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What I want to do in this videos is to make to

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parametrizations of essentially the same curve, but we're going

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to go along the curve a different rates.

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And hopefully we'll be able to use that to understand, or get

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a better intuition, behind what exactly it means to take a

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derivative of a position vector valued function.

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So let's say my first parametrization, I have

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x of t is equal to t.

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And let's say that y of t is equal to t squared.

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And this is true for t is greater than or equal to 0,

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and less than or equal to 2.

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And if I want to write this as a position vector valued

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function, let me write this.

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x1, call that y1, and let me write my position vector valued

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function; I could say r1-- I'm numbering them because I'm

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going to do a different version of this exact same curve with a

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slightly different parametrization --so r1 one of

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t, we could say is x1 of t times i-- the unit vector i

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--so we'll just say t times i plus-- this is just x of t

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right here, or x1 of t; I'm numbering them because I'll

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later have an x2 t --plus t squared times j.

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And if I wanted to graph this, I'm going to be very careful

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graphing it because I really want to understand what the

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derivative means here.

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Try to draw it roughly to scale.

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So let's say that this is one, two, three, four.

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Then let me draw my x-axis.

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That's good enough.

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And my x-axis, I want it to be roughly to scale, one and two.

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And so at t equals 0, both my x and y coordinates are at 0-- or

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this is just going to be the 0 vector, so this is where we are

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a t equals 0 --at t equals 1 this is going to be one times

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i-- we're going to be just like that --plus 1 times j.

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1 squared is j, so we're going to be right there.

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And then at t is equal to 2, we're going to be at 2i.

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So 2i-- you could imagine 2 times i would be this vector

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right there --2 times i plus 4-- 2 squared is 4 --4 times

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j, so plus 4 times j.

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If you add these two vectors heads to tails, you're going

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to get a vector that's end point is right there.

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The vector is going to look something like this.

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So this is what, just to make it clear what we're

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doing, that's r1 of 2.

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This is r1 of 0.

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This is r1 of 1.

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But the bottom line is the path looks like this:

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it's a parabola.

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So the path will look like that.

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Now that's in my first parametrization of it.

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Actually, let me draw a little bit more carefully.

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I want to get rid of this arrows, just because I want it

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to be a nice clean drawing.

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So it's going to be a parabola.

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Let me get rid of that other point, too, just because I

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didn't draw it exactly where it needs to be; it needs

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to be right there.

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And my parabola, or part of my parabola is going to

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look something like that.

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All right.

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Good enough.

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So this is the first parametrization.

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Now I'm going to do this exact same curve, but I'm going to

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do it slightly differently.

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So let's say I'll do it in different colors.

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So x2 of t, let's it equals 2t.

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And y2 of t, let's say it's equal to 2t squared.

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Or we could alternatively write that, that's the same thing as

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4t squared, just phrasing both of these guys

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to the second power.

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And then let's say instead of going from t equals 0 to 2,

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we're going to go from t goes from 0 to 1.

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But we're going to see, we're going to cover

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the exact same path.

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And our second position vector valued function, r2 of t, is

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going to be equal to 2t times i plus-- I could say 2t squared

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--4t squared times j.

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And if I were to graph this guy right here, it would look

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like-- let me draw my axes again; it's going to look the

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same, but it's I think useful to draw it because I'm going to

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draw the derivatives and all that on it later.

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One, two, three, four.

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One, two.

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And then let's see what happens when t is equal to 0-- or r of

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0; all these are going to be 0, we're just going to have the

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zero vector; x and y are both equal to 0 --when t is equal

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to 1/2 what are we going to get here?

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1/2 times 2 is 1.

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And then we're going to get the point 1/2 squared

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is 1/4 times 4 is 1.

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So when t is equal 1/2 we're going to be at the point 1, 1.

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And when the t is equal to 1 we're going to

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be at the point 2, 4.

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So notice the curve is exactly, the path we go

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is exactly the same.

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But before we even do the derivatives, these two

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paths are identical.

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I want to think about something.

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Let's pretend that our parameter, t, really is time.

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And that tends to be the most common, that's

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why they call it t.

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It doesn't have to be time, but let's say it is time.

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So what's happening here?

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In the first parametrization when we go from 0 to 2

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seconds we cover this path.

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You can imagine after 1 second the dot moves here,

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then it moves there.

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You can imagine a dot moving along this curve, and it

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takes two seconds to do so.

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In this situation we have a dot moving along the same curve,

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but it's able to cover the same curve in only one second; and

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half a second it gets here.

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It took this guy one second to get here.

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In a one second, this guy's all the way over here; this guy

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takes two seconds to go over here.

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So in this second parametrization even though the

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path is the same, the curves are the same, the

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dot is faster.

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I want you to keep that in mind when we think about

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the derivatives of both of these position vector

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valued functions.

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So just remember the dot is moving faster for every second

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it's getting further along the curve than here; that's why it

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only took them one second.

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Now let's look at the derivatives of both

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of these guys.

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So the derivative here, so if I were to write r prime, r1 prime

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of t-- let me do that in a different color, actually,

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already used the orange; so let me do it in the blue

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--r1 prime us t.

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So the is the derivative now.

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It's going to be, remember, it's just the derivative of

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each of these times the unit vectors.

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So the derivative of t with respect to t, that's just 1.

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So it's 1 times i.

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I'll just write 1i plus-- I didn't have to write the one

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there --plus the derivative of t squared with respect

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to t is 2t plus 2t j.

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And let me take the derivative over here.

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r2 prime of t.

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The derivative of 2t with respect to t is 2, so

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2i, plus the derivative of 4t squared is 8t.

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2 times 4, it is rt.

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Just like that.

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Now the question is, what do their respective derivative

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vectors look like at different points?

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So let's look at, I don't know, let's see how fast they're

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moving when time is equal to 1.

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So let's take it at a specific point.

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This is just the general formula, but let's figure

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out what the derivative is at a specific point.

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So let's take r1 when time is equal to 1.

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And I want to take this specific point on the curve,

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not the specific point in time.

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So this point on the curve here is when time is equal

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to 1, you could say second.

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This point over here, which is the exact corresponding

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point, is when time is equal 1/2 second.

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So r1 of 1 is equal to-- we're taking the derivative

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there --is equal to 1i.

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It's not dependent on t at all.

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So it's 1i plus 2 times 1j, so plus 2j.

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So at this point the derivative of our position vector function

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is going to be 1i plus 2j.

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So we can draw it like this. so if we do 1i is like this: 1i.

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And then 2j.

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Just 2j is like that.

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So our derivative right there, I'll do it in the same

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color that I wrote it in.

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It's in this green color; it's going to look like this.

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And notice it looks like, at least its direction is-- let me

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do it a little bit straighter --its direction looks tangent

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to the curve; it's going in the direction that my

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particle is moving.

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Remember my particle is moving from here to there, so it's

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going in the direction.

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And I'm going to think about, in a second, what this length

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of this to derivative vector is.

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This right here, just to be clear is, r1 prime.

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It's a vector, so it's telling us the instantaneous change in

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our position vector with respect to t, or time, when

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time is equal to 1 second.

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That's this thing right here.

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Now let's take the exact same position here on our curve.

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But that's going to occur at a different time for this guy.

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We already said it only takes him, he's here at time

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is equal to 1/2 second.

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So let's take-- --I'll do it in the same color

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--so here we have r2.

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We're going to evaluate it at 1/2 half because this is at

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time is equal 1/2 second.

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And this is going to be equal to 2i-- this isn't dependent

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at all on time --so 2i plus 8 times the time.

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So time right here is 1/2.

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So 8 times 1/2 is 4.

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So plus 4j.

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So what does this look like?

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The instantaneous derivative here.

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Oh, and this is the derivative; have to be very clear.

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So 2i-- let me draw some more --so 2i maybe gets

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us about that far.

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Plus 4j will get us up to right around there.

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Plus 4j is that factor.

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So when you add those two heads to tails, you get this thing:

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you get something that-- let me like --you get something

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that looks like that.

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I didn't draw it as neatly as I would like to.

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But let's notice something: both of these vectors are going

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in the exact same direction.

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They're both tangential to the path, to our curve.

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But this vector is going, its length, its magnitude, is

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much larger than this vector's magnitude.

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And that makes sense because I hinted at it when we first

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talked about these vector valued position functions and

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their derivatives; is that the length, you can kind of

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view it as the speed.

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The length is equal to the speed if you imagine t being

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time and these parametrizations are representing a dot

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moving along these curves.

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So in this case, the particle only takes a second to go

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there, so at this point in its path, it's moving much faster

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than this particle is.

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So if you think about it, this vector right here, if you

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imagine this is a position factor, this is velocity.

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Velocity is speed plus the direction.

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Speed is just you know, how fast are you going?

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Velocity is how fast you're going in what direction?

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I'm going this fast-- and you could calculate it using the

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Pythagorean Theorem, but I just want to give you the intuition

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right here --I'm going that fast in this direction.

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Here I'm going this fast; I'm going even faster.

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That's my magnitude, but I'm still going in

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the same direction.

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So hopefully you have a gut feeling now of what the

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derivative of these position vectors really are.

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