AP Calculus AB 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 matter0469

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 function0703

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 – numerator0954

× 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 point1201

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