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- [Instructor] Let's say we're[br]going to trace out a curve

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where our X coordinate[br]and our Y coordinate

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that they are each defined by

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or they're functions[br]of a third parameter T.

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So, we could say that X is a function of T

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and we could also say[br]that Y is a function T.

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If this notion is completely[br]unfamiliar to you,

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I encourage you to review the[br]videos on parametric equations

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on Khan Academy.

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But what we're going to think about

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and I'm gonna talk about in[br]generalities in this video.

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In future videos we're going to be dealing

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with more concrete examples

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but we're gonna think[br]about what is the path

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that is traced out

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from when T is equal to A,

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so this is where we are[br]when T is equal to A,

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so in this case this[br]point would be X of A,

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comma Y of A,

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that's this point

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and then as we increase from T equals A

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to T is equal to B,

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so our curve might do something like this,

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so this is when T is equal to B,

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T is equal to B,

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so this point right over here is X of B,

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comma Y of B.

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Let's think about how do we figure out

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the length of this actual curve,

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this actual arc length from[br]T equals A to T equals B?

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Well, to think about that

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we're gonna zoom in and[br]think about what happens

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when we have a very small change in T?

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So, a very small change in T.

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Let's say we're starting at[br]this point right over here

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and we have a very small change in T,

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so we go from this point[br]to let's say this point

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over that very small change in T.

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It actually would be[br]much smaller than this

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but if I drew it any smaller,

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you would have trouble seeing it.

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So, let's say that that[br]is our very small change

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in our path in our arc[br]that we are traveling

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and so, we wanna find this length.

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Well, we could break it down

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into how far we've[br]moved in the X direction

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and how far we've moved[br]in the Y direction.

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So, in the X direction,

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the X direction right over here,

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we would have moved a[br]very small change in X

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and what would that be equal to?

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Well, that would be the rate of change

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with which we are[br]changing with respect to T

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with which X is changing with respect to T

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times our very small change in T

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and this is a little hand wavy,

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I'm using differential notion

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and I'm conceptually using the idea

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of a differential as an[br]infinitesimally small change

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in that variable.

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And so, this isn't a formal proof

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but it's to give us the intuition

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for how we derive arc length

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when we're dealing with[br]parametric equations.

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So, this will hopefully[br]make conceptual sense

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that this is our DX.

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In fact, we could even write it this way,

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DX/DT, that's the same thing[br]as X prime of T times DT

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and then our change in Y

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is going to be the same idea.

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Our change in Y, our[br]infinitesimally small change in Y

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when we have an infinitesimally[br]small change in T,

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well, you could view that[br]as your rate of change

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of Y with respect to T

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times your change in T,

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your very small change in T

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which is going to be equal to,

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we could write that as Y prime of T DT.

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Now, based on this,[br]what would be the length

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of our infinitesimally small[br]arc length right over here?

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Well, that we could just[br]use the Pythagorean theorem.

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That is going to be the square root of,

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that's the hypotenuse[br]of this right triangle

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right over here.

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So, it's gonna be the square root

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of this squared plus this squared.

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So, it is the square root of,

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I'm gonna give myself a[br]little bit more space here

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because I think I'm gonna use a lot of it,

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so the stuff in blue squared,

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DX squared we could[br]rewrite that as X prime

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of T DT squared

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plus this squared which[br]is Y prime of T DT squared

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and now let's just try to[br]simplify this a little bit.

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Remember, this is this[br]infinitesimally small arc length

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right over here.

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So, we can actually[br]factor out a DT squared,

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it's a term in both of these

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and so, we can rewrite this as,

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let me, so I can rewrite this

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and then write my big radical sign,

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so I'm gonna factor out a DT squared here,

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so we could write this as DT squared

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times X prime of T squared

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plus Y prime of T squared

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and then to be clear

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this is being multiplied[br]by all of this stuff

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right over there.

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Well, now if we have this DT[br]squared under the radical,

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we can take it out

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and so, we will have a DT

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and so, this is all going to[br]be equal to the square root

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of, so the stuff that's[br]still under the radical

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is going to be X prime of T squared

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plus Y prime of T squared

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and now we took out a DT

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and now we took out a DT.

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I could have written it right over here

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but I'm just writing it on the other side,

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we're just multiplying the two.

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So, this is once again just[br]rewriting the expression

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for this infinitesimally[br]small change in arc length.

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Well, what's lucky for us is in calculus

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we have the tools for adding up

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all of these infinitesimally small changes

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and that's what the definite[br]integral does for us.

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So, what we can do if we wanna add up that

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plus that plus that plus that

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and remember, these are[br]infinitesimally small changes.

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I'm just showing them[br]as not infinitesimally

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just so that you can[br]kind of think about them

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but if you were to add them all up,

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then we are essentially[br]taking the integral

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and we're integrating with respect to T

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and so, we're starting at T is equal to A,

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all the way to T is equal to B

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and just like that we[br]have been able to at least

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feel good conceptually

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for the formula of arc length

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when we're dealing with[br]parametric equations.

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In the next few videos

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we'll actually apply it[br]to figure out arc lengths.