Hey correct me if I am wrong, but on 10:41 you placed positive cos(t) instead of -cos(t) as your N. I am confused because how are you getting a positive cosine from a negative force in the "j" direction. Nonetheless, based on your answer, placing a negative in front of the cosine will grant the same answer, however, it won't need to integrate sin(2t), but instead 0. anyways, thank you for the helpful video! :)
I am so blown away by your explanations and your teaching style. I have learned so much from you and I am just in my first year at uni (studying Applied Physics). Know that you are the reason me (and many others, as I am sure you know) are passing their classes and truly enjoying the subject. Thank you!
I really like the way you give qualitative illustrations, your series on vectors, calculus, and line, flow and flux integrals are very useful for me to understand electromagnetic fields better. Thank you so much prof.
About a third of the way through your playlist, I already understand more about vector calculus then I did in a semester of University. Mr Bazett, you are a legend.
I missed the final lectures in my calculus 3 class and your videos were amazing at picking up where we left off, very intuitive and in detail about the examples. Thank you x100.
@@DrTrefor If it does not obey right hand rule then it's counter clockwise curve or vice a versa for clock wise curve am I right sir please tell me and sir please tell the analogy of shadow integral perspective of surface integral and projection of circle area is converted ellipse area please tell me the reason sir iam sorry for annoying you but this is an question In which I lost my sleep and you are great sir your efforts to teach us will be for ever with us a million thanks to you sir
cool! thanks! i get it now! i was always confused why they were using cross product formula (determinant) within in the context of a dot product… ( its because the definition of the normal out vector is based on a cross formula between the normal up and clockwise tangent vectors) YAY!
Going through the course right now. If I pass, I will definitely join your channel. It isn't much but its what I can do. Meanwhile, hats off for you, very nice content.
Wow, I always thought we only needed to use the gradient operator for finding the normal vector... Awesome, I've learned a new thing here. Blessing for you Doctor.😯😯😁😁
hi thanks for your videos, excellent animations give a perspective to what I read in books. I am from Uruguay, I do not understand English but I understand the mathematical content.
Trevor, @ around 8:31 and beyond. It appears that the ds differential outside the parentheses would just cancel out both the denominators for the unit vector components leaving dotted with . Is this sound math? It seems to work.
Dr Trefor, thank you for sharing the great contents. I have a question. Can we define N(t) as normalized dT/ds? This definition does not require defining "outward pointing" (hence does not require curve to be simple) and can be easily generalized to higher dimention. Any particular reason you use T * k.
The example vector field given at 9:52 is negative of the one solved at 10:41, but flux should still be 0 in both cases? along with marcel's correction
This might be out of scope for this video, but the cross product actually generalizes to ANY number of dimensions, but not in the way you think it dose. The cross product is only a binary operator in 3D, the 2D equivalent is a unary operation which is basically computed here, and the 4D cross product is a trinary operator of three vectors. In general, it's easier to think of the cross product as a function rather than an operator, and the number of arguments to this function is N-1 where N is the number of dimensions in which we are working. This can be derived directly from the determinant definition of a cross product, where you are finding the determinant of an NxN matrix, making the inclusion of a dummy z-direction in this example, somewhat unnecessary. This means to find the normal to a 15D "surface" embedded in 16D space, you would use a 15-vector cross product. I should note that there is also a 7D binary cross product, based on octonion multiplication, which is non-associative. But trying to get an idea of what this means geometrically, breaks my brain.
Taking inspiration from this example, does the following generalisation always work: For any two "dual" vector fields (i.e. 2 fields where for every point the vector fields evaluate to orthogonal vectors) F1 and F2, and for any simple closed continuous curve C, Flux(F1, C) = Flow(F2, C) and Flow(F1, C) = Flux(F1, C)? Realistically if this holds then i guess the condition can be loosened to F1 and F2 only needing to be orthogonal for points along C.
If it does not obey right hand rule then it's counter clockwise curve or vice a versa for clock wise curve am I right sir please tell me and sir please tell the analogy of shadow integral perspective of surface integral and projection of circle area is converted ellipse area please tell me the reason sir iam sorry for annoying you but this is an question In which I lost my sleep and you are great sir your efforts to teach us will be for ever with us a million thanks to you sir
Theses videos are super helpful for visualizing what we are actually doing but I don't get how we can just add in a z component. Wouldn't that imply that we are slowly spiraling into the sky?
Well done some the way you introduced a third dimension which is converted back to 2D was a very great and revolutionary idea and the intuitive animation of that was awesome and fabulous. Now I came to know about the minus sign between dyi and dxj. Dir I have a question. If our curve is clockwise then instead of writing N=T×k , will we write N=T×(-k) ? Sir please answer me. Sir also please make a video on how to convert a gradient vector field back to a scalar function and how to make a vector field from a scalar function. 💞💞💚
Unit T = (dx/dt, dy/dt)/sqrt((dx/dt)^2+(dy/dt)^2) vs Non unit T = (dx/dt, dy/dt) It turns out dt goes into the denominator from the numerator T = (dx, dy)/(sqrt((dx/dt)^2+(dy/dt)^2)*dt) And sqrt((dx/dt)^2+(dy/dt)^2)*dt = sqrt((dx/dt)^2*dt^2+(dy/dt)^2*dt^2)=sqrt(dx^2 + dy^2)= ds Thus the unit vector T = (dx, dy)/ds P.S. Ask ChatGPT to write it in LaTeX
Unit T = (dx/dt, dy/dt)/sqrt((dx/dt)^2+(dy/dt)^2) vs Non unit T = (dx/dt, dy/dt) It turns out dt goes into the denominator from the numerator T = (dx, dy)/(sqrt((dx/dt)^2+(dy/dt)^2)*dt) And sqrt((dx/dt)^2+(dy/dt)^2)*dt = sqrt((dx/dt)^2*dt^2+(dy/dt)^2*dt^2)=sqrt(dx^2 + dy^2)= ds Thus the unit vector T = (dx, dy)/ds P.S. Ask ChatGPT to write it in LaTeX
This is confusing me: in the example you gave when we have the integral of the flux we have a dot produt with the vector n, which means the quantity that is leaving the curve c, but it is situaded in the xy plane, because n is in the xy plane, however when you search images of flux and as you mentioned too in the video, we have this sort of idea of something coming out from a surface, but in that case shouldn't we project the integral in vetor k?
Hi I'm a little confused about the restriction that the line integral must be simple. It seems like you can still compute the integral on a non simple curve, and I guess if you just ignore the part of the intersection where you previously were, then you can still find a vector normal to your current trajectory i.e. tangent vector. So why must the curve be simple?
Hey correct me if I am wrong, but on 10:41 you placed positive cos(t) instead of -cos(t) as your N. I am confused because how are you getting a positive cosine from a negative force in the "j" direction. Nonetheless, based on your answer, placing a negative in front of the cosine will grant the same answer, however, it won't need to integrate sin(2t), but instead 0. anyways, thank you for the helpful video! :)
Thanks Marcel, you are of course correct!
@@DrTrefor anytime! :)
I am so blown away by your explanations and your teaching style. I have learned so much from you and I am just in my first year at uni (studying Applied Physics). Know that you are the reason me (and many others, as I am sure you know) are passing their classes and truly enjoying the subject. Thank you!
I really like the way you give qualitative illustrations, your series on vectors, calculus, and line, flow and flux integrals are very useful for me to understand electromagnetic fields better. Thank you so much prof.
Thank you! Indeed E&M is one of the best use cases for vector calc
About a third of the way through your playlist, I already understand more about vector calculus then I did in a semester of University. Mr Bazett, you are a legend.
That's awesome, glad it is helping!
I missed the final lectures in my calculus 3 class and your videos were amazing at picking up where we left off, very intuitive and in detail about the examples. Thank you x100.
I was just waiting for this video to come out. Thanks doc!
Hope you enjoyed it!
@@DrTrefor Just finished it. You're great, i understood everything in one go
@@DrTrefor If it does not obey right hand rule then it's counter clockwise curve or vice a versa for clock wise curve am I right sir please tell me and sir please tell the analogy of shadow integral perspective of surface integral and projection of circle area is converted ellipse area please tell me the reason sir iam sorry for annoying you but this is an question
In which I lost my sleep and you are great sir your efforts to teach us will be for ever with us a million thanks to you sir
your explanations are IMMACULATE
Where were you when I needed you! Octogenarian here. Keep up the good work.
Sir you give feeling about maths
You really got what the original mathematician was thinking when he came up with these concepts
cool! thanks! i get it now! i was always confused why they were using cross product formula (determinant) within in the context of a dot product… ( its because the definition of the normal out vector is based on a cross formula between the normal up and clockwise tangent vectors) YAY!
Going through the course right now. If I pass, I will definitely join your channel. It isn't much but its what I can do. Meanwhile, hats off for you, very nice content.
Good luck, hope they help:)
thank you so much, your video really help a lot to understand the concept.
i am always on your videos.
Man, you're a saviour
Wow, I always thought we only needed to use the gradient operator for finding the normal vector... Awesome, I've learned a new thing here. Blessing for you Doctor.😯😯😁😁
Thank you!!
It really helps to have these diagrams. These are great videos.
hi thanks for your videos, excellent animations give a perspective to what I read in books. I am from Uruguay, I do not understand English but I understand the mathematical content.
Absolutely loving this series
Trevor, @ around 8:31 and beyond. It appears that the ds differential outside the parentheses would just cancel out both the denominators for the unit vector components leaving dotted with . Is this sound math? It seems to work.
Thank You Dr for interesting explanation.
You’re most welcome!
Amazing video, these geometric meanings are really outstanding :)
You are a genius!!! Thank you so much! Greetings from Uruguay!!! :D
You're welcome!
Woah 😯 so many viewers from Uruguay!
Dr Trefor, thank you for sharing the great contents. I have a question. Can we define N(t) as normalized dT/ds? This definition does not require defining "outward pointing" (hence does not require curve to be simple) and can be easily generalized to higher dimention. Any particular reason you use T * k.
Yes absolutely, this is equivalent.
The example vector field given at 9:52 is negative of the one solved at 10:41, but flux should still be 0 in both cases? along with marcel's correction
Thank you so much sir 🔥🔥🔥
nice , I want to know the history of concepts like (flux ,curl,etc) ,is there any book about them and their applications in physics?
Calculus: Early Transcendals by Stewart is a great one
This might be out of scope for this video, but the cross product actually generalizes to ANY number of dimensions, but not in the way you think it dose. The cross product is only a binary operator in 3D, the 2D equivalent is a unary operation which is basically computed here, and the 4D cross product is a trinary operator of three vectors. In general, it's easier to think of the cross product as a function rather than an operator, and the number of arguments to this function is N-1 where N is the number of dimensions in which we are working. This can be derived directly from the determinant definition of a cross product, where you are finding the determinant of an NxN matrix, making the inclusion of a dummy z-direction in this example, somewhat unnecessary. This means to find the normal to a 15D "surface" embedded in 16D space, you would use a 15-vector cross product.
I should note that there is also a 7D binary cross product, based on octonion multiplication, which is non-associative. But trying to get an idea of what this means geometrically, breaks my brain.
always wanted to do a video on octonion multiplication!
thank you
Taking inspiration from this example, does the following generalisation always work:
For any two "dual" vector fields (i.e. 2 fields where for every point the vector fields evaluate to orthogonal vectors) F1 and F2, and for any simple closed continuous curve C, Flux(F1, C) = Flow(F2, C) and Flow(F1, C) = Flux(F1, C)?
Realistically if this holds then i guess the condition can be loosened to F1 and F2 only needing to be orthogonal for points along C.
Bro these are awesome, genuinely
I had to comment again 3 mins later into the video, this is stunning
Haha thank you!!
Great video!!
So great ❤🎉🎉 Thanks a lot a lot
If it does not obey right hand rule then it's counter clockwise curve or vice a versa for clock wise curve am I right sir please tell me and sir please tell the analogy of shadow integral perspective of surface integral and projection of circle area is converted ellipse area please tell me the reason sir iam sorry for annoying you but this is an question
In which I lost my sleep and you are great sir your efforts to teach us will be for ever with us a million thanks to you sir
Perfect explanation
Thank you! Excellent!
Theses videos are super helpful for visualizing what we are actually doing but I don't get how we can just add in a z component. Wouldn't that imply that we are slowly spiraling into the sky?
Instead of calculating normal vector through cross product, we can use gradient vector as well, right? Also, love your videos prof
Well done some the way you introduced a third dimension which is converted back to 2D was a very great and revolutionary idea and the intuitive animation of that was awesome and fabulous. Now I came to know about the minus sign between dyi and dxj. Dir I have a question. If our curve is clockwise then instead of writing N=T×k , will we write N=T×(-k) ? Sir please answer me. Sir also please make a video on how to convert a gradient vector field back to a scalar function and how to make a vector field from a scalar function. 💞💞💚
Exactly, just flips the sign.
Please make a video on green functions
It's coming soon!
Thnks sir for response
Cool, thanks
Very useful video. Best wishes.
Thank you!!
I've understood every single thing till now except for one thing. Why is the Tangent Vector dr/ds?
Why "Unit" tangent vector? The derivative is not always equal to 1.
Unit T = (dx/dt, dy/dt)/sqrt((dx/dt)^2+(dy/dt)^2)
vs
Non unit T = (dx/dt, dy/dt)
It turns out dt goes into the denominator from the numerator
T = (dx, dy)/(sqrt((dx/dt)^2+(dy/dt)^2)*dt)
And sqrt((dx/dt)^2+(dy/dt)^2)*dt = sqrt((dx/dt)^2*dt^2+(dy/dt)^2*dt^2)=sqrt(dx^2 + dy^2)= ds
Thus the unit vector T = (dx, dy)/ds
P.S. Ask ChatGPT to write it in LaTeX
Unit T = (dx/dt, dy/dt)/sqrt((dx/dt)^2+(dy/dt)^2)
vs
Non unit T = (dx/dt, dy/dt)
It turns out dt goes into the denominator from the numerator
T = (dx, dy)/(sqrt((dx/dt)^2+(dy/dt)^2)*dt)
And sqrt((dx/dt)^2+(dy/dt)^2)*dt = sqrt((dx/dt)^2*dt^2+(dy/dt)^2*dt^2)=sqrt(dx^2 + dy^2)= ds
Thus the unit vector T = (dx, dy)/ds
P.S. Ask ChatGPT to write it in LaTeX
Why fo you use the cross product to find n ? I prefer use a 2×2 matrix isn't that more simple ?
is k the same as the binormal vector?
Thanks sir
This is confusing me: in the example you gave when we have the integral of the flux we have a dot produt with the vector n, which means the quantity that is leaving the curve c, but it is situaded in the xy plane, because n is in the xy plane, however when you search images of flux and as you mentioned too in the video, we have this sort of idea of something coming out from a surface, but in that case shouldn't we project the integral in vetor k?
just to get normal throught Va x Vb
nice
In the last Ndx is handled wrongly I guess it should be -cost -sint dt
Hi I'm a little confused about the restriction that the line integral must be simple. It seems like you can still compute the integral on a non simple curve, and I guess if you just ignore the part of the intersection where you previously were, then you can still find a vector normal to your current trajectory i.e. tangent vector. So why must the curve be simple?
By "simple", he means is that it is a curve that doesn't intersect itself. So a curve that looks like a 0, rather than an 8.
Very sensitive explanation to understand divergence and curl?
Hey boss of math,what is your country?
Canada:)
flux is like resistance but only 90° or normal. 180° flux and 0° flux==0😮
note : to revise
Now all you need is a Delorean....
:D
if a physicist watch this video then he or she will make some deal to make flux and flow exist together lol...