OK. Welcome to introductory calculus. I will start with some practical information,
and then I'll tell you a little bit about the syllabus and what we will cover in this
course and then give you some examples of differential equations from physical sciences.
And then a little bit of integration towards the end, so for for many of you, this might be the easiest course here that you take in Oxford.
But I think things will get progressively harder.
So maybe in a couple of weeks, it will be interesting to everybody if today's lecture might be a bit too easy for some of you.
OK, so let me tell you some practical information. So we have 16 lectures.
The lecture notes are online. Online, these are the lecture notes, these were written by Kath Wilkins.
She taught this course for a few years until last year, so we'll just follow them.
I guess I should have introduced myself at some points of the lecture in.
So you can call me down, my name is Daniel thorugh and will meet on.
So we'll meet twice a week. Today's special just because he's the first week we'll meet on Mondays and Wednesdays at 10 a.m., So not too early.
And you'll have eight problem sheets. So the first two problem sheets are online.
The problem sheets you'll cover in for tutorials in your college.
OK, so for our our tutorial.
What do I? So I said the lecture notes on a line, the reading list is also line.
So see online. The book that I like is Mary Boluses Mathematical Methods in physical sciences.
You know this book? And most of your colleagues should have a copy of of this, if not, the university does as well.
So this this book is quite concise, and it has various examples from physics and engineering and science,
and it also has the added advantage that if if unlike the other books on on the reading list,
if you drop this one on your foot, you might be able to to walk without seeing an orthopaedic surgeon.
All right, so that's any any questions about this?
OK, now, syllabus. So the first half of the course, about about seven or eight lectures will be devoted to differential equations.
So this is about seven eight lectures. So if two kinds ordinary differential equations, all these and partial differential equations.
Our is. So I'll give you I'll give you some examples very soon, we'll look at fairly easy examples of differential equations.
We'll learn some techniques. It's it's a combination solving them.
It's a combination of science and art. You have to do some educated guesses at some point, but it's it's quite an interesting and very useful subject.
And then after that, we'll talk about lying and double integrals, line integrals and double intervals.
And the reason these are useful is because we will be able to compute arc length.
Of of cars and areas.
Various regions in the plane or surfaces.
So this is maybe three lectures. And then finally, we'll do calculus of functions.
In two variables. So this should be viewed as a gentle introduction into multivariable calculus.
So amongst amongst the things that will do, we'll look at various surfaces, gradients, normal vectors.
We'll look at Taylor's theorem into variables, critical points and a little bit of Lagrange multipliers.
Which are useful for optimisation problems. OK.
Now, there is a lot of interaction between this course and other brilliant courses that you will take.
So intro calculus. Well, be directly useful.
Well, obviously multivariable calculus, as I said.
In a way, it's a little bit unfair. We we set set up the work we do some examples for in introduction calculus.
But then the really cool results and theorems you prove in multivariable calculus.
So we just do a little bit of the groundwork towards that.
You also do. These are also useful in dynamics and in B, the ease of you will do the next term for the series and please.
Now there is a lot of interaction between internal calculus and analysis, particularly analysis tool, which is what you do next term.
So there will be quite a few results. From analysis that will just fade and not prove, maybe prove some particular examples and so on.
But real rigorous proof seal doing analysis next term.
But then it all comes together when you revise or for your exams in Trinity.
OK. So that's that, of course, in part A.
There will be lots of applied mathematics options that will continue this differential equations is a big option, fluid and waves, et cetera.
So this is a very useful course. It's also mandatory. So you have you have to be.
Yes. OK. So now let me give you some examples of where all these might appear.
OK, so all these. So what what is a different ordinary differential equation?
So this is an equation. Involving an independent variable, let's call it X and a function of X.
Which we call it. Why? So why this would be the dependent variable and the derivatives of why?
With respect to X. So for example, DUI, the X D squared, y squared,
etc. So the order of the highest derivative that occurs when you call that the order of the differential equation.
So, for example, the simplest so the simplest kind of, Audie, would be something of the foreign DUI.
The X equals some function in X.
So the wide X equals four works, you can solve that by direct integration, so this can be solved.
So y equals so y you can think of Y as being the entity derivative of F of X and then we can use integration.
That's the simplest kind of differential equation that we can have,
and this is the reason why we'll start the course by reviewing a little bit of integration techniques.
But more interesting, there could be more interesting differential equations.
So let me give you some examples from. From physical sciences.
So, for example, from mechanics, this is something that you have all seen.
You can have Newton's second law.
Which says that the force is the mass times, the acceleration, so a is the acceleration.
But then the acceleration is a derivative, is the derivative of the velocity with respect to time.
So that's already a differential equation, but it could be a second or that differential equation,
if you think that V is B R B T, where R is the displacement?
Then you get, for example, a is the R v squared r d squared, which is a second order differential equation so that.
That's an easy example of how differential equations appear in mechanics.
Well, you could also have differential equations in.
Of engineering or if you have an electrical circuit.
So if I take a simple one, so a simple series circuit, which, for example, an R L C circuit,
which means that it has the following components it has R stands for resistor, so it has a registered R.
It has an inductor L with Inductance L, and it has a capacitor with the party capacitance C.
And it has a source of. Voltage, something like a battery v, so here I am, so I have a capacitor with C capacitance.
And the register. With our assistance and an inductor with L and inductance.
So here are our L and C our constants, they're independent of time.
But then I'm interested, for example, in the current across the circuit, so this is the current.
Across I of T is the current across the circuit, which is a function of the time.
So in terms of differential equations, t the time would be the independent variable.
And this i of T, for example, is a dependent variable.
I can also have Q of T, which is the charge across.
Capacitor. Sorry.
On the capacitor and the relation between the two of them is that I is the Q, the T.
So kick-offs law says that.
The total voltage is zero around the circuit, which in another way, the voltage V from the battery, which is a function of T.
Equals the voltage across the resistor, plus the voltage across the inductive plus the voltage on the capacitor,
and now you're right, each one of them. The voltage across the resistor by ohms law is R times-I of T.
The one on the capacitor is just one over three times the charge.
And for the inductor is El the constant times DADT, which is faraday's.
Hello. So now I can express, for example, so I have an equation involving V I.
And Q, but I is the Q DP so I can rewrite everything in terms of Q, for example.
So I can get a differential equation in Q, which will be simply.
So this would be the leading term DDT, so l times the IDP becomes L times d squared, humidity squared, plus our eye is our times.
Cue the 80 plus one over C times Q equals V, so that is a second order differential equation.
That appears in electrical circuits, you.
So it's second order, because the highest derivative is of second order.
It has constant coefficients because the constants are L, R and C are constant.
And it's what we'll call in homogeneous because this doesn't have to be zero.
So those are the type of differential equations that we can we can study.
And there are many other examples, so I'll leave one as an exercise for you.
And is the exercise? So I'll tell you, the problem is the rate at which a radioactive.
Substance, because.
Is proportional. To the remaining.
A number of. Number of attempts.
So I want you to as an exercise to write the differential equation that describes this situation.
OK, so we'll come back to things like this later.
So what? So the question is, what's the what's the differential equation?
OK. So as as you progress along in this course, in the mathematics course here you will encounter very,
very interesting and sophisticated differential equations in applied mathematics.
So we're just scratching the surface a little. All right, now, going back to what I what I said before,
the simplest kind of ODA is the idea X equals f of X, which you can solve by direct integration.
So let me review a couple of facts about integration.
So one of the most useful techniques, which I'm sure most of you are quite familiar with is integration by parts.
OK, so where does integration by parts come from? Well, it comes from the product rules, product or life needs rule.
If you want to sound fancy. For derivatives.
So if I have two functions, FMG and I multiply them and then I differentiate them, so I have four times.
Prime is f prime g plus f g prime.
Which means that. F g f times, g prime equals f times g prime minus f prime times G.
And if I integrate both sides. Then I end up with the integration by parts, which is.
F times, G Prime B X, if they're functions of X, equals F.
F times, g minus f prime times g, d x.
OK, so this is the version, the indefinite integrals version.
You can have a definite intervallo version where you put the limits of integration.
So let me spell it out. So this is the definite into integrals version.
All right. So let's do a couple of examples.
So the first example. So suppose I want to integrate X squared sign X.
The. Now. So this would solve so this would give this gives the solution.
To DUI, the X equals X squared sine X.
OK, so integration by parts, you have to decide which one is off and which ones?
G. Now clearly I would like to decrease. The power here, I know I can never get rid of the sign by differentiation.
So then maybe this, then I have to do this Earth and this is G Prime, which means that G is minus cos x.
So if I call this interval, I I is.
X squared times minus six and then minus the derivative of F, which is two x times minus cos x.
The ex. So this is minus x squared cos x plus two times x x x.
And now again, this should be F and this should be G Prime.
Score across X plus two times.
X Sign X. Minus two times.
Sign X the X, so please try to follow through what I'm doing and let me know if I make a mistake,
this is this is kind of my nightmare to do integrate into clubs like this while I'm being filmed.
This is not exactly what I like to do. So two x sign x and then minus korsak, then plus C.
Is this. So, so. Plus plus.
Thank you. Good. As I said.
So see here, denotes. A constant.
Because we're doing indefinite integral. All right, let's do another example.
So again, an indefinite integral to X minus one times Alan X squared plus one.
The X. OK. What do you think?
How which one should be f and which ones should be G or G Prime?
Say that again. Right.
So this I want to differentiate to get rid of the logger, so I should call this, which means that this is going to be different.
Thank you. And that makes G X squared minus six.
So this becomes X squared minus six times l n x squared plus one minus the integral.
Of X squared minus six times the derivative of the natural log of X squared plus one, which is two x over x squared plus one d x.
So I'm finally this term. What do I do here?
Good. So we do long division. So let's rewrite it first.
This is x squared minus x l n x squared plus one and then minus two x cubed minus x squared over x squared plus one d x.
So I have to remember how to do long division.
So how x cubed minus x. Now.
Depending how you learn this, you will draw the long division in different ways.
So you just do it your way, and I'll I'll do it my way.
So that's X and minus six cubed. Minus X then.
Well, minus x squared minus six, and that's minus one.
OK, so this means that x cubed minus x squared over x squared plus one equals x minus one.
Plus, minus six plus one over x squared plus one.
Did you get the same thing? Good. OK, so let's call this integral, Jay.
And now we come to you, Jay. The integral of X minus one plus or minus X plus one over x squared plus one.
The X, which equals X squared minus half x squared minus X.
And then. How do I integrate this term?
The fraction? So I should split x over x squared plus one.
The ex. Yeah.
And let me write the last term plus. The X over x squared plus one.
So this one, the last term, we should recognise that what is it?
Are 10 or 10 years, depending how you want to.
The notices are 10 of X, just 10 inverse of X and what do we do with this?
We can substitute, yeah, let let, let's do that so that we remember how to do substitutions.
You might just look at it and know what it is, right, but just to review substitution if I said you equals x squared plus one.
Then deal equals to X the X deal, the X equals to X, which means that this is one half.
But the you over you. Which is one half line of you.
She is one half a line of x squared plus one that you might have guessed.
Just because you have enough practise, some of you.
OK, so now let's put them all together, so Jay is one half x squared minus X.
Minus one half, Alan X squared plus one.
Plus. Ten inverse of X and some constant.
Which means that the original integral, the.
The integral in the beginning. Which I should have called I so that I don't have to roll down the balls but equals.
X squared minus x l and x squared.
Plus one. Minus twice this, so minus x squared plus two x plus Ellen.
X squared plus one. Minus 10 in verse six and then.
Plus to see. Thank you.
Any other mistakes? All right.
OK. So that's a that's an integral.
There are cases when integration by parts will not simplify either of the two functions FMG.
But what happens is if you do it twice, then you sort of come back to.
What you started with, so the typical example. And.
Is. I equals the to e to the X, Senex, the X.
All right. So maybe we don't need to go through the entire calculation.
This is in the lecture notes as well. But how would you solve it?
Right. So you do you do it? So for example, I can say that this is gee prime and this is off.
And then I integrate, I get calls and then I do it again, and I will end up with some expression minus.
They seem to grow and then I sold for it.
So you, you do this and you get the answer to be something like one half e to the X sign X minus six, then plus.
Constant. OK.
So another type of examples, which is which are more difficult,
are the ones which you cannot solve in just one go, but you have to find a recursive formula.
So I'll just do an example like that. You've you've seen other examples before.
So this is when we get a reduction or if you want to call it a recursive formula.
So I start I suppose, I'm looking at this interval.
Co-sign to the end, ex, the ex.
Now I want to label this integral I n because I'm going to get a formula of I n in terms of i n minus one or I n minus two, et cetera.
Now, there is not much choice here, what you should call and.
Mother, you should call. So I'm going to just do it, so this is coarse and minus one x times call six the X.
So this is Earth and this is g prime. Then we get costs and minus one x sine x minus.
The integral now I need to differentiate, if so and minus one costs and minus two x and then minus sine X.
And then another sign X the X. It equals costs and minus one x sine X.
Minus and minus one times, or maybe I'll make it a plus.
Costs. And minus two x sine squared x the X.
So if I write it like that. What do you do now?
He right sine squared as one minus cos squared X.
Which then gives you costs and minus one x sine X.
Plus and minus one, the integral, of course, and minus two x the X.
Minus and minus swan, the interval, of course, and Square X the X.
So now I recognise that this is the integral, of course, and minus two is I sob and minus two.
And the integral, of course, and this is in. So I have an equal that.
So if I sold for, I end. We got I and equals.
So I get an I and equals course and minus one x sine X plus and minus one i n minus two.
Which gives me the recursive formula. I n equals one over n course and minus one x sine x plus and minus one over n i minus.
So this is true for all and greater than or equal to two.
OK. Now, if I want to know all of these integrals, I mean, using this formula.
What else do I need to know? I zero and I one because it drops down by two.
So let's compute zero and one. So we also need I zero, and I won, so I zero would be just the integral the X, which is X plus a c.
And I Wan is the integral, of course, x the X, which is signing X plus C.
And now with this, you can you can get any integral you want.
For example, if you want to get, I don't know, ice.
I six. You just follow that and you get that it's one sixth course.
To the fifth. Sine X plus five over six times I for.
Which is one of the six course. Sine X plus five over six times.
I four is one fourth. Of course.
Cubed X Sine X Plus.
Three, four, two. Then what's eye to eye to his one half cos x sign x plus one half I zero i zero six.
So you put. You substitute this in there and they get an I six is one sixth course to the fifth sine x plus.
Five over 24 cores, cubed sine X.
Plus. Five times, three times, one over six times, four times to cause sine X plus.
Five times three over. I want over six times, four times to X Plus see.
The. So it has I think I think you can you can probably cook up a general formula using this example, you see how it goes.
So if I asked you to write. A Psalm involving all the terms, I think you can you can get the coefficients of each time inductively.
Good. OK, so this is a quick review of integration by parts if if you're not.
Fully comfortable with these examples or similar examples, then get up,
get together an integration textbook and do a few more examples with integration by parts, substitutions and so on.
Because differential equity, what we will do in solving differential equations will learn a lot of techniques.
But ultimately you will have to integrate some, some some functions, so you should be able to do that.
So what we learnt is how do you how to reduce the problem to integrating various functions?
But you'll have to be able to do that. OK, so we discussed about the simplest kind of odds.
Which can be solved just by direct integration. The next simplest.
All these. Are the so-called separable.
Oh, these. So we had the case did X equals f of X, which you can just integrate, the next case would be the I.D. X equals A of X times B of Y.
So what I mean by that is that this is a function only index.
Only, and similarly, B of Y is a function of.
Why only? If you have a situation like that, then you can reduce it to the direct integration with one simple trick.
If B y is not zero, then you divide by it and you get one over B of y d y the X equals a x.
And now you can integrate just as we did before.
So you'll get then the integral. Yes.
So the left hand side is the integral DUI over B of Y and the right hand side is a D X two.
And now you have to direct integration switch.
Hopefully we can we can solve the type of integrals that we have in this course will be the
time for which you can apply integration by parts or some other techniques and solve them.
If I if I were to write an arbitrary function there and ask you how to integrate it, then we can't do that in in a closed formula.
OK, so here here's an example. Find the general solution.
To the separable. Differential equation.
So the hint is already in the problem that this is a separable differential equation
x times y squared minus one plus y x squared minus one d y d x equals two zero.
And. X is between zero and one to avoid.
Some some issues. About continuity or what not?
OK. How do you separate this differential equation?
I'll do separate variables. Correct.
So I think you're. About two steps ahead of me, so I will first.
But it's correct. But let me do it step by step.
So what I will do is first isolate that.
So I have y x squared minus one d y x equals.
Minus x y squared minus one.
And then separate the variables, as the name suggests, you have y over.
Y squared minus one DUI, d x equals minus x over x squared minus Y.
OK. What do we do now? Correct.
So if we look at this, then so we integrate, let's integrate.
Well, let me write one one more, so we integrate this and we get why over y squared minus one d y equals minus x over x squared minus one d x.
So now we could do a substitution, as we did before, but I think we know how to do it.
This looks like the derivative of a logarithm. So if I differentiate a land of X squared minus one.
Then I get to X over X squared minus one.
So except X is between zero and one, so maybe it's better to write this as one x over one minus x squared and get get rid of the minus.
Sign. So then I'll do one minus x squared minus two x over one minus x squared.
So then this is minus 11 of one minus x squared.
And the half. And then plus, see?
Whereas here I'll have to put absolute values because I don't know, it's one half.
It's a long way squared. Minus one.
In absolute value. Right. Now, the easiest way to write this is to get rid of the logarithm by moving this to the other side,
using the properties of the logarithm an exponential rate.
So let's do that. So we have one half. If I move the logarithm in X to the left hand side, then I use the property.
Well, it doesn't matter much. Equals C.
Which means that. The equation will be y squared minus one times one minus six squared.
Absolute value equals it would be easy to see squared.
Or E to to see which I can just call.
Another kind of sea. And this would be a positive no.
So the equation then that we get is. So the answer then is.
Is that? Where a sea.
It's positive, but I can't relax that this is always positive,
because one minus x squared is always positive because I'm assuming X is between zero and one.
But I can rewrite the answer in a nicer form by dropping.
The absolute value and requiring and dropping the assumption on C.
So another way of doing this is.
Well, for you uniformity, I'll write it as one minus y squared equals C, so one minus y squared times one minus x squared equals C.
No assumption on C except. So she could be both positive or negative.
Except in this for me, it looks like it can't be zero. Right here, I got an exponential which is never zero.
So this is positive. I drove the absolute value, and the cost is that now see.
Can also be negative. But somehow zero is missing.
How is that possible? That doesn't look like solid mathematics.
Yes. That's right.
OK, so where did I lose that case? Right here.
So I divide it by white squared minus one. I have I did that.
Ignoring the case when y squared minus one is zero.
So note, so let's call this story here, so in star.
Why minus one is why squared minus one is not zero.
But if we need to allow that because it is possible for white it to be plus or minus one, for example.
If why is the cost and function one, then this is zero.
The idea is zero, so that's OK. So if we allow, if y is plus plus minus one is included.
In the solution. If we allow.
See, to be zero. You know, in the answer.
So then the bottom line is that the answer is. This implicit equation in Y and X, where a C can be any constant.
Good. So be careful the when when you divide by the functioning, why, as I said, here you can do that if you know that's not zero.
But sometimes you get solutions from it being zero, so you have to be careful there.
All right. That's the end of the first lecture, I'll see you tomorrow for for the second lecture and we'll do more differential equations.