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Contents

- 1 Notes for Lesson 34: Population, Scrolls, and Sound
- 1.1 Exponential and Logarithmic Models
- 1.2 Exponential and Logarithmic Models
- 1.3 Population Growth
- 1.4 Population Growth
- 1.5 First Prediction
- 1.6 First Prediction
- 1.7 Second Prediction
- 1.8 Second Prediction
- 1.9 7.1 Billion People
- 1.10 7.1 Billion People
- 1.11 Population in 2100
- 1.12 Population in 2100
- 1.13 Dead Sea Scrolls
- 1.14 Dead Sea Scrolls
- 1.15 Age of the Parchment
- 1.16 Age of the Parchment
- 1.17 Sound Power
- 1.18 Sound Power
- 1.19 Double the Power
- 1.20 Double the Power
- 1.21 Two Logs
- 1.22 Two Logs
- 1.23 Decibel Difference
- 1.24 Decibel Difference
- 1.25 Permanent Hearing Damage
- 1.26 Permanent Hearing Damage
- 1.27 Number of Doublings
- 1.28 Number of Doublings
- 1.29 More Power
- 1.30 More Power
- 1.31 My ears hurt!
- 1.32 My ears hurt!

In this last lesson of the course, you're going to be introduced to some real life examples. That can be described mathematically by exponential growth, exponential decay, and logarithmic growth. Now you've already seen the equations that correspond to each of these types of functions before, so I'm just going to refresh you on them. These forms I've written here may be slightly different than what we've talked about. Note for example, that our bases for our two exponential functions are e, and the base for our log function is 10. So this is just one way of expressing these types of functions. The question is, do you remember what each of these two functions looks like? So to jog your memory on that, here are three graphs. I've labeled them A, B and C and I'd like you to tell me which one corresponds to each of these types of functions.

Exponential growth is shown with graph C here. You can see that as x increases, y is also increasing. Exponential decay is shown by graph A. Here, as x increases, y decreases. And logarithmic growth is shown by graph B. Our function is also increasing everywhere it's domain, but this is not happening in an exponential way, as it was over here. Remember also that exponential functions have horizontal asymptotes, and logarithmic functions have vertical ones.

Now that you've jogged your memory a bit about these three types of models, let's talk about some specific examples that fit each of them. We're going to look at a pretty wide variety of topics. Population growth, carbon dating, and sound. Now earlier on in the course, we did talk about population growth, but we did this with pandas. Now we're going to do it with humans. Now there's pretty great data out there about how many people there are in the world. And we're going to check this out all the way from 1800 up through 2011. Let's take a look at some numbers. So here's a chart showing how the world population has changed in those 200 plus years. In 1800 there were only 1 billion people. And by 2011 there were 7 billion. Let's take a look at what this looks like on a graph. Looking at how this data is plotted, what kind of model do you think that this fits best? Does this look like exponential growth? Does it look like exponential decay? Or does it look like logarithmic growth?

This looks like exponential growth. We can tell because the graph is increasing everywhere, and it's getting steeper and steeper as x increases.

One of the most interest things about using real data and then trying to fit equations to it, is that it doesn't usually fit a single function perfectly. We can find curves that approximate it, but it's probably not going to be a completely 100 percent perfect fit. So let's see how that fit changes, depending on which points here we decide to base our function off of. Here, for example is the graph that we would get if we used the first two points on our chart to create and exponential graph. So what would this graph predict the population to be in the year 2013? Considering the scale on our graph here, it probably seems like it's pretty tricky to predict that. But luckily, Desmos is here to help. You'll find a link to this exact graph in the instructor comments on this question. Click on the link and then this graph will popup. Once it does, you can take a mouse and touch a point on the curve. From there, you can drag it, and select what x value you want. And from there, you can see what y value corresponds to it. Remember that this graph measures the year on the x-axis, and it shows the world population in billions of people on the y-axis. Now once you figured out what this graph says, the population will be in 2013, can you tell me if this is an accurate prediction? Think about what the population actually became by of year 2013, and see if this is the same thing. If you want to, you can also use Google to search for world population in 2013.

Let's see what this graph predicicts the population will be in 2013. I'll slide my point down to the corerect X value. In 2013, it says there should be 3.2 billion people. So is this an accurate prediction? The answer is no. We saw on our chart earlier that by 2011 there were 7 billion people in the world. So, this is greatly underestimating the total world population. Using only the first two points in our chart doesn't actually predict very well what the population ended up being.

Here's a new curve that we might use to estimate population at different dates. You can see from the graph that it actually looks like it fits a fair number of our points really closely. Remember, these green points are the coordinates of what appeared on our chart before, showing the actual population for these years. Once again, using the link to Desmos in the instructor comments, what does this blue curve predict the population will be in 2013? Thinking about that, and comparing that to the last number that we saw from our red curve in the last question, does this curve give us a better prediction?

Let's Click and Drag, to see what the population is predicted to be by the blue curve. So I want an x coordinate of 2013, and there it is, 2013. And it says that the population should be 7.695 billion. So I'm going to round that number up, just for simplicity, to 7.7 billion. And is this a better prediction than the last one? I would say yes, definitely. We know the population 2013 is probably greater than that of 2011, which was already 7 billion people. 7.7 billion is much closer to 7 billion than 3.2, our earlier prediction was.

So we know that this blue curve gives us a pretty good estimate of the world population. But, how good is it really? Well, current estimates right now say that the world population in 2013 is just under 7.1 billion people. So is the actually population growth of the world slower or faster than this graph predicts? In other words, are there more people or fewer people than predicted here. Slower growth means that there are actually fewer people. And faster growth means that there are actually more people than the blue graph shows.

The actual population growth of the world is slower than the blue graph shows.

Here are the two models we've been looking at, and I'd like to know what each of them estimates the world population would be by the year 2100. One of them is going to give us a lower estimate, and one of them is going to give us an upper estimate. So please tell me what each of those are. Once you figured out those numbers, I'd like you to tell me which estimate you think is more accurate and why. Or maybe both are inaccurate. Think about where the data that created each model came from.

The lower estimate, which comes from our red graph here, is 5.1 billion people. And our upper estimate, which comes from this blue curve, is 36.1 billion. There's such a huge difference between these two estimates, that I think if we think about all the different factors that affect world population, we can see why. There's so many things that influence the number of people in the world. For example, we've seen an increase in world population due to decrease in death rates, as people live longer because of better health care. In the future, we have no idea what's going to happen. Maybe a pandemic will wipe everyone out. Or maybe we'll have food shortages due to climate change. Or perhaps advances in science mean an abundance of food and longer average lifetimes. Who knows? If you're interested in this, and I personally think it's a pretty fascinating topic, go to the forums and talk about this with your fellow classmates. I think what this illustrates is that even though we can make models, making a long term predictions, especially based on just equations, can be pretty difficult. It's still, however, worth discussing.

Between 1947 and 1956 a collection of ancient Jewish texts was found in a series of caves along the Dead Sea. These 972 texts are known as the Dead Sea Scrolls. And they had a huge impact on history, linguistics and religious studies. Of course, reading the content of these texts was very important, but estimating their age was also really vital. The way that that was done was through carbon dating. So, what is carbon dating? Well, it's a technique used to find the age of any organic material that's up to 58,000 years old. This is done by considering the ratio of carbon 14 to carbon 12 in the material. Now if you're not super comfortable with chemistry, that is not a big deal at all. I'm going to give you all the information you need in this problem to figure everything out. You don't have to be an expert on carbon dating. What happens is that over time the amount of carbon 12 stays pretty constant. But the amount of carbon 14 decreases. These atoms decay. So the older something is, the smaller this ratio will be. One scrap of parchment from the Dead Sea Scrolls was found to have a ratio of carbon 14 to carbon 12, 0.795 times that found in plants alive today. Now, if this ratio for plants alive today, is 1 atom of carbon 14 for every 10 to the 12 atoms of carbon 12, what is the same ratio for the parchment?

We know that the ratio for the parchment is 0.795 times the ratio for plants today, which is 1 in 10 to the 12th. So we need to multiply 0.795 times 1 over times 10 to the negative 12.

Now, conveniently, there's an equation we can use to relate the ratio of carbon is that carbon 14 to carbon 12 ratio. And t is the age of the material we're looking at measured in years. Now you just found that R for the parchment is parchment? Please round to the nearest year. Think about the techniques you know using logs to simplify and then solve this equation.

Our first step is to substitue this value of r into the equation. Now we need to work toward isolating t so I'm going to divide both side of the equation by this coefficient out here, ten to the negative 12. That leaves us with 0.795 equals e to the negative t over 8223. Now what we need to do is get t out of the exponent. So what do we need to do? Well, we need to take a log. Since e is the base of our exponent here, our log should have the same base. That means we need to take the natural log of both sides. On the left side, that's just going to give us some number that we can calculate with a calculator. But, over here something a little bit more interesting is happening. We have two inverse operations that are going to cancel one another out. Leaving us with just the exponent as the entire quantity on the side. I'll leave the left side as is for now. So we have natural log of .795 is equal to negative t over 8223. Now it's easy. All we need is a little bit of multiplication. We need to multiply both sides by negative 8223. So that leaves us with t equals negative 8223 times the natural log of 0.795. Lets pull out our calculator to figure out what number this is equal to. Using a calculator and rounding gives us a value of t of about 1886. So this piece of parchment from the dead sea scrolls is about 1886 years old. Now, in a real situation, estimates are a bit better than this. They are calibrated using, for example, tree rings to determine a slightly more accurate amount of carbon-14 in the plant material at a given time. Still, it's pretty awesome that we can even come up with any sort of estimate.

We talked about earthquakes earlier on in the course. And now we are going to talk about that something that's measured in a very similar way. Surprisingly this is sound. Sound is also measured on a logarithmic scale. And we can use this equation to help us out. This says L1 equals 10 times log base 10 of P over B. Now here, B is some reference volume. it's just a constant that we're going to eliminate eventually, so we don't need to worry about what it's actually equal to. The two variables we have are important, though. P is what is known as the power of the sound, and L here, or I'm calling it L1 in this case, is a measure of the sound level. This is basically comparing the sound power P to B. Now, you may have heard about different sounds being dangerous to our hearing. And this equation can actually help us quantify that. L here, is measured in decibels or dB. And looking at the decibels ratings of different sounds it seems like a pretty small increase in the number of decibels. Actually amounts to a pretty big increase in potential damage. So why is this? Well, let's look at a few other questions to figure this out. So if the original sound power that we're talking about is p, as written in this equation, what will we need to use in its place if we wanted to talk about a sound with double that power

We would need to use 2P instead of P.

So let's say then that we have two sounds. One with the sound power of P and one with the sound power of 2P. What are their respective equations for sound level?

For this first one, we have l 1 equals 10 log base 10 of p over b. So this is exactly the equation we have up here. If power is 2 p instead, then we need to plug 2p in place of p. So that gives us l 2 equals 10 log base 10 of 2p over b

We now have expressions for L1 and L2, but there's a different way we could rate each of them. I'd like you to expand each of these right hand sides so that they contain two logs.

We can use the Quotient rule to do this expansion. The only tricky thing here, is dealing with this coefficient of 10, outside the log. Since we're seeing, for example, that log base 10 of P over B, is equal to log base 10 of P, minus log base 10 of B. The 10 still needs to multiply this entire expression. So we have L1 equals 10 times log base 10 of P minus Log base 10 of B. And for L2, we have the same thing, except with 2P instead of P.

Now that we have these expanded versions of our equations, I'd like you to use them to find L2 minus L1. Now assuming you simplify this properly, your answer should actually just be a number. When you find that number I'd like you to round it to the nearest integer. Your answer will be something that's measured in decimals.

I've started out by just writing L2 minus L1. And we can notice that both of these have a factor of 10. So let's just go ahead and pull that out. So factoring out that 10 and then also distributing this minus sign to both terms inside the second parenthesis, gives us this. 10 times log of 2P minus log of B. Minus log of P plus log of B. Something that pops out in this expression is that we have a minus log of B and a plus log of B. So these two things are going to cancel each other out, awesome. That's how we get rid of that B and that's why we didn't need to know its actual value. Looking at what this equations reduces to then, we can see how much simpler everything is. We still however have another variable left in here. What do we do with this P? Well I think that we have to bring back our quotient rule again. But this time we are going to use it in the reverse order from how we did before. We're going to condense this expression into just a single log. This gives us 10 times log base 10 of 2P over P. Well, and look at that, if we divide 2P by P, these Ps just cancel each other out. That's going to leave us with something super easy. evaluate this. We have 10 times log base 10 of 2. That's equal to 3.01 etc. Which I'm going to round to just 3, so our answer is about 3 decibels.

From what we just discussed, you learn that an increase of 3 decibels doubles the power of sound. So, this gives us some context for thinking about what a decibel means, in general. Now, it's been found that after about 8 hours of exposure to 85 decibel noise, you can experience permanent hearing damage. So, this is just an example, but every time the sound power doubles, the exposure time before hearing damage occurs halves. I wonder then how long it would take for you to experience permanent hearing damage, if instead of 85 decibels a sound was 100 decibels. Let's start off simple. What's the increase in decibels from 85 decibels to 100 decibels?

Remembering that sound power doubles every time we increase by 3 decibels. How many times is it going to double in this 15 decibel increase we were just talking about?

The answer is 5, since 15 divided by 3 is 5.

Thinking about the difference in decibels between 100 decibels and 85 decibels, and what this means in terms of power, how much more powerful is 100 decibels than 85 decibels? This is definitely a tricky question, but think back to the ones we've been doing just before this.

We saw before that with a 15 decibel increase, our power is going to double 5 times. So, what does that actually mean? Well, it means if we have some power P, we're going to double it, or multiply it by 2, and then do that again, and again, and again, and again. Which means we are multiplying P by 2, five times. That's the same as multiplying the power by 2 to the 5th power. And 2 to the

We've gathered a bunch of information over the last few questions. Here's what we know. After 8 hours of exposure to 85 decibel noise, permanent hearing damage can occur. When we increase the decibel level from 85 to 100, the power increases by a factor of 32. And the exposure time before damage occurs is cut in half every time the power doubles. So, taking all this into account, how long will it take for permanent hearing damage to occur if you're being exposed to noise of 100 decibels. Please give your answer in minutes.

Well, first thing's first, I wanted the answer to this question to be in minutes, but the time we have up here is 8 hours so let's convert that to minutes to start off. So 8 hours times 60 minutes in one hour gives us 480 minutes. Great, so this is the exposure time for an 85 decibel sound. We want to know what this is for a 100 decibel sound, so let's look at this last piece of information. If the exposure time before damage is cut in half every time the power doubles, we need to think about how many times the power doubles from 85 decibels to 100 decibels. We knew before that it was 5 times. So if the power doubles 5 times, that means that our exposure time is halved 5 times. So we need to take 480 minutes and multiply it by one half five times. That doesn't seem too bad. However, we know that if we want to multiply by one half five times, that's the same as multiplying by one half to the fifth power. One half to the fifth is one over 32, so we're actually dividing 480 by 32. And that gives us a final answer of 15 minutes. Now remember that every time the power doubles, we have a decibel increase of 3. Some sources state that instead of the exposure time being cut in half for every three decibels, this happens every five decibels, but I think it's better that we stay on the safe side. I bet you never thought we would end up talking about ear health by the end of this course. But that just goes to show you how incredibly applicable algebra is, in all spheres of life. In just this lesson we've talked about everything from sound to carbon dating, the Dead Sea Scrolls, and the population of the world, and this happened in just a few questions.