Wow this is a very original perspective. I was very impressed with the second part.. showing that the set of circles is the only solution. I, mistakenly, thought that all the early differentiation would have destroyed some information.. and allowed many other solutions in. But that's wrong! Excellent video. Subscribed.
Wow that second part was awesome. It really highlights how derivation is not a reversible operation due to destroying constant terms. Feels almost magical that exactly the circle constraints are retained. Does this preservation occur as you increase dimensions? If so, is there an intuitive explanation? None of my understandings of circles seem to connect here. Like I watched the video, but it also feels like some algebraic voodoo just happened.
We can derive a system of PDEs from the general equation of a sphere, but I haven't tried to solve them in full, so not sure what would happen with the constants. There are some general results like the Picard-Lindelof theorem which say that certain DEs have unique solutions, which can explain why the only solutions of the DE are our original family of curves. I wonder if the constraints are preserved because they are as general as possible, given an equation of the form (x-a)^2 + (y-b)^2 = r^2. My intuition is that solving the DE preserves the form (x-a)^2 + (y-b)^2 = r^2, and the solutions are as general as possible without allowing x or y to be complex (so the solution includes as many possible values of a, b, r, x, and y). But if we started with the set of all circles with centre in the first quadrant (a, b > 0), I think our DE would be the same, and its solution would become "as general as possible" and now include other values of a and b. If we started with quarter-circles, with the added constraint 0 ≤ x - a ≤ r, then I think the DE would still be the same, but the solution might allow more values of x: -r ≤ x - a ≤ r.
@@DrBarker I'm shook by your intuitive explanation. It'd make a lot of sense if the constraints are actually kind of encoded in just the solution, and the DE always gives that solution regardless of more superficial constraints. It seems almost obvious in retrospect, but that took some brain cycles to come around to. Thanks for this incredible reply!
@pyropulse You ok? I did mistype since derivatives come from derivation. Luckily, math isn't applied English and my question was communicated. That said, I can sympathize with your stress since I also get upset when I don't understand things I've just memorized.
Outside of math, a "derivative" is something that is "derived" from something else. It's easy to tell this since the words themselves are _derived_ from the same origin. This is why I much prefer the term "differential" since it is much more clearly related to the act of "differentiating," much like how "integrals" are obtained from "integrating" (even if those words also have multiple meanings depending on context).
Why a great solution! A physicist would have derived the differential equation and then just trailed ansatz to solve it and said that’s all the solutions, but you actually derived it properly
But why, tho? its like nowing, that the volume of a sphere is V=4/3r^3*pi and then saying, well d^3V/dr^3=8*pi is the more beautiful form of it. I mean its just trivial and doesnt help. Its interesting, but well it does nothing. Please explain.
It's interesting to have multiple ways to express something. It can connect or produce tools in different areas like algebra, geometry, and differential equations. For example, 3deep5me the video could have been "This Equation Describes all Right Triangles in R^2" and the thumbnail the equation of a circle. Here, as others have mentioned, differentiation destroys constants. You'll notice in the DE, there's not a single constant despite starting with 3 that are necessary and constrained. If differentiation was bijective, you'd be right that it'd be obvious, but differentiation is not. That's why when Dr. Barker re-derived exactly the circle equations from only the DE, we were like wat? You can imagine if someone just gave you that random DE and said this is exactly every circle. Even if you believed circles were a solution, you'd probably be like are you sure it's the only solution or what about some random thing like r
@Dr Barker: The differential equation you derived Dr Barker actually has a very nice intuitive interpretation after some reordering: We can rewrite the differential equation you derived as follows: y''' / y'' = 3y'' y' / (1 + (y')^2 ) Noticing that both sides are cases of logarithmic derivatives. we can rewrite both sides as : ( ln | y'' | ) ' = ( (3/2) ln ( 1 + (y')^2 ) ) ' Bringing the coefficient 3/2 as an exponent inside the logarithm and bringing this logarithm to the other side we obtain: ( ln( |y''| / ( 1 + (y' )^2 )^3/2 ) ) ' = 0 You may notice that the term inside the logarithm is precisely the definition of the curvature of the function y(x). Let's denote by K(x) the curvature: K(x) = |y''| / ( 1 + (y' )^2 )^3/2 ) For those interested, see: en.wikipedia.org/wiki/Curvature Then the differential equation you derived is simply: ln( K(x) ) ' = 0 From here, it is obvious that this differential equation can be understood as the family of functions for which we have positive (because of the ln) constant curvatures. i.e circles!!!! Really cool!
Thanks for yet another great video! Do you have an example in mind where this kind of transformation from a curve to a differential equation could be useful, or is actually used?
I would like to see a rigorous, precise formulation of the assertion you are making about the sets of solutions to this differential equation (I guess I mean the one expressed just in terms of y and its derivatives). There were lots of steps that used kind of freewheeling divisions by things, etc, that typically do not preserve solutions to equations, so it is hard to see the derivation in the video as a proof that "the solutions are all and only the semicircles". In particular, I have some trouble visualizing what happens at the endpoints of a semicircle. You have all kinds of constraints with
I have to agree... Many comments here are suggesting that differentiability is somehow incomplete. There is a good reason why integration has constants, no matter the function. It accounts for the information lost in differentials... This video was about taking multiple differentials of an equation, then integrating it, and introducing the exact same constants that differentials eliminate, and solved equations account for! I understand "Dr Barker"s approach but it is tautological at best. You reproved the equation of a circle in cartesian coordinates. It was fun to go to the world of third derivative substitutions, but it was ultimately unnecessary. I guess if your goal was to show a differential equation with only y terms to show a semicircle, you did really well. But.. why? It's neat, but unintuitive.
So, I have two questions: (i) why leave the differential equation in such a strange form, without cross-multiplying-out the divisors; (ii) is there a better way of structuring the differential equation, perhaps more revealing of the steps used either to derive it or to show that it reduces to the circles?
If you start cross-multiplying, you may end up with additional solutions that were previously not allowed, since it is assumed they are non-zero (otherwise we wouldn't be allowed to divide by them in the first place.)
Funnily enough, I originally planned to make a video on this problem but with spheres in R^3. Using the same method - eliminating the constants - we get a system of PDEs, but I figured it would be quite a long video if we were to attempt to solve the PDEs! So I went for the simpler problem instead. We could also use the same method to get equations which describe lines in R^2 (from ax + by = c), or planes in R^3 (from ax + by + cz = d).
This was a great video. I really wish I took diff eq before I switched majors from physics to biology. I was able to follow along for the most part and see where this was going. However, the reasons for some of the steps were completely lost on me because I have no idea what the end goal is supposed to look like as I haven’t taken differential equations.
I just checked this, and I think we need to add some extra constraints like y'' ≠ 0 and y''' ≠ 0 to avoid a few extra solutions. But in general, Wolfram does give semi-circles as solutions. The constants look different, presumably based on how Wolfram solved the equation, but they are equivalent to having a, b, and r.
You can change the playback speed, on youtube videos, by clicking the cog symbol. But you cannot demand how Dr Barker should make his vidoes--it is considered rude within the borders of (her majesty) Queen Elizabeths II. 'God Save the Queen.'
Hey Dr Barker, I loved the video, I found it intresting and clear. The only doubt that I have is if we could work around working in the cartesian plane. Maybe if we used a vector calculus our result would not have things like the absolute value or +- sign, still I enjoyed the videos and I am glad I sumbled into your channel
Thank you! Yes, I think the limitation of the approach in the video is that solving the DE can only even give a solution which is a function y of the variable x, which can't describe a whole circle. Perhaps we could make it work for a parametric setup (x(t), y(t)) = (a + r cos(t), b + r sin(t)) to avoid the splitting into cases.
The logical sequence has always been like this. Scientists (mathematicians, physicists, engineers), based on the set goal and the laws of nature, wrote down an equation, often differential, for the quantities under consideration. It was solved, and an integral regularity was found linking the quantities under consideration. For example, Newton's 2nd law implies the equation of vibrations of a material point in the presence of a resistance force: d2x(t)/dt2+2*β*dx(t)/dt +(ω0^2)*x(t)=0. (1) Its integration gives x(t)= A*exp(-β*t)*cos(ω*t+φ), (2) where ω=sqrt[(ω0^2)-β^2]. Equation (2) is subject to analysis, depending on the values of the parameters. Or, derived the catenary equation. d2y(x)/dx2= a*sqrt[1+(dy(x)/dx)^2] . (3) And integrating, we obtained the equation of a hanging chain (cable) y(x)= a*cosh(x/a). (4) There are many examples.
You do things differently. Take the equation of a simple curve - circle, and differentiate it to get a differential equation. What's the point of that? In my opinion - no. Although the field of activity is large. It is possible to differentiate the equations of an ellipse, a parabola, a hyperbola, a cycloid, a brachystotron, etc., etc. There are many famous plane curves.)) What will it give? Returning to your "example". It is possible to do, in my opinion, with a differential equation of only the second degree. (x-a)^2 +(y-b)^2=r^2. => ...(x-a)+ (y-b)*y' =0 => 1+ (y' )^2+(y-b)*y''=0. z(x)=y(x)-b. 1+ (z')^2 +z*z''=0 => 1+(z*z')'=0. It's easy to integrate.)
I haven't done the calculations, but we should be able to apply the same method - eliminate the constants by differentiating - to get an equation, or perhaps a system of equations, to describe other families of curve. The only possible challenge would be that the differential equation we get from one family of curves may also apply for a broader family of curves, which is the motivation for solving the DE at the end.
11:10 Why can we just get rid of the 1? ... Wait, I see: we can ditch it because we are looking at 'greater than' logic and we can say, r² + r²u² > r²u², because we know r² > 0.
As a computer engineering student, it slightly bothers me that you call the vertical bars "modulus". I know them as "absolute value", "magnitude", "mean". Modulus in computer science is %, which is the remainder after performing division. Then again, I am in America, so language can be different :-)
In quantum mechanics, one of the central operations is the squared modulus of a complex number: |z|^2. If the imaginary part is 0, then it’s just the absolute value of the number. So it’s not a regional difference, but a field difference.
Nice video, a very interesting way of using differential equations. Still, I'm not too sure if your assumption of implicit function theorem would be 100% correct here: for points with y = b, you wont get a function y(x) at all (as the differential in y of (x-a)^2 + (y-b)^2 - r^2 would be 0 for all y=b).
Why is it that we get a chain rule derivative at 06:20 ? If v is du/dx, then I'd think that dv/dx = d(du/dx)/dx = d^2u/dx^2, or how would that work. Ty and great video👍
You should interpret v to be a function of u, i.e. v = v(u). Therefore, the chain rule is necessary, since we want to calculate the derivitave with respect to x: dv/dx = dv(u)/dx = dv/du * du/dx. (In the first substitution, u is interpreted as a function of x, i.e. u = u(x), and no chain rule is necessary.)
He said the modulus, which is another way to call the absolute value. You may have heard it described in complex numbers (z), where the modulus is the length of z and the argument is the (CCW) angle z has with the x axis.
@@JoQeZzZ no, sry, never heard of the modulus as the absolute value; only as the remainder of integer division, but maybe that’s, because I‘m no native speaker…?
This is quite cool but I’m a bit lost on the restrictions of |y-b| < r, also at 13:11 how does second derivative determine whether we choose positive or negative solution ?
In this video Dr Barker takes you on a magical journey. A magical journey where Dr Barker takes you on a ride, in his car, completing a single revolution around a magical roundabout. Along this journey, Dr Barkers’ wheels start to fall off; but you can bet your boots that at the end of the journey, Dr Barker has fixed on the same wheels on again! I hope next years A-Level Mathematics Pure Paper questions students to derive the DE of any abstract circle-this video would be the perfect place for a quick-simple solution to the problem. (No joke, this topic fits in the UK A-Level Mathematics syllabus.)
