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AP Practice Exam: Section II, Part A 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: Part A 1:14
  • Problem #1: Part B 4:46
  • Problem #1: Part C 8:00
  • Problem #2: Part A 12:24
  • Problem #2: Part B 16:51
  • Problem #2: Part C 17:17
  • Problem #3: Part A 18:16
  • Problem #3: Part B 19:54
  • Problem #3: Part C 21:44
  • Problem #3: Part D 22:57

Transcription: AP Practice Exam: Section II, Part A Calculator Allowed

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

Today, we are going to continue with our AP practice exam.0004

We are going to start the free response section.0007

Again, this is going to be the same thing.0010

For part of it, you are going to be allowed a calculator.0012

For part of it, you are not going to be allowed a calculator.0015

Let us jump right on in.0021

In case you need to pull this up, there is a link down below or let me write it over here.0026

It is going to be www.online.math.uh.edu/apcalculus/exams.0032

It would be nice if I actually wrote it, /exams.0054

You are going to be clicking on where it says section 2, written.0058

We are going to be doing version 2.0064

Version one is the same thing, it is just different numbers, different arrangement, different order.0068

But it is essentially the same exam.0072

Let us get started.0074

Question number 1, we have a particular region that is going to be bounded by a couple of functions.0076

In this particular case, y = -x² and y = 4x², and the y axis.0086

The first thing we want to do is go ahead and graph this.0092

We see what we are looking at.0095

We got y = 4x², it is going to be some parabola, something like that.0105

This is our y = 4x².0111

Then we have our y = -x² which is going to be a bump function.0116

I will go ahead and just mark that there.0120

It is going to be something like that.0123

This is our function y = e ⁻x², they are all going to look like that.0127

These little bump functions.0134

It is bounded by the y axis.0135

The particular region that we are interested in is that region right there, that little triangular looking thing.0138

Part A, find the area of s.0147

This is are our region s.0151

Find the area of s.0157

The first thing we want to do, what we are going to do is we are just going to take the difference in the two graphs.0166

We are going to take a little vertical strip and we are going to integrate from 0 to whatever this x value is.0170

We need to find where the two graphs meet, so that we have an upper limit for our integral.0176

Let us go ahead and do that.0183

We set 4x² equal to e ⁻x².0184

We get 4x² - e ⁻x² = 0.0189

What you are going to do is, again, calculator is allowed here, you are just going to use this to find the roots of this.0195

Find the root of this equation, where does it hit the x axis.0201

It is going to be this and this, those two.0209

We end up with the first root is equal to 0.4515.0215

The second root is going to be symmetric.0221

On the other side, it is going to be -0.4515.0223

Therefore, the area of this region is going to equal the integral.0228

S, this region right here.0232

That is just going to be from 0 to 0.4515 and this is going to be the length of this strip.0236

The length of this strip is going to be, this is some x value,0245

it is going to be the height of the top function - the height of the bottom function.0248

Height of the top function - the bottom function gives me this height.0253

I’m going to add all of those little strips together.0258

The top function is e ⁻x².0261

The bottom function is 4x² dx.0264

I get an answer of 0.300, that is it.0268

Nice and straightforward application of an integral to find the area between two curves.0273

Let us now move on to part B.0285

Part B wants us to find the volume of the solid that is generated when this region is rotated around the x axis.0288

Let us go ahead and redraw real quickly, just a small portion of it.0297

We have that and we have the bump function.0301

This was our region s.0307

What we are going to do is we are going to rotate that around the x axis.0310

We are going to end up with this region roughly.0314

We are taking this region, we are rotating it about the x axis.0325

We want to find the volume of that solid.0329

I’m going to go ahead and use washers because again, I want to integrate with respect to the x axis.0334

Let us recall what our functions were.0340

We had y = our outside function, this was our y = e ⁻x².0342

Our lower function there which is y = 4x².0353

When I take a little strip like a washer, I’m going to take that, I’m going to turn it this way so that I’m looking at it.0359

I’m going to be looking at, essentially that.0369

I want to take this area and I want to add up all the little volume elements.0374

I think I will call this radius 2, I think I will call this outer radius 1.0385

The volume is going to equal the integral from a to b, it is going to be from here to here.0391

It is going to be the 0.4515 of the area of this circle - the area of this circle which is going to give you this area.0398

It is going to give me the area of that × dx.0409

It is going to be π × r1² - π × r2² dx.0415

r1 = e ⁻x², r2 is equal to 4x².0426

r1 is that radius, r2 is that radius.0432

Therefore, what we have is the volume equals the integral.0441

I pull out the π, 0 to 0.4515 of e ⁻x²² - 4x²² dx.0446

When I solve that, I get 1.059, very straightforward.0468

Let us go ahead and go on to part C.0476

Now what they want us to do is, now what they say is, this region is the base of a solid, 0484

where the cross section that is perpendicular to the x axis is an equilateral triangle.0491

Find the volume of the solid.0499

Let us go ahead and draw it again.0500

I have got a 4x² part, I have the x² part.0506

Basically, what they are telling me is this.0513

This is our region s that we have.0514

This function right here, this function, this was our y = e ⁻x².0517

This function right here was y = our 4x².0526

They are telling me that this is the base of a solid.0531

If I take a vertical slice perpendicular to the x axis, the cross section, if I were to take this and flip it,0534

in other words look at it from that direction, it is going to be an equilateral triangle.0545

What they are saying is this.0550

I’m going to take this and I’m going to turn it this way.0552

What I’m looking at is an equilateral triangle.0556

If I’m looking at it from that direction, I see a bunch of equilateral triangles getting bigger.0566

That is what is happening.0571

They want to know what the volume of the solid is.0573

This region is the base of the solid.0575

The cross section is going to be an equilateral triangle.0577

Let me draw my triangle a little bit better here.0582

In fact, I can draw my triangle over here.0587

I’m going to move over here, this, that, and that.0589

This right here is this right there.0602

That height is equal to e ⁻x² - 4x².0608

I’m going to draw out a perpendicular and I’m going to call that h. 0615

H because this is an equilateral triangle, all of the angles are 60.0619

When I draw out the perpendicular, I get a 30, 60, 90.0623

This is a perpendicular, therefore h is equal to half of this thing × √3.0626

It is equal to half of e ⁻x² - 4x, √3.0634

The area of that triangle is equal to ½ the base × the height that = ½ the base is that thing,0647

e ⁻x² – 4x² × the height × √3/2 × e ⁻x² - 4x.0661

Volume = the area of this thing integrated from here to 0.4515.0678

I’m going to add up now all of these triangles.0687

I’m going to add up all the triangles.0690

The volume = the integral from a to b of the area × dx, 0692

the integral from 0 to 0.4515 of the area which is now √3/4 × e ⁻x² - 4x²² dx.0701

I get a volume equal to 0.1037.0728

I hope that makes sense.0735

Let us go ahead and go on to problem number 2.0740

We are given, we have f(x) is equal to the integral from 0 to x of cos².0748

A function that is expressed in terms of an integral cos².0760

I’m changing my variables dt, on the interval from 0 to 3π/ 2.0769

What do we want here?0782

Number 2, they want us to approximate f(π) using a trapezoidal rule with n equal to 4.0784

Let us see what we have got.0798

f(π) = the integral from 0 to π cos² t dt.0801

The trapezoidal rule where n is equal to 4, that is going to equal Δ x/ 2 × f(x0) + 2 × f(x1) + 2 × f(x2).0811

You definitely want to write this out on your exam so they know what you are using.0832

+ 2 × f(x3) + f(x4), you see that you have five terms.0835

It is equal to 4, you have 1, 2, 3, 4, 5 terms, 0 to 4.0843

Δ x here, Δ x is equal to b - a/ 4.0850

It is going to be π - 0/4 = π/ 4.0857

Therefore your x0, x1, x2, x3, x4, are going to be 0, π/4, 2π/ 4, 3π/ 4, 1.0863

That is what is happening.0872

At x = 0, x(0) = 0, we have f(0) is equal to,0876

f is this, small f, this is small f.0890

We are finding F(π) which is an integral.0899

We are evaluating and approximating that integral with trapezoids.0904

F is that.0909

f(0) = 1² = 1, x1 = π/4, f(π/4) = 1/ √2² because the cos of π/4 is 1/ √2.0910

1/ √2² is ½, x2 = π/2, f(π/2) = 0² = 0, x3 = 3π/ 4, f(3π/4) = -1/ √2² which is ½.0927

x4 is equal to π, f(π) = -1, -1² is equal to 1.0959

Therefore, our approximation t4 is equal to π/4/ 2 × 1 + twice this, twice this, twice this, + that.0970

1 + 2 × ½ + 2 × 0 + 2 × ½ + 1 + π/8 × 4 = π/2.0985

Just a nice application of the trapezoidal rule, which I actually personally dislike very much.1008

I do not know why.1012

Part B, find f’(x).1016

f(x) = the integral from 0 to x of cos² t dt.1021

f’(x) is nothing more than cos² (x).1031

Straight application of the fundamental theorem of calculus.1035

C, average value of f’ on the interval.1040

They want the average value of f’ on the interval.1048

The average value = 1/ b – a, integral from a to b, f’ is cos² x dx 1/,1055

What was the particular interval, over the interval of 0 to π.1075

It is going to be π – 0, the integral from 0 to π, cos² x dx and at 0.5.1079

That is it, straightforward, nothing strange happening here.1088

Let us do question number 3 and we will done with the first half, very nice.1095

We have a particle that is moving along the x axis and its acceleration is given by, our acceleration function is 24t – 12.1101

Let me write this a little bit better.1114

It equals 24t – 12.1117

They are telling me that the velocity at 1 is equal to 8.1121

They are telling me that the position at 1 is equal to 10.1125

Part A wants us to find an expression for the velocity of this.1132

We know what the velocity is, the velocity of t is nothing more than the integral of the acceleration1135

which is the integral of 24t - 12 dt, which is going to give us 12t² - 12t + c.1144

They gave us a value of v1.1160

v1 which is going to equal 12 × 1², just plug in 1 for t - 12 + c. 1161

They tell us that this equals 8.1169

When we solve this, we get c = 8.1172

12 – 12, we get c = 8.1178

Therefore, we plug this back into there.1180

Our velocity function which is what we wanted is going to be 12t² - 12t + 8.1183

There we go, that is part A.1191

Part B, at what values of t does the particle change direction?1197

The particle changes direction when the velocity goes from positive to negative.1206

The particle changes direction when v(t) changes sign, positive to negative or negative to positive,1220

where it crosses the x axis completely, not touches it.1240

Find the root of v(t).1252

When I plug this in my calculator and I check the roots, there are no real roots.1259

Does not change direction, the particle does not change direction.1272

It might slow down, then speed up again, but it does not change direction.1291

Now part C, now they want us to find an expression for the position of the particle.1300

We know what that is.1307

We know that the position of the particle is nothing more that the integral of the velocity function.1309

It is going to be the integral of 12t² - 12t + 8 dt.1315

We are going to end up with 4t³ – 6t² + 8t, again + another constant.1325

They told us what x sub 1 is, what the position at t = 1 is.1337

It is going to be 4 × 1³ - 6 × 1² + 8 × 1 + c.1342

They tell me that that = 10.1351

When I solve this, I get c is equal to 4.1353

I plug that back in, therefore my position function = 4t³ – 6t² + 8t + 4.1356

That is what I want.1374

D, the total distance traveled from t = 1 to t = 4.1377

The particle does not change direction.1389

I do not know what is the best way to do this?1399

Total distance traveled.1403

It does not change direction.1413

I did it one way but now that I'm sort of sitting here and thinking what is the best way to do this.1419

The total distance traveled by the particle.1424

The velocity × time gives me total distance.1433

That is fine, I can just go ahead and take, 1436

Our total distance traveled, because it did not change direction, I do not have to split up the,1439

From t = 1 to t = 4 because it does not change direction,1455

I do not have to break it up into moving to the left, moving to the right, things like that.1460

I can just go ahead and I can find the distance at t = 4.1465

I can find the distance at = 1 which is equal to 196, which is equal to 10, 1470

and I can just take the difference which is going to be essentially the same as the total distance traveled.1478

If you want, you could have done it this way, the total distance.1489

Distance = rate × time, rate is velocity, velocity × time.1493

Because velocity is not constant, the total distance = the integral from 1 of 4 of the velocity function dt.1499

You can go ahead and put the velocity function which is this thing into here and integrate that, 1507

which is the same essentially as this.1513

Because when you integrate the velocity function, you are going to get the distance function.1515

When you evaluate the distance function at 4, this is function at 1, you are going to subtract.1518

It is the same thing, however you want to look at it.1523

That is it for this part, where the calculator was allowed.1528

The next part is going to be another three questions where no calculator is allowed.1531

Thank you so much for joining us here at www.educator.com.1536

We will see you next time, bye.1538