let's solve this calculus problem step

by step we need to differentiate the

natural logarithm of a quotient with

respect to X here's our starting

expression we need to find the

derivative of the natural log of x^2 + x

+ 1 /x^2 - x + 1 the first key step is

to use the logarithm property that

states the log of a quotient equals the

log of the numerator minus the log of

the denominator this gives us the

derivative of log of x^2 + x + 1 - log

of x^2 - x +1 now we can use the chain

rule to differentiate each logarithm

term separately remember the derivative

of natural log of U is 1/ U * the

derivative of U let's evaluate the

derivatives inside the parentheses the

derivative of x^2 is 2X and the deriv of

X is 1 so we get 2x + 1 for both terms

to combine these fractions we need a

common denominator we multiply the

numerator and denominator of each

fraction by the denominator of the other

fraction now comes the algebra let's

expand the numerator carefully first

term gives us 2x Cub - 2x^2 + 2x + x^2 -

x + 1 the second term expands to -2X Cub

- 2x^2 - 2x + x^2 + x + 1 combining like

terms in the numerator most terms cancel

out leaving us with -2X 2 + 2x^ 2 + 2 -

2x^2 which simplifies to 2 -

2x^2 finally we can factor out a two

from the numerator giving us our final

answer 2 * * 1 - x^2 / x 4 + x^2 + 1 and

there we have it we've successfully

differentiated this logarithmic

expression using the properties of

logarithms the chain Rule and careful

algebra

Mastering Logarithmic Differentiation [𝒍𝒏((𝒙^𝟐+𝒙+𝟏)/(𝒙^𝟐−𝒙+𝟏))] Step-by-Step Explanation

Question:
Differentiate the expression  [𝒍𝒏((𝒙^𝟐+𝒙+𝟏)/(𝒙^𝟐−𝒙+𝟏))] with respect to x.

Unlock the secrets of logarithmic differentiation with our comprehensive, step-by-step guide! This video breaks down the process of differentiating complex logarithmic expressions with clarity and precision. Perfect for students and math enthusiasts, our tutorial will help you master these techniques in no time. Whether you’re preparing for exams or just looking to enhance your math skills, this video is your ultimate resource.

#MathTutorial #Calculus #LogarithmicDifferentiation #MathSkills #STEM


In atomic physics, the Electron Shell Model
describes the distribution of electrons on

the »energy shells« of an atom in the periodic
table of elements.

For example, it can be used to understand
the electron distribution of an oxygen atom.

We can see from the periodic table that an
oxygen atom has 8 electrons.

But how are they distributed in the atom?

The shell model answers this question.

Each shell represents a certain energy.

The electrons can occupy these shells and
thus take on this energy.

Each shell can only hold a certain number
of electrons.

The electrons on the outermost shell of an
atom are known as valence electrons and they

play a crucial role in chemical reactions.

The shells are labeled with letters: The innermost
(first) shell is called the K-shell.

It can hold a maximum of 2 electrons.

Then it is completely filled.

The second shell is called the L-shell.

It can hold a maximum of 8 electrons.

The third shell is called the M shell.

It can hold a maximum of 18 electrons.

The fourth shell is called the N-shell.

It can hold a maximum of 32 electrons.

How do I know the maximum number of electrons
a shell can hold?

There is a formula for this.

Let's call the maximum number of electrons
that a shell can hold as "n_max".

You can then calculate this number using the
following formula: "n_max" is equal to two

n squared.

Where "n" is the shell number.

For the first shell, the K shell, "n equals
one".

For the second shell, the L-shell, "n equals
two" For the third shell, the M-shell, "n

equals three",  and so on!

Let's take a look at four examples.

First, the hydrogen atom.

If you look at the periodic table, you can
read the number of electrons from the atomic

number in the corner.

The hydrogen atom has an atomic number of
one.

This means it only has one electron.

We start with the innermost shell, the K shell,
and add an electron to it.

As the hydrogen atom has no other electrons,
it only has a half-filled K-shell.

The lithium atom has the atomic number three.

This means it has three electrons.

We start with the K shell.

The K-shell can hold two electrons, so we
fill it with two electrons.

The third electron is added to the next shell,
the L shell.

The oxygen atom has eight electrons.

Two are on the K shell.

The L-shell can hold a maximum of eight electrons,
so the remaining six electrons are added to

it.

The aluminum atom has thirteen electrons.

Two electrons are added to the K-shell.

Eight electrons are added to the L-shell,
then the K-shell is full.

The remaining three electrons are added to
the M-shell.

The Electron Shell Model: Distribution of Electrons Inside Atoms Explained

https://www.youtube.com/watch?v=Fyqp17JnyOQ&list=PLTjLwQcqQzNKzSAxJxKpmOtAriFS5wWy4

Here you will learn how an electric dipole (e.g. a molecule) behaves in a homogeneous and inhomogeneous electric field. More: https://en.fufaev.org/electron-shell-model

Books by Alexander Fufaev:
1) Equations of Physics: Solve EVERY Physics Problem
https://en.fufaev.org/physics-equations-book
2) Alexander Fufaev and His Never Ending Story:
https://en.fufaev.org/lifestory


wenn wir eine Wechselspannung an einen

Kondensator der Kapazität C anlegen dann

fließt durch den Kondensator einen

Wechselstrom die Wechselspannung

wechselt die Polarität mit der Frequenz

f mit dieser Frequenz ändert auch der

Wechselstrom seine Flussrichtung ein

Kondensator an den eine Wechselspannung

angelegt ist hat einen komplexen nicht

komischen Widerstand der als kapazitive

Reaktanz bezeichnet wird dieser wird

meistens mit XC abgekürzt das C steht

für das englische Wort gepassette was

übersetzt Kondensator heißt manchmal

wird die kapazitive Reaktanz auch als

kapazitiver Blindwiderstand bezeichnet

du kannst den kapazitiven

Blindwiderstand ganz einfach berechnen

du brauchst dafür die

wechselspannungsfrequenz F und die

kondensatorkapazität C

der Blindwiderstand ist -1 durch 2 mal

Pi mal F mal c p ist hier die Kreiszahl

mit dem Wert 3,14 das Minuszeichen sagt

aus dass die Wechselspannung u dem

Wechselstrom i hinterher hängt

die Einheit des kapazitiven

blindwiderstands ist oben

2 pi mal F wird übrigens oft zur

Kreisfrequenz Omega zusammengefasst

wenn du eine sehr große

wechselspannungsfrequenz benutzt dann

wird der kapazitive Blindwiderstand sehr

klein und der Kondensator lässt den

Strom leicht durch

wenn dagegen die

wechselspannungsfrequenz sehr klein ist

oder gar Null wenn also eine

Gleichspannung anliegt dann wird der

kapazitive Blindwiderstand unendlich

groß der Kondensator lässt keinen Strom

durch wie du an der Gleichung siehst

kannst du auch die Kapazität nutzen um

den komplexen Widerstand des

Kondensators anzupassen

wenn wir mit effektiv werden der

Spannung und des Stroms arbeiten dann

interessiert uns nur der Betrag des

kapazitiven blindwiderstands das

Minuszeichen können wir also weglassen

machen wir ein Beispiel du legst einen

Netzspannung an einen Kondensator mit

der Kapazität von 10 Nano Fahrrad an

die 230 Volt Netzspannung hat eine

Frequenz von 50 Herz setze die Werte ein

damit beträgt der kapazitive

Blindwiderstand 318 Kilo Ohm

um den effektiv Strom zu bestimmen der

durch den Kondensator fließt benutze die

uri-formel statt den ohmschen Widerstand

R einzusetzen benutzt du den kapazitiven

Blindwiderstand Stelle nach dem Strom um

setze die 230 Volt Effektivspannung und

318.000 Ohm ein dann bekommst du einen

effektiv Strom von 0,7 Ampere

[Musik]

Kondensator-Blindwiderstand berechnen

In diesem Kurzvideo lernst du, was der Blindwiderstand eines Kondensators ist und wie du diesen berechnen kannst. Mehr: https://de.fufaev.org/kapazitiver-blindwiderstand


was ist der Unterschied zwischen einem

elektrischen und magnetischen

Feld ein elektrisches Feld wird sowohl

durch bewegte als auch durch

nichtbewegte elektrische Ladungen

hervorgerufen ein magnetisches Feld

dagegen wird nur durch bewegte Ladungen

erzeugt eine positive Ladung ist die

Quelle des elektrischen Feldes eine

negative Ladung ist die Senke des

elektrischen Feldes in der Natur können

negative und positive Ladungen allein

vorkommen es existieren so elektrische

Monopole magnetische Monopole dagegen

soweit wir wissen sind in der Natur

nicht vorhanden das heißt es gibt keinen

allein vorkommenden magnetischen Nordpol

oder magnetischen Südpol es treten stets

Nordpol als Quelle und Südpol als Senke

gemeinsam

auf magnetische Feldlinien sind in sich

geschlossen elektrische Feldlinien

dagegen müssen nicht geschlossen sein

sondern können beispielsweise ins

Unendliche gehen wie die Feldlinien

einer

Punktladung eine Ladung Q in einem

elektrischen Feld E erfährt eine

elektrische Kraft FE = Q mal e eine

bewegte Ladung in einem zur Bewegung

senkrechten Magnetfeld B erfährt eine

magnetische Kraft FM = q* V* B die

Einheit des elektrischen Feldes ist Volt

pro Meter und die Einheit des

magnetischen Feldes ist voltsekunde pro

Quadratmeter oder kurz

Tesla wenn eine elektrische Ladung in

einem externen elektrischen Feld

platziert wird so wird die Ladung

beschleunigt dadurch ändert die Ladung

ihre kinetische Energie wenn eine

elektrische Ladung dagegen in ein

magnetisches Feld platziert wird so wird

die Ladung nicht beschleunigt die Ladung

ändert ihre Energie nicht kurz gesagt

ein elektrisches Feld verrichtet Arbeit

an der Ladung aber ein Magnetfeld tut

das nicht sogar wenn die Ladung im

Magnetfeld bewegt wird ein elektrisches

Feld lässt sich abschirmen z.B mit Hilfe

eines fairdaykäfigs ein magnetisches

Feld lässt sich dagegen nicht abschirmen

aber zumindest

[Musik]

ablenken

7 Unterschiede zwischen dem elektrischen und magnetischen Feld

Hier lernst du 7 wichtige Unterschiede zwischen dem elektrischen Feld (E) und dem magnetischen Feld (B). Mehr: https://de.fufaev.org/fragen/68


wenn wir eine Wechselspannung an eine

Spule der Induktivität L anlegen dann

Fließ durch die Spule ein

Wechselstrom die Wechselspannung

wechselt die Polarität mit der Frequenz

f mit dieser Frequenz ändert auch der

Wechselstrom seine

Flussrichtung eine Spule an die eine

Wechselspannung angelegt ist hat einen

komplexen nichtomischen Widerstand der

als induktive Reaktanz bezeichnet wird

dieser wird meistens mit XL abgekürzt

die induktive Reaktanz wird auch als

induktiver Blindwiderstand bezeichnet du

kannst den induktiven Blindwiderstand

ganz einfach berechnen du brauchst dafür

die wechselspannungsfrequenz F und die

Induktivität L der Spule XL = 2* pi* f*

l Pi ist hier die Kreiszahl mit dem Wert

3,14 die Einheit des induktiven

blindwiderstands ist oben 2pf wird

übrigens oft zur Kreisfrequenz Omega

zusammengefasst wenn du eine sehr große

wechselspannungsfrequenz benutzt dann

wird der induktive Blindwiderstand auch

sehr groß und die Spule lässt den Strom

kaum durch wenn dagegen die

wechselspannungsfrequenz sehr klein ist

oder gar Null wenn also eine

Gleichspannung anliegt dann wird der

induktive Blindwiderstand auch null die

Spule lässt einen beliebig hohen Strom

durch was einem chluss

entspricht wie du an der Formel siehst

kannst du auch die Induktivität nutzen

um den Blindwiderstand der Spule

anzupassen machen wir ein Beispiel du

legst die 230 Volt Netzspannung an eine

Spule mit der Induktivität von 500

mhenry an die angelegte Effektivspannung

hat eine Frequenz von 50 Hz setze

Induktivität und Frequenz in die Formel

ein damit beträgt der induktive blindwi

stand 157 Ohm um den effektivstrom zu

bestimmen der durch die Spule fließt

benutze die uriformel statt den omischen

Widerstand R einzusetzen benutzt du den

induktiven Blindwiderstand Stelle nach

dem Strom

um setze die 230 Volt Effektivspannung

und 157 Ohm ein dann bekommst du einen

effektivstrom von 1,5 Ampere

[Musik]

Spule-Blindwiderstand berechnen

Lerne, was ein induktiver Blindwiderstand (Reaktanz) ist und wie du diesen bei einer Spule (Induktivität) in einem Wechselstromkreis bestimmen kannst. Mehr: https://de.fufaev.org/induktiver-blindwiderstand


Let us consider a system with an ideal gas
in it.

The state of the gas, or more precisely, its
temperature T, its pressure Pi and its volume

V are described by the ideal gas equation:
Pi is equal to n times R times T times 1 divided

by V.  n is the amount of substance and indirectly
describes the number of gas particles and

R is the gas constant.

We refer to temperature, pressure and volume
as properties - they describe the macroscopic

state of the gas.

If we heat the gas and thus increase its temperature
T, we are talking about a thermodynamic process.

We could also compress the gas, thereby increasing
the pressure and reducing the volume - this

would also be a thermodynamic process.

The thermodynamic process can be isobaric,
isothermal, isochoric or adiabatic.

In the case of an isobaric process, the pressure
of the gas remains constant.

This means that Pi, n and R are pure constants
in the gas law and the volume V is proportional

to the temperature T.

The pressure-volume diagram results in a horizontal
straight line for an isobaric gas.

This means that a change in volume does NOT
lead to a change in gas pressure.

The volume V of the gas remains constant during
an isochoric process.

Then V, n and R are constants in the gas law,
and the pressure Pi is proportional to the

temperature T.

The pressure-volume diagram shows a vertical
straight line for an isochoric gas.

The change in gas pressure does NOT lead to
a change in volume.

In an isothermal process, the temperature
T of the gas remains constant.

This means that the pressure Pi is proportional
to the inverse volume: 1 by V.

The pressure-volume diagram shows a decreasing
curve for an isothermally behaving gas.

Increasing the volume of the gas leads to
a reduction in the gas pressure.

During an adiabatic process, no thermal energy
is transported OUT and into the system.

During this thermodynamic process, the temperature
T, volume V and pressure Pi of the gas can

change simultaneously.

The pressure-volume diagram shows a power
law for an adiabatically behaving gas.

The exponent gamma is referred to as the adiabatic
exponent.

Isobaric, Isochoric, Isothermal, Adiabatic?

https://www.youtube.com/watch?v=IqgMga4-P5I&list=PLTjLwQcqQzNKzSAxJxKpmOtAriFS5wWy4

Here you will learn how an electric dipole (e.g. a molecule) behaves in a homogeneous and inhomogeneous electric field. More: https://en.fufaev.org/thermodynamic-process

Books by Alexander Fufaev:
1) Equations of Physics: Solve EVERY Physics Problem
https://en.fufaev.org/physics-equations-book
2) Alexander Fufaev and His Never Ending Story:
https://en.fufaev.org/lifestory


separationsansatz

diese wird manchmal auch Produktansatz

genannt warum wirst du gleich sehen

diese Methode eignet sich für partielle

Differenzialgleichungen beliebiger

Ordnung diese Lösungsmethode dient quasi

nur dazu eine partielle die GL in

mehrere gewöhnliche differen

zahlgleichungen zu verwandeln und diese

dann mit anderen Methoden zu lösen

schauen wir uns diese Methode am besten

direkt an einem Beispiel an hier siehst

du die eindimensionale Wellengleichung

für das elektrische Feld die gesuchte

Funktion ist das elektrische Feld E

dieses hängt von x und von T ab also

genau von zwei Variablen da die gesuchte

Funktion von zwei Variablen abhängt und

ihre Ableitungen in der Nähe vorkommen

handelte sich um eine Parzelle

Differenzzahlung

wie gehen wir nun vor hier machen wir

einen Produktansatz für die gesuchte

Lösung e

= r von X mal u von T wir nehmen quasi

an dass die gesuchte Lösung e sich in

ein Produkt von zwei Funktionen R und U

aufspalten lässt die eine Funktion R

hängt dabei nur von X ab und die andere

Funktion u hängt nur von T ab in der

Wellengleichung kommen die zweite

Ableitung von E nach der Zeit und nach

dem Ort vor das heißt wir müssen unseren

Produktansatz zuerst ableiten die

Ableitung des produktansatzes nach X

ergibt u mal die zweite Ableitung von R

nach X warum weil u unabhängig von X ist

und somit in der Ableitung wie eine

Konstante ist

bei der Ableitung des Produkts nach der

Zeit Tee dagegen ist die Funktion R eine

konstante weil sie nicht von T abhängt

das Ziel ist es nun alles was von X

abhängt von dem zu trennen was von Tee

abhängt dazu teilen wir diese Geld durch

das Produkt einmal u

dadurch haben wir erreicht dass alles

was von X abhängt auf der linken Seite

steht und das was von T abhängt alles

auf der rechten Seite steht wenn du es

schaffst eine Parzelle die Franzi

Gleichung so zu separieren dann war der

separationsansatz erfolgreich

jetzt können wir X auf der linken Seite

variieren ohne dass sich die rechte

Seite verändert da auf der rechten Seite

ja kein X vorkommt das gleiche gilt für

die Zeit t wenn wir die Zeit auf der

rechten Seite verändern bleibt die linke

Seite unverändert da dort keine Zeit t

vorkommt

setzen wir also die linke und die rechte

Seite einer konstanten K =

damit haben wir eine partielle

Differenzialgleichung in zwei

gewöhnlichen Differenzialgleichungen

umgewandelt die eine differen

zahlgleichung hängt nur von nix ab und

die andere differen zahlgleichung hängt

nur von T ab die beiden differen

zahlgleichungen sind nicht mal gekoppelt

du kannst sie also unabhängig

voneinander lösen und die Lösungen dann

nach dem Produktansatz miteinander

multiplizieren um die Lösung der

partiellen Differenz zu erhalten diese

beiden gewöhnlichen

Differenzialgleichung kannst du mit dem

zuvor kennengelernten exponentialansatz

lösen

[Musik]

Produktansatz (aka. Separationsansatz)

Lerne, wie man partielle Differentialgleichungen (DGL) mit Separation der Variablen (Produktansatz) löst. Hier machen wir ein Beispiel mit der eindimensionalen Wellengleichung. Mehr: https://de.fufaev.org/partielle-dgl


consider a circuit consisting of a coil

characterized by the inductance L and

resistor R connected in serious then we

take a voltage source which provides a

voltage u0 as soon as we close the

circuit with the switch then a Time

dependent current I flows through the

coil and the resistor the current does

not have its maximum value immediately

but increases slowly due to Land's law

using kof's law we can set up the

following differential equation for the

current I remember that the point above

the I means the first time

derivative this is an inhomogeneous

linear differential equation of first

order the searched function y

corresponds here to the current I the

perturbation function s corresponds to

u/ L and is time independent in this

case we denote the homogeneous Solution

by

IH first we need to determine the

homogeneous solution IH we can quickly

calculate this using the solution

formula for the homogeneous version of

the differential equation that you

learned before the coefficient K in

front of the searched function I

corresponds to R over L and is also time

independent in this case so far so good

we may omit the constant CH in the

solution form formula here because we

consider it later anyway in the constant

a which we find in the other resolution

formula the coefficient R / L is

constant and integrating a constant only

introduces a variable T thus the

homogeneous solution is I is equal to e

to power of - R / L *

T let's insert it into the inhomogeneous

solution

formula note that one over the

exponential function containing a minus

in the exponent is simply equivalent to

the exponential function without the

minus sign now we have to calculate the

integral u/ L is a constant and can be

placed in front of the integral and when

integrating the exponential function the

exponential function is preserved only L

over R is added as a factor in front of

the exponential

function finally we hide the constant of

integration in the constant a and this

is our general solution we can simplify

it a bit more by multiplying out the

parentheses two exponential functions

simplify to

one to get a solution specific to the

problem we need to determine the unknown

constant a for that we need an initial

condition if we say that the time t

equal Z is the time when the current I

was Z because we have not yet closed the

switch then our initial condition is y

of 0 is equal to 0 insert it into the

general

solution e to the power of 0 is equal to

1 solve for a and you get a is equal to

- u0 / R thus we have successfully

determined the specific General

solution

RL Circuit: Solving Differential Equation with Variation of Parameters

Here you will learn how to solve the differential equation of the RL circuit with variation of parameters. More: https://en.fufaev.org/inhomogeneous-differential-equations-of-first-order

Books by Alexander Fufaev:
1) Equations of Physics: Solve EVERY Physics Problem
https://en.fufaev.org/physics-equations-book
2) Alexander Fufaev and His Never Ending Story:
https://en.fufaev.org/lifestory


What is the difference between a series and
a parallel connection of resistors?

Here we consider a series and a parallel circuit
with three resistors with resistances R1,

R2 and R3.

A voltage U is applied to the resistors.

Difference #1:
Look at how the three resistors are connected

to each other.

In a series circuit, the positive pole of
the voltage source is connected to one single

resistor and the negative pole is also connected
to one single resistor.

The resistors in a series circuit form a chain.

In a parallel circuit, on the other hand,
you connect the positive pole of the voltage

source to three ends of the three resistors,
and you do the same with the negative pole.

Difference #2:

In a series connection, the same current flows
through all three resistors.

The total current I is therefore equal to
the current I1 through the first resistor,

but also equal to the current I2 and I3 through
the other two resistors.

In a parallel circuit, however, different
currents I1, I2 and I3 flow through the resistors.

The total current I is the sum of the individual
currents: I = I1 + I2 + I3

In a series circuit, you will measure a different
voltage across each resistor.

Voltage U1 across the first resistor.

U2 across the second resistor and U3 across
the third.

The total voltage U is the sum of the individual
voltages: U = U1 + U2 + U3.

In a parallel circuit, on the other hand,
you will measure the same voltage across all

three resistors.

The total voltage is therefore equal to the
voltage U1 between the ends of the first resistor,

but also equal to the voltage U2 and U3.

Difference #4:

The total resistance R of a series circuit
is the sum of the individual resistances:

R is equal to R1 + R2 + R3.

In a parallel circuit, however, the total
resistance R is not the sum of the individual

resistances.

The following relationship applies here: The
reciprocal total resistance "1 over R" is

the sum of the reciprocal resistances "1 over
R1" + "1 over R2 + "1 over R3".

Difference #5:

For the conductance values of the resistors,
it is the other way round.

The total conductance G of a parallel circuit
is the sum of the individual conductances

G1, G2 and G3.

And in a series circuit, the reciprocal conductance
"1 over G" is equal to "1 over G1" + "1 over

G2" + "1 over G3".

The 5 Differences Between Parallel and Series Circuit

https://www.youtube.com/watch?v=9kTVNmejpco&list=PLTjLwQcqQzNKzSAxJxKpmOtAriFS5wWy4
00:00 Introduction
00:16 Difference: Connection
00:47 Difference: Current
01:21 Difference: Voltage
02:00 Difference: Resistance
02:33 Difference: Conductance
More: https://en.fufaev.org/lessons/1238


In theoretical physics, especially in the
theory of relativity and quantum mechanics,

we will regularly encounter symmetric and
antisymmetric tensors.

A symmetric tensor t_{ij} remains the same
if we swap its indices.

Specifically, swapping the indices of the
second-level tensor as a matrix means that

the matrix remains the same if we transpose
it, in other words if we mirror the rows and

columns on the diagonal.

This symmetry property of tensors is very
useful and simplifies calculations in computer

physics enormously.

Moreover, this property is crucial in quantum
mechanics, because symmetric matrices have

real eigenvalues.

They therefore represent physical quantities
(we call them observables) that we can measure

in an experiment.

So if you have a symmetric tensor in front
of you, as a theoretical physicist you should

immediately get a dopamine kick.

The Kronecker delta, for example, is an example
of a simple symmetric tensor.

We have considered a second-order tensor.

What if the tensor is of a HIGHER order?

What about its symmetry property THEN?

For example, if the tensor has four indices
and it remains EQUAL when we swap the first

two indices, then we are talking about a symmetric
tensor in the indices m and k.

However, we will also encounter tensors that
are antisymmetric.

An antisymmetric tensor changes sign when
we swap its indices.

If the antisymmetric tensor is represented
as a matrix, then it is equal to its negative

transpose.

Unfortunately, most tensors are neither symmetric
nor antisymmetric.

But the great thing is: Mathematically, we
can split every tensor T into a symmetric

and an antisymmetric part: T is equal to S
plus A.

Let's take a look at how we practically split
a general second-level tensor T into its two

parts.

The symmetric part S of the tensor T is "One
and a half times t_{ij} plus t_{ji}.

In the second term, we have swapped the two
indices.

Then we added the two tensors together.

The factor of one and a half is important
because we have counted the tensor twice here.

The antisymmetric part A of the tensor T is
"One and a half times t_{ij}, minus t_{ji}.

Here we have swapped the two indices, which
gives the swapped tensor a minus sign.

The factor of one and a half must not be missing
here either.

We then add the symmetric and antisymmetric
parts together to obtain the total tensor.

We are done.

The first summand is the symmetric part of
the tensor t_{ij} and the second summand is

the antisymmetric part.

Introduction to Tensors: Symmetric and Antisymmetric Tensor

https://www.youtube.com/watch?v=8K7_qPmYBiY&list=PLTjLwQcqQzNKzSAxJxKpmOtAriFS5wWy4

Here you will learn the most important property of tensors, namely their symmetry. More: https://en.fufaev.org/tensors

Books by Alexander Fufaev:
1) Equations of Physics: Solve EVERY Physics Problem
https://en.fufaev.org/physics-equations-book
2) Alexander Fufaev and His Never Ending Story:
https://en.fufaev.org/lifestory


let's assume you have bought some

jewelry and want to know whether it is

made of gold and is not fake how can

Physics help you with this first you

have to weigh the jewelry and determine

its massim then you have to find out the

volume v of the jewelry to do this you

use Archimedes principle this states

that the amount of water displaced is

exactly equal to the volume of the

immersed body so take a measuring cup of

water throw in the jewelry and look at

the milliliter scale to see how high the

water has

risen convert the value into cubic

met now the density row of the jewelry

material can be

calculated simply divide the mass by the

volume the gold has a density of

19,300 kg per cubic meter if you also

get this value then the jewelry is made

of real gold

How Can You Tell If Jewelry Is Made of Real Gold?

https://www.youtube.com/watch?v=7SjAUgeKZWg&list=PLTjLwQcqQzNKzSAxJxKpmOtAriFS5wWy4

Here you will learn how an electric dipole (e.g. a molecule) behaves in a homogeneous and inhomogeneous electric field. More: https://en.fufaev.org/mass-density

Books by Alexander Fufaev:
1) Equations of Physics: Solve EVERY Physics Problem
https://en.fufaev.org/physics-equations-book
2) Alexander Fufaev and His Never Ending Story:
https://en.fufaev.org/lifestory


to calculate the temperature of the Sun

from Earth we must first find out what

light the sun emits the sun emits a

polychromatic light that is infrared

light visible light and even X-rays and

gamma rays you need to figure out the

wavelengths Lambda of each of these

radiations and with what intensity they

are

emitted if we plot all wavelengths with

their intensities in a diagram we get a

spectral intensity distribution of the

Sun from this distribution we have to

read off the wavelength Lambda Max which

has the highest intensity so we have to

read off the maximum of the

function then you can calculate the

temperature of the sun with the wans

displacement law to do this we divide

28978 * 10 power of -6 M Kelvin by

Lambda

Max in the case of the Sun the maximum

is within the blue light spectrum at the

wavelength 490 nanom or 490 * 10 to the

power of -9

M this results in a temperature of

5,900 Kelvin or

5,630 de C

How to Determine the Temperature of the Sun from Earth?

https://www.youtube.com/watch?v=eVs-p3krM4c&list=PLTjLwQcqQzNKzSAxJxKpmOtAriFS5wWy4
Here you will learn how to determine the temperature of a glowing body (e.g. the sun) from a distance. More: https://en.fufaev.org/wien-displacement-law

Books by Alexander Fufaev:
1) Equations of Physics: Solve EVERY Physics Problem
https://en.fufaev.org/physics-equations-book
2) Alexander Fufaev and His Never Ending Story:
https://en.fufaev.org/lifestory


let's take a look at the one-dimensional

wave equation for an electric field the

function we're looking for is the

electric field e this depends on the

position X and on the time T since the

function depends on two variables and

their derivatives appear in the equation

it is a partial differential equation

here we make a product onzas for the

searched solution e e is equal to R * U

we assume that this solution e e can be

split into a product of two functions

that we call r and u one function R

depends only on X and the other function

U depends only on T in the wave equation

we have a second derivative of e with

respect to location X and a second

derivative with respect to time T

differentiating the product on that with

respect to X yields U * the second

derivative of r with respect to X Since

U is independent of X and thus acts like

a

constant in contrast in the derivative

of the product with respect to time T

the function R acts like a constant

because it does not depend on T the goal

now is to separate everything that

depends on X from what depends on T for

this we divide this equation by R *

U

thus we have achieved that everything

that depends on X is on the left side

and everything that depends on T is on

the right side if you manage to separate

a partial differential equation this way

then the separation unz was successful

now we can vary X on the left side

without changing the right side because

there is no X on the right side the same

is true for the time T if we vary the

time on the the right side the left side

remains unchanged because there is no

time T thus both sides must be constant

so let us set the left side and the

right side equal to a constant

K thus we have transformed a partial

differential equation into two ordinary

differential equations and the good

thing is that the two differential

equations are not coupled that means you

can solve them independently and then

multiply the solutions like in the

product onz to get the general solution

for the partial differential

equation you can solve the two ordinary

differential equations with the

exponential UNS you learned

before

1D Wave Equation: Solving Differential Equation with Variable Separation

Here you will learn how to solve the one-dimensional wave equation (partial differential equation) with separation of variables. More: https://en.fufaev.org/partial-differential-equations

Books by Alexander Fufaev:
1) Equations of Physics: Solve EVERY Physics Problem
https://en.fufaev.org/physics-equations-book
2) Alexander Fufaev and His Never Ending Story:
https://en.fufaev.org/lifestory


betrachten wir einen geladenen Leiter

wie zum Beispiel einen elektrisch

geladenen Metallkörper der geladene

Leiter erzeugt ein elektrisches Feld das

immer senkrecht zu seiner Oberfläche ist

warum

die elektrischen Ladungen nehme ich die

Elektronen im Inneren des Leiters können

sich frei bewegen Ladungen in der Nähe

der Oberfläche des Leiters werden als

Oberflächenladungen bezeichnet die

Oberflächenladungen können sich entlang

der leiteroberfläche also parallel dazu

frei bewegen im Gegensatz dazu können

sich die Oberflächenladungen senkrecht

zu Leiter Oberfläche nicht bewegen wenn

Sie das könnten dann würden sie aus der

Oberfläche Austreten der Leiter wäre

nicht mehr geladen und würde damit gar

keine elektrisches Feld erzeugen

die Elektronen im Inneren des Leiters

erzeugen eine elektrisches Feld E das

elektrische Feld ist eine vektorielle

Größe es hat also einen Betrag und eine

Richtung nehmen wir mal an dass das von

den Elektronen erzeugte elektrische Feld

E an der Oberfläche nicht senkrecht aus

der Oberfläche heraustreten würde es

zeigt so wie hier gezeigt in irgendeine

beliebige Richtung die nicht senkrecht

ist die Mathematik sagt uns dass wir

jeden Vektor also auch den elektrischen

Feldvektor e in zwei Anteile aufteilen

können

in einen zur leiteroberfläche

Senkrechten Anteil und in einen zuleiter

Oberfläche parallelen Anteil das zu

Oberfläche parallele elektrische Feld

führt zu einer elektrischen Kraft auf

die benachbarten Elektronen so dass sich

diese Elektronen parallel zu Oberfläche

verschieben die Elektronen verschieben

sich so lange entlang der Oberfläche bis

ein Kräfte Gleichgewicht ist in diesem

Gleichgewicht ist die Kraft durch die

gegenseitige Abstoßung der Elektronen

aufgehoben und damit auch das

elektrische Feld das parallel zur

Oberfläche zeigt wenn der parallele

fällt Anteil noch da wäre würden sich

die Elektronen weiter entlang der

Oberfläche verschieben

das zu Oberfläche senkrechte elektrische

Feld dagegen ist immer noch da damit es

gemeint wenn eine Ladung außerhalb des

Leiters platziert wird dann wird sie

eine elektrische Kraft genau senkrecht

zur Oberfläche erfahren diese Anteil des

elektrischen Feldes senkrecht zu

Oberfläche hebt sich nicht weg weil die

Oberflächenladungen dafür aus der

Oberfläche austreten müssten es bleibt

also im Gegensatz zum parallelen Anteil

weiterhin bestehen

damit steht das elektrische Feld eines

geladenen Leiters immer senkrecht auf

seiner Oberfläche

wenn wir die elektrischen Feldlinien

einzeichnen würden dann würden sie auch

senkrecht aus der Oberfläche

heraustreten

übrigens kannst du dir vorstellen warum

das elektrische Feld von geladenen

Isolatoren im Allgemeinen nicht

senkrecht auf der Oberfläche steht

Warum stehen Feldlinien senkrecht auf leitenden Oberflächen?

In diesem Physikvideo lernst du, warum das elektrische Feld bzw. elektrische Feldlinien IMMER senkrecht auf Leitern stehen! Mehr: https://de.fufaev.org/fragen/45

#universaldenker


was ist der Unterschied zwischen einer

partiellen Ableitung und einer totalen

Ableitung einer Funktion f hierbei hängt

die Funktion f von x und y ab und bei

beiden Ableitungen leiten wir hier nach

x ab bei der partiellen Ableitung die

mit einem geschwungenen Dell notiert

wird ist y unabhängig von X bei der

totalen Ableitung die mit einem D

notiert wird kann y von X abhängen die

Änderung der Variable x wirkt sich also

auch auf den Wert von Y aus machen wir

ein Beispiel betrachte die Funktion 3x

zum qu + 2Y berechnen wir zuerst die

partielle Ableitung dazu müssen wir 3x²

partiell ableiten + 2Y partiell ableiten

3x² nach X abgeleitet ist 6x und 2Y nach

X abgeleitet ist 0 weil wir ja bei der

partiellen Ableitung annehmen dass Y

unabhängig ist von x bilden wir jetzt

die totale Ableitung der Funktion das

geschwungene Dell ändert sich zu einem D

die totale Ableitung von 3x² ist 6x und

die totale Ableitung von 2Y ist nicht

mehr 0 weil y von X abhängen kann wir

können den Faktor 2 vor die Ableitung

schreiben fassen wir zusammen bei der

partiellen Ableitung sind die Parameter

x und y der Funktion f unabhängig

voneinander bei der totalen Ableitung

dagegen kann x von Y abhängen oder wie

im Beispiel y von

x

Unterschied zwischen partieller und totaler Ableitung

Hier lernst du anhand eines Beispiels, was der Unterschied zwischen der partiellen Ableitung und der totalen Ableitung ist. Mehr: https://de.fufaev.org/partielle-und-totale-ableitung


betrachten wir ein System mit einem

idealen Gas drin der Zustand des Gases

genauer gesagt seine Temperatur T sein

Druck PI und sein Volumen V werden durch

die ideale Gasgleichung beschrieben pi=

n* R* T* 1 dur V hierbei ist n die

Stoffmenge und beschreibt in direkt die

Anzahl der Gasteilchen und R ist die

Gaskonstante dabei bezeichnen wir die

Temperatur den Druck und das Volumen als

Zustandsgrößen sie beschreiben den

makroskopischen Zustand des Gases wenn

wir das Gas erhitzen und somit seine

Temperatur T erhöhen sprechen wir von

einer

Zustandsänderung ebenso könnten wir das

Gas zusammenpressen damit den Druck

erhöhen und das Volumen verringern das

wäre ebenfalls eine

Zustandsänderung die Zustandsänderung

kann dabei isobar isotherm is oder

adiabat sein bei einer Isobaren

Zustandsänderung bleibt der Druck des

Gases konstant damit sind PI n und R im

Gasgesetz reine konstanten und das

Volumen V ist proportional zur

Temperatur

T im

druckvolumendiagramm ergibt sich für ein

isobar verhaltenendes Gas eine

horizontale

gerade das heißt die Änderung des

Volumens führt nicht zu einer

Veränderung des Gasdrucks bei einer

isokoren Zustandsänderung bleibt das

Volumen V des Gases konstant dann sind V

n und R konstanten im Gasgesetz und der

Druck Pi ist proportional zur Temperatur

T im

isochorverhaltendes Gas eine vertikale

gerade die Änderung des Gasdrucks führt

nicht zu einer Veränderung des Volumens

bei einer isothermen Zustandsänderung

bleibt die Temperatur T des Gases

konstant damit ist der Druck PI

proportional zum inversen Volumen 1

durch v im

druckvolumeniagramm ergibt sich für ein

isotherm verhaltendes Gas eine

abfallende Kurve die Erhöhung des

Volumens des Gases führt zu einer

Verringerung des

Gasdrucks bei einer adiabatischen

Zustandsänderung wird keine thermische

Energie V und in das System

transportiert bei dieser

Zustandsänderung kann sich die

Temperatur T Volumen V und Druck PI des

Gases gleichzeitig ändern im

adiabatisch verhaltendes Gas ein

Potenzgesetz der Exponent Gamma wird

dabei als adiabatenexonent

bezeichnet

Isobar, isochor, isoterm, adiabat?

https://www.youtube.com/watch?v=fOX4FaRahIk&list=PLcjHhwL4seQjmFSLkz_R11cgq7-LPyuGA

Hier lernst du was eine thermodynamische Zustandsänderung ist und wann sie isobar, isochor, isotherm und adiabat ist. Mehr: https://de.fufaev.org/thermodynamische-zustandsaenderung

Bücher von Alexander Fufaev:
1) Die nützlichste Physik-Formelsammlung des Internets:
https://de.fufaev.org/formelsammlung
2) Alexander Fufaev und seine Lebensgeschichte:
https://de.fufaev.org/lebensgeschichte


If you know the Lagrangian of a system, the
Euler-Lagrange equations will provide you

with the equations of motion and describe
how the system will evolve over time.

Before we start, we need to talk about two
things. First, we define the so-called action

of a system as the integral over the Lagrangian.
For classical mechanics, this means the time

integral over the Lagrangian function, however
for relativistic mechanics, the Lorentz-covariant

formulation involves the spacetime integral
over the Lagrangian density. This action is

a functional, this means it takes a function
(in this case the Lagrangian) as an input

and after some calculation, the result is
a number. So every configuration of the system

yields as the corresponding action just a
different number. This brings us to the second

point: Hamilton’s principle. Hamilton’s
principle states, that the system we are considering

evolves in such a way, that the action takes
on an extremal value. In other words, if we

slightly change the path that the system takes,
the change in action will be almost zero.

A small mathematical detail here is, that
we do not change the starting and end points

of the path, which means that delta q for
some starting time t_1 and end time t_2 is

zero. Similar arguments lead to corresponding
constraints in the relativistic case, where

we do not change the fields at the beginning
and end point, just somewhere along the way.

Now we can start deriving the Euler-Lagrange
equations. We start by explicitly calculating

the change in the action, knowing that whatever
comes out in the end must be zero according

to Hamilton’s principle. For the classical
version, this means varying the Lagrangian

with respect to q and multiplying a small
change in q as well as varying in q dot and

multiplying a small change in q dot. For the
relativistic version, we do this with respect

to the fields phi and their derivatives d_mu
phi.

A change in the time derivative of q is the
same as the time derivative of the change

in q, so let’s write it like this. Similarly,
we can exchange the delta and the derivative

on the right.
Next we use the inverse product rule, which

let’s us write a function and the derivative
of another function as the derivative of their

product minus the derivative of the first
function times the second function. We now

write the second term as the time derivative
of dL over dq dot and delta q minus the time

derivative of dL over dq dot times delta q.
On the right, this is the same, but with a

d_mu.
From here on we can evaluate parts of the

integral, in particular the total derivative
which is now evaluated between the end time

and the starting time. As for the other terms,
we can factor out the delta q. On the right,

we evaluate between x^mu_2 and x^mu_1 and
also pull out the delta phi.

Due to our assumptions, the first term vanishes,
since it is exactly the change of q or phi

evaluated at the boundary points. That’s
why this term is usually called a „boundary

term“. And remember, since we started from
a small change in the action, the terms in

brackets have to vanish due to Hamilton’s
principle. And this is what leads to the Euler-Lagrange

equations.
Remember, that q here stands for all the variables

you have in your system. So if there is, for
example, an x and a y, you actually have two

Euler-Lagrange equations, one for x and x
dot and the other for y and y dot. As you

can see, instead of two first-order differential
equations in the Hamiltonian formulation,

we now have one second order differential
equation in the Lagrangian formulation for

each variable. And that’s pretty much it
for now, thanks for watching.

Derivation of Euler-Lagrange Equations | Classical Mechanics

The Euler-Lagrange equations describe how a physical system will evolve over time if you know about the Lagrange function. We assume that out of all the different paths a particle can take, it will be the one, where small deviations from the path won’t change the action too much. This is usually referred to as: „the action is stationary“.

If you want to read more about the Lagrangian formulation of #ClassicalMechanics, we can recommend the book „Mechanics“ by Landau and Lifshitz:
https://amzn.to/2NjI8gQ
(Affiliate-Link)

If you want to help us get rid of ads on YouTube, you can support us on Patreon!
https://www.patreon.com/prettymuchphysics


consider a large electrically charged

sphere with a charge capital Q and A

small sphere with a charge small q the

electric force between these two spheres

which are at the distance R from each

other is given by colum's law here 1/ 4

Pi epsilon0 is a constant preactor with

the electric field constant Epsilon 0

which provides the right unit of force

the unit Newton now what if you know the

value of the big charge and want to know

the value of the force that the big

charge exerts on the small charge

however you do not know the exact value

of this little charge or you

intentionally leave that value open and

only want to look at the electric force

exerted by the big charge so you have to

somehow eliminate the small charge Q

from colum's law to do so you simply

divide colum's law by Q on both sides

this way on the right hand side the

small charge drops out and instead lands

on the left hand side of the equation

the quotient on the left hand side F

over Q is defined as the electric field

e of the source charge Q by calling it

Source charge we imply that the charge Q

is the source of the electric field the

electric field e thus indicates the

electric force which would act on a

small charge when placed at the distance

R to the source charge

What is Electric Field?

https://www.youtube.com/watch?v=ZAw98lYRL30&list=PLTjLwQcqQzNKzSAxJxKpmOtAriFS5wWy4

Electric field tells how big the electric force is on an electric charge. The field strength is illustrated by field lines. The closer field lines are to each other, the greater is the field strength there, and the greater is the electric force acting on a charge there. From Course: https://en.fufaev.org/electromagnetism

Books by Alexander Fufaev:
1) Equations of Physics: Solve EVERY Physics Problem
https://en.fufaev.org/physics-equations-book
2) Alexander Fufaev and His Never Ending Story:
https://en.fufaev.org/lifestory


jeder Körper der eine Temperatur hat

also alle strahlt Energie ab wie viel

Energie strahlt unsere Sonne ab und wie

viel davon kommt bei uns auf der Erde an

wie viel Energie strahlt eigentlich die

Erde in den Weltraum ab und welche

Oberflächentemperatur hat unsere Sonne

derartige und ähnliche Fragen können von

dem Stefan bolzmann Gesetz beantwortet

werden dieses Gesetz beschreibt den

Zusammenhang zwischen der Temperatur T

eines Körpers und seiner abgestrahlten

Leistung P auch die Oberfläche a des

Körpers beeinflusst die abgestrahlte

Leistung das Stefan bolzmann Gesetz

lautet p= σma mal A* t hoch 4 hierbei

ist σma die Stefan bolzmann konstante

mit dem Wert

5,67 x 10

hoch-8 jou pro Meter qu mal K hoch 4 x

Sekunde das ist eine konstante die einen

exakten Wert hat der nur von anderen

Naturkonstanten abhängt nämlich von der

bolzmannkonstante von der

plankkonstante und von der

Lichtgeschwindigkeit was ist genau die

Strahlungsleistung P sie gibt an wie

viel Energie pro Sekunde die Oberfläche

des Körpers abstrahlt die Einheit der

Strahlungsleistung ist Watt oder Joule

pro Sekunde

außerdem gilt das Stefan bolzmann Gesetz

nur für sogenannte schwarze Körper oder

Körper die einem schwarzen Körper nahe

kommen die Bezeichnung schwarz hat

jedoch nichts mit der Farbe des Körpers

zu tun die Sonne ist beispielsweise ein

annähernd schwarzer Körper obwohl sie

wie du weißt nicht schwarz ist ein

schwarzer Körper absorbiert die gesamte

Strahlung die auf ihn trifft er

reflektiert keine Strahlung und lässt

auch keine Strahlung

durch Stefan bolzmann Gesetz kannst du

sehen wie stark die abgestrahlte Energie

von der Temperatur abhängt Verdopplung

der Oberflächentemperatur der Sonne

würde dazu führen dass auf der Erde das

16fache an Sonnenenergie ankommen würde

machen wir ein Beispiel wie man mit dem

Stefan bolzmann gesetzt die Temperatur

der Sonne bestimmt aus der bei uns

ankommenden Energie der Sonne können wir

auf die gesamte emittierte Energie der

Sonne schließen die Sonne tiert eine

Leistung von 3,8 x 10 hoch 26 jou pro

Sekunde der Radius der Sonne beträgt

6,96 x 10 hoch 8 m nehmen wir an dass

die Sonne nicht flach sondern

kugelförmig ist daher können wir die

Formel für die Oberfläche einer Kugel

verwenden um die Oberfläche der Sonne zu

bestimmen a= 4 PI

r² Formen wir das Stefan bolzmann setz

nach der Temperatur T um und setzen die

Flächenformel ein wenn wir nun die

Strahlungsleistung die Stefan bolzmann

konstante sowie den Sonnenradius

einsetzen bekommen wir

5760 Kelvin als Oberflächentemperatur

der Sonne heraus das entspricht

633° CS

Stefan-Boltzmann-Gesetz

https://www.youtube.com/watch?v=CQFkPtiHIMg&list=PLcjHhwL4seQjmFSLkz_R11cgq7-LPyuGA&index=27

Lerne, wie die Temperatur jeden Körpers, seine abgestrahlte Energie (Leistung) beeinflusst. Mehr: https://de.fufaev.org/stefan-boltzmann-gesetz

Bücher von Alexander Fufaev:
1) Die nützlichste Physik-Formelsammlung des Internets:
https://de.fufaev.org/formelsammlung
2) Alexander Fufaev und seine Lebensgeschichte:
https://de.fufaev.org/lebensgeschichte


Every body that has a temperature radiates
energy.

How much energy does our sun radiate and how
much of it reaches us on earth?

How much energy does the Earth actually radiate
into space?

And what is the surface temperature of our
sun?

These and similar questions can be answered
by the Stefan-Boltzmann law.

This law describes the relationship between
the temperature T of a body and its radiant

power P.

The surface area A of the body also influences
the radiant power.

The Stefan-Boltzmann law is "P is equal to
sigma times A times T to the power of four"

Here, "sigma" is the Stefan-Boltzmann constant
with the value "five point six seven times

ten to the power of minus 8, joules per meter
squared times Kelvin to the power of four,

times second".

This is a constant that has an exact value
that only depends on other physical constants,

namely the Boltzmann constant, the Planck
constant and the speed of light.

What exactly is the radiant power P?

It indicates how much energy per second the
surface of the body radiates.

The unit of radiant power is watt, or joule
per second.

In addition, the Stefan-Boltzmann law only
applies to black bodies or bodies that come

close to a black body.

However, the term black has nothing to do
with the color of the body.

The sun, for example, is a nearly black body,
although, as you know, it is not black.

A black body absorbs all the radiation that
hits it.

It does not reflect any radiation and does
not allow any radiation to pass through.

The Stefan-Boltzmann law shows how much the
radiated energy depends on the temperature.

Doubling the surface temperature of the sun
would result in 16 times more solar energy

reaching the earth.

Let's take an example of how to determine
the temperature of the sun using the Stefan-Boltzmann

law.

We can deduce the total amount of energy emitted
by the sun from the energy we receive from

the sun.

The sun emits a power of "three point eight
times ten to the power of twenty-six joules

per second.

The radius of the sun is "six point nine six,
times ten to the power of eight meters".

Let's assume that the sun is not flat, but
spherical.

We can therefore use the formula for the surface
area of a sphere to determine the surface

area of the sun: "A is equal to four pi times
r squared".

Let's rearrange the Stefan-Boltzmann law with
respect to the temperature T and insert the

surface area formula.

If we now insert the radiant power, the Stefan-Boltzmann
constant and the solar radius, we get "five

thousand seven hundred sixty Kelvin" as the
surface temperature of the sun.

This corresponds to "six thousand thirty-three
degrees Celsius".

Stefan-Boltzmann Law: How Bodies Radiate Energy Through Their Temperature

https://www.youtube.com/watch?v=od-jgaOQ8mc&list=PLTjLwQcqQzNKzSAxJxKpmOtAriFS5wWy4&index=17

Learn how the temperature of each body, affects its radiated energy (power). More: https://en.fufaev.org/stefan-boltzmann-law

Books by Alexander Fufaev:
1) Equations of Physics: Solve EVERY Physics Problem
https://en.fufaev.org/physics-equations-book
2) Alexander Fufaev and His Never Ending Story:
https://en.fufaev.org/lifestory


all right so here we have a satellite

that is located a distance of 7 r r

being the radius of the planet or Earth

in this case um and it wants us to find

the local acceleration due to gravity

for the satellite all right so this

one's pretty straightforward this is

just a ratio problem here uh here I have

my yellow chart surface and altitude or

before and after if you like one and two

A and B whatever it is you want to use

but have a reference point okay so here

my reference point is going to be at the

surface because I know the generic

function for acceleration neutral

gravity at the surface of the planet is

G mass of the Earth and then the race of

the earth square and then once I know

that I can set up the right side of the

uh chart here pretty quickly so here I

call it G new or the acceleration due to

gravity or the local acceleration due to

gravity for the satellite which is just

GM Earth over the radius

square uh but radius is being used kind

of vake here okay so really at the end

of the day it's the distance from Center

to Center but the diagram shows you

already that this Arrow here watch my

cursor please so here's planet Earth and

blue the distance given to you in the

schematic actually starts at the center

of the planet and it goes all the way to

the orbit of the satellite so in this

particular case 7R already includes the

radius of the planet so therefore the

distance from Center to where the orbit

is or where the Salah is is given a 7R

so therefore in the denominator I would

just substitute in 7R but use

parenthesis if you don't use parth

parentheses you'll forget to square the

coefficient 7 so it's important to use

parenthesis here so because there is an

exponent two I want to shoot that

exponent to the seven I'm also going to

introduce that uh excuse me um

distribute that exponent to the r as

well so essentially the numerator stays

the same except the bottom 7 square is

49 and then R square is R square here uh

hopefully you've seen my other videos

you can learn how to factor out

coefficients just to make things look a

little nicer you don't have to if you

know what you're doing but I like to

make it look a little nicer I notice the

only really thing difference between the

left side and the right side here is the

49 in the

denominator well because the 49

denominator I can just Factor It Out by

putting under a one and then gm/ r² I'll

keep that as the same term because on

the left side of the chart acceleration

due to gravity at the surface of the

planet is given by G mass of the earth

and the radius of the planet Square so I

notice here I have the same term here g

f Earth over r s so I want to highlight

the same

term therefore what I can do is I can

substitute out g m r² for lowercase G at

the surface of the planet which in this

case is Earth which is 9.8 so therefore

I can substitute out this whole entire

highlighted term g m e over r² for G but

then now it's the same thing 1 * G is

just G and then G over 49 G got

substituted out for 9.8 and then 9.8 ID

4.9 will give you an answer of 0.2 m/s

squar okay and that's it for this one

this one's pretty straightforward all

right good luck stay ready don't be back

Newton's Gravitational Law of Attraction_4

The radius of the satellite’s orbit around the center of the Earth is          r = 7R, where R is the radius of the Earth. What is the local acceleration due to gravity for the satellite?


so here we have an example of Young's

double slit experiment um the problem

states that the distance between the two

slits as indicated here in the diagram

is 03 millimeters I haded that in blue

for you and it does say that it's a

second order bright Fringe that's being

observed and then the angle between the

zero uh m equal to zero or the central

Fringe to the second or the right Fringe

is 215° again as shown here in the

diagram and it wants us to find the

wavelength of the light that's coming in

so to do uh so before you proceed uh you

should definitely scope out units uh

again I notic here this is given in

millimeters uh I would like to tell you

to convert everything to meters first

this way you know naturally things will

cancel out and then if you need to

convert at the very end to whatever

units they're asking for then do so at

the very end in this case I can see that

it wants it in nanometer so I know I

have to convert a little bit at the very

end uh so at to start I'll go ahead and

convert the distance between the two

slits in Mill millimeters to meters the

conversion factor between meters to

millimeters shown here on the screen for

every 1 meter there's 1,000 millimeters

uh so Place stuff in the right order and

then millimeters is going to cancel out

uh in the numerator denominator there

and if you do the calculator work

that'll give you 3.01 105 meters so now

that that's done uh we can go ahead and

proceed we want to go ahead and use

Young's doubleit function for

constructive interference because the

problem states that it is a bright

Fringe where m is equal to two uh so

we'll go ahead and bust this out again D

is the distance between the the slits uh

Theta is the angle between uh the or

Central bright frence and the second

order bright frence here and M we know

what m is and I'll get to that in just a

second okay but it's important to keep

the in mind that M must be 0 1 2 3 or

four according to Young's dite function

it cannot be like a 0.5 or 1.5 so M must

be any whole number uh

so as you can see highlighted there it

say the second order bright Fringe so

therefore m is two so even if the

problem doesn't give you like for

example in the parentheses m is equal to

2 uh they do expect for you to interpret

this information and know what value of

M to use so in this case it's pretty

straightforward we know that D is given

to us as 3.01 105 so we can substitute

that for there we know that from the

statement m is two so we can just

substitute uh m in the denominator uh

for value of two from here everything we

do know is calculator work time so 3.0 *

5 m Time s of 25 / 2 if you punch all

that in your calculator carefully you

should get a result of 5.6 T 107 M okay

but again you notice here the answer

here is in meters the question asked for

uh the wavelength and nanometers so we

just have to convert meters to

nanometers and that's pretty easy to do

uh so start with the uh statement 5.6 M7

meters times it again it's important

that you place the conversion factor

correctly here you want to put meters on

the bottom because you started off

meters so meter here is going to cancel

with meters in the denominator and that

leaves you with the units of nanometer

and in case you don't know for every one

meter there's 10 to the 9th nanometers

so if you do that properly the final

outcome will come out to be about

nanometers okay other than that it's

pretty straight for a problem okay all

right don't be whacked stay

nerdy

Young's Double Slit_3

The distance between the two slits is 0.03 mm.  The second-order bright fringe (m = 2) is measured on a viewing screen at an angle of 2.15o from the central maximum.  Determine the wavelength λ (in nm).


you have always heard in your childhood

that one raised to something is always

one but today I am going to show you

something crazy is it possible to have

one raised to some value like X that

ends up being two this is a tricky

question first of all take natural log

on both sides I will be using log to

denote natural log some people also use

Ln to denote the same you can use

whatever you want to so using this

property of logarithm the X will jump

right here and we get x * log of 1 = log

of 2 now divide both sides by natural

log of one but natural log of one is

zero and we cannot divide by zero so no

real solution exists for X but I told

you this is a tricky question so now how

can we find the complex roots for this

equation we will first start with the

oiler identity which is e raised to I

Theta is equal to cine Theta + I sin

Theta if you are dealing with the

complex world then everyone should know

this identity now if you put the value

of theta as 0o we get cosine 0 + I sin 0

cosine 0 is 1 and sin 0 is 0 so e raised

to i 0 becomes 1 but this also works for

Theta equal 2 pi 4 pi and so on so one

in complex world can be written as e

raised to I * 2 K Pi where k equal 0 1 2

and so on so this x becomes log of 2 /

log of e raised to I * 2 K

Pi we know that log of e raised to a

simply becomes a so this x becomes log

of 2 upon I * 2 K Pi so far so good now

I want this I to be on the top so

multiply and divide by I to get this now

this becomes -1 and therefore X becomes

minus I * log 2 upon 2 K piun but before

we wrap this up we need to make sure

that K is not equal to zero now

surprisingly if you look at Wolfram

Alpha and plug this equation in it you

find that it does not provide any

solution to it now take this X on the

other side and make it as 1 = 2 raed to

1 /x now if you put this equation in

Wolfram Alpha it works and we get the

same solution as what we have proved

before really cool right so good

How Can This Be True?

1^x = 2 | Math Trick | Harvard Entrance Exam | Math Olympiad | Complex Numbers

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today let's tackle an interesting

algebra problem that challenges our

problemsolving

skills also this is considered as one of

the difficult math Olympiad questions so

stay tuned our goal is to find the value

of x that satisfies this equation

equation is x^2 - x Cub = 36 can you

solve it okay as the very first step If

You observe it closely you can see that

36 can be written as 27 + 9 and what is

27 it is 3 Cub right and also 9 is 3 S

right so this 36 can be returned as 3 2

+ 3 Cub but we need a minus sign here

and not plus so if you think a little

bit we can rewrite this 9 as Min - 3

Square as well and this 27 can be

written as minus of minus 3 Cub because

- 3 Cub is - 27 and adding one more

minus will make it a plus excellent so

it is clear that one of the answers to

this question will be minus three just

by hit and trial now in order to find

the other two Roots we will factorize

this equation for that first take this

36 on the left hand side and rearrange

this equation to make it minus of X Cub

+ x² - 36 = 0 or multiply both sides by

-1 to get XB - x² + 36 = 0 now in order

to factorize it using x + 3 rewrite this

as X Cub + 3 x² - 4 x² + 36 = 0 3 - 4

will be - x² only now take x² common

from this to get x2 * X + 3 and from

here take -4 as common to get -4 * x^2 -

9 this x^2 - 9 can be written as x - 3 *

x + 3 so far so good now take x + 3

common from both of them to get x + 3 *

x^2 - 4 * x - 3 = 0 or x + 3 * x^2 - 4 x

x + 12 = 0 now this will give x = -3 and

for this part we will be using the

quadratic equation formula we have a

which is equal to the coefficient of x²

= 1 then B which is the coefficient of x

= -4 and C which is the constant term as

12 plug them into the formula to get x =

minus of -4 +- under < t - 4 2 - 4 * 1 *

12 upon 2 which will give -4 +- under <

TK 16 - 48 or under < TK - 32 upon 2 now

this will introduce the imaginary term I

as under root minus 32 can be written as

under < tk-1 * under < TK 16 * 2 or 4 <

tk2 so X will become -4 + -4 < tk2 I

upon 2 or -2 + - 2 < tk2

I this is the solution to our problem

that was amazing isn't it so good

[Music]

Can You Solve this Harvard Entrance Exam Question?

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this is such an amazing problem we are

given this right triangle whose length

of this part of the hypotenuse is five

and we have a circle inscribed in it

whose radius is 2 units we have been

asked to find the area of this shaded

region that is this blue region can you

solve it what do you think in how many

ways can this question be solved I'd

love to hear your thoughts in the

comments okay so first thing first since

this side of the triangle is tangent to

to this circle so this radius will be

perpendicular to this side so this will

be

90° now draw this radius it will also be

perpendicular to this side of the

triangle as it is also tangent to this

circle so it will be of length two and

this will be

90° look at this quadrilateral all four

sides are 90° and both of them are of

length two so it will be a square of

side length two thus both of them will

also be two great now here comes the

magic we will use the two tangent

theorem which states that the lengths of

two tangents drawn from the same

external point to a circle are equal now

this is a tangent and this is also a

tangent to this circle and they are

drawn from the same external point this

one so both of them will be equal

therefore this will also become five

same goes for these tangents both of

them will be equal let us label them as

X excellent now consider this right

triangle this side is of length x + 2

this is of length 2 + 5 or 7 and the

hypotenuse is x + 5 so we will use the

Pythagoras Theorem to find X we get x +

5² = x + 2 2 + 7 s this gives x 2 + 5 2

or 25 + 2 * x * 5 or 10 x = x 2 + 2 2 or

4 + 2 * x * 2 or 4 x + 49 this x² will

get cancelled out and we are left with

this 25 + 10 x = 4 + 4 x + 49 or 4x + 53

take 4X on left side and 25 on right

side to get 6 x = 28 or x = 28 / 6 which

can be simplified to 14 over 3 so far so

good thus this side length will be x + 2

or 14 by 3 plus this 2 can be written as

6 by 3 so it becomes 20

by3 now the area of this shaded region

will be same as the area of this

triangle minus the area of this circle

the area of this triangle will be 20x 3

* 7 / 2 or 140 / 6 which when simplified

becomes 70 by 3 and the area of this

circle will be Pi r^ 2ar or < * 2^ 2 or

4 piun so the area of this shaded region

will be 70 by 3 - 4 Pi cool were you

able to solve it what approach did you

use use to solve this question let me

know in the comments so good

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special thanks to Katrina EG for posting

this amazing geometry question on X the

end result is literally going to shock

you we have this big semicircle and

inside it we have two small semicircles

of diameter two and three and this

yellow one is tangent to the red

semicircle our job is to find the

diameter of the big

semicircle so can you solve it at first

glance if we pull this down here it

looks like the bigger diameter is simply

the sum of both of these diameters so

the answer we are looking for is five or

it will be approximately five but now

how to find out the exact value of this

diameter for that we will break this

diameter into smaller sections this this

and this one this piece right here is

the same as the radius of this

semicircle and thus it will be of length

one now to find the length of this part

first draw this radius and then join

both of these radi together this will be

1.5 and this will be 1 + 1.5 or

2.5 now we will use Pythagoras Theorem

to find the length of this part we can

do the math to find it and I have done

it for you in case you might want to

check but we have an easier method to

figure it out if we multiply the length

of both of these sides by two we get the

this as three and this is five and thus

we know that it is a three four five

Pythagorean triplets and therefore this

will be four or if we divide it by two

we get this side length as two excellent

next we will find the length of this

part and we are done let us label this

as y now here comes the magic this is

what I found the most intelligent way to

solve for this part complete this circle

now make a rectangle around it so the

length of this side of the rectangle

will be the same as the diameter of this

circle which will be three now extend

this line to the vertex of this

rectangle awesome now consider these two

triangles this angle is the same for

both these triangles and also since both

of these lines are parallel this angle

is also the same for both of them and

hence both these triangles are s similar

therefore this side which is 2 over this

side which is 1.5 equals this side which

is y + 2 over this side which is 3 cross

multiply to get this this becomes 6 over

1.5 or 4 = y + 2 or Y = 2 woohoo now is

the time for our final answer drums

rolling in and and it is 1 + 2 + 2 or 5

which is exactly the same as we

predicted at the beginning of this video

when we looked at this problem

visually isn't this cool if you liked

this video then please don't forget to

share it with your friends and support

this channel by joining our community

and becoming a member so good

[Music]

Crazy Math Challenge!

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#unitedstates


this is a pretty exciting problem you

know about the three four five triangles

you see everywhere in high school

geometry in sa a and even in

construction they show up all over the

place but today what we are going to

show will literally blow up your mind

because if you inscribe a circle in any

three four five triangles then you won't

believe but the area of this inscribed

circle is exactly Pi let us see the math

first let us draw the center of the

circle now draw a radius like this and

this the radius will always be at a

right angle to the tangent point and

therefore this whole thing will become a

square of side length R so this side

will also be R and this side will also

be R now this is three and this is R so

this will be 3 minus r similarly this is

four and this is R so this will be 4 - R

now let us clear out everything and just

keep this circle and these two lines

which are tangent to this circle here

comes the magic the lengths of two

tangent lines drawn from the same

external point to a circle are equal

specifically it is called the two

tangent theorem so this tangent line

will also be 3 minus r and the same

thing goes for this side if this is 4 -

R so this will also be 4 - R now let us

Focus focus on the hypotenuse the length

of the hypotenuse is 5 now this is also

equal to 3 - r + 4 - R so we have 7 - 2

R = 5 this gives 2 R = 2 and R = 1 now

that we have the radius of the circle as

one so what will be the area of this

circle it will be piun * R square and

putting r as 1 we get the area of this

circle as Pi isn't this cool so good

3-4-5 Triangle and π (Pi)

New Method Discovered to Calculate π (Pi) in 2024.

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today I am going to show you how to

approximate the Square t of 101 minus

the square < TK of 99 whenever you see

this type of question the first thing

you should do is rationalize the

numerator and in a minute I will let you

know why to do this multiply both

numerator and denominator by the square

root of

101+ the square < TK of 99 this way we

will have Plus in here instead of minus

using a + b * a - b as a s - b 2 the

numerator will be 101 - 99 or 2 and the

denominator will remain as it is so now

our job is to approximate this

denominator for that we will find the

integer numbers between which this sum

might exist let us call this sum as X so

we will try to find the integer value

which is just shorter than this X and

just larger than this x to do this we

will find the value of x² which will be

square T of 101 + theare < TK of 99

squar it will be 101 + 99 + 2 * the sare

< TK of 101 * theare < TK of 99 so using

this we have reduced these two ugly

square roots into a single square root

now 10 and 1 is 100 + 1 1 and 99 is 100

-1 so using this formula again we get

this value as 100 S -1 so x² will be 200

+ 2 * < TK of 100 S - 1 now how can we

get rid of this minus1 notice that 100 S

- 1 will be less than 100 square so x²

will be less than 200 + 2 * < TK 100 S

or it will be less than 200 plus plus 2

* 100 or 400 but 100 S - 1 will also be

much bigger than 99 Square so x² will be

greater than 200 + 2 * < TK of 99 s or

it will be greater than 200 + 2 * 99 or

398 so this part gives X less than 20

because 20 SAR is 400 and now if x is

greater than 398 then it is also greater

than 36 61 which is 19 squ so X will be

greater than 19 now our goal was to find

an approximate value of this one which

is

2/x since X is greater than 19 and less

than 20 so 2/x will be greater than 2

over 20 and less than 2 over 19 2 over

20 is

0.1 and 2 over 19 is

0.105 therefore our answer lies

somewhere between 0.1 and

0.105 so if we take their average we get

our answer as approximately

01025 now if we verify it using

calculator we find that we are really

close to our approximation so good

You Should Learn This Trick!

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this is such an amazing problem and I

did not expect it to have the outcome

that it had the problem is that we have

a semicircle right here and a quarter

Circle right here and then this red line

with the length of 5 cm which goes from

this vertex and is tangent to this

semicircle we want to find the area of

this rectangle so how can we solve it as

a first step let us assign a radius R to

this semicircle now many of the people

think that this radius is the same as

this quarter circle's radius but it is

nowhere given that they are the same and

in the later steps we will also discover

that we don't require them to be equal

therefore this is one of my favorite

problems so now let us label this radius

as X this piece will also be equal to r

as it is the radius of this semicircle

and this height will be equal to X which

will be the same as the radius of this

quarter Circle we can find the area of

this rectangle in terms of R and X the

base is x + 2 R and the height is X so

the area will be base x height or X plus

2 R * X which will be x² +

2xr now here comes the magic let us draw

a segment from this point of tangency to

the center of the semicircle and this

will be the same as the radius of the

semicircle which is R every radius that

intersects the tangent line at the point

of tangency is at right angle so now we

have a nice right angle triangle and

therefore yes you have guessed it right

we will use Pythagoras Theorem we get r

s + 5² = x +

r² the left hand side will be r² + 25

and the right hand side will be x² + 2xr

+ r² these r s will get cancelled out

and we are left with x² + 2 XR = 25 but

we know that this quantity is the same

as the area of this rectangle so the

area of this rectangle will be 25 don't

forget to put the cm square so good

Can This Question Be Solved?

Can This Question Be Solved?

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we have a quarter Circle and some shapes

inside it our goal is to find the area

of the blue region the trick here is to

solve this problem without using

integration pause this video and try to

solve it on your own we start by drawing

a radius from the center of this quarter

Circle which goes through this point of

intersection which will divide

everything in half let's say that both

of these blue shapes are represented by

X the white shapes by Y and this one by

Z so now we know that the total blue

area is represented by 2 * x + 2 * Z we

need to figure out what x and z are

we're dealing with a quarter of a full

circle the area of a quarter circle is

given by the formula < * Radius 2 / 4

the radius here is four so we calculate

it as 4 * 4 so the area of the quarter

circle is 4 * pi this total area of the

quarter circle is equal to 2 * x + 2 * y

+ 2 * Z we also have a semicircle inside

this quarter Circle the area of a

semicircle is given by the pi * radius

squar / 2 the diameter is 4 so the

radius is 2 when we calculate it we get

2 * pi as the area of the red semicircle

this 2 pi is equal to 2 * x + y next

let's focus on a triangle inside this

semicircle which is obtained by dividing

this semicircle into half the height of

this semicircle will be the same as the

radius of this semicircle which is

two now the area of the triangle is

calculated as base time height divided

by two here both the base and height are

two so the triangle's area is is 2 now

it is clear that the x + triangle's area

equals the area of the quarter Circle so

we subtract this triangle from both

sides to get the value of x the area of

the quarter circle is pi * 2^ 2 / 4

which is pi and the triangle area is 2

so X is equal to < - 2 now we substitute

X in this equation to get the value of y

as four the blue area is equal to 2 * x

+ 2 * Z substituting y = 4 we get the

area of the blue region as 4 Pi - 8

square units so good

Even Adults Cannot Solve This!

Fun Math Problems. Find the Area of Blue Shaded Region.

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this is an amazing geometry question

that includes a variety of Concepts like

symmetry Circles chords and

perpendicular lines we have a circle

first we draw this chord of length five

and after that we draw this line of

length six from this point which is

perpendicular to this chord followed by

another line which is perpendicular to

this one and it is of length four now

how can we use these details to

determine the radius of this Circle can

you solve it let us start by marking the

center of this circle now in a circle

the line segment from the center that is

perpendicular to a chord will also bict

the chord this means that the

perpendicular line from the center of a

circle to its chord divides it into two

equal segments this is called chord bis

sector theorem now draw this diameter

and then draw this rectangle whose side

lengths are five and six like this now

draw this line which is perpendicular to

this side of the rectangle what do you

think its length will be since this

diameter will be perpendicular by sector

of this chord and if these are 5x two so

both of them will also be 5x two as well

and since both of these lengths are the

same therefore this segment will also be

of length four awesome now draw this

line and label it as X now we will use

another theorem called intercept in

chords theorem here is one chord of the

circle and here is another chord of the

circle so by intersecting chords theorem

this length which is 6 times this length

which is X will be the same as this

length which is four times this length

which is 5 + 4 or 9 so X will be 6

amazing now draw this line such that we

get a right triangle here what do you

think this line will be yes you you

guessed it right it will be the diameter

of this circle this is because of

thales's theorem which states that if

you draw a triangle where one of its

sides is the diameter of a circle then

the angle opposite that side is a right

angle so this hypotenuse will be 2 * the

radius R and this is 12 and this 5 so we

will now use Pythagoras Theorem such

that 2

r² = 5 S + 12 square or 25 + 144 which

is equal to 169 or 13 Square so 2 R = 13

and R = 13 by 2 and that's it wasn't

this an interesting problem were you

aware of so many theorems before this

video and able to solve it let me know

in the comments so good

[Music]

Can This Be Solved?

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hey guys I saw this thing from Iota

Academy it says I want to believe and it

shows this really interesting equation

let's verify that this thing is true so

first step let's subtract one from both

sides on the left hand side we just have

this thing minus one and on the right

hand side we just have this fraction I

don't like fractions so I'm going to

multiply both sides by this denominator

these two things are going to cancel out

which leaves me with just pi to the E

minus one and on the left hand side we

could just bring this stuff down now

let's foil root pi to the E times root

pi to the E is just pi to the E root pi

to the E times negative 1 is negative

root pi to the e one times root pi to

the U is root pi to the E and 1 times

negative 1 is equal to negative 1. now

we can bring down the right hand side

these two middle terms are the same so

they're going to cancel each other out

or just left with pi to the E minus 1

equals pi to the E minus 1. so this up

here is a true statement how exciting

Fun Math Equation #shorts

#shorts


hey guys somebody mentioned me in this

one it looks like a fun one we're given

three squares and some measurements and

it wants us to calculate the area of the

entire shape if you want to try it on

your own pause it right now because I'm

going to solve it in 3 2 1 it says

calculate the area of the shape I

believe it means it wants us to add up

all three of the squares so we just got

to figure out the area of each of the

squares and then add them all together

let's start with the largest Square in

the middle and let's give the sides a

length of X next let's focus on this

gray square and let's find one of its

side lengths since we know that this

entire height is equal to X and this

portion is 2 that'll leave x - 2 for the

yellow part and since this is a square

we know the bottom is also x - 2 and now

we're starting to fill out this 17 so we

have the x - 2 right here let's color

this blue for this portion now we just

got to find this last bit and then we

can set it all equal to 17 now for the

last part since the whole thing is X and

this piece is five that leaves x - 5 for

this piece and then once again it's a

square so the base is going to match the

height so this orange portion is x - 5

now we know that the yellow plus the

blue plus the orange will equal 17 so we

have x - 2 + x + x - 5 = 17 we can copy

it down and then combine like terms x +

x + x is 3x and -2 - 5 is -7 next we can

add 7 to both sides of the equation on

the left hand side we're left with 3x

and on the right hand side 17 + 7 is 24

and then we can divide both sides by 3

and that'll give us x = 8 now we know

this yellow X is equal to 8 this blue X

is equal to 8 and this orange x is equal

8 and we can simplify 8 - 2 is 6 and 8 -

5 is 3 so the area of this gray square

is 6 * 6 which is 36 the area of the

green square is 8 * 8 which is 64 and

the area of the brown square is 3 * 3

which is 9 and then 36 + 64 + 9 is 109

let's give it a label of cm squ and put

a box around it and that's the answer to

our question the area of this shape is

109 cm squared how exciting here's a

problem I'm planning on doing for the

next video it says solve for x and it's

XX - 4 + 1 = 4x - 4 and when I do post

it I'll try to put a link to the video

somewhere around here

Area of 3 Squares

I thought this was a fun one, I hope you guys like it!


guys we're gonna evaluate a piecewise

function so for

x equals seven which of these two is

true what's the top one because seven is

greater than four we are not going to

use this one this is telling us

that we're gonna use this function so f

of seven

is equal to and we're going to plug a 7

into the x

7 plus 2 and that equals 9. so the

answer to this

is 9. had this 7 been something

different like a 2 we would have ended

up using this

lower one this x squared because 2 is in

this domain i posted more of these

online if you want to try some more

there's a whole bunch of samples you can

watch a lot of them

there's a lot of samples you can try and

they all give the answers

it's 21 different ones you can try find

that page by going to andymath.com

on the homepage just type something like

evaluating piecewise functions and this

page will come up and that's the page i

just showed you please check it out i'd

love people to use it you're awesome

and i'll talk to you soon bye

Evaluating Piecewise Functions Example Problem #Shorts

Check out my website!
https://andymath.com/

#Shorts


hey guys Pythagorean theorem is true for

all of these cases there's already a ton

of proofs for this version I'm not going

to worry about that in this video but I

do want to show how this version leads

to the other versions so let's do this

one how do we get from here to here

since this is an equation we can

multiply both sides by the same thing in

this case 1/2 and this 1/2 will

distribute to both of these terms and

this 1/2 will distribute here and we end

up with a^2 over 2 + b^2 2 = c^2 / 2 and

basically that's cutting all of these in

half so it ends up looking like this and

that's how to do the first one and you

can do something similar for equilateral

triangles the area of an equilateral

triangle is the < TK 3 over 4 * the base

squared so this formula would be < tk34

a 2 +34 b^2 = > 34 c^2 and you can

imagine how we get from here to there we

multiply both sides of the equation by <

tk3 over 4 this will distribute to both

of these and this distributes here and

we now have the equation we wanted it's

the same thing for regular pentagons you

just multiply both sides by a constant

the next interesting one is semicircles

the area of a semicircle is equal to PK

R 2 / 2 or you can rewrite that as < / 2

r^ 2 so this would end up being < / 2 r^

2 + < / 2 r^2 = < 2 R 2 so R sub1 is the

radius of this one that's going to be

half of the distance a so we can change

this to a over2 and then R sub 2 would

be the radius of this semicircle so

that's going to be half of B so this

will be B over2 and then the radius of

this one is going to be half of this

side length C so we'll change that to

c/2 so this is the formula for

semicircles and we can prove that this

is true we start with A2 plus b^2 = c^2

and then we'll divide both sides of the

equation by four and then on the left

side this four can be split to the two

terms we can rewrite each of these fours

as 2^ 2ar now looking at this a^ 2 over

2^ 2 we can rewrite this as a over 2 the

whole thing squared and the same thing

as b^2 over 2^2 that can be Rewritten as

B over 2^ 2 and we can do the same thing

to this it'd be C over 2^ 2 and then for

the next step we can multiply both sides

of the equation by Pi / 2 this Pi over2

will distribute to both of these terms

and on this side it will distribute here

and now we have the equation of the

semicircle Pythagorean theorem so we

went from the Pythagorean theorem that

we know to this new fun version and this

will work for any similar figures you

end up multiplying both sides of the

Pythagorean theorem by some constant how

exciting

Proving Different Pythagorean Theorems

Demonstrating why all these versions work.


guys this one says factor completely and

this completely is very important it

might be more than just one step so

first let's factor out the GCF the

greatest common factor for the two of

them or we can Factor two out of both of

them we can factor out an X out of both

of them and we can factor out a y

squared out of both of them so now let's

go after this first one two times what

gives us two that's one x times what

gives us x to the fourth that's X cubed

and then y squared times what gives us y

to the fifth that's y cubed so this

times this gives us this and now 2xy

squared times what gives us 2xy squared

well that's just one because this

matches this exactly a lot of people

might think we're done right now but we

can go further this right here is a sum

of cubes a cubed plus B Cubed is equal

to a plus b times a squared minus a B

plus b squared this first term is a

cubed that would mean that a is equal to

the cube root of this which would just

be x y and then for the B we want to

take the cube root of this one because

one is equal to B Cubed and that gives

us one so now we can just plug it right

into this equation so let's copy down

the GC F the 2xy squared and then this

is going to change into two parentheses

we put the a in the first spot which is

x y and then we do plus b in the next

spot and B was one and then for the a

squared we want to square this which

would be x squared y squared we squared

each of the terms and then minus a b so

we're going to minus these two things

multiplied which is just minus x y and

then plus b squared but B squared is

equal to one because one squared is one

and this is the answer to the question

we factored completely how exciting

Factoring Challenge


so we want to solve for x in this Isles

triangle in an isoceles triangle these

base angles are equal and so are these

two sides since these two sides are

equal we can just say 3x + 5 is equal to

5x + 1 and now we just solve for x

there's an x on both sides of this

equation so I'm going to subtract the

smaller one over to the other side 3 X's

cancel so we just have a 5 on the left

and 5x - 3x is 2x then we want to

subtract one from both sides and then we

div div both sides by 2 4id 2 is 2 and

2x / 2 is X so x = 2 is the answer to

the question comment below for what

you'd like me to solve next and I'll

talk to you guys soon bye Andy maath

Isosceles Triangles #andymath  #math #maths #sat #tutorials

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hey guys welcome to this limit problem

so as you know as X approaches 3 we

ultimately can just plug in a 3 for the

X and what happens when we do that is we

get 3 - 3 over 3^ 2 - 9 and this ends up

giving us 0 over 0 so anytime you have a

0 over 0 you should look to see if

something can cancel after you factor so

this is actually difference of two

squared so I'm going to rewrite this as

the limit is X approaches

3 of x - 3 over and I'm going to factor

this x -

3 x +

3 and now these two things can cancel

and we're left with just a one on top

now when I plug in the three for the X I

end up with 1 over 3 +

3 which is equal to to 1

6 and that is the answer to the question

so I hope that makes sense and I'll talk

to you guys soon

bye

Find the Limit (if it exists)

Limits Problem #1 AROUND A MINUTE! For more math made easy visit andymath.com.

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guys so this is a pretty legit calculus

challenge it uses very basic derivative

skills but it can be very tricky if

you're not careful if you want to try it

out pause it right now because I'm going

to give the answer in three two one what

it's asking for is the double derivative

of f of x to the four so I'm going to

put the derivative of derivative of f of

x to the fourth so we're going to solve

this inside first so this up here says

the derivative of f of x is G of X so

this is just going to be G of whatever

is inside here G of x to the fourth and

then we have to take the chain rule of

whatever's inside here so we're going to

multiply this by the derivative of

what's inside here which is 4X cubed our

first step is now done now we gotta take

the derivative of what's inside here

unfortunately it is a product of two

functions so we have to use the product

rule so we're going to do the derivative

of this first function and leave the

second function alone plus the

derivative of the second function

multiplied by the first function left

alone so this derivative of GX to the

fourth so it says the derivative of G of

X is f of x cubed so the derivative of G

of x to the fourth would be f of x to

the fourth Cube so then because is the

chain rule we have to multiply it by the

derivative of this inner function x to

the fourth which gives us 4X cubed and

then we're going to multiply by this

other 4X cubed plus the derivative of

this is just 12x squared and then this

GX to the 4 is just left alone we can

clean things up 4 times 4 is 16 x cubed

times x cubed is X to the sixth and then

bring this down and x to the fourth to

the third is X to the 12th plus and this

term can't really simplify so there it

is right there and that is the answer to

the question how exciting is that what

you got

Calculus Challenge Problem


guys this is implicit differentiation

it's saying find dydx when x squared

minus x y equals 25. step one we're

going to take the derivative of both

sides so what is the derivative of x

squared we bring the 2 in front and it

just gives us 2X and then we're going to

subtract and what is the derivative of x

times y well the function X is being

multiplied by the function y so we have

to use product rule so it's going to be

the derivative of x times y plus x times

the derivative of y and then what is the

derivative of 25 that's equal to zero so

this is our next step so from here we

clean up these primes X Prime is the

derivative of x which is one and Y Prime

is the derivative of y if we write that

as d y d x that's the thing we're trying

to solve for next we can distribute the

negative to both of these terms they're

both going to become negative so our

goal is to get dydx alone so let's move

all the terms that don't have a d y d x

to the other side of the equation we're

doing that by subtracting this one and

adding this one on the left hand side

we're just left with the negative x d y

d x and on the right hand side we have

the negative 2x plus y from here we can

divide both sides by negative X the

left-hand side just has the dydx which

is what we wanted to get alone and the

right hand side is basically going to

look just like this but these two

negatives will cancel and this will

become a negative so it looks like this

2x minus y over X

Impilicit Differentiation #shorts

https://andymath.com/implicit-differentiation-2/

Collab with Geeklyhub

#shorts


hey guys this looks like a fun one if

you want to try it on your own pause it

right now cuz I'm going to solve it in 3

2 1 let's connect the radius that goes

from here through this point to here and

that's going to cut everything in half

so we will have two of these blue things

two of these white things and two of

these blue things let's call these two x

these two Y and these two Z and now we

have a way to express our Blue Area it's

going to be equal to 2x + 2 Z so now we

got to do some work to find x and z

first let's look at the entire quarter

Circle the area of a quarter circle is <

r^ 2 / 4 and the radius of this quarter

circle is equal to 4 and 4^ s means 4 *

4 and then we can cancel out these two

fours and now we know the area of this

quarter circle is equal to 4 pi and this

4 Pi is the same thing as two of these

X's plus two of the Y's plus two of the

Z's next we can divide everything by two

on the left hand side all these twos

will cancel each other out and on the

right hand side 4 Pi / 2 is the same

thing as 2 pi and then next next let's

focus on this semicircle let's color it

red and let's find the area of the red

the area of a semicircle is < R 2 / 2

this four is the diameter of the

semicircle which means the radius would

be equal to 2 and 2^ 2 is equal to 2 * 2

and these twos will cancel each other

out so the area of this red region is

equal to 2 pi and this 2 pi is the same

thing as 1X 2x + y and now we can bring

everything else back next let's cut the

semicircle in half this forward split

into 2x two and then this radius wraps

all the way around so this height is

also two now let's focus on this quarter

Circle this x piece plus the triangle

will equal the whole quarter Circle next

we can subtract triangle from both sides

these triangles will cancel each other

out on the left hand side we'll have X

and on the right hand side we'll have

quarter Circle minus triangle the

quarter Circle which is p piun r 2 / 4

the triangle which is base * height / 2

the radius of this quarter circle is

equal to 2 2^ 2 is equal to 4 and these

fours will cancel each other out so the

quarter Circle has an area of Pi and

then for the triangle the base is equal

to 2 and the height is equal to 2 these

twos would cancel each other out so the

area of this triangle is equal to 2 and

now we know that X is equal to Pi - 2 in

the place of this x we can plug in pi -

2 this 2 is going to distribute to both

of these so it becomes 2 piun - 4 next

let's subtract 2 pi from both sides and

let's add four to both sides on the left

hand side these two pies will cancel

each other out these two fours will

cancel each other out and on the right

hand side these two pies will cancel

each other out so we have y is equal to

4 now we know X and Y so we're going to

substitute that into this equation in

the place of the X we'll plug in pi - 2

and in the place of the Y we're going to

plug in four let's combine the -2 and

the four to give us 2 and then we'll do

minus Pi minus 2 from both sides on the

left hand side all of this stuff cancels

out and on the right hand side 2 pi -

piun is 1 pi and now we have Z is equal

to piun - 2 2 now that we know the

values of all three of our variables we

should be able to find the blue area the

blue area is equal to 2x + 2 Z we don't

need the Y anymore and looking at this Z

and X are both equal to Pius 2 so we can

say x is equal to Z and let's substitute

in the place of this Z and X so now we

have 2X + 2x which is equal to 4X and in

the place of the X let's plug in pi - 2

and then this four can distribute to

both of those to give us 4 Pi - 8 so we

now know that the blue area is equal to

4 piun - 8 square units that's the

answer to our question so let's put a

box around it how exciting

Algebraic Approach to this one

I wanted to try this one with this algebraic approach. It was fun to go through.