Uploaded by MIT on 10.01.2011

Transcript:

PROFESSOR: Hi.

Welcome back to recitation.

In lecture you've been talking about implicitly defined

functions and implicit differentiation.

So one of the reasons that these are important is, or

that implicit differentiation is important, is it sometimes

you have a function to find implicitly and you

can't solve for it.

You don't have any algebraic method for computing the

function values as a formula, say.

So, for example, this function that I've written on the board

that I've called w of x is defined implicitly by the

equation that w of x plus 1 quantity times e to the w of x

is equal to x for all x.

So this function, some of its values you can guess.

Like at x equals 0, the function value is going to be

negative 1.

And the reason is that this can't ever be 0, so the only

way to get this side to be 0 is if w is negative 1 if

this term is 0.

So some of its values are easy to compute, but some of its

values aren't.

So for example, if I asked you what w of 3/2

is, it's very hard.

There's no algebraic way you can manipulate this equation

that will let you figure that out.

So in that situation you might still care about what the

function value is.

So what can you do?

Well, you can try and find a numerical approximation.

So in this problem I'd like you to try and estimate the

value w of 3/2 by using a linear approximation to the

function, to the curve--

yeah.

A linear approximation of the function w of x in order to

compute this value.

So as a hint, I've given you, so you're trying to

compute w of 3/2.

As a hint I'm pointing out to you that w of 1 is 0.

Right?

If you put in x equals 0 and w of 0 equals--

sorry-- if you put in x equals 1 and w of 1 equals 0 on the

left hand side, you do indeed get 1, as you should.

So, OK.

So, good.

So that'll give you a hint about where you could base

your linear approximation.

So why don't you pause the video, take a few minutes to

work this out, come back, and we can work it out together.

All right.

Welcome back.

So hopefully you've had a chance to work on this

question a little bit.

So in order to do this linear approximation that we want,

what we need to know are we need to know a base point and

we need to know the derivative of the function

at that base point.

And those are the two pieces of data you need in order to

construct a linear approximation.

So we have a good candidate for a base point here, which

is the point 1, 0.

So this curve, whatever it looks like, it passes through

the point 1, 0.

And that's the point we're going to use for our

approximation.

So we're going to use the linear approximation w of x is

approximately equal to w prime of 1 times x minus 1 plus w of

1 when x is approximately equal to 1.

So this is the linear approximation we're going to

use, and we have that w of 1 here is 0.

So this is, this is equal to w prime of one

times x minus one.

Just the w of 1 is 0.

It just goes away.

So in order to estimate w of x, and in particular w of 3/2,

what we need to know is we need to know the

derivative of w.

OK?

And to get the derivative of w, we need to use--

well, we have only one piece of information about w.

Which is we have that it's defined by

this implicit equation.

So in order to get the derivative of w we have to use

implicit differentiation.

OK?

So let's do that.

So if we implicitly differentiate this equation--

so let's start with the, the right hand side is going to be

really easy.

Right?

We're going to differentiate with respect to x.

The right hand side is going to be 1.

On the left hand side is going to be a little more

complicated.

We have a product and then this piece, we're going to

have a chain rule situation.

Right?

We have e to the w of x.

So, OK.

So we're going to take an implicit

derivative and on the left--

so OK, so the product rule first. We take the derivative

of the first part, so that's just w prime of x times the

second part-- that's e to the w of x--

plus the first part-- that's w of x plus 1--

times the derivative of the second part.

So the second part is e to the w of x.

So that gives me an e to the w of x times w prime of x.

That's the chain rule.

So that's what happens when I differentiate

the left hand side.

And on the right hand side I take the derivative

of x and I get 1.

OK, good.

So now I've got this equation and I need to solve this

equation for w prime.

Because if you look up here, that's what I want.

I want a particular value of w prime.

And as always happens in implicit differentiation, from

the point of view of this w prime it's only involved in

the equation in a very simple way.

So there's it multiplied by functions of x and w of x, but

not, you know, it's just multiplied by something that

doesn't involve w prime at all.

And then here it's multiplied by something that doesn't

involve w prime at all.

So you can just collect your w prime's and divide through.

You know, it's just like solving a linear equation.

So here if we collect our w prime's, this is w prime of x

times-- looks like--

w of x plus 2 times e to the w of x.

Did I get that right?

Looks good.

OK, so that's still equal to 1.

So that means that w prime of x is just--

well, just, you know--

it's equal to 1 over w of x plus 2 times e to the w of x.

OK, so this is true for every x.

But I don't need this equation for every x.

I just need the particular value of w prime at 1.

So that's, so I'm going to take this equation, then, and

I'm just going to put in x equals 1.

So I put in x equals 1--

well, let me do it over here-- so I get w prime of 1.

And I just everywhere I had an x, I put in a 1.

So actually, in this equation the only place x appears is in

the argument of w.

So this is w of 1 plus 2 times e to the w of 1.

OK.

So in order to get w prime of 1 I need to

know what w of 1 is.

But I had that.

I had it, it was right back here.

There was, that was my hint to you.

Right, this is why we're using this point as a base point,

which is we know the value of w for this value of x.

So we take that value.

So w of 1 is 0.

So this is just 1 over--

well, 0 plus 2 is--

2, and e to the 0 is 1.

So it's just 1 over 2.

Sorry, that's written a little oddly.

We can make it 2 times 1.

So 1 over 2.

OK.

So I take that back upstairs to this

equation that I had here.

And I have that w of x is approximately equal to w prime

of 1 times x minus 1.

So w of x is approximately equal to w prime of 1--

we saw is--

1/2 times x minus 1.

And that approximation was good near our base point.

So that's good when x is near 1.

All right.

And then, so this is the linear approximation.

And I asked for the linear approximation, its value at

the particular point, x equals 3/2.

So w of 3/2 is approximately 1/2 times--

well, 3/2 minus 1 is also 1/2--

so this is a quarter.

OK, so this is our estimate for w of 3/2.

w of 3/2 is approximately 1/4.

If you wanted a better estimate you could try

iterating this process.

Now you might have a, you know--

or choosing some base point even closer if you could

figure out the value of w and x near that, near this point

that you're interested in--

3/2.

So just to sum up what we did was we had this implicitly

defined function w.

We wanted to estimate its value at a point where we

couldn't compute it explicitly.

So what we did was we did our normal linear

approximation method.

Right?

So we wrote down our normal linear approximation formula.

The only thing that was slightly unusual is that we

had to use implicit differentiation.

In order to compute the derivative that appears in the

linear approximation, we implicitly differentiated.

OK?

So that happened just like normal, and then at the end we

plugged in the values that we were interested in, to

actually compute the particular value of that

approximation.

So I'll end there.