Growth and Decay (1 Viewer)

[CrashOveride]

My textbook actually just introduces the exponentional function (growth function) N = Ae^(kt) and then goes on to say dN/dt = kN. As ive seen on the net, it was the latter statement which led to the establishment of the actual equation.

Anyway, i was just having some trouble in understanding some of the meanings and logic behind it. If K is is defined as a growth or decay constant, i.e. the rate at which something is growing or decaying ...wouldnt just K be equal to dN/dt as the latter is defined as the rate of change of N ?

I know this is very basic but for some reason i cant get my head around this :mad1:

 

[CM_Tutor]

In an expression like dN/dt = kN, k is a rate constant, it is not the rate at which the growth / decay is occuring.

Consider, for example, a population of bacteria. They will reproduce at a rate proportional to their present population, and so dP/dt is proportional to P. However, not every bacteria is reproducing constantly, and so the rate is not equal to the population. Thus, we must introduce a rate constant to take account of other factors, and hence dP/dt = kP.

Depending on the system, the rate constant usually does have some physical significance, but it is not as simple as the rate. Note further that if, as you suggest, dP/dt = k, then P = kt + C, and so P depends only on time, and not on how many bacteria are present, which isn't reasonable.

Do you do Chemsitry? If so, I can give you an example of the physical significance of k, in terms of one of the year 12 reactions.

 

[CrashOveride]

What did you mean by "Depending on the system, the rate constant usually does have some physical significance, but it is not as simple as the rate."

I dont do Chemistry, but maybe if you posted up the example it may help.

Also, you said k is not the rate at which the growth/decay is occuring. But what about when you have a question where you find the growth rate of a population. This would then be K and doesnt the growth rate signify the rate at which the population is growing.

Thanks for your help, i actually did the whole chapter on this physical applications of calculus except i foudn on some of these last exercises it was just all mechanical, and i wasnt confident of the actual reasoning/logic behind it.

 

[CM_Tutor]

Originally posted by CrashOveride
What did you mean by "Depending on the system, the rate constant usually does have some physical significance, but it is not as simple as the rate."

I dont do Chemistry, but maybe if you posted up the example it may help.

OK, here goes:

Consider the radioactive decay of an element like iodine-131. It has a half-life of 8.05 days, and (for the chemists) occurs via beta decay.

What is meant by half-life? The half-life is the time taken for half of the sample to undergo decay. So, suppose I had 500 g of iodine-131. Since the half-life is 8.05 days, this means that:

after 8.05 days, I will have 250 g of iodine-131 and 250 g of decay products
after 2 x 8.05 = 16.1 days, I will have (1 / 2) x 250 = 125 g of iodine-131 and 500 - 125 = 375 g of decay products
after 3 x 8.05 = 24.15 days, I will have (1 / 2) x 125 = 62.5 g of iodine-131 and 500 - 62.5 = 437.5 g of decay products
and so on...

Now, what has this to do with maths? Well, this decay is governed mathematically by the decay equation
dM/dt = -kM, where M is the mass of iodine-131 present at time t. The rate constant k is determined by the
half life (t_1/2), as they are related via k = (ln 2) / t_1/2. This is what I meant by the rate constant having a physical significance.

We can then solve the differential equation to get M = M₀e^-kt where, in this case, M₀ = 500 g and
k = (ln 2) / 8.05 = 0.086105... day^-1. You can now check that my above figures (based on half-lives) are correct.

ie. at t = 16.1 days, M = 500 * e^{-0.086105...*16.1} = 500 * e^-1.38629... = 125 g, as expected.

Now, you only have to be able to handle the maths, but seeing an example of how this stuff relates to reality might help.

Also, you said k is not the rate at which the growth/decay is occuring. But what about when you have a question where you find the growth rate of a population. This would then be K and doesnt the growth rate signify the rate at which the population is growing.

Thanks for your help, i actually did the whole chapter on this physical applications of calculus except i foudn on some of these last exercises it was just all mechanical, and i wasnt confident of the actual reasoning/logic behind it.

The rate constant k is not the rate at which the population is growing. The rate at which the population is growing is dP/dt. This is (logically) proportional to the current size of the population for things like bacteria, and the rate constant is needed to take account of factors like birth rate (whatever that means for a bacterium) and death rates.

Perhaps you'd like to post an example where you didn't understand the logic, and I'll try to explain it?

 

[CrashOveride]

Ok here is a question which is actually an example in the textbook, so the worked solution is provided:

The city of Lismore had a population of 20904 in 1971 and this had grown to 21650 in 1975. Find the annual growth rate as a percentage (2 dec. places) assuming that the growth rate is proportional to the population.

Solution:
Since the rate of change of population is proportional to the population (i.e. dP/dt = kP) we can use the exponential growth equation P = P₀e^kt

From here they just take the data into account and using some logs they get k to be 0.0087662 i.e. approx 0.88%.

So if k, which is the growth rate, is 0.88% - doesn't that mean that the population is changing at a rate of 0.88% per year?


You earlier said "The rate constant k is determined by the
half life (t_1/2), as they are related via k = (ln 2) / t_1/2."

Is this an already established fact or did you derive this from some of the data you gave?


Also, something i had not though of before that you addressed earlier: "However, not every bacteria is reproducing constantly, and so the rate is not equal to the population. Thus, we must introduce a rate constant to take account of other factors, and hence dP/dt = kP."

What about when some vehicle is decelerating and its deceleration is proportional to its velocity. How could the bacteria not reproducing constantly analogy be applied here?

Thanks for your help.

 

[Winston]

I use to be sooo obsessed with these questions because i rocked at them... but now i forgot... i never bothered understanding the logic... in 2 Unit most of the time you practice a variety of questions and just follow the pattern, some can understand it through making out what all this crap is saying.. but just follow the pattern.


Hmmm i think this bit "20904 in 1971 and this had grown to 21650 in 1975" is the most important info

taking that 20904 is the initial value i.e. Po and that time i.e. t is equaled to 5 yrs i think i'm not sure if you include 1971, but lets just say, and they say that in 1975 it's equaled to 21650

so you do


21650 = 20904e^5k

21650/20904 = e^5k

taking the log of both sides

ln(21650/20904) = 5k

thus k = (ln(21650/20904)) / 5

voila

 

[CrashOveride]

It's from 1971 so T would actually be 4.

See i can also work through the questions, but im frustrated because i dont fully understand the concept behind it.

You started off straight away plugging stuff into the growth formula. That can only be done when rate of change of population is proportional to population, since from this is how we get the actual exponential growth formula (fair enough, it was given in the question).

The K constant is making me a bit confused in how its defined.

I feel so stupid, because i cant go on with more maths until i know this :mad1:

 

[CM_Tutor]

CrashOveride, what textbook is this from? The reason I ask is that the answer doesn't make mathematical sense.

The solution seems to be saying that, since k = 0.0087662, then the population is growing at a constant rate of about 0.87662 % per annum. Let's examine that assertion.

We know the 1971 population was 20904.
So, the 1972 population should be 100.87662 % of this, so
1972 Population = 100.87662 % x 20904 = 21087.248... = 21087 people (to the nearest whole person).
Thus, 1973 Population = 100.87662 % x 21087 = 21271.852... = 21272 people (to the nearest whole person).
And, 1974 Population = 100.87662 % x 21272 = 21458.474... = 21458 people (to the nearest whole person).
And, 1975 Population = 100.87662 % x 21458 = 21646.105... = 21646 people (to the nearest whole person).

However, the actual population in 1975 was 21650, and this is found using the actual formula,
P = 20904e^0.0087662*4 = 21649.997... = 21650 people (nearest whole person)

In case you're wondering whether the difference is because I required population to be in whole people, if I don't include this (usual) restriction, the method using 0.87662 % growth gives the 1975 population as 21646.689... - still the wrong answer.

The problem is that this method assumes that the percentage increase each year is the same, but it isn't. For example,

1971 population = 20904 (given)
1972 population = 20904e^0.0087662*1 = 21088.054... = 21088 people (nearest whole person)
So, percentage increase from 1971 to 1972 is [(21088 - 20904) / 20904] * (100 / 1) = 0.8802143... %

1974 population = 20904e^0.0087662*3 = 21461.038... = 21461 people (nearest whole person)
And, 1975 population = 21650 (given
So, percentage increase from 1974 to 1975 is [(21650 - 21461) / 21461] * (100 / 1) = 0.8806672... % - a slightly higher percentage increase than from 1971 to 1972.

Thus, you can't really talk of a percentage rate of increase in an absolute sense, as it is constantly changing, but you can talk of an AVERAGE rate of increase over a period.

Over the four years from 1971 to 1975, the percentage increase was
[(21650 - 20904) / 20904] * (100 / 1) = 3.56869... %
Thus, the average percentage increase per annum over this period was 0.8804767... %, as
4thrt(1.0356869...) = 1.008804767...

Basically, you need to be VERY careful with questions involving percentage changes in growth / decay questions, as many people have misconceptions (like the one you are struggling with) that percentage increase is somehow a constant given by k.

Does all this make some sort of sense?

You earlier said "The rate constant k is determined by the
half life (t_1/2), as they are related via k = (ln 2) / t_1/2."

Is this an already established fact or did you derive this from some of the data you gave?

I know this for a fact, but it is readily derived from the decay equation. At the half-life, the mass has decreased by half, so put M = M₀ / 2 and t = t_1/2 into M = M₀e^-kt, and you should be able to show that k = (ln 2) / t_1/2.

There is an equivalent result for growth circumstances, where (ln 2) / k is the time taken for the population to double. So, in the above Lismore example, (ln 2) / k = 79.07... years, and so the population doubles every just over 79 years. Let's check this:

At t = 79, P = 20904e^0.0087662*79 = 41782.196... = 41782 people (nearest whole person)
At t = 80, P = 20904e^0.0087662*80 = 42150.077... = 42150 people (nearest whole person)

And, 2 * 20904 = 41808

So, clearly the population will have doubled after a little over 79 years. It this link between k and time to
halve / double that is the physical significance of k, and not percentage increases / decreases.

Also, something i had not though of before that you addressed earlier: "However, not every bacteria is reproducing constantly, and so the rate is not equal to the population. Thus, we must introduce a rate constant to take account of other factors, and hence dP/dt = kP."

What about when some vehicle is decelerating and its deceleration is proportional to its velocity. How could the bacteria not reproducing constantly analogy be applied here?

Thanks for your help.

I'm not following you on this last part. Could you clarify?

 

[CrashOveride]

Thus, you can't really talk of a percentage rate of increase in an absolute sense, as it is constantly changing, but you can talk of an AVERAGE rate of increase over a period.

But the growth rate, or k, is defined as a constant right? So how can it vary ? A constant is something which never changes ?

The question actually had a second part which asked you to find the population at some time, assuming the growth rate was constant. But that part is treated separately and their answer to the first was in essence what i posted earlier.

But those two answers that you had where the population difference was just off by 4... is this an error in the question or? because shouldnt both method give u the same answer ?

Basically, you need to be VERY careful with questions involving percentage changes in growth / decay questions, as many people have misconceptions (like the one you are struggling with) that percentage increase is somehow a constant given by k.

But the percentage increase (in this case 0.88%) , or the growth rate, is defined by k ??

I'm not following you on this last part. Could you clarify?

Well you said in the bacterium case that the constant k needs to be introduced to take into account factors like birth, death rate etc. For some vehicle that is decelerating, and its decelartion is proportional to its velocity, what factors does the introduction of k address here?

Thanks for your help.
Hopefully this will click soon

 

[CM_Tutor]

Originally posted by CrashOveride
But the growth rate, or k, is defined as a constant right? So how can it vary ? A constant is something which never changes ?

The growth rate is not k. The growth rate is dP/dt. The term 'rate constant' (which is k) is perhaps unfortunate, as it is not the rate at which growth is occuring, nor is it trying to suggest that the growth rate is constant. As I said above, k is a constant that relates to how long it takes for the population to double, or in decay, how long it takes for the amount to halve. So, k is constant (unchanging), but it is not the rate of growth / decay.

The question actually had a second part which asked you to find the population at some time, assuming the growth rate was constant. But that part is treated separately and their answer to the first was in essence what i posted earlier.

Can you please identify the text book this is from? In exponential growth and decay, "assuming the growth rate was constant" doesn't really make sense, as it suggests that dP/dt is constant.

But those two answers that you had where the population difference was just off by 4... is this an error in the question or? because shouldnt both method give u the same answer ?

No, my point was that the 'assuming a constant rate of about 0.88 % per annum' is flawed, and for this reason gives different answers. The further you go into the future (ie. as t increases), the greater will be the discrepency between the methods.

For example, I showed above that at t = 79, P = 41782 people.
By the percentage increase method, this population would be about 20904(1.0087662)⁷⁹ = 41656.296... = 41656 people - ie. the discrepency is now up to 126 people.

I am trying to say that the whole percentage increase approach is flawed.

But the percentage increase (in this case 0.88%) , or the growth rate, is defined by k ??

No, the percentage increase over a year and the rate constant k are NOT the same thing. I think part of the confusion here arises from the fact that the two numbers are virtually identical in this case. This is a coincidence in this question. Let's illustrate with an example:

The population of a town is 1000 at the start of year 1 and 1075 at the start of year 2, and the rate of growth of population at any time is proportional to the population at that time.

This is clearly a case where dP/dt = kP, P = P₀e^kt, and P₀ = 1000.
At t = 1, P = 1075: 1075 = 1000e^k*1
k = ln (1075 / 1000) = 0.07232...
And the percentage increase over year 1 is [(1075 - 1000) / 1000] * (100 / 1) = 7.5 %

At the start of Year 3, P = 1000e^0.07232...*2 = 1155.62... = 1156 people
And the percentage increase over year 2 is [(1156 - 1075) / 1075] * (100 / 1) = 7.534... %

Notice that the percentage increase is not the same, and furthermore 7.5ish % is not the same as k = 0.07232...

Well you said in the bacterium case that the constant k needs to be introduced to take into account factors like birth, death rate etc. For some vehicle that is decelerating, and its decelartion is proportional to its velocity, what factors does the introduction of k address here?

Thanks for your help.
Hopefully this will click soon

I suppose it represents things like quality of brakes, extent of pressure on brake pedal, etc.

 

[CrashOveride]

The textbook is called "3 UNIT MATHEMATICS - Book 2" by S.B. Jones and K.E. Couchman.

Growth and decay is introduced on page 87 and the example i posted up is shown on page 88.

Perhaps it was a bit of coincidence or lack of clarity in this example that threw me off. Your last example also i belive steered me in the right direction and when i get some free time i'll read over all the stuff you posted again.

Also, you said that the growth rate is not k. But the solution for the Lismore example in this book clearly finds what k is, and says that the annual growth rate is this value of k (0.88%)

Also, if you have a look at this site:
http://www.math.wpi.edu/Course_Materials/MA1022A96/lab2/node3.html

it also says that k is the growth rate ?

 

[Calculon]

K can be looked at as the 'fudge factor', such as those used in physics to bridge the gap between 'proportional to' and 'equal to' (eg universal gravitational constant). If 'k' were made larger then the growth rate would increase, and if it were made smaller it would decrease. It is used because some populations are growing faster than others.

 

[CM_Tutor]

CrashOveride, I will look at Jones and Couchman tomorrow, and wiil post an answer for you then - sorry I didn't get to it sooner.

 

[Grey Council]

lol, why don't you get your teacher, sit down with him at lunch, and just throw everything you have at him?

imho, tis a rather long and arduos process to get a mathematical concept explained to you over the net.

Although you are to be commended for actually trying to understand it. lol

All I can suggest is, get the fitzpatrick or the Cambridge. ^_^

 

[Calculon]

Originally posted by Orange Council
lol, why don't you get your teacher, sit down with him at lunch, and just throw everything you have at him?

His teacher apparently isnt that helpful.

 

[CrashOveride]

While CM Tutor is away...

Anybody else wanna comment on the subject matter?

 

[CM_Tutor]

Originally posted by CrashOveride
The textbook is called "3 UNIT MATHEMATICS - Book 2" by S.B. Jones and K.E. Couchman.

Growth and decay is introduced on page 87 and the example i posted up is shown on page 88.

Perhaps it was a bit of coincidence or lack of clarity in this example that threw me off. Your last example also i belive steered me in the right direction and when i get some free time i'll read over all the stuff you posted again.

Also, you said that the growth rate is not k. But the solution for the Lismore example in this book clearly finds what k is, and says that the annual growth rate is this value of k (0.88%)

Also, if you have a look at this site:
http://www.math.wpi.edu/Course_Materials/MA1022A96/lab2/node3.html

it also says that k is the growth rate ?

Firstly, sorry it has taken me so long to get back to this.

Secondly, I've had a look at Jones and Couchman, and I see what you mean. However, if you look in the red box on page 87, you'll see that k is correctly defined as "the growth or decay constant for a particular population". I can see that they have, on page 88, converted the value of k (0.0087662) to a percentage (0.88 %), and I've seen this in other texts too, but the rate constant really shouldn't be expressed as a percentage, because it isn't a percentage - you'll see it expressed this way, even in questions, but that doesn't make it true. You just have to know if the give k as a percentage, divide it by 100 to actually get k, and move on.

As for the website you have linked to, let's look at the question it provides. It says that:

At the start of an experiment, a bacteria colony has mass M₀ = 2 * 10^-9 grams. After two hours, M = 1.5 * 10^-8 grams. Find the growth rate r of the colony. Find the mass of the colony after 10 hours.

It offers the answers r = 1.00745 ..., and that at t = 10 h, M = 4.75 * 10^-5 g.

Ignoring the fact that units and significant figures are mishandled, let's concentrate on what this answer r means. They suggest that r is the growth rate.

Surely, if we are measuring the amount of bacteria as a mass in g, and doing so in time measured in h, then the appropriate unit for the growth rate is gh^-1.

Does r = 1.00745... gh^-1 make any sense? No, it doesn't, because it would imply that the mass of bacteria increases by more than 1 g every hour - ridiculous when you consider that we know the mass after 2 hours is 1.5 * 10^-8 g. Hence, r is not the growth rate.

The growth rate is actually dM/dt = rM, and so initially, the growth rate is about 2.01 * 10^-9 gh^-1, which does make sense.

Now, what about my interpretation of k (or r, in this case)? I said above that (ln 2) / k is the time taken for the population to double. Applying this in this case, we get the time taken for the mass of bacteria to double is
ln 2 / 1.00745 ... = 0.6880 ... h. In other words, the mass of bacteria doubles roughly every 40 minutes. We'd expect the mass after 2 hours to be roughly 2³ = 8 times the original mass. This is 1.6 * 10^-8 g, which is in good agreement with the actual value of 1.5 * 10^-8 g, when you consider that 0.6880 ... h is actually a little over 41 minutes.

In other words, the websites definition of r as the growth rate, instead of the rate constant, is wrong.