Line Integrals of Vector Fields // Big Idea, Definition & Formula

Previously in the Vector Calculus playlist (see below), we have seen the idea of a Line Integral which was an accumulation of some function along a curve. In this video we're going to look at case where we begin with a vector field and want to measure the accumulation of the field tangential to the curve. A great example of this is the physics concept of work done by a field on a particle moving along a curve. In this video we will define the basic concept of the line integral of a vector field along a curve and then determine a formula in terms of a particular parametrization of the curve.
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[@lexixu3655]

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[@DrTrefor]

Sucks, but glad I could help!

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This is great... my professor explains it in such a way that she thinks we see what she sees in the first glimpse of it, but she only ends up losing us from the get-go and we have no idea what happens for the remaining 1.5 hours of our lectures... but you, in a little under 9 minutes , have made this simply intuitive and graspable! Thanks Dr. Bazett :D

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Before, I only knew how to apply the formulas, but your playlist helps really understanding what is going on and why, and that’s exactly what I need to pass my electromagnetic field theory exam. Great work

[@samb.7395]

Thanks for giving a short summary of previous lessons before jumping into the new stuff. It really helps make the connections.

[@carmelpule8493]

Congratulations for your presentations, I am an old man over 85 years old and this takes me back to my youth. I like the way you related the curl derivatives with the Green's theorem. A long time ago I tried to relate the connections between Curl and Green's theorem and the Cauchy Reiman conditions, Divergence and Greens theorems and the Cauchy Reiman conditions and dU/dx, dV.dy , dU/dy, and dV/dx where since U has a direction with dx but different from dy and V has a direction with dy but different from dx. the divergence and the curl seems to become so obvious. I should have published my work. Thank you for your submissions, they are excellent I used these vectors to expand electrical signals in three dimensions and recognition was done using Laplace Functions and Convolution and Fourier Integrals in three dimensions. Good old days, Thank you once again. It was so Long ago at the University of Newcastle upon Tyne UK. and now in my retirement, Brilliant work that you are doing,

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[@yellowbird8224]

I am a calculus professor, trying to get a better handle on Maxwell's Equations and Einstein's wave equation, and this course is helping me immensely! Thanks a million.

[@DrTrefor]

Cool! Glad it helped:)

[@meonjiworld]

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[@daisyg509]

I am having soooo much trouble with these last sections of calc 3 but I'm only half way in and this makes it make soooo much more sense 😭😭 thank you 🙏🏻😭

[@weplaysalot]

Was lost on the last two videos I watched but these concrete examples are great. Thinking of it as a particle moving across a vector field is very helpful

[@daisyg509]

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[@user-oc6lw2rd1q]

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[@yeast4529]

Great video and visuals. Line integrals over vector fields weren't given much of an explanation besides the formula in my course, but this video gives a great step-by-step understanding and doesn't miss out anything, thanks!

[@joaquintorres834]

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[@Princess-nu7vs]

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[@UrBigSisKey]

BEST MATERIAL OUT THERE FOR LINE INTEGRALS AND WORK. GO DOCTOR BAZETT GO

[@NguyenTran-sn8vt]

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[@DrTrefor]

Thank you

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[@DrTrefor]

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[@discussionfortherightway.9725]

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[@DrTrefor]

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[@Richard-coom-bucket]

amazing content

[@eamon_concannon]

I have a lot of experience of vector calculus, but still am enjoying your videos as review/entertainment. The concept of work used to confuse me until I realized that "work done" means "work done on an object by a particular force field when it moves along a particular path". If a particle moves along some path, it may be acted upon by many forces, and may have varying speeds depending on its initial speed but the work done on it by a particular fixed force field does not change even if the other forces acting on the particle changes while the path stays fixed. As an example, a ball could be constrained to move in a curved pipe of fixed shape at a fixed distance from the Earth. The work done by Earths gravity on the ball as it traverses the length of the pipe would be independent of any other force on the ball e.g. friction and also the initial speed of the ball. At 3:37 the tangent vector T is unit length, so we integrate the dot product of the force vector at each point on curve with the unit tangent vector at that point along curve.

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[@DrTrefor]

Thanks!

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[@DrTrefor]

Glad you like them!

[@SI_Sourav_Mondal]

স্যার আপনি গুরুদেব। আপনার ভিডিওগুলি খুবই উপকারী।

[@sergiolucas38]

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[@DrTrefor]

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[@albertbittencourt7686]

delightful explanation

[@DrTrefor]

Glad you enjoyed!

[@mohdalshad6620]

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[@DrTrefor]

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[@DrTrefor]

Glad it was helpful!

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[@jaffarmohiddin4258]

Thank you very much. You cleared my doubt about line integral.

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[@mdib2001]

Please make a video on surface integral! By the way, love the way you explain 😍

[@DrTrefor]

That’s coming soon for sure!

[@physicslover1950]

I also want to see a vector field which changes with time i.e., a vector field as a function of time will help us to visualize gravitational waves and electromagnetic waves. 💚

[@rohankalal4532]

Amazing explanation!! I hope you will continue this series all the way to important theorems related to integral theorem.(green ,stokes..) And also surface and volume integrals

[@DrTrefor]

Of course! The goal is to develop everything and end with Divergence/Stokes' theorem.

[@rohankalal4532]

Thank you

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@@rohankalal4532 aur aklal akisa ahi abhi?

[@josefhrdlicka2251]

Thank you a lot for this video! Helped me understand the topic more clearly. My question is: The vector field is probably infinite, so when we integrate over some curve, do we just get dot product of forces acting on points dS that are ON the curve, or is it work of some area around the curve? Thank you for your response.

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Very cool. Almost as good as John Gabriel's New Calculus channel

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Nice

[@proudaojiao1554]

This is where the intro lecture kicks in: s (parameter length) for the geometric visualization of the problem we are trying to solve, t (parameter time) gives us the function to actually solve the problem. Hope I'm right on this one, professor.

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[@mtach5509]

Good explanation 👍👍👍 but you forgot to put the metric tensor int integral - for general curve linear coordinate

[@Levelup383]

Tx for the lecture

[@DrTrefor]

You're most welcome!

[@arsalansyed4709]

you da best!

[@DrTrefor]

Thank you!!

[@youtoo_metoo]

I wonder why the vector "T" is dr/ds at 7:07. Thanks for video!

[@DrTrefor]

The proof is via Chain rule: dr/ds = dr/dt * dt/ds. The former of these is v. The latter of these is 1/|v| (see definition of arclength parameter). So we get v/|v| which is the definition of the unit tangent vector T.

[@youtoo_metoo]

@@DrTrefor Thanks again

[@discussionfortherightway.9725]

@@DrTrefor Thank you very much! I wondered why you didn't note, that T had to be unit. (At least I didn't notice when you implied it or so). That explains it! Greetings and love from Germany.

[@roshan8853]

@@DrTrefor I was slightly confused. Perhaps putting a 'hat' on top of the T helps show it's the unit tangent? I love the video!

[@robbybridges2854]

Indeed

[@miklosberes8565]

Great videos: the structure of the topic is clear, the explanations are easily perceived or understood, and the conclusions are self-evident. You show the beauty of math. Congratulations!! But my mother language is not English. Doc, Speak sometimes slower, please. (f.e. at the end of the video)

[@thehighground583]

This helped so much, thank you! One question: why don't we do F·r instead of F·dr? I know that T is the unit tangential component of the change in r (which could technically be the velocity) but if we want to see how much F is going along r, and r already changes direction constantly, do we really need to use dr instead of r? Thanks!

[@carultch]

In the special case of a straight line path through a uniform vector field, the integral F·dr reduces to F·r. Again, this is only for a special case. It's the general case where you have a vector field that could have any distribution, and a path that could be anything, that we have to turn a simple dot product, into a line integral with a dot product by the differential. This is what is necessary to account for the fact that the path isn't a simple straight line, and that the field changes at every point along the way.

[@joaocosta3506]

when we were evaluating integrals of parametric functions we would just integrate over the function with respect to arc length, which would give us the area under the curve. But now we have a function, which is a vector field, so there's no "area" to get. In intuitive terms I think we can interpret the vector field as a field of forces, being the vectors the representation, at a given point, of a certain force. Now supose a object moves through the field over a curve C. We're gonna integrate through the vector field, over the curve C. We know that the work done is F.d but actually this F it's not the total force acting on a certain object, it is a "useful force", in the sense that being the work a process in which we tranfer energy, the force in the formula of the work it's the force that is actually responsible for this transfer of energy. So in the case of the block we've the horizontal component of gravity, because even though gravity acts on the force it is only her horizontal component that is transfering energy trough displacing the object along the plane, we observe this in a form of kinetic energy. So the object did path C over the field of force, so we want the know the work, that is the total energy transfered by the forces trough the movement described by the object, and for that we do the integral along c of F.T with respect to the arc lenght. Which means we're summing trough every infinitesimal point of the arc (ds) the work done, therefore F.T because F is projected onto the tangent vector (dot product) and the tangent force is the "useful force" in every point, because the tangent vector is the linear aproximation of the curve at that specific point, which roughly means we're just applying the F.d force but in a infinitesimal point.Hopefully its clarifies thing to other viewers

[@eamon_concannon]

What is meant by "work done" is "work done by a particular force F". F can be any force on the object or the total (resultant) force on the object. The motion of the object will usually be due to several forces acting on the object. The confusion is probably because the letter "F" usually denotes resultant force on an object as we see in Newton's second law F = ma, but in this video F is any particular force on the object, not necessarily the resultant. If F is the resultant force on the object, then the work-energy theorem tells us that the work done by F on the object is the change in the kinetic energy of the object.

[@rahimahmer6744]

Hello, Great explanation however I am having trouble understanding how dr/ds is a unit vector and was wondering if someone could explain that to me.

[@amanmohamed8122]

In the formula part, i dont get the last step where you use chain rule ((ds/dt)*dt)

[@SuperDeadparrot]

I’m wondering if you can take the curl of that field and is that curl zero?

[@physicslover1950]

Will you please cover line integrals along a closed path and their scalar differential form? Also why these is no scalar differential form in line integrals of scalar function?

[@DrTrefor]

Yup will have a vid on line integrals w.r.t dx or dy two vids from now👍

[@matthewzuelke6721]

This is a really good explanation! However, I feel like this would have been clearer if the word "portion" were used instead of "proportion," for proportion can be greater than one which doesn't make sense in this scenario. Nevertheless, it is still a great video.

[@abhijeetdhabe7704]

How come we get a dome (3d) for x²+y²? Is it not an equation for a circle?

[@ogunsadebenjaminadeiyin2729]

Super, 🇫🇷🇳🇬❤️💕😎🙏

[@TheFpsPlayer01]

salve do brasil

[@alecwalker5270]

It appears that the gradient in the example is a mapping from a two-dimensional space into a two-dimensional space. (Both are the X-Y plane.) Yet, you have R^2 -> R on the blackboard. What am I missing?

[@Festus2022]

The math all makes sense, but what is causing the particle to move along the curve in the first place. Wouldn't there have to be another force moving the particle along the curve , so that it could "experience" the force of the vector field?

[@documentemail7299]

Please help me, why T (unit tangent vector) is used?

[@DrTrefor]

We want to know how much to add in the direction we are going, but the length of that vector shouldn't change our computation

[@ichigo_nyanko]

How can you get from "integral of F dot T ds" to "integral of F(r) dot dr/dt dt" ? Is there some substitution you use to get from one way of writing it to another? I tried doing it but I'm getting weird answers that aren't working.

[@eamon_concannon]

The unit tangent T is dr/ds because we can think of dr as a displacement vector connecting two infinitesimally close points on the curve whose length is the scalar ds, so dr/ds is a unit vector. The parameter s gives length of curve and is a function of the other curve parameter t. Without going into analysis proofs, we can treat the derivatives like fractions so we get F.T ds = F . dr/ds ds/dt dt = F. dr/dt dt after cancelling ds.

[@delsonmcgriffin1989]

5:37

[@adhirajdhungana3112]

Hi Dr. Bazett, I have always been confused with the idea of how (dr/dt)dt substitutes into dr. How does this process actually work? Love from Winnipeg, University of Victoria is one of my destinations for post grad. So, maybe someday I will see you in person.

[@eamon_concannon]

Consider Δr as a small vector connecting two points on the curve and Δt as the difference in the value of t for the two points. As Δt approaches 0 then Δr also approaches 0 and Δr/Δt approaches some value (assuming it exists) denoted by dr/dt. Therefore Δr = (Δr/Δt) Δt approaches (dr/dt)0 = 0. Doing the integration involves taking the limit of the sum of a very large number of the very small quantities (Δr/Δt) Δt as Δt approaches 0. Replacing (Δr/Δt) Δt with dr/dt dt = dr and integrating (i.e summing infinitely many "infinitesimally" small terms involving dr) is the notation for having taken the limit. A more complete explanation comes from real analysis.

[@750siddiq]

Isnt it the tangent vector dr/dt, instead of dr/ds ??.. Some clarification would be appreciated!!

[@DrTrefor]

dr/dt is indeed A tangent vector, but dr/ds is specifically THE unit tangent vector

[@750siddiq]

@@DrTrefor Thanks for your response, can i see it as (dr/dt)/mod(dr/dt)?

[@rajdipsarkar65]

How unit tangent vector equals (dr)/(ds)

[@carultch]

r = position vector relative to the origin dr = infinitesimal change in position vector ds = infinitesimal distance change along path. Has the same units as r, and also has the same magnitude as dr. Therefore dr/ds is a unit vector, since it is a change in position divided by an infinitesimal distance in that direction. And it's in the tangential direction, because that's the direction that the r-vector is changing, when compared to the previous r-vector.

[@ricardobautista-garcia8492]

Third.

[@TekCroach]

At around 2.35, I don’t think the explanation is right. You took the dot product of the gravity with kind of a unit vector tangential to the inclined plane. Then that resolved part of gravity is just multiplied (scaling) by the distance. Of course you get the same final answer. But technically it’s not right. You must take dot product of the force vector with the displacement vector. Right?

[@tkrobert1921]

Whilst I understand the rest of the video, I am confused about a particular point at 7:03, why is it that when we substitute in our definitions for T and ds, we go from a line integral to a definite integral, and why is it that the F goes from F to F(r(t))?, sorry, I may just be being dumb and missing the obvious lol but I have watched the video a couple times and I do not fully grasp why

[@muhhamedhanoon7252]

how is dr/ds unit vector ,pls explain

[@carultch]

r = position vector relative to the origin dr = infinitesimal change in position vector ds = infinitesimal distance change along path. Has the same units as r, and also has the same magnitude as dr. Therefore dr/ds is a unit vector, since it is a change in position divided by an infinitesimal distance in that direction. And it's in the tangential direction, because that's the direction that the r-vector is changing, when compared to the previous r-vector.

[@Hi_Brien]

Timing

[@HA-hd4yf]

Would you add the Arabic translation, please?

[@momentumclassesbyAvaneeshsir]

Include one example plz

[@DigitalOutlawed]

note: to revise this

[@aquilazyy1125]

I didn’t realize that T was supposed to be a unit vector until 6:29. I thought T was a projection from F to the tangent line especially at 3:37 (facepalm). How can a unit vector be so long. They’re always so tiny in my imaginations LOL! Beside that everything is so clear and fun. Another nice class!

[@DrTrefor]

ha quite true! I draw it big to make it obvious but perhaps this does more harm than good:D

[@sanaullahghazi6296]

i don't understand why we use tangential component...the my headache

[@peanutpotato2394]

Same he is so unclear about it

[@mohamedismail6273]

Work is just force tangent times distances

[@joaocosta3506]

when we were evaluating integrals of parametric functions we would just integrate over the function with respect to arc length, which would give us the area under the curve. But now we have a function, which is a vector field, so there's no "area" to get. In intuitive terms I think we can interpret the vector field as a field of forces, being the vectors the representation, at a given point, of a certain force. Now supose a object moves through the field over a curve C. We're gonna integrate through the vector field, over the curve C. We know that the work done is F.d but actually this F it's not the total force acting on a certain object, it is a "useful force", in the sense that being the work a process in which we tranfer energy, the force in the formula of the work it's the force that is actually responsible for this transfer of energy. So in the case of the block we've the horizontal component of gravity, because even though gravity acts on the force it is only her horizontal component that is transfering energy trough displacing the object along the plane, we observe this in a form of kinetic energy. So the object did path C over the field of force, so we want the know the work, that is the total energy transfered by the forces trough the movement described by the object, and for that we do the integral along c of F.T with respect to the arc lenght. Which means we're summing trough every infinitesimal point of the arc (ds) the work done, therefore F.T because F is projected onto the tangent vector (dot product) and the tangent force is the "useful force" in every point, because the tangent vector is the linear aproximation of the curve at that specific point, which roughly means we're just applying the F.d force but in a infinitesimal point.Hopefully its clarifies thing to other viewers

[@sanaullahghazi6296]

@@joaocosta3506 😍 thank you so much