Growth and decay

One word problem you will very commonly face in MA2051 is that of growth and decay! These problems all take on a very similar form and therefore can be solved with ease once you get used to them.

These problems are often about chemical half-lives, so let's do an example like that.

The rate of decay of a particular material varies proportionately to the amount present. After 2 years, 30% of the material remains. What is the equation that models the decay of this material, and what is the half-life of this material?

The first important bit of material that this problem gives is that this is a decay equation, not a growth equation. This means that the amount of material is decreasing, not increasing, as it might if this were a population of bacteria, for example. This means that there's a negative sign out front!

Then it tells that the rate is proportionate to the amount. This means that the change, or dy/dt, is equal to the amount, or y, times some constant, which we can call k.

dy/dt = -ky

That's the equation that needs to be solved! And that will always be the equation when we're dealing with decay. If we're dealing with growth, it's exactly the same except without the negative sign.

This is a separable equation. We covered separable equations a couple of posts ago, but we'll go through it now too. We must separate the y's from the t's. This means dividing both sides by y and multiplying both sides by dt, which results in:

dy/y = -kdt

Note that all y's now appear on the left side of the equals sign, and all t's are on the right. Now integrate both sides!

ln(y) = -kt + c

Solve for y:

e^ln(y) = e^{-kt + c}
y = e^ce^-kt
y = Ce^-kt Where C = the positive constant e^c

Okay! The differential equation has been solved. Now the only tricky part that is left is interpreting the other bits of information that the word problem gave us: an initial value, and the fact that they're searching for a half life.

Notice this: We have two unknown constants that must be found (C and k) but we only have one bit of information - the amount left after 2 years. That shouldn't usually be enough to find two constants. But! There is something else we can conclude just based on the type of problem this is. Watch and see!

We can start by defining the amount of material present at time 0 as being 100% of the material.

100 = Ce^-k*0 - one equation

And we are given that 30% of the original material is left after 2 years. Really, we can choose any number so long as it is 30% of the original amount. We chose 100 up above, so now we have to use 30.

30 = Ce^-k*2 - a second equation! Since we have 2 variables, 2 equations is exactly what we need in order to solve for both C and k.

Start with the first equation; it's easier because of the 0.

100 = Ce^-k*0
100 = C

Now sub into the second equation:

30 = Ce^-k*2
30 = 100e^-2k
.3 = e^-2k
ln(.3) = ln(e^-2k)
ln(.3) = -2k*ln(e)
ln(.3) = -2k
-ln(.3)/2 = k

Since all natural logs of numbers between 0 and 1 are negative, k will always be a positive constant.

So now we have the answer to the first question of our challenge: the equation that models the material's decay. Subbing in for C and k, that equation is:

y = 100e^ln(.3)t/2

Now we need to find the half life.

The half life, by definition, is the time it takes for the material to divide in half. This isn't hard to find: we must simply sub in for y with an amount that is 50% of the original amount. Recall that above, we chose 100 as being the original amount. Now, we must use 50.

50 = 100e^ln(.3)t/2

And solve for t!

1/2 = e^ln(.e)t/2
ln(1/2) = ln(e^ln(.3)t/2)
ln(1) - ln(2) = ln(.3)t/2*ln(e)
0 - ln(2) = ln(.3)t/2
-2ln(2)/ln(.3) = t

And that's the half life!

In fact, we just solved for the half life very formally, but the half life will always follow the formula ln(2)/k!

And that's decay! I don't think I'll do an explicit growth problem here because nothing changes very significantly. You just leave out the negative sign in all of your calculations.

Happy calculating,
Rachel