AP Calculus AB 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 concepts0005

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 points0323

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 use0968

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 number2934

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