Learning english was the greatest thing i have done to myself. Due to scarce learning resources (or rather videos so well made that i can be a slacker and get the idea fast and pass with little effort) i wouldnt normally have ever stood a chance in sciences. This video is so good, i have been looking for some 3 hours at resources in my language that are pretty much nonexistent and so bad in quality that you managed to explain much more in 5 minutes than all my compatriots combined in 3 hours. What a blessing. Great video.
James, very useful. I have a question. What does it mean if there's a line form (0,0) to (1,1) and the integral evaluated after the evaluation gives cero? Thank you.
A zero line integral means that there is an equal amount of area -- between the "curtain" formed by z=f(x,y) and the curve C -- that lies above the xy-plane and below the xy-plane.
Mr. James, in solution 5, if I use the standard process for parametrizstion, my y is 1+2t, as opposed to the vector you get in the end , please correct me or otherwise make the method be consistent
@@IfYoucanTakeitYoucanmakeit Just to show a different way to do it. It's better to have a bunch of flexible options for solving problems, since not every method can apply to every problem.
These are great. I watch them at 3/4 speed. I realize it's been a while since you published these, but (to you or anyone else) where does the "speed factor" |r'(t)| comes from, notationally and conceptually?
it comes from pythagorean theorem : the (small change in the x-coordinate)^2 + (the small change in the y-coordinate)^2 = (the small change of the length of your curve)^2 and all of that with respect to t. What i called the small change is actually the slope so (dx/dt)^2 + (dy/dt)^2 = (dS)^2. Finally when you want to replace dS in the original formula you take the square root of that expression and you get his formula (the modulus of vect r'(t)). And if you wonder why in the hell would you do that, thats because there's too many different variables in the original formula (x,y,S) and perform changes of variables in order to get a single variable integral (which is t in our case). Hope that'll help someone
Learning english was the greatest thing i have done to myself. Due to scarce learning resources (or rather videos so well made that i can be a slacker and get the idea fast and pass with little effort) i wouldnt normally have ever stood a chance in sciences. This video is so good, i have been looking for some 3 hours at resources in my language that are pretty much nonexistent and so bad in quality that you managed to explain much more in 5 minutes than all my compatriots combined in 3 hours. What a blessing. Great video.
Watched till the end, i am dazzled no way i might pass it
If i fail vector calculus i get kicked out of the uni, im of a good hope now
Bless you mr Hamblin you saved my calc 3 grade
Just what I needed. Thanks!
thank you so much for this, very clear and easy to understand your train of thought
Thank you sir for this video, this really helped me a lot
Thanks for this. Also, question: how do you parametrize more complicated curves? Thank you!
James, very useful.
I have a question. What does it mean if there's a line form (0,0) to (1,1) and the integral evaluated after the evaluation gives cero? Thank you.
A zero line integral means that there is an equal amount of area -- between the "curtain" formed by z=f(x,y) and the curve C -- that lies above the xy-plane and below the xy-plane.
Im struggling with this stuff, thank you
ruclips.net/video/Dz1a8GXohA8/видео.html
You are the best
Nice video!
Hi James,
In the third sum, why is the limit from 0 to 1?? Thank You...
When we parametrize a line segment using t and (1-t) like we did, the parameterization always goes from t=0 to t=1.
@@HamblinMath parametrization to evaluate line integral is always gonna go from 0 to 1?
@@darshiniikrishnan615 It doesn't have to, but it's a good default option.
@@HamblinMath alright, sir. Thank you.
Mr. James, in solution 5, if I use the standard process for parametrizstion, my y is 1+2t, as opposed to the vector you get in the end , please correct me or otherwise make the method be consistent
Don't feel like you have to only use the "standard" method; there are often simpler or easier parametrizations like the one I show here.
@@HamblinMath then could you please explain why did you take t instead of 2t+1 ?
@@IfYoucanTakeitYoucanmakeit Just to show a different way to do it. It's better to have a bunch of flexible options for solving problems, since not every method can apply to every problem.
In computation 2 where did you get the number -24t and 800 from that is in the integral from 0-1?? 4:18the upper half circleI
These are great. I watch them at 3/4 speed. I realize it's been a while since you published these, but (to you or anyone else) where does the "speed factor" |r'(t)| comes from, notationally and conceptually?
it comes from pythagorean theorem : the (small change in the x-coordinate)^2 + (the small change in the y-coordinate)^2 = (the small change of the length of your curve)^2 and all of that with respect to t. What i called the small change is actually the slope so (dx/dt)^2 + (dy/dt)^2 = (dS)^2. Finally when you want to replace dS in the original formula you take the square root of that expression and you get his formula (the modulus of vect r'(t)). And if you wonder why in the hell would you do that, thats because there's too many different variables in the original formula (x,y,S) and perform changes of variables in order to get a single variable integral (which is t in our case). Hope that'll help someone
Greatful
for the question at 7:10 wouldn't most prof just ask you to do in polar coords instead?
I'm not assuming that students know how to do line integrals in polar coordinates in this video.
thank you so much
11:13 2-3 is -1 no? Where did ya get -5?
helpful
Sorry sir how you determine the limit there? That 0 - 1 ? For question no 3
Limit is 0 to5
how do you get the range 0 to 1?
That's just how we chose to parameterize that curve. I can say more if you ask about a specific problem in the video.
thank u
Sir I need solution of one question related to line integral... Would you plxx help me?
Yes
How can I do millions likes in this video??????
Splendiferous