Follow-up on Green's Theorem! -> th-cam.com/video/8SwKD5_VL5o/w-d-xo.html Support me on Patreon: patreon.com/vcubingx Here's a small intuition correction: When I said "projecting the integral", that isn't particularly correct. It's more of a "stretching out" the curve so it becomes straight, without distorting its length. This is why we have the factor of sqrt(x'^2 + y'^2) in the conversion formula. (Credit to Dr. Steve Trettel for pointing this out!) MISTAKES! At 4:11, the problem should be (1,-1). I read it wrong and typed it wrong. Wow. Also in the first example the integral's function should be 3(3-2t)^2-2(6-7t). ALSO IN 8:15 I MISSED A MINUS SIGN HOLY
Just wanted to add onto this that if you calculate the problem at 4:11 to be to (1,1) then you should get an answer of 6 * (sqrt(29)) At least that's what I got lol
This kinda looks like 3Blue1Brown content. Not to say this isn't awsome still. That's more of a compliment than anything else but i'll be damned if you aren't an avid viewer of him.
When I saw the first seconds of the video my first thought was "Yes! More Manim users!" I think Manim should be a standard for math-related videos. It's such a great library and it's a shame that not that many people know about it. Sorry for my poor English :)
There is a minor mistake at 4:50, is should just be '-2(6-7t)', not '-2(6-7t)^2'. The answer of 8 × sqrt(53) is correct though if you meant a line segment to (1, -1)
How is 8 x sqrt(53) correct given the corrected '-2(6-7t)' term and the (1, -1) endpoint? Evaluating the eqn at bottom of 4:50 (with the correct term substituted) at t=1 gives 5 x sqrt(53), and for t=0 one gets -15 x sqrt(53), so isn't the correct answer -10 x sqrt(53)?
This is great! I like the balance you have between questions and explanation, 3blue1brown never uses questions which is a shame because it is a nice way to apply what you've learnt and get a breather from the learning. On the other hand the examples are quick and you don't waste time explaining basic calculus so it doesn't feel dragged out and uninformative like many other math videos who uses two thirds of the time for examples. Great work and I can't wait to see more!
I am a first year EE student studying for my vector calc exam . This is the best channel i have found this year regarding this topic. Keep it up, you're the best.
sir you have started to teach difficult math using animation it really helps to me.iam a engineering student from india. please dont stop uploding this kind of vedio
Thank you for giving such details. I don't know why these are not written in books clearly, and we have to search the internet for better understanding, especially for mathematics.
@@wduandy n order to parametrize a line, you need to know at least one point on the line, and the direction of the line. If you know two points on the line, you can find its direction. The parametrization of a line is r(t) = u + tv, where u is a point on the line and v is a vector parallel to the line. There are lots of possible such vectors u and v. To find one such vector v, find the difference between any two points on the line. In this example, we have A(3, 6) and B(1, 1) or I guess he meant to say B(1, -1) hence his result. So using the latter, the direction of the line will be AB = B - A = (1, -1) - (3, 6) = (-2, -7). Thus us our v. The line can then be written as r(t) = A + tv or B + tv, it doesn't really matter, but he used r(t) = A + tv. So r(t) = (3, 6) + t(-2, -7) which means x = 3 - 2t and y = 6 - 7t.
SInce we're working with a segment here, one better way to think about this is in terms of barycenters between the two endpoints with coefficients between 0 and 1. If A and B are the endpoints, then a point P inside the segment can be written as follows : P=t*A+(1-t)*B where 0
Also, the parameterization in the video is definitely wrong and this actually changes the value of the integral ! For a start, , it should be ds=sqrt(29)*dt and not ds=sqrt(53)*dt...
This is the first time when I started to understand the integrals, and it was great, like a schematization to what's Waiting for me, and be prepared to it
THANK YOU for helping for giving me a visual intuition of where the formula for arc length comes from at the start of the video!! That was something that I had previously lacked. Also love the background music!
Thank you for all your helpful videos. I think you're doing an amazing work. If I could suggest one thing, once your channel expands more, you could hire a professional voice-over guy. I think that would facilitate understanding, because visuals are already top notch, and I only ever get lost in the audio explanation. Best of luck with your channel!
One suggestion pal,if you could have used a normal curvy surface like a parabolic roof top kind and explain the line integral concept then it would be made easily more comprehensive than now.But it is a good explanation pal.Impressive one.
4:24 it would help us a lot from confusion if you take the extra step and correcting the point (1,1). It should say (1,-1). Maybe embed an annotation in the current video?
Thanks! What is the precise method of parameterizing C? Looks easy to figure out for a straight line but a wavy curve?...can't be done without a specific method. Can you define that method in detail please?
You can't always parametrize the curve by hand. For example if I draw any squiggle on a paper, you can't always parameterize it. But, when solving problems by hand you would probably get curves that can be easily parameterized, like circles, ellipses, lines, quadratics, etc. You can just Google "parametric curves" and that should answer your question if you're curious on how to parameterize those curves. Thanks for watching!
It is a better interpretation than textbook. But I would suggest go to the text book first after getting the concept of line integral and back to the animation video. The idea of line integral in a vector field becomes much more clear! Thank you for your nice video and interpreting
vcubingx cheers. Today’s learning resource and landscape has changed to the better than before. Digital tutorial video becomes indispensable as a learning companion. Before the rise of Khan academy , it seemed nobody predict that while learning ‘stem’, the help could be made from enthusiasts by digital technologies.
Really good video thankyou so much :) I am confused at 6:26 - I think the red and yellow vector arrows should be in the same direction? I think the red arrow in the circle is facing the wrong way; for the dot product to be negative there must be an obtuse angle between the vectors? Still the animation is amazing well done :)
3:52 the line integral is supposed to be the area of the orange colored region right? the arrow of the integral is pointing on the projected shadow on the xz-plane?
Excellent video and excellent the plotting web sites provided in the description!!! Thank you very much!!! I am a little puzzled at 6:25: my understanding is that the orange arrow inside the circle at the bottom indicates the traveling direction, which is the tangent direction of the curve, which is the yellow arrow direction on the curve. But at this moment, the yellow arrow is pointing toward between clock 12 o'clock and 1 o'clock direction, while the orange arrow within the circle is pointing toward 7 o'clock direction. I am puzzled. Besides that, I follow the video for the most part. It is fantastic video with nice explanation, same as others made by this man. Thank you! God bless!!!
I also understand that at this moment(6:25), the yellow arrow and the vector at that spot basically is in opposite direction, therefore their dot product is negative. The yellow value below the ruler near the bottom is indicating the dot product is negative at this moment. However, within the circle, the orange arrow and the green arrow is basically in the same direction. If so, their dot product should be positive.
You can make parentheses scale to the size of their contents in LaTeX by using \left( and ight) instead of plain ( and ); this might work better for things like your derivatives at 2:07
3:57 how could the projected area be tha same as the area under the graph? I mean if we'd stretch that curve to a line, the area of it would be much bigger. Isn't it like saying that given a line y = mx, on the OXY plane, and a segment a to b on the X axis, the distance between (a, y(a)) and (b, y(b)) is the same as d(a, b)? The pythagorean theorem says the exact opposite. I know it is not the same, I reduced the number of dimensions, but I'm still curious if I'm not noticing something. However great video, keep up the work.
Also, if I have a sheet of paper its shadow can be either a line or the shape of the paper depending how I keep it in the air. The area of the paper (the line integral) won't change however the area of the shadow (projection area) will.
I don’t think the area of the projection is the same as the area under the curve. I’m very much new to these topics from multivariable calculus but my best guess as to why it is being projected onto the OXZ plane is b/c the integral is being taken ‘with respect to’ x (with the dx at the end) so it is only multiplying the height of each rectangle by the x component of its respective ‘step’, and when the integral has the ds at the end it is equal to the area under the curve b/c it multiplies the rectangles by the actual magnitude of their respective ‘step’ instead of a projected version of it. Maybe if it were dy instead of a dx it would be projected onto the OZY plane. Hope this helps! (Sorry if it doesn’t or if I misunderstood your question)
There's an error in 7:15. The red vector inside the circle was supposed to be the (vec r) derivative, but it seems that the red vector was (vec r) itself
Ok, I'm stuck at about 2:10 into the video. You reduce your delta-x and delta-y to be derivitives but where did the delta-t come from? No explanation was given other than you can multiply and devide by the equivalent of 1. Please tell what it actually means. Where is it on the coordinate axis?
When the animation starts playing at 7:16, isn't the red arrow pointing both in the wrong direction and with the wrong magnitude? Shouldn't the red arrow, which is meant to be r'(t), be pointing in the same direction as the yellow arrow in the graph, which is also meant to be r'(t). The yellow arrow appears correct. And, since the dot product of two vectors that are more than 90d apart is negative, it would only make sense if the red arrow was on the opposite side of the green arrow, until after C crosses origo, when the green arrow should swap sides (as it correctly does) but the red arrow should remain pointing up-right (which it currently only does for the latter half of the animation). Am I mistaken?
Sorry, but on 4:34 you made a mistake. Parametrisation supposed to be , not the 6-7t. Maybe I didn't get the point, then could you please fix my issue. UPD. OK, got the pinned comment
Why are F(r(t)l and r'(t) at 7:16 aligned? Shouldn't r'(t) be orientied in the other direction because the displayed dot product is mit negative-it would be with a negative orientation. Therefore the (yellow) vector r'(t) in the vectorfield doesn't show the direction as the red one. Somehow I am confused 😅
At the time 1.54 you added delta t to the equation want to know the reason to add change in time to the equation and also the reason for adding f(x,y) to the equation .... i came here to under integration visually and thank you for such video... very difficult to understand such topics from a text book :)
I have the same question when dealing with more elaborate curves. But to parametrize a line is actually pretty simple. I'll try to be as clear as i can: First main idea is to use a variable t which conviniently for us goes from 0 to 1 So let's say we wanted to get from point a to b which are fixed using t as the parameter. Our parametric function will be f(t) So when t=0 the f(t=0) = a t=1, f(t=1) = b so you can think a bit about it and you can intuitively get to: f(t) = a * (1-t) + b * t So that those conditions stay true. The beautiful thing about it is that a and b can be anything, could be scalar valued, 2D vectors: f(t)= (ax,ay) * (t -1) + (bx,by) * t 3D vectors. The parametrization still linearly gets you from one to the next. More formally it is called linear interpolation look it up on Wikipedia it's awesome.
Neither is correct - the endpoint should be (1, -1) as shown on the graph, to give the answer dS = sqrt53dt. Also note the eqn at 4:50 of 3(3-2t)^2-2(6-7t)^2 should instead be 3(3-2t)^2-2(6-7t), to give the final answer given of 8 times sqrt53, These things have been pointed out by others. Please can you keep this pinned at the top?
Follow-up on Green's Theorem! -> th-cam.com/video/8SwKD5_VL5o/w-d-xo.html
Support me on Patreon: patreon.com/vcubingx
Here's a small intuition correction: When I said "projecting the integral", that isn't particularly correct. It's more of a "stretching out" the curve so it becomes straight, without distorting its length. This is why we have the factor of sqrt(x'^2 + y'^2) in the conversion formula.
(Credit to Dr. Steve Trettel for pointing this out!)
MISTAKES! At 4:11, the problem should be (1,-1). I read it wrong and typed it wrong. Wow.
Also in the first example the integral's function should be 3(3-2t)^2-2(6-7t).
ALSO IN 8:15 I MISSED A MINUS SIGN HOLY
vcubingx
You put the parametric equation such as 3(3-2t)^2-2(6-7t)^2 where the main function is (3x^2-2y).
I think you put a square by mistake.
@@md.rafiqurrahman792 sheesh you're right
Just wanted to add onto this that if you calculate the problem at 4:11 to be to (1,1) then you should get an answer of 6 * (sqrt(29))
At least that's what I got lol
This kinda looks like 3Blue1Brown content. Not to say this isn't awsome still.
That's more of a compliment than anything else but i'll be damned if you aren't an avid viewer of him.
Haha yeah I use the animation engine he created for his videos, which is why it looks so similar
When I saw the first seconds of the video my first thought was "Yes! More Manim users!" I think Manim should be a standard for math-related videos. It's such a great library and it's a shame that not that many people know about it.
Sorry for my poor English :)
Same here...and i love these types of videos
you should check out: manim.
I am a simple man, I see videos helping people to understand the concept of math, I hit like
As an electronics engineering I needed to refresh this 6 years after taking Multivariate calculus, and this was an excellent explanation. Thanks!
There is a minor mistake at 4:50, is should just be '-2(6-7t)', not '-2(6-7t)^2'. The answer of 8 × sqrt(53) is correct though if you meant a line segment to (1, -1)
I was thinking the same
@@michaeljuergen4326 I was second guessing myself
The two mistakes on one screen was really getting to me.
How is 8 x sqrt(53) correct given the corrected '-2(6-7t)' term and the (1, -1) endpoint? Evaluating the eqn at bottom of 4:50 (with the correct term substituted) at t=1 gives 5 x sqrt(53), and for t=0 one gets -15 x sqrt(53), so isn't the correct answer -10 x sqrt(53)?
@@DougChemist-oj4ev Because 6 -5 = 1, not 6-7 = 1 . (3,6) to (1,1)
Sorry I forgot to take the antiderivative when evaluating the final integral; the given answer of 8 times the square root of 53 is correct@@vashon100
This is great! I like the balance you have between questions and explanation, 3blue1brown never uses questions which is a shame because it is a nice way to apply what you've learnt and get a breather from the learning. On the other hand the examples are quick and you don't waste time explaining basic calculus so it doesn't feel dragged out and uninformative like many other math videos who uses two thirds of the time for examples. Great work and I can't wait to see more!
Bro you don't know value of basic calculus and we need to think of it every time
I am a first year EE student studying for my vector calc exam . This is the best channel i have found this year regarding this topic. Keep it up, you're the best.
This encouraged me to learn Vector Calculus.
ತುಂಬು ಹೃದಯದ ಧನ್ಯವಾದಗಳು... ಇದೊಂದು ಅದ್ಭುತ ಪ್ರದರ್ಶನ... ಸಂಪೂರ್ಣವಾಗಿ ಅರ್ಥವಾಯಿತು...
sir you have started to teach difficult math using animation it really helps to me.iam a engineering student from india. please dont stop uploding this kind of vedio
Thank you for giving such details. I don't know why these are not written in books clearly, and we have to search the internet for better understanding, especially for mathematics.
the animations really helped solidify the concept, great job.
Thanks! Glad you liked it!
8:14 isn't that integral should be -3(7+5t)-5(33+39t) ?
your parameterization for the curve C at 4:25 is incorrect. It should be 6-5t not 6-7t
How to do this parametrization?
@@wduandy n order to parametrize a line, you need to know at least one point on the line, and the direction of the line. If you know two points on the line, you can find its direction. The parametrization of a line is r(t) = u + tv, where u is a point on the line and v is a vector parallel to the line. There are lots of possible such vectors u and v. To find one such vector v, find the difference between any two points on the line.
In this example, we have A(3, 6) and B(1, 1) or I guess he meant to say B(1, -1) hence his result.
So using the latter, the direction of the line will be AB = B - A = (1, -1) - (3, 6) = (-2, -7). Thus us our v.
The line can then be written as r(t) = A + tv or B + tv, it doesn't really matter, but he used r(t) = A + tv.
So r(t) = (3, 6) + t(-2, -7) which means x = 3 - 2t and y = 6 - 7t.
SInce we're working with a segment here, one better way to think about this is in terms of barycenters between the two endpoints with coefficients between 0 and 1. If A and B are the endpoints, then a point P inside the segment can be written as follows :
P=t*A+(1-t)*B where 0
Also, the parameterization in the video is definitely wrong and this actually changes the value of the integral ! For a start, , it should be ds=sqrt(29)*dt and not ds=sqrt(53)*dt...
I think he made a mistake in the definition of C. He said in the video “C is the line segment from (3,6) to (-1,1)”.
This is the first time when I started to understand the integrals, and it was great, like a schematization to what's Waiting for me, and be prepared to it
dude... your visualization is crystal clear! thank's a lot !
THANK YOU for helping for giving me a visual intuition of where the formula for arc length comes from at the start of the video!! That was something that I had previously lacked. Also love the background music!
The instrumentals in this video are something else lol. Stylishly done!
such a great video! concise, intuitive and covers all the relevant information
If C was from (1,1) to (3,6), shouldn't the parametrization be [3-2t, 6-5t]?
Yes, vid is wrong.
@@vashon100 He forgot a minus sign. Rather than (1,1) it should have been (1,-1)
Dude this is great. It made everything so much clearer. Thank you for making this
Really good but I think you make the same mistake 3Blue1Brown made when he was starting. Talk a little bit slower. But otherwise great video!
Is it weird to watch this videos on 1.5x?
@@Michallote yes
The downside of being gifted means not being able to see what others cannot see.
It's my pleasure to watch it!! Thank's for doing it!!!
Thanks!
There are a mistake: the line segment from (3,6) to (1,1) is not C=(3-2t, 6-7t) 0
Your videos are great! Please keep the great job you have been doing so far.
Thank you so much!
top class fantastic
best leacture of all about introductionary line integeration
the lo-fi track is a nice touch
Thank you for all your helpful videos. I think you're doing an amazing work. If I could suggest one thing, once your channel expands more, you could hire a professional voice-over guy. I think that would facilitate understanding, because visuals are already top notch, and I only ever get lost in the audio explanation. Best of luck with your channel!
Finally, people are recognizing 3b1b's Pythonlib lmao. Good video!
I think your parametric representation of the line should be . Am I wrong ?
One suggestion pal,if you could have used a normal curvy surface like a parabolic roof top kind and explain the line integral concept then it would be made easily more comprehensive than now.But it is a good explanation pal.Impressive one.
underrated content very good
0:28 just pointing out this little mistake i did noticed f(x)=x^2+y^2; it should be f(x.y)=x^2+y^2. Thanks for doing such a great video, by the way.
Thanks for pointing that out!
4:24 it would help us a lot from confusion if you take the extra step and correcting the point (1,1). It should say (1,-1). Maybe embed an annotation in the current video?
Dope animation, it helps in understanding better
Hi, I am from India, that was great video sir... 👏👏👏👍
This is so well explained. I also liked the way you played the animation again so that we could get a clearer picture of whats going on. Thank you!
Thank you so much for this video! it helped it to better understand line integrals for my finals!
I really like the song choice in the background ... And offcourse the mathematics
Finally i understood the concept..........thankyou 😎😎😎😎
Thanks! What is the precise method of parameterizing C? Looks easy to figure out for a straight line but a wavy curve?...can't be done without a specific method. Can you define that method in detail please?
You can't always parametrize the curve by hand. For example if I draw any squiggle on a paper, you can't always parameterize it. But, when solving problems by hand you would probably get curves that can be easily parameterized, like circles, ellipses, lines, quadratics, etc. You can just Google "parametric curves" and that should answer your question if you're curious on how to parameterize those curves. Thanks for watching!
th-cam.com/video/TwpoLBayKYM/w-d-xo.html
This video here does a great job of explaining starting at 1:32.
I like to think about the line integral as being the total elevation graph as you take a hike (Curve C) through a mountainous region (f(xy))
It is a better interpretation than textbook. But I would suggest go to the text book first after getting the concept of line integral and back to the animation video. The idea of line integral in a vector field becomes much more clear! Thank you for your nice video and interpreting
I agree! I think I've improved in my explanations over time, might revisit this topic in the future!
vcubingx cheers. Today’s learning resource and landscape has changed to the better than before. Digital tutorial video becomes indispensable as a learning companion. Before the rise of Khan academy , it seemed nobody predict that while learning ‘stem’, the help could be made from enthusiasts by digital technologies.
Nicely made.
Really good video thankyou so much :)
I am confused at 6:26 - I think the red and yellow vector arrows should be in the same direction?
I think the red arrow in the circle is facing the wrong way; for the dot product to be negative there must be an obtuse angle between the vectors?
Still the animation is amazing well done :)
Nice video! Just a comment, for better notation, it should be used a "dt" for last shown equation at minute 8.11.
Incredibly helpful! Thank you so much
Thank you for the visualizer
Excellent explanation and awesome animations! Subscribed :)
3:52 the line integral is supposed to be the area of the orange colored region right? the arrow of the integral is pointing on the projected shadow on the xz-plane?
Even i have the same doubt so can you please help me incase if you have got your doubt cleared.
Thanks
At 4:24, for the parametrization to be correct, you missed a minus sign. Rather than (1,1) it should have been (1,-1)
Hello, you wrongly parametrized the curve C at 4:25 , it should be < 3-2t , 6-5t >
Thank you for the help! Cemented my understanding!
Glad it helped!
Excellent video and excellent the plotting web sites provided in the description!!! Thank you very much!!! I am a little puzzled at 6:25: my understanding is that the orange arrow inside the circle at the bottom indicates the traveling direction, which is the tangent direction of the curve, which is the yellow arrow direction on the curve. But at this moment, the yellow arrow is pointing toward between clock 12 o'clock and 1 o'clock direction, while the orange arrow within the circle is pointing toward 7 o'clock direction. I am puzzled. Besides that, I follow the video for the most part. It is fantastic video with nice explanation, same as others made by this man. Thank you! God bless!!!
I also understand that at this moment(6:25), the yellow arrow and the vector at that spot basically is in opposite direction, therefore their dot product is negative. The yellow value below the ruler near the bottom is indicating the dot product is negative at this moment. However, within the circle, the orange arrow and the green arrow is basically in the same direction. If so, their dot product should be positive.
0:28 you mean f(x,y) = x^2 + y^2 instead of "f(x)"
4:24 line segment is wrong. you drew (3,6) to (1,-1) instead of (1,1)
přesně tak
Awesome video. Thank you very much for your help.
Glad it helped!
Excellent presentations. Your videos helped me do much. Thank you.
Thanks for watching
Great vid, really helped me build intuition! Only critique is some of the coordinates of the curves didn't match the graphs you put
Great work 🔥🔥🔥
Just have one word to say "exquisite"
Thank you for that video! Do more such videos !!!
You can make parentheses scale to the size of their contents in LaTeX by using \left( and
ight) instead of plain ( and ); this might work better for things like your derivatives at 2:07
3:57 how could the projected area be tha same as the area under the graph? I mean if we'd stretch that curve to a line, the area of it would be much bigger. Isn't it like saying that given a line y = mx, on the OXY plane, and a segment a to b on the X axis, the distance between (a, y(a)) and (b, y(b)) is the same as d(a, b)? The pythagorean theorem says the exact opposite. I know it is not the same, I reduced the number of dimensions, but I'm still curious if I'm not noticing something. However great video, keep up the work.
Also, if I have a sheet of paper its shadow can be either a line or the shape of the paper depending how I keep it in the air. The area of the paper (the line integral) won't change however the area of the shadow (projection area) will.
I don’t think the area of the projection is the same as the area under the curve. I’m very much new to these topics from multivariable calculus but my best guess as to why it is being projected onto the OXZ plane is b/c the integral is being taken ‘with respect to’ x (with the dx at the end) so it is only multiplying the height of each rectangle by the x component of its respective ‘step’, and when the integral has the ds at the end it is equal to the area under the curve b/c it multiplies the rectangles by the actual magnitude of their respective ‘step’ instead of a projected version of it. Maybe if it were dy instead of a dx it would be projected onto the OZY plane. Hope this helps! (Sorry if it doesn’t or if I misunderstood your question)
Thanks for these beautiful videos..
indians are finally making quality content.
Yes finally something other than jee and neet
Is he indian? I thought its a foreign youtube channel
There's an error in 7:15. The red vector inside the circle was supposed to be the (vec r) derivative, but it seems that the red vector was (vec r) itself
4:25 the parametrization is incorrect! should be C=(3-2t,6-5t)
the correct answer is sqrt of 1044
Thank you so much, this was really helpful!
Ok, I'm stuck at about 2:10 into the video. You reduce your delta-x and delta-y to be derivitives but where did the delta-t come from? No explanation was given other than you can multiply and devide by the equivalent of 1. Please tell what it actually means. Where is it on the coordinate axis?
When the animation starts playing at 7:16, isn't the red arrow pointing both in the wrong direction and with the wrong magnitude? Shouldn't the red arrow, which is meant to be r'(t), be pointing in the same direction as the yellow arrow in the graph, which is also meant to be r'(t). The yellow arrow appears correct. And, since the dot product of two vectors that are more than 90d apart is negative, it would only make sense if the red arrow was on the opposite side of the green arrow, until after C crosses origo, when the green arrow should swap sides (as it correctly does) but the red arrow should remain pointing up-right (which it currently only does for the latter half of the animation). Am I mistaken?
Sorry, but on 4:34 you made a mistake. Parametrisation supposed to be , not the 6-7t. Maybe I didn't get the point, then could you please fix my issue.
UPD. OK, got the pinned comment
Great Msth problem
Why are F(r(t)l and r'(t) at 7:16 aligned? Shouldn't r'(t) be orientied in the other direction because the displayed dot product is mit negative-it would be with a negative orientation. Therefore the (yellow) vector r'(t) in the vectorfield doesn't show the direction as the red one. Somehow I am confused 😅
thanks a lot for this explanation but I have a doubt, how did u get the last question's limits as 0 to 1?
such a great explanation, thank you!
Chillhop music is my favourite and made me love this video more!
What software did you use to animate though?
I used manim: github.com/3b1b/manim. Glad you enjoyed!
Thank you very much...It helped a lot
great work mate...
Awesome!. I'm wondering if you made these videos using manim(3b1b) library.
3:55 I didn't get what is the projection pointed to by the arrow.?
Great work keep going,,👍
great work!
Nicely done! Can you suggest some good Manim tutorials?
Yep. Moar! manim tutorials.
Try Todd Zimmerman's. But the thing that works best for me was going through Grant's projects.
@@vcubingx thanks for suggesting, can you link? I'm not sure which "Todd Zimmerman" you are referring to.
talkingphysics.wordpress.com/2019/01/08/getting-started-animating-with-manim-and-python-3-7/
Mind blowing
Thanks
the first example was wrong wasn't it? the parameterization seems off
and in the animation, you had a square for the y and it wasn't in the problem, not sure if you used that for the final answer
Yeah both the examples have problems. I addressed all of them in a pinned comment. Thanks for pointing them out!
@@vcubingx oof sorry, i should have looked for that first
great video, thanks for sharing :)
Hey! A great video. Thank you
Thanks! You're welcome!
Nice explained thank you.
You're welcome!
this is betufial
Excellent Video!
Just one thing,
around 6:20 it should be "dot product indicate how aligned two vectors are" right ?
i think he meant 'how parallel two vectors are' instead of perpendicular
At 4:32, how did you parametrize the curve from 3x^2 -2y to become < 3 - 2t, 6 - 7t > ?
it's wrong. check the line starts at (1,-1) not (1,1)
At the time 1.54 you added delta t to the equation want to know the reason to add change in time to the equation and also the reason for adding f(x,y) to the equation .... i came here to under integration visually and thank you for such video... very difficult to understand such topics from a text book :)
Hello. How to parameterize a curve into a vector? Thank you for your answer ^_^
I have the same question when dealing with more elaborate curves. But to parametrize a line is actually pretty simple. I'll try to be as clear as i can:
First main idea is to use a variable t which conviniently for us goes from 0 to 1
So let's say we wanted to get from point a to b which are fixed using t as the parameter. Our parametric function will be f(t)
So when t=0 the f(t=0) = a
t=1, f(t=1) = b
so you can think a bit about it and you can intuitively get to:
f(t) = a * (1-t) + b * t
So that those conditions stay true.
The beautiful thing about it is that a and b can be anything, could be scalar valued, 2D vectors:
f(t)= (ax,ay) * (t -1) + (bx,by) * t
3D vectors. The parametrization still linearly gets you from one to the next. More formally it is called linear interpolation look it up on Wikipedia it's awesome.
Can you make a series of complex variable analysis!??
Noted! I definitely want to cover topics within complex analysis, probably not an entire series though
@@vcubingx thanks for consideration now can't wait to see complex analysis visually!!
How did you parametrize the curve?
AMAZING! KEEP IT UP!
At 4:15 you say "to -1,1", yet on the screen the end point is shown as (1,1). Which is correct?
Neither is correct - the endpoint should be (1, -1) as shown on the graph, to give the answer dS = sqrt53dt. Also note the eqn at 4:50 of 3(3-2t)^2-2(6-7t)^2 should instead be 3(3-2t)^2-2(6-7t), to give the final answer given of 8 times sqrt53, These things have been pointed out by others. Please can you keep this pinned at the top?
3:53 why did we squash it onto X-Z plane ?
u saved the education :)
is the answer to the first question is 11sqrt29?