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In these sets of videos, I introduced the concept of a tangent line to a point on a continuous
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smooth curve, the slopes of those tangent lines, and the derivative.
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We're going to use the slope of the second line to approximate the slope of the tangent
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lines, and also we'll be using the concept of limits.
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I'll be appealing to your intuition as I discuss the ideas, so my language is not always
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perfectly precise or accurate as I am trying to give you a feel for these concepts.
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In the second video, you'll see the formal definitions for the slope of a tangent line
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and the derivative of a function.
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So the ideas presented up to that point are to help you see why the definitions make sense.
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This is a video especially for Joey and Kira.
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It's an introduction to derivative, and here are some basic ideas.
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We have to understand.
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First of all, a line tangent to a curve at a certain point.
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Well, what does that mean?
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Let's say we have some curve, this black curve is some function.
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We would like to find the tangent line to the curve at a certain point.
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So I have four points here.
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The first one over here, this blue point, well, what do I mean by that tangent line?
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I mean something like this.
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That might be the tangent line.
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It just touches the curve right at that point.
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Now this is approximate, that's not exactly the correct line.
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I'm just eyeballing it.
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We're getting as close as we can.
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All right, I have another one over here.
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I have a green point, and it looks like the tangent line is something like this.
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Notice the tangent line looks like it has a slope of about zero.
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This first one, it looks like the slope is maybe negative one.
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All right, let's take this red point.
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This looks like it may be something like that.
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Again, that looks like about a slope one, just approximating.
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Here's another one.
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This looks more shallow.
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Maybe that's a slope of negative one half or something like that.
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First of all, we have to get the idea of what it means for a line to be tangent to a
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curve at a certain point.
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This is just basic idea.
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At this point, it's important to note the following.
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The graph or the curve, which could be a line, as long as it's not a vertical line, so
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it has to be a function.
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The graph has to be smooth and continuous on both sides of the point in order for the
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graph or the curve to have a tangent line at that point.
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All right, so what I mean is, see how this is a nice continuous curve.
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We'll be able to pick points and find tangent lines at that point.
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As long as it's continuous, what I mean is on both sides.
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So you look sort of to the right of the left and still part of the curve.
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Furthermore, there's only one tangent line at any of these points on the curve.
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It may look like, well, you could draw another tangent line.
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So here's an example.
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They say, hey, I think this looks like a tangent line.
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And that's the problem is that we need to be a little bit more precise to find exactly
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where it is.
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We're just eyeballing it right now.
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Okay, so let's get rid of this.
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Okay, so for instance, if you have some sort of curve like this and then it goes and stops,
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this is not continuous at that point or at this point where it stops.
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It's not continuous, but all these other points in between, you're going to be able to find
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tangent lines.
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Secondly, you can't have something with sharp curves like this.
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So here's places where you would not be able to talk about the tangent line.
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But all of the rest of the places on the curve are fine.
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All right, now we're going to go on.
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The key to finding the tangent line is to find the slope of the tangent line to the curve
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at that point.
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So imagine you could find the slope for an instance of this red line and you knew what this
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point was.
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Well, once you have a slope and a point, you could use the slope point formula and find
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the equation of the line.
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And as it turns out, using limits and algebra, we're actually able to find the exact slope
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of the tangent point, I'm sorry, the slope of the tangent line to the curve at a certain
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point, at least for some curves.
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So we're going to be working on that and that leads us to the definition of a derivative.
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So isn't that exciting?
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All right, so the key is to finding the tangent line.
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So let's take the graph y equals x squared minus 3, which is a parabola.
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But remember, when you're talking about functions, y is the same thing as f of x.
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So if we were going to write this in function notation, f of x equals x squared minus 3.
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And when you write in ordered pair, you're used to writing x, y, right?
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But using function notation, we could write x, f of x.
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So these are some things to just keep in mind some notational information.
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All right, this is a rough sketch of this parabola, f of x equals x squared minus 3.
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So you can see the various points it goes through.
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Let's just look at the very bottom, the minimum value here.
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And if we were going to draw the tangent line, that looks like it's just going to be straight
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across.
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In fact, that is the exact tangent line. This is the tangent line at the point zero negative
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three.
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And you can tell the slope is m equals zero.
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Now that's by eyeballing it, and we can prove it later.
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But for right now, hopefully that looks like the slope is zero to you.
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Now let's take another point and see what we do to try to figure out some, you know,
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the value at some other place.
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For instance, what if I wanted to know the slope here at one negative two?
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I'm not really sure.
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You can't, you know, when you eyeball it, you can't tell for sure what that slope is.
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So this is just a rough idea, but it looks like it's going to be positive, that's for
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sure.
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And it looks like it's going to be steeper than when.
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So the goal here is to find out the slope of that tangent line at the point one negative
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two.
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Try to figure out the find the slope of the tangent line at one negative two.
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One technique is to approximate it.
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So you could look at it and you could maybe give an approximate idea.
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Maybe it's a slope of one and a half, three halves.
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Maybe it's a slope two.
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You know, it's hard to tell because I'm not really graphing that line perfectly because
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this is just a sketch.
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So one technique is to draw what's called the secant line.
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All right, so what is a secant line?
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The secant line is a line passing through two points on a curve.
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All right, keep in mind, this was the graph of y equals f of x, right, which was x squared
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minus three.
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Okay, so we're doing f of x equals x squared minus three.
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And this would be a secant line.
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It's just going through two points of the curve.
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Now that is certainly not going to be the tangent line, correct?
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But I can figure out the slope between these two points.
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So let's just do that.
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How would you figure out the slope of this red line?
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You would have a difference, the difference quotient here.
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So you'd have one order pair of the y over x and the other order pair y over x.
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And that's going to give you negative two plus three over one, which is one.
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So the slope of this line happens to be one.
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And obviously that's not as steep as the real tangent line.
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So let's say we were trying to get a little bit closer.
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What about if I tried to get closer and I figured out, hmm, what is if I chose that for
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my other point?
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So I would have to figure out what order pair that is.
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I do know the y value is one half.
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And you'd have to plug in one half into the function.
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f of x equals x squared minus three.
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And if you put in one half, for x you get one fourth minus three, which is minus two
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and three fourths.
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Okay, so this is one minus two and three fourths.
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Okay, you can see it's much closer down to negative three.
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So I can get maybe a little bit better approximation.
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If I used the point one negative two and one half, negative two and three fourths, let's try
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that.
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If you take those two ordered pairs, one negative two and one half, negative two and three
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fourths, you would have, again, I'm going to do this one first.
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Let's see, I'll put this on the top, negative two over the positive one and negative two
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and three fourths over one half.
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Right?
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This is plus plus.
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So I have two and three fourths minus two.
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That's just going to be three fourths in the numerator divided by one minus one half,
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which is one half and then you multiply by the reciprocal and that gives us three
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halves.
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All right?
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So you see how it's getting a little bit steeper if I were doing this line right here.
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Still not the exact tangent line.
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Id.