Tired of professors who only teach the algebric way instead of actually trying to describe what happens when we use these concepts. Thanks for making it clear
@thethug19464 жыл бұрын
A stellar career as an educator awaits you. You are a born communicator
@hotvodka76412 жыл бұрын
"Hello welcome to multivariate calculus I'm Professor The Thug"
@whitealpha22654 жыл бұрын
Awesome video, finally understood the intuition behind the Jacobian and not just the algebra
@MGTOW-nn9ls3 жыл бұрын
The best explanation on internet. Excellent job. Thank you
@GizmoMaltese3 жыл бұрын
This was a very clear and concise explanation of a Jacobian. It's always been a bit of a mystery to me. Whenever they talked about focusing on a tiny area I felt lost. But now that you tie it to single variable calculus and Reimann sums it makes sense. Then tying into transformations from one variable to two variable transformations and the determinant. It all clicks. I feel like I finally now understand a lot of the stuff I learned in college.
@trigon70154 жыл бұрын
Funny thing: I was on the Khan Academy multivariable integrals course learning about triple integrals when this video came up. Perfect timing!
@jonathangrey63544 жыл бұрын
I took Calc AB last year and I understood u sub very well but I didn’t understand what the purpose of multiplying by the du factors were for. So last year I came across the Jacobian determinant and my mind was blown. It all makes sense that we just account for stretching caused mapping to different planes.
@dennisbrown53132 жыл бұрын
Excellent; many sites discus the 'how' but never really the why. This makes perfect sense
@vasuhardeo14183 жыл бұрын
Dude i came for the Jacobian and also learned whats going on for a single variable U substitution, thanks so much man. that scaling idea is pretty cool much less the transformation, WOW , awesome.
@NoActuallyGo-KCUF-Yourself3 жыл бұрын
The geometry really helps. I've used Jacobians before, but never really thought about what they _mean._
@nournote4 жыл бұрын
Amazingly explained.
@user-fh5km1ic2t10 ай бұрын
Man you are on a different level for real. Crazy insights!
@herewego80932 жыл бұрын
4:40 a really brilliant way to explain why multiply with g'. Amazing
@farrela67104 жыл бұрын
Thank you so much! Your explanation is very easily understandable!
@LovepreetSingh-jv4uj3 жыл бұрын
Awesome brother.. this connects real analysis linear algebra and differential equations 👍
@alejrandom65922 жыл бұрын
it's been days since I have tried to understand u-sub geometrically, thanks for this :)
@pendawarrior4 жыл бұрын
Thanks for the valuable intuition 👍
@muntedme203 Жыл бұрын
Excellent presentation. Thanks
@pd74843 жыл бұрын
amazing teacher omg
@xoppa093 жыл бұрын
More thinking on this matter... the 'backwards substitution' as you call it (nice name for it), which was x = g(u) in the one dimensional case, reminds me of the method of 'introducing a parameter' in precalculus. For example we might look at a complicated curve in the x,y plane made up of points (x,y), and tidly describe it by introducing a parameter t, e.g. (x,y) = (x(t), y(t)). For example the ellipse (a cos t , b sin t ). Similarly a change of variables can be thought of as a type of 'introduction of parameter(s)', where the same number of parameters are introduced as there are coordinates. Then in the u,v parametric space a rectangular differential area du dv gets mapped to a curvilinear differential area in the x,y space, in such a way that du dv gets stretched/flipped/punched by the factor |J|. Note that i said changing variables is *like* introducing parameters , i understand that it is not exactly the same. I would say its a special case of introducing parameters, since the number of parameters introduced does not usually match the number of coordinates. I usually think of a parameter, such as 't' , as a knob or slider off to the side of an x,y graph that will cause a point P in the x,y space to move around, like a ghost moving pieces around on a game board. For change of variables , the new variables u,v can be thought as two sliders inside its own "cartesian" or rectangular slider space. This means that if you grab a point in u,v space and trace the shape of a rectangle, in the x,y space you will simultaneously trace the shape of a circle (in the case of a polar coordinate change of variable) There are mathematical ways to formalize this more efficiently. Also your images make me think of the chain rule , the jacobian is just the chain rule for the derivative of a compositional function in 1 dimension. Thanks for posting this video!
@benastorga16152 жыл бұрын
Thank you so much... seriously!
@NexusEight3 жыл бұрын
Nice one!
@LucasVieira-ob6fx2 жыл бұрын
That was a very nice explanation!
@sam080903 жыл бұрын
Excellent
@iyadindia8624 жыл бұрын
Pls create a vedio on Jacobian of Implicit functions...
@donegal793 жыл бұрын
brilliant. Well done!!!!
@matejpavlovic34693 жыл бұрын
Hello, can you please tell me what is g(x) in those matlab pics? What is your initial integral? Ty!
@hemantbhosale30602 жыл бұрын
Nice one I understood a concept that wasn't understood
@BuddyNovinski2 жыл бұрын
I never had this in calculus -- and no geometry to visualize. I am so angry that I gave up on linear algebra in frustration when I could have used a better background and some hints to understand it in 1977.
@gamingfuneducation9934 Жыл бұрын
Thank you for a Great video, I have 1 quastion. Why the rectangle transforms into parralelogram in 2D case? Why not in some complex shape, while talking in some general case
@MrDlanglois2 жыл бұрын
Thank you
@tsunningwah34718 ай бұрын
it would be better if you could show what the horizontal and vertical axis represent (does the horizontal axis represents x or g(x) ?
@liviumircea69053 жыл бұрын
Algebraically I know why dx =g'(u)* du and hence to derive the substitution formula , but at your explanation I don't get why multiplying by g' the the squish canceled out
@scollyer.tuition3 жыл бұрын
[Note: my notation here is slightly different to that used in the video - I don't mention the function g] We take a rectangle occupying the interval [x, x+h] on the x-axis; it has width h (= x+h - x) We now map all points on the x-axis to a new u-axis by u=u(x). In this case, the interval [x, x+h] on the x-axis goes to [u(x), u(x+h)] on the u-axis; it has width u(x+h)-u(x) on our new u-axis. We want to relate the u-width u(x+h)-u(x) to the x-width x+h - h = h. In general, there will be some stretch factor (that depends on x) that maps the x-width to the u-width; call it S(x) and remember that a stretch of real numbers means "multiply" i.e. u-width = S(x) times x-width So we have u(x+h) - u(x) = S(x) h by definition of a stretch factor. Rearranging to solve for S(x) gives us: S(x) = (u(x+h)-u(x))/h Now consider that with a Riemann sum, we allow the width of the rectangles to go to 0 to ensure that we get an exact area from the sum i.e. we really want to consider S(x) for a very narrow rectangle near x i.e. we have to compute: S(x) = lim_{h -> 0} (u(x+h) -u(x))/h And what is that? Well, it's nothing other than the derivative of u(x) by definition of that limit. So we have shown that our stretch factor S(x) = u'(x) = du/dx for an infinitesimal rectangle on the x-axis being mapped to an infinitesimal rectangle on the u-axis via a mapping u=u(x). And, of course, to ensure that the Riemann sum on the u-axis spits out the same number as the Riemann sum on the x-axis, we *divide* by that stretch factor (to undo the stretch of the rectangles - remember that the y-displacement doesn't change), or equivalently, we *multiply* by dx/du. So the terms in the Riemann sum on the u-axis must look like f(u(x)) dx/du, and that is the form of the integrand that ensures both x-axis and u-axis integrals produce the same number. As a trivial example, suppose that we want: \int_0^1 1 dx = [x]_0^1 = 1 (i.e. we are integrating a constant, 1) and we let u=2x. Then this stretches the x-axis into the u-axis by a factor of 2, so on the u-axis our limits become 0 and 2. We can construct the x-axis Riemann sum with one rectangle, and so the unfixed-up u-axis integral is: \int_0^2 1 du = [u]_0^2 = 2 This is out by a factor of 2, which is our neglected stretch factor. Note that the stretch factor that we have to divide away is du/dx = 2 giving dx/du=0.5 and the corrected u-axis integral is: \int_0^2 1 dx/du du = \int_0^2 0.5 du = [0.5u]_0^2 = 1 - 0 = 1 as required.
@liviumircea69052 жыл бұрын
@@scollyer.tuition Thanks :)
@pinkykushwah417310 ай бұрын
Very nice
@Ishanka.5 ай бұрын
Fact that he writes in left tells the story.
@PatiparnPojanart Жыл бұрын
Thx a lot
@alexgodeye30316 ай бұрын
When we are working with the one-dimensional case, why do we not multiply by the absolute value of g' rather than g', if it's only squishing or stretching we care about?
@MuPrimeMath6 ай бұрын
If g' is negative, then this corresponds to "backtracking" on the graph of f(g(x)). We want to count backtracking as negative so that the backtracked region is not counted multiple times when it only occurs once in the original integral.
@alexgodeye30316 ай бұрын
@@MuPrimeMath Thank you for replying.
@AhMedmohamed-ly5td2 жыл бұрын
Is there proof of why we use the determinant specifically as the Jacobian coefficient?
@hrithiklanghi6418 Жыл бұрын
Nice one
@azmath20594 жыл бұрын
Why does the determinant correspond with how much the area is scaled?
@MuPrimeMath4 жыл бұрын
In some ways, that's actually the definition of a determinant! I have a video deriving the formula for 2x2 matrices: kzbin.info/www/bejne/pp3FZ5mQp9mSkNE
@tomaszsidorczuk77842 жыл бұрын
should not it be g(u) and g'(u) instead of g(x) and g'(x) on matlab right upper and lower drawings?
@MuPrimeMath2 жыл бұрын
The function input is a dummy variable because we graph on the same interval either way, so it doesn't matter what letter we use to denote it.
@tomaszsidorczuk77842 жыл бұрын
why do we define u and v implicitly rather then explicitly?
@MuPrimeMath2 жыл бұрын
There's no reason that we must define u,v implicitly, but doing so makes it more straightforward to compute the Jacobian, since the Jacobian is defined using the partial derivatives of x,y with respect to u,v.
@xoppa094 жыл бұрын
All this forwards and backwards transformations are confusing. Why not consider a mapping from u,v space to x,y space, where a small square in u,v space gets stretched by the scale factor J = |xu xv ... | to a parallelogram in x,y space.
@MuPrimeMath4 жыл бұрын
That is essentially what we're doing, since the area of any shape is scaled by the same amount by a linear transformation. We talk about backwards transformations to show why it makes sense to multiply by that scaling factor!
@audreydaleski1067 Жыл бұрын
Let's do tripple.
@jitendratiwari68864 жыл бұрын
Please also share your email id with us. I am learning optimization in my university will you make video on problems associated with it. this is the topic, in my first semester i studied laplace transform from your channel. 1) Optimization Examples, Concepts 2) Methods of nonlinear optimization - unconstrained and constrained case, optimality criteria, descent methods. Methods for differentiable and non-differentiable functions 3) Shape Optimization 4) Topology Optimization (ESO/BESO, SIMP, Level-Set-based) 5) From Structural Optimization Results towards Additive Manufacturing 6) Optimization under Uncertainties (Robust Design Optimization, Reliability-oriented Design Optimization) 7) Linear Optimization - Simplex Method - Operation Research (optional) 8) Optimization in Networks, e.g. Shortest Paths (optional)
@MuPrimeMath4 жыл бұрын
I haven't learned about optimization methods yet! In general, the series that I'm doing will depend on the classes that I'm taking at any given time.
@trigon70154 жыл бұрын
69,420th
@trigon70154 жыл бұрын
Yeah, you really deserve more views. Especially because of the quality of this video: it was very concise and helped me understand this topic.