They arrive on the scene a little while later, and find a body.
They take the temperature of the body at , and it is 31.5The victim died at .
There are far too many applications to discuss them all in this lesson, so instead this lesson focuses on working a variety of examples.
Exponential functions allow us to describe the growth (or decay) of a quantity whose rate of change is related to its current value.
1145 And we want to round our answer to the nearest 1000 years.1152 And we are also reminded that the carbon-14 half-life is 5,730 years.1154 So, let's just start with the general formula for half-life.1159 It is the amount that we originally start with, times 1/2 to the rate times time.1162 Now, in this case, the rate is going to be based on this 5730 years.1168 So, our amount is going to be equal to P--whatever our original amount is that we have of carbon-14--1172 to the 1/2, and we want a 1/2 to occur every 5730 years, so it is t/5730 years.1179 That way, when t = 5731, we will have an exponent of 1, and it will cause the 1/2 to go once.1191 If it is 5730 years twice, we will have an exponent of 2, and we will get 1/2 times 1/2: 1/4.1196 Great; so it makes sense what is going on there.1203 Now, what is our P going to be?
We probably want to figure out what the P that we are going to use is.1205 Well, notice: it said the level of carbon-14 in the bones is 7% of the carbon-14 ratio.1210 So, we were never given absolute values; we weren't given quantities of carbon-14; we were told ratios.1215 If that is the case, well, what would we want the amount to be when it is 0--when we start at 0--when the creature was very first alive?
[If you're interested, there are two different methods to set up and solve the problem given in the steps for this question.
The first part of the steps uses a method similar to the one in the video lesson, but after that is an alternative method where we look at the problem in a slightly different way.] Lecture Slides are screen-captured images of important points in the lecture.What is the percentage rate of the account given to three decimal places? What was the original principal in the account, rounded to the nearest whole dollar?A bank account that compounds continuously is opened with some original principal placed in it. At a lab, a population of bacteria is growing exponentially.0778 We do know what t is; otherwise we would have two unknowns to solve in one equation.0785 It is 5, because we are looking at the 5-year mark.0789 So, we work this out; we get 5457 dollars and 57 cents, over 4700 dollars, equals (1 r/4).0793 Now, at this point, we would probably be tempted--how can we use logs? 0806 We could take the natural log of both sides and bring down the 20; but then we would end up having ln(1 r/4).0810 We are actually losing sight of a much simpler way.0816 When we raise to a power, how do we get rid of powers?0819 If we are trying to figure out what is in the base, and the power is a known value, we just take a root.0822 We take a root, depending on what the power is.0827 If it is squared, we take the square root; in this case, it is to the 20, so we are going to raise both sides to the 1/20.0829 We are taking the twentieth root of this, so it is raised to the 1/20.0836 That cancels out here, and we are left with (1 r/4); we take the 1/20 power, which is also the twentieth root,0842 of 5457 dollars and 57 cents, divided by 4700; and that comes out to be 1.0075.0855 We subtract 1 on both sides; we get 0.0075 = r/4.0864 We multiply by 4 on both sides, and we finally get 0.03 equals our rate.0869 So, our rate, in this case, is a modest 3% return on investment.0875 Now, we can use this information to look at the ten years, at t = 10.0881 At this point, we have everything that we need to know, except for the amount.0886 The amount is the unknown--we don't know what it is going to be worth.0889 But we do know what the initial principal was, 00; we do know that it is 1 a rate of 0.03/4, quarterly,0892 to the 4 times..number of years is 10; so our amount is 4700(1 0.03/4).0902 We work that all out with a calculator, and it ends up coming out to 6337 dollars and 24 cents.0914 And that is the final amount in the account at the 10-year mark.0925 The second example: carbon-14 dating: first, we are going to talk about the idea, and then we will actually work on a specific problem.0931 Carbon-14 is a radioactive isotope; a radioactive isotope is an element that is an isotope,0937 one version of an element, that breaks down over time; it decays into something else.0944 It is something for a while, and then something happens; it undergoes fission; it breaks apart; and it turns into a different element.0949 It is radioactive: it radiates something away.0957 If you want to learn more about this, you can study either chemistry or physics to learn more about radiation.0960 It is a radioactive isotope with a half-life of 5,730 years.0964 It is created in the upper atmosphere by cosmic rays.0968 Cosmic rays from the sun or other stars hit the upper atmosphere of the earth, and they create carbon-14 isotopes.0971 Then, these carbon-14 isotopes are absorbed by living organisms--that is, plants and animals.0982 They get absorbed by trees through the carbon cycle, and then the carbon ends up getting into the animals that eat those trees,0990 and the animals that eat those animals that eat those trees.0998 Any plant that is absorbing carbon will end up absorbing this, so it ends up getting into the cycle of biology.1001 Now, carbon-14--there is a very small amount of this isotope, so it is going to make up a small, but consistent, amount of the carbon of any live organism.1010 So, it is a very, very tiny amount, but it is going to be a regular amount.1018 As long as an organism is alive, it is going to continue consuming things.1022 It is going to continue bringing carbon-14 into its body; so that amount will stay basically consistent throughout its life.1027 However, once the organism dies, carbon-14 stops being absorbed.1033 There will be no more carbon-14 pulled in, because now, since the creature is dead,1039 since the organism is dead, it is not pulling any more carbon-14 into its body.1042 It is not pulling anything else into its body.1046 Without the isotope being replenished...remember, carbon-14 is radioactive; it begins to break down over time.1048 So, the C-14 (another name for carbon-14) in our dead organism will begin to decline,1054 because the creature is no longer pulling in carbon-14 to keep its levels consistent.1060 So, the carbon-14 begins to slowly, slowly drip away; it begins to slowly dissipate, break down, and decay into other things.1064 This fact allows for carbon-14 dating, also known as radiocarbon dating (because it is based on radioactive carbon), or simply carbon dating.1072 By knowing how much C-14 would be in a live organism (we know how much that is in a live organism,1082 because we can just measure it on real, live organisms--we can figure out that this creature is alive;1088 it has this much C-14 in it--a live creature, something that was very recently alive1093 or is currently alive--has this consistent amount of C-14), then once the creature dies, it begins to steadily decline.1099 So, we can measure the amount in a dead organism.1106 And since it is steadily declining--it is declining by rules that we know about--it is declining by half-life rules--1109 we can figure out, based on the amount of carbon-14 left in the dead organism at this point, when the creature died--1115 when it went from being alive and keeping a steady state of carbon-14 to dead,1124 where it is starting to let its carbon-14 just sort of disappear.1128 So, by knowing how much carbon-14 remains, we can get an estimate of when the creature was originally alive.1132 All right, now we will look at a specific example.1139 A skeleton of an ancient human is discovered during an archaeological dig.1141 If the level of carbon-14 in the bones is 7% of the carbon-14 ratio present in living organisms, how long ago did the human die?A wooden spear is discovered in an archeological dig, and the wood has a C-14 ratio that is 0.34% of the ratio for living wood.