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En provenance du cours de University of Pennsylvania

Calculus: Single Variable Part 4 - Applications

notes

University of Pennsylvania

Calculus: Single Variable Part 4 - Applications

notes

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 fourth part--part four of five--we cover computing areas and volumes, other geometric applications, physical applications, and averages and mass. We also introduce probability.

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Physical Applications

There is so much more to applications of integrals than geometry! So many subjects, from physics to finance, have, at heart, the need for setting up and computing definite integrals. In this short but intense module, we will cover applications including work, force, torque, mass, and present & future value.

Rencontrer les enseignants

Robert Ghrist
Professor
Mathematics and Electrical & Systems Engineering

Welcome to Calculus. I'm Professor Greist, and we're about to

begin Lecture 38, bonus material. Of those topics covered in our main

lesson perhaps the most challenging is that of present value.

To reiterate. Let's say, that I were to give you a

thousand dollars in ten years, how much would it be worth to have right now?

As an equivalent, assuming that you took that money and invested it at a fixed

interest rate. Well, given what we know about such

investments, it would be worth that amount of money $1000 times e to the

negative rt where t is 10 years. That's how much it would be worth right

now. But instead of some fixed amount of money

to be given at some fixed future time. What happens if you have an income

stream, a certain amount of money that's coming in overtime and perhaps varying?

Well, to solve that more challenging problem, we integrate the present value

element. Let's see how that works in a specific

example. Which of the following income streams

would be better for you in terms of having greater present value?

Let's say, I give you the choice between earning income at a fixed rate.

I of t is equal to sum constant C or I say, well, you get less money at the

beginning, say 1 3rd C. But the amount of money that you're

getting increases over time linearly. Let's say I of t equals 1 3rd C times

quantity 1 plus rt. Which would you choose?

In the second case, there would be a time in the future where you're making much

more money than that fixed constant rate C.

Well, let's compute the present values and find out which is better.

We'll begin with the first income stream. I of t is a constant.

In this case, the present value element is e to the negative r t times i of t d

t. That is c e to the negative r t d t.

If we integrate that, well, lets see, what are our bounds on integration?

It wasn't said what sort of time frame we had.

So, let's assume a very, very long lifespan and integrate as T goes from

zero to infinity this present value element.

This integral is not so hard. The constant comes out, we integrate e to

the negative rt and get e to the negative rt over negative r.

When we substitute in t equals 0 and take the limit to t goes to infinity, then we

get simply that constant c over r. That is the present value of this

infinite income stream. Let's see how this compares with an

income stream that is linearly growing, but which starts off much lower.

If we have i of t equals 1 3rd C times 1 plus rt, then our present value element

is that same income stream times d t times e to the negative rt.

Now, integrating that is going to be a little bit more challenging, but not so

bad. It breaks up in to two integrals.

The first Is 1 3rd c, e to the negative rt dt, but that's really the same

integral that we just did for the constant income stream.

But the constant in front is 1 3rd c, therefore the first integral evaluates 2c

over 3r. For the second integral, it's going to

take a bit more work. Let's pull up the constant, Cr over 3 and

then integrate what is leftover using integration by parts.

We're going to let u d equal the t and dv be e to the negative rt dt.

I'm going to leave some of the details to you at this point, to show that what we

get is u times v. That is te over negative r times e to the

negative rt, evaluated from zero to infinity.

Well, what we get when we subtract the integral of vdu is adding the integral of

1 over r e to the negative rt dt. And that of course is a simple integral.

The first term, the uv term, vanishes at those limits.

And we're left with an integral that we can easily solve the net present value of

the second income strain is C divided by 3r plus Cr over by 3r squared that

simplifies to 2 3rd C over by r. When we compare that to the fixed income

stream, which had a present value of C over R, we see that this linearly

increasing stream is worth only 2 3rd as much over time.

That is not an obvious conclusion. Now, there are many, many other examples

in finance and economics that work well, when argued in terms of elements.

I'll leave it to you to think about what future value of an income stream might

be.

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