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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## Table of Contents

## Transcription

### 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
- Related Rates
- 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
- Example II: Ladder
- Example III: Water Tank
- Example IV: Distance between Two Cars
- Example V: Line-of-Sight

- 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

### AP Calculus AB Online Prep Course

### Transcription: Related Rates

*Hello, welcome back to www.educator.com 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 information*0301

*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 www.educator.com.*2729

*We will see you next time, bye.*2732

0 answers

Post by Shiden Yemane on November 12 at 12:39:19 PM

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 :))

Sincerely,

Acme