help-23

[eternal steppe]

well his gone

[safe radish]

@eternal steppe Has your question been resolved?

[eternal steppe]

.close

[safe radish]

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[minor barn]

Can someone please help me with this excersise where we have to use the method of Kirchhoff to find the currents?

[minor barn]

I've tried to do it, but I think I'm doing it all wrong

[tight void]

What are the equations

[minor barn]

the 3 equations at the bottom are the solutions

[tight void]

oaky

so let's start off with labeling the currents

[minor barn]

like this right

[tight void]

make sure you write down I_1, I_2, and I_3

on the corresponding currents

and the direction for I_1 is not an convenient one

$17-I_1\left(1\right)-I_2\left(5\right)=0$

[flat frigate]

nosqldb

[tight void]

we get equation 1

[tight void]

nope!

[minor barn]

oh

how do I do it then?

[tight void]

let me attempt to solve the question first

just in case I'm a lil too rusty 😭

[minor barn]

haha ok no worries

[tight void]

$37-I_1\left(1\right)-I_3\left(5\right)=0$

[flat frigate]

nosqldb

nosqldb

[tight void]

$I_1 - I_2 - I_3 = 0$

[flat frigate]

nosqldb

[tight void]

$I_1 = I_2 + I_3$

[flat frigate]

nosqldb

[tight void]

okay

so I got the right answer now

ur labeling is not correct

[minor barn]

How do I correct it?

[tight void]

so the second and third arrows look good

in directio

but the first one

change the direction

and then label it $I_1, I_2, and I_3$

[flat frigate]

nosqldb

[minor barn]

The one above U3 right?

[tight void]

no change the direction for the one with U_1

and then ur good

send a pic again

[minor barn]

Like this?

[tight void]

QUE PASO chico

let me fix it 😭

[minor barn]

so angry you started speaking spanish 💀

[tight void]

@minor barn

[minor barn]

Oh I see

[tight void]

now use the junction rule

here

$I_1 - I_2 + I_3 = 0$

[flat frigate]

nosqldb

[minor barn]

wait

[tight void]

yep

good

now $17 - R_1(I_1) - 5(I_2) = 0$

[flat frigate]

nosqldb

[tight void]

I'll let you find the other equation involving the other parallel component

[minor barn]

How did u know that this equation was correct? isnt I_1 negative as well?

$20 - 5(I_2 - I_1) - 5(I_2) = 0$

$(I_2) = 4$

$(I_1) = 4$

[flat frigate]

RR

[minor barn]

@tight void This is correct right?

$5(I_2 - I_1) - 5(I_2) = 20 <=> 5(I_2) - 5(I_1) - 5(I_2) = 20$

[flat frigate]

RR

[safe radish]

@minor barn Has your question been resolved?

[regal imp]

$5(I_2 - I_1) - 5(I_2) = 20 <=> 5(I_2) - 5(I_1) - 5(I_2) = 20$

[minor barn]

?

[regal imp]

How did you use the bot?

[minor barn]

you have one extra space between 20 and <=>

$5(I_2 - I_1) - 5(I_2) = 20 <=> 5(I_2) - 5(I_1) - 5(I_2) = 20$

[flat frigate]

RR

[minor barn]

$5(I_2 - I_1) - 5(I_2) = 20 <=> 5(I_2) - 5(I_1) - 5(I_2) = 20$

[regal imp]

$5(55)x(55)

/:

[minor barn]

you must use the dollar sign on the end as well

[regal imp]

Ohhhhhhhh

[minor barn]

$5(55)x(55)$

[flat frigate]

RR

[regal imp]

Thanks

[minor barn]

Do you know how to solve this excersise?

[regal imp]

Yeah

Ofc

[minor barn]

can you help me?

[minor barn]

Is this correct?

[regal imp]

Lemme check

Should be

How do I do a square root sign to the bot?

[minor barn]

So how do I get I_1 = 2A as an answer then?

[minor barn]

i think its

$sqrt(1)$

[flat frigate]

RR

[minor barn]

idk

[regal imp]

Lemme see

[tight void]

I have to go but @minor barn

here are the systems of equation

swith a = I_1

b = I_2

c = I_3

hopefully 3x3 can help explain my solution

[regal imp]

$sqrt(100)$

[flat frigate]

3x3 cuber \learnt CFOP

[regal imp]

=?

[tight void]

$\sqrt(100)$

[flat frigate]

nosqldb

[regal imp]

Ty

$\sqrt(100)=10$

[flat frigate]

3x3 cuber \learnt CFOP

[minor barn]

Wait

is my solution correct though?

[regal imp]

$\sqrt(100)=1849?$

[flat frigate]

3x3 cuber \learnt CFOP

[regal imp]

S,kxd. Xdoxfxf

XD

Xfxf

so

[minor barn]

it would be cool if someone could help me

[minor barn]

<@&286206848099549185> can anyone please help me?

[sinful zodiac]

WHAT ARROWS ARE YOU USING

SO WE'RE ON THE SAME PAGE

[minor barn]

What?

the numbers at the bottom are the solutions

[sinful zodiac]

OJ

FIRST WE SPLIT IT UP INTO LOOPS

AND FOR EACH LOOP, WE MAKE AN EQUATION USING KIRCHOFF'S VOLTAGE LAW (KVL)

YOU KNOW HOW TO DO THAT?

@minor barn

[minor barn]

I think for this excersise, we do

$17 = -(I_1) -5(I_5)$

[flat frigate]

RR

[minor barn]

and for the right loop we do

$20 = -5(I_2) - 5(I_2 - I_1)$

[flat frigate]

RR

[minor barn]

right?

[sinful zodiac]

YE

NOW WE HAVE 2 EQUATIONS WITH 2 UNKNOWN VARIABLES

SO WE CAN SOLVE SIMULTANEOUSLY

[minor barn]

Alright and to do that we multiply all this here with 5 right

so we can cancel out I_1 from both equations

[sinful zodiac]

THATS ONE WAY TO DO IT YEAH

[minor barn]

and if we do that, we get $85 = -25(I_2) and 20 = -10(I_2)$

[flat frigate]

RR

[minor barn]

and that becomes

$105 = -35(I_2)$

[flat frigate]

RR

[sinful zodiac]

85 = -5Q -25P
20 = -10P +5Q

105 = -35P

[sinful zodiac]

YEP

[minor barn]

alright

that becomes

$(I_2) = -3$

[flat frigate]

RR

[minor barn]

I_2 over here is positive 3A

but we have negative 3A

why is that?

[sinful zodiac]

YEAH YEAH, LET'S CONTINUE

[sinful zodiac]

ILL EXPLAIN LATER

[minor barn]

Ok

[sinful zodiac]

LETS FIND I_1 AND I_3

[minor barn]

to find I_1 we can use this equation

$20 = -5(-3) - 5(-3 - I_1)$

[flat frigate]

RR

[minor barn]

$20 = 15 + 15 + 5(I_1)$

[flat frigate]

RR

[minor barn]

and that eventually becomes

$-2 = I_1$

[flat frigate]

RR

[sinful zodiac]

NOW, I_3

[minor barn]

I_3 = I_2 - I_1

so -3 -2 = -1

$I_3 = -1$

[flat frigate]

RR

[sinful zodiac]

YE

ALRIGHT, NOW IMMA SAY WHY ALL OUR VALUES WERE NEGATIVE

FOR CIRCUIT DIAGRAMS, THE LONG TERMINAL OF THE BATTERY IS +VE , AND THE SHORT TERMINAL IS -VE

CONVENTIONAL CURRENT, (WHICH GIVES A VALUE OF +17V) GOES FROM POSITIVE TO NEGATIVE

[minor barn]

So if there is an excersise with the kirchhoff method

and our circuit is looped like that,

[sinful zodiac]

SAME HERE

[minor barn]

then our values become negative and we have to take the absolute value of it

[sinful zodiac]

MM, NOT EXACTLY

STILL HAVE MORE TO SAY

IF WE WEREN'T USING THE CONVENTIONAL THING, AND WENT FROM -VE TO +VE, (THE RED ARROW) , WE WOULD GET -17V FOR THE EMF OF THE BATTERY

DIFFERENT DIRECTION OF LOOP = -VE OF EMF VALUE

THESE ARE THE EQUATIONS WE STARTED WITH

AND IF WE USED CONVENTIONAL CURRENT I_1 WOULD POINT UP THERE, AND I_2 WOULD POINT DOWN
(CONSIDERING ONLY THIS LOOP)

[tight void]

no way you're still doing this 😭

[sinful zodiac]

V=IR FOR ANY PARTICULAR RESISTOR

[minor barn]

You didnt finish helping : (

[sinful zodiac]

AND WE GOT THIS, WHICH MEANS WE ASSIGNED THE CURRENT I_1 TO BE POINTING DOWNWARDS

[minor barn]

I see

[sinful zodiac]

SOO

WHEN WE GOT A -VE VALUE FOR CURRENT, THAT JUST MEANT WE ASSIGNED IT WRONG, IT WAS MEANT TO POINT UP

SAME FOR ALL THE RESISTORS

[minor barn]

And can we ever get a negative current?

Does that mean we did something wrong if we do get one?

[sinful zodiac]

NOT NECESSARILY

[minor barn]

Right

[sinful zodiac]

AT THE BEGINNING OF A QUESTION LIKE THIS, WE HAVE TO RANDOMLY GUESS THE DIRECTION EACH CURRENT IS GOING

IF WE END UP WITH A -VE CURRENT, WE JUST HAVE TO REMEMBER TO SWITCH UP OUR INITIAL ASSUMPTION WHEN WE REACH THE ANSWER

[minor barn]

Ooh I see

thank you!

[sinful zodiac]

🔥

WE ASSUMED ALL 3 , AND WE ASSUMED ALL 3 WRONG

UNLUCKY

[minor barn]

Haha

but luckily we only have to change the sign

[sinful zodiac]

YE

CHANGING SIGN MEANS CHANGING DIRECTION

JUST REMEMBER, WHENEVER UR CHOOSING THE DIRECTION FOR A LOOP

IF IT GOES +VE TO -VE USE 17V
AND IF IT GOES -VE TO +VE USE -17V

[minor barn]

Got it

[safe radish]

@minor barn Has your question been resolved?

[safe radish]

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[fluid spoke]

how to find inflection point

[fluid spoke]

already found local min and max

how find inflection

[fleet tendon]

find the second derivative

look for zeros and see if the sign changes on each side

[fluid spoke]

ohhh

is the inflection point where concavity changes

[fleet tendon]

yes

[fluid spoke]

ok

i didn't remember the name

thanks

.close

[safe radish]

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[broken nimbus]

How do you present a counter example for a proof for one-to-one / onto?

[plucky elk]

!original

[safe radish]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[broken nimbus]

Just a general question

[plucky elk]

By showing it doesn't satisfy the definition

[fickle pendant]

if the question is "prove f(x) is not inj/surjective" then you're looking for an example of f(a) = f(b) for a /= b or an x such that you have no a f(a)=x

[broken nimbus]

Gotcha thanks

.close

[safe radish]

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[lyric pilot]

having trouble with this problem in my textbook (calc 3)

[lyric pilot]

specifically the part where it says costheta = 0 is the y-axis... why would it be the y-axis?

what would the level curve for costheta = 1 be?? and costheta = -1

[hardy lion]

Solve for cos(theta)=0

What values of theta do you get and how does that correspond to the xy-plane?

[safe radish]

@lyric pilot Has your question been resolved?

[safe radish]

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[lean otter]

ive already messed something up and idk what

[main mural]

polynomial division very wrong

shhh no spoilers

i just wanted to see the factorisation not the roots lol

[delicate sphinx]

Where's your coefficient for the x^3 term?

[lean otter]

huh its synthetic division i was tryna do

[lean otter]

1?

[delicate sphinx]

Wait did you break x^3 -6x^2 + 57x - 98 to be (x - 2)(6x^2 + 57x - 98)?

[lean otter]

yeah

[delicate sphinx]

That's not how that works

[lean otter]

https://www.youtube.com/watch?time_continue=14&v=uc7sg106MGI&embeds_referring_euri=https%3A%2F%2Fwww.myopenmath.com%2F&source_ve_path=MjM4NTE

this what i was following

[delicate sphinx]

You missed the fact that it did long division with on the cubic

[main mural]

also that

[lean otter]

oooh whoops

wait ima try sec

[delicate sphinx]

It didn't do x^3 - 12x^2 + 49x - 70 to be (x - 6)(-12x^2 + 49x - 70)

[lean otter]

I know

i was in the middle of tryna find what actually goes there

thats why i was tryna do the synthetic division thing

then that answer would go after the (x-2) right?

[delicate sphinx]

Yes

[lean otter]

So ima re do it with the 1

ye thats what was wrong

i thought he had skipped the cube coefficient for some reason

but is there a better way to do this?

[delicate sphinx]

Better way how?

[lean otter]

idk i seen someone say something about ugh not that way lol

but its wtvr lol ill just do it this way thanks for catching the mistake

[delicate sphinx]

If you can't use a calculator, that's the more ideal way to do it

[lean otter]

yeah i do but teacher wants the work smh

cuz like yk we could just use program lol

and what he mean by simplified square roots

just like not get the square root of the number?

[delicate sphinx]

Like $\sqrt{20} = 2\sqrt{5}$

[flat frigate]

CaptainNova22

[lean otter]

oh thats dumb

alr

[lean otter]

so is 2-7i and 2+7i correct?

[delicate sphinx]

No

Show your work

What was the quadratic after you did the long division?

[delicate sphinx]

You applied the quadratic formula wrong

Under the sqrt, it's b^2 - 4ac

And b = -4

So b^2 is (-4)^2 = 16

[lean otter]

huh my calc saying -16

[delicate sphinx]

$-4^2 \neq (-4)^2$

[flat frigate]

CaptainNova22

[delicate sphinx]

You need parentheses

[lean otter]

god i hate math

ok ima solve it once again

3rd times the charm

[delicate sphinx]

The quadratic is fine, you just did b^2 wrong

So from that step and on, is where you need to fix

[lean otter]

so im getting a god awful ugly number now

cuz then i get 180

and square root it

13.41640786

or is this where he means leave it

[delicate sphinx]

It shouldn't be 180

b^2 - 4ac is (-4)^2 - 4 * 1 * 49 = -180

So you should have sqrt(-180)

And you can simplify sqrt(-180)

[lean otter]

yeah neg 180

i got that

but then i change it to positive right

and add i

[delicate sphinx]

Not just that

You can simplify sqrt(180)

[delicate sphinx]

Using this concept

[lean otter]

i dont get that

[delicate sphinx]

https://youtu.be/BPY7gmT32XE?si=0He1LD9Xuu4-XqkU&t=212
Try watching this, and it's like 5 minutes from where I linked

That's how you can simplify radicals

[lean otter]

4 and 45 is wrong?

[delicate sphinx]

You want the biggest perfect square that can go into 180

[lean otter]

ong he didnt teach this to us

[delicate sphinx]

4 and 45 isn't wrong but you can simplify 45 more

What's a perfect square that can go into 45?

[lean otter]

5 and 9

[delicate sphinx]

So 180 = 4 * 9 * 5

9 * 4 = 36

So the largest perfect square in 180 is 36

Then you can continue the process from the video

[lean otter]

so 6 sqrt(5)

[delicate sphinx]

Close, don't forget the i

Because it was sqrt(-180)

[lean otter]

yeah but thats right for the simplified?

[delicate sphinx]

Yes

[lean otter]

thank god

ok so when i go to devide by 2

what happens with them

divide

[delicate sphinx]

Well what do you have now?

[delicate sphinx]

Notice how you can divide by 2 and simplify that fraction

[delicate sphinx]

Kinda like what you did here

[lean otter]

but i cant devide a square root right

[delicate sphinx]

You can't divide into the sqrt but you can divide the 6 that's outside of it

[lean otter]

and the square root of 5 just doesnt change or

[delicate sphinx]

Does not change

[lean otter]

so 4+3(sqrt(5i))

is that right

[delicate sphinx]

No

First the i is outside

[lean otter]

[delicate sphinx]

And second the 4 can be simplified

[lean otter]

oh yeah duh

2

So is that final

[delicate sphinx]

I suggest writing it as $2 + 3i \sqrt{5}$

[flat frigate]

CaptainNova22

[delicate sphinx]

So that way it's clear that the i isn't under the root

[lean otter]

Alr

[delicate sphinx]

And you forgot one more thing

[lean otter]

The minus one

Right

[delicate sphinx]

It's plus and minus

[lean otter]

Ik teacher said to write it separate

Otherwise he would get a bunch of lazy answers that are right but like not done or something like that

Idk

Is this finally right lol

[delicate sphinx]

Yes

And don't forget the x = 2, since that was given at the very beginning

[lean otter]

Right

[delicate sphinx]

Because you were finding the roots to the cubic

[lean otter]

I’m so failing this lmao

Welp ima go do it all over again and hopefully get the right thing 😂

[delicate sphinx]

Definitely practice

I think khan academy should have practice on simplify radicals

[lean otter]

He never went over this whole simplify square root thing

It’s the practice test I’m praying it’s just not in the actual test

I got one last question

I always get lost on which thing i gotta do to get which answer

like one of the answer i do the whole quadratic formula thing

like this one for example

like b and c

[delicate sphinx]

It's asking for what time it will make splash down so that means you are solving for t, so you set that equation equal to 0, and use the quadratic formula

[delicate sphinx]

Max height is the vertex because the vertex is the highest or lowest point in a parabola

[lean otter]

yeah i remeber that part

[delicate sphinx]

And how far horizontally is setting the equation equal to 0 and solve for x

[lean otter]

its just so many rules on when to do one thing to get an answer and when to do another thing like

idk

just hard to keep it all straight

[safe radish]

@lean otter Has your question been resolved?

[safe radish]

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[tawdry salmon]

Yo

[tawdry salmon]

Im confused

I did this

But I did not get this

[hardy lion]

What did you get?

[tawdry salmon]

Is this what I was supposed to get

sqrt3 - 1

1 + sqrt3

[hardy lion]

Multiply the numerator and denominator by sqrt 3

[tawdry salmon]

What for

You arent rationalizing ??

Theres still square root in denominator right

[hardy lion]

Idk, thats just how they roll i guess

[tawdry salmon]

But the top doesnt match up

They have -sqrt while mines -1

[hardy lion]

(sqrt3)*(sqrt(3)-1) is?

[tawdry salmon]

Ok

Thanks

[hardy lion]

Yup np 👍

[tawdry salmon]

Again

I have cos195

I split it into cos(240 - 45)

cos240 * cos45 + sin240 * sin45

1/2 * sqrt2/2 + sqrt3/2 * sqrt2/2

Is there anything wrong with this

They expected this

Oh

Mine have a plus

Theirs a minus

Huh

Ok I see

Its at 240*

[safe radish]

@tawdry salmon Has your question been resolved?

[safe radish]

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[atomic zodiac]

If when finding the limit as x approaches infinity, and the degree of the coefficient on the numerator is greater than the degree of the coefficient on the denominator, the limit is infinity, then why is this answer - infinity?

[stoic dune]

Your memorized rule is wrong

Consider dividing top and bottom by x, then plugging in inf

[atomic zodiac]

yeah but

i know how to do it that way

but is there like a certain exception for the rule?

[stoic dune]

The wrong one?

[atomic zodiac]

wym the wrong one?

here look

https://www.youtube.com/shorts/-gqa8S3qVxQ

[stoic dune]

I'll let somebody else watch that if they need to. I'll bow out here and reiterate: Your memorized rule is wrong

[atomic zodiac]

ok

then what is the "right" procedure

[little mesa]

you need to factorise the coefficients of the leading terms out first

[little mesa]

...to get the answer you would normally get

[safe radish]

@atomic zodiac Has your question been resolved?

[atomic zodiac]

wym?

sorry

<@&286206848099549185>

[atomic zodiac]

.close

[safe radish]

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[tawdry salmon]

at the bottom

Does the minus distribute to both tana and tanb

[rotund valley]

yes

[tawdry salmon]

Shit why?

[rotund valley]

because tanA and tanB are being multiplied

[tawdry salmon]

Doesnt make sense

[rotund valley]

so the substraction sign applies to both of them

[tawdry salmon]

Why dont u multiply them first

Then do the minus

Is that same thing

[rotund valley]

that would be equivalent to 1+(-TanA*TanB)

[rotund valley]

it would be the same thing

your not subtracting just the TanA from 1

your subtracting the product of TanATanB from 1

this is the step right before converting to Tan A and Tan B

you can see that its the product of TanA and TanB subtracted from 1

[queen parcel]

= (-1)(tan A)(tan B)

[tawdry salmon]

So the way to convert from tana to sin/cos is just divide sina/cosa ok

[queen parcel]

tan A = sin A / cos A

correct

[tawdry salmon]

I know thats so obvious lmao

What do you do if you get like sin(5)

Do you just do same as usual

Reference angle

Is 85

Oh shit

Now what

We should already have triangle of a actually

so I guess we can sketch that triangle and get sin of it

[safe radish]

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[rapid delta]

$$z^7 = 1$$
(a) Find the the roots of $z^7 = 1$ in mod-arg form
\ Done. z = 1, $\text{cis}(\pm\frac{2\pi}7), \text{cis}(\pm\frac{4\pi}7), \text{cis}(\pm\frac{6\pi}7),$
\
(b) If w is a non-real root, show that $w+w^2+w^3+w^4+w^5+w^6 = -1$
\Done. Using sum of roots $(\frac{-b}a) = 0$
\ \
\textcolor{red!40}{(c) Show that the quadratic equation $z^2 + z + 2 =0$ has roots $w+w^2+w^4$ and $w^3 + w^5 + w^6$
And this is where im stuck, i tried to solve for z in qf but got $-\frac12 \pm 7i$ and have no idea what to do with that}

[flat frigate]

魔法の🌙kitty!

[drowsy karma]

you incorretly solved this quadratic function

[rapid delta]

o

[drowsy karma]

not 7 but its sqaure rrot

$-\frac{1}{2}\pm i\frac{\sqrt{7}}{2}$

[rapid delta]

oh trueee

also $-\frac12\pm\frac{i\sqrt7}2$ i think

[flat frigate]

魔法の🌙kitty!

[drowsy karma]

yes yes

)

[rapid delta]

now........

idk 😇

what to do with it

[flat frigate]

Joanna Angel

[drowsy karma]

ok you can name those w sums

like x and y

and if they are rots

roots*

they shud follow the given quadratic equation

[rapid delta]

name the w sums x and y?

[drowsy karma]

$\text{ let }x = w+w^{2}+w^{4}$

[flat frigate]

Joanna Angel

[drowsy karma]

etc

[rapid delta]

do i plug x into the equation $z^2+z+2$ to give $x^2+x+2=0$, then by solving for x its the same eq as this

[flat frigate]

魔法の🌙kitty!

[drowsy karma]

we knmow that x is the solution of the given equaiton

so x must be eual to roots you have fond before

you may also use the analogus mehtod you did in (b), using Viete'as or just sum of what you have foudn you get sum of roots = -1

form other side,, x + y = -1, too

[rapid delta]

ohhh alr ty i got it !!!

i said $x = w + w^2 + w^4$ and $y = x^3 + w^5 + w^6$, from (b) then $x+y=-1$
\ and when $x = -\frac12+\frac{i\sqrt7}2, y = -\frac12-\frac{i\sqrt7}2$, then $x+y$ also $= -1$
\ hence $x, y$ are roots of $z^2+z+2$

[flat frigate]

魔法の🌙kitty!

[rapid delta]

@drowsy karma does this seem right?

[drowsy karma]

right, because you have proved in b) that x + y = -1, usng my variables

but in same time

there exist Viete'a formuals

and sum fo -1/2 +-isqr etcc

si also equakl to -1

so both methods are the same

[rapid delta]

mmkkk tyy

[drowsy karma]

in other way,m c ) was an appplicaiton

waht you di din a ) and b)

[rapid delta]

ur typing is quickly decaying

n e way i think i got it tyty

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[restive parrot]

https://gyazo.com/98fbc9217298475493e89d8da16822ba
how do i even begin to find set and number

[main laurel]

You guess, like really, just guess and make adjustments. First guess could be A=[0,5]

[safe radish]

@restive parrot Has your question been resolved?

[restive parrot]

|A| = 5

[restive parrot]

<@&286206848099549185>

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@restive parrot Has your question been resolved?

[restive parrot]

<@&286206848099549185>

does this even ping anyone

.close

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[orchid agate]

Hi is there anyone who knows how to solve derivative of this function?

[icy lance]

probably some chainy product rule

let $u=e^x+sin(x)$
$$y=u^u$$
$$y=e^{uln(u)}$$

[flat frigate]

AℤØ

[orchid agate]

ok and than after derivating whole exponent what is the next step?

[icy lance]

you need dy/du and du/dx
dy/du is also a chain rule but you can probably just do it in your head

[orchid agate]

ok thanks

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[shy glade]

How did the conversion from the first to second equation happen?

[safe radish]

@shy glade Has your question been resolved?

[safe radish]

@shy glade Has your question been resolved?

[plucky elk]

Do you know product rule?

,tex .diff rules

[flat frigate]

riemann

[shy glade]

I know the product rule yes

[plucky elk]

$d(fg) = (df)g + f(dg)$ and then use chain rule

[flat frigate]

riemann

[shy glade]

Uhh okay but how does that d(fg)= equation work? I don't know how to do that

[plucky elk]

It doesn't make sense until you take multivariate calculus

https://math.libretexts.org/Bookshelves/Calculus/Calculus_3e_(Apex)/12%3A_Functions_of_Several_Variables/12.04%3A_Differentiability_and_the_Total_Differential

[shy glade]

I'm learning differential equation do I learn them after this?

[plucky elk]

First couple paragraphs

Uhhh depends on your university . Many people take multivariate calculus before differential equations

[shy glade]

Oh i probably did not then

[plucky elk]

Doesn't take that long to read and do an example, maybe 10 minutes

[shy glade]

and aint gonna take it probably since cs degree

[shy glade]

ok i will read it thanks

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[minor barn]

The excersise question: "The box has a weight of 550N. Decide the force in each cable which carries the box."

Can someone walk me through on how to do this please?

[frigid spruce]

what have u tried so far

[minor barn]

So I calculated the angle between A and C, which is 36.87°

Fy should be equal to 550N, so I know the vertical force

and thats it

I don't know how to continue anymore

I'm stuck

btw those 3 numbers on the right are the solutions

@frigid spruce

[frigid spruce]

5 min plj, busy rn

[minor barn]

ok

[frigid spruce]

ok so

equate thier vertical and horizontal components

because it is at equilibrium

[minor barn]

so Fx = Fy?

[frigid spruce]

no

[minor barn]

F(cos(3/5)) = 550

oh

How then?

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@minor barn Has your question been resolved?

[safe radish]

@minor barn Has your question been resolved?

[minor barn]

<@&286206848099549185>

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@minor barn Has your question been resolved?

[sinful zodiac]

THE BOX ISN'T IN MOTION, SO WE SAY THE FORCES IN THE SYSTEM ARE IN EQUILIBRIUM; NEWTON'S 1ST LAW

WHAT THAT MEANS IS:
UPWARD FORCES OF SYSTEM = DOWNWARD FORCES OF SYSTEM
LEFTWARD FORCES = RIGHTWARD FORCES

[sinful zodiac]

THIS SYTEM CAN BE THOUGHT OF AS A POINT (THE INTERSECTION OF THE CABLES) BEING ACTED UPON/PULLED BY 3 FORCES:

WE HAVE 2 UNKNOWNS

AND WE HAVE THE MEANS TO MAKE 2 EQUATIONS

[safe radish]

@minor barn Has your question been resolved?

[minor barn]

But T1 is not equal to T2 because they're both at different angles though?

[sinful zodiac]

YE, THE TENSION IN EACH CABLE IS DIFFERENT BECAUSE OF DIFFERENT ANGLES

[minor barn]

Wait, with leftward forces, do you mean Fx?

oooooh

But my question is, how do I find Fx?

[sinful zodiac]

DO YOU KNOW HOW TO SPLIT A VECTOR INTO ITS X & Y COMPONENTS

THE 550N ONLY ACTS DOWNWARD
T1 HAS A LEFTWARD AND UPWARD COMPONENT
T2 HAS A RIGHTWARD AND UPWARD COMPONENT

[minor barn]

FCos(x) = Fx
FSin(x) = Fy

[minor barn]

Oh I see

but didnt you say upward forces = downward forces?

Isn't the upward component for both 550N then?

[sinful zodiac]

BOTH OF WHAT YOU SAID IS CORRECT

[minor barn]

ok

pls continue

[sinful zodiac]

WAIT, YOU MEAN BOTH CABLES?

EACH HAVING 550N UPWARDS?

[minor barn]

Yes

the Fy for each should be 550N right

[sinful zodiac]

NOP

LEMME REPHRASE

FOR A SYSTEM IN EQUILIBRIUM:
SUM OF UPWARD FORCES= SUM OF DOWNWARD FORCES
SUM OF LEFTWARD FORCES = SUM OF RIGHTWARD FORCES

SUM OF UPWARD FORCES= SUM OF DOWNWARD FORCES
T1_Y + T2_Y = 550

[minor barn]

ooooh

[sinful zodiac]

THE FY OF BOTH SHOULD ADD UP TO 550 TO CANCEL THE DOWNWARD, THUS CANCELLING ANY VERTICAL MOTION

[minor barn]

Alright, so how do we find the Fy of one of them?

[sinful zodiac]

THE SAME YOU WOULD DO VECTORS

[minor barn]

But we don't have the F for T1 or T2

do you want to use 550N?

[sinful zodiac]

PPL MAKE FORCE-DIAGRAMS OR FREE-BODY DIAGRAMS SO WE COULD SEE ALL THE VECTORS INVOLVED

[sinful zodiac]

YE

[minor barn]

ok 550sin(36.87°) = 330N

that is Fy for T2

[sinful zodiac]

UHHHH

NOP

T2 DOESNT HAVE MUCH RELATION TO 550N BY ITSELF

[minor barn]

hm, could you show me how you do it then?

[minor barn]

but what is the F???

is it T2

[sinful zodiac]

AND WE DONT HAVE A VALUE FOR T2 SO WE JUST LEAVE IT AS T2

[minor barn]

but it's all uknown variables apart from Sin(x)?

[sinful zodiac]

YE

[minor barn]

hm

ok pls continue

[sinful zodiac]

USING THIS,
T2_X = T2 COS(36.87)
T2_Y = T2 SIN(36.87)

AND WE DO THE SAME FOR T1

THEN WE USE THIS

[minor barn]

so T2Sin(36.87) * T1sin(30) == 550N

[minor barn]

oh right

T2Sin(36.87) + T1sin(30) = 550N

and how do we continue from here?

[sinful zodiac]

WE GOT THIS FROM CONSIDERING THE VERTICAL COMPONENTS

WE NEED ANOTHER EQUATION

[minor barn]

Ok

how do we do that

sorry I know very little about this

[sinful zodiac]

ITS OJ

WE USED THE VERTICAL ONE

SO NOW LETS USE THE HORIZONTAL

550 DOESNT HAVE A HORIZONTAL COMPONENT SO WE DONT USE IT

[minor barn]

Well can't use anything then

T1cos(30) + T2cos(36.87) = x

[sinful zodiac]

IF THERES NOTHING WE JUST PUT 0

BUT THERE IS

ONE IS PULLING LEFT, ONE IS PULLING RIGHT

[minor barn]

ok?

[sinful zodiac]

SUM OF LEFTWARD FORCES = SUM OF RIGHTWARD FORCES
T1_X = T2_X

[minor barn]

hm ok

[minor barn]

So we only have to find 1 of them

T1cos(30°) = T2cos(36.87°)

[sinful zodiac]

NOW WE HAVE 2 EQUATIONS, WITH 2 UNKNOWN VARIABLES

[sinful zodiac]

WHENEVER THERE'S SOMETHING LIKE THIS I IMMEDIATELY THINK TO TRY SIMULTANEOUS EQUATIONS

[minor barn]

we can take one equation

lets take the left one

and do T2 = (550N - T1sin(30)/sin(36.87) right

and then we can substitute the T2 in the right equation with this one

[minor barn]

T1 = 402.62N

[sinful zodiac]

NOT SURE WHERE YOU GOT THAT FROM

[minor barn]

ah damn

I misread something

[sinful zodiac]

TRAGIC 🔥

[minor barn]

T1 = 434.43N

[sinful zodiac]

NOP

THIS 550 - T1SIN(30) IS SUPPOSED TO BE IN A BRACKET

SINCE THIS IS HOW WE GOT IT

[sinful zodiac]

REMOVING THE BRACKET DISRUPTS THIS MULTIPLICATION OF T2 AND COS(36.87)

I HAVE TO GO NOW, BUT THE CORRECT ANSWERS SHOULD CORRELATE WITH THESE

[minor barn]

I did it

thank you so much holy shit

.close

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[leaden scaffold]

Can someone help me with understanding this?

[leaden scaffold]

WHat does it mean to exchage the order of integration when evaluating a double integral?

[tight void]

WOAH

NOT ALWAYS!

what if you have dy dx

but your dy bounds is in terms of x

what do you do then

[fleet condor]

its always true if you ignore the times when its not

[lean thorn]

It's asking if you can always do this $\int \int f(x, y) dxdy$ you can re-write it as $\int \int f(x, y) dydx$

[flat frigate]

MellowDramaLlama

[tight void]

so true bunny

[fleet condor]

throw in some arbitrary bounds to show that the bounds swap as well?

[tight void]

https://tenor.com/view/snowball-bunny-carrot-nuf-nuf-looking-around-gif-24073283

[leaden scaffold]

Keep in mind this is not true and false, it's just a standard I was asked to understand by an exam date because I will be assessed on it.

[lean thorn]

yeah for sure. But that gets complicated without an example because the inner bounds are dependent of the outer bounds :X

[fleet condor]

$\int_a^b \int_c^d f(x, y) dxdy$ you can re-write it as $\int_c^d \int_a^b f(x, y) dydx$

[tight void]

throw in an example of a tetrahedron q

[leaden scaffold]

Do you recommend any place to find exampe=les on it? I've been searching..

[lean thorn]

Ah yeah

[south path]

this just sounds like Fubini's theorem

[lean thorn]

Hey I remember you! I used to have a llama portrait 🙂

[lean thorn]

we chatted like a year ago ahah

[lean thorn]

Oh yeah duh. Yeah if you have something like a square in the x and y plane then there is no issue with changing the bounds. It gets more complicated with tetrahedrons like nosqldb says

[fleet condor]

yeah i was just imagining {0 < x,y < 1}

[tight void]

sonny dear

did that answer ur q

[lean thorn]

my favorite example ❤️

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@leaden scaffold Has your question been resolved?

[leaden scaffold]

Yes! Thanks.

.close

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[winter roost]

Please help guys... I can't find the overlap. The surface area of both shapes combined is 243.42, and the textbook answer of the surface area but minus the overlap is 165.03...

[winter roost]

Jfjgkfigkeoog

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@winter roost Has your question been resolved?

[safe radish]

@winter roost Has your question been resolved?

[misty bay]

remember to not include the floor of the building

the overlap is just the area of the back face of the attached shed

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[oblique fjord]

I'm having trouble with this exercis: Find the inclination 𝜃 (in radians and degrees) of the line with slope m.
m = −1 edit: I keep getting
𝜃 =\frac{7\pi }{4} radians and
𝜃 = 157.5 degrees?

[safe radish]

@oblique fjord Has your question been resolved?

[safe radish]

@oblique fjord Has your question been resolved?

[plucky elk]

Inclination is measured from the positive x axis anti clockwise.

Plot a line of slope -1 and rotate until you hit the line

[safe radish]

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[rigid path]

4.34

:D

[long shore]

ok

[rigid path]

Yes

:D

[long shore]

i just need to type this out so give me a sec

[rigid path]

Lol np

[long shore]

,tex $\frac{3r^2t - 6rt^2}{6r^2t^3} \div \frac{8t^3r - 4t^2r^2}{6r^3t^4}$

[flat frigate]

suds

[long shore]

that is another way to write the equation

so you can use the same change flip way to put it into one fraction

[rigid path]

Oh ic

[long shore]

,tex $\frac{3r^2t - 6rt^2}{6r^2t^3} \times \frac{6r^3t^4}{8t^3r - 4t^2r^2}$

[flat frigate]

suds

[long shore]

do you need more with that?

[rigid path]

Uhh lemme try and solve it

And I'll tell you

[long shore]

alright

[rigid path]

Yeah I have no clue what I'm doing bruh

😭

[hazy elbow]

do you know how to factorize it?

[rigid path]

No

😭

[hazy elbow]

Just for instance:

[long shore]

alright

so first try to multiply out the equation

[hazy elbow]

r^2*t+r*t^2=rt(r+t)

$r^{2}t+rt^{2}=rt\left(r+t\right)$

[flat frigate]

B-eard

[rigid path]

Uh I was not taught factorization 😭

[long shore]

,tex $\frac{(3r^2t - 6rt^2)(6r^3t^4)}{(6r^2t^3)(8t^3r - 4t^2r^2)}$

[frigid spruce]

sorry but ur an undergrad wtf 😭

[rigid path]

Hwhwhst the haill

[rigid path]

Yes bro

[long shore]

i need to fix it gimme a sec

[rigid path]

I am indeed an undergrad

[desert iron]

🤯🤯🤯

[frigid spruce]

and u haven't been taught factorisation

[rigid path]

I am also stupid and my friend let me join this server because she gets precalc help

[rigid path]

Yes I'm in 8th grade

😭

[frigid spruce]

8th grade is undergrad?

[rigid path]

I think so

[long shore]

its pre university

[rigid path]

Now I'm getting confused bruh

[long shore]

might want to change the thingy

[rigid path]

Yeah

Oh thats what I change it to suds?

[long shore]

ok so do you know how to multiply out this equation ive done?

yes

do you want me to show the next step for that

[rigid path]

Uhh ok

[long shore]

well do you know how to multiply those brackets out?

because you do that first

[rigid path]

Uhhh

Oh ok 😭😭

Bare with me here

[flat frigate]

suds

[rigid path]

Uhhhh

So that's

..

[long shore]

,tex $ = \frac{18r^5t^5 - 36r^4t^6}{48r^3t^6 - 24r^4t^5}$

[flat frigate]

suds

[long shore]

do you know how to do that or do you want me to explain?

[rigid path]

You like use I forgot the name of the property uhh it's some type of multiplication one?

Like

[long shore]

expanding?

[rigid path]

Uhhh maybe??

[long shore]

yes

so lets say its (a - b) x c

[rigid path]

Right

[long shore]

you multiply the c value by each thing in the brackets

[rigid path]

It's ac - bc

[long shore]

so its ac - bc

[rigid path]

Right?

[long shore]

yes

[rigid path]

Ye

[long shore]

and when you have exponentials like $r^3$

[flat frigate]

suds

[long shore]

,tex $(r^x)^y$

[flat frigate]

suds

[long shore]

it will be x + y

[rigid path]

That's x times y

Oh nvm

Lol

[long shore]

its different with exponentials

[rigid path]

Ic

[long shore]

wait sorry

no you're right

[rigid path]

😭

[long shore]

i just gave you the wrong example

[rigid path]

Oh it's ok

[long shore]

it is x times y

i mean

[rigid path]

Uhhh

[long shore]

,tex $a^m \times a^n$

[flat frigate]

suds

[long shore]

is m + n

[rigid path]

Ye okk

[long shore]

which is what we do for the equation

so you see how we've expanded it now?

[rigid path]

Kk

[long shore]

are you ready for the next step

[rigid path]

Uh yeah

I think

[long shore]

or do you wanna try multiplying it out yourself without checking what i did

[rigid path]

Is it subtraction?

Uhh sure

[long shore]

alright let me know when you're ready

[rigid path]

Oh ic now

[long shore]

,tex $ = \frac{18r^5t^5 - 36r^4t^6}{48r^3t^6 - 24r^4t^5}$

[flat frigate]

suds

[rigid path]

It's like adding and multiplying ish?

[long shore]

kind of

[rigid path]

Is the best way I can explain it

😭

It's like 6*3 and then you add the exponents

[long shore]

but you cant subtract things with different exponentials

[rigid path]

Yw

Ye

[long shore]

so what you might wanna do is factorise it again

[rigid path]

So you uhh

[long shore]

can you give that a go and tell me what you get

[rigid path]

You change it to like

Kk

[long shore]

take out the like variables

[rigid path]

Uhh

Kinda like this?

[long shore]

almost but not quite

[rigid path]

Oh what did I do wrong😭

[long shore]

you want to take the ones that are the same out the front

[rigid path]

Wut

[long shore]

so 6tr(r + t) is the same as 6tr^2 + 6r^2t

[rigid path]

Ye

Wait so what did I do then in confusion

[long shore]

tbh idk what you did

[rigid path]

Thanks bro

Me neither

😭

[long shore]

but for the top line you can see that both are multiplied by 18 right?

so we can do that first nice and simple

[rigid path]

Wut?

[long shore]

,tex $18(r^5t^5 - 2r^4t^6)$

[flat frigate]

suds

[long shore]

for the top line yes?

because when you mutiply it out it is the same

[rigid path]

Uhh.... idk

Is that the full top line?

[long shore]

no im just doing is slowly so you understand

so then we want to take the rs and the ts out the front

,tex $18r^4t^5(r - 2t)$

[flat frigate]

suds

[long shore]

because thats the most of the r's and the t's you can take oyt

[rigid path]

Ic

[long shore]

ok can you do the bottom line for me then

[rigid path]

Uhhh

Maybe?

[long shore]

give it a go

take your time dont worry

[rigid path]

Ok...

Ok wait I don't understand bruhhh i feel so stupid 😭😭

[long shore]

dont worry you're not stupid

would you be comfortable with calling so i can explain it better?

im a 17yro girl

if not thats fine we can keep messaging here but it might just be easier

[rigid path]

Uhmm my parents don't allow me to call

[long shore]

thats fine dw i get it

[rigid path]

I trust you tho

👌

[long shore]

yeah i wanna make sure you know im not a weirdo i just wanna help

ok so what dont you get

[rigid path]

Factorization

[long shore]

ok let me quickly make up an equation for you

,tex $3a^2b^3 + 6ab^4$

[flat frigate]

suds

[long shore]

lets start by factorising this one

so what do they have in common?

[rigid path]

Uhh

They are both multiples of 3

[long shore]

think of it as dividing