For more information, please see full course syllabus of AP Calculus AB

For more information, please see full course syllabus of AP Calculus AB

### 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 x *0023

*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 line*0336

*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 and*1128

*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 see *1692

*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 means*1703

*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

Last reply by: Professor Hovasapian

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

Last reply by: Professor Hovasapian

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