Using a line integral to find the work done by a vector field example Using a line integral to find the work done by a vector field example
Using a line integral to find the work done by a vector field example
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- Let's apply what we learned in the last video into a concrete
- example of the work done by a vector field on something
- going through some type of path through the field.
- So let's say that I have a vector field.
- It's defined over r2 for the x-y plane.
- So it's a function of x and y.
- It associates a vector with every point on the plane.
- And let's say my vector field is y times the unit vector i
- minus x times the unit vector j.
- And so you can imagine if we were to draw-- let's
- draw our x- and y-axes.
- I'll do it over here.
- If we were to draw our x- and y-axes, this associates a
- vector, a force vector-- let's say this is actually a
- force vector-- with every point in our x-y plane.
- So this is x and this is y.
- So if we're at the point, for example, 1, 0, what will the
- vector look like that's associated with that point?
- Well, at 1, 0, y is 0, so this will be 0, i minus 1, j.
- Minus 1, j looks like this.
- So minus 1, j will look like that.
- At x is equal to 2-- I'm just picking points at random, ones
- that'll be -- y is still 0, and now the force vector here
- would be minus 2, j.
- So it would look something like this.
- Minus 2, j.
- Something like that.
- Likewise, if we were to go here, where y is equal to 1 and
- x is equal to is 0, when y is equal to 1, we have 1, i minus
- 0, j, so then our vector is going to look like
- that at this point.
- If we're to go to 2-- you could get the picture.
- You can keep plotting these points.
- You just want to get a sense of what it looks like.
- If you go here, the vector's going to look like that.
- If you go maybe at this point right here, the vector's
- going to look like that.
- I think you get the general idea.
- I could keep filling in the space for this
- entire field all over.
- You know, just to make it symmetric, if I was here,
- the vector is going to look like that.
- You get the idea.
- I could just fill in all of the points if I had to.
- Now, in that field, I have some particle moving, and let's say
- its path is described by the curve c, and the
- parameterization of it is x of t is equal to cosine of t, and
- y of t is equal to sine of t.
- And the path will occur from t-- let's say, 0 is less than
- or equal to t is less than or equal to 2pi.
- You might already recognize what this would be.
- This parameterization is essentially a
- counterclockwise circle.
- So the path that this guy is going to go is
- going to start here.
- Well, you can imagine, t in this case, you could almost
- imagine is just the angle of the circle, but you can
- also imagine t is time.
- So at time equals 0, we're going to be over here.
- Then at time of pi over 2, we're going to have traveled a
- quarter of the circle to there, so we're moving in
- that direction.
- And then at time after pi seconds, we would have
- gotten right there.
- And then all the way after 2pi seconds, we would have gotten
- all the way around the circle.
- So our path, our curve, is one counterclockwise rotation
- around the circle, so to speak.
- So what is the work done by this field on this curve?
- So the work done.
- So the work, we learned in the previous video, is equal to the
- line integral over this contour of our field, of our vector
- field, dotted with the differential of our movement,
- so dotted with the differential of our movement dr.
- Well, I haven't even defined r yet.
- I mean, I kind of have just the parameterization here, so we
- need to have a vector function.
- We need to have some r that defines this path.
- This is just a standard parameterization, but if I
- wanted to write it as a vector function of t, we would write
- that r of t is equal to x of t, which is cosine of t times i
- plus y of t times j, which is just sine of t times j.
- And likewise, this is for 0 is less than or equal to t, which
- is less than or equal to 2pi.
- And this are equivalent.
- The reason why I took the pain of doing this is so now I can
- take its vector function derivative, and can figure out
- its differential, and then I can actually take the dot
- product with this thing over here.
- So let's do all of that and actually calculate this line
- integral and figure out the work done by this field.
- One thing might already pop in your mind.
- We're going in a counterclockwise direction, but
- at every point where we're passing through, it looks like
- the field is going exactly opposite the direction
- of our motion.
- For example, here we're moving upwards.
- The field is pulling us backwards.
- Here we're moving to the top left.
- The field is moving us to the bottom right.
- Here we're moving exactly to the left.
- The field is pulling us to the right.
- So it looks like the field is always doing the exact opposite
- of what we're trying to do.
- It's hindering our ability to move.
- So I'll give you a little intuition.
- This'll probably deal with negative work.
- For example, if I lift something off the ground,
- I have to apply force to fight gravity.
- I'm doing positive work, but gravity's doing
- negative work on that.
- We're just going to do the math here just to make you
- comfortable with this idea, but it's interesting to think about
- what's exactly going on even here.
- The field is-- the field I'm doing in that pink color,
- so let me stick to that.
- The field is pushing in that direction, so it's always
- going opposite the motion.
- But let's just do the math to make everything in the last
- video a little bit more concrete.
- So a good place to start is the derivative of our
- position vector function with respect to t.
- So we have a dr/dt, which we could also write
- as r prime of t.
- This is equal to the derivative of x of t with respect to t,
- which is minus sine of t times i plus the derivative of y
- of t with respect to t.
- Derivative of sine of t is just cosine of t.
- Cosine of t times j.
- And if we want the differential, we just multiply
- everything times dt, so we get to dr is equal to-- we could
- write it this way.
- We could actually even just put the d-- well,
- let me just do it.
- So it's minus sine of t dt-- I'm just multiplying each of
- these terms by dt, distributive property-- times the unit
- vector i plus cosine of t dt times the unit vector j.
- So we have this piece now.
- And now we want to take the dot product with this over here,
- but let me rewrite our vector field in terms of in
- terms of t, so to speak.
- So what's our field going to be doing at any point t?
- We don't have to worry about every point.
- We don't have to worry, for example, that over here the
- vector field is going to be doing something like that
- because that's not on our path.
- That force never had an impact on the particle.
- We only care about what happens along our path.
- So we can find a function that we can essentially substitute y
- and x for, their relative functions with respect to t,
- and then we'll have the force from the field at any
- point or any time t.
- So let's do that.
- So this guy right here, if I were to write it as a function
- of t, this is going to be equal to y of t, right? y is a
- function of t, so it's sine of t, right? that's that.
- Sine of t times i plus-- or actually minus x, or x of
- t. x is a function of t.
- So minus cosine of t times j.
- And now all of it seems a little bit more
- If we want to find this line integral, this line integral is
- going to be the same thing as the integral-- let me pick
- a nice, soothing color.
- Maybe this is a nice one.
- The integral from t is equal to 0 to t is equal
- to 2pi of f dot dr.
- Now, when you take the dot product, you just multiply the
- corresponding components, and add it up.
- So we take the product of the minus sign and the sine of t--
- or the sine of t with the minus sine of t dt, I get-- you're
- going to get minus sine squared t dt, and then you're going
- to add that to-- so you're going to have that plus.
- Let me write that dt a little bit.
- That was a wacky-looking dt.
- dt, and then you're going to have that plus these two guys
- multiplied by each other.
- So that's-- well, there's a minus to sign here so plus.
- Let me just change this to a minus.
- Minus cosine squared dt.
- And if we factor out a minus sign and a dt, what is
- this going to be equal to?
- This is going to be equal to the integral from 0 to 2pi of,
- we could say, sine squared plus-- I want to put the t --
- sine squared of t plus cosine squared of t.
- And actually, let me take the minus sign out to the front.
- So if we just factor the minus sign, and put a minus
- there, make this a plus.
- So the minus sign out there, and then we factor dt out.
- I did a couple of steps in there, but I think you got it.
- Now this is just algebra at this point.
- Factoring out a minus sign, so this becomes positive.
- And then you have a dt and a dt.
- Factor that out, and you get this.
- You could multiply this out and you'd get what we
- originally have, if that confuses you at all.
- And the reason why I did that: we know what sine squared of
- anything plus cosine squared of that same anything is.
- That falls right out of the unit circle definition of
- our trig function, so this is just 1.
- So our whole integral has been reduced to the minus integral
- from 0 to 2pi of dt.
- And this is-- we have seen this before.
- We can probably say that this is of 1, if you want to
- put something there.
- Then the antiderivative of 1 is just-- so this is just going to
- be equal to minus-- and that minus sign is just the same
- minus sign that we're carrying forward.
- The antiderivative of 1 is just t, and we're going to evaluate
- it from 2pi to 0, or from 0 to 2pi, so this is equal to
- minus-- that minus sign right there-- 2pi minus t
- at 0, so minus 0.
- So this is just equal to minus 2pi.
- And there you have it.
- We figured out the work that this field did on the particle,
- or whatever, whatever thing was moving around in this
- counterclockwise fashion.
- And our intuition held up.
- We actually got a negative number for the work done.
- And that's because, at all times, the field was actually
- going exactly opposite, or was actually opposing, the movement
- of, if we think of it as a particle in its
- counterclockwise direction.
- Anyway, hopefully, you found that helpful.
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