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Everyday Calculus Reader Question

Reader S. Cho wrote in with the following question:

Hello! I have read your book, Everyday Calculus. It is very interesting and gives me a knowledge for math. I realized that math is all around us. But I have two questions in the book. First, I met the formula “r(V)=k sqrt(P_0 l) over V” in chapter 1. However I couldn’t find this formula in any other books. Could you give me the source of it or a reference book? Second, does P_0 mean the electric power? I’m waiting for your reply. Sincerely yours.

 

The formula r(V)=k\frac{\sqrt{P_0l}}{V} comes from the combination of the three formulas:

P_0=IV,         R=\rho \frac{l}{\pi r^2},          V=IR.

The first formula relates the voltage (V) and current (I) produced by a power source (P_0). The second is Pouillet’s Law, which relates a power line’s resistance to the flow of electric current R to the power line’s length l and radius r (I’ve assumed the power line is a long cylinder). The constant \rho is called the electrical resistivity. The last formula is Ohm’s Law.

Now, if we solve Ohm’s Law for I and plug that into the first equation we get

P_0=\frac{V^2}{R}.

Then, using Pouillet’s Law we get

P_0=\frac{V^2(\pi r^2)}{\rho l}.

Solving this for r^2 yields

r^2=\frac{\rho}{\pi}\frac{P_0l}{V^2}.

Taking the square root of both sides and letting k=\sqrt{\rho/\pi} gives

r=k\frac{\sqrt{P_0l}}{V}.

In the first chapter of Everyday Calculus I explain how this equation spelled doom for Thomas Edison. His early power grid operated at a fixed voltage of 110 Volts, which meant that as the length of the power line (l) increased, the power line’s radius (r) got bigger too. To avoid large power lines hanging over pedestrians (clearly a danger for anyone walking underneath them) Edison therefore had to build his power plants very close to his customers. This represented a huge cost, and limited the early usage of electricity. But the same equation also tells us that we can avoid this issue by using much larger voltages, which is what’s done today. These high voltages coming out of the power plants–in some cases nearly 1 million Volts–are reduced to the familiar 120 Volts we see in homes using transformers, which are the result of other neat physical principles (and math) at work.

 

 

How To Talk To Your Kids About Math (And Why You Need To)

When I started elementary school my mother, like all mothers, began asking me “what did you learn in school today?” I was always eager to share, and did. I talked about the things I was learning in all of my classes. I even remember some of those conversations–like the time I told her about cumulus clouds. After that I remember mom pointing them out to me regularly, and we’d bond over the ensuing conversation…yes, about clouds. Then middle school came. The concepts got more serious–governments and their roles, Greek tragedies, etc.–but she still asked her question, I still answered, and we still bonded over those conversations. It seemed that no matter what the subject was, I could always enjoy a nice discussion about it with my mother. That made me happy; it also made me want to keep learning. But on the horizon was one subject that would eventually drive a wedge between us: math.

Continue reading the article on the Huffington Post here.

Why You Need To Eat Pie This Saturday

Give me 3.14 minutes to explain.

This Saturday is March 14th, 2015. Every March 14th the entire world comes together to celebrate Pi Day (yeah right, we’re lucky to get, like, 5 people). The “pi” referenced here is that magic number that you may remember from math class. The picture below should jog your memory…

Read the rest of the article on the Huffington Post here.

How Much Gold Is In The Oscar Statuette?

In a recent Huffington Post article (see here) I asked how much gold the Oscar statuettes actually contain. Considering that these little men are like newborn babies (based on their official dimensions and weight), we can answer the question by calculating the statue’s surface area. One good approximation to a newborn’s surface area is the equation

BSA=0.024265 \times W^{0.5378}H^{0.3964}, \

where BSA is the body surface area in square meters, W is the weight in kilograms, and H is the height in centimeters. Using the official weight and height of 8.5 pounds and 13.5 inches, respectively, gives a body surface area of about 0.2036 square meters. To convert this to square feet, we use the fact that 1 meter is equivalent to about 3.28084 feet. So, 1 square meter is about

1 \, \text{m}^2 \, = (3.28084)^2 \, \text{ft}^2 \, = 10.7639 \, \text{ft}^2. \

Using this, we can then convert the 0.2036 square meters into square feet:

0.2036 \, \text{m}^2 \, \times \frac{10.7639 \, \text{ft}^2}{1 \, \text{m}^2} = 2.1912 \, \text{ft}^2. \

The square root of this is about 1.4803 feet, or about 17.76 inches. That means those little Oscar statues have a gold coating about equal to the area of a square of side 17.76 inches. (As I describe in the Huffington Post article, this is roughly the area enclosed by the square made up of your two forearms.)