Calculus Teacher ~ transform ~ physics teacher.

@qav_cnzo_

Ай бұрын

first time seeing him teaching phisics😅

@ridesafealways4929

Ай бұрын

​@@qav_cnzo_Because he is a mathematician. He focuses more on much much harder maths than those we use in engineering

@MrUtah1

Ай бұрын

ℒ{calculus teacher} = physics teacher

@Patrik6920

Ай бұрын

Well .. ∂F(Math)dX = Practical application aka Physics ..Usually all math was and are invented to solve real world problems...

@kingoreo7050

3 күн бұрын

A lot of the number theory that early mathematicians like euclid did never found usefulness until thousands of years later. Most high level maths done now is in that same boat of just discovering interesting things in whatever mathematical object they happen to be playing with. Maths is discovered almost always for the purpose of discovery itself and there is no obligation that it becomes useful, though it often does end up being.

bprp physics basics?

That takes me back to the 1980s when 0:06 I was playing with my Sinclair ZX81. I wrote a neat little program to find the centroid of an I beam. I then extended it to do the same for any shape as long as it was made up of rectangles. I think the initial data entry was first how many rectangles, and then for each rectangle, the location of the bottom left of each rectangle from any convenient origin, and it's width and height. The result was the coordinates from the previously defined origin. It was a nice little problem to code as I was learning the principles of simple coding .

Real centroid formulas: m = ∬ρ(x, y)dxdy Mx = ∬yρ(x, y)dxdy My = ∬xρ(x, y)dxdy Centroid: (My/m, Mx/m) Also, in 3D, m = ∭ρ(x, y, z)dxdydz, and the centroid is equal to (Myz/m, Mzx/m, Mxy/m). Using this formula, we can derive the centroid of a given function z = f(x, y) under the curve is equal to: x- = x∬(f(x0, y0) - g(x0, y0))dxdy y- = y∬(f(x0, y0) - g(x0, y0))dxdy z- = ∬(f(x0, y0))^2dxdy

@joeythreeclubs

Ай бұрын

What do you mean "real centroid formulas"?

@joeythreeclubs

Ай бұрын

Also what are m, Mx, and My?

@AlbertTheGamer-gk7sn

Ай бұрын

@@joeythreeclubs Real centroid formulas are the formulas they are derived from. The formulas used in the video are only used for EXPLICIT R -> R functions (y = f(x)), etc. However, if you have an implicit f(x, y) function in the 2D plane, you use the real centroid formulas.

@AlbertTheGamer-gk7sn

Ай бұрын

@@joeythreeclubs m = mass M = first moment of inertia (Mx = moment of inertia of x-axis, My = moment of inertia of y-axis)

Great explanation 👌

4:00 : Since you're doing a centroid rather than a centre of mass, a more direct 1-dimensional analogy is where you (arbitrarily) cut the 10-metre bar somewhere (not in the middle), find the centroid of each piece, and compare those to the (obvious) centroid of the entire bar. Then you'll see that you need to weight each piece by its length.

Perfect explanation.

I did the same thing some months ago but I used inverse function to find the y coordinate

The x coordinate of a rectangle to be integrated would be (x + 1/2 dx). The area of that same rectangle is (x + 1/2 dx) f(x) = x f(x) + x/2 f(x) dx. Integrating this we get Int (x f(x) + x/2 f(x) dx) dx = Int x f(x) dx + Int [ x/2 f(x) dx ] dx. The second integral vanishes as dx approaches 0.

”Just hold up something heavy like the two markers here”

@mhm6421

Ай бұрын

He's really strong

this is very cool

balancing the torque to find centroid in a line.

Thank u❤

Sir please make a video on how to find standard deviation

Cool! Would you made a video with the same calculations but for unevenly distributed mass/density?

Please do AP Calculus AB 2024 FRQs whenever you can, those are the ones I took. Great video 👍🏾

What if the density wasn’t uniformly distributed?

@jamescollier3

Ай бұрын

it gets more complicated lol.

@yplayergames7934

Ай бұрын

Then, integrate

@CptnWolFox

Ай бұрын

For a one-dimensional object like a rod, if you know how the density ρ varies as a function of position, you can use this: ( ∫ xρ(x) dx ) / (total mass).

@headshotgaming6808

Ай бұрын

Double integrals

@danielhinrichsen6927

Ай бұрын

You'd probably have to deal with line integrals

Were you able to slove that integral BPRP?

You could instead do x̅=∫xdA/∫dA and y̅=∫ydA and setup double integrals or integrals in terms of inverse functions as appropriate. And if an area is bounded by piecewise functions or other complexities, you can still break it apart into components and sum them.

haha @4:26: 'So what, exactly, does d1*m1 do, though? This, right here, is called the 'moment'... at the moment, we are doing moments in Calculus. heh." Love this dude lol

Hello there, can you help me with my integration question? The question is Integrate e^-x . secx

In the first example, does the centroid of the whole shape necessarily lie on the line joining the 2 centroids of the rectangles?

@ACheateryearsago

Ай бұрын

If the mass is distributed uniformly within the body

(2,4)

Shouldn't the X coordinate of bigger rectangle be 3 because 1/2 of 4 + 1/2 of 2 = 2 + 1 = 3?

@Ninja20704

Ай бұрын

No its 4 because u have to add 1/2 of 4 (which is 2. Starting from the left border of the big rectangle and not the centroid of the smaller rectangle.

Moment of force

An equation common for structural engineering

Isn’t that barycenter?

Sometimes i be feelin like the person with mass m2 lately…

7:21 hahaha