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

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

### AP Practice Exam: Section I, Part B Calculator Allowed

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
- Problem #1 1:22
- Problem #2 4:55
- Problem #3 10:49
- Problem #4 13:05
- Problem #5 14:54
- Problem #6 17:25
- Problem #7 18:39
- Problem #8 20:27
- Problem #9 26:48
- Problem #10 28:23
- Problem #11 34:03
- Problem #12 36:25
- Problem #13 39:52
- Problem #14 43:12
- Problem #15 47:18
- Problem #16 50:41
- Problem #17 56:38

### AP Calculus AB Online Prep Course

### Transcription: AP Practice Exam: Section I, Part B Calculator Allowed

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

*Today, we are going to continue our practice exam.*0005

*It is going to be Section 1 Part B, where the calculators are actually allowed.*0009

*You are going to be using the calculator regularly in this section.*0014

*Let us go ahead and get started.*0018

*Let me let you know once again where you can find this, in case you already forgot or did not see it.*0020

* Of course, there is a link down below in the quick notes.*0028

*I will go ahead and write it here.*0030

*You can find this at www.online.math.uh.edu/apcalculus/exams.*0032

*When you pull out this page, it is going to be section 1 part B.*0055

*The version that we are going through is version 5.*0064

*There are multiple exams on this page.*0069

*I would recommend just going through all of them as practice.*0070

*But we are going to be doing version 5 for this.*0073

*Let us get started, problem number 1.*0076

*I kind of like this purple, it is nice.*0080

*That is okay, I will go ahead and work in blue, I think.*0082

*It is a nice color that we usually work in.*0085

*Number 1 asks us to give a value of c that satisfies the conclusions of the mean value theorem for a given function.*0089

*Our function is f(x) = x² – x – 1.*0101

*The interval that we are concerned with is 1 to 3.*0110

*The mean value theorem basically says that, if f is continuous on a closed interval, it is differentiable on the open interval.*0116

*Then, there is some number c in that interval such that f’ at c is going to equal the actual slope from point 1 to point 3.*0123

*Let us go ahead and run through this.*0139

*Looking at this, it is definitely continuous.*0140

*Any polynomial, all polynomials are continuous, or the entire real line.*0143

*The first hypothesis is satisfied.*0148

*F is continuous on the closed interval 1,3.*0151

*F is a definitely differentiable because polynomials are differentiable everywhere.*0156

*F is differentiable on the open interval 1,3.*0163

*The conclusion that we can draw, there exists some number c such that c is between 1 and 3,*0170

*such that f’ at that value of c is equal to f(3) – f(1)/ 3 – 1.*0186

*That is what the mean value theorem says.*0196

*Let us go ahead and take the first derivative because we are going to be looking for f’.*0199

*F’ at x is equal to 2x – 1.*0204

*F’ at c is nothing more than 2c – 1.*0211

*We are going to be looking for that value of c.*0216

*Let us go ahead and calculate this value and set this f’(c) equal to this and solve for c.*0219

*That is all you have to do.*0226

*F(3), when I put it in here, I’m going to get a 9 – 3 – 1 is equal to 5.*0227

*f(1) is equal to 1 - 1 - 1 is equal to -1.*0244

*f(3) - f(1)/ 3 - 1 is equal to 5 - a -1/ 3 – 1.*0255

*This is going to be 6/2.*0267

*It is going to equal 3.*0269

*Therefore, we have, f’ at c is going to equal 3 or f’ at c is 2c – 1.*0271

*2c - 1 = 3, then we just solve for c.*0283

*2c = 4 and c = 2.*0287

*Our choice is going to be c, for this one.*0291

*Nice and straightforward, just have to make sure that the hypotheses of the mean value theorem are satisfied.*0295

*It has to be continuous on the closed interval and it has to be differentiable on the open interval.*0300

*If either of those two is not satisfied, you cannot use the mean value theorem.*0306

*Hypothesis is very important.*0311

*Number 2, it asks us to give us a function.*0317

*Let us go ahead and write this function down.*0321

*Our f(x), we have got 4x³ + 2 × e ⁺x.*0324

*It tells us that this function is invertible, it wants us to find the derivative of the inverse at x = 2.*0337

*What it is asking us to find is f inverse' at 2, that is what we want.*0346

*We are going to start off by graphing this.*0359

*Because again, you have this exponential function, you have your calculator, go ahead and graph it to take a look at it.*0361

*The graph is going to look something like this.*0370

*This over here, it is going to go something like that.*0375

*This is f, this is going to be the original function.*0379

*Now the inverse of that, we know what the inverse is.*0382

*The inverse is just a reflection about the line y = x.*0385

*Let me go back to blue here.*0394

*The inverse is going to look something like that.*0396

*This value is 2, f(0) is 2.*0404

*This graph is f inverse, what they want is the slope.*0410

*They want the derivative of the tangent line.*0417

*They want the slope of the curve at that point.*0422

*f inverse' at 2.*0426

*We actually know how to do this.*0432

*It turns out that the derivative of the inverse is nothing more than the reciprocal of the derivative of the original function.*0434

*f inverse', 1/ f’ at a given point.*0445

*We just have to be very careful at which point we are doing it.*0449

*Here is what is going to happen.*0452

*What we are looking for, f inverse' at 2 is actually going to be the inverse of 1/ f’ at 0, not 1/ f’ at 2.*0455

*Here is why, this point is 2,0, it is inverse point, if you will, is 0,2.*0469

*The x value for the inverse function is 2 but this 2, because inverse is switch x and y values, *0484

*when we go up to 2,0, the x value, this point which is the essentially the inverse point of this, its x value is 0.*0495

*Essentially, it is just going to be the reciprocal of that slope, right there.*0503

*That slope is f’ at 0.*0509

*F’ inverse at 2 is not 1/ f’ at 2.*0514

*You have to find the f inverse’ at 2.*0520

*You have to find where, that is going to be the y value of the original function.*0528

*Up here, it is going to be 1/ f’ at 0.*0534

*We need to find f’at (0), I hope that make sense.*0537

*Be very careful, that is why it is a good idea to actually graph this.*0539

*The argument for what we are looking for is going to have to be the y value of the original function,*0545

*where the inverse function is this.*0550

*Let us go ahead and do f’(x).*0559

*F’(x) is going to equal 12x² +,*0563

*I’m sorry, my real function was wrong, e ⁺2x.*0578

*There you go.*0581

*Yes, is it e ⁺2x, now I’m getting confused, 2e ⁺2x.*0590

*What function am I looking at here?*0595

*I think we are good.*0599

*4x³ + 2e ⁺2x, the derivative is going to be 12x² + 2e ⁺2x × 2, 4e ⁺2x.*0600

*We are going to find f’(0), it is going to be 12 × 0 which is 0.*0612

*e⁰ is going to be 1, this is going to be 4.*0618

*f inverse’ at 2 is equal to 1/ f’ at 0.*0625

*It is just equal to ¼, it is going to be c.*0634

*Sorry about the little confusion there.*0640

*I have written two different things on the page and I got a little confused for a moment.*0644

*Let us go on to question number 3.*0649

*What is question number 3 asks us?*0653

*Question 3 is asking us to give a value.*0664

*We actually have a graph, let me see it here.*0670

*They gave us a graph and this graph looks like that as it passes through 0 and it passes through 3.*0675

*1, 2, and 3, it is a little bit of parabola like that.*0685

*It actually passes through that point.*0690

*They tell us that this is actually a graph of f’(x).*0694

*This is the derivative of f.*0701

*What they are asking us to do is they want us to give a value of x where f actually achieves a local minimum.*0703

*A local min, we know that local min or local max, we know that f’(x) has to equal 0.*0712

*Since this is the f’ graph, it is going to be either this point or this point.*0719

*If I look at 3, if I look at the point 3, and the other point is point 0.*0732

*It is either going to be 3 or 0, which one of those is it going to be?*0742

*To the left of 3, the derivative.*0746

*This is the derivative graph, it is negative, it is decreasing.*0748

*To the right of 3, it is positive, it is above the x axis.*0753

*It is increasing.*0757

*Therefore, decreasing then increasing, our local min is at x = 3.*0760

*That is how we do with.*0767

*0, to the left you have a positive, it is increasing and it is decreasing.*0768

*Here it is actually going to be a local max there.*0771

*They want the local min, x = 3.*0776

*Our choice is a.*0778

*Number 4, basically, you are just going to be graphing the function that they gave you.*0786

*We have a graph right here and let us go ahead and say what this graph is.*0791

*I will go ahead and do over here.*0798

*f(x) =, the piece wise graph, we have got - x - 5 whenever x is less than -2.*0799

*We have x² + 1, when x is greater than or equal to -2, less than or equal to 1.*0812

*We have 2x³ -1, when x is greater than 1.*0820

*That is it, 3 piece wise function.*0826

*Now we answer some questions about this question.*0828

*Question number 1 is, is f continuous at x = -2.*0831

*-2, no, it is not continuous there.*0837

*Question 2 is asking us is f differentiable at x = 1?*0841

*At 1, we also have a discontinuity here.*0847

*The left hand limit and the right hand limit are not equal, no it is not differentiable at x = 1.*0850

*3, is x = 0 a local minimum?*0856

*Is x = 0 a local min?*0859

*Yes, it is a local min because to the left and right of that, the values of the function are higher than that.*0863

*Yes, three is definitely true.*0869

*Number 4, is x - 2 an absolute max?*0873

*x - 2 an absolute max, no because there is something higher over here.*0878

*Only 3 is true, our choice is c.*0884

*That is it, graph it, read it right off the graph.*0888

*Let us see what question 5 has for us.*0895

*Question 5, we are given some integrals and we are asked to determine another definite integral.*0903

*They give us that the integral from 0 to 50 of 4 f(x) dx is equal to 3.*0909

*They tell us that the integral from 2 to 50 of f(x) dx is equal to 2.*0921

*They want us to evaluate the integral from 0 to 2 of f(x) dx.*0930

*This is just an application of the properties of the integral.*0938

*We are going to go ahead and just write the integral from 0 to 2 of f(x) dx, which is what we want.*0943

*It is going to equal 1/4 of 4 f(x) dx which is the integral from 0 to 50 +,*0950

*this upper is going to come down here, so that they end up canceling, so to speak.*0965

*50 to 2 of f(x) dx.*0970

*This is equal to this.*0975

*Now I just plug the values in.*0977

*This is equal to ¼, I know that the integral from 0 to 50 is this one, 3.*0980

*This is -2, if I’m not mistaken.*0990

*Yes, that is -2.*0997

*The integral from 2 to 50 is -2.*0999

*The integral from 50 to 2 is going to be the negative of that.*1006

*It is going to be - -2.*1011

*When I add these together, I’m going to end up with 11/4, the choice is d.*1015

*I hope that made sense.*1022

*Just be very careful.*1023

*Here what I have done is, by flipping the integral, I’m basically taking the negative of this one, - -2.*1028

*I hope that made sense.*1039

*Number 6, what is number 6 asks us?*1047

*It wants us to find an approximate location of the local maximum for the particular function.*1051

*The particular function that they gave us is this and I have gone ahead and graphed it.*1058

*F(x) = 4x³ + 3x² - 2x.*1063

*That is it, basically, just go ahead and graph it and use your zoom feature, your trace feature on your calculator,*1071

*whatever your calculator uses or calculate, basically, to find this.*1077

*They want the value of a local maximum.*1082

*That is it, use the zoom feature, in this particular case, it turns out to be this point -0.7287, 1.502.*1088

*In this case, our choice is going to be b.*1102

*That is it, just graph it and read it right off.*1105

*Either zoom or calculate it directly, however is it that you want, go ahead and do it.*1108

*Let us move on to number 7.*1119

*Number 7 asks us to find the approximate average value of function.*1127

*The function that we are concerned with is going to be 4x ln 3x.*1132

*Let me just double check and make sure that it is correct, 4x ln 3x.*1139

*On the interval, of course, average value is always specified on an interval, on the interval 1 to 4.*1145

*Very simple, we know what the average value is, it is equal to 1/ b – a *1153

*where the first number is a and the second number is b × the integral from a to b of the function itself.*1159

*That is it, it is the same as any other average.*1167

*When you take an average, let us say of 10 numbers, you add those 10 numbers and you divide by 10.*1170

*An interval is continuous, it is not 1, 2, 3, 4, 5, 6, 7, 8, 9, 10.*1175

*You are essentially adding an infinite number of numbers.*1179

*The sum, if you remember the integral is just the sum.*1183

*Once you add the numbers, that is the integral.*1185

*And then, you divide by how many there are.*1187

*Where, how many there are, is basically the length of the interval.*1190

*This is the same average that you are used to, always.*1192

*This is nice and simple.*1198

*1/ 4 – 1/ the integral 1 to 4, 4x of ln of 3x dx.*1199

*Again, just go ahead and use your calculator to evaluate the integral numerically.*1211

*You are going to get an approximate value of 20.77.*1216

*Your choice is going to be d.*1222

*Number 8, let us see what number 8 has for us.*1227

*We have a region that is enclosed by the graphs of a couple of functions and*1235

*we rotate it around the y axis to generate a solid.*1239

*We want to find the volume of this solid.*1244

*The two functions are y = x³ - 1 and y = x – 1.*1248

*You can go ahead and graph it or if you know what it looks like, *1256

*you can graph it by hand and save the calculator for the actual evaluation of the integral.*1259

*The region is going to look like this.*1263

*1, this is going to be the x³ - 1 graph.*1269

*The x - 1 is going to be the line going this way, something like that.*1278

*Essentially, we are going to be rotating this around the y axis.*1287

*We are going to end up getting that and we are going to end up getting that.*1295

*We want the volume of that solid.*1305

*We can do this with washers, we can do this with shells.*1308

*I think we can just go ahead and do it with shells.*1312

*We are basically going to take a volume element.*1317

*And then, we are going to integrate that volume element.*1320

*Let me go ahead and work in red for this one.*1326

*A volume element, I decided I think I'm going to go ahead and go shells.*1329

*I'm going to take that and that.*1335

*Basically, a little bit of a cylindrical shell.*1341

*And then, I’m going to integrate, add them up, all over from here to here, from 0 to 1.*1344

*That is how I’m going to go ahead and do this.*1357

*Basically, we are going to take the area of a cylindrical shell.*1360

*The area of the cylindrical shell is going to be 2π x × the height.*1366

*This cylindrical shell, if you will, something like that.*1375

*We are just going to add them all up.*1386

*This is going to be a cylindrical shell.*1387

*This thickness is going to be dx.*1390

*What we want is, we want the surface area then we are going to add them up.*1392

*That is going to be our dx.*1398

*2π x × h.*1399

*X is going to be the radius, then 2π r is going to be the circumference.*1404

*And that multiply by h which is that, that is going to give me the area of the outside of the circular shell.*1410

*The dx is going to give me a volume.*1417

*The volume element is 2π x h dx.*1419

*H itself, how do we find this height?*1426

*The height is going to be the difference between that and that.*1430

*The height is going to be the function x – 1.*1436

*For any value x, I have this and I have this.*1439

*I’m going to do x - 1 - x³ – 1.*1445

*It is going to be this function - this function.*1454

*That is going to give me the difference, the height.*1457

*I hope that makes sense.*1460

*If I have a particular x, it is going to be this height - this height.*1462

*It is this height, they give me essentially just the difference between the two functions.*1471

*I get x = x - 1 – x³ + 1*1478

*The 1 cancel, I get a height equal to x - x³.*1486

*Therefore, my volume element is going to be 2π x.*1492

*My h his x - x³ and it is going to be × 2 × dx.*1500

*Now that takes care of this.*1509

*I have to multiply this by 2 because now I have the bottom part.*1513

*Therefore, I have a dv element of 4π × x² – x⁴ dx.*1525

*Now I can integrate.*1547

*Now the volume is going to be the integral from 0 to 1.*1548

*I’m going to add up all the cylindrical shells.*1552

*That is fine, I will just leave it in here, it is not a problem.*1560

*X² - x⁴ dx and when I evaluate that, I get a volume equal to 1.676 cubic units.*1563

*The choice is actually d. *1574

*I hope that made sense.*1576

*You could have done this with washers.*1578

*We take that region and we draw it like that.*1582

*You could have done it this way but then and you have to integrate along y.*1587

*You are going to have to change this, which is why I went with shells.*1592

*It was given in terms of x and y, as a function of x.*1596

*I integrated along x.*1599

*Along x, I decided to just use the shells.*1601

*Number 9, let us see, instantaneous rate of change.*1609

*They want to know the approximate instantaneous rate of change of a given function.*1618

*This function f(t) is actually expressed as an integral.*1622

*0 to 4t of cos(x) dx.*1628

*Fundamental theorem of calculus tell us that the instantaneous rate of change, *1640

*we know that an instantaneous rate of change is the derivative.*1644

*The derivative f’ t is going to be, when I have the function expressed as an integral,*1646

*I go ahead and just drop the integral sign, when I take the derivative.*1659

*It is going to be cos, I put that in here, of 4t.*1663

*But because this is not t, it is a function of t, I have to multiply by the derivative of that.*1668

*They want us to evaluate the instantaneous rate of change at a certain point.*1677

*What they want is f’ at π/4.*1682

*f’ at π/4, when I put π/4 into this expression, I get 1.662.*1688

*That is it, nice and simple, just a straight application of the fundamental theorem of calculus I’m plugging in.*1698

*Number 10, we have a particular integral and they want to know what the area is*1705

*when this integral is evaluated by a trapezoidal rule for n = 3.*1717

*The integral that they gave us is 0 to 1 of sin of π(x) dx.*1724

*Essentially, what we are going to do is we are going to calculate this integral with our calculator.*1732

*We are going to do a trapezoidal rule approximation.*1735

*We are going to subtract 1 for the other and that is going to be our area.*1739

*In this particular case, n = 3.*1742

*When n = 3, the integral from a to b of f(x) dx, the trapezoidal approximation is going to b Δ x/ 2.*1746

*This is just a formula that you have to know.*1761

*It is going to be f(x) is 0 + 2 f(x1) + 2 × f(x2) + f at x3.*1764

*n = 3, you are going to have four terms.*1779

*X0 and x3 and the n, and 2 × down below or Δ x is equal to be - a/ n.*1782

*In this case, b - a/ n.*1792

*b - a is equal to 1 - 0/ 3.*1799

*Δ x = 1/3.*1805

*We just need to find f(x) 0, f(x1), f(x2), f(x3).*1808

*When I plug those in, f(x) is 0, it is going to be this is f right here, sin(π) x. *1815

*It is going to be sin(π) × 0 which is equal to 0.*1828

*Let me write out everything explicitly here, and a little more clear.*1841

*We have x(0) = 0, that is going to give us, f(x0), π × 0 = 0.*1849

*x sub 1 is equal to 1/3 because Δ x is 1/3, it is going to be 0, 1/3, 2/3, 1.*1871

*We are breaking up this interval into that many parts.*1880

*x1 is equal to 1/3.*1884

*Therefore, f of 1/3 is going to equal the sin(π/3) which is equal to 0.86603.*1886

*x2 is equal to 2/3, therefore, the f(2/3) is going to equal the sin of 2π/ 3.*1905

*It also equals 0.86603.*1917

*x of 3 is equal to 1, f(1) is equal to the sin(π).*1923

*The sin(π) is equal to 0.*1931

*We have the trapezoidal approximation is equal to, we said Δ x/ 2.*1937

*It is going to be 1/3/ 2.*1944

*I will do it in this way.*1948

*I will actually write everything out.*1949

*0 + 2 × 0.86603 + 2 × 0.86603 + 0.*1953

*That is going to give some answer of 0.577353.*1966

*The actual value of the integral itself, when I evaluate this integral, it is going to equal 0.63662.*1975

*I take that number - that number.*1991

*I get an area = 0.63662 - 0.577353.*1995

*It is going to give me an area of 0.05927.*2010

*Our choice of is d, that is it, just got to know this formula.*2016

*That is it, Δ x/ 2, and then, however many you have.*2021

*If n is 6, you are going to have 7 terms.*2024

*If n is 19, you have 20 terms.*2026

*Always n + 1 terms in a trapezoidal approximation.*2029

*The 1 and n stay, f(x) and f(x) ⁺n, the ones in between all are multiplied by a factor of 2.*2032

*Not too bad.*2044

*What does number 11 ask us?*2047

*They are telling us that the amount of money in a bank is increasing at a certain rate.*2050

*The rate is going to be dollars per year.*2055

*They give us a time, a year where t = 0.*2057

*In this case, it turns out to be 2005.*2061

*They want to know, what is the approximate total amount of increase from 2005 to 2007?*2064

*The rate at which the money is increasing is equal to 10,000 × e⁰.06 t.*2071

*It is going to be that many dollars per year.*2086

*t is in years.*2093

*They tell us that t = 0, corresponds to the year 2005.*2099

*They want to know the increase, this is the rate of increase,*2105

*they want to know what the increase is from the year 2005 to 2007.*2113

*If t(0) is 2005, then t = 2 is 2007.*2120

*What we are going to do, the increase from 2005 to 2007.*2127

*If the rate of increases this amount per year, I multiply by the number of years.*2141

*But I integrate, because it is a function of t.*2147

*It is the integral from 0 to 2 of r(t) dt.*2150

*That is it, I just calculate that integral with my calculator and it turns out to be $21,250.*2156

*That is it, I hope that makes sense.*2163

*Dollars per year × year.*2167

*Years cancels year, you are left with dollars.*2171

*But because the rate of increase, the dollars per year is not constant.*2174

*I cannot just multiply them, I have it integrate it.*2178

*That is it, very simple.*2181

*Number 12, let us see what is number 12 asking us.*2189

*A particle is moving with a certain acceleration which they give us here and its initial velocity is 0.*2195

*For how many values of t does the particle change the direction?*2202

*A particle changes direction, when velocity goes from positive to negative or negative to positive.*2207

*When the velocity curve actually crosses the x axis, that is when it changes direction.*2213

*The acceleration that they give us here,*2222

*Our acceleration that they give us is 2t² - 4t.*2227

*They tell us that the initial velocity is 0.*2234

*Was that correct , yes, the initial velocity is 0.*2237

*For how many values of t does it change direction?*2244

*For how many values of t does the velocity curve cross the x axis?*2255

*We need to find the function for the velocity.*2260

*We know what the velocity is.*2263

*The velocity is just the integral of the acceleration.*2266

*It is going to be integral of 2t² - 4t dt.*2269

*When we do that, we get 2t³/ 3 - 2t² + c.*2279

*They tell us that the initial velocity is actually equal to 0.*2288

*v(0) is nothing more than, I put 0 into this for t.*2292

*It is going to be 0 + 0 + c which implies that c itself = 0, *2299

*which means now I can go ahead and put that into here to get my velocity function is equal to 2t³/ 3 – 2t².*2305

*Now I graph that function and I see where it crosses the x axis.*2316

*Not hits the x axis only, it have to fully cross.*2321

*It has to go from positive to negative or negative to positive.*2323

*Let us see what we do next.*2331

*This is a graph of the function that we just dealt with.*2336

*This is the velocity function 2t³/ 3 - 2t².*2340

*This is the velocity function, if something changes direction, it is what they want.*2355

*Does the particle change direction?*2359

*It changes direction when the velocity goes from positive to negative or negative to positive.*2360

*Notice here, the velocity goes to 0 but it stays negative.*2364

*It means it is moving to the left.*2367

*It stops but it keeps moving to the left.*2369

*Here it only crosses once, there is only one place where it actually changes direction.*2373

*Our choice is e, hope that makes sense.*2380

*There is only one place where it actually changes direction.*2386

*Number 13, let us see what we have got.*2395

*This is a related rates problem.*2401

*We have a sphere, they want to know fast the volume of the sphere is changing,*2403

*when the surface area is 3m² and the radius is increasing at a rate of ¼ m/ min.*2413

*Let us see what we have got.*2426

*They tell me that the radius is changing.*2428

*This is a basic related rates problem.*2430

*You may be given at least one rate, at least one, you might be given more.*2433

*You are going to asked to find another rate.*2438

*Your goal, your task is only to find the relationship between the two variables that you choose.*2440

*And then, differentiate implicitly and just plug your values in.*2446

*They tell me that the radius is changing at ¼ m/ min.*2451

*I’m going to skip the units here.*2469

*The rate of change of the radius is ¼.*2473

*It is increasing at ¼ m / min.*2476

*What they want is the rate of change of volume.*2480

*They want dv dt, that is what they want.*2482

*Our task is to find the relationship between that and that.*2484

*We know what that is, the volume of the sphere is 4/3 π r³.*2490

*Now we differentiate, dv dt is equal to 4 π r² dr dt.*2499

*They are telling me, they want to know what this value is when the surface area is equal to 3.*2516

*We know the formula for surface area, it is 4 π r².*2527

*They want to know when that is equal to 3.*2532

*It is 4 π r², the rate of change of the volume, when the surface area is 3,*2537

*I just put in the 3 where I see 4 π r², and I multiply by dr dt which I have.*2545

*3/4 m³/ min is my answer, which is going to be choice e because that is in decimal form, 0.75.*2552

*I hope that makes sense.*2564

*All related rates problems are the same.*2565

*They will give you a rate, they will ask for a rate.*2568

*Your job is to find an explicit expression between this variable and that variable.*2571

*Using whatever other information is given to you, and then differentiate that expression.*2576

*Here is the expression, differentiate the expression.*2582

*This is what we want.*2585

*We arrange it for whatever they ask, that is all.*2586

*Let us see what do we got number 14.*2597

*Rectangle, upper two vertexes, the graph, give the decimal approximation of the maximum possible area.*2604

*What we have is this situation.*2611

*We have a function which is e ⁻9x².*2619

*When I graph this function, I get something like this.*2631

*It is called a dump function, because it looks like a little dump.*2639

*They are telling me that there is this rectangle that is in here.*2643

*They want to know what the maximum area of the rectangle can be.*2649

*Let us take a look at what we have got.*2656

*This is going to be x, this is going to be f(x).*2659

*Therefore, the area of this rectangle is going to be 2x × f(x).*2666

*2x × f(x) which is going to equal 2x × e ⁻9x².*2675

*What we are going to do, we are going to graph the area.*2686

*We are going to graph the derivative because the area is what we want to maximize.*2689

*When you maximize something, we maximize the function like this one.*2696

*You take the derivative and you set it equal to 0.*2700

*Since we have our calculator, we are just going to graph the function, graph the derivative, *2703

*and we are going to see where the derivative crosses the x axis.*2707

*We are going to find where the derivative = 0.*2710

*That is the x value that we are going to use to put in our particular function.*2714

*Let us go ahead and do that.*2721

*Once again, maximization means this.*2723

*Find the x values where a’(x) = 0.*2729

*This green right here, this is our original function.*2750

*This is our area function.*2754

*This is the function that we want to maximize.*2755

*This purple function, this is a’(x).*2758

*That is the derivative of that.*2763

*Our a(x), remember we said this is 2x e ⁻9x².*2765

*It does not matter what this is.*2771

*It is the derivative of this.*2772

*I need to know where the derivative crosses the x axis, where the derivative is equal to 0.*2773

*It is equal to 0 to places here and here.*2778

*This is a negative value, I cannot have a negative length because this is a box that we are talking about.*2781

*A box whose length is this part is x.*2787

*That is the value that I'm interested in.*2792

*That value, I read it off the graph, zoom in, use my calculator, however it is that I want to use it.*2795

*I end up with this value is 0.236.*2800

*a(0.236), you can see that the area function is maximized here.*2808

*The x value is here, the derivative confirms that.*2815

*That is all we are doing.*2818

*It is a 0.236, the area of 0.236, when I put 0.236 into here, I get an area of 0.2859.*2819

*My choice is a, I hope that made sense.*2831

*Number 15, let us see what number 15 is asking us.*2839

*They are saying that a rough approximation from the natlog of 5 is going to be 1.609.*2848

*They want us to use this approximation and differentials or linear approximations to approximate.*2855

*They tell us that the natlog of 5 is equal to 1.609.*2864

*They want us to find an approximate value for the natlog of 521/ 100.*2876

*That is just the natlog of 5.21.*2884

*We are going to use a linear approximations and differentials here.*2890

*Actually linear approximations not differentials.*2892

*The linear approximation of a value is equal to that function at that value + the derivative at that value × x - x0.*2898

*In this particular case, we want to find the value at 5.21.*2913

*It is going to be the f(5) + f’ at 5 × 5.21 – 5.*2920

*f(x) is just equal to the natlog of x.*2934

*f’(x) = 1/ x, f’ at 5 = 1/5.*2939

*Our answer, the approximation of the natlog of 5.21 is equal to f(5),*2949

*which is f(5) which is ln of 5 which is 1.609 + f'(5) which is 1/5 × 5.21 – 5.*2958

*When I calculate this, I end up with 1.651.*2976

*My answer is c, linear approximation, that is it.*2980

*That is all you are doing.*2986

*This is going to be the difference between what you know and what you are looking for.*2988

*That is all that is happening here.*2993

*Just remember, linear approximation.*2998

*Let us say this is a function, the whole idea of the derivative is 2.*3001

*As long as you are staying pretty close to the curve, the line itself is actually pretty good approximation.*3005

*This point was our ln 5.*3014

*This point was going to be our ln 5.21.*3019

*These are linear approximation, we used the derivative instead of actually using the logarithm function because it is very close to it.*3027

*That is all we have done here.*3036

*Let us take a look at number 16, let us see what we have got.*3043

*What is 16 asking us, it is giving us a function and it is telling us that it is differentiable everywhere.*3050

*They want to know what n is.*3056

*f(x) is equal to, it is a piece wise function here.*3060

*We have got nx³ - x for x less than or equal to 1.*3067

*We have got nx² + 5 for x greater than 1.*3074

*Once again, they tell us that it is differentiable everywhere.*3082

*They want to know what is n.*3091

*We want to know what n is.*3095

*Let us go ahead and work in red here.*3098

*Differentiable means it is continuous.*3101

*Because differentiability implies continuity, not the other way around, continuous.*3107

*We know that this function is continuous everywhere.*3115

*The differentiability also means that the left hand derivative = the right hand derivative.*3122

*The left hand derivative, I take the derivative of this, the derivative of that.*3139

*The left hand derivative, as I approach it from the left.*3148

*I have got 3nx² - 1 is equal to 2nx.*3150

*I got myself one equation there.*3160

*Because I know that the left hand derivative is equal to the right hand derivative.*3161

*That has to be true.*3164

*The limit as x approaches 1 from the left of f(x) = the limit as x approaches 1 from the left of nx³ - x is equal to, I plug 1 in for x.*3170

*It is going to be n - 1 and the limit as x approaches 1 from above, they are differentiable, left hand and right hand derivative are equal.*3201

*What I’m saying is that it is continuous which means that*3216

*the left hand limit and the right hand limit are going to be the same because it is continuous everywhere.*3219

*Therefore, those two things have to be the same.*3226

*If f(x) = the limit as x approaches 1 from above of that function mx² + 5 = m + 5.*3232

*These are equal, precisely because f is continuous everywhere.*3252

*It is continuous everywhere because it is differentiable everywhere.*3259

*Now I have got, we have got n - 1 is equal to m + 5.*3263

*Let us go ahead and write this as n - 6 is equal to m.*3274

*I’m going to go ahead and plug n – 6, wherever I see that.*3283

*I got 3nx² - 1 = 2 × n - 6 × x.*3290

*I get 3nx² – 1 = 2nx - 12x.*3303

*If I did my math correctly and I end up with x = 1.*3311

*I know the x = 1 because I'm talking about 1,*3324

*because 1 is the point at which it is going to be both continuous and differentiable.*3329

*X = 1, I get 3n.*3334

*When I plug in x = 1 to there and to there, I get 3n - 1 = 2n – 12.*3340

*I get n = -11, my choice is e.*3351

*In this particular case, when they said it is differentiable everywhere, it implies that it is continuous everywhere.*3355

*Because it is differentiable, the left hand and right hand derivative have to be the same.*3360

*I take the derivative of both functions, to the left of 1 or the right of 1 *3366

*because it is differentiable at 1 and I set them equal to each other, that is one of my equations.*3370

*Because it is continuous, I know that the left hand limit, the original function = the right hand limit of the original function.*3375

*I set those two things equal to each other.*3381

*That gives me another equation, an m and n.*3383

*Now I just solve the two equations.*3387

*In this case, they just wanted n.*3388

*I could have done both, not a problem.*3390

*Let us see number 17, which I think is our last problem here.*3396

*They want to know which of the following functions has a vertical asymptote at -1 and a horizontal asymptote at y = 2.*3403

*They gave us some choices.*3413

*Vertical asymptote at x = -1 and they want a horizontal asymptote at y = 2.*3414

*This is what they are asking.*3435

*Basically, you are just going to check each function, that is it.*3439

*Denominator, where the denominator is 0, that is going to be your vertical asymptote.*3442

*Horizontal asymptote, it depends on the nature of the rational function, what the top is and what the bottom is.*3450

*I notice that the 2x² + 1/ x² – 1, it goes to 2 as x goes to infinity.*3457

*Horizontal asymptote means taking x to infinity and see what happens.*3471

*Essentially, as x goes to infinity, these do not really matter.*3475

*x²/ x², they cancel.*3479

*You are just left with 2/1 which is 2.*3481

*Sure enough, y = 2 is a horizontal asymptote and x² – 1, this is x – 1 × x + 1.*3484

*x cannot be -1 because it is actually going to end up making the denominator 0.*3501

*Our choice is b, nice and straightforward.*3507

*That takes care of Section 1, Part B, where the calculator was allowed for the exam.*3513

*Next time, we are going to start with our free response question.*3520

*Thank you so much for joining us here at www.educator.com.*3524

*We will see you next time, bye.*3526

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