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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### AP Practice Exam: Section II, Part B No Calculator

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 #4: Part A 1:35
- Problem #4: Part B 5:54
- Problem #4: Part C 8:50
- Problem #4: Part D 9:40
- Problem #5: Part A 11:26
- Problem #5: Part B 13:11
- Problem #5: Part C 15:07
- Problem #5: Part D 19:57
- Problem #6: Part A 22:01
- Problem #6: Part B 25:34
- Problem #6: Part C 28:54

### AP Calculus AB Online Prep Course

### Transcription: AP Practice Exam: Section II, Part B No Calculator

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

*Today, we are going to finish off our AP practice exam by doing the no calculator portion of the free response question.*0004

*Let us jump right on in.*0012

*Once again, you can find this particular test, this particular section, all of them, at the following.*0016

*The links are down below in the quick notes.*0023

*I will go ahead and write it here for you anyway.*0024

*They are at www.online.math.uh.edu/apcalculus/exams.*0027

*This is going to be section 2, written, and we are using version 2.*0052

*If you pull it up, leave it on the screen or print it out, however you want to do it.*0062

*Let us get started.*0067

*This is going to be question number 4.*0068

*That is fine, I will stick with black, it is not a problem.*0071

*Yes, that is fine, I will stick with black, at least for now.*0074

*I should change colors in between.*0077

*Number 4, it asked us to consider the graph of the following functions.*0080

*We have f(x) = 2x⁴ - 4x².*0086

*For part A, they want us to find the relative maxima and minima.*0095

*They want us to find both the x and y coordinates.*0101

*Let us go ahead and start.*0105

*Basically, nice and straightforward.*0106

*You just take, for relative max and min, local max and min.*0108

*You take the derivative, you set it equal to 0.*0112

*You solve for the x values.*0113

*f’(x) is going to equal 8x³ - 8x and we are going to set that equal to 0.*0116

*I’m going to factor out an 8x and we are going to have x² - 1 = 0.*0127

*Therefore, we have each of those factors equal to 0.*0136

*I’m just going to go ahead and factor the whole thing.*0141

*It is going to be x - 1 × x + 1 equal to 0.*0144

*We are going to have x = 0 is one point, x = 1, and x = -1.*0149

*We go ahead and we draw ourselves a little line, 0, 1, -1.*0160

*We are going to pick points over here, over here, over here, and over here.*0168

*We are going to put them into here to see whether we get something that is greater than 0 or less than 0.*0173

*Greater than 0 means the function is increasing.*0178

*Less than 0 means the function is actually decreasing because the derivative is the slope.*0180

*Let us go ahead and put some values in.*0186

*When we try -2, we are going to get a negative number × a negative number × a negative number.*0190

*That is going to give me a negative number, it is going to be negative, it is going to be decreasing.*0199

*Over here, if I try, let us say -0.5, it is going to be a negative number × a positive number ×,*0205

*Let me write this out.*0214

*For -2, it is going to be negative × a negative × a negative.*0215

*For 0.5, you are going to end up with negative, positive.*0218

*I’m sorry, negative, -0.5 - 1 is negative and this is going to be positive.*0227

*We are going to end up with positive, this is going to be increasing.*0235

*If I try maybe 0.5, this is going to be positive, 0.5 – 1 this is going to be negative, 0.5 this is going to be positive.*0238

*I will get a negative number here.*0248

*This is going to be decreasing.*0249

*If I try 2, it is going to be positive × positive × positive.*0251

*It is going to be increasing.*0257

*Decreasing, let me go to blue now.*0259

*Decreasing, increasing, this is a local min.*0262

*Decreasing, increasing, this is a local min.*0266

*Our local mins happened at -1 and 1.*0271

*Relative min at x = -1 and 1.*0279

*We have relative max.*0285

*Over here because it is increasing and decreasing, that is a relative max.*0287

*Relative max at x = 0.*0291

*Now let us go ahead and find the actual y values of this because that is what we wanted.*0296

*Again, to find the y value, you put it back into f.*0300

*You put these x values back into f not into f’.*0303

*Just be very careful which function you are using.*0306

*f(-1) it is going to end up equaling -2.*0310

*Therefore, we have -1, – 2.*0316

*f(1) is going to equal -2.*0319

*Therefore, we have 1, –2.*0323

*f(0), when I put it into the function, the original function, it is going to b 0.*0327

*I end up with 0 and 0.*0334

*My relative mins are going to be -1, -2 and 1,-2.*0336

*My relative max is going to be 0,0.*0348

*That is it, nice and straightforward.*0353

*Now we take a look at part B, part B asks us to find the coordinates of the points of inflection.*0358

*Again, the xy coordinates of the points of inflection.*0364

*Our point inflection is where the concavity changes from positive to negative or from negative to positive,*0367

*which means we are going to now look at the second derivative and set that equal to 0.*0372

*We have f, the original f was 2x⁴ - 4x².*0380

*f’ was equal to 8x³ - 8x.*0388

*f” is equal to 24x² – 8.*0396

*It is the second derivative that we set equal to 0, to find where the concavity changes.*0403

*When I solve this, I end up with x² is equal to 1/3 which gives me x equal to + or -1/ √3.*0411

*The points of inflection, the x values of the points of inflection are + and -1/ √3.*0425

*We want the coordinates so let us go ahead and find the f values.*0432

*Yes that is fine, I can do it over there.*0437

*Let us go ahead and take f(1)/ √3.*0439

*When I put 1/ √3 into this,*0444

*I will just write it all out, that is not a problem.*0448

*1/ √3⁴ - 4 × 1/ √3².*0452

*Let me see, what do I get.*0465

*I think I may have done the calculation wrong on my paper but that does not matter*0469

*because this is the answer, the rest is just arithmetic.*0471

*1/ √3 × 1/ √3 is 1/3.*0476

*1/3 × 1/3 is 1/9.*0480

*It looks like I did not do make arithmetic mistakes.*0483

*2/9 - 4/3 = -10/9.*0487

*That is the y valve for 1/ √3.*0490

*When I do f(-1/ √3), I end up with - 10/9.*0496

*The points of inflection are 1/ √3 – 10/9 and -1/ √3 – 10/9.*0506

*That is it, nice and straightforward.*0527

*Part C, they want us to find the intervals of increase and decrease for the function.*0531

*We already did that when we took care of Part A, when we found the critical values, the max's and min’s.*0536

*Let us recall what it is that we did.*0543

*We had 0, we had 1, and we had -1.*0545

*We ended up with a decreasing, increasing, decreasing, increasing.*0550

*The intervals of increase turns the intervals on which the function is increasing.*0559

*It is going to be from -1 to 0 union 1 to positive infinity.*0565

*That is it, nothing strange, just read it straight off.*0574

*The interval of the points of concavity.*0580

*Part D, I do not need to write it down.*0582

*I’m sorry, not the point of concavity, intervals of positive concavity.*0590

*Positive concavity, we said that the points of inflection are -1/ √3 and positive.*0598

*We said that we had -1/ √3 and 1/ √3.*0610

*Those are going to be in this interval right there.*0621

*Basically, we already know what the graph looks like.*0626

*The graph looks like this.*0630

*We know that it is positively concave here and we know that it is positively concave here.*0639

*For positive concavity, we really do not need to do any more work.*0646

*The interval is -infinity to -1/ √3.*0654

*From here all the way to this point, wherever it actually changes from concave up to concave down.*0659

*And then, from here to positive infinity to 1/ √3 to positive infinity.*0667

*We do not need to do any more work.*0676

*We have already done all the work necessary.*0678

*Let us move on to question number 5 here, let us see what we have got.*0683

*Question number 5, we are going to be dealing with the equation x² - 2y² + 3xy = -4.*0689

*Part A wants us to find an expression for the slope of this implicitly defined function*0703

*anywhere along the curve, at any point xy.*0710

*Basically, they are just asking for the derivative dy dx, y’, however you want to think about it.*0714

*We are just going to differentiate implicitly.*0719

*This is going to be 2x - 4yy’ + 3 ×, again, I’m somebody who likes to separate out my constant and just deal with my functions.*0721

*It is going to be this × the derivative of that, xy’ + this × the derivative of that y × 1 = 0.*0735

*We get 2x - 4yy’ + 3xy’ + 3y = 0.*0747

*2x + 3y = y’.*0759

*I’m going to move the y, in terms of y’ over to the right, factor out the y’, 4y - 3x.*0764

*I get y’ is equal to 2x + 3y/ 4y - 3x and that is the expression we were looking for.*0775

*That is dy dx, the slope along the curve at any given point xy.*0786

*Nice and straightforward, just a nice application.*0792

*Part B wants us to find the equation of the normal line to the curve at the point 1 - 1.*0795

*In other words, if that is the curve, that is the tangent line, that is the normal line.*0815

*The normal line is going to be, this is the tangent, this is the normal.*0823

*We are going to find the equation of the tangent line and we are going to basically take the reciprocal of the slope,*0826

*to get the slope of the normal line passes through the same point 1 - 1.*0831

*We have the equation for it, nice and straightforward.*0836

*The first thing we want to do is find the slope of the tangent line.*0839

*We are going to do y’ at 1 - 1.*0842

*y’ is this thing, we just put in 1 - 1 into x and y respectively.*0847

*It is going to equal 2 × 1 + 3 × -1/ 4 × -1 - 3 × 1.*0853

*It is going to be 2 – 3, it is going to be -1.*0864

*This is going to be -4 - 3 – 7, we end up with 1/7.*0868

*When we take the reciprocal of that, the slope of the normal line is -7.*0874

*We get y - y1 which is -1 = -7 × x - 1.*0885

*There you go, that is your answer, and you are welcome to leave it like that, not a problem.*0894

*You do not necessarily have to simplify it, unless it asks you to do so.*0897

*Part C, what is part C asking of us?*0905

*Part C, they want to know what the second derivative d² y dx² is at 1 – 1.*0909

*y’ we know is 2x + 3y/ 4y - 3x.*0925

*Basically, we are just going to find y”, that is what we want to find.*0940

*We want to find y”, what is that?*0943

*Let us just use the quotient rule, that is fine.*0948

*I mean, there are other ways we can do it but you go ahead and use the quotient rule.*0950

*y” = this × the derivative of denominator × the derivative of numerator – numerator*0954

*× the derivative of the denominator divided by the denominator².*0962

*This × the derivative of that.*0966

*We have got 4y - 3x × the derivative of this which is 2 + 3y’ - 2x + 3y × the derivative of this, 4y’ - 3/ 4y -3x².*0967

*Let us go ahead and multiply everything out.*0996

*This is going to equal 8y + 12yy’ - 6x – 9xy’ - 8xy’*0999

*+ 6x – 12yy’ + 9y/ 4y – 3x².*1018

*Let us see, what have I got, 12yy’ – 12yy’ – 6x + 6x.*1048

*I have got 8y 9y, that left and that left.*1059

*y” is going to equal 17y - 17xy’/ 4y - 3x².*1068

*Find y’ of 1 - 1 first, to put into here and then we will find the y”.*1094

*We already know what y’ is at 1 – 1, it = 1/7, that was the slope of the tangent line.*1108

*Once again, we have got y”, let me write it again, our expressions 17y - 17xy’/ 4y - 3x².*1121

*When I put in the 1 - 1 including this which is that,*1138

*I get y” at 1 - 1 is equal to 17 × -1 - 17 × 1/7/ 4 × - 1 - 3 × 1² = -17 – 17/7/ 49.*1143

*You end up with -1/36/ 343.*1176

*There you go, nice and straightforward.*1184

*Let us see, what does part D asks us?*1192

*At the point 3a – 2a, what is the nature of the tangent line to the curve for any point*1201

*that has the coordinates 3a and -2a.*1214

*Let us say if a = 3, then it would be the point 9 – 6, what is the nature of the tangent line at 9 - 6 or at 3 – 2, or at 6 – 4.*1218

*For different values of a, it is going to be different points.*1233

*We literally just plug it in.*1236

*They want to know what the nature of the line tangent.*1237

*It is going to be y’.*1241

*We have an expression for y’, that was 2x + 3y/ 4y - 3x.*1242

*Let us basically put in this and this into x and y respectively.*1251

*y’ at the points 3a and -2a, it is equal to 2 × 3a + 3 × - 2a/ 4 × - 2a - 3 × 3a.*1257

*You are going to get 0/-17a = 0.*1283

*Whenever you have anything that is of a nature of 3a – 2a,*1291

*the tangent line through any point whose coordinate is given by 3a - 2a is horizontal.*1295

*That is it, straight application of basic differential calculus, nothing strange about it.*1308

*Number 6, number 6 is a related rates problem.*1320

*Let us go ahead, as you can see it here, you have water being poured into a tank.*1325

*They gave you the height of the tank.*1330

*They gave you the radius.*1333

*They gave you the water level of the tank is increasing at a constant rate of 0.5 ft/s.*1334

*Find different things.*1343

*Let us go ahead and get started.*1345

*Let us draw a picture before anything else.*1346

*Let us go ahead, we have something like this.*1349

*We have an inverted conical tank.*1352

*Of course, we have the water level.*1356

*This is going to be our h, they tell me that the radius is 24.*1362

*They tell me that the height of the tank is 120.*1372

*Water is being poured in here and it is filling up.*1377

*Let me just go ahead and draw little something like this.*1381

*They tell me that the water level in the tank is increasing at a constant rate of 0.5 ft/s, dh dt.*1386

*Dh dt, the height is changing at 0.5 ft/s.*1393

*Part A says, find an expression for the volume in the tank in terms of the height.*1409

*We know that volume for a conical tank is going to be 1/3 π r² × h.*1418

*This is a function of two variables, r² and h.*1429

*We want it to be only in terms of height.*1431

*We need to find an expression relating the radius and height that I can substitute in here.*1436

*Once again, I’m going to use similar triangles here.*1444

*I’m going to take half.*1451

*This is a cross section of this thing right here.*1455

*This is 24 and this is 120.*1458

*This is h and this is r.*1465

*There is a relationship.*1470

*h is to 120, as r is to 24, similar triangles.*1472

*I get 24h = 120r.*1482

*I'm going to get r is equal to 1/5h.*1487

*I’m going to take this 1/5h, I’m going to plug it in here.*1493

*I get volume = 1/3 π × h/ 5² × h.*1498

*The volume = π h³/ 75, that is my answer for part A.*1508

*Find an expression for the volume, in terms of the height.*1524

*Part B, let us see what part B is asking us to do.*1534

*It wants us to know how fast is the water being poured in at the instant, that the depth of the water is 20 ft?*1538

*When h is equal to 20 ft, how fast is the water being poured in?*1548

*How fast water being poured in, there is a little bit of typo here.*1565

*It says feet per second, it should be cubic feet per second.*1569

*It should be cubic feet per second.*1577

*If for any reason there is not a typo, I will explain that after I finish this particular, the part B of the problem.*1579

*When we say how fast, they are asking for how fast does the volume changing?*1588

*How fast does water being poured in?*1592

*Volume is cubic feet per second, not feet per second.*1593

*What they are asking us basically do is to find dv dt, how fast does the volume changing?*1597

*Dv dt = what, when h = 20?*1604

*We have an expression volume = π h³/ 75.*1612

*Let us differentiate both sides with respect to t.*1620

*dv dt = 3π h²/ 75 dh dt, which is equal to π h².*1625

*I’m going to put the dh dt on top and I’m going to leave this down here, 25.*1643

*We already have dh dt, the rate at which the height is changing was 0.5 ft/s.*1649

*They gave us the h, now we just solve for dv dt.*1659

*Dv dt, when h = 20, is nothing more than π.*1663

*Put the 20 in, 20² × dh dt which is 0.5/ 25.*1670

*That is going to equal 200π/ 25.*1679

*You answer is 8π ft³/ s, that is your answer.*1686

*I know that there is a typo.*1694

*If for any reason there was not a typo and this is a trick question.*1696

*If they say how fast in feet per second is the water being poured in,*1700

*at the instant that the depth of the water is 20 ft, feet per second is the change in the height.*1704

*The change in the height is constant.*1710

*The answer would just be 0.5.*1712

*If you run across a situation like this, you definitely want to call your proctor and ask them is this a typo,*1714

*how can I interpret this, or do both.*1720

*Just do the problem and then write down.*1722

*If this is not a typo, the answer is this.*1725

*All your bases are covered.*1729

*Let see with c says, they want to know how fast the area of the surface of the water is changing, when the depth is 20?*1733

*Now area is π r².*1747

*We know that r is 1/5 × h.*1752

*Therefore area, in terms of the height of the water is π × h/ 5² which is just π h²/ 25.*1758

*Now that we have that expression, we differentiate with respect to t.*1770

*How was the area changing, da dt = 2 π h/ 25 dh dt.*1776

*Because h is a function of t, we are taking the derivative.*1790

*Da dt, when h = 20 = 2 × π × 20 × 0.5/25 is going to equal 20 π/ 25 = 4π/ 5, this is ft²/ s area.*1796

*That is it, straightforward application.*1822

*All related rates problems are the same.*1824

*There is a rate that they give you, at least one, maybe more.*1826

*There is a rate that they want.*1830

*There is a rate that they give you, let us say dm dt.*1833

*There is a rate that they want, let us say dp dt.*1838

*Your job is to find a relationship, an equation, that expresses one in terms of the other.*1841

*It could be this in terms of that or that in terms of that.*1847

*It does not really matter.*1849

*Once you have that expression, you differentiate with respect to t.*1850

*And then, you rearrange the equation as you see fit, plugging in the values that are given to you.*1854

*My friends, we have come to the end of our AP Calculus course.*1861

*It has been an absolute pleasure being able to present AP Calculus to you like this.*1865

*I wish you the best of luck in all that you do.*1871

*I wish you the best of luck on the AP Calculus exam.*1873

*I thank you for joining us here at www.educator.com.*1876

*We hope to see you soon, take care.*1879

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