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0:11
Okay, 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 many of you this might be the easiest course here that you'll take in Oxford,
0:46
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. Okay so let me tell you some practical information. So we have 16 lectures. The lecture notes are online.
1:24
These are the lecture notes which were written by Cath 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 point as the lecturer so you can call me Dan my name is Dan Ciubotaru and we'll meet twice a week today is special just because it's the first week. We'll meet on Mondays and Wednesdays at 10am, so not too early and you'll have eight problem sheet. So
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the first two problem sheets are online. The eight problem sheets you'll cover in four tutorials in your college. Okay so four hours for tutorials. So I said the lecture notes are online, the reading list is also online (so see online) the book that I like is Mary Boas's Mathematical Methods in Physical Sciences you know this book and most of your colleges should have a copy of this. If not the university does as well. So
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this is this book is quite concise and it has various examples from physics and engineering and science it also has the added advantage that 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 orthopedic surgeon. All right, so that's that - any questions about this? Okay, now syllabus
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so the first half of the course - about seven or eight lectures will be devoted to differential equations. So this is about seven/eight lectures so of two kinds ordinary differential equations (ODEs) and partial differential equations (or PDEs) so 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 a combination solving them - it's a combination of science and
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art. You have to do some educated guesses at some points but it's quite an interesting and very useful subject and then after that we'll talk about line and double integrals and the reason these are useful is because we will be able to compute arc lengths of curves and areas of various regions in the plane or surfaces. So this is maybe three lectures
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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 among the things that we'll do we'll look at various surfaces gradient normal vectors will look at Taylor's theorem in two variables, critical points and a little bit of Lagrange multipliers, which are useful for optimization problems. Okay, now there is a lot of interaction between this
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course and other premlim courses that you will take. So intro calculus will be directly useful in well obviously multivariable calculus as I said. In a way it's a little bit unfair as we set up the work and we do some examples for in introduction calculus but then the really cool result and theorems you prove in multivariable calculus, so we just do a little bit of the ground work towards
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that. You also do these and they are also useful in dynamics and in PDEs which you will do next term fourier series and PDEs. Now there is a lot of interaction between intro calculus and analysis particularly analysis two which is what you do next term so there will be quite a few results from analysis that will just say and not prove, maybe prove some particular examples and so on, but real rigorous proofs you'll do in analysis next term but then it all comes together when
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you revise or for your exams in Trinity. Okay so there'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 etc so this is a very useful course it's also mandatory so you have to be here. So now let me give you some examples of where all these might appear. Okay so all these DEs, so what is an ordinary differential equation? So this is an equation involving
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an independent variable, let's call it x and a function of x which we call it y. So y this would be the dependent variable and the derivatives of y with respect to x so for example dy/dx d^2y/dx^2 etc you know so the order of the highest derivative that occurs we call that the order of the differential equation. So for example the simplest so the simplest kind of ODE would be something of the form dy/dx equals some function in x so dy/dx equals f(x) - you can solve that
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by direct integration so this can be solved so y equals. So you can think of y as being the antiderivative of f(x) and then we we can use integrations 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 there can be more interesting differential equations, so let me give you some examples from physical sciences. So for example from mechanics,
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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).
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But then the acceleration is a derivative. It is the derivative of the velocity with respect to time. So that's already a differential equation but it could be a second-order differential equation if you think that v is dr/dt where r is the displacement. Then you get for example a is d^2r/dt^2 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
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engineering, or if you have an electrical circuit. So if I take a simple one, so a simple series circuit, which for example: an RLC circuit - which means that it has the following components: it has R stands for resistor, so it has a resistor R, it has an inductor L with inductance L and it has a capacitor with capacitance C and it has a source of voltage something like a battery V.
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So I have a capacitor with C capacitance and the resistor with R resistance and an inductor with L inductance. So here are R, L and C are 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(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(t)
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for example is a dependent variable I can also have Q(t) which is the charge across capacitor on the capacitor and the relation between the two of them is that I=dQ/dt so Kirchoffs law says that the total voltage is zero around the circuit. Which is in a another way saying the voltage V from the battery which is a function of t equals the voltage across the resistor plus the voltage across the resistor plus the voltage on the capacitor and
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now you write each one of them. The voltage across the resistor by Ohm's law is R I(t), the one on the capacitor is just 1/C and for the inductor is L dI/dt which is Faraday's Law. So now I can express for example so I have an equation involving V I and Q but I is dQ/dt so I can rewrite everything in terms of Q, for example. So I can get a differential equation in Q which will be simply: this would be the leading term dI/dt so L times dI/dt becomes L
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times d^2Q/dt^2 + RdQ/dt +1/C Q = V, so that is a second order differential equation that appears in electrical circuits. Yeah so it's second order because the highest derivative is of second order - it has constant coefficients because the constants R,L and C are constant and it's what we'll call inhomogeneous 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,
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so I'll leave one as an exercise for you. So I'll tell you the problem is the rate at which a radioactive substance decays is proportional to the remaining number of atoms. So I want you to, as an exercise, to write a differential equation that describes this situation. Okay so we'll come back to things like this later. So what the question is: what's the differential equation? Okay so as you progress along in this course in the mathematics course here, you will
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encounter very very interesting and sophisticated differential equations in applied mathematics, so we're just scratching the surface a little bit. All right now, going back to what I what I said before - the simplest kind of ODE is dy/dx equals f(x) which you can solve by integration so let me review a couple of facts about integration. So one of the most useful techniques which I'm sure
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most of you are quite familiar with is integration by parts. Okay so where does integration by parts come from? Well it comes from the product rule. If I have two functions f and g and I multiply them and then I differentiate them so f times g 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 dx if they're functions of
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x. It equals f times g minus f prime times g dx. Okay so this is the version indefinite integral's version, you can have a definite integral version where you put the limits of integration.
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So let me spell it out. So this is the definite integral's version. Alright, so let's do a couple of examples. So the first example - so suppose I want to integrate x^2 sin(x). So this would solve, so this would give this gives the solution to dy/dx equals x^2 sin(x) ok so in the integration by parts you have to decide which one is f and which one's g. Now clearly I would like to decrease the power here I know I can never get rid of
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the sine by differentiation so then maybe this then I have to do this f and this is g prime which means that g is -cos x so if I call this integral I I is x squared times minus cos x and then minus the derivative of f which is 2 x times minus cos x dx. This is minus x squared cos x plus 2 times x cos x dx.
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And now again this should be f and this should be g prime of course x plus 2 times x sine x minus 2 times sine x dx. So please try to follow through what I'm doing and let me know if I make a mistake. This is kind of my nightmare to integrate integrals like this while I'm being filmed. Right exactly what I like to do so 2x sin(x) and then minus cos(x) then plus C is this... so plus plus thank you, as I said so C here denotes
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a constant because we're doing indefinite integrals. All right let's do another example.
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So again an indefinite integral: 2x minus 1 times Ln (x^2+1)dx. Ok. What do you think? Which one should be f and which one should be g or g prime. Say that again? Right so this I want to differentiate to get rid of the logarithm, so I should call this f which means that this is going to be g prime, thank you and that makes g: x squared minus x so this becomes x squared minus x ln (x^2+1) minus the integral of x squared minus times the derivative of the
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natural log of x squared plus 1 which is 2x over x squared plus 1 dx so and finally this term. What do I do here? Good so we we do long division, so let's rewrite it first. This is x squared minus x ln x squared plus 1 and then minus 2x cubed minus x squared over x squared plus 1 dx so I have to remember how to do long division, so I have x cubed minus x, now depending how you learn this
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you will draw the the long division in different ways so you just do it your way and I'll do it my way. So that's x minus x cubed minus x then minus x squared minus 6 and that's minus 1. Ok so this means that x cubed minus x squared over x squared plus 1 equals x minus 1 plus minus x plus 1. Did you get the same thing? Okay so let's call this integral J and now we compute J. The integral of x minus 1 plus minus
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x plus 1 over x squared plus 1 dx which equals x squared minus 1/2 x squared minus 6 and then how do I integrate this term? The fraction so I should split x over x squared plus 1 dx. Yeah and let me write the last term plus dx over x squared plus 1. So this 1, the last term we should recognize that - what is it? It's arctan(x) or tan inverse depending how you want to denote it this is arctan of x which is tan inverse of x. Now what do we do with this we can substitute.
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Yeah 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 u equals x squared plus 1 then du equals 2x dx, so du/dx equals 2x which means that this is 1/2 d u / u which is 1/2 Ln of u which is 1/2 Ln of x squared plus 1 that you might have guessed just because you have enough practice some of you. Okay so now let's put them all together. So J is 1/2 x squared minus x minus 1/2
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Ln x squared plus 1 plus tan inverse of x + some constant, which means that the original integral, the integral in the beginning which I should have called I so that I don't have to roll down the boards, that equals x squared minus x Ln x squared plus 1 minus twice this. So minus x squared plus 2x plus Ln x squared plus 1 minus tan inverse x and then plus 2c. Thank you. Any other mistakes? Alright.
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Okay so that's a that's an intro. There are cases when integration by parts will not simplify either of the two functions f and g but what happens is if you do it twice then you sort of come back to what you started with. So the typical example is I equals the integral e to the x times sine x dx. 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 it so for example I can say that this
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is g prime and this is f and then I integrate I get cos and then I do it again and I will end up with some expression - they seem to grow and then I solve for it. So you do this and you get the answer to be something like 1/2 e to the x sin(x) minus cos x then plus constant.
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Okay so another type of example 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 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, suppose I'm looking at this integral: cosine to the n x dx. Now I want to label this
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integral I(n) because I'm going to get a formula of I(n) in terms of I(n-1) or I(n-2) etc. Now there is not much choice here what you should call f and what you should call g so I'm going to just do it. So this is cos^(n-1) x times cos x dx so this is f and this is g prime, then we get cos n minus 1 x sin(x) minus the integral. Now I need to differentiate f so (n-1) cos n minus 2 x and then minus sine x and then another sine x dx, which equals cos n minus 1 x sine x.
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- n - 1 times, or maybe I'll make it a plus, and minus 2 x sine squared x dx. So if I write it like that what do you do now? You write sine squared as 1 minus cos squared x, which then gives you cos n minus 1 x sine x plus n minus 1, the integral of cos n minus 2 x dx minus n minus 1 the integral of cos n squared x dxm so now I recognize that this is the integral of course n minus 2 is I sub n
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minus 2 and the integral of cos, and this is I(n) so I have I(n) equals that so if I solve for I(n) we get I(n) that n I(n) equals cos^(n-1)(x)sin(x)+ (n-1)I(n-2), which gives me the recursive formula
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I(n) equals 1 over n cos (n minus 1) x sine x plus n minus 1 over n. So this is true for all n greater than or equal to 2. Okay now if I want to know all of these integrals I(n) using this formula - what else do I need to know? I(0) and I(1) because it drops down by 2. So let's compute I(0) and I(1), so I(0) would be just the integral dx which is x plus C, and I(1) is the integral of cos x dx which is sine x plus C and now with this you can you can get any integral you want,
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for example if you want to get I don't know I(6),
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you just follow that and you get that it's 1/6 cos^5(x)sin(x) plus 5 over 6 times I(4) which is 1 over 6 cos^5(x) sine x plus 5 over 6 times I(4) is 1/4 cos cubed x sine x plus 3/4 I(2), then what is I(2) is 1/2 cos x sine x plus one-half I zero but I zero is x so you put your substitute this in there and I get an I six is 1/6 cos to the fifth x sine x plus five over 24 cos cubed x sine x plus 5 times 3 times 1 over 6 times 4 times 2 cos x sine x plus 5 times 3 over
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over 6 times 4 times 2 x, so it has 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 sum involving all the terms I think you can you can get the coefficients of each term inductively. Okay so this is a quick review of integration by parts. If you're not fully comfortable with these examples or similar examples, then get get an integration textbook and do a
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few more examples with integration by parts, substitutions and so on because in solving differential equations we'll learn a lot of techniques but ultimately you will have to integrate some functions so you should be able to do that. What we learn is how to reduce the problem to integrating various functions but you'll have to be able to do that. Okay so we discussed about the simplest kind of these
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which can be solved just by direct integration. The next simplest of these are the so-called separable.
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so we had the case dy/dx equals f(x) which you can just integrate the next case would be dy/dx equals a(x) b(y). So what I mean by that is that this is a function in only x and similar b of y is a function of y 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 1 over b(y) dy dx equals a(x),
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and now you can integrate just as we did before. So you'll get then the integral so the left-hand side is the integral dy over b(y) and the right-hand side is a(x)dx and now you have to direct integrations which hopefully we can we can solve, right. The type of integrals that we have in this course will be the kind for which you can apply integration by parts or some other techniques and solve them if I if I
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were to write an arbitrary function there and ask you how to integrate it then we can't do that in a closed formula. Okay so 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 1 plus y x squared minus 1 dy/dx equals to 0 and x is between 0 & 1 to avoid some issues about continuity or whatnot.
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Okay, how do you separate this differential equation? How do you separate the variables? Correct, so I think you're about two steps ahead of me. 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 dy/dx equals minus xy squared minus one and then separate the variables, as the name suggests. You have y over y squared minus one dy/dx equals minus x over x squared minus one.
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Okay what do we do now? Correct, so if we look at this then, so we integrate, let's integrate, well let me write one more, so we integrate this and we get y over y squared minus one dy equals minus x over x squared minus 1 dx so now we could do 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 ln of x squared minus 1 then I get 2x over x squared minus 1, so except x is between 0 & 1 so maybe it's
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better to write this as 1 x over 1 minus x squared and get get rid of the minus sign. So then I'll do 1 minus x squared minus 2x over 1 minus x squared so then this is minus Ln of 1 minus x squared
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and a half and then plus a C. Whereas here I will have to put absolute values because I don't know. It's 1/2 Ln so y squared minus 1 in absolute values 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 logarithmic, so let's do that. So I have 1/2. If I move the logarithm in x to the left hand side then I use the property, well it doesn't matter much, it equals C which
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means that the equation will be y squared minus 1 times 1 minus x squared absolute value equals it would be e to C squared or e to to C which I can just call another kind of C and this would be a positive number so the equation then that we get is so the the answer then is
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where C is positive. But I can relax that, this is always positive because 1 minus x squared is always positive, because I'm assuming x is between 0 & 1 but I can rewrite the answer in a nicer form by dropping the absolute value and dropping the assumption on C. So another way of this for uniformity, I'll write it as 1 minus y squared equals C so 1 minus y squared times 1 minus x squared equals C. No assumption on C except so C could be both positive or negative
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except in this formula it looks like it can't be 0, right? This here I got an exponential which is never 0 so this is positive, I drop the absolute value and now C can also be negative but somehow 0 is missing. How is that possible? That doesn't look like solid mathematics? Yes?
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That's right. Okay, so where did I lose that case? Right here. So I divided by y squared minus one. I did that ignoring the case when y squared minus one is zero, so note so let's call this star here so in star y -1 is y squared minus 1 is not 0 but if we need to allow that because it is possible for y to be plus or minus 1 for example. If y is the constant function 1 then this is 0 and dy/dx is 0, so that's okay. So if we
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allow if y is plus minus 1 is included in the solution if we allow C to be 0 in the answer.
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So then the bottom line is that the answer is this implicit equation in y and x where C can be any constant. Good. So be careful when when you divide by the function in y, 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!
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