WEBVTT mathematics/ap-calculus-ab/hovasapian
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Hello, welcome back to www.educator.com, welcome back to AP Calculus.
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Today, we are going to do some example problems for the concepts
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that we have been discussing in the last couple of lessons which is slopes of curves.
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Let us jump right on in.
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It seems like a long problem, it is not.
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It is just there is a lot of information, that is all.
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We just have to sort of keep track of the stuff.
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A large tank can hold 600 gallons of water.
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It takes 25 minutes to fill up the tank.
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At various values of time, the volume of the tank is measured and the table below gives the appropriate data.
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I think I will work in purple.
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Here we have the time in minutes, volume in gallons.
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At the beginning, there is nothing in the tank.
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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.
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This is a tabular representation of a particular function.
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A, which is the independent variable and which is the dependent?
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What is the average slope between t0 and t5, between t5 and 10, between 10 and 15, 15 and 20, 20 and 25?
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On the average slope that means two points.
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Here are our two points, 0,0, 5,30.
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It is going to be 30 - 0/ 5 – that.
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For the average slope between 5 and 10, it is going to be 100 -30, 10 – 5.
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Express the slopes as rates of change and interpret what these mean physically.
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D, what is the instantaneous slope at t = 20?
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In other words, exactly when the clock hits 20 minutes, what is the instantaneous slope not average?
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We can find an average between 10 and 15 or 15 and 20.
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We can find an average between 20 and 25.
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We want the instantaneous at 20 and what does this mean physically.
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Let us go ahead and get started.
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Part A, which is the independent variable and which is the dependent variable?
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Let us look back at the question.
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The volume in the tank is measured.
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Any time something is measured, that is the dependent variable.
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It is that simple.
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Volume is the dependent variable, time is the independent variable.
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Therefore, what you actually have is volume which is the depended, is a function of time.
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Just t here, I ended up with T here, it does not really matter.
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Volume is a function of time.
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In this particular case, this function is expressed as a table.
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You are going to see functions expressed in 3 ways, primarily as a table,
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as an actual graph, or as an actual function that they give you.
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Like, let us say, v = 13 t³, an explicit equation.
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A function can be given to you as a table, it can be given to you as a graph.
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It can be given to you as an explicit function.
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In this case, the table tells us, because we know that volume is measured, it tells us the volume is a function of t.
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We do not have the explicit function.
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But that is okay, we have the table of values, we should be able to extract some data here.
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That takes care of a.
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What is the average slope between t = 0, t = 5, and all that business.
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Let us go ahead and take care of that next.
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But I'm going to actually draw this out so you can see it.
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So we can convert this into some sort of a graph.
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Let me at least go ahead and do that.
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I’m going to take 5, 10, 15, 20, 25.
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I have 1, 2, 3, 4, 5, 600.
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This is 100, this is 300, and this is 500.
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I’m going to graph my points.
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I’m going to plot my points.
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When my points were at 0, it was 0.
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If 5 minutes later, it was 30.
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I’m going to go ahead and put my point there.
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At 10 minutes, there is 100.
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It is right there.
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At 15, they said I had 220 gallons.
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Maybe someplace like that.
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At 20 minutes, they said I have 350 gallons.
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1, 2, 3, it would be someplace like that.
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At 25, I had 600, someplace like that.
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This is the graphical representation.
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This was volume and this was time t.
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Volume was some function of time.
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Now this gives us discreet values, individual ones.
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It is pretty fair to say that, let us go ahead and do this in black.
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If I were to extrapolate, I’m looking at some sort of function like that.
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There you go, if I need to deal with anything in between.
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The idea is there is another branch of mathematics called numerical analysis,
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where you can actually find the best fitting curve that matches these data points
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because I only measured for these 5 or 6 values.
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But the idea is that, it is some function that we actually can make continuous, nice, smooth curve.
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We said part b, we want to find the average slope between here and here, here and here, here and here.
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Essentially, what we are doing, average slope means to find the slope of that line segment,
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find the slope of that line segment.
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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.
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Let us go ahead and do, average slope between 0 and 5 minutes,
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that is going to be volume final - volume initial/ time final - time initial.
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The volume here was 30, the volume here is 0.
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The time here is 5, the time here is 0.
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We have 6.
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The average slope for the time interval 5 to 10.
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Now we are finding the slope of that.
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This is going to be a certain xy value.
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This is going to be a certain xy value.
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Again, it is going to be Δ y/ Δ x.
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The Δ y is going to be, when I look at the table of values.
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At 10, the value was 100 so it is going to be 100 - the value here which was 30/ 10 – 5.
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This value is going to be 14.
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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.
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It is going to end up being 220 – 100.
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220 is the y value at 15, 100 is the y value at 10/ 15 – 10.
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I get a value of 24.
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The average slope on the time interval from 15 to 20, it is going to end up being, at 20, the value is 350.
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At 15, it is 220, this is 20 – 15 and I get 26.
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The average value from the final time interval from 20 minutes to 25 minutes.
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The y value of 25 minutes is 600, it is this line segment right here.
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The y value at 20 is 350 divided by 25 – 20 and I get 50.
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There we go, I have got 6, 14, 24, 26, 50.
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These are the average slopes from these time intervals.
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0 to 5, 5 to 10, 10 to 15, 15, 20, 20, 25.
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That is all I’m doing.
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You know what average slope mean, find the slope of the secant line, two points.
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Let us take a look at part c.
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Part c asks, it said express these average slopes as rates of change and interpret them physically.
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We said in the last lesson, let us just take 0 to 5.
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Our 0 to 5 time interval, we said that the average slope was 6.
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We know our function was volume is a function of time.
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Therefore, volume is expressed in gallons, that was the unit per 1 minute.
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The independent variable, the unit of the independent variable is the denominator.
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The unit of the dependent variable is the numerator.
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The 6 is the numerical value of the slope, physically, 6 gallons per 1 minute.
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What this says is, on average, because we calculated an average slope.
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On average, over the time interval 0 minutes to 5 minutes, the tank is being filled 6 gallons for every 1 minute that passes.
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That is what this is saying.
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6 is numerical value that I found for the average slope.
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Volume, gallons is a function of time.
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Seconds, in this case actually was in minutes that we measured it in.
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It is 6 gallons per minute or 6 gallons per 1 minute.
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In general, on average, between 0 and 5, for every minute that passes, 6 part of water is being added.
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That is the physical interpretation.
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This is a rate of change, it is the rate at which the volume changes per change in time.
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It is a rate of change, how fast something is changing per something else.
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When you hear rate, there is a per in there somewhere, it has to be.
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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.
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Let us do the 15 to 20, it is the same thing.
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The time interval from 15 to 20, the numerical value that we found for the slope, the average slope was 26.
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What that means is that, it is actually 26/1.
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It means 26 gallons per 1 minute.
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On average, during the time period between 15 and 20, for every minute that passes, 26 gallons is being pumped into that tank.
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It is a rate of change, the change in volume per unit change in time.
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That is all that is going on.
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We might as well go ahead and do the last one, 20 to 25, same thing.
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The numerical value that we ended up getting I think was 50, if I’m not mistaken.
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This is 50 which is the same as 50/1, that means 50 gallons per 1 minute.
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During the time increment from 20 minutes to 25 minutes, after I started filling up the tank,
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during that time, on average, for every minute that passes, I’m adding 50 gallons to the tank.
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The volume is changing 50 gallons per 1 minute.
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It is a rate of change.
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Notice that these rates of change are not constant.
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Between 0 and 5, it is 6 gallons per minute.
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From 15 to 20, it is 26 gallons per minute.
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20 to 25, it is 50 gallons per minute.
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The rate of change is changing.
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The average slope is changing.
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It is filling up faster, for every minute that goes by.
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More water is being pumped in.
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It is like I'm turning the faucet, for every moment, I’m actually opening up more and allowing more water to flow in.
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Let us see what we have got.
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Let us see if I got more blank page, I do.
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Part d, it asks for the instantaneous slope at t = 20.
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They also want us to interpret this physically.
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So far, what we have done is we have calculated averages, now they want an instantaneous.
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What they are asking us to find is a derivative.
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They are saying 20 minutes after I start filling the tank.
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Exactly at 20 minutes, how much water is being pumped into the tank, in gallons per minute?
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They want the instantaneous rate of change.
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They want the derivative.
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How do we find the derivative?
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In the previous lesson, we said the only way to find the derivative is to find some function, whatever this function is.
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V = f(t), some explicit function, to find the derivative, to differentiate it.
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To find that derivative and then plug in 20 into that.
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We do not have that here.
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However, there is a way to do it graphically and that is what I’m going to discuss next.
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Before I do that, I want to discuss what is it that I actually use, when I work graphically.
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When I work graphically, and I’m going to introduce a piece of software that I use
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to make my life a little bit easier and you are welcome to do it.
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Of course, you are going to be required to use your calculators for much of the work that you do.
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Certainly, when you take the AP exam, you are going to have to use your calculators.
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You are not going to have software at your disposal.
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It is up to you, if you want to work with your calculator that is fine.
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If when you are doing it, you happen to have a computer in front of you, if you would much rather work on computer,
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as long as you know how to work on your calculator, that is fine.
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I just happened to prefer working with online stuff.
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Occasionally, I will graph things by hand.
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Oftentimes, I will graph things on the computer.
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Before I continue with this problem, let me introduced this piece of software.
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When we are working graphically, I use the online grapher called the desmos.
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Many of you have probably heard of it, if not, not a problem.
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Grapher called desmos which is just at www.desmos.com.
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It is absolutely fantastic and so easy to use.
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Literally, it will take you like 3 seconds to learn how to use it.
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When you pull up www.desmos.com, on the home page, you will see a large red button that says launch calculator.
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This is a big button and it will say launch calculator.
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Press that button.
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Press this button and you will get a graphing screen.
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At this point, you can start graphing.
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Enter a function, start graphing.
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Now start graphing.
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Do not worry, I'm actually going to devote on the next few lessons to a quick tutorial on what desmos is.
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I will go ahead and I will pull it up, show you how to use it,
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show you how to enter the functions and show you the few things that you are going to need.
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It is going to be very quick.
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If you want, just go to it and you can figure it out yourself.
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That little help button actually is very short.
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One thing I love about desmos, it is very intuitive and there is not a lot of discussion, in terms of the help.
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They present only what you need, very quick, easy, manageable terms.
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You can start using it right away.
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In any case, that is that.
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Let us go ahead and return to the problem in hand.
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Again, we are concerned with trying to find an instantaneous slope, an instantaneous rate of change,
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the derivative at a value of t = 20.
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Let us see, where is it, here we go.
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Essentially, what I have done with desmos is I went ahead and plotted the values.
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This is the time t and this is the volume.
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At time 0, there is nothing in it.
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5 there is 30, at 10 there is 100 gallons.
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15 to 20, I have plotted these as points that you can see.
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What I have done is, I’m going to go ahead and connect these with individual line segments.
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When I do that, I get that.
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I have at least some sort of rough idea of what this looks like.
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Again, mind you, this is usually a smooth curve which we can use numerical analytic techniques to find it.
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For our purposes, this is absolutely fine.
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20, this is the point that we are interested in, right there.
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Let us go ahead and talk about what is that we are going to be doing here.
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The instantaneous slope that we are looking for means the slope of the tangent line through that point.
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When you have a graph, you can actually find the derivative, the slope of the tangent line.
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Here is how you do it, by doing it graphically.
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What you do is you draw the best line that you can, the best tangent line to the curve.
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That is the best one that I could come up with.
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That is the first thing that you do.
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Let us actually write that down.
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Draw the best tangent line.
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Next, what you are going to do is you are going to pick two points on that line.
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Pick two points on the tangent line.
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Three, you are going to calculate the slope between those two points.
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Use the two points to calculate the slope.
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This is a graphical technique for actually finding the slope of the tangent line, the derivative at a particular point.
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If you do not have an explicit function, then you can find the derivative of.
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You can just go ahead and draw out the best function, draw your tangent line, and then just pick two points.
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I’m going to pick a point there and a point there.
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I'm going to find out what they are.
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In this particular case, the points that I end up actually picking were the points, I think, 24.
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It ended up being 500, 24, 500, I read them right off the graph.
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Over here, I think I ended up picking 15 and 175.
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At the point 15, I go to 175, that is that point.
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Drew my best tangent, take those two points, and now I find my instantaneous slope.
00:23:00.600 --> 00:23:13.700
My instantaneous slope = 500 – 175, change in y/ change in x/ 24 – 15.
00:23:13.700 --> 00:23:23.200
I get an answer of 36.1.
00:23:23.200 --> 00:23:24.700
There is another approach that I can use.
00:23:24.700 --> 00:23:28.600
Let me go ahead and just describe it.
00:23:28.600 --> 00:23:41.100
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.
00:23:41.100 --> 00:23:51.200
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.
00:23:51.200 --> 00:23:57.800
A slope of the secant line looks almost parallel to the slope of the tangent line.
00:23:57.800 --> 00:24:04.100
What I can do, if I wanted, I can find the slope of the secant line between this point and this point.
00:24:04.100 --> 00:24:06.900
That is another way of doing it, it is going to be the same.
00:24:06.900 --> 00:24:13.400
The only difference between this line and this line, as far as our slopes are concerned is you just moved it down.
00:24:13.400 --> 00:24:17.000
Slope wise, parallel lines have the same slope.
00:24:17.000 --> 00:24:27.100
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.
00:24:27.100 --> 00:24:28.600
In general, I do not do that.
00:24:28.600 --> 00:24:31.600
In general, I just draw the best tangent line that I can.
00:24:31.600 --> 00:24:35.800
I pick two points on that line randomly and I find the slope.
00:24:35.800 --> 00:24:41.500
In this particular case, we found 36.1.
00:24:41.500 --> 00:24:57.000
As far as the physical interpretation is concerned, the slope is a rate of change.
00:24:57.000 --> 00:25:02.000
An instantaneous slope is an instantaneous rate of change.
00:25:02.000 --> 00:25:10.700
It is 36.1 gallons per 1 minute.
00:25:10.700 --> 00:25:18.400
In other words, 36.1 gallons per minute.
00:25:18.400 --> 00:25:36.300
This means that at time = 20, this means at exactly 20 minutes after I started filling up the tank,
00:25:36.300 --> 00:26:05.800
the tank is being filled 36.1 gallons per every minute that passes.
00:26:05.800 --> 00:26:13.200
Again, I mentioned once before, let me go ahead and do this in red.
00:26:13.200 --> 00:26:18.900
The particular thing that I have done with desmos is I have connected these particular data points with lines.
00:26:18.900 --> 00:26:21.600
We know, in general, that is not a line.
00:26:21.600 --> 00:26:28.100
It is going to be some sort of a smooth sort of function, like that.
00:26:28.100 --> 00:26:31.400
Again, let me return to what it is that I mentioned early on.
00:26:31.400 --> 00:26:38.300
You are going to be given a function, in this particular case, volume is a function of time.
00:26:38.300 --> 00:26:40.600
You are going to be given a function in three different ways.
00:26:40.600 --> 00:26:45.700
You could be given a table which you can graph.
00:26:45.700 --> 00:26:50.800
You can be given the graph, just given the graph as it is.
00:26:50.800 --> 00:26:53.000
That is another way that you are going to be given a particular function.
00:26:53.000 --> 00:26:56.000
The 3rd way is you are actually given an explicit function.
00:26:56.000 --> 00:27:05.800
For example, this could have been v = ½ t³, something like that.
00:27:05.800 --> 00:27:12.100
That is an explicit function of volume as a function of time that passes.
00:27:12.100 --> 00:27:18.300
You can be given a table, a graph, or an explicit function.
00:27:18.300 --> 00:27:22.200
Here we use a table to make a graph.
00:27:22.200 --> 00:27:26.400
From the graph, we physically drew a tangent line.
00:27:26.400 --> 00:27:29.400
From that tangent line, we extracted the data.
00:27:29.400 --> 00:27:35.600
Now the idea behind calculus is our preference is always to be able to find an explicit formula.
00:27:35.600 --> 00:27:38.900
Because once we have an explicit formula, we are going to develop techniques.
00:27:38.900 --> 00:27:44.900
In other words, differentiating this and coming up with a derivative for this,
00:27:44.900 --> 00:27:52.000
some function of t, some v’(t) which is the derivative of this function.
00:27:52.000 --> 00:27:58.600
So that at any place along t, I can just plug in a particular value and find what the instantaneous slope is.
00:27:58.600 --> 00:28:01.000
We are not always going to be that lucky.
00:28:01.000 --> 00:28:05.200
More often than not, we will be working with functions, because again, we are teaching calculus.
00:28:05.200 --> 00:28:07.000
We are trying to develop the techniques of calculus.
00:28:07.000 --> 00:28:13.500
But understand in the real world, oftentimes, you will be given a table of data that you have to convert to a graph.
00:28:13.500 --> 00:28:19.300
You will either work from the graph directly or use numerical techniques to approximate, find a function.
00:28:19.300 --> 00:28:24.700
From there, work your calculus magic.
00:28:24.700 --> 00:28:27.100
Let us move on to another problem.
00:28:27.100 --> 00:28:31.000
This one, let y = √x – 3.
00:28:31.000 --> 00:28:35.800
The point 4,1 is a point on this curve, we will call it p.
00:28:35.800 --> 00:28:41.700
A variable point q whose coordinate is xy also lies on the curve for various values of x.
00:28:41.700 --> 00:28:48.300
P is fixed, q is anywhere else on the curve but different values of x.
00:28:48.300 --> 00:28:52.200
q is the point that is actually moving.
00:28:52.200 --> 00:28:59.700
Using a calculator, calculate the slope of the secant line pq, for the following values of x.
00:28:59.700 --> 00:29:05.600
4 is the point that is fixed, that is the x value that is fixed, its y is 1.
00:29:05.600 --> 00:29:13.400
We are going to plug in for x, here 3, 3.5, 3.9, 3.99, 3.99.
00:29:13.400 --> 00:29:20.800
Notice we are approaching 4 from below, 3 going up to 4.
00:29:20.800 --> 00:29:24.700
If this is the x value 4, this is 3, this is 5.
00:29:24.700 --> 00:29:29.100
We are going to be taking different values of x getting close to 4.
00:29:29.100 --> 00:29:33.600
Also, we are going to start and work our way down to 4.
00:29:33.600 --> 00:29:34.900
They are different values.
00:29:34.900 --> 00:29:46.600
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.
00:29:46.600 --> 00:29:51.000
Using the calculator, calculate the slope of the secant line for all of these.
00:29:51.000 --> 00:29:57.900
And then B says, using the results of A, speculate it to the value of the instantaneous slope of this function at the point.
00:29:57.900 --> 00:30:06.500
Now they are saying use the average slopes, to see if you can come up with some idea of what the instantaneous slope is.
00:30:06.500 --> 00:30:10.700
Using the slope you get, find an equation of the tangent line through the point.
00:30:10.700 --> 00:30:16.500
We already know how to find the equation, if we are given a slope and a point.
00:30:16.500 --> 00:30:25.200
It is y - y1 = slope which is m × x - x1.
00:30:25.200 --> 00:30:28.900
We have the point, now we just need to find the slope which we are going to get from part B.
00:30:28.900 --> 00:30:31.200
We just plug it into here and get our equation.
00:30:31.200 --> 00:30:36.300
First, we need to find what is going on there.
00:30:36.300 --> 00:30:42.000
Let us draw this out and see what it is that we are actually doing.
00:30:42.000 --> 00:30:49.500
That is fine, let us draw this out.
00:30:49.500 --> 00:31:00.200
I think I’m going to go to purple because I really like purple, it is very nice.
00:31:00.200 --> 00:31:05.900
This is 1, this is 2, this is 3, 4, 5, 6.
00:31:05.900 --> 00:31:12.100
Our function is y is equal to √x – 3.
00:31:12.100 --> 00:31:15.400
This is the radical function shifted over 3 to the right.
00:31:15.400 --> 00:31:26.700
It starts there at 3 and it goes that way.
00:31:26.700 --> 00:31:30.900
Our point p is fixed at 4,1.
00:31:30.900 --> 00:31:34.200
This is our point p.
00:31:34.200 --> 00:31:39.200
Our q, here is 5, this is the value for 5.
00:31:39.200 --> 00:31:44.300
Our q is going to be different values of 3, 3, 3.5, 3.9.
00:31:44.300 --> 00:31:47.300
q is going to move from here to here.
00:31:47.300 --> 00:31:50.600
To here, it is going to get closer that way.
00:31:50.600 --> 00:31:53.300
Here, this is going to be the other set of q.
00:31:53.300 --> 00:31:56.900
It is going to be this point and then it is going to be 4.5.
00:31:56.900 --> 00:31:58.000
It is going to be this point.
00:31:58.000 --> 00:32:01.900
q is a movable point, q is going to approach p from the left.
00:32:01.900 --> 00:32:04.600
q is going to approach p from the right.
00:32:04.600 --> 00:32:12.400
We want to find the slopes of the secant lines for different values of q.
00:32:12.400 --> 00:32:14.000
That is what we are doing, we are finding average slopes.
00:32:14.000 --> 00:32:16.300
We are finding pq.
00:32:16.300 --> 00:32:20.500
This is going to be one q, this is going to be another q.
00:32:20.500 --> 00:32:29.800
Let us say q1, q2, q3, q4, q5, q6, q7, q8, q9, q10, things like that.
00:32:29.800 --> 00:32:41.700
This is that, therefore, our q is the point xy.
00:32:41.700 --> 00:32:48.500
We know what y is, it is going to be the point x × √x – 3.
00:32:48.500 --> 00:32:49.900
Those are the coordinates of the various q.
00:32:49.900 --> 00:32:55.200
It is based on whatever the value of x is.
00:32:55.200 --> 00:33:03.900
Let us go ahead and magnify this part a little bit and see what we are dealing with.
00:33:03.900 --> 00:33:09.600
I’m going to go ahead this way.
00:33:09.600 --> 00:33:17.000
I’m going to exaggerate the curve a little bit, just to make it a little bit easier for me to deal with.
00:33:17.000 --> 00:33:25.100
Let us say this is our 3, this is our 4, this is our 5.
00:33:25.100 --> 00:33:29.600
This is the point p, this is the point 4,1.
00:33:29.600 --> 00:33:43.000
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.
00:33:43.000 --> 00:33:47.100
For 5, we are going to find the slope of that line and we are going to move q over here.
00:33:47.100 --> 00:33:53.600
We are going to find the slope of that line, that is what we are going to do.
00:33:53.600 --> 00:34:08.200
The average slope = Δ y/ Δ x.
00:34:08.200 --> 00:34:14.800
p is the point 4,1, that point is fixed.
00:34:14.800 --> 00:34:18.400
That does not change, that is one of our points.
00:34:18.400 --> 00:34:28.800
q is the point x √x - 3.
00:34:28.800 --> 00:34:36.500
Our average slope for pq is going to equal y2 - y1/ x2 - x1.
00:34:36.500 --> 00:34:44.300
It is going to be √x - 3 - 4/ x.
00:34:44.300 --> 00:34:50.300
This is, sorry, -1/ x – 4.
00:34:50.300 --> 00:35:01.900
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.
00:35:01.900 --> 00:35:16.500
3, 3.5, 3.9, 3.99, 3.9999, and then 5, 4.5, 4.1, 4.01, 4.001.
00:35:16.500 --> 00:35:19.900
Notice what is happening to these lines.
00:35:19.900 --> 00:35:23.900
At some point, you are going to end up getting that.
00:35:23.900 --> 00:35:27.700
From this direction, you are going to end up getting that.
00:35:27.700 --> 00:35:31.600
You are actually going to end up getting the tangent line that we will see in just a minute.
00:35:31.600 --> 00:35:33.000
Let us go ahead and do this.
00:35:33.000 --> 00:35:38.400
Again, I went ahead and I use desmos to calculate the values for me.
00:35:38.400 --> 00:35:48.800
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.
00:35:48.800 --> 00:35:52.700
These are the values of y that I get.
00:35:52.700 --> 00:36:03.100
This is q1, this is q1, q2, q3, q4, q5, q6, q7, q8, q9, and q10.
00:36:03.100 --> 00:36:06.400
The slope was this equation right here.
00:36:06.400 --> 00:36:16.000
It is going to be y2 - y1/ x2 - x1.
00:36:16.000 --> 00:36:20.100
y2 - y1/ x2 - x1.
00:36:20.100 --> 00:36:26.400
y2 - y1/ x2 – x1, these are the various slopes.
00:36:26.400 --> 00:36:31.900
When the x value of q is 3, the slope is 1.
00:36:31.900 --> 00:36:39.200
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.
00:36:39.200 --> 00:36:43.100
This gives me the different slopes.
00:36:43.100 --> 00:36:58.300
Some things to notice, the x value of q is running from 3 to 4.
00:36:58.300 --> 00:37:06.300
It is a line segment, q is getting closer to p from below.
00:37:06.300 --> 00:37:19.200
As q goes from 3 to 4, notice what the slope approaches.
00:37:19.200 --> 00:37:54.000
The slope approaches, it looks like it is 0.5.
00:37:54.000 --> 00:38:03.200
Let me rewrite this.
00:38:03.200 --> 00:38:09.200
This is p and those are the things that we are finding.
00:38:09.200 --> 00:38:10.800
This is q, this is q.
00:38:10.800 --> 00:38:17.700
We are moving q closer to p, closer to p, find the line segment.
00:38:17.700 --> 00:38:32.800
As q approaches p from below, p is 4, from 3 to 4, from below.
00:38:32.800 --> 00:38:35.600
The line, here is the 4, here is the 3.
00:38:35.600 --> 00:38:38.000
We are approaching from less than 4.
00:38:38.000 --> 00:38:42.700
We say from below.
00:38:42.700 --> 00:38:52.000
The slope approaches 0.5, the average slope, the slope of the secant line.
00:38:52.000 --> 00:38:59.400
But notice the difference between 3.99 and 4 is really a short line.
00:38:59.400 --> 00:39:08.600
Now as q approaches p from above, if we go from 5 down to 4.
00:39:08.600 --> 00:39:16.400
Once again, 4.1, 4.4, .449, .48, .498, .499.
00:39:16.400 --> 00:39:36.900
As q approaches p from above, the slope, again, it approaches 0.5.
00:39:36.900 --> 00:39:39.300
Let us talk about what this means.
00:39:39.300 --> 00:39:51.200
Essentially, what we are looking at is this.
00:39:51.200 --> 00:39:55.700
I think you guys can already figure out what is going on.
00:39:55.700 --> 00:40:02.900
That is 4, that is 5, that is 3.
00:40:02.900 --> 00:40:10.900
I need to exaggerate it so you can actually see it.
00:40:10.900 --> 00:40:16.000
That is not very good, that is okay.
00:40:16.000 --> 00:40:24.300
I fixed p and I started with q1 over here and I moved q to 3.5.
00:40:24.300 --> 00:40:26.100
In other words, I’m moving q closer and closer.
00:40:26.100 --> 00:40:29.400
I’m finding that slope and that slope.
00:40:29.400 --> 00:40:38.000
Notice these slopes, here is the q, here is the p, from your perspective.
00:40:38.000 --> 00:40:42.200
q is approaching p along that curve right there.
00:40:42.200 --> 00:40:47.300
The slope is rising.
00:40:47.300 --> 00:40:54.700
Eventually at this point, you are going to hit, if I take 3.99, if I get infinitely close to 4,
00:40:54.700 --> 00:40:58.900
the slope of that secant line is going to be the slope of the tangent line.
00:40:58.900 --> 00:41:01.000
It is going to turn in to the tangent line.
00:41:01.000 --> 00:41:04.000
The same thing from this side, this slope.
00:41:04.000 --> 00:41:07.500
From your perspective, p is here, q was up here.
00:41:07.500 --> 00:41:11.700
The curve goes this way, from your perspective.
00:41:11.700 --> 00:41:15.000
Now I’m moving q closer.
00:41:15.000 --> 00:41:21.900
If q was closer, this line segment, the slope is going to increase.
00:41:21.900 --> 00:41:24.300
Increase until I hit this.
00:41:24.300 --> 00:41:27.300
That is what is going on.
00:41:27.300 --> 00:41:36.500
Part B, it is reasonable and we saw that this average slope approaches 0.5 from below,
00:41:36.500 --> 00:41:41.600
this average slope approaches 0.5 from above.
00:41:41.600 --> 00:42:05.700
It is reasonable to conclude at x = 4, the instantaneous slope = 0.5.
00:42:05.700 --> 00:42:15.300
That is another technique that you can use, in order to find the instantaneous slope at a given.
00:42:15.300 --> 00:42:22.200
You can take average slopes and get closer and closer and closer, move a particular fixed one point.
00:42:22.200 --> 00:42:25.500
The point where you are interested in the instantaneous slope.
00:42:25.500 --> 00:42:32.600
Take a point q and move it move it this way and calculate all of these average slopes.
00:42:32.600 --> 00:42:37.100
As you get infinitely close, 3.99, 3.999, 3.99999,
00:42:37.100 --> 00:42:40.700
you are going to find that the slope is getting close to one number and it is not moving.
00:42:40.700 --> 00:42:44.800
In this particular case, it is 0.5.
00:42:44.800 --> 00:42:47.500
In fact that is going to be the procedure that we are going to use.
00:42:47.500 --> 00:43:06.600
Earlier on, I introduce this idea of the limit as h goes to 0 of f(x) + h – f(x)/ hs.
00:43:06.600 --> 00:43:10.500
As h goes to 0 of this thing, that is the process that we are doing.
00:43:10.500 --> 00:43:14.700
We are going to find a secant line, that is this.
00:43:14.700 --> 00:43:20.600
We are going to take h, this distance to 0.
00:43:20.600 --> 00:43:22.400
We are going to make an infinitely close.
00:43:22.400 --> 00:43:26.300
At some point, a number is going to emerge.
00:43:26.300 --> 00:43:28.100
A function is going to emerge.
00:43:28.100 --> 00:43:29.900
That is going to be this.
00:43:29.900 --> 00:43:34.100
I can do it right now, I can just take limiting values.
00:43:34.100 --> 00:43:37.700
I can find secant, secant, secant, when it gets close to one side.
00:43:37.700 --> 00:43:40.600
And then, secant, secant, when it gets close to on the other side.
00:43:40.600 --> 00:43:45.700
If from below and from above, the two slopes happen to meet,
00:43:45.700 --> 00:43:49.900
if they happen to end up being the same number, we define that as the actual slope of that.
00:43:49.900 --> 00:43:52.000
We define that as the derivative.
00:43:52.000 --> 00:43:58.500
Hopefully, you understood the process that we went through here.
00:43:58.500 --> 00:44:05.400
Final, the equation of the tangent line.
00:44:05.400 --> 00:44:10.700
The equation of the tangent line.
00:44:10.700 --> 00:44:15.500
Again, if some of the stuff is not altogether that clear, do not worry about it, we will continue to discuss it.
00:44:15.500 --> 00:44:17.300
It is not a problem.
00:44:17.300 --> 00:44:28.400
We said that the equation of any line is y - y1 = the slope × x - x1.
00:44:28.400 --> 00:44:38.100
We have the slope, they said the slope is 0.5 at the point 4,1.
00:44:38.100 --> 00:44:41.700
We have the point p which is 4,1.
00:44:41.700 --> 00:44:55.100
y - 1 = ½ × x – 4, that is the equation of our tangent line.
00:44:55.100 --> 00:45:04.000
The slope of the tangent line ½, that is the instantaneous slope at x = 4.
00:45:04.000 --> 00:45:14.400
That is the derivative of the function y = √x – 3, at the point where x = 4.
00:45:14.400 --> 00:45:17.400
The numerical value of the derivative is ½.
00:45:17.400 --> 00:45:25.400
It is the slope of the tangent line to the curve at that point.
00:45:25.400 --> 00:45:46.600
Let us see what else we have got here.
00:45:46.600 --> 00:45:56.100
Let me write that out again, just in case.
00:45:56.100 --> 00:46:27.700
This was another graphical technique, another graphical method for finding the instantaneous slope at a point p,
00:46:27.700 --> 00:46:53.600
by moving a point q closer and closer to p along the curve.
00:46:53.600 --> 00:47:10.800
Calculating the various average slopes of pq.
00:47:10.800 --> 00:47:46.000
As q gets insanely close to p, the slope of pq gets insanely close to the instantaneous slope.
00:47:46.000 --> 00:47:56.700
Again, what we are doing is we are fixing the point p.
00:47:56.700 --> 00:48:00.900
We are interested in that slope.
00:48:00.900 --> 00:48:04.400
We pick a point q over here, we find that slope.
00:48:04.400 --> 00:48:07.400
We move q over here, we find that slope.
00:48:07.400 --> 00:48:11.300
We move q over here, we find that slope.
00:48:11.300 --> 00:48:14.000
We move q a little closer, a little closer, that slope.
00:48:14.000 --> 00:48:19.600
Notice how these slopes, the lines, they are coming up.
00:48:19.600 --> 00:48:25.900
The slope, the numerical value of the slope is coming up and eventually it is going to be that.
00:48:25.900 --> 00:48:32.600
As q gets infinitely close to p, the slope of pq is going to actually b, this instantaneous slope.
00:48:32.600 --> 00:48:35.300
When we do the same thing from above.
00:48:35.300 --> 00:48:37.400
That is that slope, that is another q.
00:48:37.400 --> 00:48:45.700
Then we find that slope and that slope, from your perspective.
00:48:45.700 --> 00:48:51.400
This slope is coming up, this is pq, pq, pq, pq.
00:48:51.400 --> 00:48:54.100
At some point, the slope itself is going to approach a value.
00:48:54.100 --> 00:49:01.200
If the numerical value of the slopes, as you go from below, approach a number
00:49:01.200 --> 00:49:05.100
and it happens to be the same number that it approaches as you go this way,
00:49:05.100 --> 00:49:12.600
that number that we got, in this example, 0.5, that is the slope of the tangent line.
00:49:12.600 --> 00:49:14.100
There is another graphical technique.
00:49:14.100 --> 00:49:18.500
I have two graphical techniques, you do not need an actual function.
00:49:18.500 --> 00:49:25.400
You just need a graph, you can even draw the line, pick two points, find the slope.
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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.
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In this case, it was 0.5.
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If a ball is thrown up in the air with initial velocity of 10 m/s,
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its height above the ground is a function of t, that is t seconds after I throw it.
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It is given by the function, this, explicit function.
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The height as a function of t is 10t -4.9t².
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Find the average velocity for the following time intervals.
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1 to 2, 1.5 to 2, 1.9 to 2, 1.99 to 2, 1.999 to 2.
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Notice 2 is fixed, we are shortening the time interval.
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From your perspective, 2 is here, 1 is here.
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We are going 1, 1.5, what is the average, find the instantaneous velocity at t = 2.
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We are going to do the same thing.
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We are going to find the average, average, average, average, average.
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We are going to see if the slope is approaching some number.
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That is what we are going to do.
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The first thing we want to do is let us go ahead and draw this out.
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A, I have got h(t), this I’m going to draw by hand.
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It is 10t - 4.9t².
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I have got this is equal to t × 10 - 4.9t.
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I’m going to this equal to 0 to see where it hits the x axis.
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It hits the x axis at t = 0 and it hits the x axis at 10 divided by 4.9 which is 2.04.
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0, this is 2.04, this is a parabola.
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This is a parabola 10t – 4t², it is a parabola that opens downward.
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I want to see what height is 2.04, at the halfway mark, it is going to be 1.02.
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When I put in 1.02 into h and I solve, I get that it is equal to 5.1, I think.
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If I’m not mistaken, something like that.
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Here is a graph of the function, it is a parabola.
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I'm interested in the time interval from 1 to 2.
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There to there, that is what I'm interested in.
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I know what the slope is, the slope is the change in y/ the change in x.
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It is going to be, t is time in seconds, h is height it is meters.
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This is going to be meters per second.
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The slope is a velocity, the slope is going to give me the velocity of this thing.
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From the time increment, from 1 to 2, my average slope which is my average velocity,
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because we just said that slope is dependent variable divided by the independent variable, it is going to equal from 1 to 2.
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It is going to be a slope of that line.
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It is going to be h(2) - h(1)/ 2 – 1.
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I end up getting -4.7 m/s.
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When I do 1.5 to 2, the average velocity, any variable with the line over it means average = h(2) – h(1.5).
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1.5 is here, now I’m finding that secant line.
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Sorry, not that secant line.
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2 was the one that is fixed, it is 1.5 that is moving.
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I’m finding that secant line, right there, from here to here, that secant line.
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H(2), h(1.5)/ 2 – 1.5.
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I end up with -7.15 m/s.
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I do the same for 1.9 to 2.
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I end up with an average velocity of - 9.11 m/s.
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If I do 1.99 to 2, I find an average velocity of -9.551 m/s.
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If I do 1.999 to 2, I get an average velocity of -9.595 m/s.
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I decided to go one further, 1.9999 to 2, and I found an average velocity of -9.59951 m/s.
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Clearly, I'm approaching -9.6.
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Part B, my instantaneous velocity at t = 2.
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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.
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Instantaneous velocity t = 2, as we approach 2 from below, from 1 approaching 2 from below,
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the slope approaches, that is what the arrow means.
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The slope approaches -9.6 m/s.
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Our instantaneous slope which is our instantaneous velocity, which is our derivative at 2 = -9.6 m/s.
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Here is an tabular form.
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Our function is this, at different values of x, these are my y values, my height.
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At the value of 2, it is equal to 0.4.
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The slope that I'm finding is, I'm fixing the point 2,0.4.
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It is going to be the y value of 0.4 - this value which is that/ 2 – this.
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y2 - y1/ x2 - x1, the next point, y2 - y1/ x2 - x1.
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The next point, y2 - y1/ x2 - x1, that is what this says.
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The x values are 1, 1.5, 1.9, 1.99999.
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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.
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It looks like it is approaching the -9.6.
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Therefore, I’m using average slopes, as I get really close to the point 2, where I'm interested in an instantaneous slope,
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I see where the average slope is getting close to.
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If I want, I can keep going, 1.999999999.
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You are going to find that it would not go past -9.6.
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That is going to be my answer.
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If I wanted to, I can also approach it from the other side.
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I can also approach 2 from 3.
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I can go 3 down to 2, I can go 3 to 2.5, 2.1, 2.01, 2.001, 2.0001.
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You are going to see that it approaches -9.6.
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That is what is going on here.
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Hopefully, these three problems have helped you see and hope to clarify what it is that we actually discussed in the previous two lessons.
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Thank you so much for joining us here at www.educator.com.
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We will see you next time, bye.