## How does the speed of light relate to Maxwell’s equations?

You may have seen in your physics class that the speed of light can be calculated from the permittivity of free space, $\varepsilon_0$, and the permeability of free space, $\mu_0$, as follows:

$c = \dfrac{1}{\sqrt{\varepsilon_0 \mu_0}}$

where  $\varepsilon_0 = 8.85 \; C^2/N \cdot m^2$ and  $\mu_0 = 4 \pi \cdot 10^{-7} \; T \cdot m/A$ .

Sure enough, plug the numbers into your calculator and you do get the correct value for the speed of light. It’s a very famous result, and helped convince Maxwell that light is an electromagnetic wave. But the units just don’t seem to work out, do they? Well, actually they do. Let’s do the dimensional analysis. We want to show the following:

$\dfrac{m}{s} = \dfrac{1}{\sqrt{\dfrac{C^2}{N \cdot m^2} \times \dfrac{T \cdot m}{A}}} \; \; (1)$

From the equation for the magnetic force on a charged particle:

$F = qvB \; \rightarrow \; B = \dfrac{F}{qv}$

If we express this equation in terms of its units and simplify we have

$T = \dfrac{N}{C \cdot m/s} = \dfrac{N \cdot s}{C \cdot m}$

We also note that 1 A = 1 C/s, so we substitute for T and A in equation (1) above to get:

$\dfrac{m}{s} = \dfrac{1}{\sqrt{\dfrac{C^2}{N \cdot m^2} \times \dfrac{N \cdot s}{C \cdot m} \dfrac{m \cdot s}{C}}} = \dfrac{1}{\sqrt{\dfrac{s^2}{m^2}}} = \dfrac{m}{s}$

Physics

## finding the number of diagonals in a polygon

How many diagonals are there in an octagon?

Instead of trying to count them all, use the formula for the number of diagonals in any polygon:

$d= \dfrac{n(n-3)}{2}$

where n = the number of sides and d= the number of diagonals. For an octagon, n= 8, so

$d= \dfrac{8(8-3)}{2} = 20$

This is a good formula to memorize and its derivation is pretty simple, so let’s see it. We will use the same octagon diagram, but only draw the diagonals from one vertex:

You can see that from this vertex on the upper left, a diagonal can be drawn to every vertex except the two adjacent vertices and to itself. So the number of diagonals from any vertex in an n-gon is (n-3). Since there are n vertices, the total number of diagonals is n(n-3). But we’ve just counted every diagonal twice (once from each endpoint), so to find the number of different diagonals, we divide by two. This gives us the formula above.

Geometry

## finding internal or external angles of regular polygons

How do you find the measure of an internal angle or an external angle on a regular polygon? It’s quite easy, if you memorize two formulas. All you have to do is count the number of sides on the polygon and plug it into the formula.

For internal angles, use the formula

$\alpha = \dfrac{180(n-2)}{n}$

For external angles, the formula is even easier:

$\beta = \dfrac{360}{n}$

For example, here is a regular octagon, so n = 8. The internal angle is labeled α and the external angle is β.

$\alpha = \dfrac{180(8-2)}{8}=135 \textdegree$

$\beta = \dfrac{360}{8} = 45 \textdegree$

And here’s a trick to save you some time: You can see in the figure that the internal and external angles are always supplementary. (This is true in any polygon, even if it isn’t regular.) So if you need to find the internal angle, you can find the external angle instead with the simpler formula. Then the supplement to that angle is the internal angle!

Geometry

## Calculating permutations and combinations

When counting up the number of ways an event can occur, you use the formulas for permutations and combinations. You should be familiar with the nPr and nCr commands on your calculator, and this is the easiest way to evaluate these problems. But if your calculator doesn’t have these functions, there is a fairly simple way to set up these operations. This is the way we had to calculate permutations and combinations when calculators did not have these functions built in. Practice a couple of these examples and you’ll see that you can calculate permutations and combinations almost as quickly as your calculator can do it.

Calculating nPr

To calculate nPr, you will multiply together r consecutive numbers, starting with n and counting down. For example, 12P 3 is equal to 12*11*10 = 1320. We started with 12 (the value of n) and counted down to 10 so that we had 3 numbers (3 is the value of r). As another example, 7P5 = 7*6*5*4*3 = 2520.

Calculating nCr

To calculate nCr, create a fraction. The numerator is the same as above; that is, start with n and count down r consecutive numbers. The denominator is the smaller of r! and (n-r)!. For example,12P3 is

$\dfrac{12*11*10}{1*2*3}$

Before you calculate this fraction, simplify it. All of the terms in the denominator will always cancel out with terms in the numerator, leaving you with just numbers in the numerator to multiply together. For example,

$\dfrac{12*11*10}{1*2*3} = 2 * 11 * 10 = 220$

To calculate 7C5, note that 7C5 = 7C2 . Then,

$\dfrac{7*6}{1*2} = 7 * 3= 21$

## Calorimetry calculations II—measuring heats of reactions

In my previous post, I showed you how to work a calorimetry problem when you drop a hot block of metal into the calorimeter. In this post, we will look at the heat of reaction problem. The principles are the same: all of the heat released or absorbed by the reactants is absorbed or released by the water. In formula terms, this is written as ΔH = -q. From the M-CAT equation we examined in the previous post, we can write this as

$\Delta H_{rxn} = -m_{solution} c_{solution} \Delta T_{solution}$

We use this equation to find the molar heat of reaction as shown here:

Example: The temperature of a beaker containing 150 mL of water drops by 3.85°C when 8.00 g of ammonium nitrate are added. Determine the molar heat of solution for ammonium nitrate. Assume the specific heat capacity of the solution is 4.18 J/g·°C.

Answer: The heat of solution is defined as the ΔH for the following reaction:

$NH_4NO_3 \; (s) \rightarrow NH_4^+ \; (aq) + NO_3^- \; (aq)$

The mass of 150 mL of water is 150 g. Therefore the total mass of the solution is 158 g. So

$\Delta H_{rxn} = -(158)(4.18)(-3.85) = 2540 \; J$

The molar mass of ammonium nitrate is 80.0 g, so the molar heat of solution is

$\dfrac{2540 \; J}{8.00 \; g \; NH_4NO_3} \times \dfrac{80.0 \; g}{1 \; mol} = 25.4 \; kJ/mol$

Notes: We must include the mass of the ammonium nitrate in the M-CAT equation because it cools down along with the water. We also assumed the specific heat capacity of the solution was the same as the specific heat capacity of pure water. For dilute solutions, this is a reasonable assumption.

Chemistry

## Calorimetry calculations I—dropping hot blocks into water

Calorimetry refers to the measure of how much heat is transferred in a reaction or phase change. We measure this heat with a device called a calorimeter. Although serious chemists will use serious calorimeters costing hundreds or thousands of dollars, you can duplicate much of their work by pouring some water into a Styrofoam cup. Your answers won’t be as precise as the serious chemists’ answers, but for two or three sig figs, this method is good enough.

There are two kinds of calorimetry problems; we’ll look at one here and the other in the next post. The first involves dropping a heated block of metal into the cup of water. The block cools down and the water warms up until the block and water are at the same temperature (thermal equilibrium). In the second type of problem, a chemical reaction takes place in the water. If the reaction is exothermic, the water absorbs the heat and warms up. If the reaction is endothermic, heat is absorbed from the water and the water cools down.

Let’s examine the first problem: There are two guiding principles we observe. The first is that heat will flow from the hot block into the water until the block and the water are at the same temperature. The second principle is that (we assume) there is no loss of heat to the air. In other words, the heat lost by the block exactly equals the heat absorbed by the water.

Both the heat lost by the block and the heat the water absorbs are governed by a simple equation:

$q = mc \Delta T$

Here q is the amount of heat gained or lost, m is the mass of the substance, c is the specific heat capacity of the substance, and ΔT is the change in temperature. Many teachers call this the M-CAT equation because the Δ looks like an A if you squint really hard. If this helps you remember it, you can call it that too. The value of c, the specific heat capacity, is a constant for a given substance. It measures how much heat must be absorbed by 1 g of a material to raise its temperature by 1°C. You might be asked to memorize the specific heat capacity of water. It is 1.00 cal/g·°C or 4.18 J/g·°C. You probably will not need to memorize specific heat capacity values for any other substance.

With the equation, calorimetry problems of this type are simple algebra problems. We have $q_{water} = m_{water} c_{water} \Delta T_{water}$ and  $q_{block} = m_{block} c_{block} \Delta T_{block}$. Also, since the heat lost by the block equals the heat gained by the water, we can write $q_{water} = -q_{block}$. Therefore,

$m_{water} c_{water} \Delta T_{water} = -m_{block} c_{block} \Delta T_{block}$ .

This is our magic equation. There are two kinds of questions you will answer with this equation. In one, we tell you the initial and equilibrium temperatures and you calculate the specific heat capacity of the block substance. In the other, you are given the initial temperatures and the specific heat of the block and you calculate the equilibrium temperature. Here are examples that show both types:

Example 1: A 25 g block of aluminum is heated to 85°C and placed into a calorimeter with 80 g of water at 25°C. The final temperature of the calorimeter and block is 28.8°C. Calculate the specific heat of aluminum.

Solution 1: We calculate the ΔT of the water (3.8°C) and the aluminum block (-56.2°C) and plug everything into the equation above.

(80)(4.18)(3.8) = -(25)(c)(-56.2). Solving gives c = 0.90 J/g·°C.

Example 2: A 25 g block of aluminum is heated to 85°C and placed into a calorimeter with 80 g of water at 25°C. The specific heat of aluminum is 0.90 J/g·°C. What is the equilibrium temperature of the calorimeter?

Solution 2: This problem (really just a rewording of example 1) is only a little more complicated to solve than example 1. First note that ΔT = Tf – Ti. We make this substitution and our magic equation becomes

(80)(4.18)(Tf – 25) = -(25)(0.90)(Tf – 85)

Distribute and simplify to get 334.4 Tf – 8360 = 1912.5 – 22.5 Tf. Solving gives Tf = 28.8°C.

Note: This is one of the rare times you do not need to convert the temperature from °C to kelvins. This is because the temperature term in the equation is a change in temperature and not an absolute temperature. Changing the temperature by 1°C is the same as changing it by 1 K.

Chemistry

## An algorithm for calculating the square root of a number

How do you find the square root of a number? You use a calculator, of course! But what if you can’t find your calculator? Did you know there’s an algorithm that will allow you to derive a square root of a number? My dad taught it to me a long time ago before calculators were around. It would surprise me if anyone you know under the age of 40 has ever seen it. It’s a slow, painstaking process, so only use it if you have a lot of time to waste. Frankly, I’d recommend waiting until you get a new calculator, but in case you’re interested, here it is. It’s easiest to explain with an example. Let’s find the square root of 300. Because we want to calculate some digits after the decimal point, we will write it as 300.0000

$\sqrt{300.0000}$

The first step is to separate the digits into groups of two. Starting from the decimal point, mark off each pair of digits. If there are an odd number of digits to the left of the decimal point, the leftmost digit will be a single digit and not a pair. Then start from the decimal point again and count off the digits to the right by twos. In our example, the “3” in 300 is a single digit and all the others are pairs.

Now we are ready to calculate. Our first digit is a three. We find the largest integer whose square is less than this number. Since 12 = 1 < 3 and 22 = 4 >3, our number is 1. We place a 1 above the 3, just like we are doing a long division problem.

Next, copy this digit on the line below the 300.

This next step looks a lot like a long division problem. Multiply the (red) 1 by the (tan) 1 and put the product under the 3. Then subtract, and bring down the next two digits. Our example will look like this:

Now it gets a little strange. Take the (red) number above the 300 and double it. Write this number on the next line down on the left and add an underscore. 1 doubled is 2, so our example now looks like this:

The underscore is a place holder for an unknown digit. We need to find a single digit that we will place above the line (over the “00”) and in the placeholder. We want the product of these two numbers to be as large as possible without being larger than the current remainder. Let’s say we decide the digit is 6. Then 6 ·26 = 156. If the digit is 7, then 7 ·27 = 189. If the digit is 8, then 8 ·28 = 224. This is larger than 200, so our digit is 7. We place it above the radical and in the placeholder as shown below. Do the multiplication and subtraction as before. Write the remainder and bring down the next two digits. Our example now looks like this:

Now we repeat this process over and over for each new digit. Double the number above the radical and add a placeholder. 17 ·2 = 34, so our problem now looks like this:

Again, we need a digit above the line and in the placeholder so that the product is less than the remainder. 3·343 = 1029 < 1100.   4·344 = 1376 >1100. So the digit we want is 3. Do the multiplication and subtraction and bring down the next two digits. Our example looks like this:

Let’s do it one more time. Double the number over the radical and add a placeholder:

The next digit we need is a 2. (2·3462 = 6924 < 7100; while 3·3463 = 10389 > 7100). Multiply and subtract and bring down the next two digits.

17.32 is a pretty good approximation of the square root of 300. You can repeat this process as often as you want to get even more digits in your solution. The next number we would write on the left would be 3464_. You can see that the number on the left gets bigger with each step, so the process gets pretty unwieldy. If you need more than three or four digits in your square root, make sure you have a lot of paper, or go find that calculator!

## Graphing sine and cosine functions like a pro

When graphing a sine or cosine curve, the first thing you must do is determine the amplitude, period, phase shift and vertical shift. See my previous post (Graphing Sine and Cosine Functions – Intro) if you need help with this analysis. In this post, we will graph the function

$\displaystyle f(x) = -3 \sin (2x + \frac{\pi}{2}) -1$

We quickly determine the four values we need:

Amplitude = |-3| = 3

Period = $2 \pi /2 = \pi$

Phase shift = $-(\pi /2)/2 = - \pi/4$ (that is, $\pi /4$ units to the left)

Vertical shift = -1

This is all the information we need in order to complete the graph. Just follow this procedure step-by-step.

 1. Put values on the coordinate axes. On the y-axis, you typically make each square equal to one unit, but you can change this if you want. To determine the scale on the x-axis, take the period and divide by 4. This will be the scale on the x-axis. In our example, the period is $\pi$, so each square will be $\pi /4$. The vertical axis will be one unit per square. What do you do if your teacher gives you a grid with the numbers already in place? You should get a blank piece of graph paper and do your own grid! 2. Use the vertical shift to draw a dashed line across the figure. This is the location of the midline of your graph. In our example, the vertical shift is -1, so we draw a dashed line at y= -1. 3. Use the amplitude to draw two more dashed lines—one above the midline and one below. These represent the maximum and minimum values of your function. In our example, the amplitude is 3. Three units above -1 is 2—that’s our maximum dashed line. Three units below -1 is -4—that’s where our minimum is located. 4. Plot the starting point of your graph, using the vertical shift and phase shift as a guide. Our function is a sine curve, which starts at the midline. The phase shift is $\pi /4$ to the left, so our initial point is $\pi /4$ units left of the y-axis. If our function had been a cosine curve, our initial point would be plotted on the maximum line instead of the midline (or on the minimum line if A is negative). It’s hard to see, but note that I’ve placed a green dot at the “start” point; the coordinates are $( - \pi /4, -1).$ 5. Moving one square to the right at a time (because each square is one quarter of a period), plot points at the maximum, midline, minimum and midline. This is one period of your function. If you want to graph more than one period, continue the process. In our example, we’ve plotted points for two complete periods. Note that because A is a negative number (-3), our first point after the starting point is at the minimum instead of the maximum. Look closely, and you will see that I’ve placed a green dot every square to the right of our first point. 6. Connect the dots with a nice smooth curve. You’ve graphed the sine curve like a pro!

## Graphing sine and cosine functions– an Intro

One of the most complicated skills you need to learn in your trig class is how to graph sine and cosine functions. This scares a lot of students, but you can tame this process if you make one simple observation: Every sine and cosine curve has exactly the same shape! No matter the amplitude or period or phase shift, the curve looks just like this:

You only need to place the graph in its proper position on the coordinate axes. This is (mostly) easy to accomplish if you can remember only two things about the sine and cosine parent curves:

1)     The sine curve y = sin x “starts” at the origin and goes up to its maximum, while the cosine curve y = cos x “starts” at its maximum.

2)     For either curve, you can break one period into four equal intervals. At each interval, the curve moves from its midline to the maximum to the midline to the minimum to the midline to the maximum to… over and over again. So all you need to do is find the starting point, and plot the points on the curve at each ¼-period interval.

We will always write our functions in standard form:

$f(x) = A \; sin(Bx+C) + D \; or \; f(x)= A \; cos(Bx+C) + D$

(Note that some textbooks prefer to write the formula in a slightly different form:

$f(x) = A \; sin(B(x+C)) + D \; or \; f(x)= A \; cos(B(x+C)) + D$

We will discuss how that affects your work below.)

Each of the constants A, B, C & D affects the position of the curve and you need to analyze this before you graph the curve. Let’s look at each of them in turn:

A: The absolute value of this number tells you the amplitude of your curve.

B: The period of your curve is determined by dividing $2\pi$ by B.

C: The phase shift is found by dividing -C by B. A positive value means the phase shift is to the right. A negative value means the phase shift is to the left. (If your class uses the version of the equation above with the B factored out, then the phase shift is equal to C.)

D: The vertical shift is equal to D.

Here’s an example to show how you would calculate all these values.

$\displaystyle f(x) = -3 \sin (2x + \frac{\pi}{2}) -1$

Here, A = -3; B = 2; C = $\pi/2$; and D = -1. Therefore,

Amplitude = |-3| = 3

Period = $2 \pi /2 = \pi$

Phase shift = $- (\pi /2)/2 = - \pi /4$ (that is, $\pi /4$ units to the left)

Vertical shift = -1

When you need to graph a sine or cosine curve, always determine these four values first. Then you are ready to graph the function. We’ll do that in our next post.

## Using tree diagrams to find conditional probabilities

Those problems that ask you to find the probability of a series of events “without replacement” can be scary because the probabilities of each event keep changing. (These are known as conditional probability problems.) If the number of possible outcomes isn’t too large, you can tame these problems by using a tree diagram to simplify your calculations.

1. For the first event, draw a tree branch for each possible outcome.
2. At the end of each branch, draw a tree branch for each possible outcome of the second event.
3. Continue until you have a column for every event.
4. For every branch on the tree, write down the probability of that event occurring at that location.
5. Then multiply all the branches from first event to last event to find the probability of any one outcome.
6. Add various events together to get the probability of any compound outcome.

Here’s a simple example that shows how this process works. Let’s say you have a candy dish with 10 red candies, 15 green candies and 20 blue candies. You want to know the probability that you draw at least two red candies or at least two blue candies. There are a lot of different possibilities here, but a tree diagram simplifies everything greatly. Start by drawing a tree with every possible outcome (R, G and B in this example). Then from each outcome, draw another tree representing each outcome for the second draw. Repeat for the third draw. Your tree will look like this:

(You can see that this process will get pretty unwieldy if there are too many outcomes or too many events.)

Next, label each branch with the probability for that outcome. Note that the probabilities change depending on which outcomes have already occurred. For our example, the tree would now look like this:

Finally, for each of the branch ends at the right, multiply together all the probabilities leading to that endpoint. For example, the very top branch, which represents RRR, you would multiply 10/45*9/44*8/43 to find the probability of getting a red candy on all three draws. The final table looks like this (to make the table easier to read, we have calculated only those branches that represent at least two reds or at least two blues):

The probability of our desired event is then the sum of all of listed probabilities: 0.5352.

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