INSTRUCTORS Raffi Hovasapian John Zhu

Enter your Sign on user name and password.

• Follow us on:
Start learning today, and be successful in your academic & professional career. Start Today!
This is a quick preview of the lesson. For full access, please Log In or Sign up.
For more information, please see full course syllabus of AP Calculus AB

• ## Transcription

 2 answersLast reply by: Peter FraserSat Nov 18, 2017 4:22 PMPost by Acme Wang on April 2, 2016Hi Professor,Actually in Example III, 40 cm/min = 0.4 m/min but you wrote into 0.04 m/min so may be the final answer should be 12500 cm^3/min instead of 53,000 cm^3/min?Sincerely,Acme

### More Related Rates Examples

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
• Example I: Shadow 0:14
• Example II: Particle 4:45
• Example III: Water Level 10:28
• Example IV: Clock 20:47
• Example V: Distance between a House and a Plane 29:11

### Transcription: More Related Rates Examples

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

Today, I thought it would be nice to do some more related rates examples.0005

I never get enough practice with those.0010

Let us jump right on in.0013

Our first example says, a light is mounted on top of a 20ft pole.0015

A man who is 6ft tall walks away from the pole at 2 m/s.0021

How fast does the tip of his shadow move away from him, when he is 50ft from the pole?0025

Let us draw this out, let me go ahead and use black here.0034

I have my pole, I have my light on top, and I have my man.0039

Let me put him right over here.0049

The light beam is going to cast a shadow.0054

Typically, his shadow is going to be over there.0062

They say that the pole is 20ft and they say that he is 6ft tall.0066

He walks away from the pole at 2 m/s.0072

He is walking that way at 2 m/s.0075

Let us assign some values here.0079

I’m going to call this x, from here to the man is going to be x.0082

How fast does the tip of his shadow moving away from him?0093

I will call the distance between him and the tip of the shadow, I’m going to call that y.0098

They tell me that he is walking away from the pole at 2 m/s, that is a rate.0105

In other words, x is changing at 2 m/s.0109

Therefore, dx dt = 2 m/s, that is the rate that they give us.0114

How fast does the tip of his shadow moving away from him?0122

In other words, how fast is y changing?0127

What they want is dy dt, I need to find the relation between x and y.0130

Let me write that down.0140

We want a relation between x and y.0142

I have one, these are similar triangles.0155

There is this triangle, small one, and there is the bigger triangle.0159

Those are similar.0164

I’m going to go ahead and keep this in black, actually.0165

I have got y/6 is equal to x + y/20.0169

I will go ahead and solve this.0182

I get 20y = 6x + 6y.0183

I end up with 14y = 6x, this is my final equation.0199

Now I differentiate, once I have my final equation in the form that I want.0210

Let me actually go to the next page here.0217

I have got 14y = 6x.0219

When I differentiate, I get 14 dy dt = 6 dx dt.0225

They want dy dt, I solved for that.0236

Dy dt = 3/7 × dx dt.0239

Dx dt = 2.0248

This equals 3/7 × 2, I get the dy dt is equal to 6/7 ft/s.0252

There we go, nice and simple.0266

Again, the hardest part is drawing it out.0268

A particle moves along the curve y = x².0278

As the particle passes through the point 3,9, its x coordinate is changing at 2 cm/s.0281

What is the rate of change of the particles distance from the origin, at this point?0288

Let us draw ourselves a picture here.0296

Let us go ahead and take the first quadrant, that little parabola like that.0303

Let us say this is the point 3,9.0309

They say that a particle is moving along this curve.0316

As the particle passes through, its x coordinate is changing 2 cm/s.0322

Therefore, dx dt = 2 cm/s.0326

What is the rate of change of the particle’s distance from the origin?0335

The distance from the origin is this distance.0339

This distance is going to be, if this is x and this is y, we need a relation between d and x only.0348

We have a distance formula.0368

The distance formula is going to be √x1 - x2² + y1 - y2².0369

In this particular case, we are going to get d is equal to,0392

in other words, what is happening here, we are not going to use 3,9 just yet.0400

We have to use the general equation.0404

It is always going to be the case, you want the most general equation.0406

In this case, the general equation for distance is in terms of x and y because this particle is moving.0409

It just happens that at certain point 3, 9, that is where we are going to plug those in because that is what they are asking about.0416

But we are using the general equation.0423

In this case, at some random point on here is just going to be xy.0424

The other point is 0,0.0429

The distance is actually going to be, x - 0² + y - 0².0432

I’m going to go ahead and write this to the ½.0439

I end up with d is equal to x² + y² ^ ½.0444

I have a relationship between d and x but I also have a y in there.0454

I have to get rid of that y.0458

I need to find a relation between y and x, such that, so I can plug it in.0460

I already have a relation between y and x.0465

Y = x², therefore, y² = x⁴.0467

This d becomes (x² + x⁴) ^½.0472

That is my relation right here, this is my relation between d and x.0486

The variable for the rate they want and the variable for the rate they give me.0492

Now I can differentiate.0496

I have dd dt is equal to ½ × x² + x⁴⁻¹/2 × 2x + 4x³ × dx dt, chain rule.0499

Let us go ahead and plug in the rest.0532

Let me move on to the next page, let me rewrite this.0537

Dd dt is equal to ½, x was 3.0541

We wanted it at x = 3 and y = 9.0549

3² + 3⁴⁻¹/2 × 2 × 3 + 4 × 3³ × dx dt which was 2.0554

Dd dt = 1/ 2 √9d × this is 6, this is 108.0576

It is going to be × 114 × 2.0592

When I do all my math, dd dt is going to equal 12 cm/s.0598

When I work out all of the mathematic, it is going to be 12 cm/s.0609

That is it, nice and straightforward.0614

Just need to keep everything calm, cool, and collected.0618

It does all this work.0623

Example number 3, water is draining out of an inverted conical tank at a rate of 8000 cm³/min,0628

while water is also being pumped in at the other end at a constant rate.0636

The tank has a height of 10, a radius of 3m.0641

If the water level is rising at a rate of 40 cm/m, when the water level is 2m, find the rate at which the water is being pumped in.0646

There is a lot going on here, whole lots of information.0657

Let us see if we can make sense of some of this.0660

Let us go ahead and draw us a picture of an inverted water tank.0664

I have got this and I have got my water level.0669

They say that my radius is 3.0675

They say that the height of my water tank is 10, these are in meters.0682

Also be careful, we have some mixed units.0686

We have centimeters and we have meters, cm³, m, and cm.0689

We are going to have to choose a unit and work in that or we can convert afterwards.0696

We do want to mix units within the same problem.0701

Let me see, we have this.0708

Let us go ahead and call this the height of the water level.0710

Water is being pumped in.0717

If the water level is rising at a rate of 40 cm/min.0720

That is the rate that they give us.0725

We have dh dt is equal to 40 cm/min which is also equal to 0.040 m/min.0727

In case, we work in meters.0744

Now, I have water coming in, pumped in at a constant rate.0749

That is the rate that they are looking for.0755

However, water is leaving at 8000 cm³.0757

The change in volume is actually going to be, the dv dt it is going to be the water in.0762

The rate at which the water is coming in - the rate at which the water is leaving.0770

I’m going to call that d in dt.0775

And then, it is going to be –d out dt.0783

The rate at which water is coming in - the rate at which water is leaving.0789

This is equal to d in dt - 8000 cm³ min.0794

They want the rate at which the water is being pumped in.0808

What they are looking for here is actually this d in dt.0813

Let us go ahead and write that.0819

We have dn dt, I’m going to move this 8000 over.0820

That is actually going to be the dv dt + 8000.0826

Our task here is to actually find this dv dt and then add the 8000 to it,0836

and that will give us our final answer which is the rate at which water is being pumped in.0844

Let me write it again.0859

We need to find dv dt.0865

We need a relation between v and h.0869

We have something close, we have the volume of a cone = 1/3 π r² h.0872

This is in terms of r and h.0880

I need to find a relation between r and h, to substitute in there.0882

We want h only on the right.0887

That is when I’m going to use my similar triangles.0899

I’m going to go ahead and take half my tank.0902

Half my tank has a height of 10.0908

Let me make it a little more appropriate here.0911

If this is 3, that is 10.0917

This is my water level.0922

This length is r and this is h.0926

There is a relationship, 3 is to 10 as r is to h, which implies that I have 3h = 10r.0931

Therefore, r = 3h/10.0944

Perfect, now I have my volume = we said 1/3 π r² h = 1/3 π × 3h/10² × h.0948

When I solve this, I get that my volume is equal to 3 π/100 h³.0969

That is my relation between v and h.0982

Now I differentiate.0989

I get dv dt = 9 π/100 h² dh dt.0991

Let me see, I just need to plug in my h and dh dt.1011

We know those already.1019

They said when the water height is 2, dh dt they said is 0.040 m/min, because I decided to work in meters.1020

3m, 10m, I decided to use the meters per minute version.1034

I get that dv dt is equal to 9 × π × 2²/100 × 0.040.1038

When I solve this, I get the dv dt is equal to 0.045 m³/min.1057

I found the dv dt, that is this.1074

We said that the d in dt was equal to the dv dt + 8000 cm³/min.1077

But this is m³/min, I need to decide whether I'm going to change the 8000 to m³ or this m³ to cm³.1097

I think I'm going to go ahead and convert my 0.045 m³ to cm³.1107

We convert 0.045 m³/min to cm³/min.1118

Let us go ahead and do that real quickly.1132

Here is how that works.1138

It is not just one conversion factor, I have cm³.1139

Here is how I write this, 0.045, m³ is going to be m × m × m/min.1142

I just broke it up into its meters, ×, there 100 cm in a meter, 100 cm in a meter, 100 cm in a meter.1154

I have to account for every single meter.1169

It is not just one conversion.1171

M cancels m, m cancels second m, third m cancels third m.1173

Now I have cm × cm × cm.1179

This is going to be cm³/min.1182

You are going to end up with 45,000 cm³/min.1184

This is dv dt.1193

D in dt is equal to dv dt + 8000 = 45000 + 8000.1199

The rate at which the water is being pumped in is equal to 53000 cm³/min.1215

There we go.1230

This is all based on the fact that the change in volume is going to be the amount of water.1232

The volume of water being coming in - the volume of water coming out.1236

I hope that made sense.1243

The minute hand of the clock is 10cm long, while the other hand is 5cm long.1249

How fast does the distance between the tips of the hands changing at 2 pm?1256

Let us draw this out.1262

We have a clock, this is the minute hand, this is the hour hand.1266

This is 10 and this is 5.1275

The distance between them is this thing right here.1277

I’m going go ahead and call that c.1280

How fast does the distance between the tips of the hands changing?1286

What we are looking for is dc dt = what?1289

Notice, in this particular problem, they did not give me a rate.1298

They did not give it to me explicitly.1301

They did not actually say it in the problem.1303

I told you earlier that there is a rate that they are going to give you and there is a rate that they are going to ask for.1305

They are asking for the dc dt, where is the rate that they gave you?1310

They did not write it down but we do have it.1314

It is the rate at which the minute hand is actually moving around the circle.1317

In other words, this θ right here, I’m going to call that angle θ.1321

I know what dθ dt is.1326

Dθ dt is equal to 2 π rad, it makes the minute hand goes around in 60 min.1332

It goes 2 π rad in 60 minutes, that is equal to 0.1047 rad/min.1346

This is the rate that I am given, but I’m not given explicitly.1361

It was part of the problem, it is there.1365

I just needed to recognize it, this is the rate that I know.1368

I know how fast the minute hand is going around.1374

It goes one revolution which is 2 π rad in 60 min.1376

I do the 2π divided by 60, I get 0.1047 rad/min.1381

I have the dθ dt, that is the rate that I know.1385

I'm looking for dc dt.1388

I need a relationship between c and θ.1391

Is there a relationship between c and θ?1395

Yes, there is, we are going to use the law of cosign.1397

C² is equal to 10² + 5² - 2 × 10 × 5 × cos of θ.1402

C² = 125 - 100 cos θ.1417

This is my relation between the two variables.1426

I differentiate, I have 2c dc dt = 100 × sin θ dθ dt.1431

Dc dt = 100 sin θ/c × dθ dt.1452

We need θ and we need c at 1pm.1473

The clock is divided into 12 hours.1488

Sorry about that, I get a little confused between looking here and looking at what I have written.1511

Also, a clock is divided into 12 hours, we know what θ is.1514

Each hour is 2 π rad divided by 12.1525

Each hour is π/6 rad.1544

2pm is 2 π/6 which is π/3.1554

2 π/ 6 = π/3 rad, that is θ.1562

Θ = π/3.1569

Let us find c².1577

Now we need c, as well, since we have θ.1584

We have c² is equal to 125 - 100 × cos θ.1587

That is equal to 125 - 100 × cos π/3.1595

C² = 125 – cos π/3.1607

What is the cos of 60°, that is going to be ½.1613

½ × 50 is c² = 125 -50, that is equal to 75.1621

C = √75.1631

I will just go ahead and leave it that.1636

That is fine, I will just go ahead and write 5 √3.1641

The calculation that I actually did on my paper, I end up doing for the angle θ = π/6.1644

I did it for 1 pm, this problem asks for 2 pm.1651

I’m just going to leave mine in radical form, instead of decimal form.1654

C = 5 √3.1657

We said that dc/dt is equal to 50 × sin θ, sin π/3 / 5 √3 × dθ dt which was 0.1047.1660

I will go ahead and leave the answer that way.1696

If you want, you can go ahead and plug it in your calculator and see what you get.1699

That was it because the dc dt, remember, dc dt we said was equal to 50 × sin θ/ c dθ dt.1705

Dθ dt is 0.1047 θ π/3.1720

C we just found was 5 √3.1725

That is it, the only difficulty with this problem was realizing that the rate that they were supposed to give us was not explicit.1729

It was in the problem, we have to extract it based on drawing the picture and seeing what relations exist.1741

That was the only difficulty with this one.1747

A plane flying at a constant speed of 350 km/h passes over a house 1.5 km below.1754

Just as it passes over the house, it begins to climb at an angle of 25° from the horizontal.1761

How fast does the distance between the house and the plane changing 1 minute after the plane starts climbing?1768

Let us draw a picture here.1775

The plane is flying this way.1778

And then, we have our house, I will go ahead and put it over here.1781

It is flying this way.1788

The minute it passed over the house, it actually starts to climb.1789

The angle at which it is going to climb is going to be 25°.1798

You know what, I think I ended up doing something slightly wrong here.1820

I’m going to make a little change.1829

I’m going to say 30°.1830

There we go, I think it will make the math a little bit easier and also points out what order it is that I have written.1838

Because there is a little bit discrepancy between what I have here and what I have here,1842

but it does not change the nature of the problem.1846

1.5 km below, this height right here is 1.5 km, at constant speed of 350 km/h.1850

They want to know, how fast is the distance between the house and plane is changing?1865

They want this.1871

I'm going to call that c, I’m going to call this b.1875

How fast does the distance between the house and the planes changing 1 minute after the plane starts climbing?1885

They want dc dt, that is what they want.1891

What rate did they give us?1899

They gave us 350 km/h, that is this distance.1900

Db dt, they gave us db dt, that is 350 km/h.1907

Db dt dc dt, we need a relation between c and b.1918

Do we have one, the answer is yes.1926

Once again, we are going to use the law of cosign.1929

I have got c² = 1.5².1933

Let me write it out, just in case.1945

I have got c² = a² + b² -2 ab × cos C.1949

That is this one right here.1959

I will make this a c, I will make this a bigger C.1962

The angles we usually have capitals and the lengths we have smaller.1965

C² = a, I will take a to be this side of the triangle.1970

It is going to be 1.5² + b² - 2 × 1.5 × b × cos.1975

I know this angle, this is just 90° + 30.1985

This is just the cos of 120.1988

I'm going to get c² = 2.25 + b² +, when I do the 2 × 1.5 × cos 120, I get 1.5 b.1993

There we go, I have my relation between b and c from the law of cosign that I have extracted from this.2010

I differentiate, I have got 2c dc dt = this is 0.2019

This is going to be 2b db dt + 1.5 db dt.2029

Therefore, dc dt is equal to, it is going 2b + 1.5/2c × db dt.2042

I just factored out the db dt and wrote the 2b + 1.5 as one unit.2061

I just need to know what b and c are.2069

B, the length is going to equal how far the plane travelled in 1 minute.2087

It is travelling 350 km/h, how far is it going to be after 1 minute?2093

Let us go ahead and do that.2101

B is going to equal 350 km/h × 1hr is 60 min × 1 min.2103

Hour cancels hour, minute cancels minute.2124

I end up with 5.8 km.2127

B is equal to 5.8 km.2132

I need to find c.2136

Now that I have b, I can find c.2138

C² = 2.25 + 5.8² - 2 × 1.5 × 5.8 × cos of 120.2141

I get c² is equal to 44.59.2158

I get c is equal to 6.7 km.2164

I have everything that I need.2169

My dc dt, the rate of change at which that is going to be 2b + 1.5/ 2c × db dt.2172

Therefore, dc dt is equal to 2 × 5.8 + 1.5/ 2 × c which was 6.7 × db dt which was 350 km/h.2194

Therefore, the rate of change of the distance between the house and the plane is equal to 342.2 km/h.2216

There we go, thank you so much for joining us here at www.educator.com.2232

We will see you next time, bye.2236