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Algebra 2
Course: Algebra 2 > Unit 7
Lesson 2: Constructing exponential models according to rate of changeConstructing exponential models: half life
Sal models the decay of a Carbon-14 sample using an exponential function.
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If carbon-14's mass gets halved every 5730 years, then wouldn't carbon-14 never disappear? If the current mass gets halved, there would always be something to half for the next 5730 years.(7 votes)
When we say that a mass is "halved", we don't mean that, after 5730 years, half suddenly disappears. We're saying that atoms of carbon-14 (let's say there are 1000) are disappearing at a certain rate (we don't know the exact rate) which is decelerating (so we lose less every 5730 years). So, after 5730 years, we will have in total lost 500 atoms of carbon-14, leaving us at 500. But this happened gradually. Maybe we lost a couple atoms every 10 years or so.
Eventually, we get down to the point where we have very few atoms, say 3. If atoms are gradually disappearing, can you really say that after 5730 years we'll have 1.5 atoms (which doesn't make sense)? No, you'd say that maybe we lost 2 atoms over all that time. 1/2 is not a perfect rate. It's just really, really close.
it's also worth noting that, eventually, when you reach only 1 atom, you can't really divide that anymore, so you're stuck at 1 atom. It would be silly to keep halving that, because then we wouldn't have carbon-14 anymore.(21 votes)
Sal's answer is 741*(1/2)^t/5730. I got 741*(2)^-t/5730. They are the same thing, right?(1 vote)
Yes! If you factor the -1 out of your exponent, you would get 741*((2)^-1)^t/5730. 2^-1 simplifies to 1/2, so your whole expression would then be the same as Sal's: 741*(1/2)^t/5730(4 votes)
I was so curious how can we use Exponential function in our real life?(2 votes)
One common use for exponential models is calculating compounding interest. When you leave money in a bank, the bank may give you interest, which is additional money based on the amount you already have. When the amount of interest you get increases as you gain interest, then we call it compounding interest.
CI = P * (1 - r/n) ^ (n * t) - P
where CI is the compound interest, P is the original amount of money (principal amount), r is the interest rate, n is the number of times the interest is calculated (compounded), and t is the length of time the money is left in the bank.(3 votes)
Just saying that at2:16Sal should have put parentheses around the 1/2.(3 votes)
On2:18Sal wrote 1⁄2^2 but shouldn't he do it with parentheses?(2 votes)
Yes, Sal should be using parentheses. Technically, his version only applies the exponent to the 1, not the whole fraction.(2 votes)
Hi! At3:15Sal said that 1/2 can be called common ratio. But the "step" for the common ratio should be 1. Here t goes from 0 to 5730. Shouldn't the common ration be (1/2)^(1/5730)? As always thanks for the grate job. Greeting from Greece!(2 votes)- Yes. He doesn't stress this, in the interest, maybe, of making this easy to learn. Every year the C14 is reduced by a factor of "the 5,730th root of 1/2". But a step of 5730 yrs is just as valid.(1 vote)
- At1:29why didn't you multiply 741 by 1.5 instead of 0.5?(0 votes)
- Read the problem carefully. It tells you that the carbon-14 loses half its mass. Your method of multiplying by 1.5 would mean that its mass is increasing by 1/2.
Hope this helps.(4 votes)
- At1:29why not multiply 741 by 1.5 instead of 0.5(1 vote)
- When we multiply by 1.5, it means we are adding an additional 0.5. Multiplying by 0.5 just outputs a half of the value we multiplied by.(1 vote)
I was trying to expand this question little bit and think further and got stuck ... need help ... I was asking my self when will carbon -14 sample loose 1/8 of its mass or when it will loose 12,5% which is same as multiplying by 0,875 right ? - after how many years? so if M(t) = 741*(1/2)^t/5730 what I was asking my self is actually for what t will this equation be true (1/2)^t/5730 = 0,875 ( because for that ''t'' it would be same as multiplying by 0,875) right ? did I setup everything correct for my problem ? - If yes - how can I solve for ''t'' in the equation?(0 votes)- I am 100% clueless on how to answer this question.
"Mercury-194 has a half-life of 444 years. What is the constant value for mercury 194? What equation would give us the amount of mercury remaining after t years?
Please help(0 votes)- There are several ways to set up this kind of decay problem. First some terminology we will need:
The amount of a radioactive substance at a given time (after t years have passed) would be N(t)
The amount of the radioactive substance at the beginning of the time period (0 years have passed) would be N₀
The half-life is the amount of time for the material to decay enough to lose 1/2 of its radioactive nuclei. The multiplier is 1/2. We will raise that by the number of years divided by the number of years in the half-life.
In this problem, we are given that it takes 444 years for the substance to lose 1/2 of its radioactive nuclei, so in each year, it will tick through only one-444th of its half-life. So our exponent is t/444
We then can say that N(t) = N₀ (1/2) ^ (t/444)
You asked what the constant value is for mercury 194. In this equation, the original amount is considered to be a constant. In the chemistry of radioactivity, there are other constants, but if you are just trying to answer the question without going into nuclear-level details, the one I gave you should do!
Better get a good book to wait around for those 444 years to pass!(1 vote)
2:16Video transcript
- [Voiceover] We're told
carbon-14 is an element which loses exactly half of
its mass every 5730 years. The mass of a sample of carbon-14 can be modeled by a function, M, which depends on its age, t, in years. We measure that the initial mass of a sample of carbon-14 is 741 grams. Write a function that models the mass of the carbon-14 sample remaining t years since the initial measurement. Alright, so, like always, pause the video and see if you can come
up with this function, M, that is going to be a function of t, the years since the initial measurement. Alright, let's work through it together. What I like to do is, I
always like to start off with a little bit of a table
to get a sense of things. So let's think about t, how much time, how many years have passed
since the initial measurement, and what the amount of
mass we're going to have. Well, we know that the initial, we know that the initial mass of a sample of carbon-14 is 741 grams, so at t equals zero, our mass is 741. Now, what's another interesting
t that we could think about? Well, we know at every 5730 years, we lose exactly half of
our mass of carbon-14. Every 5730 years. So let's think about what
happens when t is 5730. Well, we're going to
lose half of our mass, so we're going to multiply this times 1/2. So this is going to be 741 times 1/2. I'm not even gonna calculate
what that is right now. And then let's say we have
another 5730 years take place, so that's going to be, and I'm just gonna write two times 5730. I could calculate what it's going to be. 10,000, 11,460 or something like that. Alright, but let's just
go with two times 5730. Is it 10? Yeah. 10,000 plus 1400 so 11,400 plus 60. Yeah. So 11,460. But let's just leave it like this. Well, then, it's gonna be this times 1/2. So it's gonna be 741 times 1/2 times 1/2. So we're gonna multiply by 1/2 again. And so this is the same thing
as 741 times 1/2 squared. And then, let's just think about if we wait another 5730 years, so three times 5730. Well, then it's going
to be 1/2 times this. So it's going to be 741, this times 1/2 is gonna
be 1/2 to the third power. So you might notice a little
bit of a pattern here. However many half-lifes we
have, we're gonna multiply, we're gonna raise 1/2 to that power and then we multiply it
times our initial mass. This is one half-life has gone by, two half-lifes, we have
an exponent of two, three half-lifes, we multiply by three. Sorry, we multiply by 1/2 three times. So what's going to be a
general way to express M of t? Well, M of t is going to be
our initial value, 741, times, and you might already be identifying this as an exponential function, we're going to multiply times this number, which we could call our common ratio, as many half-lifes has passed by. So how do we know how many
half-lives have passed by? Well, we could take t, and we could divide it by the half-life. And try to test this out. When t equals zero, it's gonna
be 1/2 to the zeroth power, which is just one, and
we're just gonna have 741. When t is equal to 5730, this exponent is going to be
one, which we want it to be. We're just gonna multiply our
initial value by 1/2 once. When this exponent is two times 5730, when t is two times 5730, well then the exponent is going to be two, and we're gonna multiply by 1/2 twice. It's going to be 1/2 to the second power. And it's going to work
for everything in between. When we are a fraction
of a half-life along, we're gonna get a non-integer exponent, and that, too, will work out. And so this is our function. We are, we are done. We have written our function, M, that models the mass
of carbon-14 remaining t years since the initial measurement.