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

### 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

### AP Calculus AB Online Prep Course

### 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

2 answers

Last reply by: Peter Fraser

Sat Nov 18, 2017 4:22 PM

Post by Acme Wang on April 2, 2016

Hi 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