Your next math quiz.

M508 · Oct 5, 2020

If your neighbor has 9 oranges and 3 cars all adding up to the same length what is the distance from the white house to the Sun?

1.Yes
2.19

Space · Oct 5, 2020

uh am gonna go for all of the above.

DeerFat · Oct 5, 2020

If you are driving 30km/h for 2 hours, how big is the car?

Wavie · Oct 5, 2020

If you have 3 busses and two apples is it true or false that the empire state building is taller then a normale apple % by the dubi state tower and how many cars can you fit in on house yes or no?

EchoFX · Oct 5, 2020

Math question worth all the marks: "What is the principal sum of a bond purchased at 120% of it's initial cost if the first three years of interest were respectively: 3.8%, 4.5% and 6%." (Don't forget that interest is compounded annually) Good luck, it's a real scenario.

The answer is a percentage, FYI.

heidi · Oct 5, 2020

I saw math in the title and my brain just shut off lol

EchoFX · Oct 5, 2020

Here's one more:

Ebba paid $110,000 for a segregated fund with 75% maturity guarantee. At maturity, in addition to the market value of the fund, she received $5,000 as a top-up. Calculate the market value of the fund at maturity.

a) $77,500

b) $82,500

c) $105,000

d) $110,000

snn · Oct 6, 2020

y=mx+b

florinev · Jul 4, 2021

As soon as the word mathematics appears in my field of vision, I turn off all the ability to think calmly. I remember numerous additional classes that I took to prepare for exams. It was an absolute horror. All the problems and equations seemed terribly complicated to me. And your question, in general, would cause me depression if I got caught on the exam. Haha, but I understand that this is a fake test. It's better to already take tests like am I pretty or ugly on websites than to remember mathematics, you know. You can't change me.

Exvious · Jul 7, 2021

EchoFX said:
Math question worth all the marks: "What is the principal sum of a bond purchased at 120% of it's initial cost if the first three years of interest were respectively: 3.8%, 4.5% and 6%." (Don't forget that interest is compounded annually) Good luck, it's a real scenario.

The answer is a percentage, FYI.

found the business major what a fuckin nerd

TDestroyer28 · Aug 21, 2022

All of the above.

TinyMuffinLord · Aug 21, 2022

Simple, 42

Hero · Aug 22, 2022

this is easy,

$\frac{\partial\mathcal{L}}{\partial x_i}=\frac{d}{dt}\frac{\partial\mathcal{L}}{\partial \dot{x}_i}$

This is called the Euler-Lagrange equations (plural) because this is actually several equations. Each different variable (x1=x, x2=y, x3=z) tells you something different. In regular ol’ calculus, if you want to find the value of x that extremizes a function f(x), you solve

$\frac{df}{dx}=0$

for the value x. Using the Euler-Lagrange equations is philosophically similar: to find the path that extremizes S, you solve

$\frac{\partial\mathcal{L}}{\partial x_i}=\frac{d}{dt}\frac{\partial\mathcal{L}}{\partial \dot{x}_i}$

for the path

$\vec{x}(t)$

.

The Lagrangian from earlier, for a free-falling object near the surface of the Earth, is:

$\mathcal{L}=\frac{1}{2}m\left|\dot{\vec{x}}(t)\right|^2-mgz(t)=\frac{1}{2}m\left[\left(\dot{x}(t)\right)^2+\left(\dot{y}(t)\right)^2+\left(\dot{z}(t)\right)^2\right]-mgz(t)$

For z:

$\begin{array}{l}\frac{\partial\mathcal{L}}{\partial z}=-mg\\[2mm]\frac{\partial\mathcal{L}}{\partial \dot{z}}=m\dot{\vec{z}}(t)\\[2mm]\frac{d}{dt}\frac{\partial\mathcal{L}}{\partial \dot{z}}=m\ddot{\vec{z}}(t)\end{array}$

So the E-L equation says:

$m\ddot{\vec{z}}(t)=-mg$

or

$\ddot{\vec{z}}(t)=-g$

In other words, “everything accelerates downward at the same rate”. Doing the same thing for x or y, you get

$\ddot{\vec{x}}(t)=\ddot{\vec{y}}(t)=0$

, which says “things don’t accelerate sideways”. Both good things to know.

You wanna be even slicker, note that this Lagrangian is independent of time. That means that

$\frac{\partial\mathcal{L}}{\partial t}=0$

. Therefore, applying the chain rule:

$\begin{array}{rl}\frac{d\mathcal{L}}{dt}=&\frac{\partial\mathcal{L}}{\partial t}+\sum_i\left(\dot{x}_i\frac{\partial\mathcal{L}}{\partial x_i}+\ddot{x}_i\frac{\partial\mathcal{L}}{\partial \dot{x}_i}\right)\\[2mm]=&\sum_i\left(\dot{x}_i\frac{\partial\mathcal{L}}{\partial x_i}+\ddot{x}_i\frac{\partial\mathcal{L}}{\partial \dot{x}_i}\right)\end{array}$

But we have the E-L equations! Plugging those in:

$\begin{array}{rl}=&\sum_i\left(\dot{x}_i\frac{d}{dt}\frac{\partial\mathcal{L}}{\partial \dot{x}_i}+\ddot{x}_i\frac{\partial\mathcal{L}}{\partial \dot{x}_i}\right)\\[2mm]=&\sum_i\frac{d}{dt}\left(\dot{x}_i\frac{\partial\mathcal{L}}{\partial \dot{x}_i}\right)\end{array}$

And therefore:

$\frac{d}{dt}\left(\sum_i\dot{x}_i\frac{\partial\mathcal{L}}{\partial \dot{x}_i}-\mathcal{L}\right)=0$

This thing in the parentheses is constant (since it never changes in time). In the case of

$\mathcal{L}=\frac{1}{2}m\left[\left(\dot{x}\right)^2+\left(\dot{y}\right)^2+\left(\dot{z}\right)^2\right]-mgz$

we find that this constant thing is:

$\begin{array}{rl}&\sum_i\dot{x}_i\frac{\partial\mathcal{L}}{\partial \dot{x}_i}-\mathcal{L}\\[2mm]=&\left[\dot{x}\frac{\partial\mathcal{L}}{\partial \dot{x}}+\dot{y}\frac{\partial\mathcal{L}}{\partial \dot{y}}+\dot{z}\frac{\partial\mathcal{L}}{\partial \dot{z}}\right]-\mathcal{L}\\[2mm]=&\left[\dot{x}(m\dot{x})+\dot{y}(m\dot{y})+\dot{z}(m\dot{z})\right]-\left[\frac{1}{2}m\left[\left(\dot{x}\right)^2+\left(\dot{y}\right)^2+\left(\dot{z}\right)^2\right]-mgz\right]\\[2mm]=&\frac{1}{2}m\left[\left(\dot{x}\right)^2+\left(\dot{y}\right)^2+\left(\dot{z}\right)^2\right]+mgz\end{array}$

Scarox93 · Aug 24, 2022

Lunar said:
this is easy,

$\frac{\partial\mathcal{L}}{\partial x_i}=\frac{d}{dt}\frac{\partial\mathcal{L}}{\partial \dot{x}_i}$

This is called the Euler-Lagrange equations (plural) because this is actually several equations. Each different variable (x1=x, x2=y, x3=z) tells you something different. In regular ol’ calculus, if you want to find the value of x that extremizes a function f(x), you solve
$\frac{df}{dx}=0$
for the value x. Using the Euler-Lagrange equations is philosophically similar: to find the path that extremizes S, you solve
$\frac{\partial\mathcal{L}}{\partial x_i}=\frac{d}{dt}\frac{\partial\mathcal{L}}{\partial \dot{x}_i}$
for the path
$\vec{x}(t)$
.

The Lagrangian from earlier, for a free-falling object near the surface of the Earth, is:

$\mathcal{L}=\frac{1}{2}m\left|\dot{\vec{x}}(t)\right|^2-mgz(t)=\frac{1}{2}m\left[\left(\dot{x}(t)\right)^2+\left(\dot{y}(t)\right)^2+\left(\dot{z}(t)\right)^2\right]-mgz(t)$

For z:

$\begin{array}{l}\frac{\partial\mathcal{L}}{\partial z}=-mg\\[2mm]\frac{\partial\mathcal{L}}{\partial \dot{z}}=m\dot{\vec{z}}(t)\\[2mm]\frac{d}{dt}\frac{\partial\mathcal{L}}{\partial \dot{z}}=m\ddot{\vec{z}}(t)\end{array}$

So the E-L equation says:

$m\ddot{\vec{z}}(t)=-mg$
or
$\ddot{\vec{z}}(t)=-g$

In other words, “everything accelerates downward at the same rate”. Doing the same thing for x or y, you get
$\ddot{\vec{x}}(t)=\ddot{\vec{y}}(t)=0$
, which says “things don’t accelerate sideways”. Both good things to know.

You wanna be even slicker, note that this Lagrangian is independent of time. That means that
$\frac{\partial\mathcal{L}}{\partial t}=0$
. Therefore, applying the chain rule:

$\begin{array}{rl}\frac{d\mathcal{L}}{dt}=&\frac{\partial\mathcal{L}}{\partial t}+\sum_i\left(\dot{x}_i\frac{\partial\mathcal{L}}{\partial x_i}+\ddot{x}_i\frac{\partial\mathcal{L}}{\partial \dot{x}_i}\right)\\[2mm]=&\sum_i\left(\dot{x}_i\frac{\partial\mathcal{L}}{\partial x_i}+\ddot{x}_i\frac{\partial\mathcal{L}}{\partial \dot{x}_i}\right)\end{array}$

But we have the E-L equations! Plugging those in:

$\begin{array}{rl}=&\sum_i\left(\dot{x}_i\frac{d}{dt}\frac{\partial\mathcal{L}}{\partial \dot{x}_i}+\ddot{x}_i\frac{\partial\mathcal{L}}{\partial \dot{x}_i}\right)\\[2mm]=&\sum_i\frac{d}{dt}\left(\dot{x}_i\frac{\partial\mathcal{L}}{\partial \dot{x}_i}\right)\end{array}$

And therefore:

$\frac{d}{dt}\left(\sum_i\dot{x}_i\frac{\partial\mathcal{L}}{\partial \dot{x}_i}-\mathcal{L}\right)=0$

This thing in the parentheses is constant (since it never changes in time). In the case of
$\mathcal{L}=\frac{1}{2}m\left[\left(\dot{x}\right)^2+\left(\dot{y}\right)^2+\left(\dot{z}\right)^2\right]-mgz$
we find that this constant thing is:

$\begin{array}{rl}&\sum_i\dot{x}_i\frac{\partial\mathcal{L}}{\partial \dot{x}_i}-\mathcal{L}\\[2mm]=&\left[\dot{x}\frac{\partial\mathcal{L}}{\partial \dot{x}}+\dot{y}\frac{\partial\mathcal{L}}{\partial \dot{y}}+\dot{z}\frac{\partial\mathcal{L}}{\partial \dot{z}}\right]-\mathcal{L}\\[2mm]=&\left[\dot{x}(m\dot{x})+\dot{y}(m\dot{y})+\dot{z}(m\dot{z})\right]-\left[\frac{1}{2}m\left[\left(\dot{x}\right)^2+\left(\dot{y}\right)^2+\left(\dot{z}\right)^2\right]-mgz\right]\\[2mm]=&\frac{1}{2}m\left[\left(\dot{x}\right)^2+\left(\dot{y}\right)^2+\left(\dot{z}\right)^2\right]+mgz\end{array}$

I refer you, sir, to the first line of your signature as my response.

Your next math quiz.

M508

I made one post

Space

eeeee

DeerFat

EGO Is My Life!

Wavie

soundcloud.com/waviethekid

EchoFX

Echo Effects

heidi

Mindless Self Indulgence

EchoFX

Echo Effects

snn

Poster

florinev

I made one post

Exvious

Poster

TDestroyer28

Rookie

TinyMuffinLord

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Hero

:3

Scarox93

The (Retired) Guardian of 2Fort's Sewers

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