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## Multivariable calculus

### Course: Multivariable calculus > Unit 4

Lesson 3: Line integrals in vector fields- Line integrals and vector fields
- Using a line integral to find work
- Line integrals in vector fields
- Parametrization of a reverse path
- Scalar field line integral independent of path direction
- Vector field line integrals dependent on path direction
- Path independence for line integrals
- Closed curve line integrals of conservative vector fields
- Line integrals in conservative vector fields
- Example of closed line integral of conservative field
- Second example of line integral of conservative vector field
- Distinguishing conservative vector fields
- Potential functions

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# Using a line integral to find work

Using a line integral to find the work done by a vector field example. Created by Sal Khan.

## Want to join the conversation?

- If a particle return back to same initial point then displacement is 0, thus workdone is 0(work done =force*displacement)...

for any close path displacement is 0 thus the work done is also 0... but in the above vedio we got -2pi. how it is possible?? can u pls explain me.

thanks in advance.(11 votes)- What work really depends on is the field. If you have a conservative field, then you're right, any movement results in 0 net work done if you return to the original spot. With most vector valued functions however, fields are non-conservative. In a non-conservative field, you will always have done work if you move from a rest point.

In the video, the field is non-conservative, because as Sal showed, moving around a circle nets you some work.(20 votes)

- why did the i and j (unit vectors) disappeared after the dot product at9:25?(14 votes)
- yes the result is a scalar, and the way you calculate the dot product two vectors is to multiply each element from each vector and add them like this V1 = a1i+b1j and V2= a2i+b2j then V1 dot V2 = a1*a2+b1*b2 which is a scalar without the i and J components.(5 votes)

- So when we say the work done was -2pi, that is the work done by the field. Would the work done by us to keep the particle moving in that path then be 2pi?(8 votes)
- Like you stated, if the field does -2pi units of work, then the force pushing against the field must do +2pi units of work. Pavle, the logic you use is slightly skewed. Recall, work is force times distance (and as we have learned now, it is actually the tangential component of force in the direction we are moving times the distance moved), so if the particle doesn't move, then there is no work done by anything, the field or us.

If the field did -2pi units of work, then the particle definitely moved. the amount of work the field "absorbed" (negative) is the same amount that we "put in" (positive).

Pavle, you would be correct in saying that the particle does not move if we applied the same Force as the vector field but in the negative direction (in this case -yi+xj), but remember force is different than work.

Another way to think about it is through the law of conservation of energy, energy can not be created or destroyed, just change forms. Therefore, if the field does negative work to us (take energy from us) then we must do positive work to the field (supply energy to it). So yes, if the field does -2pi units of work, then we do +2pi units of work. If the field does positive work, then we do the same amount of negative work, etc.

(assuming that we are the ones moving the particle in this instance)(10 votes)

- Is it possible to evaluate the line integral without using a parameter?(8 votes)
- Convert the parametric function into single variable function and find the integral of the real-valued function.(1 vote)

- At1:43, how is he graphing the vectors? at3:37, how does he know whether it's clockwise or counterclockwise? thanks(5 votes)
- f(x,y)=yi+xy ---> f(1,2)=2i+1j ---> means plot the vector with the components of 2i+1j at the point (1,2). Make sense now?

Also he graphed ccw because when you plug in t values, the x and y values correspond to a ccw motion.(4 votes)

- Why does Sal not write the integral sign with the circle at3:58? It
*is*a closed loop.(5 votes)- You could write the closed-loop integral symbol here, but it is not necessary. It is normally used for mathematic definitions, when you want to say "this is only true for closed loops".(3 votes)

- Doesn't r(t) = cos(t) i + sin(t) j point outward, rather than tangentially along the circular path?(3 votes)
- Yes, r(t) will point outward from the origin to that circle. However, r'(t) points tangentially along the circle.(4 votes)

- How are you sure that it's a counter clockwise curve?(3 votes)
- If we take the 1st quadrant (where x and y are positive), we see that as t gets larger the cos will get smaller, so x starts at 1 when t=0 and moves to 0 when t=pi/2. We also see that as t gets larger sin gets larger, thus y=0 when t=0 and y=1 when t=pi/2. If we graph these as points (x,y) marking a new point for each value of t we see that a quarter of a circle is drawn in the counter clockwise direction. Extending this to the other 3 quadrants we see it's a counter clockwise curve.(3 votes)

- So if you integrate over C and C is the circumference of the circle, then the limits of integration should be 2*pi*r, where r is the radius of the circle in meters. So in this case it is 1 meter to make it simple, so r =1 meter. And the force opposing the movement of the particle is in N, so you get 2 pi Nm. Correct? The reason I mention this is if you replace the 1 with a variable r, then the answer is generalized for all Work done by a particle moving CCW in a CW field, that is - 2 pi r. So you could use this formula instead of going through the derivation each time.(3 votes)
- What are the units of 2 pi in the answer?(2 votes)
- we could only figure that out if we knew what units of force and distance were being used in the question ...

without that information it could be some bizzarre archaic unit (particularly here in the semi-medieval US ;-)(1 vote)

## Video transcript

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 straightforward. 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.