help-10

[wild swallow]

they're very much different

[daring steppe]

because (a+b)v=av+bv

in both cases

[wild swallow]

yeah and a and b are scalars here

[daring steppe]

its just that <0,0,0> is wrong

[wild swallow]

?

<0,0,0> is wrong
is not a valid statement

how can a single element be wrong

[daring steppe]

<x1,y1,z1> + <x2,y2,z2> = <0,0,0>

doesnt make sense

[wild swallow]

what about it doesn't make sense

it's a definition

[daring steppe]

what we pointed out earler

[wild swallow]

which is what?

[daring steppe]

(1+1)<1,0,0> = <2,0,0> !=<0,0,0>

but

[wild swallow]

that is just saying that equality doesn't hold

[daring steppe]

that doesn't mean (1+1)<1,0,0> != 1<1,0,0>+ 1<1,0,0>. These, to me see like they are equivalent statements

[daring steppe]

yeah

i apologize its 4 am, im exhausted and ,my math in general is not very good haha

[wild swallow]

i'm not sure what you're trying to say

[daring steppe]

algebraically this statement makes sense to me: (1+1)<1,0,0> != 1<1,0,0>+ 1<1,0,0>

[wild swallow]

<x1,y1,z1> + <x2,y2,z2> = <0,0,0>
is a perfectly fine definition

[daring steppe]

yes

but we're talking about vectors

[wild swallow]

it just doesn't define a vector addition operation

[daring steppe]

and its not a vector space

[wild swallow]

there's nothing wrong with it as a function

[daring steppe]

yeah

but im confused 1. what does that mean if its not a vector

[wild swallow]

a vector is an element of a vector space

this is highly context dependent

[daring steppe]

ok

[wild swallow]

in your case, the "tuples" you are working with do not belong to a set which forms a vector space

so they are not called vectors

[daring steppe]

i guess im also trying to ask why they say that (1+1)v != 1v+1v rather than saying v + v != <0,0,0>

[wild swallow]

well you can't conclude immediately that v + v can never equal <0,0,0>

the axioms don't allow for that

[daring steppe]

but i can say (1+!)v != 1v+1v

why?

[wild swallow]

because of the distributivity axiom

[daring steppe]

ok

i think ive figured out the issue im having

[wild swallow]

the distributivity axiom requires that

(a+b)v = av + bv for all vectors v, scalars a, b

[daring steppe]

yeah

a = 1

b= 1

[wild swallow]

your operations fail this

[daring steppe]

v = <x1,y1,z1>

out of those values which fail?

[wild swallow]

you set a = 1, b = 1, v = <1,0,0> for example

this not the only example

[daring steppe]

yeah ik

[wild swallow]

but in this case, the equality does not hold

so distributivity fails

[daring steppe]

does the equality not hold

[wild swallow]

it does not

[daring steppe]

or v1+v2 != <0,0,0>?

[wild swallow]

no you are doing your own thing now

you must check distributivity exactly as written

[daring steppe]

ok

[wild swallow]

(a+b)v = av + bv for all vectors v, scalars a, b

LHS = (1+1)<1,0,0> = 2<1,0,0> = <2,0,0>

[daring steppe]

the other confusion i have

is that in the og equation we have 2 different vectors

but in the distributive rule there is only 1

[wild swallow]

RHS = 1<1,0,0> + 1<1,0,0> = <1,0,0> + <1,0,0> = <0,0,0>

[daring steppe]

is the equation suggesting that the rhs would have to = 1v+v2 but doesnt

[wild swallow]

which equation?

all of the equations written here are actually definitions

[daring steppe]

the <x1,y1,z1>+<x2,y2,z2>=<0,0,0>

[wild swallow]

you just have to be able to parse that from reading the question

[daring steppe]

whats the difference then between an equation and definition

[wild swallow]

well the definitions prescribe what the LHS of the equations in the picture should evaluate to

[daring steppe]

yeah

[wild swallow]

so you are not trying to solve for x1 x2 x3 or whatever

the equations are just definitionally true for all inputs

[daring steppe]

right

[daring steppe]

what does this mean tho

ok i think i see

[wild swallow]

it's a prescription, not a hypothesis

[daring steppe]

if I adjusted the definition to <x1,y1,z1> + <x2,y2,z2> = <x3,y3,z3> would this then be a vector space

or is the addition stil scalar

I think im just gonna leave it there. I'm eating up this channel and i need to get some rest. Thanks for the help and sorry for hacking such a thick skull.

.close

[obtuse pebble]

Channel closed

Closed by @daring steppe

Use .reopen if this was a mistake.

[wild swallow]

i'll just say that this does not give a definition

[fossil crag]

yep, because what are x_3,y_3 and z_3

[obtuse pebble]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• 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.

[timid silo]

I dont understand how to do it😭😭

[acoustic harness]

kif 7alek

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• 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.

[distant rampart]

I'm working on an exercise involving the functions ( g(x, y) = (x^2, xy) ) and ( f(u, v) = (uv, u-v) ), and I'm supposed to find the Jacobian ( J_{f \circ g} ) in two different ways: a) Determine the composition ( f \circ g ). b) Find the Jacobian ( J_{f \circ g} ) directly from the expression for ( f \circ g ). c) Find the Jacobian ( J_{f \circ g} ) using the chain rule.

[warm shale]

dghf

dghf

[distant rampart]

For c) I'm unsure.

[kind hawk]

what is the chain rule

[distant rampart]

Using jacobi?

[kind hawk]

yes

[distant rampart]

$$g: R ^m \rightarrow R ^n, g(x)=y$$
$$f: R ^n \rightarrow R ^p, f(y)=z$$

Then $h=f \circ g$ is a function $h: R ^m \rightarrow R ^p$, $h(x)=f(g(x))$.

[warm shale]

dghf

[distant rampart]

$J_h(x)=J_f(g(x)) \cdot J_g(x)$

[warm shale]

dghf

[distant rampart]

So would this be the chain rule? If we were to take the derivative of h(x)?

@kind hawk

[kind hawk]

yes

[distant rampart]

That is, outer derivative * inner derivativee

[kind hawk]

can you see how that is similar to the "normal" chain rule in one dimension?

[distant rampart]

J_f' * j_g'

[distant rampart]

I suppose yes, though I am not sure if I wrote it using the right notation, shouldn't I use a prime somewhere?

[kind hawk]

the J is the prime basically

in one dimension a derivative is a '

[distant rampart]

Ah true

[kind hawk]

in more than one dimension a derivative is J

[distant rampart]

The J is the partial derivative for each component right?

[kind hawk]

the matrix of all of those, yes

so now, compute J_f and J_g and then plug those into the formula

and then multiply the two matrices

[distant rampart]

Ok, the question in my problem states, what's the difference between the J_fog of the chain rule and the Jacobian of the J_fog that coms from the expression of fog.

And I don't know concretely what they mean.

Which method is this (from the answer they provide)?
$$\text { c) } J_{ f \circ g }=J_{ f } J_{ g }=\left(\begin{array}{cc}
v & u \
1 & -1
\end{array}\right)\left(\begin{array}{cc}
2 x & 0 \
y & x
\end{array}\right)=\left{u=x^2, v=x y\right}=\cdots$$

[warm shale]

dghf

[kind hawk]

thats exactly the chain rule

with bad notation

[distant rampart]

Ok, and determining the jacobian J_fog directly from the expression fog?

What does that mean

[kind hawk]

what you did. first compute fog, then differentiate

[distant rampart]

So just the jacobian matrix of this $f \circ g(x, y)=\left(x^3 y, x^2-x y\right)$

[warm shale]

dghf

[kind hawk]

yes

[distant rampart]

Ok thanks!

[distant rampart]

Btw what was the bad notation

[kind hawk]

the ={u=...}=...

[distant rampart]

Oh, why is it bad

@kind hawk

[kind hawk]

cause the product of two matrices is not equal to some set

[obtuse pebble]

@distant rampart Has your question been resolved?

[obtuse pebble]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• 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.

[lone bobcat]

Yo

[timid silo]

What do you need help with

Graphing?

[lone bobcat]

so so would it be

a straight line from bottom left to the -1

and then from top right to the 1

[timid silo]

Show your work

[random burrow]

yeah

[lone bobcat]

k ty

[random burrow]

nicee

[lone bobcat]

.close

[obtuse pebble]

Channel closed

Closed by @lone bobcat

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• 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.

[vivid knot]

Double checking if this is correct.

Did I use the correct ratio?

<@&286206848099549185>

[obtuse pebble]

@vivid knot Has your question been resolved?

[vivid knot]

<@&286206848099549185>

[vivid knot]

<@&286206848099549185>

.close

[obtuse pebble]

Channel closed

Closed by @vivid knot

Use .reopen if this was a mistake.

[obtuse pebble]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• 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.

[civic karma]

hi, i am having trouble understanding the notion of the double integral //f(x,y)dS whereas S is a random region not the xy-plane

say f(x,y) is a function above the region S in an enclosed space, then is //f(x,y)dS equals to the volume in the enclosed space between f(x,y) and the region S?

[dark stirrup]

Yes

Oh. S is not contained in the xy plane?

[civic karma]

no

[dark stirrup]

How does this work as a double integral then if S has a z component. You'll need to parameterize S so that it depends on two variables u,v, and then express f in terms of u and v, and the double integral would represent that volume

[civic karma]

in my textbook it says
says S= z(x,y)
dS=sqrt(1+z'(x)^2+z'(y)^2)dxdy
so they used the surface area formula to get dS and i dont understand why they did that

well, here is the question:
calc //sqrt(1+x^2+y^2)dS whereas S is z+(x^2+y^2)/2, 0<=x,y<=1

i dont know how is this expressed visually

[obtuse pebble]

@civic karma Has your question been resolved?

Channel closed

Closed by @civic karma

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• 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.

[timid silo]

how do I solve this

[mystic jetty]

have you tried already

[timid silo]

I don't know how to start

[mystic jetty]

have you ever solved a system of equations

[timid silo]

Yeah but I can't remember

[mystic jetty]

alright

so the general idea is to manipulate the 2 equations so that you only have 1 variable in one equation

for that, you can add and subtract the equations from each other

[timid silo]

And how do I do that

[mystic jetty]

by adding/subtracting every piece. For example, if you wanted to subtract the first equation from the second, you would have (19x-15x)+(8y-(-y))=(236-40)

[timid silo]

Would that result in 4x+9y=196

[calm prism]

just you must delete the "y"

[mystic jetty]

its 4x instead of 14x, but yes

[timid silo]

Yeah, I just realized

[mystic jetty]

you can also do multiples, like subtracting an equation 3 times, which you of course get my multiplying every piece by 3 and then subtracting

with those tools, you should find a way to get rid of the 8y in the second equation

[calm prism]

like that

[timid silo]

wait

How

I am kinda confused

[calm prism]

you can multiple first equation

by 8

[timid silo]

so x=536/139?

[calm prism]

yes

then x = 4

if you have x

you can calculate y

[mystic jetty]

by inserting into one of the equations

[timid silo]

Wait, 536/139=4?

[calm prism]

I think so

[mystic jetty]

theres a command

, 536/139

that it?

no

damn

[calm prism]

shit wait

139x = 556

[mystic jetty]

,w 536/139

[warm shale]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by ilith

[mystic jetty]

there you go

[calm prism]

I wrong calculate 320 + 236

[timid silo]

Oh

[calm prism]

it's 556

[timid silo]

Yeah

[calm prism]

not 536

[timid silo]

I just realized again

[calm prism]

wait I'm gonna smoke

[timid silo]

oh

is that a slang or are you really gonna smoke

[mystic jetty]

bro needs a break

[timid silo]

oh

[mystic jetty]

so you get the idea?

[timid silo]

so x=4 y=20

[timid silo]

yeah I did

better than I did on class

thank you guys

[mystic jetty]

,w system of equations: 15x-y=40 19x+8y=236

[warm shale]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by ilith

[mystic jetty]

not what i wanted

bad bot

,w system of equations: 15x-y=40 ;19x+8y=236

[warm shale]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by ilith

[mystic jetty]

there we go

yea youre right

congratulations

[timid silo]

thank ya

[calm prism]

nope

I smoked seriously

[mystic jetty]

.close

type that

@timid silo

[timid silo]

yeah, thank you guys

.close

[obtuse pebble]

Channel closed

Closed by @prime ridge

Use .reopen if this was a mistake.

[obtuse pebble]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• 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.

[fresh copper]

find y' of x^2/x+y = y^2 + 1

[obtuse pebble]

@fresh copper Has your question been resolved?

[ruby wyvern]

the derivative of x^2/x+y?

or the derivative of the whole equation?

[fresh copper]

find y'

so dy/dx

[ruby wyvern]

so its d/dx (x^2/x+y)?

use quotient rule maybe

[fresh copper]

no

i tried

i failed

its implicit differentiation

[fresh copper]

dy/dx of x^2/x+y

[ruby wyvern]

ok, give me a sec

wait

[ruby wyvern]

first you have to isolate y

then find the derivative

[fresh copper]

no

you cant do that

the function's behaviour is similar to a circle

you cant find the derivative of a circle by isolating y

even if you do, its going to be +-, which defeats the whole purpose

[ruby wyvern]

then quotient rule migth be the solution

[fresh copper]

ok

thx

.close

[obtuse pebble]

Channel closed

Closed by @fresh copper

Use .reopen if this was a mistake.

[obtuse pebble]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• 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.

[night lodge]

Status!

[echo burrow]

hi

[rough stump]

hello

I know tangent form

x^2 = xx1 and y^2 = yy1

this one i try to do with this

do you know how to this with this method

[night lodge]

Its for x² and y² if i remember correctly

[rough stump]

ohh ya soory

typo mistake

[night lodge]

Ok and u know for xy

[rough stump]

yup

xy = x1y + yx1 /2

[night lodge]

Yes

[rough stump]

i put y = -1 in equation and get value of x

[night lodge]

Replace them with these u will get ur eqn of tangent

[night lodge]

Yess

[rough stump]

i do but i do not getting it so can you try if you do not mind

[night lodge]

U will get ur x1 and y1

[night lodge]

Value of x?

[rough stump]

x = 0 and x = 5

put x = 5 cuz hint from option

[night lodge]

Yes we will be getting two values of x

As u mentioned

[rough stump]

now we put x = 5 and y = -1 right?

[night lodge]

Yeah right

[rough stump]

do you put values

[night lodge]

Yes put those values in place of x1 and y1

[rough stump]

i do

but i did not get correct answer

[night lodge]

What r u getting

Can u show me ur work

[rough stump]

hmm

soory

but do not have phone currently

i am on laptop

x^2 + 5xy - 4y^2 + 4 = 0

[night lodge]

Okk

[rough stump]

xx1 + 5(xy1 + y1x/2) - 4yy1 + 4 = 0

5x + 5(-x + 5y/2) + 4y + 4 = 0

5x - ( 5x + 25y ) / 2 + 4y + 4 = 0

correct

10x - 5x + 25y + 4y + 4 = 0

5x + 29 y + 4 = 0

this is what i got

did i do something worng

[night lodge]

Here multiplied whole eqn by 2 but u didnt multiply 2 with 4y and 4

[rough stump]

but result is same

we did not get required eq

[night lodge]

Whats the ans

[rough stump]

Alpha option

5x + 33y + 8 = 0

their is some thing wrong with our 33y term

[night lodge]

Yeah let me check

Ohh check with x=0

There are two tangents

Forgot about it

Sorry i cant help u further i need to go

[rough stump]

ok bye

.close

[obtuse pebble]

Channel closed

Closed by @rough stump

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• 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.

[lone bobcat]

here first

lol

[lone bobcat]

part b and below, the final solution is correct, that the function is continuous, but is my working out correct?

also, is there a quicker way to check for the continuity of a function between intervals

i dont want to check each point individually

do i just check for the limits/ each endpoint of the interval?

then assume between them its continuous

wait lemme pull the intermediate value theorem from the last crevice of my smoothed brain

[sour granite]

For continuity at point $a$ you need $\lim_{x\to a} f(x) = f(a)$

[warm shale]

π=√g

[sour granite]

In other words $\lim_{x\to a^-} f(x) = \lim_{x\to a^+} = f(a)$

[warm shale]

π=√g

[lone bobcat]

yes

[sour granite]

So actually yeah that looks good

[lone bobcat]

ty

vm

[sour granite]

I would explicitly write $2+a^2 = 2-b^2$ somewhere though

[lone bobcat]

ok

[warm shale]

π=√g

[lone bobcat]

.solved

[obtuse pebble]

Channel closed

Closed by @lone bobcat

Use .reopen if this was a mistake.

[obtuse pebble]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• 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.

[wicked gull]

s

[wicked gull]

hey guys i need help with this limit

[timid silo]

what have you tried?

[wicked gull]

im trying to solve the limit of these two only cuz theyre exponential

and dominates n

ok so how about this

idk how to use bot commands youll have to do with my hand writing

[obtuse pebble]

@wicked gull Has your question been resolved?

[slim cove]

looks good to me

assuming you don't need to be super rigorous about proving your result is correct

[obtuse pebble]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• 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.

[forest shell]

can someone help

[wet moss]

you could just plug points in to see if the equation is true or not

[forest shell]

oh i did that but i still got it wrong 💀 maybe calculation then

[wet moss]

probably

[forest shell]

is that the only way though?

cuz this is study prep for a test that is timed

[wet moss]

you just need to verify that the LHS on one of these points will be 25

[wet moss]

i mean its probably the quickest

[forest shell]

okay icc ty

is it D then

[wet moss]

one way you could make it quicker would be to realise that A is the centre, so it wont be that, and so it probably wont be B either since its a similar answer, so test C and D and it basically cuts the work in 2

[wet moss]

yeah

[forest shell]

yay thx <3

[wet moss]

np

[obtuse pebble]

@forest shell Has your question been resolved?

Channel closed

Closed by @forest shell

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• 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.

[forest shell]

pls 🤲

[empty maple]

What does "the equations exactly has one solution" mean to you

[forest shell]

like discriminant ig?

[empty maple]

yes

[forest shell]

ohhhh

wait i get it now

thank you ily <3

[empty maple]

no problem

[obtuse pebble]

@forest shell Has your question been resolved?

[reef tendon]

just solve it like you solve a quadratic formula

[reef tendon]

a=18 and b/24

b=24

[obtuse pebble]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[obtuse pebble]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• 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.

[next widget]

How do I continue

[obtuse pebble]

@next widget Has your question been resolved?

[next widget]

<@&286206848099549185>

[obtuse pebble]

@next widget Has your question been resolved?

[tranquil summit]

@next widget Hi. Your first two lines are looping, so they do nothing. Then on your third line you wrote 2.5, where did you get this value?

[warm shale]

BH

[obtuse pebble]

@next widget Has your question been resolved?

[obtuse pebble]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• 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.

[undone crown]

Let A = {1, 2, 3}. The relation E = {(1, 1),(2, 2),(3, 3),(2, 3),(3, 2)} is an equivalence relation on A. F = {(1, 1),(2, 2),(3, 3),(1, 2),(2, 1)} is another equivalence relation
on A. Compute the composition F ◦ E. Is F ◦ E and equivalence relation on A?

[undone crown]

I said the composition was {(1,1),(1,2),(2,1),(2,2),(2,3),(3,1),(3,2),(3,3)}

i simply just added the c's from F wherever the b's matched in both relations

E being the ordered pair (a,b) and F being the ordered pair (b,c)

is this correct?

[elfin burrow]

looks correct to me

[undone crown]

so its just treated like a regular composition right?

[elfin burrow]

what do you mean "regular composition" ?

[undone crown]

that was actually poor wording on my part

[undone crown]

so what I did was that if, say (1,1) in relation E, ill look in relation F that has the same value b so in this case b=1.
So with those two relations we have the composition ordered pairs (1,2) and (1,1)

is that the right way to think about it

[tulip prawn]

that is, I think, the definition of composition

(stated in a very informal way)

[obtuse pebble]

@undone crown Has your question been resolved?

[undone crown]

so you can have multiple ordered pairs coming from the first relation?

also, this would be reflexive, symmetric but not transitive right?

so the composition is not an equivalence relation

[tulip prawn]

yep

[tulip prawn]

not symmetric, look closely

[tulip prawn]

yep

[undone crown]

oh

no(1,3) but theres a (3,1)

makes sense

that answers my questions

thank you

.close

[obtuse pebble]

Channel closed

Closed by @undone crown

Use .reopen if this was a mistake.

[obtuse pebble]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• 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.

[agile hornet]

ayy help me please

[agile hornet]

the first photo is the question and the second on eis the graph

[brazen gorge]

!status

[obtuse pebble]

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

[agile hornet]

1

[shut lagoon]

Do you know what h and k correspond to in that form?

[agile hornet]

no i honestly dont remember the form being used when i was school

[shut lagoon]

The couple (h,k) corresponds to the coordinates of the vertex of the parabola.

[agile hornet]

ohh ok

i get it

so which one is x and which one is y

is h=x

?

[shut lagoon]

Yes

[agile hornet]

dude omg im starting to remember

cool

is the a just a place holder

[shut lagoon]

a can be any number, but in this case they specify that it it either -1 or 1.

Positive a means the parabola opens upwards, negative means downwards.

[agile hornet]

damn ok

boom thats it?

[tulip prawn]

yah pretty much

[agile hornet]

somehow i got it wrong

i did everything correct but got it wrong

nevermind i got what i got wrong

[obtuse pebble]

@agile hornet Has your question been resolved?

Channel closed

Closed by @agile hornet

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• 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.

[sinful frigate]

hello

[hot drum]

hi

[iron edge]

Do you have any questions?

[gloomy vector]

are we being raided or smth lol

[iron edge]

Imma close the channel

[trim portal]

weird

[iron edge]

.close

[obtuse pebble]

Channel closed

Closed by @iron edge

Use .reopen if this was a mistake.

[trim portal]

3 accounts named completely randomly, joined today, created 3 days ago are saying hello in help channels

[sour sun]

hello

[gloomy vector]

exactly

their names are also spam

[iron edge]

Idk the point

For their purpose

[obtuse pebble]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• 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.

[weak apex]

hi can someone pls double check my work

[tulip prawn]

I think you should by using pythagorean theorem for the magnitudes of number 2

[weak apex]

oh ya i wasnt sure ab that bc the mag is technically given i think

idk if he wants us to use pyth theorem

or just like use the given info that each square is 5 lb

also for the graphing part in problem 1

would the left graph or right graph be correct

[tulip prawn]

how?

[weak apex]

it says the three forces A, B, and C are drawn to scale of 1 square = 5lb

and i thought mag meant size

so like A is 3 squares, making it 15lb

is that wrong bc i honestly dk

[tulip prawn]

A is 3 squares horizontal, 2 squares vertical

so Pythag says

[warm shale]

nameless individual

[weak apex]

so its mag would be roughly 18lb?

if yes, then why does the problem state that each square is 5lb. like just out of curiosity, what would i do w that info

[tulip prawn]

its the $\times5$

[warm shale]

nameless individual

[weak apex]

ohhh im actually slow

you're awesome ty for helping

i'll be back in a bit lmao

ty

.close

[obtuse pebble]

Channel closed

Closed by @weak apex

Use .reopen if this was a mistake.

[weak apex]

.reopen

[obtuse pebble]

✅

[weak apex]

wait actually sorry forgot one more thing

for the graphing part in problem 1
would the left graph or right graph be correct

[tulip prawn]

left

[weak apex]

ok ty

.close

[obtuse pebble]

Channel closed

Closed by @weak apex

Use .reopen if this was a mistake.

[obtuse pebble]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• 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.

[weak apex]

ok sry back again

[weak apex]

can someone pls check if this is correct

also am i crazy or are questions 2 and 5 the same thing

[tulip prawn]

looks like it

[weak apex]

oh ok good

i didnt know if he wanted smthing else

also im confused ab problem 4

[tulip prawn]

decompose $F$ along $a$ and $b$.

[warm shale]

nameless individual

[weak apex]

uhh

wdym

[obtuse pebble]

@weak apex Has your question been resolved?

[weak apex]

im still a bit confused

i remember the prof doing smthing like this

for a similar problem

[hidden compass]

Yeah that's it

[weak apex]

this is correct?

[hidden compass]

Yes, ynot?

[weak apex]

idk

wait

do u mind double checking if this is also right

[weak apex]

the other angle is throwing me off

[hidden compass]

Magnitudes are always positive, by definition

[weak apex]

i get that

but the drawing

i dont know why it's right

[obtuse pebble]

@weak apex Has your question been resolved?

[weak apex]

.close

[obtuse pebble]

Channel closed

Closed by @weak apex

Use .reopen if this was a mistake.

[obtuse pebble]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• 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.

[inland plaza]

Can someone please give the second ii bit

[crude coral]

just find sin and cos and use those values in it

[inland plaza]

I did that

[marsh geyser]

Are you talking about the 1/sqrt(7) exercise?

[crude coral]

sin^2 theta = (sin theta)^2

[inland plaza]

Yes

[runic badger]

it should simplify to cos2theta

[marsh geyser]

You can rewrite the rhs like (1-tan^2x)/(1+tan^2x)

[runic badger]

or just put value of theta ig

[crude coral]

u can simplify if u want tbh

[inland plaza]

Well I did the cosec = cot² + 1

Is that correct?

[runic badger]

use the reciprocal relation

[crude coral]

id just switch into sin and cos

[marsh geyser]

You have the value of tan

Use it

[inland plaza]

I'm getting numerator as a fraction

So I'm a but confused

[marsh geyser]

What fraction

[inland plaza]

Wait I didn't see that message sorry

Let me try that

Tbh I don't know how to apply that

Should we substitute sec value?

[marsh geyser]

What sec?

[inland plaza]

-sec²theta

[marsh geyser]

At this point im not sure if we are looking at the same exercise

[obtuse pebble]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[marsh geyser]

(cosec^2x-sec^2x)/(cosec^2x+sec^2x)=(1-tan^2x)/(1+tan^2x)

[inland plaza]

Ohh

Tysm!

.close

[obtuse pebble]

Channel closed

Closed by @inland plaza

Use .reopen if this was a mistake.

[inland plaza]

.reopen

[obtuse pebble]

✅

[inland plaza]

1st question

[swift stream]

!status

[obtuse pebble]

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

[inland plaza]

1

A+b confused me

Sorry if that's inconvenient

[swift stream]

Take the dot product between a+b and a

Cross product will also work

[inland plaza]

Uhhh

Wait

I'm confused tho doesn't a+ b and a have a same term

[swift stream]

So?

[inland plaza]

Is a the vector??

And a +b along x

Just asking

[swift stream]

(a+b)•a=|a|²+a•b and
(a+b)•a=|a+b||a|cos(angle between a+b and a).
By this you can find out the angle between a+b and a

[inland plaza]

Wait so just resultant formula?

37costtheta is 4/5

[swift stream]

Im not understanding what you are saying

[inland plaza]

Mb I'm stupid please wait a min

I need to draw it

[swift stream]

Do you understand why doing this will give us the angle between a+b and a?

[inland plaza]

Nope

I'm struggling in this chapter

I skipped 3 grades

I'm 13

If you're still there I got tan alpha is 3/4

[obtuse pebble]

@inland plaza Has your question been resolved?

[obtuse pebble]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[obtuse pebble]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• 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.

[lime rampart]

Can someone help me I need math videos I basically skipped g6, g7 g8 and g9 I didn't listen at all

[shy ruin]

ig you could try khan academy

well also the level of math you did in those grades can depend on what area you are in and your schools curriculum

[lime rampart]

Ok thanks

[obtuse pebble]

@lime rampart Has your question been resolved?

[lime rampart]

Yes

.close

[obtuse pebble]

Channel closed

Closed by @lime rampart

Use .reopen if this was a mistake.

[obtuse pebble]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• 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.

[echo burrow]

a probability riddle my teacher gave for extra credit: if there are three events, A, B, and C. and P(A)=1/100, if two random events B and C occur, what is the probability that one of them is A

[grizzled shore]

Two random events B and C?

[echo burrow]

like a coin toss or dice roll

@grizzled shore

[grizzled shore]

If 2 random events like B and C happen

That’s already 2 events

How can one of them be A

[plain stag]

this question is pretty nonsense

!original

[obtuse pebble]

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

@echo burrow Has your question been resolved?

[obtuse pebble]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• 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.

[snow herald]

Hi, I'm reviewing some basic vector calculus stuff
I was solving a problem where a paramterisation C was given
$$\begin{cases}
{x = t^3}\{y=t}\{z=t^2}
\end{cases}$$
and a surface
$$S = z^3+xyz-2=0$$
The question is to calculate the angle between the derivative of C and the normal to S in their intersection point p. (Domain is limited to the first quadrant)\
I found p to be $\begin{bmatrix}{1}\{1}\{1}\end{bmatrix}$\
after taking the derivative of C and plugging in those values I got $\begin{bmatrix}{3}\{1}\{4}\end{bmatrix}$ which should be the vector pointing along the derivative of C in p.
Now for the normal vector to S in p, I couldn't quite find this one, so I looked up a solution in my book. And apperantly you need to take $\del S$ in p, I'm really confused, because the gradient of a surface shouldn't be the normal vector... can anyone explain?

[warm shale]

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

[slim cove]

If you have a surface given by f(x,y,z) = 0, then the gradient of f is ALWAYS perpendicular to the surface

Intuitively, this is because the gradient of f represents the direction that you increase the fastest, so you'll be "getting away from f=0" in the fastest possible direction

i.e. perpendicularly

[slim cove]

(Such a surface is known as a "level set" because it's the level 0 of f)

[snow herald]

I'm not sure I can visualise this one

[slim cove]

Think of f(x,y,z) = z for example

Then f(x,y,z) = 0 is just going to be the xy-plane right

[snow herald]

yeah

[slim cove]

And the gradient of f will be the direction in which z increases the fastest, so (0,0,1)

And you can see that (0,0,1) is perpendicular to the xy-plane

It turns out this method works for really any choice of f though, not just f(x,y,z) = z

[snow herald]

oh wow, this is pretty usefull to know

I wanted to paramterise into u-v ant take their crossproduct

[slim cove]

That method will also work

This one is faster though

[snow herald]

yeah, wayy faster

do you know any proofs of this one by any chance?

[slim cove]

(Thinking of it in 2D might also be an easier visualization. For example, consider the function f(x,y) = x^2 + y^2. Then the level sets of f look like circles. For example f(x,y) = 9 is a circle of radius 3. Then, the gradient of f is (2x, 2y) which points radially outwards, perpendicular to the circle.)

[slim cove]

Lemme think about it, it should be pretty straightforward I think

[snow herald]

I might've missed it during my lectures and it seems like a pretty nice property

[snow herald]

oh so the gradient is normal to the level set circles

but not to the surface

[slim cove]

There is no surface in this example

it just shows that the gradient is perpendicular to the level sets

In your original q tho, the surface was a level set

so the gradient will be perpendicular

[snow herald]

ah okay, I usually interpret f(x,y) = x^2 + y^2 as a paraboloid

[slim cove]

Yea, similarly you could interpret f(x,y,z) as a shape in 4D, and f(x,y,z) = 0 as one slice of it, but 4D is impossible to visualize which is why I took it a dimension down haha

[snow herald]

ah alr, makes sense

[slim cove]

Here's one argument that's pretty simple, but I'm not sure it's 100% correct lol. But if you have a surface f(x,y,z) = 0 that you can parameterize in terms of x and y and express z as a function z(x,y), then the tangent vectors of the surface will be (1, 0, dz/dx) and (0, 1, dz/dy). I think dz/dx should be given by -df/dx / df/dz, and dz/dy should be given by -df/dy / df/dz. And then when you take the dot product of the gradient (df/dx, df/dy, df/dz) with these vectors, you get zero, showing it's perpendicular.

The reason why dz/dx = -df/dx / df/dz is because if you take the derivative w.r.t. x of f(x,y,z) = 0 where z is a function of x and y, you get df/dx + df/dz dz/dx = 0 by the chain rule, and then solving gives you dz/dx = - df/dx / df/dz.

There's probably a neater proof, but that's the first thing that popped into mind.

[snow herald]

ohhh okay

so z becomes a function of the other two variables, so we can use chain rule

one more question, are those tangent vectors normalised?

ah nvm, the 1 just gives it the right direction

oh yeah indeed the dot product equals 0

[slim cove]

No, you can normalize them easily though

Btw is your username a reference to Faust?

[snow herald]

I meant cross product, not dot product woops

[slim cove]

Oh, you can also check that the cross product of those two vectors is a multiple of the gradient

The dot product just seemed more straightforward to me, plus it works in any number of dimensions

[snow herald]

ah no, just one of the biblical demons ig

[slim cove]

Ah

[snow herald]

demon*

is Faust a book ?

[slim cove]

Yes, the only issue is that sometimes you cannot write z as a function of the other variables, e.g. if the surface is x=0. But in that case you could write x as a function of the other variables, and the same proof should work fine. See the "implicit function theorem" for more details.

[slim cove]

It's a legend where Faust is the hero that makes a deal with the demon Mephistopheles

A lot of books were written about this story, including a famous one by Goethe

I only know of it well because there's some haunting pieces of music called the Mephisto Waltzes (and one called the Mephisto Polka) based on the story

[snow herald]

ohh, that's pretty interesting, I'll look into it

sadly my summer vacation is a bit cluttered with studying for the re-examinations

so I don't have much time to do other stuff

[snow herald]

where does the negative sign come from?

[obtuse pebble]

@snow herald Has your question been resolved?

[slim cove]

See the next message

It comes from solving for dz/dx in the equation df/dx + df/dz dz/dx = 0

[obtuse pebble]

Channel closed

Closed by @snow herald

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• 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.

[zenith orbit]

$log(n) \leq c*log(log(n^4))$

the base of log is unknown, how do i know if this holds for a constant c?

[warm shale]

Ayanokoji

[zenith orbit]

as in, how do i prove/disprove (go from here)

[slim cove]

Two questions:

1. Do you mean as n → ∞? Or for all positive integers n?
2. Do you mean ≥ instead of ≤?

[zenith orbit]

1. yes, n is a natural and this is a sequence a_n=n type inequality
2. no i do mean <=, i believe this needs to be disproven by showing c cant be a constant in this context, but i cant do it with double logs

[slim cove]

Ah, you're trying to disprove, I see

[zenith orbit]

i might be wrong but i think this is incorrect, otherwise i'll need to find the c contstant or find an estimated value c>=...

[slim cove]

Are you allowed to use calculus? It should be fairly easy to show that
[\frac{\log(\log(x^4))}{\log(x)} \to 0]
as (x\to\infty), so no constant (c) will work for all values of (n)

[warm shale]

biased_estimatERIC

[slim cove]

Using something like L'Hospital's rule

[zenith orbit]

we are, but we didnt learn log limits, can you explain?

[zenith orbit]

also didnt learn this, whats this say?

[zenith orbit]

here this is true if $x > 4*log(x)$, which is not obvious

[warm shale]

Ayanokoji

[zenith orbit]

.tc dark

,tc dark

[warm shale]

I don't recognise the option dark. Use ,texconfig to see the list of options.

Jul 5, 2024, 7:33 PM

[zenith orbit]

,tc

[warm shale]

Personal LaTeX configuration.

​ ​ ​ ​ ​ ​ autotex: Disabled (may be overriden by guild settings)
keepsourcefor: Don't delete source (may be overriden by the guild)
​ ​ ​ ​ ​ ​ ​ colour: Using the light colourscheme
​ ​ ​ alwaysmath: Disabled
​ ​ ​ alwayswide: Disabled
​ ​ ​ ​ namestyle: DISPLAYNAME
autotex_level: WEAK

To see more detailed information use ,texconfig <option>.
To set an option use ,texconfig <option> <value>.

Preamble

No personal or guild preamble, using the global default preamble.
Use ,preamble to view or modify your preamble!

Jul 5, 2024, 7:33 PM

[zenith orbit]

,tc colour dark

[warm shale]

You have switched to the dark colourscheme.

[obtuse pebble]

@zenith orbit Has your question been resolved?

[obtuse pebble]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[obtuse pebble]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• 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.

[timid silo]

There exist integers x,y in Z s.t. c = ax+by > 0 for some integers a,b <=> gcd(a,b)=c

Here's my proof. I'm stuck on how to show it in the (<=) path, like how can I proceed from saying d|c(gu + hv) to d|c? It would require gcd(d,gu+hv) = 1 which I have no idea how to show

[timid silo]

If I can show d|c then I can show c = a(uk) + b(vk) = ax+by and finish the proof

[kind hawk]

well it doesnt help that the claim is false

well the => direction is false

[timid silo]

It's from David M Burton, (a) here

[smoky vigil]

that's because of the >0 restriction?

i think?

[timid silo]

Oh, well while I was reading the chapter, it assumed the common divisors to be positive since their negatives are also common divisors

[kind hawk]

well the image is correct but you didnt write down the question correctly

[timid silo]

Because of >0?

Let me go through the chapter once again

It assumed gcd to be positive hence I assumed c to be positive as well

[kind hawk]

look at the question you wrote down and the actual question asked in the book

there is a very significant difference

you arent supposed to show gcd(a,b)=c

[timid silo]

Oh my god I read | as = the whole time

I'm really sorry

.close

[obtuse pebble]

Channel closed

Closed by @dusk knot

Use .reopen if this was a mistake.

[obtuse pebble]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• 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.

[mild bay]

Apparently the answer is 2/3 and I have literally no idea how to get that answer

[dapper bloom]

Did factoring help?

[wet moss]

have you tried using l'hopitals rule?

if you are allowed to

[mild bay]

l'hopitals rule is in the next chapter

[wet moss]

ah fair

[mild bay]

I mean, I just ended up with 1/x-1

[dapper bloom]

I think you can factor both and have the x-1 factors cancel

Isn't x^3-1 = (x-1)(x^2 + x + 1) or something?

[mild bay]

Lemmie try

[dapper bloom]

,w factor x^3 - 1

[warm shale]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by dootdooter

[dapper bloom]

Double checking..

[meager gale]

XD

[mild bay]

Wtf

Yea I don't think I'm ready for calc 😭😭

My algebra is so much worse than what I thought it was

[meager gale]

this actually reassures me soo much bcz other ppl second guess themselves too i literally do this 😭 especially someone as based as doot

[dapper bloom]

This is a single formula

One that is also easily forgotten

I would not worry about it.

Then general approach of "hmm weird ratio of polynomials that I can't evaluate. What if I factor and cancel?" Is the more important take away here.

[thin pawn]

I'm so mad I fell for your bio Doot

[mild bay]

Mhm

[meager gale]

pinging grass rn

[coarse wren]

Since we know x-1 is a factor of x^3 - 1 due to the factor theorem you can just use long division or synthetic division or whatever division to get the other factor

[dapper bloom]

So long a you know this you can always review factoring a little bit to figure out stuff like this

[meager gale]

in general tho

[dapper bloom]

Yeah true.

[meager gale]

like when you see a problem like this

its usually factoring so

you can just mess around with factors and see what works

[mild bay]

Mhm, I see now, thanks

I'm glad I have to take college algebra before precalc cuz my algebra is absolute shit

[viscid gull]

U just get it from finite geometric series formula

[meager gale]

you will get better at algebra from doing calc too

as you are right now.

[viscid gull]

and for that I just remember if S is a geometric series then xS - S does some nice telescoping and work fron there

Oops not telescoping

Just cancels out terms

[mild bay]

.close

[obtuse pebble]

Channel closed

Closed by @mild bay

Use .reopen if this was a mistake.

[obtuse pebble]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• 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.

[brazen smelt]

for the bottom one... how?

[brazen smelt]

.close

[obtuse pebble]

Channel closed

Closed by @brazen smelt

Use .reopen if this was a mistake.

[wet moss]

just from what x, y and z equal to in spherical coords

[obtuse pebble]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• 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.

[fathom schooner]

A particular coin is biased so that P(H) = 2.5 x P(T)

State in simple fraction form P(head) and P(tail)

[fathom schooner]

why is P(head)5/7 and P(tail) 2/7

[quaint wing]

Remember that P(H) + P(T) = 1

That gives you two equations with two unknowns

[fathom schooner]

ffs

thanks

.close

[obtuse pebble]

Channel closed

Closed by @fathom schooner

Use .reopen if this was a mistake.

[obtuse pebble]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• 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.

[jaunty mica]

I was watching this video thats visualizing the detivative of trigonometric functions startong with sin(x), while it showed difference im angle and resulting difference in sinx I couldnt understand this part in the diagram. How is it that the little triangle thats formed has dϴ as the length, thats the little difference in input angle at the centre right? So how is the length at circumference being measured in radians

[gritty geyser]

Think of the relationship between radians and theta

[jaunty mica]

What does that mean

[gritty geyser]

Define a radian

What does a radian mean in terms of the radius

[jaunty mica]

An angle that creates an arc as long as the radius on circumference

[gritty geyser]

Okay, so in a unit circle we know the radius is 1 right

[jaunty mica]

Oh wait so its just theta times 1

[gritty geyser]

The arc length is equal to theta in radians here

[jaunty mica]

Ye i get it now

[gritty geyser]

s=rtheta

[jaunty mica]

Ye

[gritty geyser]

Alrighty

[jaunty mica]

Wait i have one more question

[gritty geyser]

shoot

[jaunty mica]

Hold up

[jaunty mica]

Ok so

Right after it says that the little triangle is similar to the larger of the unit circle but how do we know for sure

[gritty geyser]

Think about how the legs of the large triangle change as theta changes

[jaunty mica]

Sin gets longer cos smaller

Adj smaller opp longer

[gritty geyser]

Sure, but since we know that theta and dtheta are directly related, we also know these changes in sine and cosine (the sides of the large triangle) and dsine and dcosine in the small triangle are proportional

To each other

[jaunty mica]

I dont get it suppose theya for large triangle gets smaller, the dtheta would be same

[patent owl]

both triangles are right angled triangles. The hypotenuse of the large triangle forms a 90° angle with the side marked dθ. the unmarked side of the small triangle forms a 90° angle with the right edge of the large triangle. Hence can you see that

sorry for the scuffed image, I'm drawing on my phone

[jaunty mica]

The angle with dtheta can only be roughly 90 right since its not a proper tangent at hypotenuse edge

[exotic tiger]

I need help with this calc question

[jaunty mica]

Still sont see the similarity tho

[gritty geyser]

#❓how-to-get-help

[jaunty mica]

Hows it for sure that unmarked side and right edge make a right angle

[jaunty mica]

Oh wait no i get that now

[gritty geyser]

Because that’s how sine and cosine are represented

[jaunty mica]

Ye ik the unit circle

But this proportionality is dont get

dtheta will always be constant for whatever theta (and by that I mean same as it approaches 0)

So for a smaller/larger theta dtheta we’ll take the same and so dsin(theta) and unmarked side also have to be the same length

As any other theta

What I mean is this triangle will be the same shape for any theta right? ( the small one) but I think dsin(theta) and unmarked might change a bit

[gritty geyser]

We aren’t comparing theta to dtheta here

We are comparing it to the radius

1

Also a constant

Two constants hypotenuses

[jaunty mica]

What I mean is that as theta gets smaller dtheta will be steeper right?

[robust hound]

any helpers can help about my quick math? #help-17

[jaunty mica]

But say we take 0.001 as the tiny difference so that will remain same right

The length is same but steeper at some points

And so dsin(theta) and unmarked side vary proportionally

[gritty geyser]

We aren’t directly comparing theta to dtheta here, the slope of the circle isn’t relevant at this moment, remember, it is in terms of radians, dtheta is a infinitesimal value, not a slope

[jaunty mica]

Ok

[gritty geyser]

It is also constant

Like the radius

And we know that a small change in theta will produce a small change in sine

[jaunty mica]

Approaches 0 so how constant

[gritty geyser]

Right?

[gritty geyser]

When we say dtheta the value isn’t gonna change so just think of it as some infinitely tiny number

[jaunty mica]

If we compare it to dtheta elsewhere in unit circle that way?

[gritty geyser]

There’s no reason to think it changed

[gritty geyser]

Wdym

[jaunty mica]

Like we basically wanna know what happens to rate of change as the difference approaches 0 right? For any function

So we get the accurate value

Thats basically ruling out the negligibly small values while differentiating like we do in power rule for dx^2 and greater

Suppose I have a function x^n so to differentiate it i use the first principle, so in the numerator I have
x^n +nx^(n-1)dx+ (n(n-1))/2 X x^(n-2)dx^2…. - x^n

And that whole thing over dx

so factorize dx, cancel and whatever powers of dx are left are negligibly small

[gritty geyser]

I don’t see the relevance

All we need to prove is that the two triangles are similar

And then the problem is solved

[jaunty mica]

Right

[gritty geyser]

Did you watch the full segment

To the end

[jaunty mica]

Yes

[gritty geyser]

So you understand what comes after proving the similarity right

[jaunty mica]

But latgely sisnt get it so tryna understand strp by strp

[jaunty mica]

Ye

[gritty geyser]

Okay so let’s start with dtheta

[jaunty mica]

dsimtheya is badically same as cos ratio

[gritty geyser]

dtheta is a constant, we are comparing it to 1, the hypotenuse of the big triangle

These are both constants

Let’s move on

The angle between dsin and dtheta is theta

[jaunty mica]

So dtheta is the difference in the angle of triangle we take

[gritty geyser]

Precisely

[gritty geyser]

?

Well kind of

I guess

[jaunty mica]

We take a small difference roght? To find derivative

[gritty geyser]

Just think of dtheta as a very very very small value of theta (and since the radius is 1, that value translates directly into arc length, adding this tiny value to the rest of the arc)

[gritty geyser]

Yes

[jaunty mica]

Yes that part i get

[gritty geyser]

Okay so now let’s look at the angles in the two triangles

[jaunty mica]

And the upright little side is the difference caused in sin

[gritty geyser]

the large triangle has an angle of theta right

[jaunty mica]

Yes

Both have 90 so wr just have to prove one is theta

[gritty geyser]

The angle in the tiny triangle is also theta, we know this because of the way these triangles are generated.

[jaunty mica]

Ye this part i dont get

[gritty geyser]

Remember the relationship of theta and the arc length

[jaunty mica]

I get how wr can say that if rotated the sides line up

When we make it bigger

So they’re similar but hows that theya

[jaunty mica]

So are you saying we made it small through a ratio such that hypotenuse, from 1 becomes the small dtheta and other are also made smaller same way

Using the same ratio

[gritty geyser]

Because the same theta that gave us the big sine will give us the tiny sine, because the radius of this triangle is dtheta not 1

but the values are still proportional

We have two seperate radius values

For two imaginary circles

[jaunty mica]

Ok yes if we draw a second circle we get dsintheta

[gritty geyser]

Yes

With the same angle

SAS similarity

[jaunty mica]

But that still doesnt prove that theta is equal to bigger triangle’s theta

[jaunty mica]

Is this correct?

[gritty geyser]

Yes imagine we scale down the radius of the unit circle to some value dtheta, everything else will be proportional to the unit circle in the same way, just like how trigonometric functions of the same angles in triangles of varying sizes are all proportional to each other, otherwise the field of trigonometry would be useless if they weren’t proportional

[jaunty mica]

So first we pick a little difference in theta of the unit circle

So at the corcumference dtheta is formed

Now we scale back the big triangle’s hyp to that dtheta

[gritty geyser]

are you familiar with rcosx = x and rsinx =y

[jaunty mica]

And so the other sides also scale back

And now we have the triangle

[gritty geyser]

Yes

[jaunty mica]

No dont think so

[gritty geyser]

Imagine it as a scaled down unit circle

[jaunty mica]

Think i underatand it better now

[gritty geyser]

Everything is proportional to the actual unit circle

This is fundamental to trigonometry

[jaunty mica]

Does it have to be the unit circle i imagine? That makes it a bit complicated to imagine

[gritty geyser]

That’s why the sine of an angle, whether it’s made of sides of 10 and 11 or 20 and 22, is still the same

[jaunty mica]

Ye

[gritty geyser]

The unit circle is important because it is a basic reference

Every other circle is proportion to it

[jaunty mica]

The triangle from the unit circle alone doesnt suffice?

[gritty geyser]

That’s how we derive sine in triangles who hypotenuse is not 1

[jaunty mica]

To scale down to dtheta

[gritty geyser]

Because of the relationship of the triangle sides, the unit circle tells us how how sine behaves on a simple unit scale that can be scaled up or down depending on what the hypotenuse is, in this case it’s dtheta

[jaunty mica]

Yes

The unit citcle helps to understand the ratios

[gritty geyser]

Yes

[jaunty mica]

And the triangle of any angles can then be scaled and ratios stay same

[gritty geyser]

Yes

[jaunty mica]

So thats what happens woth the little and bigger triangle

[gritty geyser]

The length of the sides doesn’t matter because of the ratio

[jaunty mica]

And thats why they havr theta

[gritty geyser]

Exactly

[jaunty mica]

We just scale it back to dtheta, other sides scaled back too and now we rearrange to find dsin(theta)

[gritty geyser]

well we find dsin(theta)/dtheta

[jaunty mica]

Yes

[gritty geyser]

Which is costheta

[jaunty mica]

So we have two marked sides

So derivative is

dsin(theta)/dtheta

Lime u said

And regardless of length we’ll get same ratio

[obtuse pebble]

@jaunty mica Has your question been resolved?

[obtuse pebble]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• 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.

[dusky arrow]

i asked this question before but went afk so couldnt see the answer

[dusky arrow]

how to get ab in the underlined eq to the lhs

or is there another way to solve this eq