INSTRUCTORS Raffi Hovasapian John Zhu

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• ## Transcription

 1 answerLast reply by: Professor HovasapianThu Jul 7, 2016 7:10 PMPost by Haleh Asgari on July 6, 2016Hello Professor Raffi, I really enjoy your lectures but I do not get what you did with the absolute value at the end of Ex 2. How did you get < (5-x)^(1/3) + 0.1 and > (5-x)^(1/3) + 0.1?Thanks in advance, HA 2 answersLast reply by: Professor HovasapianThu Apr 7, 2016 2:02 AMPost by Acme Wang on April 1, 2016Hi Professor,I am little confused in Example IV. Why dr means the error in radius? And why dA is the error in Area?Also, What is the graph of L(x)? Is L(x) the tangent line at x sub0?P.S: I really appreciate your classes and they are really really super useful! Thank you very much!Sincerely,Acme

### Linear Approximations & Differentials

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

• Intro 0:00
• Linear Approximations & Differentials 0:09
• Linear Approximations & Differentials
• Example I: Linear Approximations & Differentials 11:27
• Example II: Linear Approximations & Differentials 20:19
• Differentials 30:32
• Differentials
• Example III: Linear Approximations & Differentials 34:09
• Example IV: Linear Approximations & Differentials 35:57
• Example V: Relative Error 38:46

### Transcription: Linear Approximations & Differentials

Hello, welcome back to www.educator.com, and welcome back to AP Calculus.0000

Today, we are going to talk about linear approximations and differentials.0004

Let us jump right on in.0008

Essentially, what we are going to do is we are going to have some function f(x).0011

Instead of f(x), we are going to come up with something called a linearization of x0023

which we are just going to call l(x).0027

It is going to allow us to use this, instead of this, when we are not too far away from a particular point whose value we do know.0028

Let us jump right on in.0039

I’m going to start off by drawing a picture here.0040

Let me go like this and something like that.0044

Let us see we have got some curve, some function, and then, we have the tangent line.0049

This is going to be our x0 and let us say over here is our x.0059

If we go up, we got this point and this point.0066

This point is going to be our x0, y0.0072

This point over here, this is going to be our xy.0080

It is going to be the y value at another point along the curve.0083

This is our f(x), it is the actual f(x).0088

Now this line here, this line, it is y - y0 = m × x - x0.0093

m is the slope of the line, the derivative.0110

y0 and x0 those are the points that it actually passes through.0112

We know what this is.0117

This height right here, this height is just y0.0120

y is the same as that right there.0128

This height, this is m × x - x0.0131

x - this difference right here, this is x - x0.0138

If I multiply, I have a bigger triangle here.0142

If I'm here, if I move a distance x - x0, the height, this part is nothing more than the slope × the distance.0149

I hope that makes sense.0162

Remember from the slope is Δ y/Δ x =, the Δ y,0163

in other words how high you move is nothing more than the slope × the Δ x.0170

The slope × this is your Δ x, you have your slope.0176

This height right here.0180

Let me write this down now.0183

The whole idea of the derivative and the tangent line which this is to a curve,0185

is to be able to approximate the curve, approximate values along the curve.0208

I will explain what I mean using the picture, after I finish writing this,0222

is to be able to approximate the values along the curve by using the tangent line instead.0226

As long as we do not move to far away from our x0.0244

If we know some point on the curve, this tangent line gives me an approximation.0261

I know if I do not move too far away from x0, either in this direction or in this direction,0266

If I do not move too far away, the values of x along the curve are going to be all these values right here.0273

But if I do not move too far away, then this value right here is a pretty good approximation to this value right here.0283

It gets better and better, the closer I actually stay to x0.0290

That is what this is all about.0295

The whole idea of the derivative is to approximate a complicated curve near a certain point.0297

If I want the value at a point near a point that I know, I do not have to use the function itself.0304

I can just go ahead and use the derivative of the function.0310

It gives me an approximation and that is what this is.0312

Instead of finding this value, I find this value.0316

This value comes from this.0321

If I just move this y0 over to this side, I get this y value is actually equal to y0 + this height.0324

That is what this equation is actually saying.0334

If I rearrange this equation, rearranging the equation of the tangent line0336

gives my y value is actually equal to my y0 value + m × x – x0.0354

This is the value that I really like, but I’m going to approximate it by that.0364

y0 is this height, m × x - 0 is this height.0370

If I add those two heights which is what this says, I get this y value which is a pretty good approximation to that.0376

That is the whole idea, this is the linearization.0386

Now let me do this in terms of f(x).0389

Let us do it again.0392

I'm going to change the labels but it is still the same thing.0395

I have got a graph, I have a tangent line.0402

This is my x0, this is my x.0409

This point right here, this is x sub 0, f(x) sub 0.0418

Let me write this a little bit better.0426

Instead of y, I’m going to call it f(x) sub 0.0427

This point is x sub 0, f(x) sub 0.0431

This point right here, this is nothing more than x f(x).0441

Now this height, let me go ahead and go this height.0453

This is nothing more than f(x0), right.0461

If I go straight across that means this height is nothing more than f(x0).0467

This height is f’ at x0 × x - x0 because the equation of this line is f(x) - f(x0) = f’ at x0 × x – x, y - y0 = m × x - x0.0480

I just replaced y, y0 with f(x) and f(x0).0508

Therefore, this point which is an approximation to this point is this height + this height gives me an approximation.0522

Now we have f(x) rearranging this, moving this over to that side, = approximately f at x0 + f’ at x0 × x - x0.0540

That will take me to that point right there.0559

It is almost equal to that point right there.0562

Again, this is greatly magnified.0565

We call this the linear approximation.0568

We call this the linear approximation of f(x) at x sub 0.0575

It is symbolize with the l.0592

The linear approximation is equal to f(x0) + f’ at x0 × x - x0.0595

It is also called the linearization of f(x) at x0.0607

It is also called the linearization of f(x) at x = 0.0611

When you are given an x0, you are going to find the f(x0), you are going to find the f’.0625

At x0, you are going to put those numbers here and here.0630

You are going to form this thing.0633

Now that you have a linearization, now you can replace f(x).0636

You can replace, now l(x) replaces f(x).0642

That is what is going on here.0660

When you create this linearization, this function, which is going to be some function of x,0661

you can use that instead of f(x) directly.0667

It is an approximation.0671

Instead of dealing with the function directly, you are going to deal with the linear approximation to the function.0674

You are going to deal with the equation of the tangent line, that is what is happening here.0679

Let us do an example, I think it will make sense.0685

Example 1, use linear approximations to find the natlog of 5.16.0689

What it is that is actually happening here?0716

I will draw it out real quickly.0718

We know that the natlog function is something like that, it crosses at 1.0719

We want to know what the natlog is at 5.16.0726

Here is 5, let us just say that 5.16 is right there.0730

This is 5.16.0735

We want to know this value.0738

We will use linear approximation not the actual function itself.0743

5.16 is close to 5, we go to 5.0747

We draw a tangent line at 5.0752

I’m going to use this equation, the linearization.0756

In other words, the equation of the tangent line to find that value.0759

If I magnify this area, it is going to look something like this.0764

The curve is going to be like that, then it is going to be like this.0768

Here is where the actual point is, I’m going to find this one that is really close to it.0773

Let us go ahead and do it.0779

Let us recall our linearization is equal to f(x0) + f’ at x0 × x - x0.0780

5.16 is close to 5, our x0 is 5.0792

Our x that we are interested in is 5.16, we have that.0798

Again, it needs an x0.0802

Our x0, in this case is 5, just based on what the problem is asking.0805

5.16 is closest to 5.0809

The function itself is the natlog of x because the natlog function is what we are dealing with.0815

The derivative f’(x) = 1/x.0824

f(5) is equal to 1.609, f’ at 5 = 1/5.0831

Therefore, the linear approximation, the substitution that I can make for f(x) for the natlog is f(x0).0845

X0 was 5, f5 is 1.609 + f’ at x0.0855

F’x0 is 5, f’ at 5 is 1/5 + 1/5 × x – x0.0863

There you go, this is my actual function of x that I can substitute now for the log of x.0871

This is the linearization function.0882

Now I want to solve it.0884

I want log of 5.16.0886

Therefore, the l(5.16) is equal to 1.609 + 1/5, x is 5.16 – 5.0888

When I solve this, I get 1.6414, that is my answer.0906

I identify my function, log(x).0915

I identify my x0, I have the linearization equation.0918

I calculate the f at x0, I calculate f’ at x0.0925

I put it in, now I have my linearization function.0929

This is a substitute for the original function log(x).0932

l(x) replaces f(x).0937

l(x) approximates, that is why we can replace it, approximates f(x).0943

This function is an approximation.0950

When I'm close to 5 for ln(x).0952

That is what is happening here.0958

By the way, the actual value of the natlog of 5.16 is equal to 1.6409.0963

Not bad, 6409, 6414, that is very good approximation.0975

As long as we do not deviate too far from x0, either in this direction or this direction,0982

the tangent line is a good approximation to the curve.0987

Clearly, if that is your curve, as you get further and further away from the x value,0995

it is going to deviate from the curve.1006

Now you are here and here as opposed to here and here.1008

The idea is to stay as close as possible.1011

Here are our f(x), once again, is the natlog of x.1019

The linearization was 1.609 + 1/5 × x – 5.1029

We asked for the natlog of 5.16 which is the l(5.16).1041

In this case, the absolute value of the linearization, the actual f(5.16) – l(5.16),1058

the actual value - the linear approximation.1077

This is the actual - the linearization was 1.6409 - 1.6414, actually gave us 5 × 10⁻⁴.1082

The error was on the order of 10⁻⁴.1101

In general, what you can do is, if you know what error you want to be within, you can adjust your choice of x.1106

Here we just happen to pick 5.16.1116

We said it is not that far from 5, let us choose x0 = 5.1120

If you want your error to be within a certain number, if you want to keep your error,1124

let us say to a 0.1 or 0.01, 0.056, whatever it is that you want, you can actually adjust and1128

it will tell you how far you can deviate from the x0 that you choose.1134

In general, if you want to, I should say if you want,1139

the error between f(x) and l(x) to be within a certain bound,1152

then you have to solve the following.1177

You have to solve that.1182

You have to put in, in this case it would be ln(x) - 1.609 + 1/5 x – 5 under the absolute value sign.1188

You have to solve the certain bound b.1197

I should say a certain bound b.1203

You have to solve this equation either analytically or graphically.1208

As you will see in a minute, graphically is probably the best way to do this.1212

Let us go ahead and do an example, I think it will make sense.1216

Our example number 2, we are going to let f(x) equal the 3√5 - x.1224

The question is for what values of x will the linearization at x sub 0 equal 1 be within 0.1?1237

We are saying the x is equal to 1.1266

How far can I move to the right of 1 and to the left of 1, to make sure that the error,1269

the difference between the actual value of the function and the linear approximation of the function stays less than 0.1?1275

Now I’m specifying the error that I want, now I need to know how far I can move away.1284

Let us go ahead and do it.1290

f(x) is equal to the 3√5 - x which is equal to 5 - x¹/3.1293

l(x), we know that l(x) is f(x0) + f’ at x0 × x - x0.1305

Let us find f(x0) first.1320

Let me go ahead and do this in red.1324

I have got f(x0).1326

x0 is 1, f1 is equal to 5 - 1 is 4, 4¹/3.1328

It is going to be the √4 which is going to equal 1.5874.1340

I have taken care of that, now I need to find f’ at 1.1348

f’ at x is equal to 1/3 × 5 – x⁻²/3 × -1.1358

When I put in 1, I will write it this way, -1/3 × 5 – x⁻²/3.1368

Therefore, f’ at x0 is f’ at 1.1387

I put 1 into this and I end up with, -1/3 × 5 – 1⁻²/3 = -0.1323.1392

I have taken care of that.1404

I will go back here, I put this number and this number into here and here.1407

Now I have my linearization, my l(x) is equal to 1.5874 - 0.1323 × x - x0 which is 1.1413

This function can now replace this function, if I'm not too far from 1.1426

If x is 1.1, 1.2, 1.3, whatever.1436

If I’m not too far away, I can replace this function with a linearization.1439

It is saying, how can I keep this error between the two?1443

How can I keep the error less than 0.1?1449

We have to solve f(x) – l(x) less than 0.1.1451

We want the absolute value, we want the difference.1463

We do not care about the positive or negative.1466

That is why there is an absolute value sign.1467

What we want is this, f(x) is 5 - x¹/3 - 1.5874 - 0.1323 × x – 1.1471

We want that to be less than 0.1.1491

This is the same as 1.5874 - 0.1323 × x - 1 less than greater than 5 – x¹/ 3 + 0.1, 5 - x¹/3 - 0.1.1494

I hope that makes sense why that is the case.1524

We have an absolute value sign, get rid of the absolute value sign.1526

You put a -0.1 here.1530

Then, I just move this function over that side, move this function over to that side.1531

I think the best thing to do is graph this.1540

You are going to graph the function, you are going to graph the function -0.1.1546

You are going to graph the function + 0.1.1550

You want this thing to be between those two graphs.1554

This is a straight line, this is the equation of the tangent line to this function at the point 1.1560

What you get is the following.1572

When we graph this, we get this.1575

The middle graph is the actual graph itself.1586

Let me write that down.1590

The middle graph that this is one, that one.1592

The middle graph, that is the function itself, 5 – x¹/3.1598

The lower graph, that one, that is 5 - x¹/3 - 0.1.1607

The upper, exactly what you think, it is 5 - x¹/3 + 0.1, that is that one.1615

The line, this, that is the linearization, that is the tangent line at the point 1.1624

Notice it touches the graph at x = 1.1634

The black line is the linearization.1640

The black line is l(x).1645

In order for the difference between l(x) to be between 0.1 above and 0.1 below,1649

we need to make sure that this tangent line,1661

we see where it actually touches either the upper or the lower graph and we read off the x values.1664

As long as the linearization stays between the upper and lower which comes from what we just did in the previous page,1671

remember we solve the absolute value, we rearranged it.1679

We put 0.1, -0.1, we move the function over.1684

That inequality has to be satisfied, that inequality is this graph.1688

Instead of solving analytically, just do it graphically and you can just see1692

where it touches the upper and lower, and read off the x values.1698

Let us write that down.1702

Wanting the absolute value of f(x) – l(x) to be less than 0.1 which is what this graph says means1703

we want the tangent line which is l(x) between the upper and lower curves, the upper and lower graphs.1722

Just read off the x values.1746

When you read off the x values, you have -2.652.1755

That means as long as x, here is 1, this is our x sub 0.1762

As long as x goes that far and goes this far, as long as x is between there and there,1769

my linear approximation will give me an approximation to the actual function itself, the 5 – x³, to an error of 0.1.1779

That is what this means.1792

As long as I stay within this and this, the tangent line itself does not go past 0.1.1793

By specifying an error, I solve this thing.1800

Graphing this thing, I see where the tangent line touches and I read off the x values, less than 3.398.1803

This is there.1818

As long as x is in this interval, the difference between f(x) and the actual linearization is less than 0.1.1819

I hope that made sense, that is what you are doing.1828

Now let us talk about differentials.1833

Essentially the same thing except in calculus notation.1835

It is really simple.1842

Let y equal x³.1846

We know that dy/dx is equal to 3x².1849

Let us move this over, this is just a number, a small number.1854

dy = 3x² dx, this is the differential.1858

The differential of x³ at a particular x, at a particular x sub 0 is 3 × x sub 0² dx.1865

This says, if I change my x value by a small amount, and the small amount is dx,1875

then the y value changes by this small amount.1902

The y value changes by this small amount, in terms of graphs.1912

Same thing that we did, except now we are talking about really tiny motions.1927

We have our function, we have our tangent line, this is our x0.1931

Now this distance is dx, this distance is dy.1940

We know what the distance dy is.1953

It is just of the slope × dx.1955

There you go, the slope at a given point.1961

X0 is that, the slope is the derivative.1964

This is just notation, that is all it is.1967

It is the exact same thing.1970

There you go.1972

If you want to know how the y value changes when you make a small change in x,1974

just move this x over and multiply it by the slope, the derivative at that point.1980

From a given point, your change in y is going to be this much, if you change x by this much.1984

That is all this says, the differential.1992

If I said what is the differential of the function x³ at x0 = 2, I will in plug 2 to here.1995

2 × 2 is 4, 3 × 4 is 12.2007

We have got dy = 12 dx.2010

If I move away from 2, a distance of 0.1, y is going to change by 12 × 0.1.2015

That is what this is saying.2025

If I change x by this much, how much is that going to change?2027

It is the same thing here right.2030

It is a rate of change.2032

If I change x by a certain amount, how much is y going to change?2033

Except now I have a dx here, all I have done is actually move it over.2037

There is nothing different than what it is that we are doing.2041

We are just looking at it slightly differently.2044

Let us do an example, very simple.2049

Example 3, what is the differential of f(x) = e ⁺x³ + sin x?2055

Very simple, just take the derivative.2075

This is just y = e ⁺x³ + sin(x), dy dx = e ⁺x³ + sin(x) × 3x² + cos(x).2078

Therefore, the differential dy is equal to 3x² + cos(x).2103

I just decided to move it over to the left, × e ⁺x³ + sin(x) dx.2110

For a particular x value, at a particular x value, from that x value, from that x0,2122

I should say from that particular x0 value, if I move away to the left or to the right by dx,2128

the value of my function, the vertical movement is going to be that.2135

That is all this is, the differential.2140

If I have a differential movement in x, what is my differential change going to be in y, that is all this is.2142

We actually call that the differential, when we have moved the dx over.2148

Let us try another example here.2157

Let us try example 4, use differentials to evaluate e⁰.02.2162

Linear approximation, differential, it is essentially the same thing.2180

We are just different notation.2183

-0.02 is close to 0, I’m going to take x0 = 0.2188

That is going to be my base point.2200

The function that I’m interest in is, clearly y = e ⁺x.2202

Dy dx = e ⁺x.2209

Therefore, dy = e ⁺x dx.2214

At the point it is equal 0, we get the differential is equal to e⁰ × dx.2223

e⁰ is 1, dy = just 1 dx.2233

Our dx from 0, our 0 point, we are at -0.02, we are moving this way.2241

Now we are at 0.02.2249

Our dy is equal to 1 × -0.02.2252

Therefore, e⁰.02 is equal to e⁰ + dy + the differential from that point.2265

I’m going to evaluate it at that point and if I change by dx, dy is going to change this much.2292

From 0, if I go to 0.2, it is going to be e⁰ +, in this case I calculated that the differential is -0.02 =, e⁰ is 1 + -0.02 = 0.90.2299

I hope that makes sense.2323

Let us finish off with a nice little problem here.2326

The radius of a circular table is measured to be 30 inches with a maximum error and measurement of 0.4 inches.2330

We have a table and measure the radius to be 30.2338

At a possible error 0.4 inches which means it could be 29.6 or could be 30.4.2346

It might actually be a slightly smaller or slightly bigger.2355

What is the possible error in the calculated area of the table between the 30 + 0.4, the 30 - 0.4?2362

What is that possible error?2371

In other words, how much of a difference, what is the outside extra area or if it is smaller what is the inside extra area?2373

What is the error in the actual area, if I have a possible error of 0.4 away from 30?2385

Express the error as relative error as well.2394

We will do both of those.2397

The area, we know that the area of the circle is π r².2399

The differential, if I change 30 to 30.4, or to 29.6, that 0.4 is my differential, that is my dr.2405

Dr here equal 0.4 + or -.2419

The change in the area, the differential of the area, da dr = 2π r.2424

Therefore da = 2π r dr.2434

I change my area by a certain amount.2440

My area is going to change by a certain amount.2442

This dr is the error in the radius.2446

Da is going to be the error in the area.2450

That is what they are asking, estimate the possible error in the calculated area.2453

We are actually going to express it as a differential.2458

Here r = 30, dr = 0.4.2462

Therefore, da is equal it 2 × π × 30 × 0.4.2469

da = 75.4 in², that is the error.2482

The error is 75.4 in².2494

If I made a measurement of 30, if they are telling me that my error is off by 0.4 inches + or -,2501

that means the area that I calculated, the error in the area is going to be +75.4 or -75.4.2511

That is how big of a difference my error in the area is going to be.2523

It is just using differentials, that is all it is going on here.2527

Relative error, let us go to blue.2530

Relative error is the error divided by the calculated value.2534

A couple of ways that we can do this.2550

The error itself we calculated, that was 2π r dr.2552

The actual value, the area is π r².2558

The π cancels π, the r cancels r.2562

You get 2 × dr/ r.2565

Here the relative error is 2 × dr 0.4/ r which is 30, 0.027.2571

If I express this is a percentage, I will move this over and that would be 2.7%.2587

That means that if the error that I'm making measurement from 30 is 0.4,2592

that means the difference in area is going to be 2.7% of the area calculated at 30.2602

If it is 30.4, that means I'm going to be over by 2.7% of my actual area measured at 30.2609

If I’m at 29.6, if the error is under the 30, that means I'm going to be short by 2.7% of my total area calculated at 30.2618

That is what this means.2629

Relative error is the error itself that you calculate divided by the actual area of the value that you measured.2630

We could have done this directly.2639

Here we actually did relative error.2641

We use da/a, the error/ the actual value.2642

We get it in terms of variables and then we plug the variables in.2647

I could just have done it directly.2651

We could also have just done it directly.2658

In other words, the definition of relative error is equal to the actual error/ the actual value.2669

Sorry, I should say the error/ the actual value, da/a.2679

We found da, we found 75.4.2687

Area is equal to π r², area = π × 30².2692

It is equal to 2827.43.2700

da/a, error/ actual value, error/ actual value = 75.4/ 2827.43 = 0.027.2707

Again, this can give, if I want to speak in terms of percent, 2.7%.2725

That is what a differential does.2731

Anytime you have a given function, go ahead and differentiate, move the dx over.2732

And that tells you, if I change x by a certain amount, how much does y change?2736

That is it, it is really what we have been doing all along.2743

Now we just want to think about it that way.2746

Once again, if I have y = some f(x).2749

I think it would be the best if I just use actual functions here.2761

An example, if y = sin(x), dy dx, the rate of change of sin(x) is equal to cos(x).2768

If I just want to concentrate on how much does y change when I change x, I move the dx over.2779

If I change dx by 0.2 away from the point π,2786

If I’m at the point π, and if I go 0.2 away from π to the left or 0.2 away from π to the right, y changes by cos(π) × 0.2.2794

That is the differential, that is the whole idea of the differential.2808

In this case, the differential actually gives me some error.2811

If I have a measured value, and if I’m not quite sure about that measured value,2815

let us say it is + this way and + this way, my measured value is my x0.2820

The difference positive or negative is my dx.2827

The y gives me the change in the overall function that I'm dealing with, in this case it was area.2832

Area = π r².2838

Here it is cos x dx.2839

This is the dx, this is the dx.2843

The differential itself gives me the total change of whatever it is that I’m calculating.2844

I hope that makes sense.2849

Thank you so much for joining us here at www.educator.com.2852

We will see you next time, bye.2853