help-13

[quick flume]

?

[dire geode]

What

[quick flume]

But a=x no?

Holy shit

.close

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[tall eagle]

I am trying to solve the following vector problem:

A pilot is flying a Cessna airplane at 180 mph (airspeed). He would like to fly in the direction N45◦W, but there is a 32 mph wind in the S60◦W direction. What direction should the pilot set his course for in order to fly along his desired track? What will his speed relative to the ground be?

I know that I need to break things up in components and solve a system of equations but am struggling to create the system and then solving it.

[cedar kiln]

@tall eagle Has your question been resolved?

[tall eagle]

<@&286206848099549185>

[slate plover]

Oh that is a simple Galileo's transformation $\vec V_trans = \vec V_ref + \vec V_obs $

$$\vec{ V_trans} = \vec{ V_ref} + \vec{ V_obs} $

the wind vector is your reference vector while the fly direction is your desired transformed vector

the vector you are looking for ( the pilot vector is then $ \b V_pilot = \b V_wind - \b V_fly $

the speed relative to the ground is the same exceot that now the v_ref is now zero, so V_pilot = - V_fly

Correction V_pilot = V_fly - V_wind

and V_ground = V_fly

I messed up with the signs earlier

[cedar kiln]

@tall eagle Has your question been resolved?

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[hexed vortex]

A rocket has an initial velocity vi =(-27.2 m/s)i+(12.8 m/s) j. The thrusters are fired, and the rocket undergoes constant acceleration for 21.6 s resulting in a final velocity of ve =(353 m/s)i + (34.4 m/s) j.

It's asking for the magnitude of the acceleration

[hexed vortex]

I did v - vi / t

But got it wrong somehow

I got .890m/s^2

[tepid sail]

jessa

is this your channel?

[hexed vortex]

Yes it says my name on it

I think the other guy posted right after I posted

Posted at the same time

[tepid sail]

calculate $a = \frac{v_f-v_i}{t}$

[wraith dagger]

nosqldb

[tepid sail]

and then calculate the magnitude of a

[hexed vortex]

That's what I did yeah but I got the wrong answer somehow

[tepid sail]

√(((34.4-12.8)/21.6)^2 + ((353+27.2)/21.6)^2)

I got sm like this

[hexed vortex]

This is my calculation

[tepid sail]

if I did my paranthesis correctly

[hexed vortex]

Wait

How come you divide by t inside parenthesis and not the whole thing by t

[tepid sail]

wym

again

calculate $a = \frac{v_f-v_i}{t}$

[wraith dagger]

nosqldb

[tepid sail]

THEN calculate magnitude of a

[hexed vortex]

Not sure if I'm understanding, are you getting the components of a and then finding the magnitude

[tepid sail]

yes

[hexed vortex]

Why is the above equation not the acceleration? Why can't we do
(Magnitude of V - Magnitude of Vi)/t

[tepid sail]

you can take out the t

and calculate the magnitude of $(v_f-v_i)$

[wraith dagger]

nosqldb

[hexed vortex]

Hmm I don't still don't quite understand what is wrong with

$\frac{sqrt({35.3}^2 + {34.4}^2) - sqrt({-27.7}^2 + {12.8}^2)}{21.6}$

[tepid sail]

jessa

alr

hmm

let's just consider a diff situation to see why

there are two drunk people on like frat row and

[wraith dagger]

Jessa

[tepid sail]

if you use ur method

the acceleration has magnitude 0

[hexed vortex]

How could we get the acceleration for one person from two people

What does [-1,0] mean I haven't encountered that notation before

-1 x direction and 0 y direction?

[crimson sedge]

real

[tepid sail]

ye lol

[hexed vortex]

I am confused still lol

[tepid sail]

jessa uh

let's say you have two vectors in space

[hexed vortex]

Hmm the finding accerlation components make sense but why the other method doesn't work doesn't make sense

[tepid sail]

$x_1, x_2$

[wraith dagger]

nosqldb

[hexed vortex]

Ok

[cedar kiln]

@hexed vortex Has your question been resolved?

[cedar kiln]

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[wraith trellis]

Is this proof valid

[ornate hearth]

I don't think so

you need to define 6 is congruent to 4

[cedar kiln]

@wraith trellis Has your question been resolved?

[wraith trellis]

i did

[ornate hearth]

oops I missed that

it seems correct otherwise

[wraith trellis]

alr cool thanks

.close

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[native lion]

hello I need someeone to check this for me

can someone let mee know if this is right or not?

[cedar kiln]

@native lion Has your question been resolved?

[spring rivet]

It looks correct to me

[native lion]

ok

[cedar kiln]

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[royal loom]

Would anyone know any resources that do a good job with this? I'm not really following the book

Or if anyone is able to explain

I don't understand the first approximation why it is any good, or why we can justify that it becomes "better and better" as r->0

[tepid sail]

@royal loom I liked this exlpanation: https://www.youtube.com/watch?v=WfEQabCGAqI

[royal loom]

MIT OCW goat

@tepid sail I just scrolled through that a bit, I'm not seeing the limit described above, are you sure it is in there?

I'm less asking about the divergence theorem and more so just about this specific corollary of it

[flint plinth]

isn't this just another continuity argument

[royal loom]

Well why is the first approximation true?

[flint plinth]

if div F is continuous at a, then when you're within a small radius of a, div F approximately equals its value at a

[flint plinth]

isn't that what the squiggly-equals part is saying?
LHS = divergence at a itself
RHS = average of divergence on a small ball of radius r

[royal loom]

I don't see how it is saying that though

[royal loom]

oh

the integral is giving the average divergence over that ball

[flint plinth]

yep

the scale factor is just 1/vol(ball)

[royal loom]

and as the ball gets smaller then the average approaches the value at the center supposedly by continuity

[flint plinth]

yep

similar to the argument in your other problem last night

[royal loom]

using the integral mean value theorem?

[flint plinth]

yea, or just that abs(integral(blahblah) - divF(a)) <= delta
like we argued last night
here we assume that r is small enough that |divF(x) - divF(a)| < delta holds for all |x-a| < r

[royal loom]

So would something like this be good reasoning

$$\lim_{r\to 0}\frac{3}{4\pi r^3}\iint_{|x-a|=r} F\cdot n dA$$
By the integral MVT there is an $\vec{a}\in |x-a|=r$ such that
$$\lim_{r\to 0}\frac{3}{4\pi r^3} \text{div }(\vec{a})\cdot \iint_{|x-a|=r}1dA$$
$$\lim_{r\to 0} \text{div }(\vec{a})=\text{div }(\vec{a})$$

[wraith dagger]

Austin

[royal loom]

and this a becomes the a at the center of the ball in order for it to be in all of these intervals as r->0

[flint plinth]

probably you want to apply the integral MVT to the volume integral, not the flux integral

also i'm not sure what the distinction is between your arrow-a and ordinary a

[royal loom]

they should all be arrowed I guess

book had a bolded

[flint plinth]

well a is not in |x - a| = r, the latter is a circle around a, and does not include a

(or sphere or whatever)

[royal loom]

I see

[flint plinth]

i think you can just argue like before...
since F is C1, its divergence is continuous, hence given delta > 0 we have for sufficiently small r that |divF(x) - divF(a)| < delta for all |x - a| < r

[royal loom]

yes it makes more sense to do it to the volume integral

[flint plinth]

now use that estimate in the vol integral

(or you can use MVT if you prefer)

both ways you'll need to use continuity of divF at a

[royal loom]

$$\frac{3}{4\pi r^3}\iiint_{B_r} \text{div }\vec{F} dV$$
by the IMVT there is an $\vec{a}\in B_{r}$ for all $r$ such that $$=\text{div }(\vec{a})$$
So,
$$\lim_{r\to 0} \frac{3}{4\pi r^3} \iiint_{B_r} \text{div }\vec{F} dv= \lim_{r\to 0} \text{div }\vec{F}(\vec{a})$$

[wraith dagger]

Austin

[royal loom]

Where $\vec{a}$ is in $B_r$

[wraith dagger]

Austin

[royal loom]

so as r->0 by continuity then, this is going to be just the center of the ball

and we get our proof

?

I don't know if I am wording the last part good enough

but I get the idea

[flint plinth]

the point a is already in use, i don't think it's a good idea to reuse it

[royal loom]

Right

[flint plinth]

the point you're getting from the MVT might not be the center of the ball

(which is the a in the text)

[royal loom]

but it's always in the ball

[flint plinth]

yea

[royal loom]

and as we limit the balls radius to 0

it is the center

[flint plinth]

and by continuity, divF at your a is close to the divF at the center

[flint plinth]

well again, by continuity

[royal loom]

yes

thanks again Bungo

idk why this one trips me up a bit

[flint plinth]

the kind of limit is actually a derivative in disguise

[royal loom]

oh god

[flint plinth]

or maybe i guess you'd call it a generalization of a derivative

but that's not really relevant to understand this part of the textbook haha

[royal loom]

I have another smaller thing you might know about

is this unit normal

[flint plinth]

yes

[royal loom]

are you sure

[flint plinth]

100%

[royal loom]

I didn't think it had to be

[flint plinth]

well if you scale n you're gonna scale the result

they can't all be right

[royal loom]

but

given a parametrization of S

there will be one normal vector

now the question is do you scale that down to unit

or not

[flint plinth]

dA is the "magnitude"
n is the direction

together they give you the surface area of a small patch, along with the orientation

if you scale either one of them by say 2, then you're inflating the area by 2

dA is supposed to be the one that captures the area

so leave n's magnitude alone as 1

right there in folland

unit normal vector

that's at the top of page 233

[royal loom]

that is for a different theorem

[flint plinth]

doesn't matter

[royal loom]

that's for greens

right?

[flint plinth]

all the surface integrals use the same convention

i thought green's was the 2d version

[royal loom]

it is

[flint plinth]

i'm in the same chapter as you, surface area and surface integrals

[royal loom]

chapter 5.5

is stokes

[flint plinth]

sure but it's using surface integrals

which are used throughtout this chatper

and they all have n dA

[royal loom]

1

isn't this a non-unit normal?

[final crag]

I will note that when using parametrizations (and not the explicit one), the n finds a way of cancelling out often in that it feels like you didn't use a unit normal

[royal loom]

just a random example of stokes theorem I found

yes and also here this one is from my class

[flint plinth]

yea that vector is n dA
not just n

[royal loom]

the normal is (0,r,r)

which will have non-unit magnitude

not trying to like be picky I just don't think it is the unit normal, although now I feel less sure

[royal loom]

it does look like the dA is seperate in the image though?

[final crag]

just to add to this as a side note but I don't think this is the issue here exactly, if you look at the lines leading up to equation 16.6.5 it kinda shows you how the unit normal tends to cancel out https://math.libretexts.org/Bookshelves/Calculus/Calculus_(OpenStax)/16%3A_Vector_Calculus/16.06%3A_Surface_Integrals#mjx-eqn-equation_5
just a plug to use differential forms so you don't even have to worry about it 😉

[flint plinth]

check here

see how the n has a denominator

but it cancels with the factor next to it

[royal loom]

what is that formula

[flint plinth]

https://en.wikipedia.org/wiki/Surface_integral#Surface_integrals_of_vector_fields

[royal loom]

so

if something is cancelling

then what is going on in this example

where n=(0,r,r)

here's the rest of it

" Let C be the curve of intersection of y+z = 0 and x^+y^2=a^2 oriented in the counterclockwise direction when viewed from a point high on the z-axis"

so it's like we parametrize S which is the graph parametrization in polar

take r partial, cross theta partial

to get the normal

(0,r,r)

which points up so it's orientation preserving (but not unit)

and we dot it with the curl and the answer is still fine

if we normalized it our answer would be wrong

so I'm confused

[final crag]

this is a line integral so far

[royal loom]

yeah because it's stokes theorem

[final crag]

yes

I understand

but if you show the work for the surface integral, you'll see that when you do it

if you did the unit normal

and then you figured out dS
you'll get dS = ||(0,r,r)|| dr dtheta

[royal loom]

Isn't this the work for the surface integral?

[final crag]

no

oh I see

sorry Im misreading

[royal loom]

No worries

[final crag]

Okay, so let's do it the "correct" way
You have G_r x G_t= (0, r, r), |G_r x G_t| = √2r right? So n = (0, 1/√2, 1/√2) so far so good
now what is dS? dS = |G_r x G_t| dr dtheta
so your integrand is (-r sin t, r cos t, 2)*(0, 1/√2, 1/√2) √2r dr dt

which ends up giving be (-r sin t, r cos t, 2)*(0, r, r) dr dt all over again

the idea is that when you normalize the normal, you divide by G_r x G_t coming from the parametrization, by the surface area element dS = |G_r x G_t| dr dt so this normalization cancels out

that's what the like I put above and the equation stuff that Bungo sent both show

[royal loom]

I think that I understand now

thank you @final crag and @flint plinth

okay but actually maybe that was a little preemptive I have one final clarifying question

in practice for stokes theorem, we don't have to actually normalize it right?

[final crag]

if you want to make this all better I would highly suggest differential forms instead, it unifies it all and makes it much easier! at the cost of some work up front for sure

in practice for flux integrals you don't have to normalize

yes

[royal loom]

Amazing

[final crag]

stokes has flux integralss so yes in that case too!

[royal loom]

I might get a book on differential forms to read sometime soon

but this is for my class right now so I can't pick the way I learn it XD

Ty very much

[final crag]

hahah I know! but just some side reading that helps you gain fluency and actually shows you that Stokes, Green, Divergence theorem are all actually the same thing

no worries!!

[royal loom]

yes we've talked about this a bit

we went over the statement of generalized stokes but it uses lots of differntial forms like you're saying

so it kind of flys over my head

i don't know multi linear algebra

[final crag]

totally fair!

[royal loom]

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[thick portal]

What am i doing wrong?

[surreal cave]

Have you drawn a picture or the triangle?

[upper abyss]

What you've put in is not an exact value. That's a decimal approximation.

[upper abyss]

Oh but you can't put in exact values, can you

[thick portal]

might be because it should be -3?

since its in 3rd quadrant

[upper abyss]

Basically, make sure to check your signs. The method you've shown above doesn't always capture the signs.

[thick portal]

ah i see

it was -sqr(5)/3 while i was putting sqr(5)/3

.close

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[loud ibex]

what is the equation i need to use to solve this

[flint plinth]

!status

[cedar kiln]

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

[loud ibex]

2

[flint plinth]

can you show where you're stuck?

[loud ibex]

i know how to find slope but im not sure what numbers im suppossed to plug in

[flint plinth]

well you want (change in y) / (change in x)
between the given points P and Q

P is (5, -4)
and Q is given by that formula, where you have to plug in the given value of x for each part

[cedar kiln]

@loud ibex Has your question been resolved?

[loud ibex]

is this what it should look like

[cedar kiln]

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[pseudo trellis]

Is there a mistake on the answer key? I got 35, but the answer key says 45?

[pseudo trellis]

I got:

[wraith dagger]

Someone

[raven shard]

10(2) not 10(3)

[tepid sail]

^^

[pseudo trellis]

oh my goodness

my bad

[tepid sail]

the boy is a liar

[raven shard]

the boy

[pseudo trellis]

thank you 😂

[raven shard]

🙂 welc

[pseudo trellis]

.solved

[cedar kiln]

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[crimson sedge]

i got (-x-7) log 16 = -7x log 11

then whats next?

[drifting marlin]

log16 and log11 are just numbers, so treat them as such

they're just gonna be constants like that -7

[crimson sedge]

so log -16x - log-112?

that doesnt make sense

[clear umbra]

the -x is outside the log

-(log (16))x

[crimson sedge]

how so?

im confused

[clear umbra]

like the person above said, log 16 is just a number

maybe if you dont like looking at it we'll just call it A for now

so now on the left you have (-x-7)A

do you know how to expand that?

[crimson sedge]

-Ax -7A

[crimson sedge]

right?

[clear umbra]

yes

same thing on the right side

if you dont like looking at log(11), just call it B for now

[crimson sedge]

okay

now what?

[clear umbra]

rearrange

solve for x

put back A and B

[crimson sedge]

-Ax-7A = B

take out A

A(-x-7) = B-7x

A = (B)/(-x-7)

right?

[clear umbra]

wheres your -7x on the right

[crimson sedge]

A(-x-7) = B-7x

?

im so lost ngl

[clear umbra]

whats log(11^-7x)

[crimson sedge]

-7x log 11

[clear umbra]

so what should this be

[crimson sedge]

A(-x-7) = -7x + B

[clear umbra]

no...

[crimson sedge]

log 11?

[clear umbra]

B = log 11

[crimson sedge]

A(-x-7) = -7xB

[clear umbra]

ok good

now expand and rearrange everything

[crimson sedge]

log 16(-x-7) = -7x log 11

im so lost lmaoo

[clear umbra]

ok now put that log16 into the brackets

what will you get

[crimson sedge]

log -16x - 112

[clear umbra]

did you mean log(-16x) or?

[crimson sedge]

yes

sorry

[clear umbra]

no..

[crimson sedge]

whaaa

[clear umbra]

you dont put that -x into the log

[crimson sedge]

then what is it lol

[clear umbra]

its just -log(16)x

[crimson sedge]

ahh i see

okay so -log(16)x - 112 = -7x log 11

[clear umbra]

-112?

where did that come from

[crimson sedge]

16x - 7

[clear umbra]

again

you dont put that 7 inside the log

[crimson sedge]

oh

[clear umbra]

its just simply -7log(16)

[crimson sedge]

but doesnt thjat mean log 16^-7

[clear umbra]

yes

but we dont need to do that

just leave it as it is

[crimson sedge]

-log(16)x -7log (16) = -7x log 11

[clear umbra]

ok good

[crimson sedge]

add -7log(16)?

to both sides

[clear umbra]

yep

[crimson sedge]

-log(16)x = -7x log 11 + 7log(16)

[clear umbra]

then what do we do

[crimson sedge]

i have no idea :/

[clear umbra]

you have x on the left and x on the right

[crimson sedge]

add

[clear umbra]

ok good

[crimson sedge]

-7xlog 11

-log(16)x + 7xlog 11 = 7log 16

[clear umbra]

7x*

[crimson sedge]

yes

[clear umbra]

ok now both terms have an x

what do we do with it

[crimson sedge]

7xlog 11/ log 16x?

i have no idea

[clear umbra]

no..

[crimson sedge]

i appreciate your patience, im teaching myself this as of rn

[clear umbra]

since both terms have an x

[crimson sedge]

take out x

[clear umbra]

ok good

[crimson sedge]

but idk how

[clear umbra]

what happens if you take an x out of -log(16)x

[crimson sedge]

x( -log16)

[clear umbra]

and what about 7xlog(11)

[crimson sedge]

x(7log(11)

[clear umbra]

ok good

so what are you left with after taking x out on both terms on the left

[crimson sedge]

x(-log16)+(7log11)

[clear umbra]

no..

[crimson sedge]

😦

[clear umbra]

both of those things are added together

[crimson sedge]

?

[clear umbra]

• should be inside the brackets

[crimson sedge]

how so?

[clear umbra]

remove the brackets in the middle

[crimson sedge]

x(-log16+7log11)

[clear umbra]

ok good

now one last step

what do you do now

[crimson sedge]

divide the right by (-log16+7log11)

[clear umbra]

ok good

[crimson sedge]

x = (7log16)/(-log16+7log11)?

[clear umbra]

yes

[crimson sedge]

how do u compute this on a calculator

[clear umbra]

just type it like how it is

[crimson sedge]

can you do it for me i dont have a calculator like that

i dont have a calculator option

on the hw

[clear umbra]

didnt the question ask for answer in logs

so you dont need to type it in a calculator

just leave it like that for your answer

[crimson sedge]

okay thank you 🙂

arigato

mucho arigato

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[royal loom]

I'm really not considering the proof sketch for this theorem at all

it references the proof of the last theorem, which made a lot of sense, but I don't see how it's applicable (I can send if anyone wants to see )

actually nvm, I see i misread and never saw the part that says "and satisfies *"

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[upper laurel]

no idea what that means but at the very least you need to show that this happens for every not-divisible-by-3 p - 1 instead of just p = 3

by default a proof applies for anything and everything that fits

[forest flame]

hey

oh

[upper laurel]

really could see the order you read those messages in

[forest flame]

wym

[forest flame]

hi can u help me out a bit

what approach should i be taking here?

[upper laurel]

sorry though I have no idea what any of that means so youll have to ask someone else

[forest flame]

instead of mod 3 i have to do all mod p where p -1 is not divisible by 3

oh ok

[cedar kiln]

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[split crest]

A) If the derivative of f at 0 exists in R, does it imply that there exists some delta > 0 such that f is bounded on the interval (-delta, delta)?

B) If the derivative of f at 0 exists in R* (with +-infinity), does it imply that there exists some delta > 0 such that f is bounded on the interval (-delta, delta)?

[split crest]

<@&286206848099549185>

[wraith dagger]

spq_64_t

[cedar kiln]

@split crest Has your question been resolved?

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[sharp sequoia]

ummmmmm

[sharp sequoia]

is this vector projection?

[wraith dagger]

spq_64_t

[sharp sequoia]

what does the brackets mean?

is the u time v?

[keen nest]

Brackets are a dot product

[sharp sequoia]

is this formula wrong?

[keen nest]

What does the hat mean above a?

[sharp sequoia]

unit vector i believe

[keen nest]

Ah, if a is a unit vector, then it is true, I believe, the formula would simplify to $proj_a(b) = (a,b)a$ indeed

[wraith dagger]

spq_64_t

[sharp sequoia]

am just having an issue understanding this

what is it conveying

like i need a picture in my head

thats why i tried using desmo

[keen nest]

Are you confused why proj_a(b) is defined as it is in general, or how do you get a formula with a^hat from it?

[sharp sequoia]

no am confused as to what is its use

like what is it giving me

whats the meaning of the asnwer

like projab

[keen nest]

I mean pictorially you kinda draw a perpendicular from b to a if that helps

[sharp sequoia]

and that perpendicular line

is the projection?

[keen nest]

No, where it intersects the line in the direction of a would be the projection

[sharp sequoia]

so the green thing here?

[keen nest]

The green one is collinear with u, not v

It should be (3, 1, 0) times some factor

Well, the factor is (u,v)/(v,v)

[sharp sequoia]

so the first one?

[keen nest]

Yeah, just the scale isn't right

the projection is (18/5, 6/5, 0) -- try drawing this one

Then you may notice that u - proj_v(u) is orthogonal to v -- hopefully, this helps with the intuition

[sharp sequoia]

the projection is longer than

[keen nest]

Yeah, that's alright

[sharp sequoia]

so what does this mean

is that how much of the red vector lies on the orange ?

or what

like this is so abstract whats the significance

[keen nest]

Pretty much

[keen nest]

The significance of the projection? I mean it can be useful, one very famous example where it's used is gram-schmidt orthogonalization process

[sharp sequoia]

oh wait that made sense since its the unit vector of the orange times the red

[keen nest]

Yeah

[sharp sequoia]

okay thanks for everything and thanks for having the patience dealing with my stupidity

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[finite raven]

If A,B have measure zero, then A X B has measure zero.

[finite raven]

A,B are subsets of R^n and R^m

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[unborn pond]

i can try it and then ill lyk if im capable of doing it

packaged food =32 hourly

frozen food = 6

32:6

which is 16:3

is that right

[crimson sedge]

@unborn pond your answer to the other question was incomplete

just as an fyi

[unborn pond]

huh

[crimson sedge]

you werent finished yet lmao

[unborn pond]

OH LOL

MY BAD

[crimson sedge]

my yes wasnt a yes towards that being the answer

[unborn pond]

hahahaha

ok lets continue that

[crimson sedge]

yeah so the net increase is 4%

[unborn pond]

ys

yes

[crimson sedge]

but now you have to find the percentage of the increase

so like

% Increase=Amount of increase/Original Whole amount

[unborn pond]

where'd you get that equation from

[crimson sedge]

idk uh, that's just how percentages work

[unborn pond]

oh ok!

so 4/33?

oh wait

no

[crimson sedge]

it was 29% originally so

[unborn pond]

ok!

so 4/29?

[crimson sedge]

yes but multiply by 100

[unborn pond]

which is 14%

[crimson sedge]

yeah

[unborn pond]

okay i see!

[crimson sedge]

thats it

[unborn pond]

tysm

wait so whats the difference of % net increase and percentage of increase

[unborn pond]

^

[crimson sedge]

idk how to explain it. Like the 4% isn't weighted it's just how much the difference was

[unborn pond]

like the 4% is essentially the difference but not truly the percentage from 29 to 33???

[crimson sedge]

yes

[unborn pond]

ok

thats a little confusing

ok can we do my next set of questions 🙂

[crimson sedge]

i gotta go but just post them again someone else will come

[unborn pond]

ok!

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[cobalt plover]

Let ABC be the isosceles triangle, with base BC s, and the inscribed circle Ω. The center of the circle Ω is I again the tangent points of Ω with the sides BC, AB, respectively AC are X, Y, respectively Z. The circumscribed circle of the triangle XZC is denoted Γ and the tangent in Z to the circle Γ intersects XY in T. Prove that the points C, I, T are collinear.

[cobalt plover]

I really think I need to use Menalaus' Theorem but I'm not sure where...

Any help please?

[cedar kiln]

@cobalt plover Has your question been resolved?

[cobalt plover]

<@&286206848099549185>

[cedar kiln]

@cobalt plover Has your question been resolved?

[cobalt plover]

<@&286206848099549185>

[cedar kiln]

@cobalt plover Has your question been resolved?

[cobalt plover]

<@&286206848099549185>

[cedar kiln]

@cobalt plover Has your question been resolved?

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[mighty rune]

are there any good rules of thumbs for determining what a decent order of integration is for 3 axes in triple integration?

[cedar kiln]

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[severe elm]

dont i not have enough information to answer?

[stable edge]

is it known that LN is a diameter?

[severe elm]

it wasnt mentioned but that might b true

[stable edge]

it looks like one but idk if that's just the diagram

[severe elm]

it was

they just never mentioned it

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[grizzled shale]

hi can i get help with a problem

[grizzled shale]

i want to see the work for how these trig identities are equal

[lyric galleon]

https://tenor.com/view/billy-bounce-fortnite-emotes-gif-4553861362510094707

[grizzled shale]

i dont know where to start

[lyric galleon]

hahah me neither

[rapid elk]

bro

[grizzled shale]

yeah

hard

[clear umbra]

which one exactly do you need help with

[grizzled shale]

33 and 34 and 36 specifically

[clear umbra]

for 33 have you tried expanding the right side

[grizzled shale]

yes and i got tan theta/2 squared

[cedar kiln]

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[cinder shard]

Hello, I am looking with some help understanding the following:

Distance from point to plane
Distance from origin to plane
Distance between planes
Intersection of lines and planes
Intersection of three planes
Angle between planes

[cinder shard]

Could anyone please assist?

[cedar kiln]

@cinder shard Has your question been resolved?

[flint plinth]

what is your specific question

[cinder shard]

Its more assistance with understanding the processes

But I can share the examples I have if it helps?

[flint plinth]

sure, pls focus on one question at a time tho

the above list is the content of like a full lecture or more

[flint plinth]

ok

[cinder shard]

Do you mind if I DM you?

[flint plinth]

do you know which two are parallel?

why, let's just use the channel

[cinder shard]

ok

i can guess

but i dont really know why

2/3 are

[flint plinth]

do you know about normal vectors to planes

[cinder shard]

ahaha

ok

lets move on

i see

[flint plinth]

ok

[cinder shard]

i can tell that their normal vectors are parallel, and since they are both orthogonal to each plane, then each plane is parallel to each other

thank you!

[flint plinth]

ok, any thoughts on how to do that one?

[cinder shard]

previous one has given me an idea, thought I am not certain how to resolve it yet

[flint plinth]

yea, think about normal vectors

[cinder shard]

i am thinking that the "step" vectors must be dot product 0

so yeah

n dot step = 0

e.g. i-2j for line 1

[flint plinth]

step = direction vector of the line?

[cinder shard]

yeah

[flint plinth]

yes that's true

[cinder shard]

i dont know the terms cuz im picking this stuff back up

[flint plinth]

if you're given two vectors (the direction vectors for the two lines in this case), do you know how to find a third vector that is orthogonal to both?

[cinder shard]

is that the easiest way to prove it though? I feel like if i had more planes/lines it would be a HUGE question manually testing each

cross product

[flint plinth]

yep

[cinder shard]

gotchu

[flint plinth]

that's what you want here

[cinder shard]

thats much better

cross product parallel to a normal

cheers

legend

[flint plinth]

and they want the plane itself

so you need two things

normal vector

which you get from the cross product

and a point on the plane

which you can get using any point on either line

since both lines are contained in the plane

[cinder shard]

got you, cheers

i have done it before with intersecting lines (rather than lines in the plane)

it was just the cross product idea i didnt immediately see

which is quite easy to use

Do you mind if we keep going?

[flint plinth]

sure

[cinder shard]

smashing these out so quick

thank you!

okay so, for F

is it

just a simultaneous application?

gaussian?

or similar

just throwing out ideas can't really see it quite yet

[flint plinth]

well that line is in both planes

so its' orthogonal to both of their normal vectors

so...

[cinder shard]

right

cross

[flint plinth]

yep

[cinder shard]

then

point

to start the line position

using a position vector

[flint plinth]

yea you just need a point in both planes

[cinder shard]

in the new line eqn

[flint plinth]

do you know how to get such a point?

[cinder shard]

just trying to think of a way rn, one sec if thats okay

[flint plinth]

sure

[cinder shard]

going through these ideas super quick, which I love, but I gotta contain it all haha. Thanks again

All I can see

is that if I apply a simultaneous approach I can eliminnate one variable

and then i have a bi-variate equation which describes a line in cartesian form

[flint plinth]

yea

[cinder shard]

but i feel like that is not the way

lol

[flint plinth]

you have one degree of freedom (since the intersection is a line)

you can freely choose one of the coordinates

say take z = 0

[cinder shard]

GOTCHU

right

thats really cool

[flint plinth]

then the plane equations become 2 eq, 2 vars

[cinder shard]

yeah got ya

just not used to finishing with a solution like that

effectively i have a line equation

in 2 dimensions

which i take at face value

sub anything i want

[flint plinth]

there are probably other equivalent ways, that's just the first that occurred to me

[cinder shard]

get the other

use that with one of the original equations

and then i have the full 3 coord

plus the direction vector from before

[flint plinth]

yes

[cinder shard]

amazing

okay cool

[flint plinth]

distance from a point to a plane

is along a line in what direction?

[cinder shard]

i have done the point to a plane now

specifically origin

i have a textbook in front of me with a method i dont quite understand but it appears to be a shortcut so i am trying to get my head around it

[flint plinth]

my first thought would be, take the normal vector to the plane

the shortest line between the origin and the plane is in that direction

the origin is a point on the line

find the equation for the line using those two pieces of info

and find where the line intersects the plane

but there's probably a shorter way

[cinder shard]

okay do you mind trying to work out this alternate way with me if I post a screenshot?

[flint plinth]

sure

[cinder shard]

its a short explanation/proof im just not seeing it

legend thanks, one sec

[flint plinth]

,rotate

[cinder shard]

so let me clarify here

i can see the process

if i have ax+by+cz = constant
I can do n given by ai+bj+ck

Divide the constant by the magnitude of n i mean

and that apparently gives distance

I can't picture the connection to the distance

[flint plinth]

yea that's valid, did you work through the logic involving the dot product?

[cinder shard]

i tried to but i just couldnt see it

[flint plinth]

what's the first statement that you don't get

[cinder shard]

(m x unit-n) dot n = k

[flint plinth]

you agree that (m n_unit) is the point in the plane that is closest to the origin?

[cinder shard]

yeah

i think fundamentally

i dont understand the meaning of k

in a vector plane eqn

[flint plinth]

that's just a measure of how much the plane is "shifted"

[cinder shard]

so

if its 0

it crosses the origin?

yeah

[flint plinth]

if k is zero then (0,0,0) is in the plane, right?

[cinder shard]

it must okay

yeah so like

[flint plinth]

and otherwise it's not

[cinder shard]

if k is 2

does that mean there is a distance of 2

no because there are coefficients

so its related but not direct...

[flint plinth]

yea it depends on the coefficients

you can scale the whole equation by some constant and that will change the coefficients and k by the same factor

[cinder shard]

okay so like is there a conceptual way to picture k exactly or because of tri-variate coefficients its hard

[flint plinth]

so in some sense only their "ratio" matters

well one way to view it is this

suppose we normalize the plane equation so that the coefficients (which are a normal vector) correspond to a unit normal vector

then the equation (in vector form) is:
n . (x,y,z) = k

now recall that hte dot product can be expressed in terms of lengths and cosine of angle:

n . (x,y,z) = |n| |(x,y,z)| cos(theta)

= |(x,y,z)| cos(theta)
since |n| = 1

and this all equals k

rearrange this:

cos(theta) = k / |(x,y,z)|

[cinder shard]

where is that theta

[flint plinth]

it's the angle between the (x,y,z) vector and the normal vector

now recall cos = adj/hyp

so viewing |(x,y,z)| as the hypotenuse of a triangle

then k is the adjacent

which is the distance (in the normal direction) to the plane

[cinder shard]

oh okay

[flint plinth]

so basically a geometric explanation of what your book says

[cinder shard]

so i think i was struggling because of not having a concrete picture of that

but i may be there

[flint plinth]

now i assumed that n was unit length

if not then you have to scale by |n|

hence your book's formula

[cinder shard]

so for any arbitrary point in the plane

[flint plinth]

k/|n|

[cinder shard]

x,y,z

i can basically see the magnitude of the position vector to x,y,z

as the hypotenuse of a triangle

where the base of the triangle is the shortest distance

vector/line

from origin to the plane

[flint plinth]

yes

right

[cinder shard]

ok let me just re-wrap my head around that

and connect it to process

[flint plinth]

oh just one small technicality

k may be negative

so the distance is actually |k|

[cinder shard]

got u

[flint plinth]

(the plane could be on either "side" of the origin)

[cinder shard]

right so

correct me if i am wrong here

my understand seems to be a little different to the form that the book has it in. It seems to relate to the idea of a normalised plane

I am effectively finding the distance by simply "normalising" k in isolation

using k/mag(n)

[flint plinth]

yea

think of it as normalizing the whole plane equation

by dividing by |n|

then the coefficients are just the unit normal

and by my argument above,

[cinder shard]

right

[flint plinth]

the new constant, which is k/|n|

[cinder shard]

100%

[flint plinth]

is the distance

[cinder shard]

you are incredible

cheers so much

[flint plinth]

sure, glad it helped!

[cinder shard]

do you mind looking at the last 3 or gtg?

[flint plinth]

gtg, gonna have dinner now

but you got this

[cinder shard]

all good, i really appreciate the help

[flint plinth]

sure, gl

[cinder shard]

dm me if you want a tip tbh

idk if allowed haha

[flint plinth]

haha nw

cheers

[cinder shard]

thank you again

.close

[cedar kiln]

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[cedar kiln]

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[cinder shard]

hey

[severe elm]

i got to (x-4)^2 + (y-3)^2 = 13

[cinder shard]

okay

[severe elm]

but r needs to be a whole number

[cinder shard]

why does R need to be a whole number?

ill check the working to there, one sec

[formal prism]

it should be x-2)^2

[severe elm]

part of the question is graphing a point on the circle and it only accepts whole numbers

ohhhh

[cinder shard]

yeah

thats it

[severe elm]

ty one esc

[cinder shard]

haha

[severe elm]

ok ty

.close

[cedar kiln]

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Send your question here to claim the channel.

Remember:
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• Show your work, and if possible, explain where you are stuck.
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[remote lantern]

for this question, shouldnt there be 3 critical points

in the given interval?

[abstract lotus]

i believe it is only 2

[remote lantern]

hm

because this is how i see it

because thats still less than 2pi correct

5pi / 3 = 5.23

which is less than 6.28 = 2pi

[brazen wigeon]

i dont think youre doing this right

[remote lantern]

hm

[brazen wigeon]

can you please explain $$\pi - \frac{\pi}{3} = \frac{2\pi}{3} = 2.09$$

[wraith dagger]

@brazen wigeon

[brazen wigeon]

like how did you get these numbers

[remote lantern]

im finding the critical points right

[brazen wigeon]

yep

[remote lantern]

so pi/3 is my xra which the reference angle

[brazen wigeon]

do you know about derivatives

[remote lantern]

ik those 2 are right because my teacher got it too

im following her example for it

[brazen wigeon]

wait just checking

are you in USA?

[remote lantern]

canada

[brazen wigeon]

ok what math class are you in

[remote lantern]

so the given interval is 0 - 2pi

this is calc

[brazen wigeon]

ok

then i think you should use derivatives

[remote lantern]

i did

i used the first derivative and made it equal 0 and solved for x

because its a trig funciton u need to find the acute reference angle

[abstract lotus]

d/dx = 1 + 2cosx = 0 : x = cos^-1(-1/2) :x = 2pi/3, 4pi/3

[remote lantern]

and that reference angl is pi/3

[brazen wigeon]

,tex $$\frac{\dd }{\dd x} = 1 + 2 \cos(x) = 0$$

[wraith dagger]

@brazen wigeon

[remote lantern]

yes

[brazen wigeon]

so that means:

[remote lantern]

cosx = -1/2 now

[brazen wigeon]

yep

[abstract lotus]

find where x is -1/2

[brazen wigeon]

ohhh i think i see

i think i kind of understand what you're saying

[remote lantern]

yea

then u do

cos^-1 (1/2) = pi/3

cos inv(1/2)

its negative but u make the ref always postive right

[brazen wigeon]

,tex $$\text{Wrong: }\arccos(\frac{1}{2}) = \frac{\pi}{3}$$
$$\text{Correct: }\arccos(-\frac{1}{2}) = \frac{\pi}{3}$$

[remote lantern]

so thats why

[brazen wigeon]

i think that is your mistake

[remote lantern]

i get the first two

but shouldnt there be a thiurd

[brazen wigeon]

the 1/2 should bee negative

[remote lantern]

where?

no

it has to be positive

[abstract lotus]

look at unit circle

[wraith dagger]

@brazen wigeon

[brazen wigeon]

what is it

[remote lantern]

im following the teachers video atm as wel

she stated that when finding the ref angle it has to be postive

when u arc cos or arc sin it

no matter what

i get how to get the first two

and that it fits between the interval of 0 - 2pi

but i thought there would be another crtical point which is 2pi - p/3 becaue if u do that it still fits in that given interval

we were taught to use the CAST rule to find the CP

[abstract lotus]

look at unit circle of 5pi/3

the x is 1/2 and u are looking for -1/2

[remote lantern]

hm im confused

[abstract lotus]

im not sure why u think there is a third

[remote lantern]

5pi /3 = 5.23?

thats less than 6.28

because the 5.23 is still less than than the given interval

[abstract lotus]

but cos(5pi/3)=1/2 ur looking for where cosx = -1/2

[remote lantern]

it cant go beyond 2pi which is 6.28 right

[brazen wigeon]

right now, the interval is not your problem

[brazen wigeon]

,tex $$\cos(\frac{5\pi}{3}) \neq -\frac{1}{2}$$

[wraith dagger]

@brazen wigeon

[remote lantern]

hm

i thought we would keep going around the cast rule in the quadrants

until we hit a point where it extends the interval

[abstract lotus]

im not sure what you are referring to

[remote lantern]

like im using the cast rule to find the CP

here

[brazen wigeon]

one sec lemme research this cast rule

[remote lantern]

ok

thaks

are u familar witht eh cast rule tuna

[abstract lotus]

ive never heard of that

plug in 5pi/3 to derivative - 1+2cosx=0

[remote lantern]

because our value for cos x = -1/2 we have to draw the lines in quadrant S and T because thats where Cos is negative

so 2 and 4^

like I have

[brazen wigeon]

but 5pi/3 is in quadrant 4

[remote lantern]

yes

because half a circle is pi

[brazen wigeon]

and in quadrant 4, cosine is positive

[remote lantern]

so from quadrant 1 to 2 is pi

[abstract lotus]

ur looking for angles in 2 and 3 not 2 and 4

[remote lantern]

il draw it out