I am currently taking Calc. 3 online and your videos have saved me this semester. Thank you so much for breaking these concepts down in such a graspable way. You are the reason I am going to pass and with an A.
Even your shirt is teaching! Great explanation as always. Your videos have been of great help to me lately. I will always be grateful. Thanks Dr. Trefor
@@DrTrefor thanks! I'm sitting here with the videos on one side and my textbook on the other side so I can do the practice problems after each section is finished. working out great so far and I'm going to do some the chapter review problems at the end after I finish all the sections I need to. My teacher's homework guide goes through this chapter even though we didn't get to it in class, so that's helping me gauge how much practice I need to do.
Strange connection I made that helps me to remember the concept is to think of the endpoints as the initial and final letters of a word, with the rest of the letters jumbled in between. As long as those two letters are retained, I can still recognize what the word is supppesd to be, like in that one graphic with many misspelled words that congradulates you for being able to read it despite the rampant spelling errors.
1:20 the work done by gravity will be zero. As the work done is F.dr and in x axis there is no field and in y axis there is no displacement and DR u gave the formula to be F.dx + F.dy from a to b. Wrote this for my revision. Thank You Dr. For this lovely playlist
The ending kinda explains as to why in Physics, we say Conservative Fields are the ones who has the power to create some sort of a Potential Energy in a body. For example, Gravitational Potential, Spring Potential etc. Apart from that, we do learn that F = ∇U where F is any field and U is its Potential
In the gravity example, the work done is unaffected by the velocity field due to no change in the gravitational potential energy. So a conservative field is one where the line integral is unaffected by the velocity field?
Velocity fields and force fields are two separate fields altogether. Even non-conservative force fields are not necessarily affected by velocity fields. A conservative field means that a closed-loop path has a line integral of zero. Applied to reality, this means that if the field represents a force, that it is a zero sum game to take a closed loop path through the field, as it does no net work on you. There is an exchange of energy between potential and kinetic, but no sources or sinks of energy to allow it to enter or leave the mechanical domain. Assuming all other forces are zero work forces. In other words, it is impossible to walk uphill both ways. One way is has a net uphill, and the other way has a net downhill, or both ways are net neutral. If you could walk uphill both ways, in a hypothetical alternate reality, this means that there is an alternative path that would allow you to walk downhill both ways.
@@carultch In what way is the line integral equal to zero; do you mean the magnitude is equal to zero? Or do you mean if you treat it as a vector and partially differentiate it, it will be equal to zero because ▽F = 0? And if so (if there is a change in potential energy), wouldn’t the gradient be of a non-zero value? But I understand the concept of a conservative field that there is no energy leakage in the system.
@@oliverrosol2706 To recap: Vector field: F(x, y) = Path: circle of radius 1, starting at the (0, 1) and proceeding clockwise Potential function: f(x, y) = x^3 - 2*y^2 + K Line integral outcome, prior to applying bounds: W = [sin(t)^3 - 2*cos(t)^2] from t1 to t2 Notice the similarity between the potential function, and the indefinite line integral? I don't even need to evaluate both of them, because they both ultimately are the same function. One is of t, and the other is of x and y. I'd consider this to be luck that it turned out exactly this way, but it is expected that the two will be related. Onward with the evaluations: For t = 0 to pi/2, which is from the 12 O'clock to 3 O'clock position: W = (-1) - (-2) = +1 For t=pi/2 to pi, i.e. 3 O'clock to 6 O'clock: W = (-2) - (-1) = -1 For t=pi to 3*pi/2, i.e. 6 O'clock to 9 O'clock: W = (-3) - (-2) = -1 For t=3*pi/2 to 2*pi, i.e. 9 O'clock to 12 O'clock: W = (-2) - (-3) = +1 Add it all up, and as expected, we get zero. During the first quadrant of this path, the field does positive work. The next two quadrants, it does negative work. Finally, the final quadrant has positive work, returning us to net work of zero. The field is pushing on the object during the parts when it does positive work, and an agent would need to push the object against the field when it does negative work. The line integral is ultimately the integral of the dot product, of the vector field and the infinitesimal segment of the path. So a dot product of the field and the velocity, that we then integrate with respect to time t. If the field is aligned with our movement, it does positive work, if the field opposes our movement, it does negative work. This field does both, and on net, the work adds up to zero.
@@oliverrosol2706 My comment disappeared. Anyway, what I did was derive how a line integral would be set up, for the given F(x, y) and a circular path of radius 1, that is clockwise from (0, 1). Here's some key information. Radius vector of the path: r = Corresponding velocity vector: dr/dt = The line integral is: W = integral F(x, y) dot dr/dt * dt Thus: W = integral dot dt W = integral (3*x^2 * cos(t) + -sin(t)*(4*y)) dt Substitute in x and y terms from original r-vector function, and simplify: W = integral(3*sin^2(t)*cos(t) - 4*sin(t)*cos(t)) dt
Actually, electric fields are non-conservative if they're time-variant. Static charges (zero current) and constant currents (DC circuits in steady-state) are associated with conservative electromagnetic fields, while time-varying currents (such as AC circuits) are associated with non-conservative electromagnetic fields. Besides that, good video! :)
This might be stupid questions, but as math student i dont understand much about physics. Could some one explain why gravitation has no curl? Should the gravitation pull towards the center of the mass of an object? For example people can stand at the north and south pole of the globe.
The curl or the circulation density for a conservative field is zero....which is obvious from the green's theorem....but what significance does this have physically?
This is mostly another way to rephrase the concept of path independence. From stokes theorem you can plug in the curl of the conservative field being zero to get the circulation around any closed loop is zero
second comment. in video about flux and circulation you imagin field F=X+Y as velocity field and as fountain ..my question is what is the physical quantity that flux measure? its unit is m^2/s however flow in physics have a unit of m^3/s?
The issue is we've been doing thing in 2D for now, largely to build intuition in the simpler situation, but will upgrade to 3D in a few videos from now. You can still think of it as m^3/s if you think the surface of water where the field is just constant in the z direction, i.e. all moving in the same way.
@@DrTrefor So I take it that's a "I'm not too sure myself" ah, it's so much of the experience. We need someone who's whole schtick on youtube is helping other youtubers fix audio issues with worked examples.
I got little confused or lost in this video. I think we have to assume that an object or particle moves under the influence of the field on its own and takes different paths from one point to another rather than me moving it. am I missing something?
I think over large scale the field may not be conservative because the gravitational force varies in different points so there will be no perfect negative components to cancel and just give work done for shortest distance
In Newtonian gravity, as long as all bodies are moving slow, relative to the speed of light, gravity is conservative at any given snapshot in time. If the sources of gravity are moving, you may find apparent closed paths through gravity with net work done, but it really isn't a closed path, since the energy came from the KE of the source bodies of the field, and the initial conditions are not perfectly restored. Look up gravitational slingshot. With relativistic effects, there is a non-conservative behavior to gravitational fields called gravitomagnetism.
I am currently taking Calc. 3 online and your videos have saved me this semester. Thank you so much for breaking these concepts down in such a graspable way. You are the reason I am going to pass and with an A.
Even your shirt is teaching! Great explanation as always. Your videos have been of great help to me lately. I will always be grateful. Thanks Dr. Trefor
Haha, I love that T-shirt. Glad I've been helping:)
My calc 3 teacher didn't cover the last few sections of chapter 16 so I'm here cramming for my final that's in 3 days. This series is a life saver!
good luck!
@@DrTrefor thanks! I'm sitting here with the videos on one side and my textbook on the other side so I can do the practice problems after each section is finished. working out great so far and I'm going to do some the chapter review problems at the end after I finish all the sections I need to. My teacher's homework guide goes through this chapter even though we didn't get to it in class, so that's helping me gauge how much practice I need to do.
We will be learning this mid November. Perfect timing.
Nice!
What major are(were) you in?
@@oximas-oe9vf Mechanical. You?
Engineering, I am a first year student so I haven't specialized yet
thinking of Mechatronics or Computer Engineering though.
OMMG (gosh) I love your shirt! IT is so clever and conveys everything AMAZINGLY! The video is also very imformative and I'm learning so much!!
Strange connection I made that helps me to remember the concept is to think of the endpoints as the initial and final letters of a word, with the rest of the letters jumbled in between. As long as those two letters are retained, I can still recognize what the word is supppesd to be, like in that one graphic with many misspelled words that congradulates you for being able to read it despite the rampant spelling errors.
thanks for this!!
@@Dina-he1uc yruoe wloceme
So today, I guess, I discovered a goldmine.
Hope it helps!
Brilliant and intuitive!
Thank God you exist. ❤
I loved the shirt and this great explanation. Thanks a lot!!!!
Took my cal3 with this guy before. Fantastic teacher!
Thanks Eric! Hope your semester is going well:)
1:20 the work done by gravity will be zero. As the work done is F.dr and in x axis there is no field and in y axis there is no displacement and DR u gave the formula to be F.dx + F.dy from a to b. Wrote this for my revision. Thank You Dr. For this lovely playlist
I keep getting distracted by your shirt and having to rewatch the video lol, love that shirt.
Sir, Please upload a complete course of this series as soon as possible.
I will! Should be 2-3 vids per week coming out in vector calculus:)
The ending kinda explains as to why in Physics, we say Conservative Fields are the ones who has the power to create some sort of a Potential Energy in a body. For example, Gravitational Potential, Spring Potential etc. Apart from that, we do learn that
F = ∇U where F is any field and U is its Potential
I hope you get rewarded as much benefit as you provide, god bless and thank you.
Thank you!!
In the gravity example, the work done is unaffected by the velocity field due to no change in the gravitational potential energy. So a conservative field is one where the line integral is unaffected by the velocity field?
Velocity fields and force fields are two separate fields altogether. Even non-conservative force fields are not necessarily affected by velocity fields.
A conservative field means that a closed-loop path has a line integral of zero. Applied to reality, this means that if the field represents a force, that it is a zero sum game to take a closed loop path through the field, as it does no net work on you. There is an exchange of energy between potential and kinetic, but no sources or sinks of energy to allow it to enter or leave the mechanical domain. Assuming all other forces are zero work forces.
In other words, it is impossible to walk uphill both ways. One way is has a net uphill, and the other way has a net downhill, or both ways are net neutral. If you could walk uphill both ways, in a hypothetical alternate reality, this means that there is an alternative path that would allow you to walk downhill both ways.
@@carultch In what way is the line integral equal to zero; do you mean the magnitude is equal to zero? Or do you mean if you treat it as a vector and partially differentiate it, it will be equal to zero because ▽F = 0? And if so (if there is a change in potential energy), wouldn’t the gradient be of a non-zero value? But I understand the concept of a conservative field that there is no energy leakage in the system.
@@oliverrosol2706 To recap:
Vector field: F(x, y) =
Path: circle of radius 1, starting at the (0, 1) and proceeding clockwise
Potential function: f(x, y) = x^3 - 2*y^2 + K
Line integral outcome, prior to applying bounds: W = [sin(t)^3 - 2*cos(t)^2] from t1 to t2
Notice the similarity between the potential function, and the indefinite line integral? I don't even need to evaluate both of them, because they both ultimately are the same function. One is of t, and the other is of x and y. I'd consider this to be luck that it turned out exactly this way, but it is expected that the two will be related.
Onward with the evaluations:
For t = 0 to pi/2, which is from the 12 O'clock to 3 O'clock position:
W = (-1) - (-2) = +1
For t=pi/2 to pi, i.e. 3 O'clock to 6 O'clock:
W = (-2) - (-1) = -1
For t=pi to 3*pi/2, i.e. 6 O'clock to 9 O'clock:
W = (-3) - (-2) = -1
For t=3*pi/2 to 2*pi, i.e. 9 O'clock to 12 O'clock:
W = (-2) - (-3) = +1
Add it all up, and as expected, we get zero.
During the first quadrant of this path, the field does positive work. The next two quadrants, it does negative work. Finally, the final quadrant has positive work, returning us to net work of zero. The field is pushing on the object during the parts when it does positive work, and an agent would need to push the object against the field when it does negative work.
The line integral is ultimately the integral of the dot product, of the vector field and the infinitesimal segment of the path. So a dot product of the field and the velocity, that we then integrate with respect to time t. If the field is aligned with our movement, it does positive work, if the field opposes our movement, it does negative work. This field does both, and on net, the work adds up to zero.
@@oliverrosol2706 My comment disappeared. Anyway, what I did was derive how a line integral would be set up, for the given F(x, y) and a circular path of radius 1, that is clockwise from (0, 1). Here's some key information.
Radius vector of the path:
r =
Corresponding velocity vector:
dr/dt =
The line integral is:
W = integral F(x, y) dot dr/dt * dt
Thus:
W = integral dot dt
W = integral (3*x^2 * cos(t) + -sin(t)*(4*y)) dt
Substitute in x and y terms from original r-vector function, and simplify:
W = integral(3*sin^2(t)*cos(t) - 4*sin(t)*cos(t)) dt
Great presentation!
Actually, electric fields are non-conservative if they're time-variant. Static charges (zero current) and constant currents (DC circuits in steady-state) are associated with conservative electromagnetic fields, while time-varying currents (such as AC circuits) are associated with non-conservative electromagnetic fields.
Besides that, good video! :)
Are you an electrical engineer ;o
@@Michael748159263 Yep. 🤔
@@altuber99_athlete You smart, I quit that shit within 2 months. Now I'm doing mechanical engineering studies (;
@Michael Chen I’m sure you’ll find out that mechanical engineering is by no means any less difficult. Source: I am a mechanical engineer.
What is dt in the integration is it time? Please help me I am very confused
Super vidéo.
You are a great explainer!
Where can I buy your shirt??
How did F(r(t)) reduce to just ?
Shouldn't it be -h(t) instead?
The vector feild is -1 j not -y j
can you talk about liberal vector fields?
Wut?
@@maxwellsequation4887 the opposite of conservative, politically, is liberal
Dr Trefor i love your videos but i do have to ask if there is a way i can get a tshirt like that
Where did you buy your shirt from?
This might be stupid questions, but as math student i dont understand much about physics. Could some one explain why gravitation has no curl? Should the gravitation pull towards the center of the mass of an object? For example people can stand at the north and south pole of the globe.
The curl or the circulation density for a conservative field is zero....which is obvious from the green's theorem....but what significance does this have physically?
This is mostly another way to rephrase the concept of path independence. From stokes theorem you can plug in the curl of the conservative field being zero to get the circulation around any closed loop is zero
Sir where can I buy ur shirt ....waaah did someone notice that ??
Hello, how can I have video slides?
second comment.
in video about flux and circulation you imagin field F=X+Y as velocity field and as fountain ..my question is what is the physical quantity that flux measure?
its unit is m^2/s however flow in physics have a unit of m^3/s?
The issue is we've been doing thing in 2D for now, largely to build intuition in the simpler situation, but will upgrade to 3D in a few videos from now. You can still think of it as m^3/s if you think the surface of water where the field is just constant in the z direction, i.e. all moving in the same way.
@@DrTrefor thanks can you draw a diagram that concider z constant ,please?
btw, what's up with the audio on recent vids? Kinda hurts my ears
Sigh, I hate audio:(
@@DrTrefor So I take it that's a "I'm not too sure myself" ah, it's so much of the experience. We need someone who's whole schtick on youtube is helping other youtubers fix audio issues with worked examples.
where is the t-shirt from!!!??? love IT!
Where do you buy those T shirts? Thanks for the amazing videos btw!
Thanks a lot sir 🔥🙏🔥
I got little confused or lost in this video. I think we have to assume that an object or particle moves under the influence of the field on its own and takes different paths from one point to another rather than me moving it. am I missing something?
Where did you get that shirt!?
I think over large scale the field may not be conservative because the gravitational force varies in different points so there will be no perfect negative components to cancel and just give work done for shortest distance
In Newtonian gravity, as long as all bodies are moving slow, relative to the speed of light, gravity is conservative at any given snapshot in time. If the sources of gravity are moving, you may find apparent closed paths through gravity with net work done, but it really isn't a closed path, since the energy came from the KE of the source bodies of the field, and the initial conditions are not perfectly restored. Look up gravitational slingshot.
With relativistic effects, there is a non-conservative behavior to gravitational fields called gravitomagnetism.
great video! btw where is the shirt from?
Where can I get that shirt?
where can i find that t shirt
Not an exact match, but check out my amazon store link in description of the videos, I found a pretty close match
You are great !
Thank you!:)
You missed a negative sign: Given F( x, y ) then work = -\int( F . dr ) .
Where'd you get that shirt?
I linked a similar - but not identical - one in my amazon store, check out description:)
@@DrTrefor Thanks, G. Excellent content. Excellent response time. Excellent customer support.
But how do I get that shirt?
Thank you very much
Respect sir
need help on discrete math thanks like your t-shirt!
Great video :)
ty
Hello Sir, what does ‘ net flow is zero’ mean in vector calculus?
That the integral of the curve is zero through the vector field
Great T shirt ❤
i love your shirt
Thank you very much :)
A good example to explain potential function. Thank you, Sir.
Too good
great shirt
I want ur T shirt
Poor audio quality
What about a socialist field?