help-27

[undone chasm]

yea

[iron holly]

How is that related though?

[undone chasm]

wdym

[iron holly]

Why is angle ACB just angle AOB divided by 2?

[undone chasm]

its a circle theorem

ok so like

take two points on a circle

the arc subtended by the two points from the center of the circle has twice the angle of the same arc from the edge

[iron holly]

Is it the theorem that “the angle at the center of a circle is twice the angle at the circumference when both are subtended by the same arc”?

[undone chasm]

yes

this is one case of that

the other is when c is actually between a and b

please please pleaseeeeeeee never ask math to AI

[iron holly]

So angle AOB is double angle ACB because of this theorem

[undone chasm]

yea

[iron holly]

I’m trying to visualize this better

[iron holly]

I see that angle AOB is found through its isosceles properties with angles OAB and OBA

[undone chasm]

we can prove it

do you see AOP and AOQ are also isoceles?

because again both radii

[iron holly]

Ya

[undone chasm]

do u know the exterior angle property

BOP = OPA + OAP

BOQ = OQA + OAQ

the exterior angle is the sum of the two other interior angles

since OPA=OAP and OQA=OAQ, BOP=2*OAP, and BOQ=2*OAQ

soo

BOP+BOQ=2*(OAP+OAQ)

therefore POQ=2*PAQ

[iron holly]

Very interesting

[undone chasm]

I’d recommend drawing each step out

it helps u visualize

[iron holly]

Wait wait wait so one last time in context of the original problem, the whole premise is that angle ACB is half of angle AOB due to this theorem?

[iron holly]

Unlike this though ACB is such an awkward angle in relation to AOB

Like it has part of its angle sort of cut off for lack of better words

[undone chasm]

its just an example figure

you can try it with any angle!

[iron holly]

Is this part of no concern since triangle ACB encompasses triangle AOB?

[undone chasm]

upp

i mean it’s just lines, not like

encompassing per se

[iron holly]

ya true

sorry for the instantiation of 3D terminology

wait wait wait so for any problems like these

if an angle lies along the circumference of a circle in a triangle

a triangle “inside” the larger triangle with a point on the circumference is half the inner angle not on the circumference?

[undone chasm]

what

hold on i have to process

[iron holly]

np

[undone chasm]

wdym by along

[iron holly]

like on the circumference of a circle

[undone chasm]

like

for example like point C?

[iron holly]

ya

[undone chasm]

ok

“with a point on the circumference”

?

[iron holly]

for example point C

maybe my interpretation was lost somewhere along the way

the solution is based off the exterior angle theorem right

WAIT WAIT WAIT

so just as AOB is double x

nvm

OH MY GOSH

ANGLE ACB IS SUBTENDED BY AB

AND THE INSCRIBED ANGLE THEOREM STATES THAT ANGLE AT CIRCUMFERENCE IS HALF THE ANGLE AT THE CENTER

AND WE FOUND THE ANGLE AT THE CENTER USING X IN RELATION TO AOB

@undone chasm

[undone chasm]

sorry

uhh

yea

sorry 😭

[iron holly]

this is revolutionary

[undone chasm]

you should publish it

[iron holly]

IMMEDIATELY

so in sum

[undone chasm]

Mary

[iron holly]

Angle AOB is found by noticing that triangle AOB is an isosceles triangle in which AO and BO are radiuses extending from the center. Noticing that triangle AOB is an isosceles triangle due to the fact that two sides of the triangle are equal to the circles radius, we can understand that the angles opposite to those sides (X and ABO) are equal. From this, we can conclude that AOB is equal to 180-2x. Now looking at angle ACB, we can see using the inscribed angle theorem which states that an inscribed angle on a circle is half the measure of the central angle that subtends or forms the same arc (which in this case, is AB, which angle ACB forms, and which triangle AOB forms). From this, we can deduct that angle ACB is half of 180-2x, or 90-x. We know that w + 90 - x is equal to 180 because ACD is a straight line. Rearranging for w: 180=w+90-x turns into w = x + 90.

I just needed to realize this

That the angle is related to the arc AB

thank you for this @undone chasm

this has bothered me for a hour

What math are you in now?

I see your pre uni

[undone chasm]

i mean

like

[iron holly]

IS IT WRONG 😱

[undone chasm]

technically im in middle school (im >13 tho)

no it’s correct

but i like doing advanced math 🥰🥰🥰

[iron holly]

are you taking high school math in middle school?

[undone chasm]

is this high school math?

[iron holly]

I think the normal track for US high school students is to take geometry in 10th grade

[undone chasm]

oh what

[iron holly]

yeah lol

I never took geometry 💀

I’m passing calculus 2 with flying colors now

But I’ve never seen this stuff before

I really like it though

I haven’t had a formal Euclidean geometry class though

[undone chasm]

my favorite thing in geometry was 2 column proofs

[undone chasm]

what are you learning rn?

[iron holly]

Never heard of it

Series

I love series

[undone chasm]

my class is doing things in a crazy order

see that’s normal

[iron holly]

wait what class are you in

[undone chasm]

somehow im in lhopitals rule after improper integrals 💀

[iron holly]

we did that too 😭

[undone chasm]

im taking calc again to be able to take the AP

[iron holly]

I believe it’s because limits are an important preface to take before series evaluation

They should’ve gone over lhopital before improper though

[undone chasm]

oh i see

[iron holly]

Improper integrals are used a lot in series

[undone chasm]

we haven’t gotten to integral test yet lol

[iron holly]

What country are you from 💀

[undone chasm]

US

murica 🇺🇸🇺🇸🇺🇸🇺🇸🏈🏈☕️🦅🦅🦅🦅🦅

[iron holly]

Bro and you’re of middle school age JEEZ

What do you want to do after high school

I better see you at Wall Street

Or in the astrophysics department at mit

[undone chasm]

lmao

ig my real dream is to become a professor

[iron holly]

what do you want to specialize in?

[undone chasm]

either analytic geometry or arithmetic dynamics

[iron holly]

being a professor seems awesome

[iron holly]

very interesting

have you heard of differential geometry

I might take it in 3 semesters

I have to take real analysis first though

I didn’t take geometry in high school but I don’t think that’ll be an issue

[undone chasm]

i hateeeeee real analysis

complex analysis is so much better tbh

[iron holly]

I don’t have enough knowledge to be able to make a comment on this

[undone chasm]

complex is the only complete algebra and it shows

so many edge cases in RA

CA is just like

wow

a path

[iron holly]

Like as in no question is left unanswered?

[undone chasm]

wdym

[iron holly]

By this I’d assume you mean there’s some areas of RA that are ambiguous and that this doesn’t exist in CA

[undone chasm]

yea

[iron holly]

I see

They should take some ideas from physics and introduce ideas sort of like dark matter to explain the uncertainties

lol

I gtg

tysm btw

I wish u the best I’m sure you’ll go very far

[undone chasm]

omgg thank you!!!

[iron holly]

have a good day!!!!

[undone chasm]

you too!!!!!!!!!!

(you should type .close)

[iron holly]

oh ok sorry 😭

.close

[devout snow]

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[iron holly]

cya and good luck again!! :))

[devout snow]

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[dusk tide]

hii

[dusk tide]

technically, would the answer be 5?

[supple knot]

well use pen and paper to check your work

[devout snow]

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[uneven thistle]

i was thinking like adding up two seperate integrals? one from 0-4 and one from 4-6 but thats not right. also im confused abt the equation for f(x) bc it doesnt look like the equation is just x?

[winter patrol]

from the graph that looks like sqrt(x)

[uneven thistle]

it is

but how do ik that from the question

is it just a typo ?

oh yeah i looked it up and the college board question is sqrt

[devout snow]

@uneven thistle Has your question been resolved?

[uneven thistle]

umm no

[devout snow]

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[safe nova]

could anyone calc 1-2

[faint gorge]

this looks like something with arcsin(y)

[safe nova]

how did u get arcsin y out of that

[faint gorge]

i said something

besides that iknow the derivative of arcsin

you can derive it using trigonometry

[safe nova]

hmmmmm

what about the limits

do i do anythign with those

[solid hull]

integrate then substitute and compute the upper and lower bounds

[safe nova]

so i do what then what then what

intergrate what

the bounds?

[solid hull]

you compute the integral of the expression

then you subtract the lower bound substituted by the upper bound substituted

[safe nova]

bro im acc brain dead

how do u subsitue the lower bound by the upper bound

[devout snow]

@safe nova Has your question been resolved?

[proven salmon]

I dont
understand what gizmic is saying

[devout snow]

@safe nova Has your question been resolved?

[proven salmon]

No it hasnt and i wanna know too!

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[glacial sable]

Just for fun I was looking at this math problem where you take a tetrahedron shaped crystal consisting of one atom on the top layer, three on the next-to-top layer, six on the third layer, ten on the fourth layer, and so forth If there are exactly 1,000,000 layers, specify the total number of atoms in the entire crystal. Pretty straight forward however, I was told it could be solved on a pen and paper and without the use of a computer. My first thought was to just to take a cube with sides of a million (volume of a quintillion) and then find the volume of the tetrahedron in the cube but thats only aprox

[undone chasm]

so

do you know how to count the atoms in each layer?

[glacial sable]

I am trying to find a solution without using a calculator.

[undone chasm]

yes

do you know how to count the atoms in each layer without a calculator?

[glacial sable]

yes however theres a million layers?

[undone chasm]

hmm count isn't a great word

like

if i asked for the number of atoms in the 896,562th layer could you give me an expression that counts how many atoms there are?

[glacial sable]

it would be 896,562 + every number before it

[undone chasm]

yes

ok so the first step would be to find an expression for that

that doesn't involve 896,561 plus signs

[glacial sable]

n

[undone chasm]

what's n?

the nth layer doesnt have n atoms

[glacial sable]

the nth layer has the nth layer as a number + every number before it right?

then you do that for every layer

[undone chasm]

yes but we can find a concrete mathematical formula for the atoms in the nth layer

like without "and so on" or a ridiculous amount of addition

[glacial sable]

n * (n + 1) / 2

[undone chasm]

yea

howd u get that?

[glacial sable]

i googled it lol

[undone chasm]

i just ask bc everyone gets it in a diff way and i wanted to

oh ok lmao

so thats the formula

[glacial sable]

im not great at math but i like the theoretical side of it what really interested was finding a solve you can do on paper

[undone chasm]

now we need to just sum that for the million layers in the tetrahedron

oh ok

so then we can separate this into n^2/2+n/2

we use the formula above on the right one

sum(n^2/2)+n(n+1)/4

do u wanna figure out the sum for n^2 or should i tell you lol

[glacial sable]

isint it just n * n

[undone chasm]

i mean

that's what n^2 is

but the sum of 1^2+2^2+3^2+...+(n-2)^2+(n-1)^2+n^2

[glacial sable]

hey boss I skipped school to drag hoses around for a living like i said im bad at math

[undone chasm]

soooo is that a no

its n(n+1)(2n+1)/6

[glacial sable]

gotcha

[undone chasm]

so the sum is n(n+1)(2n+1)/12+n(n+1)/4

now to simplify this

$\frac{n(n+1)}{4}\cdot\left(1+\frac{2n+1}{3}\right)$

[woven radish]

blahaquil

[glacial sable]

ok and i just plug 1 million into n ?

[undone chasm]

$\frac{n(n+1)(n+2)}{6}$

[woven radish]

blahaquil

[undone chasm]

yes

[glacial sable]

1 sec lemme open up desmos

[undone chasm]

it's 166,667,166,667,000,000 fyi lol

there exist sum formulae for all n^k

and thence you can predict your tetrahedrel crystals in any dimension!!!!

[glacial sable]

and theoretically this can all be done on paper

[undone chasm]

not just theoretically lol

[glacial sable]

i meant theoretically more in the sense of someone smarter than me can

[undone chasm]

oh

yea

so basically this case (3d)

[glacial sable]

hey i actually have to know some math for my job so im trying to get a little better but its mostly pressure stuff

[undone chasm]

so like technically

wait do you know sigma notation?

[glacial sable]

yes

[undone chasm]

yea

so the tetrahedral case is like $\sum_{k=1}^n\sum_{j=1}^k j$

[woven radish]

blahaquil

[atomic condor]

the sum of triangular numbers

[undone chasm]

ya

[glacial sable]

welp thx

i just smashed my finger in a car door ima go eat dinner now

[undone chasm]

(you should type .close)

[gleaming socket]

Ouch

[gleaming socket]

.close

Nothing happened

[undone chasm]

@glacial sable opened it

so they gotta close it

[atomic condor]

xd

[devout snow]

@glacial sable Has your question been resolved?

[glacial sable]

defenitly broke it good thing it was my middle finger and not a important one

ahhh shit i cant play guitar

[devout snow]

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[molten birch]

Deez questions

[molten birch]

Answers:
First: D
Second: C
Third: C

In the first question I can only prove the first statement correct

[devout snow]

@molten birch Has your question been resolved?

[molten birch]

<@&286206848099549185>

[devout snow]

@molten birch Has your question been resolved?

[molten birch]

[gray summit]

😅 I'm getting ac > 3

[warped hearth]

For question #20, ACE + ECD will equal AFE. ECD is 60 degrees because thats an equilateral triangle. 60 + 20 is 80
Question #39, you should get 48 because the similar lines at AE and AB make two similar triangles kind of mirrored if you know what mean, and FCB = ADB if that's true

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[inner ibex]

HOW DO i do part a?

[pseudo basin]

surely it's just that the resultant force on the plank is 0 bc it's in equilibrium

[inner ibex]

so i do force of boy down + force of girl down + force n = 0?

[pseudo basin]

...yes

[inner ibex]

oh

do i make gravity negative then?

.close

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[placid barn]

!gethelp

[placid barn]

Hello, I'm studying to take a G.E.D test, I'm curious if anyone could help me learn some of whats on this practice test.

These were the questions I don't understand

[pseudo basin]

so the topics you're missing then are:

• volume of a cone (and perhaps volume in general)
• forming linear equations from an x-y table
• surface area of a cylinder (and perhaps surface area in general)

which one do you want to start with

[placid barn]

Let's start with 2 & 3, I made a Google Doc to keep detailed notes, may I DM you editing access? The doc has been very helpful.

[pseudo basin]

you want me to edit your notes...?

well, ok i guess i can at least take a look at them

[placid barn]

DM's sent

May we speak in a VC?

[pseudo basin]

ok yeah sure you've DM'd me anyway

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[terse bear]

Are my choosing true ? Not a test just a bonus exercise

[terse bear]

<@&286206848099549185>

[frozen aurora]

all good

[wheat totem]

hey anyone here to help

hey does anyone know how to derive the magnitude of resultant of relative velocity of b wrt a my head hurts at this point i cant make any sense out of that formula and how it was derived

[devout snow]

@terse bear Has your question been resolved?

[radiant hare]

Open your own channel

Also give more context

[devout snow]

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[abstract bane]

how do i do 1 (b)

[visual hazel]

do you have PR

[abstract bane]

no

I know (a) i) and ii)

[visual hazel]

well you kinda need to know PR to do b

[abstract bane]

idk then

.close

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[visual hazel]

?? ok

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[smoky gyro]

i know z must be between 0 and 2 inclusive but im struggling to justify why using algebra and a geometric arguement. i was told that that i couldnt use the triangle inequality becasue u cant order complex numbers and then told that i cant separate the absolute value into cases so i dont know what im supposed to do??

[pseudo basin]

i was told that i couldn't use the triangle inequality
says who?

like, yes, with complex numbers the breakdown into cases doesn't work anymore -- but the triangle inequality still does, and i would even argue it's much more visual in the complex case!

@smoky gyro

[smoky gyro]

well i wasnt directly told but for my inequality arguement a helper said i cant do it like that because complex numbers cant be ordered

[pseudo basin]

do you remember who that helper was

cause i think they were misguided

|z+w| <= |z| + |w| DOES hold for z and w complex

because, crucially, this inequality does NOT involve comparing complex numbers

the magnitude of a complex number is real

[smoky gyro]

not sure

[pseudo basin]

anyway, your geometric argument essentially is that |z-2| + |z-0| >= |0-2|, considering the triangle with vertices at 0, 2 and z

[pseudo basin]

yes exactly

only if z falls on the line segment between 0 and 2 will there be equality

[smoky gyro]

right

[pseudo basin]

(and the "less than" case happens precisely never)

[smoky gyro]

we have this from the triangle inequality right

[pseudo basin]

yes

[smoky gyro]

but the original question is asking me for a<=b

but only way for a<=b is if a=b

[pseudo basin]

we know this holds for every single z

[pseudo basin]

correct!

[smoky gyro]

yes

but how can i justify that z must be between 0 and 2? or is the diagram enough

[pseudo basin]

the diagram is enough

[smoky gyro]

okay

i asked austin how he would do it and he suggested that my inequality is in the form of an ellipse but itll be a line segment instead between 0 and 2

[pseudo basin]

@weak cove is he talking about you

[smoky gyro]

yes

[pseudo basin]

ellipse with foci 0 and 2 sure yeah

[smoky gyro]

yes

this is what he said

[pseudo basin]

yeah he's right

[smoky gyro]

but ive never rlly worked with ellipses before so is the triangle inequality arguement enough

[pseudo basin]

yes

[smoky gyro]

ok thank you

.close

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[inner ibex]

what is linear dependenced?

[smoky gyro]

hello bum chicken

i wish i knew

[hollow ice]

Do you know mutually independent? Its the opposite of that

Some linear combination of a and b gives you the vector c

if they are linearly dependent

[inner ibex]

whats mutqually indepdnent

i did that in my probaqbility test last year i think

[hollow ice]

something like ax + by = c, where x and y are some constants

[hollow ice]

Its similar, but somewhat tough if you wanna relate it to this. Just go with ax+by=c

[inner ibex]

o

[hollow ice]

This is a formal definition if you want one

[inner ibex]

so like this

[hollow ice]

yea

[inner ibex]

hmm but then what

[hollow ice]

but m and n are real numbers

[inner ibex]

how do I prove it

[hollow ice]

If you manage to find the real numbers m,n then thats that

you proved them dependent

[inner ibex]

wow dis is really hard

can I use simultaneous equations

[hollow ice]

Yea

sure

[inner ibex]

or am I just supposed to look at it and know

[hollow ice]

separate the components, and then solve. Its not tough at all

[rough nova]

Yeah this is basically simultaneous equations with i,j,k as variables

[inner ibex]

now wat

[hollow ice]

You only got 2 variables

Use two of the equations to find m and n

[inner ibex]

do I do solve simultaneous

[hollow ice]

ye

[inner ibex]

for 2 equations for m and n

and then I sub into remaining one to see if the solution matches?

[hollow ice]

Yea

[inner ibex]

oh

ok

makes sense

ok thank u !

[hollow ice]

np

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[bleak arrow]

conceptual question, when would you parametrize in cylindrical unit vectors as opposed to cartesian ones?

r(t) = (p(t), 0 , z(t))
r'(t) = (dp/dt, p*dphi/dt, dz/dt)

[bleak arrow]

use another channel, this channel is occupied

[marble tree]

Ok

[devout snow]

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[bleak arrow]

<@&286206848099549185>

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[novel drift]

I have no clue how to solve this, I have the answer but I just dont see how to get to it

<@&286206848099549185>

[hazy cradle]

have u made diagram?

[novel drift]

yeah

[hazy cradle]

show?

[novel drift]

a is the whole BC length, same for b

[hazy cradle]

,rccw

[novel drift]

B1 and A1 are mid points

[novel drift]

where the circle touches

i have to find AB

in expression with a and b

You got anything?

[hazy cradle]

nope

[novel drift]

sheet

[hazy cradle]

trying to see if there is any property

[novel drift]

theres one but i dont see how i can apply it

when the sum of the top and bottom angles is 180 and is equal to the sum of the side angles in the quadrilateral only then can there be a circle around it

[hazy cradle]

ye the basic case for a cyclic quad

[novel drift]

yeah

the answer is root of (a^2+b^2)/2

[hazy cradle]

idk bro

imma have to go

[novel drift]

alr bro thanks for trying

[faint prawn]

I got something see if it helps

Draw all three medians
And try power of point on points A and B

with respect to circle

[novel drift]

that could work

it does not work

[devout snow]

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@novel drift Has your question been resolved?

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@novel drift Has your question been resolved?

[hazy cradle]

ig it comes with cosine rule

but its too long

and messy

[devout snow]

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[rough hatch]

1

[rough hatch]

how do i find the bounds

[cinder bobcat]

since the circle is on the right half of the xy plane, your angle needs to cover the right half of the xy plane

so you should do -pi/2 to pi/2

[rough hatch]

ohh i see

so its in terms of the quadrents

rather than the circle itseldf

[cinder bobcat]

it's easier to at least think about the quadrants to find the bounds

eyeballing it usually works luckily

but sometimes you need to do more work to find your bounds if the shape is really complicated

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[rough hatch]

i need help with this

[rough hatch]

Is this the right track

Wait

There

[pseudo basin]

waaa

,rcw

[rough hatch]

I forgot the ^2

[pseudo basin]

almost

the bounds of the integral are correct

though i do recommend replacing them with like $-\alpha$ and $\alpha$ or something and note to the side that $\alpha = \arccos(1/18)$ or sth

[woven radish]

Ann

[pseudo basin]

anyway the integrand should be $(18 \cos(\theta))^2 - 1^2$

[woven radish]

Ann

[rough hatch]

Yea i noticed and added the^2

[pseudo basin]

yeah ok correct integral now

[rough hatch]

Rewritting this right

[pseudo basin]

-1 integrates to -θ

oh er

you didnt integrate yet

my bad

[rough hatch]

Yea

[pseudo basin]

this is correct thus far i think yes

[rough hatch]

For integrating

@pseudo basin

[pseudo basin]

seems ok but why b and a in the last line

[rough hatch]

u said to use that

the alpha and beta

[pseudo basin]

where what

where did i mention a beta

i said alpha and -alpha

[rough hatch]

oh

[pseudo basin]

did you read something that didnt exist

[rough hatch]

no i just forgot
since they are the same number just pos and neg

@pseudo basin

[pseudo basin]

how did arccos(1/16) happen

also more importantly

you're asked for a decimal answer

why are you not using a calculator

or hell even desmos at this point

[rough hatch]

thats suppose to be a 18, the circle was just too small

[rough hatch]

im trying to put it in desmos rn

[pseudo basin]

it's very straightforward to put it in desmos lol

[rough hatch]

wait

oh is that the graohing one

[pseudo basin]

what other desmos thing were you thinking of

are you just using its standalone calculator

desmos graphing calculator is very powerful

[rough hatch]

how do i do this

[rough hatch]

the scientific oen?

[pseudo basin]

do what

anyway yeah ig the scientific one doesnt have the integration stuff

[rough hatch]

i was trying to figure out how to enter it, but then i saw how u did it

[pseudo basin]

entering stuff is fairly straightforward in desmos

[rough hatch]

i think i did something wrong

[pseudo basin]

thats odd...

try 505.870 maybe?

probably some rounding bullshit...

[rough hatch]

didnt work

[blazing jetty]

maybe it rounds off to nearest integer

[pseudo basin]

it says 3dp

.... i would flag this question somehow tbh

im not seeing where we are both going wrong here

[supple knot]

where does 323 come from

[pseudo basin]

OH

WAIT

[blazing jetty]

oh mb lol

[pseudo basin]

OH

MY FUCKING

GOD

WE BOTH FORGOR TO HALVE IT

theres a half

that we both just completely forgor about

divide that shit by 2 lol

252.935 should be it

[rough hatch]

💀

🥲

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[worldly tartan]

Maybe a small question, but i can't seem to understand how i am supposed to explain how the projection of vectors formula can be used to find the direction of a vector, when it is reflected of a blank surface
i have proved the formula, but i can not seem to understand how it is used. When googling i keep getting redirected to other formulas that i highly doubt that im supposed to know for my assignment
and help is appreciated

[worldly tartan]

I thought about using the orthogonal vector instead to find it, but i did not come far with that

my vector is (1,2,3) and the plane im given is 2x-3y+z=4
and so i figured the orthogonal vector would be (2,-3,1)
which i then substituted into the projection of vectors formula ,but im not sure of where to go from here

I mean that im supposed to explain(in text) how it its possible to use the vectorprojection formula to find the orthogonal line that is reflected of a blank surface

but it seems as if the vectorprojection formula can only describe when the vector is going to hit the blank surface

not the vector after it is reflected orthogonally

i hope that makes more sense

kind of. I have to use the projection formula to explain how it can be used to find the orthogonal line

that's what i did up earlier

but im not sure of where to go from there

Yeah

no

the thing is im not allowed ot use the reflection formula in the wording of the problem

hmm ive drawn it out on paper

I agree

my teacher has been known to make mistakes, but she is nice so it's alright

the question is unfortunately in Danish and hard to translate, but at least im glad im not the only one who finds it vague

Oh nice 😄

Swedish+

?

Ah alright

not quite its more like the projection comes down from up top and lands and is then reflected out orthogonally

It's a final project for Danish A level math

Im trying to draw a beautiful rendition in paint

yeah i do get it

my problem is just how to use the projection of vectors formula to calculate the angle of the green reflection

ey thanks anyways

ill see what i can do

https://tenor.com/view/warcraft-gaming-gif-27702118

<@&286206848099549185>
i am still stuck on my problem that is mentioned above^
any help is greatly appreciated

[zenith star]

!status

[devout snow]

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

[worldly tartan]

2

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@worldly tartan Has your question been resolved?

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@worldly tartan Has your question been resolved?

[worldly tartan]

https://tenor.com/view/worried-monkey-worried-monkey-gif-5992280527583416561

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@worldly tartan Has your question been resolved?

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[worldly tartan]

https://tenor.com/view/monkey-gif-18890916

[devout snow]

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[worldly tartan]

.solved

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[bleak arrow]

Trying to find the volume with the following constraints. If I wanna start with dz, the limits would be -sqrt((x-y^2)/2) to sqrt((x-y^2)/2), to find the 2D region R when the triple integral reduces to a double integral, can I just set z = 0? or do I have to set the two z's equal to each other?

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[gray summit]

Triple integral of what?

[bleak arrow]

1dV

The volume

[gray summit]

Oh

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[summer charm]

suppose i want to calculate the integral of sinz/z over a unit circle

[devout snow]

Please don't occupy multiple help channels.

[summer charm]

suppose i want to calculate integral of sinz/z over a unit circle

i guess that is called residue

[pseudo basin]

$\oint_{|z|=1} \frac{\sin(z)}{z} \dd{z}$?

[woven radish]

Ann

[summer charm]

Yes

but the radius of circle wont effect outcome

so i take radius very small

then use sinz=z-z³/3!+z⁵/5!-z⁷/7!...

[pseudo basin]

i mean sure

sin(z)/z has only a removable discontinuity at 0

you will find that the residue is just zero and so is your integral, won't you

[summer charm]

but integral is 2π

[pseudo basin]

?!

says who

[summer charm]

Orginal question was to integrate sinx/x from 0 to infinity

[pseudo basin]

...

[summer charm]

i do it from -infty to infty

[pseudo basin]

what an XY 💀

[summer charm]

then i take appropriate curves

what is XY?

[pseudo basin]

http://xyproblem.info/

[summer charm]

oh

sorry

Is this right thinking?

or am i wrong?

[pseudo basin]

so the integral over that top contour should still be 0 yes but you can't really take the limit as R -> infty here

because $|\sin(x + iy)| \approx \sinh(y)$ for big $y$

[woven radish]

Ann

[wet rune]

ummmmmmmmmmmm

[wet rune]

sintaxis error

[summer charm]

oh yes i didnt consider that |sinz| could be > 1

Thanks

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[faint gorge]

maybe that helps
https://www.youtube.com/watch?v=8x_UjFUySRg

[acoustic leaf]

one thing to be careful about is that different conventions swap the role of theta and phi

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[undone chasm]

can we derive a method to tell if a given cubic is monotone by analyzing the sign of the general form of its derivative?

[undone chasm]

like is there some relation 3a, 2b and c must have that means 3ax^2+2bx+c > or < 0 for all x?

[lunar harbor]

Recall the discriminant (specifically for a quadratic)

[undone chasm]

negative determinant lol

thats what I just found too

ok I think I got it

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[slate pawn]

if anyone is free I could use help figuring out mathematica code for this line integral. I missed a week of class and my professor is out rn, so I need help learning the theory/correct way as well.

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[frozen horizon]

where do I go from here and is this right so far

[strange axle]

You got your signs mixed up in the first/second step

Oh wait nvm I misread

[frozen horizon]

oh okay

[strange axle]

Looks right so far

First term looks like a trig sub

Second I would say integration by parts and then a u sub

[frozen horizon]

i cant just do trig sub for the second one directly?

[strange axle]

3rd is probably also a trig sub, but Idk, I would check

Is there a trig sun that works well there?

[frozen horizon]

idk i kinda forgot how it works lol

[strange axle]

Lemme see

Actually it probably would work there

But I think you'd end up with some sort of product of trig functions, you should check first though

Do you need a reminder on trig subs?

[frozen horizon]

yeah i dont remember how it works

ill just watch a video or something

thanks

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[bleak arrow]

This is the question

[fossil locust]

I don't understand why you are trying to do this

just set up the triple integral first

[bleak arrow]

i already did

for dz
-sqrt((x-y^2)/2) to sqrt((x-y^2)/2)

i'm trying to get the projection

[fossil locust]

z ranges from -sqrt((x-y^2)/2) to sqrt((x-y^2)/2)

so then what do your y and x range from?

[bleak arrow]

can i just set z = 0 to find the domain D?

[fossil locust]

no

set up $\int \int \int 1 \ dz \ dy \ dx$ with the appropriate bounds

[woven radish]

south

[bleak arrow]

so I set them equal to each other?

[fossil locust]

no

[bleak arrow]

if i do that, I get x >= y^2, and there's y>=0 and x+y^2<=8

so that's my region R

[fossil locust]

well how I would proceed is $y^2 \le 8 - x \implies 0 \le y \le \sqrt{8 - x}$

[woven radish]

south

[bleak arrow]

I already know how to do double integrals. I'm just trying to get the region R here.

from the constraints

[fossil locust]

I know, and I'm helping you with that
I've bounded y for you

I guess if this helps you

pictures help a lot

wait hold on sorry

I assumed x >= 0 but like

[fossil locust]

the region in the original question is unbounded I bet

hopefully I'm not making some silly error

oh wait, any sum of squares must always be positive, so 0 <= y^2 + 2z^2

so yeah that forces x >= 0

my pattern recognition lit up and assumed something without me consciously realising why

$y^2 + ( \sqrt{2} z)^2 \ge 0$ to be extra clear for all $y, z \in \mathbb R$

[woven radish]

south

[bleak arrow]

do you agree that the region is given by the following equations:

x + y^2 <= 8
y^2 <= x
y >= 0

My question is, how did they arrive to these equations. x + y^2 <= 8 and y >= 0 are given, but how did they arrive to y^2 <= x?

[fossil locust]

so that you have 0 or positive <= 0 or positive for y^2 + 2z^2 <= x

[fossil locust]

should be y^2 <= 8 - x

that's a typo in your book

[bleak arrow]

how did they get y^2 <= x

it seems they set z = 0

[fossil locust]

oh right, well you can make z range from 0 to sqrt((x - y^2)/2) then don't forget to multiply the original interval by 2

it's an argument by symmetry

case a) with the function being 1

[bleak arrow]

if u set the two bounds for z equal to each other, u also get y^2 <= x

[fossil locust]

but I think worrying about such tricks is confusing you

[bleak arrow]

u just graph it right? u don't use these tricks 😆

[fossil locust]

yes

just graph it and $0 \le y \le \sqrt{8 - x}$ and $0 \le x \le 8$

[woven radish]

south

[bleak arrow]

I hate graphing in 3D.

[fossil locust]

you don't need to graph in 3D!

[bleak arrow]

then how do u get ur region

[fossil locust]

your whole point about graphing the 2D region after the z-bounds was correct!

[bleak arrow]

so let's say I wrote my z bounds

now I wanna get the 2D region

[fossil locust]

z ranges from -sqrt((x-y^2)/2) to sqrt((x-y^2)/2) is correct

then you just need to bound x and y

[bleak arrow]

how do i get the 2D region

[fossil locust]

so you need to graph the region where $x + y^2 \le 8$ and $y \ge 0$

[woven radish]

south

[bleak arrow]

what about x>=y^2

u didn't include that

[fossil locust]

forget about it

[devout snow]

@bleak arrow Has your question been resolved?

[devout snow]

@bleak arrow Has your question been resolved?

[bleak arrow]

why? don't we wanna get the 2D region? isn't it a constraint? in the solution they use it

[fossil locust]

oh they've done a different order

[fossil locust]

AH okay so $y^2 + 2(0) \le y^2 + 2z^2 \le 8$ and that's how

[fossil locust]

then if you plot the region you will get that y ranges from 0 to sqrt(8), oh but there's another restriction cause of z arghh

[bleak arrow]

not -sqrt(8) to sqrt(8)?

[fossil locust]

fair enough it's good to think of it from multiple perspectives

[fossil locust]

nope, look at the graph again

wait

the graph isn't the only story

[woven radish]

south

[fossil locust]

yes because if $z = 2$, you get $y^2 + 8 \le x$ but you also have $x + y^2 \le 8$ remember

so that's the limiting case, $y^2 + 2z^2 = 8$ exactly

[woven radish]

south

[bleak arrow]

i got this

[fossil locust]

nearly correct

should be the upper half of that region

[bleak arrow]

oh ok

my bad

[fossil locust]

ah so yeah when you combine all the restrictions you should use your graph not mine

[bleak arrow]

so u basically set z = 0

[fossil locust]

yes so any time you see z^2, z^4, z^6, z^8 and so on

those are all squares

[bleak arrow]

so i set z = 0 in these cases?

[fossil locust]

z^2 >= 0, z^4 >= 0 for all real z and so on

[fossil locust]

yes!!

well you set z^2 = 0 or z^4 = 0 which is just z = 0

so same thing

yes so there's a hidden limiting case and that's why

it's so messed up

[bleak arrow]

what if it's only defined for positive square root

do i still use z = 0

[fossil locust]

oh yeah absolutely

so z^(1/2) and z^(1/4) and z^(1/6)... too

[bleak arrow]

because there was one problem where it's wrong to use z = 0, do u wanna see that example?

"Find the volume between x^2 + y^2 + z^2 = 9 and z = 1/8 (x^2 + y^2)"

Here we can't just plug in z = 0.

even though our z is squared

[fossil locust]

oh that's best done in spherical coordinates

[bleak arrow]

i know but if we did it in rectangular we can't plug z = 0 to get the region

[fossil locust]

I think it just comes from subbing in no?

wait

basically you should try to replace x^2 + y^2 here

so you get 8z + z^2 = 9 and that's very different now

why I tell you not to use Cartesian is because r^2 + z^2 = 9 and z = 1/8 * r^2 is much easier to see, how to proceed

[bleak arrow]

yes but how do u get ur region

u just combine them

[fossil locust]

yes

[bleak arrow]

why can't we just set z= 0

when can we just do that

[fossil locust]

like of course you must have z >= 0 because z = 1/8 * (never negative)

but there's an even stricter condition

now you know what values z is exactly

from subbing in

[fossil locust]

try to sub in whenever you can, random things

if you can solve for something, solve it

otherwise you're just left with the default assumption that z^2 >= 0

[bleak arrow]

i think here for z = 1/8 (x^2 + y^2), we can't just plug in 0 as x^2 + y^2 = 0 doesn't make sense

[fossil locust]

so in that example, z >= 0 overlaps with z = -9, 1

you reject the -9 and you're left with z = 1 only

[fossil locust]

yeah that's not a limiting case, z = 0

you already know what z can be in that situation

so you have to instead figure out which z values are possible

[bleak arrow]

i see

[fossil locust]

I lied obviously knowing this now use cylindrical

[bleak arrow]

.close

[devout snow]

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[devout snow]

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[glad peak]

Yoo

[glad peak]

Can someone help me with basic calc

[blazing jetty]

sure

[lime plaza]

Send the question first so bot pins it

[glad peak]

It's not a question tho. I mean ok

[lime plaza]

That's a long question

[blazing jetty]

frfr

[glad peak]

Yo Sorry, give me some times I'll make the question and be right back

[lime plaza]

Lol he's been typing for 5 minutes

[glad peak]

It's my personal question actually..

[blazing jetty]

😭😭

[lime plaza]

:(

[glad peak]

and I can't think how I the fck I ask it

[lime plaza]

..

[glad peak]

It's related to financial..

I'll be brb you can help someone else if needed

[devout snow]

@glad peak Has your question been resolved?

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[leaden star]

I tried all kinds of rationalisation

What should I do

No lhopital

[upper ermine]

$\lim_{x \to 2} \frac{\sqrt{x} - \sqrt{2} + \sqrt{x + 2}}{\sqrt{x^2 -4}}$

[woven radish]

MOD BY ZERO ERROR

[leaden star]

Yee

[upper ermine]

hmm

im assuming u cant do a cheap graphing or numbers close to 2 being plugged in thing right

[leaden star]

Yes

[upper ermine]

gimme a sec

[leaden star]

I cant

Its supposed to be some kind of rationalization

[mild comet]

So no calculator, right?

[upper ermine]

isnt the function undefined for x < 2

so then shouldnt the lim not exist

[leaden star]

Yes

[upper ermine]

cuz its defined for 2+ but not for 2-

[leaden star]

Yes

[leaden star]

Hmmm

[strange axle]

Have you tried doing difference of squares on the bottom then partial fractions

[leaden star]

Yes

[upper ermine]

id just say dne cuz lhs != rhs

[strange axle]

Wait wait wait

I'm being silly

The bottom has a higher power of x in it

So it has to go to Infinity quicker

[upper ermine]

oh wait yeah lol

[upper ermine]

wait no thatd mean it goes to 0..?

[strange axle]

Meaning this will asymptote

[strange axle]

Oops yeah

[leaden star]

Ans is 1/2

[upper ermine]

???

[strange axle]

@leaden star yeah idk about that one

[leaden star]

Photomath also says that

[upper ermine]

the answer is def not 1/2

[leaden star]

But it does it with lhopital

[strange axle]

Ummmmmmmm

[upper ermine]

photomath probably copied it wrong

[upper ermine]

,ask $\lim_{x \to 2} \frac{\sqrt{x} - \sqrt{2} + \sqrt{x + 2}}{\sqrt{x^2 -4}}$

[woven radish]

Wolfram Alpha doesn't understand your query!
Perhaps try rephrasing your question?
Click here to refine your query online

Requested by hailinqs

[upper ermine]

oh wait

its minus

i copied fn wrong

$\lim_{x \to 2} \frac{\sqrt{x} - \sqrt{2} + \sqrt{x - 2}}{\sqrt{x^2 -4}}$

[woven radish]

MOD BY ZERO ERROR

[upper ermine]

yes it's 1/2

[leaden star]

What did u do diff

[upper ermine]

lemme find an algebraic manipulation

[upper ermine]

i put a plus instead of minus

[leaden star]

Z

Ah

[upper ermine]

you do partial fractions yeah

[leaden star]

Ok but why does the limit exist

[leaden star]

Im not sure what that is

We usually rationalize and then evaluate for the denominator part of the diff of squares

Ill write it out

[upper ermine]

\[
\frac{\sqrt{x} - \sqrt{2} + \sqrt{x - 2}}{\sqrt{x^2 -4}} = \frac{\sqrt{x} - \sqrt{2} + \sqrt{x - 2}}{\sqrt{x - 2}\sqrt{x + 2}}
\]

\[
\frac{\sqrt{x} - \sqrt{2} + \sqrt{x-2}}{\sqrt{x-2}\,\sqrt{x+2}} = \frac{\sqrt{x} - \sqrt{2}}{\sqrt{x-2}\,\sqrt{x+2}} + \frac{\sqrt{x-2}}{\sqrt{x-2}\,\sqrt{x+2}}.
\]

[woven radish]

MOD BY ZERO ERROR

[upper ermine]

the first addend approaches 0 and the second addend approaches 1/2

do you see that?

$$\lim_{x \to 2} \frac{\sqrt{x} - \sqrt{2}}{\sqrt{x-2},\sqrt{x+2}}$$
Let $h = x - 2 \implies x = 2 + h$. $x \to 2$, $h \to 0$

[woven radish]

MOD BY ZERO ERROR

[upper ermine]

and i think u can work out the rest

[upper ermine]

second addend goes to 1/2 by cancelling sqrt(x-2)

[leaden star]

Can u check this out

What am I doing wrong

@upper ermine

[upper ermine]

do not rationalize

[upper ermine]

you need to split your fraction like this

[leaden star]

Right but we havent done partial fractions

I know what it is , you do A/... + B/... and solve

But is it possible with just rationalization?

I figured it out

Really simple

Want me to send you a pic

With just difference of squares

.close

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[primal elm]

I am unable to get the correct answer, maybe it's brain fog,can smone help me?

<@&286206848099549185>

Idk if whatever I have done is bs.
But can someone guide me, please?😭

[devout snow]

@primal elm Has your question been resolved?

[primal elm]

Just figured it out.

.close

[devout snow]

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[quaint ether]

Saying that $|e^{-x} | < 1 $ is incorrect

[devout snow]

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[winged oyster]

Hello, Need determinate value of g(0), Please and ty for the help

[safe fractal]

How would you approach this problem?

[winged oyster]

I really don't have much of an idea, I'm a bit desperate :(

I don't just want the answer, I also want how to solve it.

[low coyote]

Look at the ones right there

And try to think what they mean

[safe fractal]

Well you need to read the value of graph g when x is 0

[pseudo basin]

!noans

[devout snow]

The purpose of this server is to help you learn, not to hand out answers. Do not ask someone to give you the answer directly.

[pseudo basin]

oh

i can't read 💀

but yeah uh

you need to look at the green curve and specifically at the point whose x coordinate is 0

[safe fractal]

Can you identify where x = 0 is on the x axis?

[winged oyster]

I promise I'll try, but I don't know. Can you be more specific with the question?

I'm very sorry, I'm not very good at calculus.

Or could you explain it to me in another way?

[winter patrol]

this isn't calculus