Calculus is one of the grandest achievements of human thought, explaining everything from planetary orbits to the optimal size of a city to the periodicity of a heartbeat. This brisk course covers the core ideas of single-variable Calculus with emphases on conceptual understanding and applications. The course is ideal for students beginning in the engineering, physical, and social sciences. Distinguishing features of the course include: 1) the introduction and use of Taylor series and approximations from the beginning; 2) a novel synthesis of discrete and continuous forms of Calculus; 3) an emphasis on the conceptual over the computational; and 4) a clear, dynamic, unified approach. In this third part--part three of five--we cover integrating differential equations, techniques of integration, the fundamental theorem of integral calculus, and difficult integrals.
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Из курса от партнера University of Pennsylvania
Calculus: Single Variable Part 3 - Integration
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Integrating Differential Equations
Our first look at integrals will be motivated by differential equations. Describing how things evolve over time leads naturally to anti-differentiation, and we'll see a new application for derivatives in the form of stability criteria for equilibrium solutions.
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Robert GhristProfessor
Mathematics and Electrical & Systems Engineering
Welcome to Calculus.
I'm Professor Greist, we're about to begin Lecture 19 on
More O.D.E.S. In this lesson we'll consider some new families
of ordinary differential equations, their solutions, and what they're good for.
Along the way we'll learn some new methods and some new applications.
Among the many differential equations that you may encounter,
a few families are worthy of note.
Some ODEs are autonomous, meaning that the right-hand
side depends on x alone and does not depend on t.
Those equations that do depend on t are called non autonomous equations.
They can be very difficult to solve.
Some, however, are solvable by the following separation method.
If dx/dt is in the form of a function of x alone times a function of t alone,
then following the pattern of what we did for the linear differential equation,
we can separate out all of the x terms on the left.
All of the t terms on the right then it makes sense to
integrate both sides if these integrals are computable and
if you can solve for x then you're in good shape.
Let's look at an example.
If you consider a body that is falling under the action of gravity,
you'll notice different behavior depending on its shape.
This is due to drag force.
If we assume a drag force model where that force is proportional
to the velocity of the body then, what do we have?
Well, our body is being acted on by a gravitational force of magnitude m,
the mass, times g, gravitational constant pointed down.
Pointed upwards is the drag force that is assumed proportional to the velocity, v.
By some constant of proportionality, kappa the drag coefficient.
Newton's law says that the mass times the acceleration,
dvdt, is equal to mg minus kappa times v.
Here we have an autonomous and separable, ordinary differential equation.
By doing a little bit of factoring and
algebra, we can move to the left hand side,
the quantity v minus mg over kappa, all in the denominator.
On the right hand side, what is left over is negative kappa
divided by m dt integrating both sides.
What do we do?
On the left, you have to think a moment to see that the entire
derivative of dv over v minus constant is in fact,
the logged of quantity v minus that constant.
On the right-hand side the integral is very easy.
It is negative kappa over m times t.
We'll put our constant of the integration on the right side.
Now, to solve for v, what do we do?
We exponentiate both sides.
On the left we have v- mg over kappa.
On the right we have E to the negative kappa over M times T,
that's times E to the constant of integration we're
just going to call that a new constant with the same name C.
Now solving for V what do we have V is constant E to the negative kappa over MT.
Plus mg over kappa.
Let's think about this for a moment.
If we take a limit as t goes to infinity, what happens?
Well the exponent in the exponential is negative.
That means that term goes to zero.
And goes to zero rapidly.
What remains is a constant velocity, equal to m times g over kapa.
This is called the terminal velocity.
Depending on the value of kappa.
It may be very fast or rather slow.
Separable ODEs also arise in analysis of a fluid leaking from a tank.
This is regulated by Torricelli's Law which relates h,
the height of the liquid with A, the cross sectional area at the top.
The law states that the rate of change of h with respect to t is proportional
to the square root of h, and inversely proportional to A.
If we consider say a cone, or
anything where the cross sectional area is quadratic in h by some constant.
Maybe pi, maybe not.
Then we obtain h prime is proportional to h to the negative three halves.
This is separable.
And moving the h to three halves over to the left side.
And then integrating both sides, yields on the left,
h to the five-halves times two-fifths.
And on the right, a constant times t plus an integration constant.
Solving for h,
gives a linear function of t raised to the two-fifths
power where the constants will depend on the physics in your initial condition.
What this implies is that h approaches zero very rapidly at the end.
How rapidly?
It is in O of the final time minus t raised
to the two-fifths power, that comes into zero very very sharply,
when you're looking at something to a power of two-fifths.
Let's move on to a different class of nonautonomous ODEs.
These are the linear nonautonomous ODEs.
There of the form dx dt equals A of t times x
times B of t where A and B are functions of t.
For example, dx dt equals tx plus one half t squared is of this form.
We say that it's a linear equation because if you suppress the t
dependence you get something of the form dxdt equals ax plus b.
That right hand side should remind you of the equation.
The line, well, how do we solve such a problem?
The method that we will introduce is that of the integrating factor.
It's a little tricky so hang on.
Let us denote that integrating factor by I.
We are going to multiply both sides of this equation by an as yet
unknown function I.
Rearranging terms a little bit and multiplying through gives us
I times dx dt minus AI times x.
And on the right I times B.
Now what is this good for?
Well if you look at that left hand side and keep in mind the product rule
then you see something that should remind you a little bit of the product rule if,
if we had something that looked like I times dx dt + dI dt times x.
Well then, we would recognize that that's just the derivative of I times x.
And that's something that's gonna be helpful to us.
So what would we need to be true in order to have this?
We would need to have dI dT = -AI.
Now, focus on that because what we're doing is deriving
a differential equation for this unknown integrating factor I.
That is separable.
That is solvable by the standard techniques that we've been using.
When we solve that equation, what are we going to get?
I is e to the integral of negative A dt,
which we could write in shorthand.
As E to the negative integral A.
If we choose such a function I and multiply both sides of our original
equation by this then what we obtain is on the left the derivative of I times X.
On the right I times B.
Now we can integrate both sides, because differentiation and
integration undo each other.
On the left we have I times X.
On the right, the integral of I times B.
We can solve for x and obtain, after substituting in our formula for I.
A scary looking but very useful equation.
The solution is x is e to the integral of a times the integral
of b times e to the negative integral of a.
Let's demonstrate this formula in the context of a mixing formula.
Assume you have a tank that holds a 1000 gallons and is 90% full.
An additive is injected at a rate of 10 gallons per minute.
The solution is mixed well and
then drained out at a rate of 5 gallons per minute.
The question is what is the concentration of the additive
at that moment when the tank is full?
There are a couple of features to this problem.
We're gonna have to analyze.
First of all, we're talking about when the tank is full so
we've gotta be concerned with volume.
Also, we are told the rate of change of the volume,
that coming in and that going out.
Lastly, what we're asked about is the concentration of the additive in the tank.
Let's set a few variables.
First of all, V is a function of t is our volume.
Next comes Q(t), that's the quantity of the additive that is put into the tank.
So that when we talk about the concentration, C(t),
that is simply Q(t) over V(t).
Let's derive the differential equation and then solve it.
By considering flow rates coming in and out,
its easy to see that the volume is the initial volume, 900 plus 5T,
meaning that the tank will be full at precisely T = 20.
If we consider the time rate of change of the amount of additive, Q,
well that additive comes in at a rate of 10 gallons per minute,
but the fluid is drained out at a rate of 5 gallons per minute.
That drained out fluid is not pure additive,
it depends on a concentration, C.
Thus we derive a differential equation on Q,
as follows, dQ dt + Q /180 + t = 10.
This is a linear, nonautonomous ODE,
where the function a is -1 /180 + t.
That means our integrating factor is e to
the integral of dt/ 180 + t Of course that
integral is simply the log of 180 + t,
and when we exponentiate that we obtain simply 180 + t.
If we therefore multiply the differential equation by
this integrating factor we obtain Very simply 180
plus t times dQ/dt plus Q equals 10 times quantity.
180 plus t, the left hand side is the derivative of Q times quantity 180 plus t.
Integrating both sides gives us on the left 180 plus t plus q.
On the right 1,800 t plus five t squared, plus a constant.
You can show based on the initial condition
that there's no additive at the beginning, that that integration constant is zero.
Now we almost have the solution to our problem.
What do we need to do?
We solve for Q as 1800 t plus 5 t squared over 180 plus t.
But we're asked about the concentration at the time when the tank is full.
At T equals 20 we get the amount of additive to be 190.
Therefore the concentration at that time is 190.
Divided by the volume of the tank, that means that
the additive is present at 19% when the tank is full.
There are many similar types of problems for which integrating factors are helpful.
Some of them are like this involving mixing and tanks.
Others come from entirely different applications.
Indeed, this is just the beginning
of the wonderful subject of differential equations.
Where there many more families of differential equations that require
specialized solution.
A few of these are doable with the tools that you now have in hand.
But many of these differential equations require multi variable calculus.
Some can be very difficult to solve indeed.
We now have in hand several types of ODE's,
their solutions, and their applications.
That's the good news.
The bad news is most differential equations are hard to solve.
Some seem impossible.
In our next lesson,
we'll consider what to do when the solution seems to slip from our grasp.
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