# Lessons from teaching gifted elementary students (Part 7b)

Every so often, I’ll informally teach a class of gifted elementary-school students. I greatly enjoy interacting with them, and I especially enjoy the questions they pose. Often these children pose questions that no one else will think about, and answering these questions requires a surprising depth of mathematical knowledge.

Here’s a question I once received after a really big hailstorm:

How big would a 1000-pound hailstone be?

The guesses from the students ranged from the size of a small car to the size of a large pick-up truck. Here’s how I gave a reasonably accurate answer using only mental arithmetic. This kind of describes the way that I try to size up things when only an approximation (and not an exact answer) is necessary.

First, I had some real-world experience that quickly told me the answer was going to be deceptively small. When I was young — maybe 10 or 12 years old — I was getting ready for a picnic, and I was assigned cut a block of ice — maybe a cubic foot of ice, if memory serves — into smaller chunks. (The party organizer bought a block of ice instead of a bag of ice to economize.) I remembered how incredibly heavy that block of ice was even though it wasn’t much larger than a basketball… several of us kids had a lot of trouble lifting the block of ice as we prepared to chop it into pieces. So, for the sake of argument, if that cubic foot of ice weighed about 100 pounds, then 8 cubic feet would weigh 800 pounds. So, based on that chance encounter with a block of ice when I was a kid, my guess would have been that the hailstone would measure 2 feet across.

Back to the problem at hand.

First, I converted to metric. I knew that there are about 2.2 pounds in a kilogram, and so I knew that the block would weigh something like 400 or 450 kilograms. I knew that I would be making plenty of crude approximations, so I just went with 400 kilograms and didn’t worry too much about immediately calculating $1000/2.2$.

Next, I knew that metric units were originally defined so that a cubic centimeter of water weighs a gram, so that a 10 cm-by-10 cm-by 10 cm cube of water weighs one kilogram. Ice (hail) is slightly less dense than water (after all, ice floats in water), but for crude approximation purposes, I ignored this.

So, if a cube of ice with a side length of 1 decimeter (10 cm) weighs 1 kilogram, then a cube of ice with a side length of $\sqrt[3]{400}$ decimeters would weigh about 400 kilograms.

How big is $\sqrt[3]{400}$? Well, I have memorized that $7^3 = 343$ and $8^3 = 512$, so it’s between 7 and 8 someplace… say 7.5. So the answer would be a cube of side length 7.5 decimeters. Also, I have memorized that 1 decimeter (10 cm) is approximately 4 inches, so the cube would have side length $7.5 \times 4 = 30$ inches.

Finally, hailstones are more spherical in shape than cubic, and a sphere of diameter $d$ has less volume than a cube of side length $d$. So the answer should be a bit larger than 30 inches, so I just rounded up to a nice even number: 36 inches (one yard).

This calculation took me about a minute to do in my head and another half-minute to re-do to make sure I didn’t botch the arithmetic. So I held my hands about a yard apart (perhaps the crudest part of this calculation), pretending to hold a ball of diameter a yard across, and announced, “The hailstone would be about this big.”

Of course, a more thoughtful analysis produces the actual answer. The density of ice at the freezing point is 0.9167 grams per cubic centimeter, and 1 pound converts to 0.453592 kilograms. So:

$\displaystyle \frac{4\pi}{3} r^3 \times 0.9167 \frac{\hbox{grams}}{\hbox{cm}^3} = 1000 \hbox{pounds} \times \displaystyle 0.453592 \frac{\hbox{kilograms}}{\hbox{pounds}} \times 1000 \frac{\hbox{grams}}{\hbox{kilograms}}$

$r^3 \approx 118127 \hbox{cm}^3$

$r \approx 49.1 \hbox{cm}$

$r \approx 49.1 \hbox{cm} \times \frac{1 \hbox{inch}}{2.54 \hbox{cm}}$

$r \approx 19.3 \hbox{inches}$

Therefore, the sphere would have a diameter of twice that, or 38.6 inches.

# Lessons from teaching gifted elementary school students (Part 7a)

Every so often, I’ll informally teach a class of gifted elementary-school students. I greatly enjoy interacting with them, and I especially enjoy the questions they pose. Often these children pose questions that no one else will think about, and answering these questions requires a surprising depth of mathematical knowledge.

Here’s a question I once received after a really big hailstorm:

How big would a 1000-pound hailstone be?

My head hurts thinking about hail that large. After about a minute of thinking, without using a calculator or even a pencil, I gave my answer: about a yard across.

I’ll reveal how I got this answer — which turns out to be a lot close than I had any right to expect — in tomorrow’s post. In the meantime, I’ll leave a thought bubble if you’d like to think about it on your own without using a calculator.

# Sphere Packing Solved in Higher Dimensions

I enjoyed reading this bit of mathematical news: https://www.quantamagazine.org/20160330-sphere-packing-solved-in-higher-dimensions/

The opening paragraphs:

In a pair of papers posted online this month, a Ukrainian mathematician has solved two high-dimensional versions of the centuries-old “sphere packing” problem. In dimensions eight and 24 (the latter dimension in collaboration with other researchers), she has proved that two highly symmetrical arrangements pack spheres together in the densest possible way.

Mathematicians have been studying sphere packings since at least 1611, when Johannes Kepler conjectured that the densest way to pack together equal-sized spheres in space is the familiar pyramidal piling of oranges seen in grocery stores. Despite the problem’s seeming simplicity, it was not settled until 1998, when Thomas Hales, now of the University of Pittsburgh, finally proved Kepler’s conjecture in 250 pages of mathematical arguments combined with mammoth computer calculations.

# Engaging students: Finding the volume and surface area of spheres

In my capstone class for future secondary math teachers, I ask my students to come up with ideas for engaging their students with different topics in the secondary mathematics curriculum. In other words, the point of the assignment was not to devise a full-blown lesson plan on this topic. Instead, I asked my students to think about three different ways of getting their students interested in the topic in the first place.

I plan to share some of the best of these ideas on this blog (after asking my students’ permission, of course).

This student submission comes from my former student Avery Fortenberry. His topic, from Algebra: finding the volume and surface area of spheres.

How does this topic extend what your students should have learned in their previous courses?

The topic of volume and surface area of spheres is building upon the students’ knowledge of area and circumference of a circle.  A sphere is similar to a circle in that a circle is a closed shape with all points equidistant from the centerpoint (the distance is the radius) and a sphere is a closed object with all points at an equal distance from its centerpoint (the distance is also r).  Students will be familiar with the area of a circle formula, which is A=πr2 and will be able to easily use and understand the formula for volume of a sphere V=(4/3)πr3.  The same is true for circumference in relation with surface area.

What interesting things can you say about the people who contributed to the discovery and/or the development of this topic?

Archimedes was the first mathematician to discover the most important ratio in all of mathematics, π.  He did this by finding the area of a circle using shapes that were incrementally closer and closer to the same size as that circle.  In other words, he would start with a circle and enclose it within a square, then a pentagon, then a hexagon, and so on until he came extremely close to the same shape.  He used this same method to find the volume of a sphere by enclosing it within a cylinder of a known volume and cutting out piece by piece and measuring until he found the parabolic segment is 4/3 that of an inscribed triangle.

What are the contributions of various cultures to this topic?

This topic had many cultures contribute to the understanding of it.  These contributions came from Greek, Chinese, and Arabic mathematicians.  The Greek contribution came mainly from Archimedes, which I discussed in D1.  The Arithmetic Art in Nine Chapters is a Chinese book written in the 1st century that gave a formula that was close but not exact to finding the volume of a sphere.  The author of the book calculated pi as being equal to 25/8 or even as just 3 at times.  Ancient Arabic mathematicians submitted very similar ideas to the Chinese in terms of the volume of a sphere.  While it is known the Chinese derived some ideas from the Greeks, it is still unclear today how the ideas were spread to the Arabic mathematicians.

Source: http://www.muslimheritage.com/article/volume-sphere-arabic-mathematics-historical-and-analytical-survey#sec2.2

# Surface area

Source: http://www.xkcd.com/1389/

# Area of a Triangle and Volume of Common Shapes: Index

I’m using the Twelve Days of Christmas (and perhaps a few extra days besides) to do something that I should have done a long time ago: collect past series of posts into a single, easy-to-reference post. The following posts formed my series on different ways of finding the area of a triangle as well as finding the volumes of common shapes.

Part 1: Deriving the formula $A = \displaystyle \frac{1}{2} bh$.

Part 2: Cavalieri’s principle and finding areas using calculus.

Part 3: Cavalieri’s principle and finding the volume of a pyramid and then the volume of a sphere.

Part 4: Finding the area of a triangle using the Law of Sines.

Part 5: Finding the area of a triangle using the Law of Cosines.

Part 6: Finding the area of a triangle using the triangle’s incenter.

Part 7: Finding the area of a triangle using a determinant and the coordinates of the vertices.

Part 8: Finding the area of a triangle using Pick’s theorem.

# Lessons from teaching gifted elementary school students (Part 3b)

Every so often, I’ll informally teach a class of gifted elementary-school students. I greatly enjoy interacting with them, and I especially enjoy the questions they pose. Often these children pose questions that no one else will think about, and answering these questions requires a surprisingly depth of mathematical knowledge.

Here’s a question I once received:

Suppose

$A \times A = B$

$B \times B \times B = C$

$C \times C \times C \times C= D$

If the pattern goes on, and if $A = 2$, what is $Z$?

In yesterday’s post, we found that the answer was

$Z =2^{26!} = 10^{26! \log_{10} 2} \approx 10^{1.214 \times 10^{26}}$,

a number with approximately $1.214 \times 10^{26}$ digits.

How can we express this number in scientific notation? We need to actually compute the integer and decimal parts of $26! \log_{10} 2$, and most calculators are not capable of making this computation.

Fortunately, Mathematica is able to do this. We find that

$Z \approx 10^{121,402,826,794,262,735,225,162,069.4418253767}$

$\approx 10^{0.4418253767} \times 10^{121,402,826,794,262,735,225,162,069}$

$\approx 2.765829324 \times 10^{121,402,826,794,262,735,225,162,069}$

Here’s the Mathematica syntax to justify this calculation. In Mathematica, $\hbox{Log}$ means natural logarithm:

Again, just how big is this number? As discussed yesterday, it would take about 12.14 quadrillion sheets of paper to print out all of the digits of this number, assuming that $Z$ was printed in a microscopic font that uses 100,000 characters per line and 100,000 lines per page. Since 250 sheets of paper is about an inch thick, the volume of the 12.14 quadrillion sheets of paper would be

$1.214 \times 10^{16} \times 8.5 \times 11 \times \displaystyle \frac{1}{250} \hbox{in}^3 \approx 1.129 \times 10^{17} \hbox{in}^3$

By comparison, assuming that the Earth is a sphere with radius 4000 miles, the surface area of the world is

$4 \pi (4000 \times 5280 \times 12) \hbox{in}^2 \approx 8.072 \times 10^{17} \hbox{in}^2$.

Dividing, all of this paper would cover the entire world with a layer of paper about $0.14$ inches thick, or about 35 sheets deep. In other words, the whole planet would look something like the top of my desk.

What if we didn’t want to print out the answer but just store the answer in a computer’s memory? When written in binary, the number $2^{26!}$ requires…

$26!$ bits of memory, or…

about $4.03 \times 10^{26}$ bits of memory, or…

about $latex 5.04 \times 10^{25} bytes of memory, or … about $5.04 \times 10^{13}$ terabytes of memory, or… about 50.4 trillion terabytes of memory. Suppose that this information is stored on 3-terabyte external hard drives, so that about $50.4/3 = 16.8$ trillion of them are required. The factory specs say that each hard drive measures $129 \hbox{mm} \times 42 \hbox{mm} \times 167 \hbox{mm}$. So the total volume of the hard drives would be $1.52 \times 10^{19} \hbox{mm}^3$, or $15.2 \hbox{km}^3$. By way of comparison, the most voluminous building in the world, the Boeing Everett Factory (used for making airplanes), has a volume of only $0.0133 \hbox{km}^3$. So it would take about 1136 of these buildings to hold all of the necessary hard drives. The cost of all of these hard drives, at$100 each, would be about $1.680 quadrillion. So it’d be considerably cheaper to print this out on paper, which would be about one-seventh the price at$242 trillion.

Of course, a lot of this storage space would be quite repetitive since $2^{26!}$, in binary, would be a one followed by $26!$ zeroes.

# Sphere

Source: http://www.xkcd.com/1248/

# Engaging students: Finding the volume and surface area of spheres

In my capstone class for future secondary math teachers, I ask my students to come up with ideas for engaging their students with different topics in the secondary mathematics curriculum. In other words, the point of the assignment was not to devise a full-blown lesson plan on this topic. Instead, I asked my students to think about three different ways of getting their students interested in the topic in the first place.

I plan to share some of the best of these ideas on this blog (after asking my students’ permission, of course).

This student submission again comes from my former student Allison Myers. Her topic, from Geometry: finding the volume and surface area of spheres..

How could you as a teacher create an activity or project that involves your topic?

Show students pictures of the Personal Satellite Assistant (PSA). Tell students they are going to investigate how the surface area and volume of a sphere change as its radius changes.

Explain that they will also determine how big the PSA is in real life.

Remind students that NASA engineers have created a 30.5-centimeter

(12-inch) diameter model of the PSA, but they want to shrink it to 20 centimeters (8 inches) in diameter.

Use a 30.5-centimeter (12-inch) diameter globe and let students know the globe is roughly the size of the current PSA model.

Ask students how the PSA might look different if its surface area were reduced by half.

Ask how the function of the PSA might be different if its volume were reduced by half.

Ask students what information they need to calculate its surface area and volume.

If they appear confused, draw three circles of different sizes and ask students how to calculate the area of each of the circles.

The only information they need is the radius of the sphere. Review the properties of a sphere.

Ask students what formulas are necessary to calculate the surface area and volume of the sphere. Write these formulas on the board:

Show students a baseball, softball, volleyball, and basketball. Ask them if they think the surface area and volume of a sphere change at equal rates as the spheres increase from the size of a baseball to the size of a basketball.

Ask students how they will verify their hypotheses.

Curriculum

How can this topic be used in your students’ future courses in mathematics or science?

In calculus students will learn that you can revolve a curve about the x or y-axis to generate a solid. For example, a semicircle [f(x) = √(r2-x2)] can be revolved about the x-axis to obtain a sphere with radius r. From this, the different formulas for calculating the volume of a sphere can be derived.

In calculus, students will also learn how to find the surface area of a sphere by integrating about either the x or y axis.

At some point, students may also extend their knowledge of spheres into higher dimensions (hyperspheres), where they will learn how volume changes according the dimensions they are working in.

What interesting (i.e., uncontrived) word problems using this topic can your students do now?

For Volume of a Sphere:

Recent Hubble Space Telescope studies of Pluto have confirmed that its atmosphere is undergoing considerable change, despite its frigid temperatures. The images, created at the very limits of Hubble’s resolving power, show enigmatic light and dark regions that are probably organic compounds (dark areas) and methane or water-ice deposits (light areas). Since these photos are all that we are likely to get until NASA’s New Horizons spacecraft arrives in 2015, let’s see what we can learn from the image!

Problem 1

– Using a millimeter ruler, what is the scale of the Hubble image in kilometers/millimeter?
Problem 2

– What is the largest feature you can see on any of the three images, in kilometers, and how large is this compared to a familiar earth feature or landmark such as a state in the United States?
Problem 3

– The satellite of Pluto, called Charon, has been used to determine the total mass of Pluto. The mass determined was about 1.3 x 1022 kilograms. From clues in the image, calculate the volume of Pluto and determine the average density of Pluto. How does it compare to solid-rock (3000 kg/m3), water-ice (917 kg/m3)?
Inquiry:

Can you create a model of Pluto that matches its average density and predicts what percentage of rock and ice may be present?
Resource: http://spacemath.gsfc.nasa.gov/weekly/6Page143.pdf

# Volume of pyramids, cones, and spheres (Part 3)

I’m in the middle of a series of posts concerning the area of a triangle. Today, however, I want to take a one-post detour using yesterday’s post as a springboard. In yesterday’s post, we discussed a two-dimensional version of Cavalieri’s principle. From Wikipedia:

Suppose two regions in a plane are included between two parallel lines in that plane. If every line parallel to these two lines intersects both regions in line segments of equal length, then the two regions have equal areas.

In other words, if I have any kind of shape that has cross-sections that match those of the triangles above, then the shape has the same area as the triangles. Geometrically, we can think of each triangle a bunch of line segments joined together. So while the positioning of the line segments affects the shape of the region, the positioning does not affect the area of the region.

There is also a three-dimensional statement of Cavalieri’s principle, and this three-dimensional version is much more important than the above two-dimensional version. From MathWorld:

If, in two solids of equal altitude, the sections made by planes parallel to and at the same distance from their respective bases are always equal, then the volumes of the two solids are equal.

Pedagogically, I would recommend introducing Cavalieri’s principle with two-dimensional figures like those from yesterday’s post since cross-sections in triangles are much easier for students to visualize than cross-sections in three-dimensional regions.

This three-dimensional version of Cavalieri’s principle is needed to prove — without calculus — the volume formulas commonly taught in geometry class. Based on my interactions with students, they are commonly taught without proof, as my college students can use these formulas but have no recollection of ever seeing any kind of justification for why they are true. When I teach calculus, I show my students that the volume of a sphere can be found by integration using the volume of a solid of revolution:

$V = \displaystyle \int_{-R}^R \pi \left[ \sqrt{R^2 - x^2} \right]^2 \, dx = \frac{4}{3} \pi R^3$

Without fail, my students (1) already know this formula from Geometry but (2) do not recall ever being taught why this formula is correct. Curious students also wonder (3) how the volume of a sphere (or a pyramid or a cone) can be obtained only using geometric concepts and without using calculus.

For the sake of brevity, I only give the logical flow for how these volumes can be derived for students without using calculus. I’ll refer to this excellent site for more details about each step.

• Using a simple foldable manipulative (see also this site), students can see that $V = \displaystyle \frac{1}{3} Bh$ for a certain pyramid — called a yangma — with a square base and a height that is equal to the base length.
• Enlarging the yangma will not change the ratio of the volume of the pyramid to the volume of the prism.
• Cavalieri’s principle then shows that $V = \displaystyle \frac{1}{3} Bh$ for any square pyramid.
• Cavalieri’s principle then shows that $V = \displaystyle \frac{1}{3} Bh$ for any pyramid with a non-square base or even a cone with a circular base.
• Finally, a clever use of Cavalieri’s principle — comparing a sphere to a cylinder with a cone-shaped region removed — can be used to show that the volume of a sphere is $V = \displaystyle \frac{4}{3} \pi R^3$.

I note in closing that there are other ways for students to discover these formulas, like filling an empty pyramid with rice, pouring into an empty prism of equal base and height, and repeating until the prism is filled.