Follow-up on Green's Theorem! -> ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-8SwKD5_VL5o.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 :)
@@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...
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!
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!
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?
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 :)
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.
I really appreciate this video, but I will allow myself 3 remarks may the one who has never made mistakes throw the first stone at you ;-) (first remark):there was a mess on the choice of the limit of the second point ptL=[[3,6,0],[-1,1,0]] # first choice (at 4min10sec in the video) but then integral is zero ( bad luck !! ) ?? ptL=[[3,6,0],[1,1,0]] # displayed choice ?? ptL=[[3,6,0],[1,-1,0]] # computed choice !! (second remark):a mistake in the video, the integrand is displayed with a Y squared (4:50 in the video) ?? (third remark): there is no explanation how to parametrize, so I give explanation below Line vector= first point vector -second point vector Segment(t)=vector(first point)+ t * Line vector thus, all the points of the segment can be traversed by varying t from zero to 1 S is along line segment between : [[3, 6, 0], [-1, 1, 0]] www.usna.edu/Users/math/rmm/_files/documents/parametrization.pdf click on the link below to see code SageMath which compute the two video integrals beware that the link below is ephemeral(youtube does not accept the very long permanents SageMath URL sagecell.sagemath.org/?q=smokeo below a link to the code in google doc because of the ephemeral SageCell link docs.google.com/document/d/1t2Q4OY5agt5LNXowD6IeU883JxhT39mYnEBcii2T0RY/edit?usp=sharing
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)
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
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?
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.
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
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.
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
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?
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 😅
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?
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.
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 :)
No you're not! I didn't entirely explain how I did it in this video. Here's an article to help you out: tutorial.math.lamar.edu/classes/calciii/eqnsoflines.aspx
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
really good, really really good. but please, go slower. at 1;20 you started going really fast and changing the animations really fast in my opinion and that can make some people loose focus, otherwise your content has so much quality, don't stop doing these please!
