The topic for today is how to change variables.
So, we're talking about substitutions and differential
equations, or changing variables.
That might seem like a sort of fussy thing to talk about in the
third or fourth lecture, but the reason is that so far,
you know how to solve two kinds of differential equations,
two kinds of first-order differential equations,
one where you can separate variables, and the linear
equation that we talked about last time.
Now, the sad fact is that in some sense, those are the only
two general methods there are, that those are the only two
kinds of equations that can always be solved.
Well, what about all the others?
The answer is that to a great extent, all the other equations
that can be solved, the solution can be done by
changing the variables in the equation to reduce it to one of
the cases that we can already do.
Now, I'm going to give you two examples of that,
two significant examples of that today.
But, ultimately, as you will see,
the way the equations are solved is by changing them into
a linear equation, or an equation where the
variables are separable. However, that's for a few
minutes. The first change of variables
that I want to talk about is an almost trivial one.
But it's the most common kind there is, and you've already had
it in physics class. But I think it's so important
in the science and engineering subjects that it's a good idea,
even in 18.03, to call attention to it
explicitly. So, in that sense,
the most common change of variables is the one simple one
called scaling. So, again, the kind of equation
I'm talking about is a general first-order equation.
And, scaling simply means to change the coordinates,
in effect, or axes, to change the coordinates on
the axes to scale the axes, to either stretch them or
contract them. So, what does the change of
variable actually look like? Well, it means you introduce
new variables, where x1 is equal to x times
something or times a constant. I'll write it as divided by a
constant, since that tends to be a little bit more the way people
think of it. And y, the same.
So, the new variable y1 is related to the old one by an
equation of that form. So, a, b are constants.
So, those are the equations. Why does one do this?
Well, for a lot of reasons. But, maybe we can list them.
You, for example, could be changing units.
That's a common reason in physics.
Changing the units that he used, you would have to make a
change of coordinates of this form.
Perhaps the even more important reason is to,
sometimes it's used to make the variables dimensionless.
In other words, so that the variables become
pure numbers, with no units attached to them.
Since you are well aware of the tortures involved in dealing
with units in physics, the point of making variables,
I'm sorry, dimensionless, I don't have to sell that.
Dimensionless, i.e.
no units, without any units attached.
It just represents the number three, not three seconds,
or three grahams, or anything like that.
And, the third reason is to reduce or simplify the
constants: reduce the number or simplify the constants in the
equation. Reduce their number is self
explanatory. Simplify means make them less,
either dimensionless also, or if you can't do that,
at least less dependent upon the critical units than the old
ones were. Let me give you a very simple
example which will illustrate most of these things.
It's the equation. It's a version of the cooling
law, which applies at very high temperatures,
and it runs. So, it's like Newton's cooling
laws, except it's the internal and external temperatures vary,
what's important is not the first power as in Newton's Law,
but the fourth power. So, it's a constant.
And, the difference is, now, it's the external
temperature, which, just so there won't be so many
capital T's in the equation, I'm going to call M,
to the forth power minus T to the forth power.
So, T is the internal
temperature, the thing we are interested in.
And, M is the external constant, which I'll assume,
now, is a constant external temperature.
So, this is valid if big temperature differences,
Newton's Law, breaks down and one needs a
different one. Now, you are free to solve that
equation just as it stands, if you can.
There are difficulties connected with it because you're
dealing with the fourth powers, of course.
But, before you do that, one should scale.
How should I scale? Well, I'm going to scale by
relating T to M. So, that is very likely to use
is T1 equals T divided by M.
This is now dimensionless because M, of course,
has the units of temperature, degrees Celsius,
degrees absolute, whatever it is,
as does T. And therefore,
by taking the ratio of the two, there are no units attached to
it. So, this is dimensionless.
Now, how actually do I change the variable in the equation?
Well, watch this. It's an utterly trivial idea,
and utterly important. Don't slog around doing it this
way, trying to stuff it in, and divide first.
Instead, do the inverse. In other words,
write it instead as T equals MT1, the reason being that it's
T that's facing you in that equation, and therefore T you
want to substitute for. So, let's do it.
The new equation will be what? Well, dT-- Since this is a
constant, the left-hand side becomes dT1 / dt times M equals
k times M to the forth minus M to the forth T1 to the forth,
so I'm going to factor
out that M to the forth, and make it one minus T1 to the
forth, okay?
Now, I could divide through by M and get rid of one of those,
and so, the new equation, now, is dT1 / dt,
d time, is equal to-- Now, I have k M cubed out
front here. I'm going to just give that a
new name, k1. Essentially,
it's the same equation. It's no harder to solve and no
easier to solve than the original one.
But it's been simplified. For one, I think it looks
better. So, to compare the two,
I'll put this one up in green, and this one in green,
too, just to convince you it's the same, but indicate that it's
the same equation. Notice, so, T1 has been
rendered, is now dimensionless. So, I don't have to even ask
when I solve this equation, oh, please tell me what the
units of temperature are. How you are measuring
temperature makes no difference to this equation.
k1 still has units. What units does it have?
It's been simplified because it now has the units of,
since this is dimensionless and this is dimensionless,
it has the units of inverse time.
So, k1, whereas it had units involving both degrees and
seconds before, now it has inverse time as its
units. And, moreover,
there's one less constant. So, one less constant in the
equation. It just looks better.
This business, I think you know that k1,
the process of forming k1 out of k M cubed is
called lumping constants. I think they use standard
terminology in physics and engineering courses.
Try to get all the constants together like this.
And then you lump them there. They are lumped for you,
and then you just give the lump a new name.
So, that's an example of scaling.
Watch out for when you can use that.
For example, it would have probably been a
good thing to use in the first problem set when you were
handling this problem of drug elimination and hormone
elimination production inside of the thing.
You could lump constants, and as was done to some extent
on the solutions to get a neater looking answer,
one without so many constants in it.
Okay, let's now go to serious stuff, where we are actually
going to make changes of variables which we hope will
render unsolvable equations suddenly solvable.
Now, I'm going to do that by making substitutions.
But, it's, I think, quite important to watch up
there are two kinds of substitutions.
There are direct substitutions. That's where you introduce a
new variable. I don't know how to write this
on the board. I'll just write it
schematically. So, it's one which says that
the new variable is equal to some combination of the old
variables. The other kind of substitution
is inverse. It's just the reverse.
Here, you say that the old variables are some combination
of the new. Now, often you'll have to stick
in a few old variables, too.
But the basic, it's what appears on the
left-hand side. Is it a new variable that
appears on the left-hand side by itself, or is it the old
variable that appears on the left-hand side?
Now, right here, we have an example.
If I did it as a direct substitution,
I would have written T1 equals T over M.
That's the way I define the new variable, which of course you
have to do if you're introducing it.
But when I actually did the substitution,
I did the inverse substitution. Namely, I used T equals T1,
M times T1. And,
the reason for doing that was because it was the capital T's
that faced me in the equation and I had to have something to
replace them with. Now, you see this already in
calculus, this distinction. But that might have been a year
and a half ago. Just let me remind you,
typically in calculus, for example,
when you want to do this kind of integral, let's say,
x times the square root of one minus x squared dx,
the substitution you'd use for
that is u equals one minus x squared,
right? And then, you calculate,
and then you would observe that this, the x dx,
more or less makes up du, apart from a constant factor,
there. So, this would be an example of
direct substitution. You put it in and convert the
integral into an integral of u. What would be an example of
inverse substitution? Well, if I take away the x and
ask you, instead, to do this integral,
then you know that the right thing to do is not to start with
u, but to start with the x and write x equals sine or cosine u.
So, this is a direct substitution in that integral,
but this integral calls for an inverse substitution in order to
be able to do it. And notice, they would look
practically the same. But, of course,
as you know from your experience, they are not.
They're very different. Okay, so I'm going to watch for
that distinction as I do these examples.
The first one I want to do is an example as a direct
substitution.
So, it applies to the equation of the form y prime equals,
there are two kinds of terms on the right-hand side.
Let's use p of x, p of x times y plus q of x
times any power whatsoever of y.
Well, notice, for example,
if n were zero, what kind of equation would
this be? y to the n would be
one, and this would be a linear equation, which you know how to
solve. So, n equals zero we already
know how to do. So, let's assume that n is not
zero, so that we're in new territory.
Well, if n were equal to one, you could separate variables.
So, that too is not exciting. But, nonetheless,
it will be included in what I'm going to say now.
If n is two or three, or n could be one half.
So anything: even zero is all right.
It's just silly. Any number: it could be
negative. n equals minus five.
That would be fine also. This kind of equation,
to give it its name, is called the Bernoulli
equation, named after which Bernoulli, I haven't the
faintest idea. There were, I think,
three or four of them. And, they fought with each
other. But, they were all smart.
Now, the key trick, if you like,
method, to solving any Bernoulli equation,
let me call another thing. Most important is what's
missing. It must not have a pure x term
in it. And that goes for a constant
term. In other words,
it must look exactly like this. Everything multiplied by y,
or a power of y, two terms.
So, for example, if I add one to this,
the equation becomes non-doable.
Right, it's very easy to contaminate it into an equation
that's unsolvable. It's got to look just like
that. Now, you've got one on your
homework. You've got several.
Both part one and part two have Bernoulli equations on them.
So, this is practical, in some sense.
What do we got? The idea is to divide by y to
the n. Ignore all formulas that you're
given. Just remember that when you see
something that looks like this, or something that you can turn
into something that looks like this, divide through by y to the
nth power, no matter what n is. All right, so y prime over y to
the n is equal to p of x times one over y to the n minus one,
right, plus q of x.
Well, that certainly doesn't look any better than what I
started with. And, in your terms,
it probably looks somewhat worse because it's got all those
Y's at the denominator, and who wants to see them
there? But, look at it.
In this very slightly transformed Bernoulli equation
is a linear equation struggling to be free.
Where is it? Why is it trying to be a linear
equation? Make a new variable,
call this hunk of it in new variable.
Let's call it V. So, V is equal to one over y to
the n minus one.
Or, if you like, you can think of that as y to
the one minus n. What's V prime?
So, this is the direct substitution I am going to use,
but of course, the problem is,
what am I going to use on this? Well, the little miracle
happens. What's the derivative of this?
It is one minus n times y to the negative n times y prime
In other words,
up to a constant, this constant factor,
one minus n, it's exactly the left-hand side
of the equation. Well, let's make N not equal
one either. As I said, you could separate
variables if n equals one. What's the equation,
then, turned into? A Bernoulli equation,
divided through in this way, is then turned into the
equation one minus n, sorry, V prime divided by one
minus n is equal to p of x times V plus q of x.
It's linear.
And now, solve it as a linear equation.
Solve it as a linear equation. You notice, it's not in
standard form, not in standard linear form.
To do that, you're going to have to put the p on the other
side. That's okay,
that term, on the other side, solve it, and at the end,
don't forget that you put in the V.
It wasn't in the original problem.
So, you have to convert the problem, the answer,
back in terms of y. It'll come out in terms of V,
but you must put it back in terms of y.
Let's do a really simple example just to illustrate the
method, and to illustrate the fact that I don't want you to
memorize formulas. Learn methods,
not final formulas. So, suppose the equation is,
let's say, y prime equals y over x minus y squared.
That's a Bernoulli equation.
I could, of course, have concealed it by writing xy
prime plus xy prime minus xy equals negative y squared.
Then, it wouldn't look instantly like a Bernoulli
equation. You would have to stare at it a
while and say, hey, that's a Bernoulli
equation. Okay, but so I'm handing it to
you a silver platter, as it were.
So, what do we do? Divide through by y squared.
So, it's y prime over y squared equals one over x times one over
y minus one.
And now, the substitution, then, I'm going to make,
is for this thing. V equals one over y.
It's a direct substitution.
V prime is going to be negative one over y squared
times y prime. Don't forget to use the chain
rule when you differentiate with respect-- because the
differentiation is with respect to x, of course,
not with respect to y. Okay, so what's this thing?
That's the left-hand side. The only thing is it's got a
negative sign. So, this is minus V prime
equals, one over x stays one over x, one over y.
So, it's V over x minus one.
So, let's put that in standard form.
In standard form, it will look like,
first imagine multiplying it through by negative one,
and then putting this term on the other side.
And, it will turn into V prime plus V over X is equal to one.
So, that's the linear equation
in standard linear form that we are asked to solve.
And, the solution isn't very hard.
The integrating factor is, well, I integrate one over x.
That makes log x. And, e to the log x,
so, it's e to the log x, which is, of course,
just x itself. So, I should multiply this
through by x squared, be able to integrate it.
Now, some of you, I would hope,
just can see that right away, that if you multiply this
through by x, it's going to look good.
So, after we multiply through by x, which I get?
(xV) prime for the-- maybe I shouldn't skip a step.
You are still learning this stuff, so let's not skip a step.
So, it becomes x V prime plus V equals x,
okay? After I multiplied through by
the integrating factor, this now says this is xV prime,
and I quickly check that that, in fact, is what it's equal to,
equals x, and therefore xV is equal to one half x squared plus
a constant. And,
therefore, V is equal to one half x plus C over x.
You can leave it at that form,
or you can combine terms. It doesn't matter much.
Am I done? The answer is,
no I am not done, because nobody reading this
answer would know what V was. V wasn't in the original
problem. It was y that was in the
original problem. And therefore,
the relation is, one is the reciprocal of the
other. And therefore,
I have to turn this expression upside down.
Well, if you're going to have to turn it upside down,
you probably should make it look a little better.
Let's rewrite it as x squared plus 2c,
combining fractions, I think they call it in high
school or elementary school, plus 2c.
How's that? x squared plus 2c divided by 2x.
Now, 2c, you don't call it
constant 2c because this is just as arbitrary to call it c1.
So, I'll call that, so, my answer will be y equals
2x divided by x squared plus an arbitrary constant.
But, to indicate it's different from that one,
I'll call it C1. C1 is
two times the old one, but that doesn't really matter.
So, there's the solution. It has an arbitrary constant in
it, but you note it's not an additive arbitrary constant.
The arbitrary constant is tucked into the solution.
If you had to satisfy an initial condition,
you would take this form, and starting from this form,
figure out what C1 was in order to satisfy that initial
condition. Thus, Bernoulli equation is
solved. Our first Bernoulli equation:
isn't that exciting? So, here was the equation,
and there is its solution. Now, the one I'm asking you to
solve on the problem set in part two is no harder than this,
except I ask you some hard questions about it,
not very hard, but a little hard about it.
I hope you will find them interesting questions.
You already have the experimental evidence from the
first problem set, and the problem is to explain
the experimental evidence by actually solving the equation in
the scene. I think you'll find it
interesting. But, maybe that's just a pious
hope. Okay, I like,
now, to turn to the second method, where a second class of
equations which require inverse substitution,
and those are equations, which are called homogeneous,
a highly overworked word in differential equations,
and in mathematics in general. But, it's unfortunately just
the right word to describe them. So, these are homogeneous,
first-order ODE's. Now, I already used the word in
one context in talking about the linear equations when zero is
the right hand side. This is different,
but nonetheless, the two uses of the word have
the same common source. The homogeneous differential
equation, homogeneous newspeak, is y prime equals,
it's a question of what the right hand side looks like.
And, now, the supposed way to say it is, you should be able to
write the right-hand side as a function of a combined variable,
y divided by x. In other words,
after fooling around with the right hand side a little bit,
you should be able to write it so that every time a variable
appears, it's always in the combination y over x.
Let me give some examples. For example,
suppose y prime were, let's say, x squared y divided
by x cubed plus y cubed.
Well, that doesn't look in that form.
Well, yes it is. Imagine dividing the top and
bottom by x cubed. What would you get?
The top would be y over x, if you divided it by x
cubed. And, if I divide the bottom by
x cubed, also, which, of course,
doesn't change the value of the fraction, as they say in
elementary school, one plus (y over x) cubed.
So, this is the way it started
out looking, but you just said ah-ha, that was a homogeneous
equation because I could see it could be written that way.
How about another homogeneous equation?
How about x y prime? Is that a homogeneous equation?
Of course it is: otherwise, why would I be
talking about it? If you divide through by x,
you can tuck it inside the radical, the square root,
if you remember to square it when you do that.
And, it becomes the square root of x squared over x squared,
which is one, plus y squared over x squared.
It's homogeneous.
Now, you might say, hey, this looks like you had to
be rather clever to figure out if an equation is homogeneous.
Is there some other way? Yeah, there is another way,
and it's the other way which explains why it's called
homogeneous. You can think of it this way.
It's an equation which is, in modern speak,
invariant, invariant under the operation zoom.
What is zoom? Zoom is, you increase the scale
equally on both axes. So, the zoom operation is the
one which sends x into a times x,
and y into a times y.
In other words, you change the scale on both
axes by the same factor, a.
Now, what I say is, gee, maybe you shouldn't write
it like this. Why don't we say,
we introduce, how about this?
So, think of it as a change of variables.
We will write it like that. So, you can put here an equals
sign, if you don't know what this meaningless arrow means.
So, I'm making this change of variables, and I'm describing it
as an inverse substitution. But of course,
it wouldn't make any difference.
It's exactly the same as the direct substitution I started
out with underscaling. The only difference is,
I'm not using different scales on both axes.
I'm expanding them both equally.
That's what I mean by zoom. Now, what happens to the
equation? Well, what happens to dy over
dx? Well, dx is a dx1.
dy is a dy1.
And therefore, the ratio, dy by dx is the same
as dy1 over dx1.
So, the left-hand side becomes dy1 over dx1,
and the right-hand side becomes F of, well, y over x is the same
as y over, since I've scaled them equally,
this is the same as y1 over x1.
So, it's y1 over x1, and the net effect is I simply,
everywhere I have an x, I change it to x1,
and wherever I have a y, I change it to y1,
which, what's in a name? It's the identical equation.
So, I haven't changed the equation at all via zoom
transformation. And, that's what makes it
homogeneous. That's not an uncommon use of
the word homogeneous. When you say space is
homogeneous, every direction, well, that means,
I don't know. It means, okay,
I'm getting into trouble there. I'll let it go since I can't
prepare any better, I haven't prepared any better
explanation, but this is a pretty good one.
Okay, so, suppose we've got a homogeneous equation.
How do we solve it? So, here's our equation,
F of y over x. Well, what substitution would
you like to make? Obviously, we should make a
direct substitution, z equals y over x.
So, why did he say that this was going to be an example of
inverse substitution? Because I wanted to confuse
you. But look, that's fine.
If you write it in that form, you'll know exactly what to do
with the right-hand side. And, this is why everybody
loves to do that. But for Charlie,
you have to substitute into the left-hand side as well.
And, I can testify, for many years of looking with
sinking heart at examination papers, what happens if you try
to make a direct substitution like this?
You say, oh, I need z prime.
z prime equals, well, I better use the quotient
rule for differentiating that. And, it comes out this long,
and then either a long pause, what do I do now?
Because it's not at all obvious what to do at that point.
Or, much worse, two pages of frantic
calculations, and giving up in total despair.
Now, the reason for that is because you tried to do it
making a direct substitution. All you have to do instead is
use it, treat it as an inverse substitution,
write y equals zx. What's the motivation for doing
that? It's clear from the equation.
This goes through all of mathematics.
Whenever you have to change a variable, excuse me,
whenever you have to change a variable, look at what you have
to substitute for, and focus your attention on
that. I need to know what y prime is.
Okay, well, then I better know what y is.
If I know what y is, do I know what y prime is?
Oh, of course. y prime is z prime x plus z
times the derivative of this factor, which is one.
And now, I turned with that one
stroke, the equation has now become z prime x plus z is equal
to F of z. Well, I don't know.
Can I solve that? Sure.
That can be solved because this is x times dz / dx.
Just put the z on the other side, it's F of z minus z.
And now, this side is just a
function of z. Separate variables.
And, the only thing to watch out for is, at the end,
the z was your business. You've got to put the answer
back in terms of z and y. Okay, let's work an example of
this. Since I haven't done any
modeling yet this period, let's make a little model,
differential equations model. It's a physical situation,
which will be solved by an equation.
And, guess what? The equation will turn out to
be homogeneous. Okay, so the situation is as
follows. We are in the Caribbean
somewhere, a little isolated island somewhere with a little
lighthouse on it at the origin, and a beam of light shines from
the lighthouse. The beam of light can rotate
the way the lighthouse beams. But, this particular beam is
being controlled by a guy in the lighthouse who can aim it
wherever he wants. And, the reason he's interested
in aiming it wherever he wants is there's a drug boat here,
[LAUGHTER] which has just been caught in the beam of light.
So, the drug boat, which has just been caught in a
beam of light, and feels it'd a better escape.
Now, the lighthouse keeper wants to keep the drug boat;
the light is shining on it so that the U.S.
Coast Guard helicopters can zoom over it and do whatever
they do to drug boats, --
-- I don't know. So, the drug boat immediately
has to follow an escape strategy.
And, the only one that occurs to him is, well,
he wants to go further away, of course, from the lighthouse.
On the other hand, it doesn't seem sensible to do
it in a straight line because the beam will keep shining on
him. So, he fixes the boat at some
angle, let's say, and goes off so that the angle
stays 45 degrees. So, it goes so that the angle
between the beam and maybe, draw the beam a little less
like a 45 degree angle. So, the angle between the beam
and the boat, the boat's path is always 45
degrees, goes at a constant 45 degree angle to the beam,
hoping thereby to escape. On the other hand,
of course, the lighthouse guy keeps the beam always on the
boat. So, it's not clear it's a good
strategy, but this is a differential equations class.
The question is, what's the path of the boat?
What's the boat's path? Now, an obvious question is,
why is this a problem in differential equations at all?
In other words, looking at this,
you might scratch your head and try to think of different ways
to solve it. But, what suggests that it's
going to be a problem in differential equations?
The answer is, you're looking for a path.
The answer is going to be a curve.
A curve means a function. We are looking for an unknown
function, in other words. And, what type of information
do we have about the function? The only information we have
about the function is something about its slope,
that its slope makes a constant 45° angle with the lighthouse
beam. Its slope makes a constant
known angle to a known angle. Well, if you are trying to find
a function, and all you know is something about its slope,
that is a problem in differential equations.
Well, let's try to solve it. Well, let's see.
Well, let me draw just a little bit.
So, here's the horizontal. Let's introduce the
coordinates. In other words,
there's the horizontal and here's the boat to indicate
where I am with respect to the picture.
So, here's the boat. Here's the beam,
and the path of the boat is going to make a 45° angle with
it. So, this is the path that we
are talking about. And now, let's label what I
know. Well, this angle is 45°.
This angle, I don't know, but of course I can calculate
it easily enough because it has to do with, if I know the
coordinates of this point, (x, y), then of course that
horizontal angle, I know the slope of this line,
and that angle will be related to the slope.
So, let's call this alpha. And now, what I want to know is
what the slope of the whole path is.
So, y prime-- let's call y equals y of x,
the unknown function whose path, whose graph is going to be
the boat's path, unknown graph.
What's its slope? Well, its slope is the tangent
of the sum of these two angles, alpha plus 45°.
Now, what do I know? Well, I know that the tangent
of alpha is how much? That's y over x.
In other words,
if this was the point, x over y, this is the angle it
makes with a horizontal, if you think of it over here.
So, this angle is the same as that one, and it's y over,
its slope of that line is y over x.
So, the tangent of the angle is y over x.
How about the tangent of 45°? That's one, and there's a
formula. This is the hard part.
All you have to know is that the formula exists,
and then you will look it up if you have forgotten it,
relating the tangent or giving you the tangent of the sum of
two angles, and you can, if you like,
clever, derive it from the formula for the sign and cosine
of the sum of two angles. But, one peak is worth a
thousand finesses. So, it is the tangent of alpha
plus the tangent of 45°. Let me read it out in all its
gory details, divided by one,
so you'll at least learn the formula, one minus tangent alpha
times tangent 45°.
This would work for the tangent of the sum of any two angles.
That's the formula. So, what do I get then?
y prime is equal to the tangent of alpha, which is y over x,
oh, I like that combination, plus one, divided by (one minus
y over x times one).
Now, there is no reason for doing anything to it,
but let's make it look a little prettier, and thereby,
make it less obvious that it's a homogeneous equation.
If I multiply top and bottom by x, it looks prettier.
x plus y over x minus y equals y prime.
That's our differential equation.
But, notice, that let step to make it look
pretty has undone the good work. It's fine if you immediately
recognize this as being a homogeneous equation because you
can divide the top and bottom by x.
But here, it's a lot clearer that it's a homogeneous equation
because it's already been written in the right form.
Okay, let's solve it now, since we know what to do.
We're going to use as the new variable, z equals y over x.
And, as I wrote up there for y
prime, we'll substitute z prime x plus z.
And, with that, let's solve.
Let's solve it. The equation becomes z prime x
plus z is equal to z plus one over one minus z.
We want to separate variables,
so you have to put all the z's on one side.
So, this is going to be x, dz / dx equals this thing minus
z, which is (z plus one) over (one minus z) minus z.
And now, as you realize,
putting it on the other side, I'm going to have to turn it
upside down. Just as before,
if you have to turn something upside down, it's better to
combine the terms, and make it one tiny little
fraction. Otherwise, you are in for quite
a lot of mess if you don't do this nicely.
So, z plus one minus z, that gets rid of the z's.
The numerator is one minus z squared over one minus z,
I hope, one, is that right,
(one plus z squared) over (one minus z).
And so, the question is dz,
and put this on the other side and turn it upside down.
So, that will be (one minus z) over (one plus z squared) on the
left-hand side and on the right-hand side,
dx over x. Well, that's ready to be
integrated just as it stands. The right-hand side integrates
to be log x. The left-hand side is the sum
of two terms. The integral of one over one
plus z squared is the arc tangent of z,
maybe? The derivative of this is one
over one plus z squared.
How about the term z over one plus z squared?
Well, that integrates to be a
logarithm. It is more or less the
logarithm of one plus z squared.
If I differentiate this, I get one over one plus z^2
times 2z, but I wish I had negative z
there instead. Therefore, I should put a minus
sign, and I should multiply that by half to make it come out
right. And, this is log x on the right
hand side plus, put in that arbitrary constant.
And now what? Well, let's now fool around
with it a little bit. The arc tangent,
I'm going to simultaneously, no, two steps.
I have to remember your innocence, although probably a
lot of you are better calculators than I am.
I'm going to change this, use as many laws of logarithms
as possible. I'm going to put this in the
exponent, and put this on the other side.
That's going to turn it into the log of the square root of
one plus z squared.
And, this is going to be plus the log of x plus c.
And, now I'm going to make,
go back and remember that z equals y over x.
So, this becomes the arc tangent of y over x equals.
Now, I combine the logarithms.
This is the log of x times this square root, right,
make one logarithm out of it, and then put z equals y over z.
And, you see that if you do
that, it'll be the log of x times the square root of one
plus (y over x) squared,
and what is that? Well, if I put this over x
squared and take it out, it cancels that.
And, what you are left with is the log of the square root of x
squared plus y squared plus a constant.
Now, technically,
you have solved the equation, but not morally because,
I mean, my God, what a mess!
Incredible path. It tells me absolutely nothing.
Wow, what is the screaming? Change me to polar coordinates.
What's the arc tangent of y over x?
Theta. In polar coordinates it's
theta. This is r.
So, the curve is theta equals the log of r plus a constant.
And, I can make even that little better if I exponentiate
everything, exponentiate both sides, combine this in the usual
way, the and what you get is that r is equal to some other
constant times e to the theta.
That's the curve. It's called an exponential
spiral, and that's what our little boat goes in.
And notice, probably if I had set up the problem in polar
coordinates from the beginning, nobody would have been able to
solve it. But, anyone who did would have
gotten that answer immediately. Thanks.