## Separable Differential Equations & Growth and Decay Model Calculus 1 AB

BAM!! Mr. Tarrou. In this lesson, we are going
to use our newfound skills of solving separable differential equations and apply those to
growth and decay models. We will do 4 examples, first one working with variables that are
inversely proportional to each other. The second example is going to be a direct application
of the rules and definitions we have up here which is having to work with 2 variables which
have direct proportionality… and are directly variable. And our third example is going to
be a problem where we have a graph of what is an exponential function- it will be growth,
we will be going through 2 points and we will be using that information to come up with
an equation of an exponential growth function- just knowing 2 points it passes through. And
our last example is going to be dealing with Carbon-14 dating.
Okay so, exponential growth and decay models or ‘growth and decay models’. Continuous
compound interest, population growth, Carbon 14 dating, etc. are examples of exponential
growth and decay. With these type of problems, the rate of change of y is proportional to
the..value of y. If y is a function of time t..( I just put quotes around these), we can
write the derivative of y with respect to t or dy/dt=ky. Now we are going to expand
that, we are going to create our growth and decay model from that. If y is a differentiable
function of t such that y is greater than zero and y prime is equal to k y, for some
constant k, then we have y is equal to c times e to the kt power. And first coming out of
our Algebra and geometry or algebra and algebra II, we might recognize this as the “PERT
“ formula, continuous compound growth and decay. Now we are just going to look at it
with a little twist of calculus, see where this comes from. So our C is going to be the
initial value in our application problems, initial value period, but like in application
problems like investing money, bacteria growth, carbon-14 dating, that will be our initial
value. And k is our proportionality constant or constant of proportion. If k is greater
than zero, then we have exponential growth and if k is less than zero then we have exponential
decay. This y prime is equal to ky, hopefully at this point in calculus we understand that
it is just another version, another way of writing dy/dt, so it is the same equation,
just extending it on to the exponential growth and decay model. How does this become this?
Let me get out my high definition, picture and picture, voila. Okay, sometimes squeaks
a little here, so hopefully it is not too bad.
We have dy/dt is equal to ky. Now you know what if I am taking a test and I go back and
look at the concept later in the book and I have forgotten this rule, we do understand
that calculus has quite a bit of memorization but we forgot this, maybe, y prime=ky; y=
c times e to the kt power, how do you come on your own. Some of these derivations are
useful. So we are going to recognize, or hopefully
we can recognize that is a separable differential equation so we are going to divide both sides
by y, and move the dt over to the other side. we have 1 over y dy is equal to k dt. Alright,
now of course it looks like we have a lot of unknowns, but k is not an unknown, it is
a constant, but we do have other unknowns, which is t and y, so now we have separated
these two variables of interest and we take the indefinite integral of both sides of our
equation with respect to that isolating variable. So, the indefinite integral of 1/y dy is the
natural log of the absolute value of y. Now our definition says that y must be positive,
you know it is a habit now, just make sure you realize or your notation shows that you
can’t take the log of a negative number, so remember that absolute value symbols are
kind of like a habit, I guess. And, the indefinite integral of a constant with respect to t is
a constant times t and with an indefinite integration, we have to add in a constant
of integration on one side, technically it is on both sides, but you subtract them to
get a new unknown constant. Now, to continue trying to get this equation
to this form, we need to get rid of the natural log function, so you make both sides, an exponent
of e, base e and log base e, those are inverse math operations so they cancel out and we
have y is equal to, now over here, we have one base with two exponents being added together,
well no when do we add exponents together? When we multiply like bases. so this can get
re written as e to the kt power times e, which hopefully we know is an approximation for
2.718.., that is an irrational number, so it is just an approximation like pi, so we
round it off to three decimal places, so it is approximately 2.718 and that raised to
a constant is simply going to be another constant. SO we have y is equal to c, if you want to
be really technical, maybe put a subscript there that these constants are not necessarily
the same as that one, or won’t be; time e to the kt power. And that is how we go from
á change in y with respect to t, or dy/dt is equal to ky where basically pert formula
comes from that we would have known from maybe algebra but definitely algebra 2, to our continuous
compound interest, or growth and decay model. Now let’s get to our first example. Write
and solve the differential equation that models the verbal statement. The rate of change of
Q with respect to ’t’ is inversely proportional to the cube of t. Okay now this is not going
to match the rule that just I had at the beginning of the video. I almost put this example at
the end of my previous video of introducing solving separable differential equations.
But, I guess I kind of didn’t because, we don’t just have direct variation, we have
inverse variation as well so when you see that proportion change like this, you can’t
just think and apply that rule I just had up there… The derivative of y with respect
to t is equal to c times e to the kt power. So let’s just take this and see what we
have here. the rate of change of Q, the rate of change, so Q is changing and it is with
respect to t and that rate of change is inversely proportional. Remember, maybe I should have
wrote that. Direct variation is of course is… Let’s put it over here.
From algebra even, you could have recognized that direct variation is y=some constant
k times x and inverse variation is y=k/x. SO inverse proportion type situation we have
going on here, k divided by the cube
of t. dQ/dt=k/t^3. Now we were asked to write the expression that goes along this verbal
statement and solve the differential equation so let’s solve and that mean we are going
to… This is going to be separable, that is the only skills that we have right now.
We are going to move the dt over here so we have dQ=k times 1/t cubed dt. I just want
to show all the steps. This time, you might not need to do this, but I am going to write
this as t to the -3 power. And now that we have isolated our 2 variables of interest,
Q and t, we find the indefinite integral of both sides with respect to the isolated variable.
So the integral of dQ is simply Q=this is going to be, let’s see here we have…we
are going to take this power and increase it by one, so t to -2 power and then we have
to divide it by the power that was increased by one and then not forget about our constant
of integration. To clean this up a little bit, I am going to write this as -k/2. Now
that t is going to drop down right. If I want to write that in a way…in a form that gets
rid of negative exponents…we are going to take that t to the negative two and drop it
down into the denominator. We get -k/2(t^2) +c. If you are not getting any points to simplify
your answer, maybe just stop right here. That is how I am going to write my final answer
for this first example. Next we are going to move on to direct variation. BAMM! The
rate of change of n is proportional to n. The rate of change of n is proportional to
n, assuming, of course that the rate of change is of course with respect to time. When t=
0, n=300 and when t=1, n=400. What is the value to n when t is 3? SO clearly I need
an equation that is not just about the rate of change; I need to solve this differential
equation. I can go through the process of solving this separable differential equation
like I did in a lot of previous examples, but this a pattern, rate of change dn/dt=kn,
so when I solve this differential equation I am going to get that n=c times e to the
kt power and if I know that rule, I just save myself a fair amount of work. There is my
equation, that is going to be some kind of growth and decay equation. When t=0, n=300,
so n(0)=c times e to the kt power. So when t=0, n is equal to 300, I don’t know what
my unknown constant is, that what I am trying to figure out, k times zero is zero and anything
to the zero power is equal to 1. As I said, c was the initial value, when t=0, we are
talking about change with respect to time when time=0, I actually kind of just said
that 300 was our initial value, but showing the work here how to figure out, we get c=300.
There is no reason to write that again. So now we have n (t)=300 time e to the kt power.
I found 1 constant but I don’t actually know the other, we don’t know what k is
and if you remember from algebra even as far back as that, when we worked with direct variation
and inverse variation, we almost had enough information to find our constant, then we
were at the equation and then we continued on to actually solve some kind of question
like this, that is what this is here for now- when t=1, n=400, that is given to us so we
know what k is equal to and we can actually write an equation and we can use it for any
other question that we are asked for. So n=400, when t=1 so, 400=300 times e to the k, which
I don’t know yet, t power, and now t is equal to 1.
Okay so this is going to be e to the k, because that’s just 1, and we are going to take
300 over to the other side. 400 divided by 300 is equal to e to the k power, of course
that is just going to be 4/thirds and we are trying to solve for k so we can’t solve
for variables in exponents so we need to take the natural log of both sides, or take any
log for that matter, any base log you like, but for base e, natural log is easier fit
there, so we have the natural log of 4 thirds=natural log of e to the k power. Well that
is log base e, natural log so those are going to cancel out. So the natural log of 4 thirds
is equal to k. So. now my equation
is n (t)=300e, now k is natural log of four-thirds t power. I would wrap either that natural
log of 4/3 in parentheses or actually maybe just the four-thirds just to make sure it
doesn’t look like t is in the natural log function. If there is no parenthesis it really
is not but just for clarity of work. Now starting to kind of cram myself into a corner here,
so let me get out of the way and clean off this board. Now that we have some room to
work, what is the value of n when t=3. What is n (3). Well that’s going to be 300 times
e to the natural log of 4 thirds times 3. That is not the natural log of four-thirds
times 3, it is the natural log of 4 thirds…times 3. Well how can I do this without jumping
straight to the use of a calculator? 300…let me move this 3 out front just for teaching
purposes. I would like to …maybe resort to the use of a calculator at the end but
do I have to immediately go. Well that’s base e and natural log
So there is no way I can do this without a calculator. Well you can get that base e and
that log of base e to cancel out in the exponent if you take the 3 and move it up as an exponent.
Okay. Well now that the base e and the log of base e or natural log are stacked against
each other, those inverse math operations will cancel out and make it 300 .Sometimes
I skip steps, clean that out, drop this down. We have got four-thirds to the third power.
Now four to the third power, four times four is sixteen times four is 64. 3 to the third
power is 27. And I think this point, I think I may just go off the top of my head and oh
my! Look at that….300 times 64 over 27 is 6400 over 9 and that is n (3) and the end
of my second example. A cheat little bit at the end but that’s okay. You know I like
to make these videos in pieces so I can edit them together or if I make a mistake more
easily correct them, but in the fear of making a mistake on camera, did I really need to
resort to some notes to go oh that divided by 3 is a 100 and that divided by 3 is 9,
and oh look 6400 over 9. Well anyway. Next Example. Find the exponential function y=c
times e to the kt power that passes through the given points. Okay so we have what looks
like, well we are told that it is exponential, and this is increasing as x increases. So
it’s an exponential growth function. And we see that it passes through a couple of
points, so let’s put these points, these x and y values into this equation and see
what we get. We have y=one half, we have an unknown constant. SO 1/2=c times e to the
k power, we don’t know what that is, we hardly know anything at all, but t, we are
talking about with respect to time, so x or t here is=2. Okay that’s one point. Now
the other point works out pretty much the same way. We have 7=c times e to the k times
6 power. It is probably going to be simpler here to get these equations, or bring these
equation together cause we have 2 equation and 2 unknowns c and k, and we have been doing
2 equations and 2 unknowns for quite a long time, when we talk about linear functions
back in algebra and even now we did it with linear combinations and substitution. We are
going to do substitution. So we have c=7 over e to the 6k power. Now we have c. Now we have
½ and bring that here. Now we have a variable we can do substitution through. SO we have
1/2=the value of our unknown constant c=7/e to the 6k power times e to the 2k power.
Now I have an equation that is written in the terms of only 1 variable not 2 and thus
we can solve for k or at least try to. Now we have e to the 2k power in the numerator
basically and e6k power in the denominator and when we divide like bases we subtract
the exponents, 2-6 is -4 and we have to move that base down to the denominator, so basically
we are going to have… I say basically then I skip a step and then somebody asks me how
did you do that so… This becomes 7 times e to the 2k-6k which becomes -4k and then
to get rid of that negative exponent, we are going to take the base and move that down
to the denominator. So we have 1/2=7 over e to the 4k power. Now again we are trying
to solve for k…and settle down the dogs. With the front blinds shut and the dogs not
barking, we are going to multiply both sides of this equation by e to the 4k power and
multiply both sides of the equation by 2 or you can call it cross multiplication. And
now we have e to the 4k power=14. Now of course we know that you can’t solve for
the variable when it is in exponent, so we are going to take the natural log of both
sides and…we will just fill up this board and erase all this work when we get to the
next step… We get natural log of e to the 4k power is equal to the natural log of 14.
And the natural log or log base e and that base of e are going to cancel out, we get
4k=ln14 and then we can divide both sided of the equation by 4 and get k=(ln 14) / 4.
Just checking my notes to make sure that I didn’t say something silly. Now that comes
in to a point where you keep working in the exact form or decimal approximation and I
did this by decimal approximation, and off the top of my head, k is going to be approximately
equal to 0.65976 if you want to carry it off that far. And now that we know what k is,
we can take that value and plug it into this expression for c and we can write the equation
for the function that we have drawn over here. C is going to be equal to, plugging this up
over here, 7 over e to the 0.65976, that is our k times 6 and that comes out to be approximately
equal to 0.13363. Okay. Now if we can get through this without my
dog going nuts again because he is starting up:) The final answer, we have y=c which is
approximately equal to 0.13363 times e to the k which is approximately 0.65976 times
t and that is a pretty good approximation cause we did good with decimals and round
off with decimals, but approximation of exponential functions where k is greater than 1, which
an exponential growth function that goes through these 2 points.
One more example coming right up involving again carbon-14 dating. BAM! Here we go. Carbon
14 dating assumes that carbon dioxide on earth today has the same radioactive content it
did centuries ago. If this is true then the amount of carbon 14 absorbed by a tree that
grew hundreds of years ago should be the same as the amount of carbon 14 absorbed by a tree
growing today. A piece of charcoal has about 19% as much carbon-14 as a new piece of charcoal.
Knowing that the half-life of carbon-14 is 5715 years, how long ago was an ancient tree
burned to create the charcoal? Now this is not a science class, it is a calculus
class, and our solving of these separable differential equations, led us into our work
here with exponential growth and decay models. So there is a probably a nice, easy formula
for doing at a time that streamlines the process, but for now because we are not in a science
class, we just hopefully recognize we are talking about here exponential decay, so going
to that model, a=c times e to the kt power, and I have to come up with a formula so I
can finish the problem. So we have an initial amount, a final amount, we have e which is
of course approximately 2.718 and I don’t know my constants, that’s what we need to
figure out right now using the fact that half-life is 5715 years. Half-life, you lose 50%. When
you talk about percentage of change, if you don’t care about the amount of change but
the percent of change, the initial amount is arbitrary. So, let’s just say that we
start off with 2 units of this carbon 14 and we end with 1. I am saying units, because
again I am not a science teacher and I don’t want to say the wrong unit and have someone
mad at me. So we start off with 2 of something and end up with 1. We have that times e to
the kt power, we actually know what t is, it is of course 5715. Divide both sides by
2 and you will see that 50% come up. I could have just said that the initial amount was
1 and our final amount was 0.5. So dividing both sided by 2, you get 0.5 is equal to e
to the 5715 times k power. I am trying to keep my board from rattling there, in case
you were wondering what I was doing. You can’t solve for variables when it is exponent, so
we take the natural of of both sides, that’s the easiest with the base of e here. Of course
we are using the natural log or log base e because that will cancel out with the base
of e on this exponential function. And we get the natural log of ½=5715k. And finishing
this up by dividing both sides of the equation by 5715, k=1 over 5715 times the natural
log of ½. You can make that a decimal right now, I am going to keep this in this form
for a little bit longer. I would caution you with rounding off very much at all is going
to cause our final answer to be off by A LOT. If you are going to take that to decimal approximation,
leave it off to 4-5 decimals, but I am going to leave it off in exact format to minimize
the round off errors by as much as possible. Okay so now… I am putting this in the way
of my problem, so let me put this down here:) My advanced picture in picture. Now that we
know the constant, we can write the equation. Our final amount is going to be our initial
amount times e to the1/5715 times the
natural log of ½ t. I got a lot of information here right? I know that I am looking for t,
how old this piece of charcoal, our final amount is 19% of what we started with so we
can let our initial amount be a 100 and let A be 19 and that would add an extra step of
dividing both sides by 100; or I could start with an initial amount of 1 and say that the
final amount is 0.19 units of this carbon-14. 1 e to the 1/5715 times the natural log of
½ t. Again, you can’t solve for variables when it is exponent, so to get rid of that
base e we take the natural of both sides. That’s going to cancel out. So we get ln
(0.19)=1/5715 times the natural log of ½ t. We are going to solve for t, so we are
going to multiply both sides of the equation by 5715, we are going to divide both sides
by the natural log of ½ to undo the multiplication and now with our scientific calculators, and
rounding it off to 2 decimal places, t comes out to be approximately 13,692.73 years. That’s
one old chump of charcoal! That’s also the end of my last example. I am Mr. Tarrou. BAMM!

## 39 Replies to “Separable Differential Equations & Growth and Decay Model Calculus 1 AB”

1. ProfRobBob says:

My lesson on solving separable differential equations and how they relate to exponential growth and decay models, direct, and inverse variation.

2. Michael Cadigan says:

That shirt is awesome! You should sell them, lol.

3. timmy brodbeck says:

You're the man

4. Văn Hồng says:

thanks for all the video lessons! I watched most of your cal 1 videos and i made a 4 on ap exam. I'm looking forward on your cal 2 videos and i hope it will guide me through my freshman fall semester

5. 2055g says:

So, Mr. Tarrou! I've been going through a bit of 'Math drama' I guess you could say, but my schedule for Math classes has been weird. But now, I'm in Calculus AB and BC (I can afford to be in both of these thanks to a course I took last year, Pure Math 1 & 2, where I learned Calculus basics) and your videos have been a lifesaver! Thanks to them, I've gotten a 95 on my Limits test, and plan on getting 100 on my next few tests and 5's on my Maths and Physics AP Exams! Thanks for all you do, it really means a lot to the students that watch these.

6. Matthew Sinnott says:

Can you do a video on convergent and divergent series?????

7. The Rea one says:

8. Oberberg says:

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9. zubiya qureshi says:

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10. zubiya qureshi says:

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11. Eljona 26 says:

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12. Eggs Benedict says:

32:00
One day his hands are going to explode

13. Jojo Jenkins says:

Thank-you so much! I really appreciate you taking the time to help us. The best thing about your videos is that they are diverse, which is truly needed because a lot of others videos will demonstrate the same kind of concepts over and over. Thanks-again!

14. Mohamed Ali says:

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15. Anthony Araujo says:

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16. Nathan Panggabean says:

32:03 onwards, you can see him crawling to left of screen 😛

17. Jaxx M says:

I need an example of calculating population growth with the DE method but using multiple time and population values to define k. I know I have to find individual k values for each population value and time. However, I don't know how to find the best k value for the final equation out from my list of k1, k2, k3, …

Example:

P = ce^kt
t=0 , p=1
t=10, p=5
t=20, p=10
c=1 ===> P=e^kt
k1 = (ln5)/10
k2 = (ln10)/20

… with k1 and k2 (more in my real questions), how do I find the final equation with only one final k value?

18. ProfRobBob says:

Closed Captioning brought to you by my YouTube and new friend from India, Jigyasa.  Thank you so much for the help:) #math

19. ProfRobBob says:

New Closed Captions brought to you by my YouTube student and friend in India, Jigyasa. Thank you so much for the help:) ‪#‎math‬﻿

20. Jordan Perez says:

Do you have a video on estimating the sum of a series?

21. Colton Woods says:

You literally got me through Calc because my professors can't teach.

22. Edu Tonaco says:

BAM! You rock! Thank you professor!

23. Martha D. says:

Are you in a green house? 🙂

24. Sandler Tossone says:

At 14:14, couldn't you just sub the {4/3)=e^k into the original n=300e^kt and use n=(300)(4/3)^t as you model for finding values of n and t? That way you avoid doing the ln's.

25. Trevor Yilk says:

At 8:08 you said that y is equal to some constant times k. Didn't you mean ye is equal to some constant times x?

26. David Fox says:

3 weeks of 'BAM!' = An entire year of University math revision. Why am I paying them when you are doing the work? Do you accept student loans?

TY Prof.

27. Gabriel Fernandes says:

Professor, thank you very much, you're amazing. Do you have any classes about Laplace tranformations?

28. Kedar Neopaney says:

Hello profesor Rob, you have the best calc tutorials on YouTube. Thank you for helping students who are struggling with math, or perhaps poor teacher. Also wondering if you do Discrete Maths tutorials or not! Again thank you very much!!!

29. Julieta Ayala says:

Hey Mr. Tarrou! Your videos helped me out a lot when I was taking Cal 1 and Cal 2. Thanks a lot!

30. Shae Cloud says:

Thank You SOOO MUCH ProfRobBob!!! // IMO you make some of the best C∆LC videos available. I was with you through C∆LC | and now I'm here for C∆LC || — Also, shout out to Math BFF / Patrick JMT / Khan / The Lazy Engineer and Krista King * You guys are the teachers of the information age!

31. Kolby Dinh says:

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32. Yousef Alamri says:

lovely shirt!

33. Kyle b says:

Hello Prof, do you have any videos on Logistic functions? Enjoy your videos great work and best vids I have seen on math since viewing!

34. Patheresa McCamy says:

Could you direct me to a video discussing 16t^2?

35. akhtar ali says:

36. akhtar ali says:

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37. Nat135 H26 says:

BAM! MR. TERRUEAU…… wait every time for that part lol

38. Anina Nazirah says:

Nice handwriting

39. Rithik Boindala says:

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