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, 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. 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 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 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 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 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 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 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 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 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 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 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, 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. 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 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. 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) 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 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 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, 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 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 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 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. 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 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. 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 a constant because we're doing indefinite
integrals. All right let's do another example. 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 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 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 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. 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 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. 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 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. 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 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. - 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 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 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, for example if you want to get I don't know I(6), 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 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 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 which can be solved just by direct
integration. The next simplest of these are the so-called separable. 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), 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 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. 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. 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 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 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 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 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 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? 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 allow if y is plus minus 1 is included in the
solution if we allow C to be 0 in the answer. 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!