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.

## 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.)

## More Snow for Boston, Says Calculus

In a recent Huffington Post article I calculated the probability that Boston will get even more snow than its current historical record. The function I ended up with was

$P(s)=e^{-0.079s}, \$

where is the additional inches of snow beyond the 45.5 inch record set on Saturday (February 15th). Let me explain how I got that function.

### First, Some Background in Probability

Let’s start with some basics. The total snowfall in any given month is a variable. But it’s not your “usual” kind of variable, it’s what we mathematicians call a “continuous random variable.” These type of variables have a function associated with them—called a “probability density function” or “PDF”—that tells us the probability that the random variable’s value will fall within a certain range (the “continuous” part means that the values of the random variable can be any real number). In our case the random variable of interest is the total snowfall in February in Boston, which I’ll denote by S, and it’s associated PDF, which I’ll denote by p(x). Then the probability that S is between, say, 45.5 inches and 45.5+h, where h is small (like, way less than 1), is approximately

$P(S=45.5) \approx p(45.5)h. \$

To approximate the probability that S is between, say, 45.5 and 46.5 we’d need to add up many terms like the one on the right-hand side. One such approximation is

$P(45.5 < S \leq 46.5) \approx p(45.5)(0.1)+p(45.6)(0.1)+\ldots+p(46.4)(0.1). \$

(This is the case when all h’s are equal to 0.1). If we want the exact answer we’d need to add up infinitely many terms, each corresponding to an h-value that is infinitesimally small. The way we do that in calculus is by integrating. So, in calculus speak,

$P(45.5

### Now Back to Boston’s Snowy Month

If you’ve made it this far one thing is clear: we can’t calculate anything without the PDF. That’s where the data linked above comes in. By downloading the total snowfall column in the data into a spreadsheet we can create the histogram below.

This histogram tells us how frequently (in the 95 years between 1920 and 2014) the Feburary snowfall total was between 0 and 5 inches, 5 and 10 inches, etc. (For example, the first bar says that the total snowfall was between 0 and 5 inches 26% of that time.) The black curve is the exponential function

$f(x)=0.511e^{-0.079x}, \$

where is the total snowfall (in inches). This curve is Excel’s best fit to the data. It’s not perfect, but it does a better job than other fits (like a linear function).

To make f(x) into a PDF we need to make sure that the probability that S is between zero and infinity is 1. (Roughly speaking, this expresses the fact that all probabilities must add to 1.) In calculus jargon, this means that we first need to calculate the integral of f(x) between zero and infinity and then divide f(x) by that value. And since

$\displaystyle\int_{0}^{\infty}0.511e^{-0.079x}\,dx=\frac{0.511}{0.079} \approx 6.416. \$

Our PDF—which I’ll call p(x)—is therefore

$p(x)=\frac{f(x)}{6.416}=0.079e^{-0.079x}. \$

(This is the PDF of an exponential distribution, a well known PDF from probability theory that has many applications to business, physics, and engineering—see here for more.)

We’ve found our PDF…woohoo! We can now calculate the probability that S—the total snowfall amount in February in Boston—will be between zero and some other number y. As before, that’s just the following integral:

$P(0 \leq S \leq y) = \displaystyle\int_{0}^{y}0.079e^{-0.079x}\,dx. \$

This is a pretty straightforward integral to calculate using a technique called u-substitution. The answer is

$1-e^{-0.079y}. \$

We’re almost done, I promise. The last step is to use the fact that 45.5 inches have already fallen. So, given that 45.5 inches of snow have already accumulated, what is the probability that s more inches will fall? Well, if we denote by y the total snowfall (i.e., y=s+45.5), then in math-speak we want to calculate

$P(S>y|S>45.5). \$

This is an example of a “conditional probability,” and by the laws of probability, this simplifies to

$1-P(S \leq y|S>45.5)=1-\frac{P(45.545.5)}, \$

and in terms of integrals becomes

$1-\frac{\displaystyle\int_{45.5}^{y}0.079e^{-0.079x}\,dx}{e^{-0.079(45.5)}}. \$

Finally, calculating and simplifying gives

$1-\frac{e^{-0.079(45.5)}-e^{-0.079y}}{e^{-0.079(45.5)}}=e^{-0.079(y-45.5)}=e^{-0.079s}. \$

This is the P(s) formula I gave at the start of the article. It’s amazing what the internet, math (oh, and Excel) can accomplish. Pretty neat huh?