Video Transcript
In many natural phenomena,
quantities grow or decay at a rate proportional to their size. For example, investing money in a
bank account which pays compound interest is a good example of exponential
growth. Though perhaps not quite at the
rate most savers would like. At the other end, atmospheric
pressure decreases exponentially with increasing height above sea level, an example
of exponential decay. In this video, we’ll learn how to
model exponential growth and decay arising from differential equations and use these
models to make predictions about future behavior.
In general, we say that if 𝑦 of 𝑡
is the value of a quantity 𝑦 at time 𝑡, and if the rate of change of 𝑦 with
respect to 𝑡 is proportional to its size, 𝑦 of 𝑡, then d𝑦 by d𝑡 is equal to 𝑘
times 𝑦 for real constants 𝑘. This is sometimes called the law of
natural growth, for values of 𝑘 greater than zero, and the law of natural decay,
for values of 𝑘 less than zero. And we can solve this differential
equation generally. Remember, whilst d𝑦 by d𝑡 isn’t
really a fraction, under certain circumstances we do treat it a little like one, and
we rewrite our differential equation as one over 𝑦 d𝑦 equals 𝑘 d𝑡.
We now integrate both sides of this
equation. Using the standard result that the
integral of one over 𝑦 is the natural log of the absolute value of 𝑦, we see that
we have the natural log of the absolute value of 𝑦 plus 𝐴 equals 𝑘 times 𝑡 plus
𝐵, where 𝐴 and 𝐵 are constants of integration. We combine these constants by
subtracting 𝐴 from both sides. And let’s call this new constant
𝐶. So, the natural logarithm of the
absolute value of 𝑦 is equal to 𝑘 times 𝑡 plus 𝐶. We now raise both sides of this
equation as a power of 𝑒, such that 𝑒 to the power of the natural log of the
absolute value of 𝑦 equals 𝑒 to the power of 𝑘𝑡 plus 𝐶.
Well, 𝑒 to the natural log of 𝑦
is 𝑦. And we don’t need the absolute
values here because we’re talking about a population. It must be greater than zero by
definition. Then, using the laws of exponents,
we see that we can rewrite 𝑒 to the power of 𝑘𝑡 plus 𝐶 as 𝑒 to the power of
𝑘𝑡 times 𝑒 to the power of 𝐶. But of course, 𝑒 to the power of
𝐶, given that 𝐶 is already a constant, is itself a constant. So, let’s call that 𝑃. And we find that the general
solution to the differential equation d𝑦 by d𝑡 equals 𝑘 times 𝑦 is 𝑦 equals 𝑃
times 𝑒 to the power of 𝑘𝑡.
Now, in fact, this 𝑃 has a special
significance. We can identify what this
significance is by imagining the scenario when 𝑡 is equal to zero. Clearly, since 𝑡 equals zero is
the initial time, we know that 𝑦 equals 𝑃 times 𝑒 to the power of zero to be the
initial value of 𝑦. However, 𝑒 to the power of zero is
one. So, we find that when 𝑡 is equal
to zero, 𝑦 is equal to 𝑃. This means that 𝑃 is the initial
value of the function. Now, we redefine a little and we
say that, in general, the solution to the initial value problem d𝑃 by d𝑡 equals 𝑘
times 𝑃 for 𝑃 of zero equals 𝑃 nought is 𝑃 of 𝑡 equals 𝑃 nought times 𝑒 to
the 𝑘𝑡th power. In order to see how to work with
these equations, let’s have a look at a specific example.
The population of rabbits on a farm
grows exponentially. If there are currently 245 rabbits
and the relative growth rate is 23 percent, find a function 𝑛 of 𝑡 to describe the
number of rabbits after 𝑡 years.
In this question, we have our
variables defined. We’re told that 𝑡 is the number of
years, and 𝑛 of 𝑡 is the function that describes the number of rabbits. We know that since the population
grows exponentially, we can use the equation d𝑃 by d𝑡 equals 𝑘 times 𝑃 to
describe it. In fact though, we’ll rewrite this
using our variables such that d𝑛 by d𝑡 equals 𝑘 times 𝑛. We could have written this using
function notation such that 𝑛 prime of 𝑡 equals 𝑘 times 𝑛. But personally, I prefer to work
with Leibniz notation.
Now, in fact, we have a value for
𝑘. We know that the relative growth
rate is 23 percent. That’s 0.23 as a decimal. And we can therefore say that d𝑛
by d𝑡 equals 0.23 times 𝑛. Now remember, once d𝑛 by d𝑡 is
absolutely not a fraction, there are certain circumstances where we treat it a
little like one. Here, we can rewrite our
differential equation as one over 𝑛 d𝑛 equals 0.23 d𝑡. We’re then going to integrate both
sides of this equation. The integral of the left-hand side
is the natural log of 𝑛 plus some constant of integration 𝑎. And the integral of the right-hand
side is 0.23𝑡 plus a different constant of integration 𝑏.
Now, note that usually the integral
of one over 𝑛 would be the natural log of the absolute value of 𝑛. But we know the number of rabbits
can’t be negative, so we don’t actually need to include that here. Combining our constants of
integration by subtracting 𝑎 from both sides, and we see the natural log of 𝑛 is
equal to 0.23𝑡 plus 𝑐. We then raise both sides to the
power of 𝑒 to find that 𝑒 to the power of the natural log of 𝑛 is equal to 𝑒 to
the power of 0.23𝑡 plus 𝑐. Well, 𝑒 to the natural log of 𝑛
is simply 𝑛.
And then, we use the laws of
exponents to rewrite the right-hand side as 𝑒 to the power of 0.23𝑡 times 𝑒 to
the power of 𝑐. But of course, 𝑒 to the power of
𝑐 is itself a constant. Let’s call it 𝑃. So, 𝑛 is equal to 𝑃 times 𝑒 to
the power of 0.23𝑡. Now, we’re almost done. There were initially 245 rabbits on
the farm. In other words, when 𝑡 is equal to
zero, 𝑛 is equal to 245. So, 245 equals 𝑃 times 𝑒 to the
power of zero because 𝑒 to the power of zero is one. So, we find 𝑃 to be equal to
245. And we found our function for
𝑛. We can say that 𝑛 of 𝑡 is equal
to 245𝑒 the power of 0.23𝑡.
Now, it’s probably quite clear that
we didn’t need to set up and solve a differential equation here. We could have simply stated the
general form for an equation that describes exponential growth and decay. It’s the general solution to d𝑃 by
d𝑡 equals 𝑘 times 𝑃 for the initial value of 𝑃 is equal to 𝑃 nought. It’s 𝑃 of 𝑡 equals 𝑃 nought
times 𝑒 to the power of 𝑘𝑡. Essentially, 𝑃 nought is the
initial value of the function, and 𝑘 is the growth rate. Of course, it’s always useful to
know how to derive this formula, when necessary.
We’ll now look at an example of a
question where we can use the general solution to the differential equation without
fully deriving it.
A doctor injected a patient with 13
milligrams of radioactive dye that decays exponentially. A doctor injected a patient with 13
milligrams of radioactive dye that decays exponentially. After 12 minutes, there were 4.75
milligrams of dye remaining in the patient’s system. Which of the following is an
appropriate model for the situation? Is it A) 𝑓 of 𝑡 equals 13 times
0.0805 to the power of 𝑡? B) 𝑓 of 𝑡 equals 13 times 𝑒 to
the power of 0.9195𝑡. Is it C) 𝑓 of 𝑡 equals 4.75 over
one plus 13 times 𝑒 to the power of negative 0.83925𝑡? Or D) 𝑓 of 𝑡 equals 13 times 𝑒
to the power of negative 0.0839𝑡.
Remember, we can model exponential
growth and decay by using the formula 𝑃 of 𝑡 equals 𝑃 of nought times 𝑒 to the
power of 𝑘𝑡, where 𝑃 nought is the initial value of 𝑃, and 𝑘 is the rate of
growth or decay. So, let’s list what we actually
have. Since the answers in this question
are listed in terms of 𝑓, we’re going to change this equation to 𝑓 of 𝑡 equals 𝑓
nought times 𝑒 to the power of 𝑘𝑡, where 𝑓 is the amount of dye that’s actually
in the patient’s system after 𝑡 minutes. Now, we’re told that the doctor
injected the patient with 13 milligrams of radioactive dye. In other words, that’s the initial
amount of dye in the patient’s system. So, 𝑓 nought is 13. And we can therefore say that 𝑓 of
𝑡 is equal to 13 times 𝑒 to the power of 𝑘𝑡.
But we also know that after 12
minutes, there were 4.75 milligrams of dye in the patient’s system. In other words, when 𝑡 is equal to
12, 𝑓 is equal to 4.75. So, let’s substitute these values
into our function and solve for 𝑘. That’s 4.75 equals 13𝑒 to the
power of 12𝑘. We’ll solve for 𝑘 by first
dividing both sides by 13. And we see that 4.75 over 13 must
be equal to 𝑒 to the power of 12𝑘. Next, we find the natural logarithm
of both sides of this equation. This is useful as it’s the inverse
operation of raising something to the power of 𝑒. And so, these two operations will
essentially cancel one another out. In other words, the natural log of
𝑒 to the power of 12𝑘 is just 12𝑘.
So, we find that the natural log of
4.75 over 13 equals 12𝑘. We’ll solve now by dividing both
sides by 12. And so, 𝑘 is equal to a 12 of the
natural log of 4.75 over 13. Typing this into our calculator and
we obtain 𝑘 to be equal to negative 0.083900, and so on. Correct to three significant
figures, that’s 𝑘 is approximately equal to negative 0.0839. And substituting this back into our
equation gives us that 𝑓 of 𝑡 equals 13 times 𝑒 to the power of negative
0.0839𝑡. And we see that that corresponds to
the answer D. 𝑓 of 𝑡 is 13 times 𝑒 to the
power of negative 0.0839𝑡.
Once we have these models, we can
use them to make predictions about the future behavior. Let’s have a look at an example of
this form.
At the start of an experiment, a
scientist has a sample which contains 250 milligrams of a radioactive isotope. The radioactive isotope decays
exponentially at a rate of 1.3 percent per minute. a) Write the mass of the isotope
in milligrams, 𝑚, as a function of the time in minutes, 𝑡, since the start of the
experiment. And b) Find the half-life of the
isotope, giving your answer to the nearest minute.
Remember, we can model exponential
growth and decay using the formula 𝑃 of 𝑡 equals 𝑃 nought times 𝑒 to the power
of 𝑘𝑡, where 𝑃 nought is the initial value of 𝑃, and 𝑘 is the rate of growth or
decay. In our case, there are 250
milligrams of the sample at the start of the experiment. So, 𝑃 nought must be equal to
250. We said that 𝑘 is the rate of
growth or decay. Since the isotope is decaying, this
value is going to be negative. And 1.3 percent as a decimal is
0.013. So, 𝑘 is negative 0.013. We can therefore say that the
function that describes the mass in terms of minutes is 𝑚 equals 250 times 𝑒 the
power of negative 0.013𝑡.
We’re now going to look at part
b). Here, it helps us to understand the
definition of the phrase half-life. It’s the time taken for the
radioactivity of the isotope to fall to half of its original value. In our case, that’s half of
250. That’s 125 milligrams. We need to work out for what value
of 𝑡 𝑚 is equal to 125. We therefore set 𝑚 equal to 125
and solve for 𝑡.
We see that 125 equals 250 times 𝑒
to the power of negative 0.013𝑡. We then divide both sides of this
equation by 250. 125 divided by 250 is 0.5. So, we see that 0.5 equals 𝑒 to
the power of negative 0.013𝑡. We then take the natural logarithm
of both sides of this equation to get the natural log of 0.5 equals the natural log
of 𝑒 to the power of negative 0.013𝑡. But of course, the natural log of
𝑒 to the power of negative 0.013𝑡 is just negative 0.013𝑡. And so, our final step is to divide
both sides of this equation by negative 0.013. That gives us 𝑡 equals 53.319
minutes which, correct to the nearest minute, is 53 minutes.
In this video, we’ve learned that
we can solve differential equations of the form d𝑃 by d𝑡 equals 𝑘𝑃 by writing
them in the form one over 𝑃 d𝑃 equals 𝑘 d𝑡 and integrating both sides. When we do, we see that the
solution is of the form 𝑃 equals 𝑃 nought 𝑒 to the 𝑘𝑡, where 𝑃 of zero equals
𝑃 nought. We saw that whilst it’s useful to
be able to derive the formula, it’s absolutely fine to quote it and apply it to
problems involving exponential growth and decay.
