# Applications of First Order Differential Equations: Exponential Decay Part 1

– [Narrator] Welcome back for more Applications of First Order
Differential Equations. In this video, we’ll
cover exponential decay. The differential equation
dP dt equals k times P or P prime of t equals k times P of t models exponential decay,
which means the rate of decline or decay is proportional to the population or amount and nothing else. So taking a closer look at this, this is telling us that the change of P with respects to time is
equal to some constant k times our function P. And because we have exponential decay, k is going to be negative. We’ll actually solve this
differential equation in the first example to
find our function P of t, but the only solution to this is the exponential function given here, P of t equals p sub zero times e
raised to the power of kt. Where P sub zero is the initial amount, k is the exponential
decay rate, t is the time, and P of t is the amount after time t. And this function here should remind you of exponential decay
problems that you solved in a previous algebra class. Let’s take a look at our first example. Polonium-218 decays exponentially. The decay rate is 23.1% per minute and can be modeled by
the differential equation dP dt equals negative 0.231 times P. So again, because we
have exponential decay, notice how this derivative is
equal to this constant here which is our decay rate,
times the function P. Part a, we want to find an
equation that satisfies the differential equation. We’re going to let P of
zero equal P sub zero. Now we already know
from the previous slide and from algebra that the
solution to this is going to be this function here, but
we’re actually going to verify this by solving
this differential equation for part a. Part b, if 1000 grams of
polonium-218 is present at time t equals zero, how much is
present after 90 seconds? And then what is the
half-life of polonium-218? So let’s start with our first question. We can solve this differential equation by using separation of variables. So we’re first going to write
this in differential form. So we would have dP
equals negative 0.231Pdt. And now we’re going to
divide both sides by P, or multiply both sides by one over P. So we’d have one over P dP equals negative 0.231 dt. And now from here,
we’re going to integrate both sides of the equation. The anti-derivative of one
over P with respects to P would be natural log absolute value of P. But since P is never going to be negative, we can just write natural log P. We’ll put the constant of
integration on the right side. So this is going to be equal to the anti-derivative of
0.231 with respects to t. It’s going to be negative 0.231t plus a constant of integration. Since our goal is to
solve this equation for P, e raised to the power of natural log P must be equal to e raised to
the power of this sum here. Looking at the left side,
because this is base e and this is log base e, your natural log, this simplifies nicely to P. And because it’s going to be a function of t, we can write P of t equals, now on the right side, notice how we’re adding the exponent
so we can write this as e to the negative 0.231t times, this would be e to the c. But e to the c is just a constant. So we’re going to let e to the power of c equal
the initial value P sub zero. So now we have our general solution to the given differential equation which models an exponential decay. P of t is equal to P sub zero, again this is sum constant,
which we’re letting equal e to the power of c times
e raised to the power of negative 0.231t. And again, this should be
the solution that we expect from our previous classes in algebra. And now we should be
able to use this function to answer the two remaining questions. If 1000 grams of polonium-218
is present at t equals zero. So they’re telling us that
P of zero equals 1000, how much is present after 90 seconds? You’ll need to be careful here because remember the time was given in minutes, so 90 seconds is going to
be equal to 1.5 minutes. So we know that t is equal to 1.5. Let’s go ahead and verify that
if P sub zero equals 1000, P sub zero would be 1000. So if P of zero equals 1000, this would have to be equal to P sub zero times e raised to the power of, well if t is zero, the exponent is zero. E to the zero is equal to one. We’re quickly verifying that P sub zero, our initial amount would be 1000. So now we have the particular solution P of t must equal 1000 times e raised to the
power of negative 0.231t. And now to find how much is
present after 1.5 minutes, we just need to evaluate P of 1.5. So we have 1000 times e raised
to the power of negative 0.231 times 1.5. We’ll have to determine this
value in the calculator. So we have 1000 times e raised to the power of negative 0.231 times 1.5. We’ll go and round this to the hundreds. This would be 707.16, and
the units would be grams. And then for the last question, they’re asking us to
determine the half-life of polonium-218. So starting with our function, P of t is equal to P sub zero
times e raised to the power of negative 0.231t. We can let the initial amount
P sub zero equal anything. Let’s go ahead and assume
that it’s equal to 1000 as it was in part b. So if we start with 1000,
half life is the time it takes for half of that to decay. That means P of t would
be 500, or half of 1000. So we want to solve the equation 500 equals 1000 times e raised to
the power of negative 0.231t. So now we’ll divide both sides by 1000 to isolate the exponential part. So this simplifies to one, the left side simplifies to one half, which is always the case for half life. And on the right side we have e raised to the power of negative 0.231t. To solve for t we’ll
now take the natural log of both sides of the equation. And then we can apply
the power property here. So we can move this exponent to the front of the natural log. So we have the natural log of one half must equal negative 0.231t
times natural log e. But natural log e is equal to one. And we don’t have to
include the factor of one, so the last step here is
to divide both sides by our exponential decay
rate, or negative 0.231. So this simplifies to t. And on the left side, we’ll
go back to the calculator. So we have natural log one half, or .5 divided by our constant,
which is negative 0.231. And our time is in minutes,
so we’ll say that the time is approximately three minutes. Now that we determined the
half-life the long way, there is a shortcut. Whenever we want to find the half-life for exponential decay,
let’s call it capital T, it’s going to be equal to natural log one half divided by the constant of
the exponential decay rate. And if you remember, when
we had exponential growth, the shortcut to determine doubling time was natural log two divided by k. I think we’ll go ahead and
stop here for this video. We’ll take a look at a second example of exponential decay in part two.

## 7 thoughts on “Applications of First Order Differential Equations: Exponential Decay Part 1”

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