For more information, please see full course syllabus of AP Calculus AB

For more information, please see full course syllabus of AP Calculus AB

### Example Problems for Slopes of Curves

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
- Example I: Water Tank
- Part A: Which is the Independent Variable and Which is the Dependent?
- Part B: Average Slope
- Part C: Express These Slopes as Rates-of-Change
- Part D: Instantaneous Slope
- Example II: y = √(x-3)
- Part A: Calculate the Slope of the Secant Line
- Part B: Instantaneous Slope
- Part C: Equation for the Tangent Line
- Example III: Object in the Air

- Intro 0:00
- Example I: Water Tank 0:13
- Part A: Which is the Independent Variable and Which is the Dependent?
- Part B: Average Slope
- Part C: Express These Slopes as Rates-of-Change
- Part D: Instantaneous Slope
- Example II: y = √(x-3) 28:26
- Part A: Calculate the Slope of the Secant Line
- Part B: Instantaneous Slope
- Part C: Equation for the Tangent Line
- Example III: Object in the Air 49:37
- Part A: Average Velocity
- Part B: Instantaneous Velocity

### AP Calculus AB Online Prep Course

### Transcription: Example Problems for Slopes of Curves

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

*Today, we are going to do some example problems for the concepts *0005

*that we have been discussing in the last couple of lessons which is slopes of curves.*0008

*Let us jump right on in.*0011

*It seems like a long problem, it is not.*0015

*It is just there is a lot of information, that is all.*0016

*We just have to sort of keep track of the stuff.*0019

*A large tank can hold 600 gallons of water.*0021

*It takes 25 minutes to fill up the tank.*0025

*At various values of time, the volume of the tank is measured and the table below gives the appropriate data.*0028

*I think I will work in purple.*0035

*Here we have the time in minutes, volume in gallons.*0037

*At the beginning, there is nothing in the tank.*0040

*5 minutes later, there is 30 gallons, 10 minutes later a 100, at 15 minutes there is 220, at 20 there is 350, at 25 that is 600.*0042

*This is a tabular representation of a particular function.*0051

*A, which is the independent variable and which is the dependent?*0056

*What is the average slope between t0 and t5, between t5 and 10, between 10 and 15, 15 and 20, 20 and 25?*0060

*On the average slope that means two points.*0070

*Here are our two points, 0,0, 5,30.*0074

*It is going to be 30 - 0/ 5 – that.*0078

*For the average slope between 5 and 10, it is going to be 100 -30, 10 – 5.*0081

*Express the slopes as rates of change and interpret what these mean physically.*0089

*D, what is the instantaneous slope at t = 20?*0093

*In other words, exactly when the clock hits 20 minutes, what is the instantaneous slope not average?*0097

*We can find an average between 10 and 15 or 15 and 20.*0104

*We can find an average between 20 and 25.*0108

*We want the instantaneous at 20 and what does this mean physically.*0110

*Let us go ahead and get started.*0116

*Part A, which is the independent variable and which is the dependent variable?*0120

*Let us look back at the question.*0124

*The volume in the tank is measured.*0126

*Any time something is measured, that is the dependent variable.*0130

*It is that simple.*0133

*Volume is the dependent variable, time is the independent variable.*0136

*Therefore, what you actually have is volume which is the depended, is a function of time.*0140

*Just t here, I ended up with T here, it does not really matter.*0148

*Volume is a function of time.*0151

*In this particular case, this function is expressed as a table.*0156

*You are going to see functions expressed in 3 ways, primarily as a table,*0163

*as an actual graph, or as an actual function that they give you.*0167

*Like, let us say, v = 13 t³, an explicit equation.*0171

*A function can be given to you as a table, it can be given to you as a graph.*0176

*It can be given to you as an explicit function.*0179

*In this case, the table tells us, because we know that volume is measured, it tells us the volume is a function of t.*0182

*We do not have the explicit function.*0189

*But that is okay, we have the table of values, we should be able to extract some data here.*0192

*That takes care of a.*0196

*What is the average slope between t = 0, t = 5, and all that business.*0201

*Let us go ahead and take care of that next.*0206

*But I'm going to actually draw this out so you can see it.*0210

*So we can convert this into some sort of a graph.*0215

*Let me at least go ahead and do that.*0218

*I’m going to take 5, 10, 15, 20, 25.*0225

*I have 1, 2, 3, 4, 5, 600.*0237

*This is 100, this is 300, and this is 500.*0242

*I’m going to graph my points.*0248

*I’m going to plot my points.*0250

*When my points were at 0, it was 0.*0251

*If 5 minutes later, it was 30.*0254

*I’m going to go ahead and put my point there.*0259

*At 10 minutes, there is 100.*0261

*It is right there.*0264

*At 15, they said I had 220 gallons.*0264

*Maybe someplace like that.*0269

*At 20 minutes, they said I have 350 gallons.*0271

*1, 2, 3, it would be someplace like that.*0274

*At 25, I had 600, someplace like that.*0281

*This is the graphical representation.*0285

*This was volume and this was time t.*0291

*Volume was some function of time.*0297

*Now this gives us discreet values, individual ones.*0300

*It is pretty fair to say that, let us go ahead and do this in black.*0305

*If I were to extrapolate, I’m looking at some sort of function like that.*0310

*There you go, if I need to deal with anything in between.*0316

*The idea is there is another branch of mathematics called numerical analysis, *0319

*where you can actually find the best fitting curve that matches these data points*0323

*because I only measured for these 5 or 6 values.*0330

*But the idea is that, it is some function that we actually can make continuous, nice, smooth curve.*0333

*We said part b, we want to find the average slope between here and here, here and here, here and here.*0343

*Essentially, what we are doing, average slope means to find the slope of that line segment, *0353

*find the slope of that line segment.*0362

*That is what we are asking, average slope, secant line, the slope of the secant line, the line that connects two points on the graph.*0370

*Let us go ahead and do, average slope between 0 and 5 minutes, *0381

*that is going to be volume final - volume initial/ time final - time initial.*0399

*The volume here was 30, the volume here is 0.*0409

*The time here is 5, the time here is 0.*0415

*We have 6.*0422

*The average slope for the time interval 5 to 10.*0429

*Now we are finding the slope of that.*0432

*This is going to be a certain xy value.*0435

*This is going to be a certain xy value.*0438

*Again, it is going to be Δ y/ Δ x.*0440

*The Δ y is going to be, when I look at the table of values.*0444

*At 10, the value was 100 so it is going to be 100 - the value here which was 30/ 10 – 5.*0449

*This value is going to be 14.*0462

*The average slope from 10 to 15, I take a look at the y values and I divide by the difference of the time values.*0468

*It is going to end up being 220 – 100.*0479

*220 is the y value at 15, 100 is the y value at 10/ 15 – 10.*0484

*I get a value of 24.*0494

*The average slope on the time interval from 15 to 20, it is going to end up being, at 20, the value is 350.*0499

*At 15, it is 220, this is 20 – 15 and I get 26.*0514

*The average value from the final time interval from 20 minutes to 25 minutes.*0523

*The y value of 25 minutes is 600, it is this line segment right here.*0529

*The y value at 20 is 350 divided by 25 – 20 and I get 50.*0536

*There we go, I have got 6, 14, 24, 26, 50.*0551

*These are the average slopes from these time intervals.*0555

*0 to 5, 5 to 10, 10 to 15, 15, 20, 20, 25.*0558

*That is all I’m doing.*0561

*You know what average slope mean, find the slope of the secant line, two points.*0563

*Let us take a look at part c.*0569

*Part c asks, it said express these average slopes as rates of change and interpret them physically.*0571

*We said in the last lesson, let us just take 0 to 5.*0599

*Our 0 to 5 time interval, we said that the average slope was 6.*0605

*We know our function was volume is a function of time.*0614

*Therefore, volume is expressed in gallons, that was the unit per 1 minute.*0622

*The independent variable, the unit of the independent variable is the denominator.*0634

*The unit of the dependent variable is the numerator.*0639

*The 6 is the numerical value of the slope, physically, 6 gallons per 1 minute.*0643

*What this says is, on average, because we calculated an average slope.*0651

*On average, over the time interval 0 minutes to 5 minutes, the tank is being filled 6 gallons for every 1 minute that passes.*0659

*That is what this is saying.*0695

*6 is numerical value that I found for the average slope.*0700

*Volume, gallons is a function of time.*0704

*Seconds, in this case actually was in minutes that we measured it in.*0707

*It is 6 gallons per minute or 6 gallons per 1 minute.*0711

*In general, on average, between 0 and 5, for every minute that passes, 6 part of water is being added.*0716

*That is the physical interpretation.*0724

*This is a rate of change, it is the rate at which the volume changes per change in time.*0725

*It is a rate of change, how fast something is changing per something else.*0732

*When you hear rate, there is a per in there somewhere, it has to be.*0736

*Let us not do all of this, let us just do like to 0 to 5, I will skip 5 to 10, I will skip the 10 to 15.*0747

*Let us do the 15 to 20, it is the same thing.*0753

*The time interval from 15 to 20, the numerical value that we found for the slope, the average slope was 26.*0758

*What that means is that, it is actually 26/1.*0764

*It means 26 gallons per 1 minute.*0768

*On average, during the time period between 15 and 20, for every minute that passes, 26 gallons is being pumped into that tank.*0777

*It is a rate of change, the change in volume per unit change in time.*0788

*That is all that is going on.*0796

*We might as well go ahead and do the last one, 20 to 25, same thing.*0798

*The numerical value that we ended up getting I think was 50, if I’m not mistaken.*0804

*This is 50 which is the same as 50/1, that means 50 gallons per 1 minute.*0810

*During the time increment from 20 minutes to 25 minutes, after I started filling up the tank, *0830

*during that time, on average, for every minute that passes, I’m adding 50 gallons to the tank.*0837

*The volume is changing 50 gallons per 1 minute.*0842

*It is a rate of change.*0848

*Notice that these rates of change are not constant.*0849

*Between 0 and 5, it is 6 gallons per minute.*0852

*From 15 to 20, it is 26 gallons per minute.*0854

*20 to 25, it is 50 gallons per minute.*0857

*The rate of change is changing.*0862

*The average slope is changing.*0866

*It is filling up faster, for every minute that goes by.*0871

*More water is being pumped in.*0874

*It is like I'm turning the faucet, for every moment, I’m actually opening up more and allowing more water to flow in.*0877

*Let us see what we have got.*0887

*Let us see if I got more blank page, I do.*0894

*Part d, it asks for the instantaneous slope at t = 20.*0896

*They also want us to interpret this physically.*0912

*So far, what we have done is we have calculated averages, now they want an instantaneous.*0920

*What they are asking us to find is a derivative.*0923

*They are saying 20 minutes after I start filling the tank.*0926

*Exactly at 20 minutes, how much water is being pumped into the tank, in gallons per minute?*0929

*They want the instantaneous rate of change.*0934

*They want the derivative.*0937

*How do we find the derivative?*0938

*In the previous lesson, we said the only way to find the derivative is to find some function, whatever this function is.*0940

*V = f(t), some explicit function, to find the derivative, to differentiate it.*0947

*To find that derivative and then plug in 20 into that.*0952

*We do not have that here.*0956

*However, there is a way to do it graphically and that is what I’m going to discuss next.*0958

*Before I do that, I want to discuss what is it that I actually use, when I work graphically.*0963

*When I work graphically, and I’m going to introduce a piece of software that I use*0968

*to make my life a little bit easier and you are welcome to do it.*0975

*Of course, you are going to be required to use your calculators for much of the work that you do.*0979

*Certainly, when you take the AP exam, you are going to have to use your calculators.*0985

*You are not going to have software at your disposal.*0986

*It is up to you, if you want to work with your calculator that is fine.*0989

*If when you are doing it, you happen to have a computer in front of you, if you would much rather work on computer, *0993

*as long as you know how to work on your calculator, that is fine.*0998

*I just happened to prefer working with online stuff.*1001

*Occasionally, I will graph things by hand.*1005

*Oftentimes, I will graph things on the computer.*1007

*Before I continue with this problem, let me introduced this piece of software.*1011

*When we are working graphically, I use the online grapher called the desmos.*1015

*Many of you have probably heard of it, if not, not a problem.*1029

*Grapher called desmos which is just at www.desmos.com.*1033

*It is absolutely fantastic and so easy to use.*1041

*Literally, it will take you like 3 seconds to learn how to use it.*1044

*When you pull up www.desmos.com, on the home page, you will see a large red button that says launch calculator.*1048

*This is a big button and it will say launch calculator.*1078

*Press that button.*1084

*Press this button and you will get a graphing screen.*1092

*At this point, you can start graphing.*1110

*Enter a function, start graphing.*1112

*Now start graphing.*1116

*Do not worry, I'm actually going to devote on the next few lessons to a quick tutorial on what desmos is.*1123

*I will go ahead and I will pull it up, show you how to use it, *1130

*show you how to enter the functions and show you the few things that you are going to need.*1132

*It is going to be very quick.*1136

*If you want, just go to it and you can figure it out yourself.*1138

*That little help button actually is very short.*1141

*One thing I love about desmos, it is very intuitive and there is not a lot of discussion, in terms of the help.*1144

*They present only what you need, very quick, easy, manageable terms.*1149

*You can start using it right away.*1155

*In any case, that is that.*1158

*Let us go ahead and return to the problem in hand.*1159

*Again, we are concerned with trying to find an instantaneous slope, an instantaneous rate of change,*1165

*the derivative at a value of t = 20.*1172

*Let us see, where is it, here we go.*1179

*Essentially, what I have done with desmos is I went ahead and plotted the values. *1182

*This is the time t and this is the volume.*1187

*At time 0, there is nothing in it.*1190

*5 there is 30, at 10 there is 100 gallons.*1191

*15 to 20, I have plotted these as points that you can see.*1193

*What I have done is, I’m going to go ahead and connect these with individual line segments.*1200

*When I do that, I get that.*1206

*I have at least some sort of rough idea of what this looks like.*1211

*Again, mind you, this is usually a smooth curve which we can use numerical analytic techniques to find it.*1213

*For our purposes, this is absolutely fine.*1219

*20, this is the point that we are interested in, right there.*1221

*Let us go ahead and talk about what is that we are going to be doing here.*1229

*The instantaneous slope that we are looking for means the slope of the tangent line through that point.*1233

*When you have a graph, you can actually find the derivative, the slope of the tangent line.*1245

*Here is how you do it, by doing it graphically.*1253

*What you do is you draw the best line that you can, the best tangent line to the curve.*1255

*That is the best one that I could come up with.*1264

*That is the first thing that you do.*1267

*Let us actually write that down.*1269

*Draw the best tangent line.*1273

*Next, what you are going to do is you are going to pick two points on that line.*1285

*Pick two points on the tangent line.*1292

*Three, you are going to calculate the slope between those two points.*1298

*Use the two points to calculate the slope.*1306

*This is a graphical technique for actually finding the slope of the tangent line, the derivative at a particular point.*1317

*If you do not have an explicit function, then you can find the derivative of.*1322

*You can just go ahead and draw out the best function, draw your tangent line, and then just pick two points.*1325

*I’m going to pick a point there and a point there.*1331

*I'm going to find out what they are.*1333

*In this particular case, the points that I end up actually picking were the points, I think, 24.*1336

*It ended up being 500, 24, 500, I read them right off the graph.*1357

*Over here, I think I ended up picking 15 and 175.*1363

*At the point 15, I go to 175, that is that point.*1368

*Drew my best tangent, take those two points, and now I find my instantaneous slope.*1374

*My instantaneous slope = 500 – 175, change in y/ change in x/ 24 – 15.*1380

*I get an answer of 36.1.*1394

*There is another approach that I can use.*1403

*Let me go ahead and just describe it.*1405

*I notice that I have data points for 15, that is 15 to 20, and I have a data point for 25, that is 25 and 600.*1408

*If I draw a secant line, I notice that the secant line, in this particular case, it looks reasonably symmetric from this point to this point.*1421

*A slope of the secant line looks almost parallel to the slope of the tangent line.*1431

*What I can do, if I wanted, I can find the slope of the secant line between this point and this point.*1438

*That is another way of doing it, it is going to be the same.*1444

*The only difference between this line and this line, as far as our slopes are concerned is you just moved it down.*1447

*Slope wise, parallel lines have the same slope.*1453

*If I want, for a particular point, a tangent at line at that point, I can pick points that flank it, that I have values for, if it is reasonably parallel.*1457

*In general, I do not do that.*1467

*In general, I just draw the best tangent line that I can.*1468

*I pick two points on that line randomly and I find the slope.*1471

*In this particular case, we found 36.1.*1476

*As far as the physical interpretation is concerned, the slope is a rate of change.*1481

*An instantaneous slope is an instantaneous rate of change.*1497

*It is 36.1 gallons per 1 minute.*1502

*In other words, 36.1 gallons per minute.*1511

*This means that at time = 20, this means at exactly 20 minutes after I started filling up the tank,*1518

*the tank is being filled 36.1 gallons per every minute that passes.*1536

*Again, I mentioned once before, let me go ahead and do this in red.*1566

*The particular thing that I have done with desmos is I have connected these particular data points with lines.*1573

*We know, in general, that is not a line.*1579

*It is going to be some sort of a smooth sort of function, like that.*1581

*Again, let me return to what it is that I mentioned early on.*1588

*You are going to be given a function, in this particular case, volume is a function of time.*1591

*You are going to be given a function in three different ways.*1598

*You could be given a table which you can graph.*1600

*You can be given the graph, just given the graph as it is.*1606

*That is another way that you are going to be given a particular function.*1611

*The 3rd way is you are actually given an explicit function.*1613

*For example, this could have been v = ½ t³, something like that.*1616

*That is an explicit function of volume as a function of time that passes.*1626

*You can be given a table, a graph, or an explicit function.*1632

*Here we use a table to make a graph.*1638

*From the graph, we physically drew a tangent line.*1642

*From that tangent line, we extracted the data.*1646

*Now the idea behind calculus is our preference is always to be able to find an explicit formula.*1649

*Because once we have an explicit formula, we are going to develop techniques.*1655

*In other words, differentiating this and coming up with a derivative for this,*1659

*some function of t, some v’(t) which is the derivative of this function.*1665

*So that at any place along t, I can just plug in a particular value and find what the instantaneous slope is.*1672

*We are not always going to be that lucky.*1678

*More often than not, we will be working with functions, because again, we are teaching calculus.*1681

*We are trying to develop the techniques of calculus.*1685

*But understand in the real world, oftentimes, you will be given a table of data that you have to convert to a graph.*1687

*You will either work from the graph directly or use numerical techniques to approximate, find a function.*1693

*From there, work your calculus magic.*1699

*Let us move on to another problem.*1705

*This one, let y = √x – 3.*1707

*The point 4,1 is a point on this curve, we will call it p.*1711

*A variable point q whose coordinate is xy also lies on the curve for various values of x.*1716

*P is fixed, q is anywhere else on the curve but different values of x.*1722

*q is the point that is actually moving.*1728

*Using a calculator, calculate the slope of the secant line pq, for the following values of x.*1732

*4 is the point that is fixed, that is the x value that is fixed, its y is 1.*1740

*We are going to plug in for x, here 3, 3.5, 3.9, 3.99, 3.99.*1745

*Notice we are approaching 4 from below, 3 going up to 4.*1753

*If this is the x value 4, this is 3, this is 5.*1761

*We are going to be taking different values of x getting close to 4.*1765

*Also, we are going to start and work our way down to 4.*1769

*They are different values.*1773

*q is going to be 3, f(3), 3.5, f(3.5), 3.9, f(3.9), 3.99, f(3.99), and so on.*1775

*Using the calculator, calculate the slope of the secant line for all of these.*1786

*And then B says, using the results of A, speculate it to the value of the instantaneous slope of this function at the point.*1791

*Now they are saying use the average slopes, to see if you can come up with some idea of what the instantaneous slope is.*1798

*Using the slope you get, find an equation of the tangent line through the point.*1806

*We already know how to find the equation, if we are given a slope and a point.*1811

*It is y - y1 = slope which is m × x - x1.*1816

*We have the point, now we just need to find the slope which we are going to get from part B.*1825

*We just plug it into here and get our equation.*1829

*First, we need to find what is going on there.*1831

*Let us draw this out and see what it is that we are actually doing.*1836

*That is fine, let us draw this out.*1842

*I think I’m going to go to purple because I really like purple, it is very nice.*1849

*This is 1, this is 2, this is 3, 4, 5, 6.*1860

*Our function is y is equal to √x – 3.*1866

*This is the radical function shifted over 3 to the right.*1872

*It starts there at 3 and it goes that way.*1875

*Our point p is fixed at 4,1.*1887

*This is our point p.*1891

*Our q, here is 5, this is the value for 5.*1894

*Our q is going to be different values of 3, 3, 3.5, 3.9.*1899

*q is going to move from here to here.*1904

*To here, it is going to get closer that way.*1907

*Here, this is going to be the other set of q.*1910

*It is going to be this point and then it is going to be 4.5.*1913

*It is going to be this point.*1917

*q is a movable point, q is going to approach p from the left.*1918

*q is going to approach p from the right.*1922

*We want to find the slopes of the secant lines for different values of q.*1924

*That is what we are doing, we are finding average slopes.*1932

*We are finding pq.*1934

*This is going to be one q, this is going to be another q.*1936

*Let us say q1, q2, q3, q4, q5, q6, q7, q8, q9, q10, things like that.*1940

*This is that, therefore, our q is the point xy.*1950

*We know what y is, it is going to be the point x × √x – 3.*1962

*Those are the coordinates of the various q.*1968

*It is based on whatever the value of x is.*1970

*Let us go ahead and magnify this part a little bit and see what we are dealing with.*1975

*I’m going to go ahead this way.*1984

*I’m going to exaggerate the curve a little bit, just to make it a little bit easier for me to deal with.*1989

*Let us say this is our 3, this is our 4, this is our 5.*1997

*This is the point p, this is the point 4,1.*2005

*We are going to find that q and we are going to find the slope of that line and another q, the slope of that line and another q.*2009

*For 5, we are going to find the slope of that line and we are going to move q over here.*2023

*We are going to find the slope of that line, that is what we are going to do.*2027

*The average slope = Δ y/ Δ x.*2033

*p is the point 4,1, that point is fixed.*2048

*That does not change, that is one of our points.*2055

*q is the point x √x - 3.*2058

*Our average slope for pq is going to equal y2 - y1/ x2 - x1.*2069

*It is going to be √x - 3 - 4/ x.*2076

*This is, sorry, -1/ x – 4.*2084

*We are going to find the slope of that line, that line, and then another one, and another one, for all the different values of x.*2090

*3, 3.5, 3.9, 3.99, 3.9999, and then 5, 4.5, 4.1, 4.01, 4.001.*2102

*Notice what is happening to these lines.*2116

*At some point, you are going to end up getting that.*2120

*From this direction, you are going to end up getting that.*2124

*You are actually going to end up getting the tangent line that we will see in just a minute.*2128

*Let us go ahead and do this.*2131

*Again, I went ahead and I use desmos to calculate the values for me.*2133

*p is 4,1, for the different values of x, 3, 3,5, 3,9, 3,99, 3,99, 5, 4.5, 4,1, 4,01, 4,001.*2138

*These are the values of y that I get.*2149

*This is q1, this is q1, q2, q3, q4, q5, q6, q7, q8, q9, and q10.*2153

*The slope was this equation right here.*2163

*It is going to be y2 - y1/ x2 - x1.*2166

*y2 - y1/ x2 - x1.*2176

*y2 - y1/ x2 – x1, these are the various slopes.*2180

*When the x value of q is 3, the slope is 1.*2186

*When the slope of the next line segment, when the x value of q is 3.5, the slope is this, the slope is this.*2192

*This gives me the different slopes.*2199

*Some things to notice, the x value of q is running from 3 to 4.*2203

*It is a line segment, q is getting closer to p from below.*2218

*As q goes from 3 to 4, notice what the slope approaches.*2226

*The slope approaches, it looks like it is 0.5.*2239

*Let me rewrite this.*2274

*This is p and those are the things that we are finding.*2283

*This is q, this is q.*2289

*We are moving q closer to p, closer to p, find the line segment.*2291

*As q approaches p from below, p is 4, from 3 to 4, from below.*2298

*The line, here is the 4, here is the 3.*2313

*We are approaching from less than 4.*2315

*We say from below.*2318

*The slope approaches 0.5, the average slope, the slope of the secant line.*2323

*But notice the difference between 3.99 and 4 is really a short line.*2332

*Now as q approaches p from above, if we go from 5 down to 4.*2339

*Once again, 4.1, 4.4, .449, .48, .498, .499.*2348

*As q approaches p from above, the slope, again, it approaches 0.5.*2356

*Let us talk about what this means.*2377

*Essentially, what we are looking at is this.*2379

*I think you guys can already figure out what is going on.*2391

*That is 4, that is 5, that is 3.*2396

*I need to exaggerate it so you can actually see it.*2403

*That is not very good, that is okay.*2411

*I fixed p and I started with q1 over here and I moved q to 3.5.*2416

*In other words, I’m moving q closer and closer.*2424

*I’m finding that slope and that slope.*2426

*Notice these slopes, here is the q, here is the p, from your perspective.*2429

*q is approaching p along that curve right there.*2438

*The slope is rising.*2442

*Eventually at this point, you are going to hit, if I take 3.99, if I get infinitely close to 4,*2447

*the slope of that secant line is going to be the slope of the tangent line.*2455

*It is going to turn in to the tangent line.*2459

*The same thing from this side, this slope.*2461

*From your perspective, p is here, q was up here.*2464

*The curve goes this way, from your perspective.*2467

*Now I’m moving q closer.*2472

*If q was closer, this line segment, the slope is going to increase.*2475

*Increase until I hit this.*2482

*That is what is going on.*2484

*Part B, it is reasonable and we saw that this average slope approaches 0.5 from below,*2487

*this average slope approaches 0.5 from above.*2496

*It is reasonable to conclude at x = 4, the instantaneous slope = 0.5.*2501

*That is another technique that you can use, in order to find the instantaneous slope at a given.*2526

*You can take average slopes and get closer and closer and closer, move a particular fixed one point.*2535

*The point where you are interested in the instantaneous slope.*2542

*Take a point q and move it move it this way and calculate all of these average slopes.*2545

*As you get infinitely close, 3.99, 3.999, 3.99999,*2552

*you are going to find that the slope is getting close to one number and it is not moving.*2557

*In this particular case, it is 0.5.*2561

*In fact that is going to be the procedure that we are going to use.*2565

*Earlier on, I introduce this idea of the limit as h goes to 0 of f(x) + h – f(x)/ hs.*2567

*As h goes to 0 of this thing, that is the process that we are doing.*2586

*We are going to find a secant line, that is this.*2590

*We are going to take h, this distance to 0.*2595

*We are going to make an infinitely close.*2600

*At some point, a number is going to emerge.*2602

*A function is going to emerge.*2606

*That is going to be this.*2608

*I can do it right now, I can just take limiting values.*2610

*I can find secant, secant, secant, when it gets close to one side.*2614

*And then, secant, secant, when it gets close to on the other side.*2618

*If from below and from above, the two slopes happen to meet, *2620

*if they happen to end up being the same number, we define that as the actual slope of that.*2626

*We define that as the derivative.*2630

*Hopefully, you understood the process that we went through here.*2632

*Final, the equation of the tangent line.*2638

*The equation of the tangent line.*2645

*Again, if some of the stuff is not altogether that clear, do not worry about it, we will continue to discuss it.*2651

*It is not a problem.*2655

*We said that the equation of any line is y - y1 = the slope × x - x1.*2657

*We have the slope, they said the slope is 0.5 at the point 4,1.*2668

*We have the point p which is 4,1.*2678

*y - 1 = ½ × x – 4, that is the equation of our tangent line.*2682

*The slope of the tangent line ½, that is the instantaneous slope at x = 4.*2695

*That is the derivative of the function y = √x – 3, at the point where x = 4.*2704

*The numerical value of the derivative is ½.*2714

*It is the slope of the tangent line to the curve at that point.*2717

*Let us see what else we have got here.*2725

*Let me write that out again, just in case.*2746

*This was another graphical technique, another graphical method for finding the instantaneous slope at a point p, *2756

*by moving a point q closer and closer to p along the curve.*2788

*Calculating the various average slopes of pq.*2813

*As q gets insanely close to p, the slope of pq gets insanely close to the instantaneous slope.*2831

*Again, what we are doing is we are fixing the point p.*2866

*We are interested in that slope.*2877

*We pick a point q over here, we find that slope.*2881

*We move q over here, we find that slope.*2884

*We move q over here, we find that slope.*2887

*We move q a little closer, a little closer, that slope.*2891

*Notice how these slopes, the lines, they are coming up.*2894

*The slope, the numerical value of the slope is coming up and eventually it is going to be that.*2899

*As q gets infinitely close to p, the slope of pq is going to actually b, this instantaneous slope.*2906

*When we do the same thing from above.*2912

*That is that slope, that is another q.*2915

*Then we find that slope and that slope, from your perspective.*2917

*This slope is coming up, this is pq, pq, pq, pq.*2926

*At some point, the slope itself is going to approach a value.*2931

*If the numerical value of the slopes, as you go from below, approach a number*2934

*and it happens to be the same number that it approaches as you go this way,*2941

*that number that we got, in this example, 0.5, that is the slope of the tangent line.*2945

*There is another graphical technique.*2952

*I have two graphical techniques, you do not need an actual function.*2954

*You just need a graph, you can even draw the line, pick two points, find the slope.*2958

*Or you can fix the point, approach it from below, approach it from above, and see if the slopes actually converge on a single number.*2965

*In this case, it was 0.5.*2974

*If a ball is thrown up in the air with initial velocity of 10 m/s,*2979

*its height above the ground is a function of t, that is t seconds after I throw it.*2983

*It is given by the function, this, explicit function.*2987

*The height as a function of t is 10t -4.9t².*2991

*Find the average velocity for the following time intervals.*2997

*1 to 2, 1.5 to 2, 1.9 to 2, 1.99 to 2, 1.999 to 2.*3000

*Notice 2 is fixed, we are shortening the time interval.*3005

*From your perspective, 2 is here, 1 is here.*3011

*We are going 1, 1.5, what is the average, find the instantaneous velocity at t = 2.*3013

*We are going to do the same thing.*3020

*We are going to find the average, average, average, average, average.*3020

*We are going to see if the slope is approaching some number.*3025

*That is what we are going to do.*3032

*The first thing we want to do is let us go ahead and draw this out.*3034

*A, I have got h(t), this I’m going to draw by hand.*3040

*It is 10t - 4.9t².*3043

*I have got this is equal to t × 10 - 4.9t.*3052

*I’m going to this equal to 0 to see where it hits the x axis.*3059

*It hits the x axis at t = 0 and it hits the x axis at 10 divided by 4.9 which is 2.04.*3063

*0, this is 2.04, this is a parabola.*3080

*This is a parabola 10t – 4t², it is a parabola that opens downward.*3089

*I want to see what height is 2.04, at the halfway mark, it is going to be 1.02.*3094

*When I put in 1.02 into h and I solve, I get that it is equal to 5.1, I think.*3103

*If I’m not mistaken, something like that.*3112

*Here is a graph of the function, it is a parabola.*3115

*I'm interested in the time interval from 1 to 2.*3122

*There to there, that is what I'm interested in.*3130

*I know what the slope is, the slope is the change in y/ the change in x.*3136

*It is going to be, t is time in seconds, h is height it is meters.*3145

*This is going to be meters per second.*3158

*The slope is a velocity, the slope is going to give me the velocity of this thing.*3161

*From the time increment, from 1 to 2, my average slope which is my average velocity,*3168

*because we just said that slope is dependent variable divided by the independent variable, it is going to equal from 1 to 2.*3182

*It is going to be a slope of that line.*3198

*It is going to be h(2) - h(1)/ 2 – 1.*3200

*I end up getting -4.7 m/s.*3208

*When I do 1.5 to 2, the average velocity, any variable with the line over it means average = h(2) – h(1.5).*3221

*1.5 is here, now I’m finding that secant line.*3238

*Sorry, not that secant line.*3246

*2 was the one that is fixed, it is 1.5 that is moving.*3253

*I’m finding that secant line, right there, from here to here, that secant line.*3256

*H(2), h(1.5)/ 2 – 1.5.*3265

*I end up with -7.15 m/s.*3269

*I do the same for 1.9 to 2.*3275

*I end up with an average velocity of - 9.11 m/s.*3281

*If I do 1.99 to 2, I find an average velocity of -9.551 m/s.*3293

*If I do 1.999 to 2, I get an average velocity of -9.595 m/s.*3308

*I decided to go one further, 1.9999 to 2, and I found an average velocity of -9.59951 m/s.*3319

*Clearly, I'm approaching -9.6.*3336

*Part B, my instantaneous velocity at t = 2.*3342

*As we approach 2, as t runs from 1 to 2, 1, 1.5, 1.9, gets closer and closer and closer to 2, my slope is approaching -9.6.*3349

*Instantaneous velocity t = 2, as we approach 2 from below, from 1 approaching 2 from below,*3365

*the slope approaches, that is what the arrow means.*3384

*The slope approaches -9.6 m/s.*3389

*Our instantaneous slope which is our instantaneous velocity, which is our derivative at 2 = -9.6 m/s.*3395

*Here is an tabular form.*3415

*Our function is this, at different values of x, these are my y values, my height.*3418

*At the value of 2, it is equal to 0.4.*3426

*The slope that I'm finding is, I'm fixing the point 2,0.4.*3431

*It is going to be the y value of 0.4 - this value which is that/ 2 – this.*3441

*y2 - y1/ x2 - x1, the next point, y2 - y1/ x2 - x1.*3455

*The next point, y2 - y1/ x2 - x1, that is what this says.*3461

*The x values are 1, 1.5, 1.9, 1.99999.*3468

*The slopes 4.7, 7.5, -9.11, -9.55, -9.595, notice the jumps, all of a sudden, are now in the 9 range, now in the 9.5 range.*3473

*It looks like it is approaching the -9.6.*3488

*Therefore, I’m using average slopes, as I get really close to the point 2, where I'm interested in an instantaneous slope,*3491

*I see where the average slope is getting close to.*3505

*If I want, I can keep going, 1.999999999.*3509

*You are going to find that it would not go past -9.6.*3513

*That is going to be my answer.*3518

*If I wanted to, I can also approach it from the other side.*3519

*I can also approach 2 from 3.*3521

*I can go 3 down to 2, I can go 3 to 2.5, 2.1, 2.01, 2.001, 2.0001.*3524

*You are going to see that it approaches -9.6.*3532

*That is what is going on here.*3537

*Hopefully, these three problems have helped you see and hope to clarify what it is that we actually discussed in the previous two lessons.*3539

*Thank you so much for joining us here at www.educator.com.*3549

*We will see you next time, bye.*3551

2 answers

Last reply by: Professor Hovasapian

Fri Aug 12, 2016 6:16 AM

Post by Kitt Parker on August 11 at 03:42:07 PM

at 52.12 I'm very curious about how you derived 2.04. Right before that, you mention plugging in 0 which equal 0 and then the equation is re-written with an answer of 2.04. I'm unsure where this came from.

1 answer

Last reply by: Professor Hovasapian

Sat Feb 27, 2016 5:53 AM

Post by Acme Wang on February 27 at 05:12:10 AM

Hi Professor Hovasapian,

So according to this lecture, as Q insanely close to P(the fixed point), the slope of the secant line(average slope)between Q and P would be the instantaneous slope at point P, right?

Thank you very much!

Sincerely,

Acme

1 answer

Last reply by: Professor Hovasapian

Wed Dec 30, 2015 1:00 AM

Post by Isaac Martinez on December 28, 2015

Hello Professor Hovasapian,

I was wondering if you would please help me understand what is h in example 3 part A? For example, when you're looking for the slope of [1,2], to solve you entered

h(2)-h(1)/2-1 = -4.7

How do you arrive to this answer?

Thank you!