be ‘big brown spider'”. He asked why. Not only did I not know the rule involved, I had never even imagined that anyone would ever say it the other way until that moment.

Tutoring has been exposing my own brown big spiders – the little steps and bits of knowledge that I take for granted – for years. I’ve rarely stopped to notice it.

Just to follow each step in the Archimedes argument, you must make an enormous number of mathematical connections behind the scenes in your mind. Here’s a partial list:

• A “sphere” is a round three-dimensional object like, a ball
• Every point on the surface of a sphere is the same distance from the center
• The “surface” of the sphere means its outside edge, or skin
• A “point” is a little dot with no size at all. It simply marks a place.
• You can represent three-dimensional figures in two dimensions with certain types of drawing.
• The point of doing this drawing is to make things easier to visualize.
• A “hemisphere” is half a sphere – the top half in this case
• A “cylinder” is basically a tube with constant width.
• The center of the bottom of the hemisphere is the same point as the center of the sphere it came from.
• The height of the hemisphere is the same as the distance from the center to the edge horizontally.
• This means that the cylinder drawn is twice as wide as it is tall.
• The volume of a cone is one third the area of its base times its height.
• The volume of a cylinder is its base times its height
• The area of a circle is $\pi$ times the square of its radius

And so on. I only stopped writing so that I’d eventually finish the rest of this post. Each item I added to that list sparked off several new ones I hadn’t considered.

Try writing your own list and you’ll quickly be overwhelmed by the exponentially-proliferating leaves on your conceptual tree. We didn’t even get close to things like the Cavalieri’s principle.

The items on my brown big spider list are not supposed to be mathematical facts so much as cognitive patterns the reader is required to have. For example, mathematically a point is not, “a little dot with no size at all,” as I called it. It’s a primitive notion and has no definition. The list still calls a point a dot, though, because the mathematically-accurate description isn’t helpful to a student, and isn’t they way most people think of it even when they’ve already learned geometry well.

When I started writing the list, I found myself wanting to say, “A sphere is a set of all points equidistant…”, but that’s no good. It uses the significant brown big spiders of “set” and “equidistant”, as well as the general idea of giving mathematical definitions, something most high schoolers don’t yet understand well. Then I wanted to say, “A sphere is a shape that’s symmetric with respect to rotations about any axis…” but this has all the same problems.

Ultimately, I chose “a sphere is a ball.” It’s imprecise, but it’s the way you think about a sphere before you’ve packaged the concept away so tightly you don’t need to think about it any more. Anyone who tells you a sphere is the two-dimensional manifold $S^2$ is someone who has forgotten how much they actually know about spheres. They’ve forgotten it in the good way, of course – the way David was supposed to forget the notes to Rachmaninoff. Unfortunately, I experience a crippling side effect when I forget things this way. I forget that other people haven’t yet forgotten them.

This forgetting is the psychological phenomenon of “chunking“. The most famous example involves chess players. Give expert chess players a position from a game between grandmasters and they can easily memorize the positions of thirty pieces. Give them pieces strewn randomly about the board and they’ll remember just a few – no more, in fact, than your average Joe who knows little more about chess than what the real name of the horsey is.

A position from a real game has lots of meaning, if you’re an expert. If you’re an expert you extract order from the position automatically, without consciously processing every detail. The entire task must seem quite simple to a grandmaster. Similarly, the experienced mathematician sees all the important properties of the sphere and the cylinder and the cone without having to list them out one by one, and the process is so automatic they don’t even realize they’re doing it.

In “Simple” Isn’t “Easy”, I learned not to judge the difficulty of new ideas by how simple they are, but by how familiar to the student. Despite this, I have continued to make a similar mistake when dealing with ideas the students have already learned.

“Learned” isn’t “chunked”. My student understood the meaning of “hemisphere” and the formula for the volume of a cone, but she still needed conscious effort to recall and wield those bits of knowledge. Each sat in its own corner in her mind, accessible only by dint of concerted effort, and certainly not ready to flow into a flood of beautiful ideas.

I was trying to dictate a soliloquy for her to transcribe, but I was assuming that because she could see the letters on her keyboard, should could touch-type. It turned out that the effort to hunt-and-peck was so great, all the artistry of the speech was lost.

I want to watch out for my brown big spiders in the future. I want to be more patient when they are discovered and more studious in cataloging, remembering, and working with them. Most of all, I want to look back later, and remember my students forgetting them.

### Viete’s Formula and Spinning Pizza

September 17, 2010

Have you seen Viete’s formula?:

It’s a special case of a trig identity found by Euler:

If you plug in $\pi/2$ to the trig identity and use the half-angle formula for cosine over and over, you get Viete’s formula.

But why would you want to consecutively cut angles in half and multiply their cosines? Well, you might be eating pizza.

You have a slice of pizza that is too hot to hold, so you want to balance it on your fork and gnaw at it instead. There’s a precise spot on the underside of the slice where the fork should go.

This point is called the slice’s center of mass, and we’re going to find it. By symmetry, it must be on the line cutting the slice in half lengthwise, but we don’t yet know how far down. It depends on the shape of the slice, which we measure by $\theta$, the angle its edges make.

The center of mass of the slice of pizza is a green dot. It lies on the line cutting the slice in half vertically.

A bigger slice will have its center of mass closer to the tip. We would like to know $r(\theta)$, the distance from the tip to the center of mass as a function of $\theta$.

We want to know the distance to the center of mass.

There are two limiting cases – a very skinny slice and a whole pie. A very skinny slice is basically an isosceles triangle. Its center of mass is 2/3 the way from the tip to the edge1, so

$\lim_{\theta \to 0} r(\theta) = \frac{2}{3}R$.

Let’s choose the radius of the pizza as unit of length, so $R = 1$ from here on.

In the other limiting case, an entire pizza has its center of mass right at the tip (i.e. center), so

$r(2\pi) = 0$.

To investigate intermediate cases, we start with a slice of angle $\theta$ and imagine cutting it in half lengthwise, creating two skinny pieces of angle $\theta/2$. These have their own centers of mass at $r(\theta/2)$.

The center of mass of the big piece is on the line connecting the smaller pieces’ centers of mass.

A bit of trigonometry tells us

$r(\theta) = r(\theta/2)\cos(\theta/4)$

If we take this formula and divide all angles by $2$, we get a formula for $r(\theta/2)$. We substitute this for where $r(\theta/2)$ appeared in the original. We obtain

$r(\theta) = \left[r(\theta/4)\cos(\theta/8)\right]\cos(\theta/4)$

Repeat the process ad infinitum. Rearranging the order of the terms and substituting the limiting value of $r$ for small $\theta$, we get

$r(\theta) =\frac{2}{3} \cos(\theta/4)\cos(\theta/8)\cos(\theta/16)\ldots$

It involves one half of Euler’s trig identity. If we find $r(\theta)$ by a different method and get a different expression for it, we can set our two expressions for $r(\theta)$ equal to each other, and prove Euler’s identity. We’ll do this by invoking some physics ideas.

Suppose you’re spinning some pizza dough in the air. You know, like this:

If the pizza is spinning, each little bit of dough undergoes centripetal acceleration. Where there’s acceleration, there’s force. The pizza isn’t touching anything, so the force on any one piece of pizza must be coming from the rest of the pizza.

Let’s again examine a slice of size $\theta$, this time still attached to the spinning pizza. It has two forces of size $F$ acting on it; one force is exerted by the slice to its left and one by the slice to its right.

There are two forces on the slice - one from the pizza to the left and one from the pizza to the right. They're both drawn originating from the center of mass. The slice is accelerating towards its tip (red arrow).

The sum of these forces is the mass of the slice times the acceleration of its center of mass. That acceleration is $\omega^2 r(\theta)$. Hence, if we determine the forces we can deduce $r(\theta)$.

Some trigonometry shows that the net force is $2F\sin(\theta/2)$.

Equating this to mass times acceleration, we get

$2F\sin(\theta/2) = \frac{m \theta}{2\pi} \omega^2 r(\theta)$

We might as well let $m = \omega = 1$ and solve for $r$ to get

$r(\theta) = 2F\sin(\theta/2)\frac{2 \pi}{\theta}$.

We still need to determine $F$, but we can do that because we know $r(\theta) \to 2/3$ as $\theta \to 0$. After a little algebra, we get

$r(\theta) = \frac{4}{3} \frac{sin(\theta/2)}{\theta}$

This gives us the sought two expressions for $r$. We can now equate them and simplify to

$\frac{sin(\theta)}{\theta} = \cos(\theta/2)\cos(\theta/4)\cos(\theta/8)\ldots$

1) To see why an isosceles triangle’s center of mass is 2/3 up the altitude, first show it’s true for an equilateral triangle. Then explain why all isosceles triangles have their center of mass the same fraction of the way down the altitude.