For more information, please see full course syllabus of AP Calculus AB
For more information, please see full course syllabus of AP Calculus AB
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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
AP Calculus AB Online Prep Course
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
in this case, let us just worry about this direction.0271
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
Instead of using f(x), you can use l(x) instead.0656
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
Choose your x0 wisely.1013
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
That is it, nothing particularly strange about this.2316
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
1 answer
Thu Jul 7, 2016 7:10 PM
Post by Haleh Asgari on July 6, 2016
Hello 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 answers
Thu Apr 7, 2016 2:02 AM
Post by Acme Wang on April 1, 2016
Hi 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