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## Constructing exponential models according to rate of change

Current time:0:00Total duration:4:18

# Constructing exponential models: half life

CCSS Math: HSA.CED.A.2, HSF.BF.A.1, HSF.BF.A, HSF.LE.A.2

## Video 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.