Gradient and contour maps 

Wow!!

Like knowledge can’t stop spilling into my brain😂

I have been searching for this for days! Your comment truly made it click. Thank you!

I'm glad I'm not the only one who noticed. Making these sorts of vids is difficult so I'm sure stuff like this is bound to slip through.

Thank you for this comment. Haha

Yeah I think that's right; the closer together contour lines far from the origin mean that the gradient is steep and the vectors should be large, while the more spaced out contour lines near the origin mean that the gradient is more gentle/shallow and the vectors should be small

Yea... I noticed too

​@@harry_pageneat explanation

didn't realize until I read this comment...

You probably know this by now but he actually started out in Khan Academy.

I wondered, but I thought they didn't sound the same, or the name didn't match?

I was searching for this comment

I thought the same thing. Grant... yeah right

Nir Gutman xxxx

Indeed. The relationship between electric field and electric potential is the same concept. A vector field that is the gradient of a scalar function, and that scalar function is a shortcut for evaluating line integrals of the field. Electric potential is like the "hill" of the electric field, while electric field is its gradient, which is more formally called a potential function. The relationship between electric field and electric potential involves a negative sign, so that potential can correspond to stored energy, but in pure math, the negative sign is omitted.

Only if you continue to stay on the contour line. If you move tangential, then locally, your f(x,y) value will remain the same, but eventually, you will move off the contour line, since the contour line is not necessarily a straight line.

Yo' Equations

@@maxwellsequation4887 lmao this thread is amazing

Its corresponding to f(x,y)=xy, go plot it into a 3D plotter online

I agree, vectors [y, x] are shorter near the origin

Now Consider that for a contour map value of say 25 in f, all the possible combianation of x and y are mapped that produce the value 25. Let's say that the function is f(x, y) such that f=x+y. So all the possible numbers that when added together produces 25, and yes that includes decimals, end up being quite a lot. You got 20+5, but you also got 15+10 and 15.5+9.5 and 10.9+14.1, so all the x and y combination are mapped inside that 2d map that produces the value 25. Now let's take a small step in the f and name it df. That df is very small, so small that the curve connecting all the possible combination of x and y looks like a line. Now if you are at line of f=25, the shortest way to jump towards f=25+df is the line perpendicular to the f=25+df line, as if we apply pythagorian theorem we will find out that the hypothesis is always the lingest line(h2+b2=p2). Any line else than the perpendicular will act as the hypostheis, with its own base determined by how far away it is from the shortest line. Now the other thing we found out was that the gradient is always perpendicular as gradient has its x component, that is how much it is in the x direction as the partial derivative of f with respect to x and vice versa. What really happen when we take the vector of the derivatives is illustrate how much the rate of change is changing in the x and y direction. If the gradient is steep but not long then the rate of change in the x direction is much more than in the y direction and vice versa. But you also have to remember that the rate of change is nothing but a ratio between the slope between x and dx to the chance in the input value itself. So a small change in the x direction and the small change in the y direction lands us at f+df, as df is comprised of dx in the x direction and dy in the y direction. So just by taking the pd of the function, we ended up at f +df as dx and dy was necessary for any taking of derivative. And dx and dy ended us at f+df. Now what we have done it to make a path at from f to f+df, but how is it the shortest path you may ask and justifiably so. The way that we know its the shortest path is that it always end up being perpendicular. How is this always perpendicular, idk I am not that smart but hopefully this has been helpful to you.

And yes from machine learning when we try to figure out how to make a model better by finding a gradient descent to quamten mechanics

@@satyamprakash7030 nice. I'm late, but thanks for taking the time

because contour lines by definition are basically lines that the value of the function stays the same on it, each value of the function correspond to a line. so if two lines intersect each other at a point then the function will have two values at that point, which is impossible since functions by definition only have one value at each point

@@hieudang1789 ...good answer, except what you answered was why two different contour lines can't intersect, not why contour mapping to different outputs have to be parallel. After all, two lines COULD be NOT parallel and still not intersect. Look, draw a vertical line, and a (idk, the outline of a cat or something. Yea, sure, the outline of a cat) right next to it, on its right. So we've got a vertical line passing through our input space, and the outline of a cat on the right of that vertical line passing through our input space. And say that these are both contour lines of our function (that is, the vertical line passes through a whole bunch of (x,y) points that map to some specific output when we apply the function, and the outline of the cat passes through a whole bunch of other (x,y) points that map to another specific output). Why isn't it possible for the outline of the cat (which is definitely NOT PARALLEL to the vertical line, otherwise that would mean you have a really freaky flat cat) to map to an output that's really, really close to the output that the vertical line mapped to?

This is not a general rule - however, when you zoom in, two adjacent contour lines SEEM like they are parallel (he says ‘roughly parallel’ at some point in the video)

watch the whole playlist to understand ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-ZTbTYEMvo10.html

vector calculus > makeup