Omg your an absolute legend Dr.Trefor. I have spent hours trying to understand each 1.5 h lecture only to realize this man here just took a 1.5 h lecture to 12 min and made it 5 times easily to understand. I salute you for you time and effort to make student life to learn much more enjoyable.
When I first saw line integrals I was really scared of it. After watching your videos on vector calculus series seems like a piece of cake now. One of the best teacher ever. Thanks a lot for all your hardwork.
This man is the goat! I practically passed my semester by watching @Dr. Trefor Bazett videos (specially the playlist of multivariable calculus and linear algebra). I have a lot of respect for this man; the time and work it takes to record and animate videos like this is not short nor easy.
Absolutely superb explanation - I am returning to this area of study after many years. How I could have done with TH-cam and Dr Bazett back in the 1980's! It would have saved me hours and loads of stress! Now it is my turn to communicate aspects of this subject to engineers and computer scientists - After a gap of some 30 years, I have caught up in just a few hours what I was expecting to take a few weeks of study - great resource Trefor- thanks
Dr! you have more subscriders (students) than the total no of students at the university where you teaches, and sooner you will have more students on this channel than the total students in all universities across the Canada.
i knew i would learn things quicker here , so skipped the online college lecture and came here to learn it the easy way instead of wasting 1 hour in the class
I really enjoyed this video and the explanation was superb and I would love to see you explain the method of computing a line integral of F.dr to some value x1, y1 , z1 as dummy variables and finding the function from there.
Interesting suggestion. I've accepted your challenge, to show that you get the same result, by performing a line integral. I've opted to start from the origin, but you can really start from anywhere and still get *a* function whose gradient is the given vector field. Another starting point, will produce a different potential function that only differs by an arbitrary constant. Using Trefor's example: F(x, y, z) = Let P, Q, and R indicate the three component functions of this vector field, which are all functions of x. Set up a path, starting at the origin, and aiming toward your destination point. To keep it simple, make this a straight line path. Call the path r(t), and let t be a parameter that goes from 0 to 1 r(t) = Take the derivative to find r'(t): r'(t) = The line integral (call it W, for work) is given by: W = integral F(x, y, z) dot dr Since r'(t) = dr/dt, this means we can substitute r'(t) dt for dr: W = integral F(x, y, z) dot dt This means: W = integral dot dt The dot product integrand evaluates as: P(x, y, z)*x1 + Q(x, y, z)*y1 + R(x, y, z)*z1 dt Substitute components of r for x, y, & z: P(x1*t, y1*t, z1*t)*x1 + Q(x1*t, y1*t, z1*t)*y1 + R(x1*t, y1*t, z1*t)*z1 dt Replace functions with their given functions: W = integral ([y1*t*cos(x1*t) + y1*t]*x1 + [sin(x1*t) + x1*t]*y1 + 1*z1) dt Distribute: W = integral [x1*y1*t*cos(x1*t) + 2*x1*y1*t + y1*sin(x1*t) + 1*z1) dt Evaluate each sub-integral, and pull constants out in front: x1*y1*integral t*cos(x1*t) dt = y1*t*sin(x1*t) + y1/x1*cos(x1*t) 2*x1*y1*integral t dt = x1*y1*t^2 y1*integral sin(x1*t) dt = -y1/x1*cos(x1*t) z1*integral 1 dt = z1*t Notice that y1/x1*cos(x1*t) exists for both positive and negative terms, which cancel. Thus: W = [y1*t*sin(x1*t) + x1*y1*t^2 +z1*t] eval from 0 to 1 Evaluate from 0 to 1: W = y1*sin(x1*t) + x1*y1 + z1 And this is our general potential function for this vector field. Notice that it matches Trefor's result at 8:15, when you drop the subscripts, which is exactly what we expect.
going through examples like this is great! would be cool maybe to see more of these. Maybe like a playlist on doing difficult questions from each topic and going through them slowly...
Just out of curiosity: It’s also possible to find the potential by computing a line integral from a reference point to an arbitrary point (x, y, z) along an easy path (like a straight line) since a conservative field is path independent and then use the fundamental theorem of line integrals. Do you think it’s easier than computing a bunch of integrals and partial derivatives?
You could try it, but it will be a lot easier to integrate each component, reconcile the common terms, and match the gaps left by the partial-constants of integration. Setting up a generalized path to all points, would be a lot more challenging.
We can prove that the constant of integration that is a function of the other two variables (say y and z) than the variable of partial integration (x) is zero if differentiate phi again wrt another variable (say y) and equate it to the y component of the conservative vector field and use the property that curl is zero derive and equality between dFx/dy and dFy/dx we get Fy =Fy + C(y,z) Hence C(y,z) = 0 Similarly other constants of integration are also zero.
You aren't adding up the results. Instead, you are looking to fill in the gaps, left by the "constants" of integration. I put constants in quotes, because they really are functions of the other variables. Given a vector field: F = Integrate M(x,y,z) with respect to x. You will get a function in the form of m(x,y,z) + A(y, z) Integrate N(x,y,z) with respect to y. You will get a function in the form of n(x,y,z) + B(x, z) Integrate P(x,y,z) with respect to z. You will get a function in the form of p(x,y,z) + C(x, y) Any terms in common among m(x,y,z), n(x,y,z) and p(x,y,z), you should only have one copy of them in your final result. If there are terms in the n-function and/or p-function that only depend on y and z, then they are part of the function A(y,z). If there are terms in the m-function that only depend on x, then they are part of the functions B(x,z) and C(x,y). Any terms depending on all three variables, will be in all three anti-derivative functions. Only one copy of them, should be in your final answer for the potential function. You then complete it with a final constant of integration, which I generally call +K.
I find hilarious how for some degrees on some countries this is taught during the 3rd or 4th semester while. In another ones is not even taught but assumed to already master it. (My situation: first semester, they assume I already know the whole course of Calculus IV) 😅
Because if all we care about, is finding *a* potential function, and not necessarily any specific potential function, or the general family of all possible potential functions, it's good enough to just keep it simple and let the final constant of integration be zero. The final constant of integration will cancel anyway, if using the potential function to evaluate a line integral.
The way you sat.... 🙄🙄. But nice content. Why didn't you use software writing this time sir.? This video is without animations. Once the concept has been defined, Math Simulations allow us to produce many more visual examples quickly. I hope you understand.
I clicked "like", not for the sake of the big TH-cam algorithm in the dark sky, but because you are good teacher.
Haha thank you
@@DrTrefor I am only in around 5-7th grade yet I can understand your explanations. Keep up the good work sir!
Omg your an absolute legend Dr.Trefor. I have spent hours trying to understand each 1.5 h lecture only to realize this man here just took a 1.5 h lecture to 12 min and made it 5 times easily to understand. I salute you for you time and effort to make student life to learn much more enjoyable.
Wow, thanks!
It's incredible the difference in teaching skill when someone is just there to "get their paycheck" vs someone who is passionate about the material.
1.5 hrs for scalar potential!! then your teacher must also be covering the origin of scalar potential
When I first saw line integrals I was really scared of it. After watching your videos on vector calculus series seems like a piece of cake now. One of the best teacher ever. Thanks a lot for all your hardwork.
I swear to God your videos explain in 15min things which I was trying to understand for the last 7 weeks of lectures. Thanks for your work
I spent two days in frustration trying to understand this, and you taught it to me in twelve minutes, thank you.
So glad it helped!
This man is the goat! I practically passed my semester by watching @Dr. Trefor Bazett videos (specially the playlist of multivariable calculus and linear algebra). I have a lot of respect for this man; the time and work it takes to record and animate videos like this is not short nor easy.
Absolutely superb explanation - I am returning to this area of study after many years. How I could have done with TH-cam and Dr Bazett back in the 1980's! It would have saved me hours and loads of stress! Now it is my turn to communicate aspects of this subject to engineers and computer scientists - After a gap of some 30 years, I have caught up in just a few hours what I was expecting to take a few weeks of study - great resource Trefor- thanks
Subscribed!
No more words.
Thank you!!
Thank you for explaining this in an easy language, I literally wasted 1-2 hrs in study material while you explained it in 10 mins!
I have a habit.I click like before i see the whole video,because i know that it's gonna be a perfect one!Thank you!
Thank you, Dr. Bazett! Currently taking Calc V and this series of videos has helped a lot with my broader conceptual understanding.
Calc V???
Glad they have been helping!
what the hell is calc v??? lol
I thought Calc 4 (Vector calc) was the max? ( Of course you sre never finished with calculus as it is boundless, but in terms of courses)
Love and respect from INDIA.
Thank you sir.
This really helped me see the process after much confusion thanks.
Very awesome video. It made calculus concept clearer that I ve been studying since 1 , 2 years
Your lectures are just amazing and quite helpful, thankyou so much for your lectures ...
Beautifully explained!!!
Dr. Trefor Bazett, you do not put a line through your z???
Dr! you have more subscriders (students) than the total no of students at the university where you teaches, and sooner you will have more students on this channel than the total students in all universities across the Canada.
That would be quite something!
This series is so helpful! Thanks so much
You're so welcome!
If you just do all three integrals and find their union you gonna end up at the same place. Great video by the way
Thank you so much for this sir
U did a great job
Now it is too easy to understand potential function.💕
Glad it helped!
I love this playlist!
i knew i would learn things quicker here , so skipped the online college lecture and came here to learn it the easy way instead of wasting 1 hour in the class
Man!!! You're a legend.🙌🙌🛐🛐
Thank you so much!!!
Thank you very much sir 🙏🙏🙏
Thank you
I really enjoyed this video and the explanation was superb and I would love to see you explain the method of computing a line integral of F.dr to some value x1, y1 , z1 as dummy variables and finding the function from there.
Interesting suggestion. I've accepted your challenge, to show that you get the same result, by performing a line integral. I've opted to start from the origin, but you can really start from anywhere and still get *a* function whose gradient is the given vector field. Another starting point, will produce a different potential function that only differs by an arbitrary constant.
Using Trefor's example:
F(x, y, z) =
Let P, Q, and R indicate the three component functions of this vector field, which are all functions of x.
Set up a path, starting at the origin, and aiming toward your destination point. To keep it simple, make this a straight line path. Call the path r(t), and let t be a parameter that goes from 0 to 1
r(t) =
Take the derivative to find r'(t):
r'(t) =
The line integral (call it W, for work) is given by:
W = integral F(x, y, z) dot dr
Since r'(t) = dr/dt, this means we can substitute r'(t) dt for dr:
W = integral F(x, y, z) dot dt
This means:
W = integral dot dt
The dot product integrand evaluates as:
P(x, y, z)*x1 + Q(x, y, z)*y1 + R(x, y, z)*z1 dt
Substitute components of r for x, y, & z:
P(x1*t, y1*t, z1*t)*x1 + Q(x1*t, y1*t, z1*t)*y1 + R(x1*t, y1*t, z1*t)*z1 dt
Replace functions with their given functions:
W = integral ([y1*t*cos(x1*t) + y1*t]*x1 + [sin(x1*t) + x1*t]*y1 + 1*z1) dt
Distribute:
W = integral [x1*y1*t*cos(x1*t) + 2*x1*y1*t + y1*sin(x1*t) + 1*z1) dt
Evaluate each sub-integral, and pull constants out in front:
x1*y1*integral t*cos(x1*t) dt = y1*t*sin(x1*t) + y1/x1*cos(x1*t)
2*x1*y1*integral t dt = x1*y1*t^2
y1*integral sin(x1*t) dt = -y1/x1*cos(x1*t)
z1*integral 1 dt = z1*t
Notice that y1/x1*cos(x1*t) exists for both positive and negative terms, which cancel.
Thus:
W = [y1*t*sin(x1*t) + x1*y1*t^2 +z1*t] eval from 0 to 1
Evaluate from 0 to 1:
W = y1*sin(x1*t) + x1*y1 + z1
And this is our general potential function for this vector field. Notice that it matches Trefor's result at 8:15, when you drop the subscripts, which is exactly what we expect.
Genius teaching ❤️❤️❤️👍
Absolutely amazing explanation......👏👏
going through examples like this is great! would be cool maybe to see more of these. Maybe like a playlist on doing difficult questions from each topic and going through them slowly...
Absolute legend
You taught it so well. 👍
friggin mad lad thanks for the help
Namaste sir ji 🙏🏻
Why it is necessary that 'C' is a function of (y,z) . it may be a single constant
you nailed this one i give you that'
Thank you!!!
You're welcome!
this was awesome
How do you do it the other way around? Determining if a scalar potential has a corresponding conservative vector field and finding that vector field?
Finding the field is easy if you have a potential function, just differentiate each component.
Hey, isn't this suspiciously like solving for an exact differential equation?
Yes indeed! The context is quite different but the method almost exactly the same.
I didn't realize how terrible my profs are at teaching. Thank you for al your work. Uni is a scam.
Happy to help!
Just out of curiosity:
It’s also possible to find the potential by computing a line integral from a reference point to an arbitrary point (x, y, z) along an easy path (like a straight line) since a conservative field is path independent and then use the fundamental theorem of line integrals.
Do you think it’s easier than computing a bunch of integrals and partial derivatives?
You could try it, but it will be a lot easier to integrate each component, reconcile the common terms, and match the gaps left by the partial-constants of integration.
Setting up a generalized path to all points, would be a lot more challenging.
We can prove that the constant of integration that is a function of the other two variables (say y and z) than the variable of partial integration (x) is zero if differentiate phi again wrt another variable (say y) and equate it to the y component of the conservative vector field and use the property that curl is zero derive and equality between dFx/dy and dFy/dx we get Fy =Fy + C(y,z) Hence C(y,z) = 0 Similarly other constants of integration are also zero.
4:54 I thought you need to check for boundary conditions in a PDE
I am confused as to how you added the numbers in the last example. How did you get (1*0+1*pi+3-0-0-1) from f(pi,1,3)-f(0,2,1)/
you just saved my life and you don't even know it
Super
i dint get how you finalize the integrals of M,N,and P...i thought you had y sin x in both M and N TO MAKE 2[Y SIN X]
You aren't adding up the results. Instead, you are looking to fill in the gaps, left by the "constants" of integration. I put constants in quotes, because they really are functions of the other variables.
Given a vector field:
F =
Integrate M(x,y,z) with respect to x. You will get a function in the form of m(x,y,z) + A(y, z)
Integrate N(x,y,z) with respect to y. You will get a function in the form of n(x,y,z) + B(x, z)
Integrate P(x,y,z) with respect to z. You will get a function in the form of p(x,y,z) + C(x, y)
Any terms in common among m(x,y,z), n(x,y,z) and p(x,y,z), you should only have one copy of them in your final result. If there are terms in the n-function and/or p-function that only depend on y and z, then they are part of the function A(y,z). If there are terms in the m-function that only depend on x, then they are part of the functions B(x,z) and C(x,y).
Any terms depending on all three variables, will be in all three anti-derivative functions. Only one copy of them, should be in your final answer for the potential function.
You then complete it with a final constant of integration, which I generally call +K.
at 6:43 how is it sinx + x partial derivative of c(y,z) with respect to y is not zero. c can be y^2 and all
Sir,we can find scalar potential by f.dr only when our curl f is zero ???..if not zero then how we can find our potential?
You can only find the scalar potential when the curl is zero
I find hilarious how for some degrees on some countries this is taught during the 3rd or 4th semester while. In another ones is not even taught but assumed to already master it. (My situation: first semester, they assume I already know the whole course of Calculus IV) 😅
What book I should follow for vector calculus???I have thomas calculus.
Why can we assume the remaining constant after integrating C(z) is zero?
Because if all we care about, is finding *a* potential function, and not necessarily any specific potential function, or the general family of all possible potential functions, it's good enough to just keep it simple and let the final constant of integration be zero. The final constant of integration will cancel anyway, if using the potential function to evaluate a line integral.
Good luck
There is that small pink dot in the middle of the screen. At first I tried cleaning it because I thought the screen is dirty :D
youtube should give university degrees as well cause u learn the most here
11:19 Engineers: Answer is 5
its "zee" not "zed"... fight me
🇨🇦 ftw:D
The way you sat.... 🙄🙄. But nice content. Why didn't you use software writing this time sir.? This video is without animations.
Once the concept has been defined, Math Simulations allow us to produce many more visual examples quickly.
I hope you understand.
I love animations, but sometimes I want to go at the same "pace" that students would work through a problem if they were doing an actual test problem.
Whoever disliked this video i will find you!
thank you