help-19

[forest yoke]

For a moment

Now we know sin 60 degrees is root 3/2

Right?

From the trignometry table

U know the table right?

[mystic saffron]

Okie

[forest yoke]

Ok so

[forest yoke]

Here the input is 60 degrees

And the output is root3/2

Right?

So in an inverse function the
Output and input switch

There fore in the inverse function

[forest yoke]

The INPUT is root3/2

[mystic saffron]

so the input now is (sqrt)3/2

and output is 60

[forest yoke]

Exacy

[forest yoke]

Exactly

Ur right

[mystic saffron]

How does that sovle the equation?

[forest yoke]

We need to find the out put

[forest yoke]

And u just did

So therefore

The solution to that equation is 60 degrees

[mystic saffron]

huh??

[forest yoke]

The solution is just the output

The solution of the equation

Is the OUTPUT

[signal crown]

never forget to work in radians in this kind of exercise

[forest yoke]

Ok bro

60 degrees is pi/3 radians

So ur answer is pi/3 radions ( or just 60 degrees)

The solution of that sum

Is just equal to the out put

[forest yoke]

And u found that your self

Do u follow?

There fore the soln is pi/3 ( or 60 degrees)

Now you can use this ligic in all the other questions

Its simple if you calmly think about it

Did this help u understand how we got the answer?

[mystic saffron]

I dont get it

[forest yoke]

Which part did u not get?

[mystic saffron]

because all im seeing is
sin-1 ((sqrt)3/2)
= 60 degrees

Like I don't think nothing happened

I just need to input 60 degrees

Is that it?

[forest yoke]

Yeah

Thats it

[mystic saffron]

Soo...

all answers are 60 degrees?

[forest yoke]

Yeah?

[forest yoke]

Not all

Only the 1st one

Im trying to explain to you how to think about these sums

I suggest you watch a youtube video on the topic " how to find solutions to inverse functions" from yt

To get a deeper understanding

I have to go eat some food rn

[mystic saffron]

Okie

Imma go do that

Okayyy

Soo

I kinda got it

okay now I don't know

1. 60 degrees

3. 240 degrees

4. 180 degrees

[fiery prawn]

@mystic saffron Can you help me do math hw ? I’m in 8th grade btw

[mystic saffron]

uhhh

1 + 1 is 8

there

[fiery prawn]

<@&286206848099549185>

[fiery prawn]

Wdym ?

[mystic saffron]

Well

This is MY CHANNEL

[molten pine]

Waduh

[mystic saffron]

Obviously no :3

Shoo

6. 135 degrees

1. 60 degrees
2. ?
3. 240 degrees
4. 180 degrees
5. ?
6. 135 degrees
7. ?
8. ?
9. ?
10. ?

WAIT

WHAT THE HECK

[tacit haven]

for 2,3 find the values inside the inverse functions

[mystic saffron]

Well I don't know what's cos

I only know sine

[tacit haven]

whats your definition of sine?

[mystic saffron]

in between -90 and 90 degrees

only follows Q1 and Q2

[tacit haven]

huh?

sinx is the ratio opp/hypotenouse, in a right triangle that has x degrees/radians

[mystic saffron]

ok

[tacit haven]

also its the y value in this diagram

cosx is the ratio adjacent/hypotenouse

which is also the x value in the diagram

[mystic saffron]

https://tenor.com/view/bakugo-katsuki-bakugou-funny-anime-gif-23616036

SO MOST OF THE ANSWERS I DID IS WRONGG

DANG IT

This is why I don't trust youtube

[tacit haven]

you should honestly learn trig functions first😭

[mystic saffron]

I WAS JUST FOLLOWING WHAT THIS YOUTUBE LINK SAID

https://www.youtube.com/watch?v=jt7p-mCC0ng

[tacit haven]

theyre correct...

[mystic saffron]

but???

[tacit haven]

your answers are correct too..

why do you think youre wrong

[mystic saffron]

force of habit

[tacit haven]

real

[mystic saffron]

usually people here would start saying "Oh you didn't answer this you're cringe and stupid""

LIKE HOLD ON

IM SOLVING

[tacit haven]

yeah you could like use this diagram to finish this qn

[mystic saffron]

SCREW IT

1. 60 degrees
2. 135 degrees
3. 240 degrees
4. 180 degrees
5. ?
6. 135 degrees
7. 135 degrees
8. 210 degrees
9. ?
10. 135 degrees

Done :3

<@&286206848099549185>

[autumn bolt]

for the second one, can you figure out what cos(3pi/4) is?

[mystic saffron]

uhhhh 135 degrees?

[autumn bolt]

135 degrees is 3pi/4

[mystic saffron]

yes?

[autumn bolt]

now can you find what cos 135 degrees is

[mystic saffron]

uhh

3pi/4

<@&286206848099549185>

[signal crown]

for the second one, use that : cos(a) = sin(pi/2 - a)

[mystic saffron]

<@&286206848099549185>

oh-

[mystic saffron]

I don't get it

[signal crown]

complementary angle

but well its a well known formula

[mystic saffron]

I mean

how do I use it

[signal crown]

cos(3pi/4) = sin(pi/2 - 3pi/4)

we are switching to sin because we have arcsin

[mystic saffron]

cos(3pi/4) = sin(pi/2 - 3pi/4)
= sin(-2)

[signal crown]

?

$\frac{\pi}{2} - \frac{3\pi}{4} = \frac{2\pi - 3\pi}{4} = -\frac{\pi}{4}$

[clever fjord]

Herels

[mystic saffron]

I don't get it

[signal crown]

what dont you get ?

[mystic saffron]

Because I remembered cross multiplication?

you didn't apply cross multiplication

why?

[signal crown]

ah

$\frac{\pi}{2} - \frac{3\pi}{4} = \frac{4\pi - 6\pi}{2 \times 4} = \frac{-2\pi}{8} = -\frac{\pi}{4}$

[clever fjord]

Herels

[mystic saffron]

That's better :3

[signal crown]

anyway

we got this

cos(3 pi/4) = sin(-pi/4)

and we have : arcsin(cos(3 pi/4))

then we are left with : arcsin(sin(-pi/4))

since arcsin is the inverse of sin, we are only left with -pi/4 for the result

[mystic saffron]

Okie

1. 60 degrees
2. -pi/4
3. 240 degrees
4. 180 degrees
5. ?
6. 135 degrees
7. 135 degrees
8. 210 degrees
9. ?
10. 135 degrees

[signal crown]

same logic with the other

[forest yoke]

Omg ull are still going at it

[mystic saffron]

so same logic as number 3?

[signal crown]

just look for an angle such that its tan is equal to -sqrt3/3

[mystic saffron]

OMG. I CAN'T BELIEVE IM SAYING THIS BUT I MISS YOU SO MUCH

[forest yoke]

Lol is ok

Whats up

[mystic saffron]

I don't get that but okay

so it applies with number 3?

[signal crown]

trig circle

[forest yoke]

Damn well done bro u got most the answers

[mystic saffron]

what

[mystic saffron]

I feel like there are all wrong because Herels is telling me what to do in number 2

[forest yoke]

Oh

Its ok , your trying sums trust me this always happens

[signal crown]

unit circle I meant

[forest yoke]

Circle with radius 1

[signal crown]

ah you were talking about the n°3

[forest yoke]

Atb @signal crown

[signal crown]

arccos(sin(4pi/3))

[signal crown]

atb ?

[forest yoke]

All the best

[mystic saffron]

Soo

How do I do number 3

[signal crown]

same logic as the second

you transform the sin into a cos

[mystic saffron]

4pi/3 - 3pi/4 = - pi/pi = -pi?

[signal crown]

sin(a) = cos(pi/2 - a)

therefore : sin(4pi/3) = cos(pi/2 - 4pi/3)

[mystic saffron]

This is confusing

[signal crown]

hmm

how would I explain this

[mystic saffron]

How about

give me the formulas of sin and cos?

Maybe I got confused about those

[signal crown]

there is none, they are functions
but in geometry they are linked to the sides of a triangle

[mystic saffron]

Okay this is not workig

[mystic saffron]

At first I thought, okay the question is around cos so the formula of sin is pi/2

[signal crown]

you are confusing angles and function

[mystic saffron]

Dang it

I don't know anymore!

[signal crown]

sin and cos are functions and take angle as their input

that means, the sin of an angle will give you a value

[mystic saffron]

So what pi/2 suddenly pop up out of no where

[signal crown]

but sin and cos are also linked through some formula

[signal crown]

from the fact that the cos and sin of two complementary angles are the same

im too lazy to do a proof of this

in trigonometry class maybe they gave you this formula

[mystic saffron]

This is an activity that was given with no discussion.

[signal crown]

if we have two angles a and b, and their sum gives pi/2 (or 90 degree), then we say they are complementary

[signal crown]

sadly
your exercises need a lot of past knowledge

function, trigonometry...

[mystic saffron]

so for both it's pi/2?

[signal crown]

both what

[mystic saffron]

cos and sin?

[signal crown]

sin(a) = cos(pi/2 - a)
cos(a) = sin(pi/2 - a)

remember that at least

[mystic saffron]

so they are both

okay

[mystic saffron]

pi/2 - 4pi/3 = 2pi - 3pi/pi - 3 = pi/3pi = 3?

<@&286206848099549185>

[tacit haven]

so cah TUAH!!

[mystic saffron]

What

https://tenor.com/view/mission-yozakura-family-mission-yozakura-family-yozakura-family-gif-35564743320074738

1. π/3
2. -π/4
3. 5π/6
4. -π/2
5. -π/6
6. π/4
7. (sqrt)2/2
8. 5π/6
9. π/3
10. -π/4
    ​

!status

[odd edge]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[mystic saffron]

4

[autumn bolt]

all good to go

[mystic saffron]

Sooo are the correct?

[autumn bolt]

yes

checked

[mystic saffron]

YIPPEE

[odd edge]

@mystic saffron Has your question been resolved?

[mystic saffron]

1. sin^-1 (cos (-pi/4))
2. arcos (sin (5pi/2))
   3)cos^-1 (-tan (3pi/4))

1. pi/4
2. 0
3. 0

!status

[odd edge]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[mystic saffron]

4

<@&286206848099549185>

[paper onyx]

@mystic saffron if you are asking about these, then they are all correct

[mystic saffron]

Okie

[paper onyx]

Is that it?

I just realized you have had this channel for 6 hours 😶

@mystic saffron if that's all for today, please .close the channel

[mystic saffron]

There's a lot

im just solving it

!status

[odd edge]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[mystic saffron]

4

[paper onyx]

Can you send the answers?

[mystic saffron]

okay lemme send a photo

[mystic saffron]

Test 1:

1. 2sec^2 x
2. sin x/1−sin x
3. cos x/sin x (1 − cos x)
4. 1 + sin x/cos^2 x − sin x − 1
5. cos x/sin x −1

[paper onyx]

Hmm

1. Is wrong, also Isn't even terms of sin and cos

2. Can be simplified furthur

@mystic saffron

[mystic saffron]

Dang okay

[paper onyx]

Try 1st one again

[slim saddle]

Can someone please tell me the formulas for sin(x+y) and cos(x+y)

[paper onyx]

!occupied

[odd edge]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[mystic saffron]

2 tan^2 x+1?

[paper onyx]

It is right tho

Turn it into sin and cos

[mystic saffron]

number 4 can be round "−cscx"?

[paper onyx]

-1/sinx

[mystic saffron]

sin^2 x+1/cos^2 x
for number 1?
​

[paper onyx]

Yup

Good job

Any issues in the second part?

[paper onyx]

The prove identities part

[mystic saffron]

wait lemme solve them

[odd edge]

@mystic saffron Has your question been resolved?

[odd edge]

Channel closed

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[odd edge]

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[mystic saffron]

wow

[halcyon lava]

Can you send a picture instead of a video?

[silent comet]

Sure

i just have a problem with the angles

i need the full angle between a and b

So i can use a parallelogram

and find the rest

[halcyon lava]

Angle between a and b is theta

[silent comet]

Yeah

I need theta

[halcyon lava]

Wdym full angle?

[silent comet]

They only gave me 30 degrees

[halcyon lava]

That's the angle between the resultant and a

[silent comet]

i need the angle between a and b so i can use a parallelogram and find B angle

[silent comet]

Yes and i need theta

I dont need the angle between a and r

[halcyon lava]

You could also think of B as R - A

[mystic saffron]

are you familiar with dot product ?

[silent comet]

they gave it in physics but i cant use it here cause the question is about parallelograms

Whenever i try to use the sin law i always have a missing component

Like sina/a=sinb/b

[mystic saffron]

$\tan \alpha = \frac{A_1 \sin \theta}{A_2+A_1 \cos \theta}$

[silent comet]

whats that

[mystic saffron]

this is derived from the parallelogram law of addition

(i think i got the name wrong)

[silent comet]

I have never seen this in my life

[mystic saffron]

and $\theta$ is the acute angle between $A_1$ and $A_2$

[mystic saffron]

oh

then idk

pretty sure the methods i mentioned is the only to solve this

[silent comet]

Hmm.. is it possible to solve this using the components method

[mystic saffron]

oh

yes

ig

$F_A +F_B \cos \theta = u \cos 30$

@silent comet

[silent comet]

Yup

I got the magnitude

But still i used the wrong method

Now that i have the magnitude i can find the angle

Using the sin law

[silent comet]

Uhm how can i find it using the dot product

[mystic saffron]

dot product
$$\vec a \cdot \vec b = \norm{\vec a} \norm{\vec b} \cos \theta$$

sheesh

lets taking the two vectors of which we know the magnitudes and the angle between them

that would be Fa and u

agreed?

[silent comet]

Yea

[mystic saffron]

substitute them here

[mystic saffron]

and instead of u

you can write $F_a + F_b$ since u is the resultant of those both

[mystic saffron]

try to do it from here

[silent comet]

Uhm imma go finish my bio and come back to this

thanks for helping tho

.close

[odd edge]

Channel closed

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[odd edge]

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[mystic saffron]

p

[mystic saffron]

how did we go from logx=0.5 to x = square root of 10

[narrow crypt]

raise 10 to both sides

and 10^0.5 = 10^(1/2) = sqrt{10}

[mystic saffron]

0.5^10?

[narrow crypt]

10^{log_x} = 10^{0.5}

10^{log_x} simplifies to x

[chilly ravine]

10^0.5

[mystic saffron]

oh okay

[green ledge]

this is the main thing

the basic property of logarithms

[mystic saffron]

alright thanks

.close

[odd edge]

Channel closed

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[serene radish]

hey im having trouble with this problem with line integrals... i tried doing this but it kind of trips me up that the vector field is three dimensional while the curve is 2D so could someone possibly explain how this problem is done? thanks!

[serene radish]

is this just FTC where because it starts and ends at the same place, it would be 0?

ok so i've denoted the line segments as C1, C2, and C3 respectively and parameterized them

parameterize C1 to be <3t, 5t> 0<t<1

[nocturne belfry]

this only works if the vector field is conservative

is it

[serene radish]

its in my notes somewhere 🙏

lemme try to find

F = deltaf

wait

the upside down

gradient

gradientf

[nocturne belfry]

$\text{curl}(F) = \nabla \cross F = \mdet{ i & j & k \ \pdv x & \pdv y & \pdv z \ P & Q & R }$

lets see if that works

its a lil tight

but thats okay

[serene radish]

oh wow

cross

[nocturne belfry]

can you do this? i mean the computation here

[serene radish]

wait i learned it if the partials are equal not the cross product

[nocturne belfry]

thats the easier case, in 2d

[serene radish]

oh yea i can 😭

[nocturne belfry]

in 3d its more complicated

[serene radish]

isnt it just if all the partials are equal...

6

[nocturne belfry]

6?

[serene radish]

nah its like

[nocturne belfry]

its not all of them equal, its specific partials

you want this curl to be 0 everywhere

[serene radish]

yea yea

[nocturne belfry]

you dont want any local areas of vorticity

but thats not just all the partials equal

if you take the det

its certain partials

[serene radish]

i was on some crack when i was doing this problem istg but i had the specific partials equal

so it was conservtive

but idk i was using M and N and W and P and idk where W came from

[nocturne belfry]

for example the first component you would want $$\pdv{F_3}{y} = \pdv{F_2}{z}$$

[clever fjord]

jan Niku

[serene radish]

righttt yea that thing

i think i found it online

this thing right

[nocturne belfry]

yea, i can rewrite mine to use this language if you like it more

[serene radish]

nah its all good ima try that rq

[clever fjord]

jan Niku

[serene radish]

ITS CONSERVATIVE

THANK THE HEAVENS

SO I CAN USE FTC

RIGHT?

[nocturne belfry]

this is usually some theorem

[serene radish]

f(b)-f(a)

[nocturne belfry]

fundamental theorem of line integrals

well, but in any case, yea

[serene radish]

ok wait so it's a closed thing starting and ending at (0,0)

would that mean that f(r(b)) = f(r(a)) = 0

so integral = 0

[nocturne belfry]

yea

[nocturne belfry]

doesnt have to start and end at the origin

just the same point

we place some requirements on the curve it follows

and the vector field has to be conservative

[serene radish]

no yea but in this case it does

[nocturne belfry]

it does, yea

[serene radish]

bruh my teacher trolling me fr

okk thanks for ur help

[nocturne belfry]

its a common exercise to confirm that this works out

if youre not tasked with just proving it as a special case of the theorem

[serene radish]

no proving

alr alr thanks

.close

[odd edge]

Channel closed

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[lunar prawn]

Can someone help me solve this logic question?

One of the star athletes you see on the field has committed about 400 felonies. The lad wanted to figure out exactly how many he's committed. He pulled up his criminal record and started counting them in clusters to save time but completely lost track by the end. All he can recall is that when he counted them 3 at a time, he had 1 left over. When he counted them 4 at a time, he had 2 left over. When he counted them 5 at a time, he had 4 left over. When he counted them 7 at a time, he had 2 left over. How many felonies did the star athlete commit?

[warped urchin]

i thought this was a joke until i read further and realized it was a system of linear congruences lmao

[signal yacht]

oh no modular arithmetic

it's even, ends in 4, not divisible by 4

id write python code

[odd edge]

@lunar prawn Has your question been resolved?

[lunar prawn]

.close

[odd edge]

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[untold pike]

Few days ago I were thinking about another method of wrtiting down positive integrers. I had idea which has many parameters and variables and it looks like this
I ve been talking it with my math teacher and she said she doesnt have any clue for solving it. Can anyone help me with it? where abcd are positive integrers

[orchid torrent]

This is equivalent to saying that $$5^{a+b+c+d}-3^{b+c+d}+2^{c+d}-6^{d} \equiv 0 \pmod{4}$$ or simply $$1-(-1)^{b+c+d}+2^{c+d}-2^{d}{ \equiv 0 \pmod{4}$$ atp, just a bit of annoying casework.

[clever fjord]

Civil Service Pigeon
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[untold pike]

Ok but does it doesnt show the thing I am looking about

I am sorry i didnt say about ut

can I write that way all positive integrers @orchid torrent ?

[orchid torrent]

My intuition tells me that a bounding argument should suffice to derive some sort of counterexample

In general, if you just make stuff up out of thin air, it's not going to work out "nicely" (which makes sense because the "nice" stuff is a small fraction of all the "possible" stuff you could create)

[untold pike]

ok but waht if I checked few numbers for counter example and it looks possible to write every positive integrer/?

[orchid torrent]

🤷‍♂️

[untold pike]

ok but do you know who should i have ask?

[orchid torrent]

idk I doubt a lot of people would be willing to seriously try this

[orchid torrent]

due to this

hold up

[untold pike]

thank you very much

[orchid torrent]

``def find_extended_integer_solutions(max_val=25):
solutions = []
for a in range(1, max_val + 1):
for b in range(1, max_val + 1):
for c in range(1, max_val + 1):
for d in range(1, max_val + 1):
if 5**(a + b + c + d) + 2**(c + d) - 3**(b + c + d) - 6**d == 4:
solutions.append((a, b, c, d))
return solutions

find_extended_integer_solutions(max_val=25)``

bashing it up to a, b, c, d \leq 25 already fails to yield a solution for 1

[orchid torrent]

so I feel like this should hold

[untold pike]

ok

but what explanation is behind it

?

@orchid torrent

and doesnt it looks like a=1 and bcd=0

?

@orchid torrent ?

[orchid torrent]

Show that the minimum value is obtained when a=b=c=d=1

that yields the minimum to be 149 upon substitution of said values

hence values from 1 to 148 are already not attainable

[orchid torrent]

?

[orchid torrent]

Also, as a quick note, a lot of helpers find it annoying when you repeatedly ping them like this, so chill out a bit on that lol

[orchid torrent]

Oh also, small additional remark on this - a lot of the time when you make stuff with no additional insight, it's typically very easy to invoke small counterexamples and bounding arguments, especially in the face of a very general question like this

Anyway, if you don't have any other questions, imma dip 👋

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[untold pike]

thx

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[cyan socket]

I tried applying snells law but it resulted in an undefined angle

[forest sky]

can you show your work?

[odd edge]

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[cyan socket]

In order to calculate theta_2 I wrote the equation 1 * sin(40) = 1.7 * sin(theta_2). (everything is in degrees). this means theta_2 is around 22.3 degrees. Then to find theta_3, I did 60 - 22.3 since the angles in a triangle add to 180. Applying snell's law again results in the equation 1.7 * sin(37.7) = 1 * sin(theta_4). However this results in an undefined value of theta_4

I believe this means this is total internal reflection, however the diagram says otherwise

[forest sky]

could you explain more how you got theta_3? what triangle are you referring to?

[cyan socket]

If you extend the perpendiculars on both sides

they intersect somewhere near the middle

and a quadrilateral with angles 60, 90, 90 and 120 is formed

therfore theta_2+120+theta_3=180

so theta_3 = 60-theta_2

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[barren coral]

Yeah that's right

Waait

Yeah that does look right

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[thin willow]

unsure how this works

[quartz trellis]

<@&268886789983436800>

bruh

[chilly ravine]

Dumb ahh don't even know how to advertise a server

At least tell us what the server is about then I'll consider joining

[paper onyx]

<@&268886789983436800>

[chilly ravine]

A kid's game is NOT math bro😭🙏

[thin willow]

im scared

i was looking for help

[mystic saffron]

[paper onyx]

This is annoying

[quartz trellis]

bro mods not responding

[chilly ravine]

Says the one advertising a kid's game😂

[paper onyx]

Really sorry for the inconvenience

.close

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[mystic saffron]

💀💀

[thin willow]

i feel violated....

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[gaunt fiber]

how do i do this

[wicked kestrel]

!occupied

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[wicked kestrel]

That may come in handy later

What information does F(x) tell us?

It's not frequency

It's the cumulative distribution function

How do you interpret F(1) = 1/6?

If you don't know what it is then you can't solve the problem

If they've not taught you what it is, you can't be expected to answer this

If they have, now's the time to refresh your memory

It would almost surely have been introduced earlier

You're using the cdf of a discrete distribution after all, which I imagine is in an earlier chapter

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[odd edge]

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[hexed patrol]

find out a new channel

[wide drift]

Brother I posted half a second after him

[eager yacht]

No problem

I moved to a new channel

[hexed patrol]

mb g i jus got in the channel

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[dim dew]

my guess is since its a sphere, we use spherical coordinates?

[timid sky]

yes you could

set up the parameterisation and find the differential surface element

[dim dew]

so here would it be x = 5sinphicostheta,
y = 5sinphi sin theta
z = 5cosphi
r(theta,phi) = < 5sinphicostheta,5sinphi sin theta, 5cosphi>

then do i plug that in F(r)?

[timid sky]

well yes but you need to substitute for each value of x,y,z that you wrote into F

[dim dew]

okok

so F(r) = 15sinphisintheta + 4, -25cosphi, 10sinphicostheta -1>

[timid sky]

ye

now find the differential surface element

[dim dew]

how do i do that?

get paritial derivitive for each component?

oh i get partial derivtive with respect to theta for each,
Then again with respect to phi?

[timid sky]

It was $$d\vec S = \left(\pdv{\vec r}{\phi}\cross \pdv{\vec r}{\theta}\right)d\phi d\theta$$ if i recall correctly

[clever fjord]

Aero

[dim dew]

hm okok

[timid sky]

you need to cross them but i think you got the idea down

[dim dew]

okok

dr/dtheta = <15sinphicosthetea, 0 , -10sinphisintheta>

dr/dphi = < 15cosphisintheta,25sinphi,10cosphicostheta>

doesnt cross product order matter how do i know which one to put first

[timid sky]

The normal vector to your surface is always taken to be outward-facing for your surface conventionally, and you can use the right-hand rule to see that with phi and theta

[dim dew]

for right hand rule, we usually connect our index to the start of the vector. how would it work here

[timid sky]

Think of it this way; dr/dphi points in the direction of increasing phi, "downwards" if you look at the sphere in the conventional perspective. dr/dtheta points "around" the sphere. If you pick a point on the sphere's surface, and you point out your index finger in the direction of dr/dphi and middle finger in the direction of dr/dtheta, your thumb will point outwards from the sphere

[dim dew]

😭 im confused

so middle finger is towards theta and index is towards phi? and thumb is Z axis then?

[timid sky]

This is a better visual of what i am trying to say

Ok so the above actually flips your theta and phi, but the same logic applies. Try to put your fingers in the same directions as e_theta and e_phi and you should get e_r with your thumb

[dim dew]

if er is thumb

what is etheta and ephi?

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[mystic saffron]

whys "infection or cold", P(I U C)?

i think it should be intersect

not union

wait

interesect is infection and cold?

but still

union would be infection or cold OR infection and cold

because union will include the intersect as well

infection or cold should be union - intersection

correct?

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[bronze nova]

Can yall please help me with a to c I'm having a hard time with it

[steep mantle]

what are you struggling with?

[proven cape]

Can someone help me with lim x -> 0 sin x/x

@bronze nova pls help

[steep mantle]

!occupied

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[proven cape]

I dont know what work show

I dont have my work

[muted oxide]

use l'hopitals rule

its 1

[bronze nova]

Idk how to solve it I don't understand my profs explenation

[proven cape]

How to use It?

[muted oxide]

go to a help channel

this one is taken

sorry lobster

[proven cape]

Ah ok

Sorry

[steep mantle]

heres for a diffrent problem, the way is the same

[bronze nova]

It's blurry for me

[steep mantle]

lemme find a diffrent one on google

heres one

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[upbeat roost]

can someone help me with these questions?

[upbeat roost]

it would be great if someone could first tell me how to calculate the interquatile range

<@&286206848099549185>

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[edgy creek]

Can someone explain to me why this proof says anything about limits?

[edgy creek]

except for however this happend I think I know what's going on.

But I really don't understand how this prooves anything about what is supposed to be proven

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[ember meteor]

Circle $\omega$ circumscribes square $ABCD$. $M$ is the midpoint of $AB$. Circle $\Gamma$ passes through $M,O,D$. If $AD$ meets $\Gamma$ at $P$, $\omega$ meets $\Gamma$ at $Q$, how to prove $PA=PQ$?

[clever fjord]

babario

[ember meteor]

i cant really use ptolemy so i tried proving $\angle PQA=\angle PAQ$, but the points are like from 2 different circles so i cant even use circle theorems, i can get $\angle PQC$ from cos rule where C is the center of $\Gamma$ but reckon it wont be useful

[clever fjord]

babario

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<@&286206848099549185>

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[mystic saffron]

in this question i can find angle of A when sin (A) equals sqrt 2/2 and B when sin (B) equals sqrt 2/2 then convert them to radian and get their difference to know the length of AB and just multiply by height to get the area then do same thing with XL, but what i don’t understand is how subtracting angles leads to us getting a length?

[cold sage]

'b' is from 0 to B
'a' is from 0 to A
if i do b-a i get the separation of A to B

youre finding the angle between two angles that start at the axis basically here

[mystic saffron]

yeah but for me to get an area i need lengths and im not sure how subtracting angles leads to me getting a length

[cold sage]

consider it an angular length if you want

its still an angle

on the graph its a unit

[mystic saffron]

but its not a unit length its just the difference between two angles meanwhile something like sqrt 2/2 is a unit length

[cold sage]

the sqrt2/2 is dimensionless if you want to get overly specific

i think youre just overcomplicating it slightly

i wouldnt pay mind to the dimensions of the axes, at least not here

[mystic saffron]

yeah but im a little confused because here its like angle dimensions, its just the difference between two angles, how can the difference between two angles be equivalent to a dimension of a rectangle?

[cold sage]

i could say the y axis is dimensionless, the x is in radians, then area on that graph is measured in radians

its not literally a 'length'

but you can still find its area on the graph

like we can find the area of graphs where we dont know the dimensions whatsoever

[cold sage]

on this graph all angles lie on the same line

angular separation becomes a straight line

[mystic saffron]

by saying dimensionless do you mean that in the coordinate plane magnitudes all go by the same measuring unit (aka length unit) regardless of their original measuring unit (cm, m, km etc. on the coordinate plans all measurements have the same measuring unit) ?

[cold sage]

yeah im saying the y axis could be considered as having no unit at all

like the number 2 by itself

[mystic saffron]

aha i didnt know angles counted too i thought it was just distance quantities

[cold sage]

and normally we would take these things into account when measuring area, its true, like here area would be measured in radians

in your answer all units will cancel out anyway though so you dont have to mind that so much

[mystic saffron]

okay so in summary its because in the coordinate plane none of the quantities have any measuring units so it doesnt matter ?

[cold sage]

they do, but because you have Area/Area the units all disappear

eg if i had a graph where the y axis was Joules and the x was time
my Area unit would be Js

but you just have to pay mind to the fact that angle is an axis here, so they lie on a straight line, thats why you can have a rectangle

[mystic saffron]

what if it wasnt area/area? what if it was just asking for the area of one shape?

[cold sage]

id give it the unit radians

since its radians * dimensionless

to my knowledge anyway, unless y has some contextual dimensions

[mystic saffron]

so its fine if i multiply two totally different quantities (time and speed for example) as length and width ?

[cold sage]

yeah

[mystic saffron]

since quantities are dimensionless

ah okayy i never really knew this

[cold sage]

the product would be in m

like a speed time graph

the area is distance

[mystic saffron]

ahaa i get it

but in this case its multiplying angle by length and those two dont really have a relation like speed and time

[cold sage]

you can graph things with no relation

[mystic saffron]

so if the area is distance in a case where speed and time are the dimensions what would the area be in the case where angle and length are dimensions?

[cold sage]

you can have rad m if that was the case

[mystic saffron]

alright so since this is the coordinate measuring units dont really matter since measurements are dimensionless unless stated otherwise?

[cold sage]

yeah

if we didnt know units we may just literally write 'unit'

[mystic saffron]

alright thank you ^^

.close

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[molten sparrow]

For 10: how do I know which one it’s increasing at

[turbid compass]

A function is increasing when the y value is increasing (getting bigger)

At what interval is the y value of the function going up?

[molten sparrow]

infinity

right cuz you can’t see the stopping point

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[fair depot]

How do I visualize this?

[fair depot]

.close

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[wanton sierra]

I dont know if anyone here knows anything about deterministic finite automatas, but:

[wanton sierra]

this works right ?

[latent scaffold]

Yes

[wanton sierra]

ty

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[mystic anchor]

Probably my least favorite math topic as of yet

[quasi sparrow]

Did you use double angle for cos?

,tex .double angle

[clever fjord]

riemann

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[soft basin]

To simplify a rational expression, both the numerator and denominator must be written in factored form. Then, look for factors that make a ______ ___and simplify

it must be 2 word and 8 letter wth is this crosspuzzle😭😭
can someone plz help

[soft basin]

1 Down

[narrow crypt]

,rotate

[soft basin]

<@&286206848099549185>

[soft basin]

u figure it out yet?😭

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[soft basin]

.reopen

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✅

[soft basin]

<@&286206848099549185>

[spiral quarry]

hi everyone i need treding chanel who can join me

[sand horizon]

<@&268886789983436800>

[viscid flint]

not in this server

[soft basin]

can somone help with my question😭

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[ancient crest]

hey if anyone could help it would be greatly apprechiated. Im having trouble understanding how to solve this problem, and this is as far as I think i can get.

hold on ill reorientate it

oh crazy i didnt know that was a thing

[boreal crag]

ye

[modest nacelle]

Oh no, it's an imperial system. 🤮

[boreal crag]

I'm guessing first you need an equation for the velocity over time

[ancient crest]

i have my profs done but im ngl i dont understand it

this is his work but i still dont understand.

[boreal crag]

He integrates acceleration to obtain velocity

Then figures out what the constant of integration (starting velocity) should be

Then integrates velocity to obtain position

Then figures out again what the constant of integration (starting position) should be

[ancient crest]

hmm alright

so then im assuming that using that solved position, he plugs in the time where x' (velocity) is 0?

[boreal crag]

Yeah

[ancient crest]

ohhhh

i understand it now

ight i got the right answer now

thanks for the help

not too sure how to close this

[forest sky]

you can use .close or .solved

[boreal crag]

.close

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[graceful parrot]

Do i make x equal to y?

[graceful parrot]

like

both x to one side

[amber schooner]

what

[graceful parrot]

like

do i move over

[amber schooner]

you mean solve for x in terms of y

[graceful parrot]

yes

sorry

[amber schooner]

since you’re revolving around the y axis yes you would if you’re doing the disk method

$\pi \int_c^d R^2(y) dy$

[clever fjord]

knief

[graceful parrot]

what was the point of telling me the volume that its V= pi/3 h^3/2?

[graceful parrot]

here

if i just use that

[amber schooner]

"show that"

[graceful parrot]

ah

skull

i cant read

ty

.close

[odd edge]

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[graceful parrot]

got it

[odd edge]

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[deep mason]

I'm trying to numerically track eigenvectors of the matrix $M(t)$ as it changes smoothly with time ($t$). Are there any established methods for doing so?

[clever fjord]

Ginger

[deep mason]

I have two methods: 1) calculate the eigenvectors at each time step, and then match those eigenvectors to those in the previous time step, and 2) integrate the evolution of the eigenvectors via the differential equation that gives their evolution

neither are working well, and i believe it is because the eigenvalues become (nearly) degenerate at some point in time

[odd edge]

@deep mason Has your question been resolved?

[low locust]

there are already simple 2x2 examples of matrices with eigenvalues for example +-sqrt(t) and you can imagine the trouble you then get for letting t go over 0

[deep mason]

yes!

[deep mason]

but i thought method 2 should do ok here

well

really i'm just after suggestions for what to do

we can imagine that the eigenvalues become nearly degenerate but not exactly

and the 2x2 matrix example is nice

well, in practice, it turns out that, with method 2, the step size of the integrator becomes infinitesimally small and the integration fails

(maybe this means that the eigenvalues are becoming exactly degenerate?)

[low locust]

I would need to think about it more and actually look at the matrix and stuff and frankly I dont have the time for that right now

[low locust]

but this sounds plausible

[deep mason]

ok makes sense

is it worth posting the whole matrix? it's kind of a complicated 6x6 one

even for others if you have to go

[low locust]

definitely do it

[deep mason]

ok sure

[-ik,  Az, -Ay,   0, Bz,  -By],
[-Az, -ik,  Ax, -Bz,  0,   Bx],
[ Ay, -Ax, -ik,  By, -Bx,   0],
[  0,  Cz, -Cy,  ik,  Dz, -Dy],
[-Cz,   0,  Cx, -Dz,  ik,  Dx],
[ Cy, -Cx,   0,  Dy, -Dx,  ik]

A, B, C, and D are 3D vectors. the four 3x3 matrices that make this up (the four "corners" of this 6x6 matrix) are the matrices that represent A cross ..., B cross..., etc., and then finally, ik's are added on the diagonals

i suppose the eigenequation is then equivalent to

$$\lambda \vec u = - \vec A \times \vec u -ik \vec u - \vec B \times \vec v$$
$$\lambda \vec v = - \vec C \times \vec u - \vec D \times \vec v + i k \vec v$$

[clever fjord]

Ginger

[deep mason]

to repeat the problem:

i'm wondering how to track the eigenvectors of this matrix (or any matrix) numerically as the matrix changes smoothly in time (e.g. since A, B, C, and D change in time), which becomes difficult when the eigenvalues become nearly degenerate

[meager juniper]

An interesting observation is that your trace is 0, thus the sum of eigenvalues is also zero, thus this is matrix which is expressible as the lie bracket of two other matrices P and Q

I don't know if this helps at all though

[deep mason]

interesting!

and hello again

[meager juniper]

Hello again

Seeing as you have some rather nice looking block matrices, I was hoping to find an easy way to find the eigensystem using them

Though I haven't found anything yet.

[deep mason]

yeah, you're right the form looks kinda nice

i've been starting a bit at the vector form of it without luck so far

($\vec u$ and $\vec v$ are 3D vectors that, when stacked on top of each other, are the 6-component eigenvector)

[clever fjord]

Ginger

[deep mason]

i'm also curious about the Lie bracket form, i'll keep that in mind at the very least

this question is still open so anyone feel free to jump in

[odd edge]

@deep mason Has your question been resolved?

[odd edge]

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[eternal rune]

hello

[eternal rune]

Im having trouble in the 2nd step of this solution

how did they proceed after the 2nd step

after using the property of conjugate

[south plume]

$(\overline{z})^n = \overline{(z^n)}$

[clever fjord]

artemetra

[quartz trellis]

everything is given in the picture that you posted

[eternal rune]

oops

.close

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[odd edge]

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[quick needle]

Q: Prove that \lim \limits{k\to \infty} a_k =\lim\limits{k\to \infty} \frac1{\sqrt[k]{k^3+2k}}=1.

Consider the denominator

$$\sqrt[k]{k^3+2k} = \sqrt[k]{k(k^2+2)}= \sqrt[k]{k} \cdot \sqrt[k]{k^2+2}$$.

I know that $\sqrt[k]{k}\to 1$ and $\sqrt[k]{a}\to 1$ for all $a\in \mathbb{R}^+$, but am unsure how I can apply the result here

[quick needle]

wtucck

[proven harbor]

Consider the denominator $\sqrt[k]{k^3 + 2k} = \sqrt[k]{k(k^2 + 2)} = \sqrt[k]{k} \cdot \sqrt[k]{k^2 + 2}$.

[quick needle]

YEAY

[spiral basalt]

You can do something very brutal here
k^3 <= k^3 + 2k <= 2k^3
Then apply known formulas, as you have the cube of sqrtk

This will generally work for any polynomial

[clever fjord]

Kakaka
Compile Error! Click the reaction for more information.
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[quick needle]

Q: Prove that \lim \limits_{k\to \infty} a_k =\lim\limits_{k\to \infty} \frac1{\sqrt[k]{k^3+2k}}=1.

Consider the denominator

$$\sqrt[k]{k^3+2k} = \sqrt[k]{k(k^2+2)}= \sqrt[k]{k} \cdot \sqrt[k]{k^2+2}$$.

I know that $\sqrt[k]{k}\to 1$ and $\sqrt[k]{a}\to 1$ for all $a\in \mathbb{R}^+$, but am unsure how I can apply the result here.

[clever fjord]

Kakaka
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[quick needle]

what makes it "brutal"

[spiral basalt]

2k <= k^3 is quite the brutal upper bound

[quick needle]

yeah in what sense is it "brutal"?

first time I'm hearing of "brutal" used to refer to an inequality

as it in only works for some k?

[spiral basalt]

Sometimes you use fairly tight upper bounds
Here for k = 10 you say "yeah 20 <= 1000 so this works"

[quick needle]

"What" will generally work?

bounding all terms by an integer multiple of the dominant term?

[spiral basalt]

This trick of bounding it above by a monomial
Shows that sqrtn -> 1 for any polynomial with positive leading coefficient

[quick needle]

erm wot

still confused as to what yo umean by "brutal"

here, your 2k≤ k^3 bound only works for k≥2

not for k=1

[spiral basalt]

You can always easily fix that by saying <= 2k^3

[quick needle]

so it doesn't work for all k \in N

[spiral basalt]

Otherwise such bounds are indeed only asymptotically true

(which is all we need)

[quick needle]

ok

so "brutal" bound = asymptotic bound?

i.e. only holds for some k-tail?

[spiral basalt]

It's that the bound is basically a whole order of scale/magnitude higher

So it's the opposite of tight

[spiral basalt]

No we just say asymptotic

[quick needle]

How?

both are O(k^3)

you haven't increased a linear factor

[spiral basalt]

Theta(k) vs theta(k^3)

[quick needle]

ok

so in like applications, that's a really shitty ub

but for the purpose of grinding out the maths, it's ok

right?

[spiral basalt]

You'd be surprised how much theory relies on presumably egregiously untight upper bounds to get results

[quick needle]

wowow ok

[spiral basalt]

Were if you were to guesstimate how much higher you upper bound is, in the end it's off by a huge factor, so you might expect your bound to be a very innacurate predictor in practice
Then you see that actually the worst case reaches that and a better study shows the average case isn't that far off

[quick needle]

wait

hollup

$\sqrt[x]{x}$ isn't monotonically increasing or decreasing for $x\geq 0$ though

[clever fjord]

Kakaka

[quick needle]

so how do I bound $\sqrt[k]{k^3+2k}$?

[clever fjord]

Kakaka

[spiral basalt]

It's between sqrtk^3 and sqrtk sqrtk^3

[odd edge]

@quick needle Has your question been resolved?

[odd edge]

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[lucid magnet]

Helo

[odd edge]

Please don't occupy multiple help channels.

[lucid magnet]

Why

Bro bot

[quartz trellis]

@lucid magnet stick to one channel

[lucid magnet]

Howww

[quartz trellis]

close one channel

[lucid magnet]

Nobody here

Howw

Bro help me i cant read

Close channel

[quartz trellis]

.close

[lucid magnet]

.close

[odd edge]

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[quartz trellis]

aight stick to one channel

[odd edge]

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[silver stump]

hi, i need to know how people think mathematically, for example this is a question i got on the test, and had no clue how to even approach it:
a class of n students is going on a school trip and the teacher must separate the class into two groups. the first group must have 3 students while the second group must have at 3 students. Then, two students approach the teacher and ask her to make sure that they aren't in the same group. Upon adhering to this request, the amount of ways in which the teacher can assign the groups halves. find n

[vale vortex]

so

the 2nd group

has atleast 4 students

[silver stump]

whoops sorry

let me edit it

[vale vortex]

wut

[silver stump]

the second group must have at least 3 students

the first group must have 3 students

[vale vortex]

so

the wording is still weird

while the second group must have at 3 students'

atleast?

[silver stump]

yea

first group must have EXACTLY 3

the second group has AT LEAST 3

so number of students in the second group is greater than or equal to 3

[silver stump]

this is correct

[vale vortex]

its a interesting thought problem

how do you think

your suppose to do it

no idea?

[silver stump]

i mean

i figure that

the second group will automatically arrange

after you pick the first group

[onyx cloak]

"Upon adhering to this request, the amount of ways in which the teacher can assign the groups halves." is smth missing here?

[silver stump]

so its just n choose 3

to arrange the groups at first?

before the request

[silver stump]

nothing is missing

[onyx cloak]

then what does that line mean

[thin kelp]

it sounds fine to me

keep going

[silver stump]

it just means that the teacher agrees

and the number of arrangements is halves

halfed

of combinations*

[silver stump]

that's all i got

[onyx cloak]

ohk

[silver stump]

i can copy the exact question

in case my wording is weird

[onyx cloak]

nah its fine, i just misread

[thin kelp]

u want to create a group of 3 that includes one of the 2

[vale vortex]

$W = \binom{n}{3}$

[clever fjord]

Diamond

[silver stump]

A teacher takes n students on a field trip. The students are assigned randomly into two groups.
For safety reasons, there must be exactly three students in the first group and at least three students in the second group
The teacher will randomly assign three students to the first group and the other students to the second group
a) Write down an expression for the number of ways that the students could be arranged
b) Two of the students ask the teacher not to work in the same group. The teacher agrees and now finds that the number of ways to assign the students is halved.
Determine the value of n

that's the exact question

[silver stump]

wdym

[silver stump]

this is the answer for part a

[vale vortex]

$W_{\text{limited}} = 2\binom{n-2}{2}$

[silver stump]

now for part b i am just so confsud

[clever fjord]

Diamond

[silver stump]

i'm pretty sure thats what the mark scheme said

something like