Line integrals in vector fields
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Line Integrals and Vector Fields
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Using a line integral to find the work done by a vector field example
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Parametrization of a Reverse Path
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Scalar Field Line Integral Independent of Path Direction
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Vector Field Line Integrals Dependent on Path Direction
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Path Independence for Line Integrals
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Closed Curve Line Integrals of Conservative Vector Fields
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Example of Closed Line Integral of Conservative Field
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Second Example of Line Integral of Conservative Vector Field
Path Independence for Line Integrals Showing that if a vector field is the gradient of a scalar field, then its line integral is path independent
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- What I want to do in this video is establish a reasonably
- powerful condition in which we can establish that at vector
- field, or that a line integral of a vector field is
- path independent.
- And when I say that, I mean that let's say I were to take
- this line integral along the path c of f dot d r, and let's
- say my path looks like this.
- That's my x and y axis, and let's say my path looks
- something like this: I start there and I go over
- there to point c.
- My end point, the curve here is c.
- And so I would evaluate this line integral, this victor
- field along this path.
- This would be a path independent vector field, or we
- call that a conservative vector field, if this thing is equal
- to the same integral over a different path that has
- the same end point.
- So let's call this c1, so this is c1, and this is c2.
- This vector field is conservative if I start
- at the same point but I take a different path.
- Let's say I go something like that: if I take a different
- path-- and this is my c2 --I still get the same value.
- What this is telling me is that all it cares about to evaluate
- these integrals is my starting point and my ending point.
- It doesn't care what I do in between.
- It doesn't care how I get from my starting point
- to my end point.
- These two integrals have the same start point and same end
- point, so irregardless of their actual path, they're
- going to be the same.
- That's what it means for f to be a conservative field, or
- what it means for this integral to be path independent.
- So before I prove or I show you the conditions, let's build
- up our tool kit a little bit.
- And so you may or may not have already seen the
- multivariable chain rule.
- And I'm not going to prove it in this video, but I
- think it'll be pretty intuitive for you.
- So maybe it doesn't need to have a proof, or I'll prove it
- eventually, but I really just want to give you the intuition.
- And all that says is that if I have some function-- let's say
- I have f of x and y, but x and y are then functions of, let's
- say a third variable, t, so f of x of t and y of t --that the
- derivative of f with respect to t is multivariable.
- I have two variables here in x and y.
- This is going to be equal to the partial of f
- with respect to x.
- How fast does f change as x changes times the derivative of
- x with respect to t-- This is a single variable function right
- here, so you to take a regular derivative.
- So times how fast x changes with respect to t.
- This is a standard derivative, this is a partial derivative,
- because at that level we're dealing with two variables.
- And we're not done. --plus how fast f changes with respect to
- y times the derivative y with respected t.
- So d y d t.
- And I'm not going to prove it, but I think it makes
- pretty good intuition.
- This is saying, as I move a little bit d t, how
- much of a d f do I get?
- Or how fast this f change with respect to t?
- It says, well there's two ways that f can change: it can
- change with respect to x and it can change with respect to y.
- So why don't I add those two things together as they're both
- changing with respect to t?
- That's all it's saying, and if you kind of imagined that you
- could cancel out this partial x with this d x, and this partial
- y with this d y, you could kind of imagine the partial of f
- with respect to t on the x side of things, and then plus the
- partial of f with respect to t in the y dimension.
- And then that'll give you the total change of f
- with respected to t.
- Kind of a hand-wavy argument there, but at least to me this
- is a pretty intuitive formula.
- So that's our tool kit right there; the multivariable
- chain rule.
- We're going to put that aside for a second.
- Now let's say I have some vector field f-- and it's
- different than this f, so I'll do it in a different color;
- magenta --I have some vector field f that is a
- function of x and y.
- And let's say that it happens to be the gradient of
- some scalar field.
- I'll call that capital F.
- And this is gradient which means that capital F is also
- function of x and y-- so I don't want to write it on a new
- line, I could also write up here; capital F is also a
- function of x and y --and the gradient, and all that means is
- that the vector field f of xy-- lower-case f of xy, is equal to
- the partial derivative of upper-case F with respect to x
- times the i-unit vector plus the partial of upper-case F
- with respect to y times the j-unit vector.
- This is the definition of the gradient right here.
- And if you imagine that upper-case F is some type of
- surface-- so this is uppercase F of xy --the gradient F of xy
- is going to be a vector field that tells you the direction of
- steepest descent at any point.
- So it'll be defined the xy plane.
- So on the xy plane it'll tell you-- so let me draw; that's
- the vertical axis, maybe that's the x axis, that's the y axis
- --so the gradient of it, if you take any point on the xy plane,
- it'll tell you the direction you need to travel to go
- into the deepest descent.
- And for this gradient field it's going to be
- something like this.
- And maybe over here it starts going in that direction because
- you would descend towards this little minimum
- point right here.
- Anyway, I don't want to get too involved.
- And the whole point of this isn't to really get the
- intuition behind gradients; there are other videos on this.
- The point of this is to get other a test to see whether
- something is path independent; whether a vector field is path
- independent, whether it's conservative.
- And it turns out that if this exists-- and I'm going to prove
- it now --if f is the gradient of some scalar field,
- then f is conservative.
- Or you could say it doesn't matter what path we follow when
- we take a line integral over f, it just matters about our
- starting point and our ending point.
- Now let me see if I can prove that to you.
- So let's start with the assumption that f can be
- written this way, as the gradient, that lower-case f can
- be written as the gradient of some upper-case F.
- So in that case our integral-- well, let's define
- our path first.
- So our position vector function-- we always need one
- of those to do a line integral or a vector line integral --r
- of t is going to be equal to x of t times i plus y of t times
- j 4t going between a and b.
- You've seen this multiple times; this is a definition
- of pretty much any path in two dimensions.
- And then we're going to say f of xy is going to be equal to
- this: it's going to be the partial derivative of uppercase
- F with respect to x-- so we're assuming that this exists, that
- this is true --times i plus the partial of upper-case F
- with respect to y times j.
- Now, given this what is lower-case f dot dr going
- to equal over this path right here?
- This path is defined by this position function right there.
- Well, it's going to be equal to, we need to figure out
- what dr is, and we've done that in multiple videos.
- I'll do that on the right over here.
- dr, we've seen it multiple times.
- Actually, I'll solve it out again.
- dr over dt by definition was equal to dx over dt times i
- plus-- I don't know why it got all fat like that
- --dy over dt times j.
- That's what dr over dt is.
- So if we want to figure out what dr is, the differential of
- dr, if we want to play with differentials in this way,
- multiply both sides times dt.
- And actually I'm going to treat dt, I'll multiply
- it; I'll distribute it.
- It's dx over dt times dti plus dy over dt times dtj.
- So if we're taking the dot product of f with dr, what
- are we going to get?
- So this is going to be the integral over the curve from--
- I'll write the c right there; we could write in terms of the
- end points as t once we feel good that we have everything in
- terms of t --but it's going to be equal to this dot that,
- which is equal to-- I'll try to stay color consistent --the
- partial of upper-case F with respect to x times that, times
- dx over d t-- I'm going to write this st in a different
- color --times dt plus the partial of upper-case F with
- respect to y times-- we're multiplying the
- j components, right?
- When you take the dot product, multiply the i components, and
- then add that to what you get from the product of the j
- components --so this j component's partial of
- upper-case F with respect to y, and then we have times-- switch
- to a yellow --dy over dt times that dt right over there.
- And then we can factor out the dt.
- Or actually, so I don't have to even write it again, right now
- I wrote it without, well let me write it again.
- So this is equal to the integral.
- And let's say we have it in terms of t; we've written
- everything in terms of t, so t goes from a to b, and so this
- is going to be equal to-- I'll write it in blue --the partial
- of upper case F with respect to x times dx over dt plus-- I'm
- distributing this dt out --plus the partial of uppercase
- F with respect to y.
- dy over dt.
- all of that times dt.
- This is equivalent to that.
- Now you might realize why I talked about the
- multivariable chain rule.
- What is this right here?
- What is that right there?
- You can do some pattern matching.
- That is the same thing as the derivative of upper-case
- F with respect to t.
- Look at this; let me let me copy and paste that just
- so you appreciate it.
- So this is our definition, or this is our-- I won't say
- definition; one can actually prove it.
- You don't have to start from there ---but this is our
- multivariable chain rule right here.
- The driven of any function with respect t is the partial of
- that function with respect to x times dx over dt plus the
- partial of that function with respect to dy over dt.
- I have the partial of upper-case F with respect to
- x times dx over dt plus the partial upper-case F
- with respect to y.
- This and this are identical if you just replace this
- lower-case f with an upper-case F.
- So this in blue right here, this whole expression is equal
- to the integral from t is equal to a to t is equal to b of-- in
- blue here --the derivative of f with respect to dt.
- And how do you evaluate-- let me just the dt in green
- --how do you evaluate something like this?
- I just want to make a point: this is just this from the
- multivariable chain rule.
- And how do you evaluate a definite integral like this?
- Well, you take the antiderivative of the
- inside with respect to dt.
- So what is this going to be equal to?
- You take the antiderivative of the inside, that's just f.
- So this is equal to f of t.
- And let me be clear.
- We wrote before that f is a function.
- So our upper-case F is a function of x and y, which
- could also be written, since each of these are functions
- of t, could be written as f of x of t of y of t.
- I'm just rewriting it in different ways.
- And this could be just written as f or t.
- These are all equivalent, depending on whether you want
- to include the x's or the y's only, or the t's
- only, or them both.
- Because both of the x's and y's are functions of t.
- So this is the derivative of f with respect to t.
- If this was just in terms of t, this is the derivative
- of that with respect to t.
- We take its antiderivative, we're left just with f, and we
- have to evaluate it from t is equal to a to t is equal to b.
- And so this is equal to-- and this is the home stretch
- --this is equal to f of b minus f of a.
- And if you want to think about it in these terms, this
- is the same thing.
- This is equal to f of x of b over y of b-- let me make sure
- I got all the parentheses --minus f of x of
- a over y of a.
- These are equivalent.
- You give me any point on the xy plane, an x and a y, and
- it tells me where I am.
- This is my capital F, it gives me a height.
- Just like that.
- This associates a value with every point on the xy plane.
- But this whole exercise, remember this is the
- same thing as that.
- This is our whole thing that we were trying to prove: that is
- equal to f dot dr. f dot dr, our vector field, which is the
- gradient of the capital F-- remember F was equal to the
- gradient of F, we assume that it's the gradient of some
- function capital F, if that is the case, then we just did a
- little bit of calculus or algebra, whatever you want to
- call it, and we found that we can evaluate this integral by
- evaluating capital F at t is equal to b, and then
- subtracting from that capital F at t is equal to a.
- But what that tells you is that this integral, the value of
- this integral, is only dependent at our starting
- point, t is equal to a, this is the point x of a, y of a, and
- the ending point, t is equal to b, which is x of b, y of b.
- That integral is only dependent on these two values.
- How do I know that?
- Because to solve it-- because I'm saying that this thing
- exists --I just had to evaluate that thing at those two
- points; I didn't care about the curve in between.
- So this shows that if F is equal to the gradient-- this
- is often called a potential function of capital F, although
- they're usually the negative each other, but it's the same
- idea --if the vector field f is the gradient of some scale or
- field upper-case F, then we can say that f is conservative or
- that the line integral of f dot dr is path independent.
- It doesn't matter what path we go on as long as our starting
- and ending point are the same.
- Hopefully found that useful.
- And we'll some examples with that.
- And actually in the next video I'll prove another interesting
- outcome based on this one.
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At 5:31, how is the moon large enough to block the sun? Isn't the sun way larger?
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