Tensor Calculus 2b: Two Geometric Gradient Examples (Torricelli's and Heron's Problems)

[@marciomunizdefarias7181]

The geometric intuition for the solution of these two problems using gradients is so beautiful that it justifies the name of the channel (MathTheBeautiful). I wish the KZbin algorithm would allow me to give 10 thumbs up for Professor Pavel.

[@MathTheBeautiful]

Much appreciated!

[@GagandeepSingh-rz7ue]

The solution to Torricelli problem using the concept of gradients was just brilliant. Blown my mind.

[@MathTheBeautiful]

Thanks! Always important to have as many perspectives as possible!

[@uditg]

seemed like some dark magic. solution was over before it even began. incredible!

[@mickodillon1480]

Well said

[@markkennedy9767]

I agree. It felt like a whole new way of doing maths

[@inigoloyola1869]

Wow that Torricelli solution was amazing -- it really makes calculus feel like a cheat code because I got this solution in a flash, whereas it took much longer for me from the first lecture's solution.

[@amirkhan355]

Wow! I'm speechless. Absolutely beautiful!

[@MathTheBeautiful]

That's how I felt when learning this stuff.

[@adarshkishore6666]

Beautiful! The simplicity and beauty of this argument is mind-blowing!

[@MathTheBeautiful]

Agreed!

[@user-pt-au-hg]

I just seen the lengths corrected on the drawing(11:50) so I think that partially answers the question. Still wondering if the magnitudes of the gradients can be different; I will have to read up on it.

[@happyhayot]

If the gradient is not orthogonal to the constraint, that means you can move along the constraint and find a smaller value, that is so intuitive, I love how you haven't even mentioned the concept of Lagrange multipliers and made total sense of it already. 🤯👏🏼

[@KeysToMaths]

9:14 Just to clarify, if the total gradient grad T at D ( where D is the point on the horizontal line that minimizes T = AP + PB for any point P on the line) has a horizontal component pointing to the right, then the directional derivative, (grad T) dot n

[@MathTheBeautiful]

Yes, that is a useful insight.

[@pabloagsutinnavavieyra2308]

Thanks a lot for this explanation. I was struggling to make sense on how the function decreased if we took the opposite side to the horizontal proyection of the sum of the gradients. I was just taking the path that that should be the case based on that I know the correct answer from the first lecture, but this logic was a little disconnected to what has been explored so far in this lectures. Thanks again!

[@DouglasTimes]

This is a topic I've wanted to study in depth for a few years now, thank you for your work on this. At 11:10 you state the angles between the 2 gradients and the river must be equal. I am having trouble seeing this as it seems that they would not necessarily be equal if the magnitudes of the 2 gradients were different.

[@dhritimanroyghatak2408]

he is saying they are unit vectors to begin with

[@styx4947]

I meant "that one proof" in the previous comment. Also I thought the trace of the Ricci tensor, R was the mean curvature, referring to the symbol of your channel. Is B meant to represent that?

[@bernandb7478]

Absolutely beautiful solution!

[@MathTheBeautiful]

I agree. Simplicity is irresistible.

[@Gamma_Digamma]

You could totally study optics way more conveniently with this application of gradients

[@timurbiktibaev4251]

I might be missing something, but it looks like in the 1st problem you already consider the correct answer (with certain angles chosen) and then just prove that this answer is indeed correct. Basically you only consider the change of length due to stretching, but don’t look into possible rotation of these lines. This solution doesn’t seem to prove that if these 3 lines are chosen at different angles, the solution would not exist

[@snnwstt]

Consider that each gradient is unitary. Now, if one is vertically down, and the two others are at 45 degree (one at the left and one at the right of the vertical), but upward, then their sum is zero in the horizontal direction, but NOT zero in the vertical direction. So that is not the optimum solution. The only way to get the vertical sum equal to zero (keeping the symmetry to keep the horizontal sum equal to 0) is to have an angle such that 2 cos(a) = 1, ( each upward gradient contributes for cos (a) upward, and the third gradient contribution is 1, downward), so a = 60 degree.

[@snnwstt]

And the only assumption is that one of the gradient has to be downward (the other two have to be upward with the same angle in magnitude on each side of the vertical to keep the symmetry). Well, you could always draw the triangle (afterward) so that it is the case. Also note that this does not strictly gives you the point itself, it does not gives you EXPLICITLY WHERE is the point. It just gives you an additional property. For an initial triangle ABC, you now look at the point M where the three triangles MAB, MAC and MCA, each have an angle of 120 degree at M, (plus the conditions to have MA the same length in MAB and MCA, etc. )

[@rkpetry]

But not if the triangle has one > 120°: you don't get negative distance from a Torricelli point outside. (Mathematicians always check the problem boundaries, as well as their solutions.) This will become important when we start checking Einstein-Minkowski Relativity solutions.

[@MathTheBeautiful]

Yes, you're right.

[@CJMilsey]

Just wondering, at 9:17, is there any reason we define the gradients in those directions? Why would we not use say the path of a person walking as the choice of gradients, which could then give us zero?

[@MathTheBeautiful]

Well, we don't have any discretion in choosing gradients. The gradient of "length of segment" is the "unit vector that points along the segment" so that's what we drew. If you are thinking about moving in a direction chosen by you, then we would talk about the directional derivative in the direction that you chose, and that would be another (interesting, but different) discussion.

[@andrewtannenbaum1]

In the second example in addition to the angles being the same, the gradient vectors need to be the same length. For if they deviate, it is impossible for their projections to cancel on the given path. This is the case as we observe that the cosine increases with an increase in the component vector. Is that correct?

[@user-pt-au-hg]

For Heron's problem (the shortest distance from A to B) it was stated the angle would have to be the same for the two gradients to cancel the x-components of the total gradient, but can the gradients be of different magnitudes where then you would have different angles so the x-components of the two individual gradients cancel each other out?

[@miloradowicz]

We are talking about gradients of two functions that have the meaning of the distance from the points A and B respectively along some path, so they grow with the same rate s(s) = s (as length of a line segment as a function of its length. Hence their gradients have the same magnitudes. The gradients could be of different magnitudes if, say, the surface on which we travel weren't flat, so for either or both of segments the distance we traveled wouldn't be equal to the length of our displacement.

[@bonbonpony]

Do we still need a gradient for that, though? Couldn't we just use the unit vectors along the distance lines directly?

[@synthelse2358]

This is probably the most amazing lesson I ever seen ... Just mind-blowing stuff. I wish you were my professor at university (10 years ago xD)... Thank you 😊

[@MathTheBeautiful]

I'm glad you feel that way. Much appreciated.

[@snnwstt]

Technically, knowing that at the point we are looking for is such that it creates three triangles (MAB, MBC, MCA) each ones having an angle of 120 degree does not EXPLICITLY define the point M: each of the new triangle is not trigonometrically defined ( we only have a side, say AB for the new triangle MAB, and the angle opposing it, 120 degree; missing either another angle or another side). So, at best, we are left with six unknowns ( the three sides MA, MB and MC, and an angle in each new triangle, such as angle MAB) with a set of six non linear (trigonometric) equations.

[@xueqiang-michaelpan9606]

this lecture is amazing!

[@starcalibre]

Outstanding examples. Thanks for posting these lectures.

[@abajabbajew]

Lovely insights. Thank you.

[@pedzsan]

At 9:05, you say the gradient of the length and draw an arrow. Then you repeat it and the second arrow is bigger. But you stated earlier that the gradient is always 1. So, the two yellow arrows should be the same length. I would have emphasized this by using your "unit" length chalk. If they are the same length and the two angles you later draw arcs for are equal, then the horizontal components will be equal and opposite and cancel out. But I had to go back and revisit part of the video to remember that the magnitude of the gradient in this case is always equal to 1.

[@MathTheBeautiful]

You're right, the drawing should have been better.

[@saadhassan9469]

Brilliant Lecture

[@noib358]

In Heron's problem if we fixed a point X in the plane, evaluated the gradient at that point, took a small step in the direction orthogonal to the gradient, and repeated the process, we would flow along the ellipse with focal points A and B passing through X.

[@ScoffMathews]

What is the name of the textbook you keep mentioning? I'm watching your lectures to try to catch up with my own Tensor Calculus course, but we don't have "a book".

[@suluclacrosnet61]

We use Dr. Grinfeld's Introduction to Tensor Analysis and the Calculus Of Moving Surfaces: goo.gl/IaCQC2

[@ScoffMathews]

Great, I'll check it out, thanks!

[@sdu28]

Can someone explain why should the combined gradient in Heron's problem be orthogonal to the constraint?

[@SplendidKunoichi]

because the problem is asking you to minimize the combined distance walked both towards the constraint before you get there and away from it after you leave at the constraint the towards and away paths go in opposite directions no matter how you look at it, but given the path should be minimized at least one of the derivatives will always be increasing, approaching the same magnitude they simultaneously increase until the combined gradient reaches a maximum; ie their sum is a unit vector which is tangent to the constraint at the solution point, the minimized sum of the two distance vectors also being unit length

[@galshaviner7724]

This is pretty awesome.

[@styx4947]

That truly is a beautiful proof.

[@MathTheBeautiful]

You mean vector-based proof?

[@styx4947]

Yes, the vector proof. But in general, the fact that you encapsulated your whole paradigm of tensor analysis as the place where geometry and calculus are married in thats one proof.

[@kingplunger1]

absolutely amazing

[@BleachWizz]

why no ones use these examples to explain lagrangian?

[@radsimu]

so when a function would grow in any direction for a given point, the gradient in that point is 0? How about the gradient of a point for which the function would decrease in any direction?

[@MathTheBeautiful]

Same! It's kind of unexpected that if you know the rate of increase in two directions, then you know it in all directions.

[@pi5549]

gosh I had to rewind 3 times to grok that first proof. I will still need to sleep on this to assimilate a new way of thinking. The second example it might be nice to complete the vector-rhombus to get the resultant.

[@AbhishekSachans]

Plain Beautiful!

[@glaucopassalacqua]

Grande!

[@adolfninh23]

hahaha 2:40 that was funny

[@rewtnode]

I don't buy the solution of the point in the triangle at minimal total distance to the vertices being the point at which the three gradients have 120 degrees angles between them: take the example of a symmetrical triangle in which the middle angle is very large, say 178 degrees.

[@RMLLcrazy]

I believe in the original problem there is a constraint on the angles of the triangle.

[@pabloagsutinnavavieyra2308]

Yes, as +Rui Lourenço mentioned, on the first lecture is mentioned that the triangle must be acute.

[@jolez_4869]

What the hell? This feels like cheating!