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Lecture Comments (3)

0 answers

Post by Shiden Yemane on November 12, 2017

1 answer

Last reply by: Professor Hovasapian
Thu Apr 7, 2016 1:48 AM

Post by Acme Wang on April 2, 2016

Hi Professor,

In Example III when I solved the problem and wrote V = 1/3 ?r^2h, I simply treated r as a constant as from the title I knew the r equals 2.5 and wrote dV/dh = 1/3?r^2, which led me to a wrong answer. So my question is how could I distinguish variable from constant? (I know this sounds a little stupid but just want to figure out everything clearly :))



Related Rates

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
  • Related Rates 0:08
    • Strategy for Solving Related Rates Problems #1
    • Strategy for Solving Related Rates Problems #2
    • Strategy for Solving Related Rates Problems #3
    • Strategy for Solving Related Rates Problems #4
    • Strategy for Solving Related Rates Problems #5
  • Example I: Radius of a Balloon 5:15
  • Example II: Ladder 12:52
  • Example III: Water Tank 19:08
  • Example IV: Distance between Two Cars 29:27
  • Example V: Line-of-Sight 36:20

Transcription: Related Rates

Hello, welcome back to and AP Calculus.0000

Today, we are going to be talking about related rates.0004

Related rates, the mathematics is actually quite simple.0008

The difficult part with related rates is setting up the problem from the verbal description.0012

Let me write this down here.0019

Related rates, I’m not going to actually talk about it very much.0022

I’m just going to launch right into the examples because the best way to explain it is by doing the examples.0025

Related rates is best explained by example.0032

The strategy for solving the related rates problems is always going to be the same for every single problem.0047

The general outline of the problem is always going to be the same.0053

Here is what it looks like.0057

The strategy for solving these problems is always the same.0062

One, it is always going to be a rate that you are given.0085

It may be more than one but generally, it is one.0089

But there is a rate or two that you are given.0093

There is a rate that you are given.0099

Two, there is a rate that you are asked for.0110

Three, your job is to find an equation that relates the variable of the rate that they give you,0130

to the variable of the rate that they ask for.0136

And the variable is going to be your choice.0138

In other words, you can label it any way you want.0141

It does not have to be x and y.0143

It could be whatever it is that you happen to be discussing.0144

You will see an example in just a second.0147

Find an equation relating the above two variables, whatever the variables you chose.0150

This is always the case.0165

The 4th and final part is differentiate the entire equation with respect to time t.0172

Because again, here we are going to be talking about a real rate, something per second,0193

something per minute, something per hour, something like that.0200

The denominator, for example, if we have like dy dt, it is going to be with respect to time.0204

Once again, let me go ahead and write this down.0217

I’m sorry, the last part is solve, solve for the rate requested.0222

The rate they ask for.0233

That is it, you just rearrange the equation if you need to, and solve for the rate that they ask for.0235

Again, the difficulty in these problems is translating the verbal description into viable mathematics.0241

I bet that has been the problem with word problems,0279

ever since you guys have started doing them in junior high school, something like that.0281

Let us launch right into the examples and we will use all of these examples to actually explain what is going on.0286

But it is the same basic strategy.0292

They are going to ask you, they are going to give you a rate.0294

They are going to ask for a rate.0296

You have to find an equation that relates those two variables.0298

Sometimes it will be direct, sometimes you are going to have to use other information0301

that is in the problem to come up with an equation that is a direct.0305

In other words, one variable on one side, the other variable on the other.0309

And then, you differentiate with respect to t and then you solve.0311

Let us see, our first one here.0317

Air is being pumped into a spherical balloon at a rate of 15 cm³ /s.0320

How fast does the radius of the balloon change, when the volume is 70 cm³?0329

Air is being pumped into a spherical balloon.0335

For all of these problems, you always have to draw a picture.0337

You might get to a point where you do not have to draw a picture,0344

when you understand what is happening, especially for the more simple ones.0346

But draw a picture, see what is going on, use your physics.0349

There is nothing in here that is going to ask you to use something that is counterintuitive.0353

You just have to trust that you understand the natural world well enough,0357

after being on earth for these many years, that you can actually figure out the rest.0362

You can come up with some physical model for what is going on.0368

We have a spherical balloon, nice and easy.0372

A spherical balloon, air is being pumped into the spherical balloon at a rate of 15 cm³/s.0377

This is the rate that they give you.0384

This is dv dt that equals 15 cm³/s.0388

The reason I chose v for my variable is volume.0395

Cubic centimeters is a unit of volume.0398

Air is being pumped in, it is a 3 dimensional object, it is a volume.0401

This is my rate, a rate of change, a derivative.0405

Volume is changing per time, it is 15 cm³/s.0409

That is the rate they give you.0414

How fast is the radius of the balloon changing?0415

The rate that they want is dr.0417

Dr dt that is the rate that they want.0422

This is r, how fast when the volume is 70 cm³.0430

This is another bit of information that we are going to use, when we finally solve.0435

This is the rate that they give us, this is the rate that they want.0441

My two variables are, let me go ahead and do this in red, are v and r.0443

I need to find the relationship that relates v and r.0450

Fortunately, I have one.0454

Volume of a sphere is 4/3 π r³.0457

Now that I have this, I differentiate with respect to t.0464

This is going to be dv dt = 4/3 π × 3 r².0467

The 3 is canceled, I'm left with 4.0480

I’m sorry, I forgot something.0485

I’m differentiating with respect to t.0488

This is like implicit differentiation except now both the v and r are functions of t.0491

It is dv dt = 4/3 π × 3 r² dr dt.0498

Dr dt, the 3 is canceled and I'm left with 4π r² dr dt.0504

We are looking for dr dt, I’m just going to solve for that.0515

Solve for dr dt.0520

Let me write this out a little bit better here.0540

I have got dv dt = 4π r² dr dt.0548

Now I’m going to isolate dr dt by dividing by 4π.0558

I get, let me write it over here, dr dt = dv dt divided by 4π r².0561

We said dv dt is equal to 15.0578

This is going to be 15/ 4π r².0581

The question is, what do we put in for r?0587

They said the volume was 70cm³.0600

I will use the equation that I have.0606

Volume = 4/3 π r³.0608

They said the volume is 70 4/3 π r³.0613

Now I solve for r.0623

That will tell me what r is, when the volume happens to be 70.0625

I get r³ is equal to 16.71.0629

I get r is equal to 2.56.0634

Dr dt, we said it is equal to dv dt divided by 4π r², that is going to be 15 divided by 4 × π × 2.56².0647

If I did my arithmetic correctly which I often do not, 0.183 and it is going to be cm/s dr.0665

R is a length, therefore, it is centimeters per second.0677

It is an area which would be square centimeters per second or volume which is cubic centimeters per second.0682

In case you are not sure about that, dv dt, the numerator is in cubic centimeters per second.0689

Down here, we have r which is in centimeters.0702

It is squared so this is going to be cm².0705

Cm² cancels this, I’m left with the unit of cm/s.0710

If you want to carry the units, that is fine.0717

Or you can go back at the end to elucidate what the units are.0719

There you go. The radius is growing at 0.183 cm every second, when the volume hits 70.0726

Notice, the rate of change actually depends on what the radius is.0735

As the radius changes, the rate at which the radius is growing changes.0741

They are two different things.0747

The radius and this is the rate of change of the radius, per unit time.0749

I hope that makes sense.0757

Again, related rates, we use the rate of the volume change to find the rate of the radius change.0760

Let us try example number 2.0772

A ladder 20 ft long rests against the vertical wall.0775

If the bottom of the ladder was pulled away at a rate of 1.2 ft/s,0779

how fast does the top of the ladder slide down the wall, when the bottom is 7 ft from the wall?0783

We go ahead and we draw our picture, always.0790

Let me go to black actually.0794

I have got a wall here and I have this 20ft ladder, this is 20ft.0797

They are pulling this bottom away.0809

They are pulling it away.0811

This top of the ladder is actually going to be going down.0814

This is growing, the bottom is growing.0819

I’m going to call it x and I’m going to call this y.0822

If the bottom of the ladder is pulled away at a rate of 1.2 ft/s, x is changing at 1.2 ft/s.0829

It is growing 1.2 ft/s.0837

Dx dt, the rate of change of x per unit time is 1.2 ft/s.0840

How fast is the top of the ladder is sliding down the wall?0849

They want to know dy dt.0853

The rate they give me, rate of change of x, the rate they want is y.0856

Now I have to find the relationship between x and y.0861

It is a triangle, I have a relationship between x and y.0869

I have x² + y² = 20².0872

I have x² + y² = 400.0879

I differentiate with respect to t.0884

This becomes 2x dx dt.0887

This becomes 2y dy dt.0892

The derivative of 400 is 0.0897

I have got 2y dy dt = -2x dx dt.0901

The 2 is cancel, I divide by y to get my final dy dt, which is why I’m isolating, = -x/y × dx dt.0911

I just plug in my values, I need to plug in an x, plug in a y.0927

My dx dt was already given to me, that is the 1.2 ft/s.0931

Let me go to the next page here.0940

Dy dt we said was equal to –x/y × dx dt, that is equal to,0944

They said that x, they wanted to know what it was.0959

Sorry about that, this is a ladder, it does not go through the wall or through the floor.0966

This was x, they wanted to how fast this is going down, when the ladder is 7 ft from the wall.0972

X is 7 and dx dt was 1.2.0980

The question is what is y, would we put in for y?0989

We have a relationship, we have x² + y² = 400.0994

X is 7 so 7², we have 7² + y² = 400.1003

We have 49 + y² = 400.1012

We have y² = 400 -49 is 351, I think.1018

Y ends up being 18.73 ft, there you go.1025

Dy dt, the rate of change of y is equal to -7/ 18.73 × 1.2.1038

We get that dy dt is equal to -0.448 ft/s, there you go.1053

Notice that dy dt is negative.1066

Negative means that, this is y, y is getting smaller.1071

This ladder is going down.1080

This distance right here is actually getting smaller, that is why this is negative.1082

This negative tells me that it is decreasing.1087

In other words, for every unit change in time 1 second, when the ladder is 7 ft from the wall,1091

the vertical distance is changing by 0.448 ft every second.1100

Notice, the rate of change depends on what x is and what y is.1107

It is going to change depending on what those things are.1116

It is not a constant rate, it is a variable rate.1119

Notice that dy dt is negative because y is decreasing.1126

Example number 3, a water tank has the shape of an inverted circular cone.1149

The base has a diameter of 5ft while the height is 7 ft.1155

It is being emptied at a rate of 1.8 m³/min.1162

The water is leaving this tank at 1.8 m³/min.1166

How fast does the water level dropping, when the water level is 3.5 ft?1171

We are looking at something like this.1176

We have an inverted cone, something like that, and it is full of water.1178

This is the water.1187

Let us go ahead and do just a full side view of this.1193

It looks something like this.1198

This is the water level.1202

They said the base has a diameter of 5 ft.1206

This is going to be 5 ft, the height is 7 ft.1209

It is being emptied at a rate of 1.82 m³/min.1216

This is the rate that they give us.1220

This is dv dt, it is the water is being emptied.1222

It is -1.8 m³/min.1229

In other words, this is how much is leaving, it is decreasing.1233

How fast does the water level dropping?1238

The water level is this height right here, from here to here.1241

Let us go ahead and call it h.1245

What they want is dh dt.1247

The rate that they give us is dv dt.1254

The rate that they want is dh dt.1256

We need to find a relationship between the volume and the h.1260

Let me write this out.1273

We do not have a direct relation between volume and height.1277

But we do have a relation among three variables, among volume, height, and radius.1292

That equation is the volume of a right circular cone is equal to 1/3 π r² × h.1311

Let me go to the next page here.1323

I have volume = 1/3 π r² h.1326

We want only h on the right side of the equality.1333

This is single variable calculus.1337

I’m relating two variables.1339

Volume, I have two variables on the right, r² and h.1342

I need just h, I need to find a relation between r and h, that I can substitute in for r.1346

We want only h on the right side.1355

In other words, we want volume to be only a function of h.1368

We must find a relation between r and h, that is some r = some function of h,1378

and substitute in for r in our equation.1406

I’m going to take half of my cone.1423

We said that this height is 7.1435

The diameter was 5, the radius is 2.5.1442

The height of the water level is this thing right here, that is h.1448

This is r, notice how as the water level drops, r actually gets smaller.1454

It goes toward the tip of the cone.1461

These are similar triangles, this one and that one.1463

I have a relationship here.1470

I can write 7 is to 2.5.1472

Let me go back to blue here.1478

I have 7 is to 2.5, as h is to r.1480

This gives me 7r = 2.5 h.1489

I solve for r, r = 2.5 h/7.1496

I stick this value into here and I will get an equation only with h on the right side.1502

V = 1/3 π r² h, becomes v = 1/3 × π × 2.5 h/7² × h.1516

I do all of my math here and I end up with v = 0.1336 h³.1539

I have my equation that relates the two variables, v and h.1551

Now I differentiate with respect to t.1557

I have got dv dt = 0.1336 × 3 h² dh dt.1566

I get dv dt = 0.4007 dh dt.1585

I end up solving for dh dt.1596

Therefore, dh dt is equal to the dv dt divided by 0.4007 h².1602

I think I forgot the h² on the previous page.1618

They want this for a water level that is 3.5.1623

H = 3.5, I stick that in there.1642

Therefore, I get dh dt is equal to, the dv dt that was -1.8.1648

And then, we have 0.4007 × 3.5².1657

That leaves me with a dh dt is equal to -0.3667.1667

I think it was m/s because the height is a distance.1679

Notice that it is negative, that means the water level is decreasing,1686

which is what you expect when a tank has been emptied with some water.1690

The negative sign means the height is decreasing, which makes physical sense.1697

Again, if you ended up with something like a positive here, hopefully, you will stop and say that does this make physical sense?1715

No, it does not.1722

If water is being emptied, water level is not dropping, it is rising.1724

Somewhere along the way, there was a minor arithmetic mistake.1727

You can use your final answer to make sure you ended up in the right place.1730

It is decreasing, in other words, water level is dropping.1736

Clearly, the most difficult aspect of these is just arranging it, drawing the picture, seeing what is going on, picking the variables.1745

Finding with is changing, what relation do I use?1752

Sometimes, it will be obvious.1756

When you clearly have a triangle, sometimes it is not going to be so obvious.1758

Let us see, example number 4.1766

One car is traveling east of 40 mph toward a certain intersection, while another is traveling north at 50 mph towards the same intersection.1770

How fast does the distance between the cars changing,1781

when car 1 is 0.5 miles from the intersection and car 2 is 0.6 miles from the intersection?1783

Let us draw this out.1791

Here is how our intersection, I will just put it right over here.1793

We have one car over here, that is actually moving at 40 mph.1797

I’m going to draw a little line towards the intersection.1808

That is one rate.1816

Another is traveling north at 50 mph.1819

I put the other car here and it is traveling this way at 50 mph.1822

How fast does the distance between the cars changing?1833

They want this.1836

Let us assign some variables.1841

I’m going to call this x, I'm going to call this y, and I’m going to call this z.1843

The rate they gave me is, they gave me dx dt and they gave me dy dt.1855

They told us it is travelling east of 40 mph towards a certain intersection.1859

Dx, how fast is x changing?1864

It is getting smaller at a rate of 40 mph.1867

Therefore, dx dt = -40 mph.1872

Why, this car is traveling 50 mph this way.1886

Therefore, y is getting smaller.1889

Dy dt = -50 mph.1892

I will do it as mph.1900

What they want is how fast does the distance between them is changing?1903

They want dz dt.1907

Dz dt equals what?1911

I need a relation between x, y, z.1915

Dx dt, dy dt, what they give me, what they want is dz dt.1923

These are the variables involved, I need relation between those variables.1928

I have one, I have x² + y² = z².1932

Let us differentiate.1940

Once I have the equation, I can go ahead and differentiate.1945

I have got x² + y² = z².1949

This is going to be 2x dx dt because we are differentiating with respect to time, + 2y dy dt = 2z dz dt.1955

Let us go ahead and cancel the 2.1971

I end up with dz dt is equal to x dx dt + y dy dt/ z.1976

Let us plug in what we know.2007

Dz dt =, they said when x is 0.5 miles away from the intersection, that is going to be 0.5.2011

Dx dt that is the -40.2021

They said when y is 0.6 miles from the intersection, that is 0.6.2025

The rate of change of y is -50/ z.2029

What is z?2039

We have a relation for z.2045

We have z² = x² + y².2047

Z² = 0.5², I will do it this way, + 0.6².2052

I get z² = 0.61 which means that z is equal to 0.781.2065

That number, I put in there, and then I solve this.2077

I get dz dt = 0.5 × -40 + 0.6 × -50 all divided by 0.781.2083

I get that my rate of change of distance between them, per unit time, is equal to -64 mph distance over time.2101

The distance between those cars.2121

We have this going this way.2130

This distance right here is changing when this guy is 0.5 miles away and when this guy is 0.6 miles away.2134

This distance is changing at 64 mph.2143

It is negative, it is decreasing because they are getting closer and closer, therefore, this is getting shorter.2147

The negative sign tells me that the distance between them is decreasing, and that is the rate.2155

Again, notice it is not constant, it depends on x, it depends on y, and it depends on z.2160

The rate at which z is changing depends on x and y and z.2170

A man 6ft tall on the ground watches a bird flying horizontally at a speed of 7 m/s and an altitude of 200ft above the ground.2183

What is the rate of change of the angle the man’s line of sight with the bird makes with the horizontal,2192

when the bird is 300ft from the man?2199

Let us go ahead and draw this out.2202

I can go back to blue here.2204

Here is the ground and I have a man who is 6 ft tall.2207

They say he is watching a bird flying horizontally.2217

Let us put the bird over here, he is flying at 7 m/s.2221

The bird is over here.2228

They tell me that its altitude is 200ft off the ground.2231

We said that this guy is 6ft tall.2239

What is the rate of change of the angle the man’s line of sight with the bird?2244

The man’s line of sight with the bird is this thing right here.2248

What is the rate of change the angle of the man’s line of sight the bird makes with the horizontal?2255

The horizontal is right here.2260

The line of sight, the horizontal, this is the angle.2267

We will call that θ, when the bird is 300ft from the man.2270

This distance is, I’m just going to go ahead and call that z.2279

Let us see, I have got myself a little triangle here actually.2287

This is a fixed distance, this is going to be 200ft - the height of the man.2293

This height right here is going to be 194ft.2302

Let me stick with the variable that I use.2315

Let us call this x.2318

Here, I have got this triangle and what they want to know is dθ dt.2322

That is what they want.2330

What rate do they actually give us?2334

I’m going to leave that as z.2338

The rate that they give us is horizontally at a speed of 7 m/s.2342

Therefore, this distance right here, we will call that x.2349

It is x because the bird is flying horizontally.2354

It is flying this way, that means this distance between the man directly below the bird, that changes at 7 m/s.2357

That is going to be dx dt, that is 7 m/s.2369

This is the rate that they give us, this is the rate that they want.2378

Therefore, I need to find a relation between x and θ.2383

What relation exists between x and θ?2394

This triangle, with other information that I have, I know this height.2399

Actually, what I can do is I can write the following.2405

I can write tan θ.2408

I can use tan θ = 194 divided by x.2409

That is a relationship between θ and x, two variables, perfect.2419

We have a relation, time to go ahead and go to the next page here.2425

Let us write our relation again.2429

We have the tan θ = 194/x.2430

Now that we have a relation, we can go ahead differentiate.2435

The derivative of tan θ is sec² θ dθ dt.2439

The derivative of this is -194/ x² dx dt.2449

I solve for dθ dt because that is what I want.2460

Dθ dt = -194 divided by x² sec² θ × dx dt.2463

I have the dx dt, they gave us dx dt.2481

They said that that was 7 m/s.2488

Therefore, this is going to be -194 × 7/ x² sec² θ.2492

Now we just need to find x².2508

Let me go back to blue here, actually, let me try black.2511

We need to find what x is and we need to find what sec² θ is, so that we can plug it in and solve for that.2514

Our questions now are, what is x and what is θ?2523

Rather, what is sec² θ?2535

Let us take care of the x first.2545

We have this triangle, remember, this was 194.2551

Here was the man was right here and the bird was right here.2559

They said they wanted this rate, how fast is this θ changing when the bird is 300ft away?2563

This is z, I have a relationship.2573

I have z² = x² + y².2575

Z is 300 = x² + 194².2580

I can actually find this.2587

X² is going to equal 52,364, which means x is actually going to equal to 228.8.2588

That takes care of x, that we can plug in here.2602

Or better yet, we already have x², we can just plug that in there.2607

I want sec² θ.2614

I finished my triangle, this is 228.8.2617

Here, the sec θ is equal to 300/ 228.8.2623

Therefore, the sec θ = 1.31.2634

Perfect, now I can go ahead and plug everything in.2640

I have got dθ dt is equal to -194 × 7 which was the dx dt.2642

We are going to divide that by x².2660

This is going to be the 228.8² × sec² θ, 1.31².2663

When I solve that, I end up with -0.015.2676

This is an angle, it is in radians per second.2684

I hope that made sense.2691

Again, there is a rate that you are given, a rate that you want.2696

Identify those and try to figure out some relationship between them.2700

Once you have the relation between them, exclusively with those whatever is given, whatever is asked for, then differentiate.2704

Rearrange the equation and then use the other information that you have, in order to fill in the rest.2713

In this case, we needed the x and we needed the sec θ.2720

We found the x first and then we found the sec θ.2723

And then, we plug in and we solve.2727

Thank you so much for joining us here at

We will see you next time, bye.2732