help-27

[winter patrol]

graph isn't accurate enough

many key coordinates are off

make a table of values as recommended

[ebon coyote]

Seems fine to me

[winter patrol]

and also write in the scale for the axes

[late scaffold]

nvm it's not fine

[fierce ruin]

why

[late scaffold]

but that also depends on the scale used, yeah

well, at x = 2, your graph is already off

[late scaffold]

if your graph has a scale of 1:1 on the y-axis, at x = 2, your graph shows a value of about y = 1.3

[ebon coyote]

So, here's the thing that's probably separating me from the others here

[late scaffold]

at best I can do you 1.5

[ebon coyote]

The question is asking for a sketch

[fierce ruin]

it does the same thing on desmos, im confused cuz shouldnt it cross at x=-2?

[ebon coyote]

To this end, I'd label the intersection point with the axes and draw in an asymptote

[late scaffold]

if they give you a grid, at least try to match coordinates

[ebon coyote]

...fair

[ebon coyote]

And it should cross there why?

[fierce ruin]

i dont understand ngl

[winter patrol]

and they also mention to use a table of values,
so its implied they want a bit more than just intercepts

[fierce ruin]

cuz it shifts down twice right?

[late scaffold]

2 units, not twice

[fierce ruin]

it says table values may help

but how

[winter patrol]

more accurate

[late scaffold]

it helps you determine key points

[fierce ruin]

this unit went by too fast idrk what im doing

[late scaffold]

try making a table first

then plot the points you get from the table

[winter patrol]

the more stuff you have in your table,
the more points you'll have
the better your graph will be

atm your graph is decent up until x=1

but the stuff to the right of that is quite inaccurate

[fierce ruin]

so for the table i write -1,-2,-3,0,1,2,3?

[late scaffold]

for safety you can also do 3

but any further is not necessary

[fierce ruin]

got it

now what do i do with it

[winter patrol]

evaluate the equation at those values of x, to get the y-coordinates
and hence your points on the graph

[fierce ruin]

can u give me an example? or like im trying to visualize it through desmos rn

where do i plug the table numbers in the equation

[frigid shale]

just do a table indicates x and y

and set value like when x =0, what is y

[fierce ruin]

0? or 1

[ebon coyote]

You're trying to graph the function y = 2^x - 2

Plug in x = 0, what do you get for y?

[fierce ruin]

0

[ebon coyote]

What's 2^0, first?

[fierce ruin]

1?

[ebon coyote]

ye

Now subtract 2...?

[fierce ruin]

-1

[ebon coyote]

ye

Now we can do the same with other x-values

https://youtu.be/OrCAXsBCjEc is a good video on this

[fierce ruin]

ill watch that rn

that was a good vid but ik this

how do i apply this here now with the equations i have

[winter patrol]

evaluate the equation at those values of x, to get the y-coordinates
and hence your points on the graph

[fierce ruin]

so the first point is 2^0= -1 right?

[winter patrol]

bad

no

[fierce ruin]

ok so for x we +1 and y +2?

[winter patrol]

y = 2^x - 2
x = 0, gives
y = 1 - 2 = -1

giving you the point (0,-1)

x = 1, gives
y = 2 - 2 = 0
giving you the point (1,0)

those y-values would be under your respective x-values
which should be easy to read

[fierce ruin]

-3,-2, -1, 0, 1, 2,3

[winter patrol]

you can choose whatever x-values you want,
ideally you'd want key values and/or values that are easy to evaluate

[winter patrol]

set up a table like in the above image,
evaluate the equation at those values,
writing the y values in the respective boxes

[fierce ruin]

what i meant by this is if were at point +2 , we minus 2 for y values?

[winter patrol]

i don't udnerstand what you mean by at the point +2

[fierce ruin]

positive 2

[winter patrol]

you calculate what the equation says

-2 will be part of all your calculations

at x=1, 2^x will be 2^1 = 2
thus at that value of x, y = 2^x - 2 = 2^1 - 2 = 2 - 2
if that's what you mean...

[fierce ruin]

like this?

[late scaffold]

you can now plot those same points on your grid

and then connect them

that being said, why is everything above the x-axis despite the -2?

this looks like 2^x without the -2

the method is correct though. just need to apply the -2 to all of the points

[fierce ruin]

so is it looking alright?

i need to fix my line 😭

[late scaffold]

no, for reasons I mentioned above

[fierce ruin]

ohh

i need to make it start from y=-2?

[late scaffold]

well see, just take an example value like x = 0. at x = 0, 2^x = 1, but then you have the -2 bringing y down to -1 (2^0 - 2 = 1 - 2 = -1)

your original graph was correct in this regard

[winter patrol]

you should have had the relevant points from completing your table

asymptotes should also be marked with a dashed line

can you show the table you made

[late scaffold]

ok see, x = 0 definitely isn't y = 1.

[late scaffold]

again, I have mentioned what I think you're doing

[winter patrol]

evaluate the whole thing

not just the 2^x

y is 2^x - 2

not just 2^x

[late scaffold]

what's going on here

which color should we look at

[fierce ruin]

the one thats most accurate

[ebon coyote]

You see... that's not clear WHICH one you mean by that

[fierce ruin]

i made 2 tables

[late scaffold]

I wish I could write multiple answers on a test and tell the examiner to pick the one that's most correct lol

[ebon coyote]

For one, this here explicitly means that for x-value 0, we have y-value -1

[fierce ruin]

LMFAOO

[late scaffold]

anyway the black one is confusing

[fierce ruin]

PLEASE

😭

[late scaffold]

why do you have two different x = 2 rows

and both having different y values

blue is correct for those three points. will need a little more points though

[ebon coyote]

So in your table, you should have something like

x | y
...
0 | -1

as an entry

[late scaffold]

oh yeah the first table is just out

[fierce ruin]

alright ignore that ss im gonna start over

[late scaffold]

as a reminder

the second table is indeed correct for those three points

just need more points, esp on the -ve side of x

[devout snow]

@fierce ruin Has your question been resolved?

[devout snow]

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[jaunty kettle]

Someone explain me equivalence and bi-conditinal or either help by providing a short video to understand.

This is a reference image about what my professer on maths course is teaching.

[lapis quarry]

Are you familiar with what it means if it is said that p implies q?

[devout snow]

@jaunty kettle Has your question been resolved?

[jaunty kettle]

Yup, but give me a short brief

I have feel of it

But don't know by definition what is the meaning of "implies" or "implication"

[upbeat isle]

Implication is like a promise

[lapis quarry]

You've got true/false for each variable so let's take p and q. Thus we got 4 different combinations - tt tf ft ff. If p->q it means out of a true p I can follow that q must be true. Additionally youve gotta get used to that out of something false (so if p is false) you dont get any information about q. So q can then be true or false, it doesnt matter. So the only case when the implication then is false is if p is true and q is false since then you cant follow q being true out of p being true.

[jaunty kettle]

Alr

But you didn't told me what implies or implication mean?

I knew this practical approach already

[lapis quarry]

Implication is the "->" in your screenshot and implies is just the verb to it so hwo you say p implies q (= p->q)

[upbeat isle]

Imagine a politician makes a promise like if I become the mayor I will lower the taxes. After he is elected then if he actually lowers taxes then he has kept his promise and if he hasn’t then it’s a false meaning

In case he becomes the mayor (A) and lowers the taxes (B)

A-> B, A is true and B is also true so the entire implication becomes true

In case he didn’t become the mayor but the taxes are still lowered

A is false but B is true so the entire implication is true

Now if he becomes the mayor and doesn’t lower the taxes

A is true but B is false so the entire implication becomes false as the result was false

So basically the entire implication depends upon the final result

[lapis quarry]

Yep - nice example ⬆️

[upbeat isle]

Hehe thanks

I failed this subject last sem

[jaunty kettle]

Oh got it

Thanks both

Now should we ahead for equivalence?

[lapis quarry]

Equivalence then is just that both - so p and q - imply each other. So the only options then are tt (p true and q true so out of p is true follows q is true ✅ AND out of q is true follows p is true ✅) and the trivial solution ff (p false so it doesnt matter what q is ✅ AND q false so it doesnt matter what p is ✅) - the other two possibilities with ft (p false so it doesnt matter what q is ✅ BUT q true so p must be true too ❌) and tf (p true so q must be true too ❌) also false here (even though q true and p false ✅ alone would be valid).

So to check for equivalence, you have to check for the first one to imply the second and the other way around.

[jaunty kettle]

Ahhh

Confused

Lol I am going on youtube. I got hint of what's going on

Thanks for the help @lapis quarry!

.close

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[jaunty kettle]

someone explain please. i am confused between this necessary and sufficent case. in the previous two questions [professor had given the cas eof necessary and sufficent respectively and i stucked their also. So now this question with both conditons intersecting each other is even a further hell.

[drifting sierra]

If A is a necessary condition for B, that means B implies A

If A is a sufficient condition for B, that means A implies B

Both at the same time means A and B are equivalent

@jaunty kettle what exactly are you confused about?

[jaunty kettle]

i understood nothing what u said @drifting sierra

these so many words and conditions have left mw confused

[drifting sierra]

Not sure what to tell you

Look at each option one by one and try to figure out whether it implies the statement or is implied by it

[devout snow]

@jaunty kettle Has your question been resolved?

[wheat pawn]

A is necessary for B means that if A does not happen, B cannot happen.
Example: Me studying is necessary for me to pass the subject. I dont study, so me passing cannot happen

A is sufficient for B means that A happening is enough to make B happen.
Example: If it rains, the ground gets wet. It raining is sufficient for the ground to get wet

neither imply the opposite.
On the first example, me studying is necessary to pass, but me studying doesnt guarantee that i will pass.
If i study, i may or may not pass. If i do NOT study, i will NOT pass

On the second example, raining is enough for the ground to get wet. But if it's not raining, the ground can also get wet for other reasons, like for example someone using a hose

[warm karma]

e.g. if you have A * B = 1, then A MUST be non-zero. But you still care about what B is before you know for sure. That is necessary.

But if you have A * B = 0, then if A = 0, you know this equality holds. You don't even need to worry about B at all anymore. That is sufficient.

[devout snow]

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[restive river]

Hi, I joined this to hopefully find anyone who can help me. All my life I have loved math but I am genuinely horrible at it. It’s not like I’m not good or struggle to understand, I can understand and I can learn fast. But my education so far has been horrible. My dad took me out of school when I was 4 and I didn’t get back into school till I was 8 and a half, I caught up quickly to the other kids but I was 2 grades behind and I did “homeschooling” at 12 which was a horrible decision because I did not do any schoolwork whatsoever. I get back into school at 13 and a half but I only had the basics down of elementary school math if you can even say that basically only anything under multiplication down, not dividing or square root or exponents nothing. They put me in math tutoring which helped because the math teacher I had was very cruel and mean so I struggled learning with her but I’d still try. Doing the math tutoring helped me understand the beginning of algebra. I became better in my class but, I decided to go back to homeschooling and then I struggled with doing any schoolwork till they forced me to meet on call with my teachers every now and then. So I did, I ended up actually learning and that was the only way I was doing some schoolwork. But specifically when I started doing math with my math teacher on call (algebra) I fell in love with it again. I remembered my passion for it which I was always obsessed with astrophysics, science and just mathematics in general. I always told myself I couldn’t though, that I wasn’t smart enough or didn’t have the money. But that teacher made me understand it and I figured out that I could do it. Sadly when I moved I had to switch schools and now I have no one to teach me math or help me learn it. That’s why I’m asking if anyone is offering to teach me. I won’t waste too much of your time I just wanna learn man. If anyone can just help me at least learn algebra or math in general that would be very much appreciated. Ty 🙏

[tender cobalt]

you could just try khan academy

[neat solstice]

Im sorry for you, schooling is so important

I can also recommend khan academy
Besides learning, i can recommend 3blue1brown, mathologer and blackpenredpen because they are interesting. But to learn, you have to do stuff, which is why khan is good

[devout snow]

@restive river Has your question been resolved?

[native gyro]

!done

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[devout snow]

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[ebon coyote]

https://tenor.com/view/furina-absolutecinema-furinaabsolutecinema-gif-1488900600823137103

[devout snow]

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[prisma garnet]

i need help

[vale raft]

and about what??

[prisma garnet]

wait

Rolle's theorem

?

[uneven coral]

Can you please send your question

[vale raft]

oh cant help with that havent seen it for the moment

[uneven coral]

I probably also might not be able to help but at least with the question up someone who might be able to help would see it

[prisma garnet]

please explain to me any example of your question

[sand dove]

?

[uneven coral]

No no, can you send a question you were stuck in?

[sand dove]

Did you have a specific question in mind when you opened the channel

!xy

[devout snow]

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.

No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question!

@prisma garnet Has your question been resolved?

[wind mason]

[supple knot]

.close

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[vestal ferry]

I love cats

[vestal ferry]

.close

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[cold haven]

uhhh

this is not the purpose of a help channel.

next time please ask an actual question.

[devout snow]

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[topaz beacon]

ya sabes que necesitas traducir para obtener las resultos mejores

[mellow cloud]

I think they want to find the minima and maxima of f in the region of A arent they?

[spring oasis]

yes

[spring oasis]

yes

[spring oasis]

yes

[mellow cloud]

First, you may want to sketch a little bit to see how the region A looks like.

You may want to find the minima and maxima of f (if exists) inside the region first, then find the extremas in the boundary of A with two conditions (not together) as $\dfrac{x^2}{2} + \dfrac{y^2}{4} = 1$ and $y = \sqrt{2}x$

[woven radish]

Queintik

[devout snow]

@spring oasis Has your question been resolved?

[spring oasis]

how does it look

to me it looks like a ellipse

I could be wrong i am bad at conics

[mellow cloud]

yes, it is an ellipse but the line $y = \sqrt{2}x$ intersect it

[woven radish]

Queintik

[mellow cloud]

you can use desmos to sketch it yourself

[spring oasis]

is it possible to get intuition by hand

also, I am not sure if I need to have a geometrical understanding of the situation

@mellow cloud

[mellow cloud]

it just help you to understand how would we get this problem step by step

[spring oasis]

if we can proceed with the geometrical info I think we would be fine

[mellow cloud]

this is the region A, both colored blue and red

if the region "looks" beautiful, then you can straightly using Lagrange multiplier, but it does not isn't it

so, another way to do it is, you may want to find the extrema inside the region first, it's straight up derivative

then, you find the extrema in the boundary, in the boundary it satisfies, either $\dfrac{x^2}{2} + \dfrac{y^2}{4} = 1$ or $y = \sqrt{2}x$, then you can substitute y in f(x;y) with x

[woven radish]

Queintik

[mellow cloud]

then, compare both extremas you found

[devout snow]

@spring oasis Has your question been resolved?

[spring oasis]

care to elaborate on that beautiful terminology

[spring oasis]

how

[spring oasis]

this whole boundary and interior confuses me

[stable storm]

well there are two possibilities, either the min/max are in the region or on the boundary of the region

you need to use two different tools to check each of them

the “interior” can be checked using the second partial derivative test whereas the boundary can be checked using lagrange multipliers

as queintik noted the boundary is not necessarily smooth

so we would need to check each section of the boundary individually

[devout snow]

@spring oasis Has your question been resolved?

[granite venture]

try to check the interior of the boundary first, see what the candidate extrema are

the numbers you get might not be extrema, but they are candidates to be one

[spring oasis]

oh I see

[spring oasis]

care to elaborate on that?

[stable storm]

so this is the region we have

see how there are two corners where the red and blue curves intersect

[spring oasis]

ye

[stable storm]

this isn't a smooth curve so we would just need to check each boundary one by one

mainly cuz we can't really find one general equation (which isn't piecewise) that encapsulates the information of both curves

[spring oasis]

how do you know if something is smooth or not

[spring oasis]

why is that

[devout snow]

@spring oasis Has your question been resolved?

[fiery marten]

Just use Lagrange multipliers

$L = \lambda_0(x^3-y^3) + \lambda_1 \Bigl(\frac12 x^2 + \frac14 y^2 - 1\Bigr) + \lambda_2 (\sqrt2 x - y)$

[spring oasis]

can I get some help

[fiery marten]

Sure

[woven radish]

spindle

[spring oasis]

the issue is

can we start from scratch

[fiery marten]

So generally the method is as follows.

We're given a function $f(x_1, \dots, x_n)$ to minimize, and a set of constraints of the form $\varphi_k(x_1, \dots, x_n) \le 0$, where $k=1, \dots, m$. Then the method prescribes to write down the following function called Lagrangian:
$$L = \lambda_0 f + \lambda_1 \varphi_1 + \dots + \lambda_m \varphi_m$$

[woven radish]

spindle

[fiery marten]

Here $\lambda_k \ge 0$

[woven radish]

spindle

[fiery marten]

Now you consider two cases: $\lambda_0 = 0$ and $\lambda_0 = 1$. In both cases solve the following system of equations: $$\frac{\partial L}{\partial x_j} = 0$$

[woven radish]

spindle

[fiery marten]

Oh and since you want all extrema you might also want to consider λ0 = -1

[spring oasis]

can I ask u a favor

there is too much generality in the way you are explaining this , is for my intro to calculus class, we always have functions frol R2 -> R

[fiery marten]

Yep, so case n=2 in my notes

[spring oasis]

and we almost always just have two constraimts, two inequalities

[fiery marten]

Okay, then m=2 as well

[fiery marten]

.

[spring oasis]

the lagrangian?

[fiery marten]

Yes

[fiery marten]

Sorry here there's no dL/dλ = 0, that's only when m=1 and the constraint is φ(x,y)=0

In general you have λ_k φ_k = 0 on the extrema

[spring oasis]

how did that happen

[spring oasis]

i thought there was two constraints

[spring oasis]

wait

can we start from scratch with a less generalized algorithm

Just keep it simple

[fiery marten]

I made a mistake. I corrected myself, by saying that you have λ_k φ_k(x, y) = 0 instead of just φ_k(x, y) = 0

[spring oasis]

what does each of this symbols mean

[fiery marten]

λ_k is so called Lagrange multiplier, it's a non-negative unknown number

[spring oasis]

can we start from scratch with a less generalized algorithm

[spring oasis]

what

[fiery marten]

φ_k(x, y) is a constraint function

[spring oasis]

we have two constraints

[fiery marten]

It comes from the method I described earlier

[spring oasis]

why are you equating them two zero

the constraints

[fiery marten]

Yep, so φ1(x, y) = 1/2 x² + 1/4 y² - 1, φ2(x, y) = sqrt(2)x - y

[fiery marten]

I don't, instead I equate λ_k φ_k to zero

[spring oasis]

what

[spring oasis]

no

the second constraint doesnt even have y

ohh

what??

[fiery marten]

Write down the second constraint here

From the task

[spring oasis]

1/2 x2 + 1/4 y2 - 1 <= 0

0 >= sqrt2x - y

[fiery marten]

Yep, so after rearranging in such a way that we have (something) <= 0 we get

Yes, we get sqrt2 - y <= 0

[spring oasis]

whereisalltheinequaloties goingto

you got rid of all of them

what the fuck is going on

@fiery marten u here?

[fiery marten]

Yes, sorry, 1 sec

[fiery marten]

Each φ_k(x, y) <= 0

I wrote the method for constraints of the form φ_k(x, y) <= 0, so we rewrote our constraints from the exercise in this form

For example, the inequality y >= sqrt(2)x doesn't look like <something> <= 0, right?
So we rearranged, and we got sqrt(2)x - y <= 0

That's why our lagrangian contains a term λ2 (sqrt(2)x - y)

[spring oasis]

no

[spring oasis]

no

[fiery marten]

Please elaborate

[spring oasis]

yes

we have y >= sqrt2x

we want to make it like, SOMETHING <= 0

[fiery marten]

Exactly

You earlier wrote 0 >= sqrt(2)x - y

It's the same inequality as sqrt(2)x - y <= 0

[spring oasis]

,, y \geq \sqrt{2}x \ y - \sqrt{2}x \geq 0

[woven radish]

Renato

[fiery marten]

Okay, now multiply both sides by -1

[spring oasis]

I see

yes is just that the >= is flipped

it confused me for a bit there

[fiery marten]

That's okay

So now we can write the lagrangian

[spring oasis]

I mean to be honest you skipped a couple of steps here and there and I needed to fill the gaps

[spring oasis]

what is that

[fiery marten]

Sorry I assumed that you will understand

[spring oasis]

what is a lagrange

[fiery marten]

1 sec

[fiery marten]

this

Let's figure out what is f and φ_k

[spring oasis]

wtf

[fiery marten]

I will try to carefully explain step by step

We want to find min and max of x³ - y² in some region

x³ - y² is our f(x, y)

[fiery marten]

We figured out the constraints:
1/2 x² + 1/4 y² - 1 <= 0
sqrt(2)x - y <= 0

[spring oasis]

notreallyno, x3-y2 sir

[fiery marten]

So we know what is φ1 and φ2

[fiery marten]

Oh my bad

[spring oasis]

yes

phi1 and phi2

[fiery marten]

Now we can write the following function:
L(x, y) = λ0 f(x, y) + λ1 φ1(x, y) + λ2 φ2(x, y)

This is lagrangian ^

λ0 (x³ - y²) + λ1 (1/2 x² + 1/4 y² - 1) + λ2 (sqrt(2)x - y)

[spring oasis]

wtf with the lambdas?

[fiery marten]

Just positive mubers, we don't know them yet

[spring oasis]

how to find

[fiery marten]

We solve a system of equations, I will elaborate in a minute

The system is $$
\begin{cases}
\frac{\partial L(x, y)}{\partial x} = 0, \
\frac{\partial L(x, y)}{\partial y} = 0, \
\lambda_1 \varphi_1(x,y)=0, \
\lambda_2 \varphi_2(x,y)=0.
\end{cases}
$$
Both $\lambda_1, \lambda_2$ must be non-negative, and $\lambda_0$ is either $0, 1$, or $-1$, we must consider all cases

[woven radish]

spindle

[fiery marten]

Sorry it's a bit hard to type on the phone

But yeah here we go

Last 2 equations in the system are called orthogonality conditions

[spring oasis]

what

[spring oasis]

help please

[fiery marten]

Of course, just let me know what exactly is unclear

[spring oasis]

what do you mean that lambda0 is random bumber

[fiery marten]

It's an unknown real number, but note that the system of equations doesn't change if we divide the entire L by any positive number

[spring oasis]

what

[fiery marten]

So when λ0 ≠ 0, we can divide L by |λ0| and the lagrangian will look like (x³ - y²) + ... if λ0 was positive, and -(x³ - y²) + ... if it was negative

That's why it's enough to consider λ0 = 0, 1, -1

[spring oasis]

we always equate this to 0?

[fiery marten]

Yes

[spring oasis]

shit looks tuff

too many possibilities

I might need some handholding

[fiery marten]

I can do that but unfortunately I will be offline in like 5 minutes for 1 hour or so

We can return to this later or maybe someone else will continue the explanation

[spring oasis]

no worries I will be waiting you until the end of time

I understand people are busy, yesterday I was dishwashing it took me like 1 hour then I cooked some meat

so sorry if I responded like shit a couple hours ago

I will watch some videos on lagrange multipliers in the meantime

[fiery marten]

That's a good idea

Because I can explain the algorithm in text but it's hard to give a good intuition and the idea behind the method using just text

[gritty terrace]

wassup

[spring oasis]

hey miss

whats up

[gritty terrace]

hru doing with this Q

[spring oasis]

like dogshit

[gritty terrace]

what've you tried so far?

[spring oasis]

spindle is making me realize idk anything about lagrange multi

[gritty terrace]

ahhh right

yes, lagrange multipliers are an alternative way to approach these kinds of Qs

though they're not strictly necessary imo

[spring oasis]

I think i like using them

my friends were saying they used langrange multi for this

[gritty terrace]

yeah they're definitely a cool approach

[spring oasis]

sure, but it's the standard

don't know if you miss, know lang multi

I was watching some videos about them

[gritty terrace]

yes i know lagrange multipliers!

[spring oasis]

in spanish they are called multiplicadores de lagrange

anyways, any idea how to do this ? 🤔

[gritty terrace]

you want me to explain lagrange multipliers to you?

[spring oasis]

sure if that's possible and some handholding

if we can start from scratch that would be perfect

[gritty terrace]

ok lemme grab lunch first

[spring oasis]

yep, ping me miss

[fiery marten]

I prepared some solution steps in the meantime

[woven radish]

spindle

[fiery marten]

Take your time to verify these, ask any questions

[woven radish]

spindle

spindle

[fiery marten]

Let's stop here for now

I'm expecting questions so yeah

@spring oasis ^

[spring oasis]

how

am I supposed to fill the gaps by myself here? how did you went from slide 1 to slide 2

@fiery marten

[fiery marten]

First equation is a quadratic equation with respect to x

That's the quadratic formula

Oh sorry I didn't write x=

[spring oasis]

,w 3ax^2 + bx+sqrt(2)c = 0

[woven radish]

Results provided by WolframAlpha

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Requested by renatoxbz

[spring oasis]

,w -2ay+(1/2)by-c=0

[woven radish]

Results provided by WolframAlpha

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

Requested by renatoxbz

[spring oasis]

how did the second one happen

[fiery marten]

It's a linear equation in y

[spring oasis]

I see

[fiery marten]

Wolfram wrote the same thing but slightly rearranged

[spring oasis]

yes

[fiery marten]

Alright

[spring oasis]

how did this shit happen

wtff

[fiery marten]

Look here and suppose λ1 > 0, λ2 > 0

Last two equations must hold

We can divide one of them by λ1 and the other one by λ2

[spring oasis]

yes of course

everything is clicking as of now

@fiery marten

[fiery marten]

Nice, so we currently have λ0 ≠ 0, λ1 > 0, λ2 > 0, and (x, y) is either (1, sqrt2) or (-1, -sqrt2)

[fiery marten]

Let's look at the first two equations here

Here's an important note

What if we substitute x=1, y=sqrt2, solve for λ1, λ2, and suddenly they end up not positive? We have to check

Because λ1 and λ2 must be positive

[spring oasis]

holy cow

where do I plug and when

where do i plug

where do you want me to plug

[fiery marten]

Let's do that. First equation gives us
$$3\lambda_0 + \lambda_1 + \sqrt{2}\lambda_2=0,$$
and the second one gives
$$-2\sqrt2\lambda_0 +\frac{\lambda_1}{\sqrt2}-\lambda_2=0$$

[fiery marten]

First and second equations in our system

[spring oasis]

why

[woven radish]

spindle

[gritty terrace]

-# btw @spring oasis i am back now

[fiery marten]

Because we're searching for (x, y, λ1, λ2) such that the system of 4 equations holds
We found x and y in the case when λ1 > 0 and λ2 > 0, but we have to find λ1 and λ2 as well

[spring oasis]

we have three unknowns and 2 eqs

[fiery marten]

And ensure that the values we found are positive

[spring oasis]

help

[fiery marten]

We already discussed that it's enough to consider λ0 = 0, 1, -1
Currently we're considering λ0 ≠ 0

[spring oasis]

ok so consider 1

[fiery marten]

Yep

[fiery marten]

Solve these equations in the case when λ0=1

[spring oasis]

b = sqrt(2)c- 3
c = -2sqrt(2) + b/sqrt(2)

c = -2sqrt(2) + (sqrt(2)c-3)/sqrt(2)

c = -2sqrt(2) +c - 3/sqrt(2)

@fiery marten

where is the mistake

[fiery marten]

b = -sqrt(2)c - 3

[spring oasis]

damn

yes I just saw it

b = -sqrt(2)c- 3
c = -2sqrt(2) + b/sqrt(2)

c = -2sqrt(2) + (-sqrt(2)c-3)/sqrt(2)

c = -2sqrt(2) - c - 3/sqrt(2)

2c = -2sqrt(2) - 3/sqrt(2)

@fiery marten

2c = -2sqrt(2) - 3/sqrt(2)
c = -sqrt(2) - 3/(2sqrt(2))

[fiery marten]

,w 3+b+sqrt(2)c=0, -2 sqrt(2) + b/sqrt(2) - c=0

[woven radish]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by equenos

[fiery marten]

Correct

[spring oasis]

wtf?

[fiery marten]

As you can see, we got λ2 = -7 / (2 sqrt(2)) < 0, which can't be, because λ2 > 0 as prescribed by the method of Lagrange multipliers

[fiery marten]

Wolfram can solve systems of equations, I used it to verify your answer

[fiery marten]

So λ0=1, λ1>0, λ2>0 lead to no solutions

We also have the case λ0=-1

[spring oasis]

,w b = -sqrt(2)c- 3 where c = -sqrt(2) - 3/(2sqrt(2))

[woven radish]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by renatoxbz

[spring oasis]

,w simplify -sqrt(2) - 3/(2sqrt(2))

[woven radish]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by renatoxbz

[fiery marten]

Yep

[spring oasis]

ok your shit looks correct to me

bur idk how to go from here?

[fiery marten]

My shit it just the system itself

[spring oasis]

any handhold please?

[spring oasis]

ohhh

[fiery marten]

Did you understand this conclusion? #help-27 message

[spring oasis]

.

[spring oasis]

why

[fiery marten]

The method tells us that λ1, λ2 must be non-negative
We considered λ1>0, λ2>0

We arrived at λ2 = -7/(2 sqrt(2)), contradiction

[spring oasis]

oh so they must be non negative b and c

[fiery marten]

Yes

[spring oasis]

either in a = 0 or a > 0 or a < 0

[tawny patrol]

The reason why λ0 equals 1 is ?

[spring oasis]

b and c are positive

[fiery marten]

We have 3 possibilities
λ0=0, λ0=1, λ0=-1
We're currently considering λ0=1

[fiery marten]

Exactly

[tawny patrol]

Ok

[spring oasis]

let's go back a Lil bit

here either b = 0 or b > 0 or b < 0

[fiery marten]

But the system holds if we divide by |a| when a≠0 so it's enough to consider a=±1 when a≠0

[fiery marten]

No, b>=0

That's what the method of Lagrange multipliers gives automatically

[spring oasis]

wdym

[fiery marten]

λ1 >= 0, λ2 >= 0, ...

[spring oasis]

that's from the theo

[fiery marten]

Yes

[spring oasis]

so either b > 0 or b = 0

[fiery marten]

Yes

[spring oasis]

or c > 0 or c = 0

[fiery marten]

And same for c

Yep

[spring oasis]

bur we also have a = 0 1 -1

help shit is getting confusing

[fiery marten]

Yes

So we have 12 cases

a=1, b>0, c>0,
a=-1, b>0, c>0,
etc

[spring oasis]

uff ok

why does

ahh I see now

[fiery marten]

First one we already discussed and arrived at no solutions

[spring oasis]

a = 1 and b > 0 and c < 0 leads to nothing

[spring oasis]

ye

[spring oasis]

help

[spring oasis]

I mean in reality we just need to check a = 1 and a = -1 and a = 0

then solve for b and c

[fiery marten]

Remember how we immediately found x and y when b>0 and c>0?

[spring oasis]

if both b and c >= 0 we good

[spring oasis]

what?

[fiery marten]

Well, yes

That's essentially what we're doing

[tawny patrol]

Is there any particular method to simplify solving this system of equations, especially for larger values of λ?

[fiery marten]

By larger values of λ you mean more constraints, right?

Generally the system can be arbitrarily complicated and might not have a closed form solution in elementary functions. I'm not aware of any easy method in general

[spring oasis]

you are trolling?

[tawny patrol]

Yes right

[fiery marten]

When equations are polynomial you can use Grobner basis I suppose

But other than that idk

[spring oasis]

we dont know the lambda values until we solve the system

[tawny patrol]

NO why?

[fiery marten]

Yeah

We found one solution of this system:
x=1, y=sqrt(2), λ0=1, λ1=1/2, λ2=-7/(2 sqrt(2))

But as you can see λ2<0 so we discard it

[tawny patrol]

I meant solving the equations in genera

[spring oasis]

like the system can be simple and still give you large lambdas after you solve it if the coefficients of the linear equations have a bad linear diophantine, isn't this equivalent to solving multiple linear diophantine equations

we dont know how large the lambdas can be until we solve the system is what I mean

[fiery marten]

Rasa probably asked about m being large, not lambdas

m being the amount of constraints

We have m=2

[spring oasis]

I seeeeeee

yeah I might have misinterpreted his message

[tawny patrol]

Given that λ0 = 1 and λ1 = 1/2, this solution might still be valid under the constraints, so we need to check the conditions further to determine whether this solution is optimal or not

[spring oasis]

c is already less than 0 so we cant use this

[spring oasis]

I agree, but don't know lagrange by heart

[tawny patrol]

Right

[fiery marten]

Okay okay

But yeah now we should also check λ0=-1 (spoiler: we will get λ1=-1/2<0 so we also discard this case)

And then check λ0=0

[tawny patrol]

I understand! It's a bit tricky, but we can work through it step by step

[fiery marten]

Important note: the method of Lagrange multipliers also tells that λ0, λ1, λ2 can not be zero simultaneously

[spring oasis]

do you mind we also go through dat

[fiery marten]

Okay, we're here, but λ0=-1: #help-27 message

(These are the equations when x=1, y=sqrt(2))

[tawny patrol]

It makes sense that λ values should not all be zero at the same time, as it could lead to an invali
d solution

[spring oasis]

how did this happen

[spring oasis]

oh my

we still have a long way to go

[spring oasis]

.

[fiery marten]

Yes, but other cases will take less time

[tawny patrol]

Sorry Renato Where are you from?

[spring oasis]

I'm from Argentina wbu

[fiery marten]

We're following this long and tedious path but at the end we will learn Lagrange multipliers, at least to some degree

[spring oasis]

can we do a little summary of the steps taking over as of now

[fiery marten]

Oh and I'm afraid I have to pause for 2 hours, maybe a bit less

[tawny patrol]

I from Iran

[fiery marten]

I can do that later

[spring oasis]

that's the important part

next time I will have a solid understanding I suppose

all of this exercises is always the same

as long as I know how to do one I will get this point right in the exam

[spring oasis]

no worries feel free to ping me

[spring oasis]

is it warm there?

its like 30 degrees Celsius rn here in buenos aires

[tawny patrol]

Yes, some areas are warm, but in some cities, it snows

[willow helm]

Oh damn and here it's 1

[tawny patrol]

How's the weather in Argentina?

[spring oasis]

is like dog shit

too fucking warm all the time

[tawny patrol]

But in the city where we live, the weather is almost mild, although some days it can get really hot

[tawny patrol]

Maybe it's because of the greenhouse gases and the many cars

[tawny patrol]

The air in Iran is very polluted. How is it there?

[tawny patrol]

Is it right or wrong ?

Renato Do you like physics?

[spring oasis]

maybe

[spring oasis]

its okay, I wouldn't say its very polluted here but I am not a climate expert

[spring oasis]

I wouldn't say I dislike them, I am just bad at them, I still prefer math over physics

[spring oasis]

where u from mate?

[willow helm]

Slovenia

[fiery marten]

Here's a requested recap. Our goal is to find all extrema of $x^3 - y^2$ in the region given by these two constraints:
$$\frac12 x^2 + \frac14 y^2 - 1 \le 0, \quad \sqrt{x}-y \le 0.$$
As prescribed by the method of Lagrange multipliers, we have the following lagrangian:
$$L(x,y)=\lambda_0(x^3-y^2)+\lambda_1\Bigl(\frac12 x^2+\frac14 y^2-1\Bigr)+\lambda_2(\sqrt2 x-y).$$
The method says that for any extremum there exist $\lambda_1 \ge 0, \lambda_2 \ge 0$ and $\lambda_0 \in {0,1,-1}$, such that $\lambda_0, \lambda_1, \lambda_2$ are not simultaneously zero, and the following system of equations holds:
$$
\begin{cases}
3\lambda_0 x^2+\lambda_1 x + \sqrt2\lambda_2=0, \
-2\lambda_0 y+\frac12\lambda_1 y-\lambda_2=0, \
\lambda_1\Bigl(\frac12 x^2+\frac14 y^2-1\Bigr)=0, \
\lambda_2(\sqrt2 x-y)=0.
\end{cases}
$$

[woven radish]

spindle

spindle

[spring oasis]

theres a typo in the second constraint but sure

[fiery marten]

Oops true

Glad you noticed

sqrt(2)x - y <= 0, of course

[spring oasis]

im in the shower let me ping u in 5 min

[woven radish]

spindle

spindle

spindle

[fiery marten]

Hopefully this gives an idea how to investigate the next candidate: x=-1, y=-sqrt(2)

Fell free to ask any questions though

[spring oasis]

how did we go from the lagrangian to the system

[fiery marten]

Here's the method in our case (2 variables x, y, 2 constraints):
$$L(x,y) = \lambda_0 f(x,y) + \lambda_1 \varphi_1(x,y) + \lambda_2 \varphi_2(x,y),$$
and the system is given by $$\begin{cases}
\frac{\partial L(x,y)}{\partial x}=0, \
\frac{\partial L(x,y)}{\partial y}=0, \
\lambda_1\varphi_1(x,y)=0, \
\lambda_2\varphi_2(x,y)=0.
\end{cases}$$
The system on the previous image is this exact thing but I subbed $f, \varphi_1, \varphi_2$ and calculated the partial derivatives

[woven radish]

spindle

[fiery marten]

You probably remember when we did this

#help-27 message

[spring oasis]

alright got it

[spring oasis]

yes

[spring oasis]

I dont understand why for the 1), why are we discarding them

[spring oasis]

the initial idea was that both λ1 ans λ2 are positive then (x,y)=(...,..)

[fiery marten]

Again, the theorem says that $\lambda_1 \ge 0$ and $\lambda_2 \ge 0$ and here we're considering the case when they're both strictly positive. And we've just concluded that they have different signs. So they can't be both positive

[woven radish]

spindle

[fiery marten]

yes

[spring oasis]

ok lets sub (x,y)=(-1,-sqrt(2))

[fiery marten]

These guys can't be positive at the same time:

[spring oasis]

here

[fiery marten]

yep go on

[spring oasis]

3λ0 - λ1 + √2 λ2 = 0
2√2 + (1/2)(λ1)(-√2) - λ2 = 0

are you sure this system of equations are right?

the first one only contains X and the second one only contains Y

[fiery marten]

yes

[fiery marten]

correct, now let's solve it wrt lmabda1, lambda2
UPD: yeah the second eq should have 2\sqrt(2)\lambda_0

[spring oasis]

how

[fiery marten]

It's a system of 2 linear equations in 2 variables

[spring oasis]

3λ0 - λ1 + √2 λ2 = 0
λ0(2√2) + (1/2)(λ1)(-√2) - λ2 = 0

[fiery marten]

After some rearrangements it looks like $$\begin{cases}
-\lambda_1 + \sqrt2\lambda_2=-3\lambda_0, \
-\lambda_1 - \sqrt2\lambda_2=-4\lambda_0.
\end{cases}$$

[woven radish]

spindle

[fiery marten]

yeah now it's correct

[spring oasis]

wait

3λ0 - λ1 + √2 λ2 = 0
λ0(2√2) + (1/2)(λ1)(-√2) - λ2 = 0

---

λ1 = λ0(3) + λ2(√2)
λ2 = λ0(2√2) + λ1(-√2)(1/2)

[spring oasis]

why u always put λ0 in the other side wtf

[fiery marten]

Because I want to solve for lambda1, lambda2

[spring oasis]

my equations look different

did you multiplied by something

[fiery marten]

yeah, by \sqrt(2)

the second eq

[spring oasis]

3λ0 - λ1 + √2 λ2 = 0
λ0(2√2) + (1/2)(λ1)(-√2) - λ2 = 0

λ1 = λ0(3) + λ2(√2)
λ2 = λ0(2√2) + λ1(-√2)(1/2)

λ1 = λ0(3) + λ2(√2)
λ2(√2) = λ0(4) + λ1(-2)(1/2)

λ1 = λ0(3) + λ2(√2)
λ2(√2) = λ0(4) + λ1(-1)

λ1 = λ0(3) + λ0(4) + λ1(-1)
λ2(√2) = λ0(4) - λ0(3) - λ2(√2)

λ1(2) = λ0(7)
λ2(2√2) = λ0(1)

[fiery marten]

What happened in the last step?

[spring oasis]

substitution

[fiery marten]

ah okay

yeah you're very close to solving this, although I don't understand why you're not using the Gauss method for solving linear systems, here it works super well

[spring oasis]

3λ0 - λ1 + √2 λ2 = 0
λ0(2√2) + (1/2)(λ1)(-√2) - λ2 = 0

λ1 = λ0(3) + λ2(√2)
λ2 = λ0(2√2) + λ1(-√2)(1/2)

λ1 = λ0(3) + λ2(√2)
λ2(√2) = λ0(4) + λ1(-2)(1/2)

λ1 = λ0(3) + λ2(√2)
λ2(√2) = λ0(4) + λ1(-1)

λ1 = λ0(3) + λ0(4) + λ1(-1)
λ2(√2) = λ0(4) - λ0(3) - λ2(√2)

λ1(2) = λ0(7)
λ2(2√2) = λ0(1)

[fiery marten]

Immaculate

Hence λ1 = 7/2 λ0, and λ2 = λ0 / (2√2)

Now you can probably tell that cases λ0=0 and λ0=-1 can be discarded

And we finally arrived at our first solution: λ0=1, λ1=7/2, λ2 = 1/(2√2), x=-1, y=-√2

[spring oasis]

why -1 discarded

[fiery marten]

Because in that case λ1<0 and λ2<0 and we don't want that

[tawny patrol]

We used the method of Lagrange multipliers , which involves taking the partial derivatives of the Lagrangian with respect to the variables and setting them equal to zero. This gives us a system of equations that we can solve for the unknowns

[spring oasis]

I forgot about that yeah

[spring oasis]

yeah

[spring oasis]

what do I do now

[tawny patrol]

Can you give me a point about Lagrange ?

Because I have A little information about solving the Lagrange

[fiery marten]

We fully finished the case λ1>0 and λ2>0 for any λ0, and now we can move on to the next set of cases:
λ1>0, λ2=0. Similarly, we once again look at our system of 4 equations from the theorem and get information from it

[fiery marten]

What exactly are you interested in?

[spring oasis]

this one?

[fiery marten]

These are the first 2 equations from this system, yes. But there're 2 more

[tawny patrol]

This doesn't mean that I'm bad at math or that I'm not interested in it

.

But I am very strong in solving ambiguous and explicit functions and equations.

As you know, Math is a big world

[fiery marten]

Yes, of course

Not knowing something doesn't make you bad at math

[spring oasis]

can we finish this spindle

λ1x + √2 λ2 = 0
(1/2)λ1 y - λ2 = 0

this is what I get for λ0=0

[tawny patrol]

EXACTLY

[fiery marten]

Okay, but we have 2 more equations to deal with

[spring oasis]

the other ones are just 0=0

[fiery marten]

Why though?

[spring oasis]

oh my bad

they are not

oof

[fiery marten]

Okay, suppose λ1>0, λ2=0

The 4th equation indeed becomes 0=0

[fiery marten]

The 3rd equation yields 1/2 x^2 + 1/4 y^2 = 1 when we divide by λ1>0

[spring oasis]

sure then what

[fiery marten]

Now we sub λ2=0 into first 2 equations

[spring oasis]

λ1x + √2 λ2 = 0
(1/2)λ1 y - λ2 = 0

[fiery marten]

We get 3λ0 x^2 + λ1 x = 0, and -2λ0 y + 1/2 λ1 y = 0

[fiery marten]

That's when λ0=0, yes

[fiery marten]

Indeed, let's consider the case λ0=0

We then have λ1 x = 0 and 1/2 λ1 y = 0

Since λ1>0, we can divide by λ1 and get x=y=0

This can't be though, because we want 1/2 x^2 + 1/4 y^2 = 1 (see #help-27 message) but we got 1/2 x^2 + 1/4 y^2 = 0

So we can now safely assume that λ0 is not 0

[fiery marten]

Okay, we're now here and we know that λ0 is not 0.
3λ0 x^2 + λ1 x = 0 holds when x = 0 or 3λ0 x + λ1 = 0 (or both), so x=0 or x = -λ1 / (3λ0)

-2λ0 y + 1/2 λ1 y = 0 is the same as (1/2 λ1 - 2λ0) y = 0, which holds when y = 0 or 1/2 λ1 - 2λ0 = 0 (or both)

Hopefully you can continue from here

[spring oasis]

is this an algorithm you are following?

this doesn't seem too systematic

[spring oasis]

this whole thing is driving me insane

like there is too many case work

[fiery marten]

There's no way to be more systematic than that, I'm afraid.
We pick which lambdas are not zero, we use this knowledge to simplify the system of equations from the theorem, we solve the system and get x, y, and lambdas

That happens in each case

[spring oasis]

ok do you mind i try drawing this whole situation because there is a lot of things going on, let me bring my tablet

[fiery marten]

It is what it is. An extremum can be

1. Inside the region
2. On the boundary
3. On the intersections of any pair of bounding curves
   The cases we're considering correspond to these things

[spring oasis]

i seeee

[fiery marten]

Perhaps, not sure if I'm capable of being more clear than now

[spring oasis]

just give me a sec, because i am getting lost in the case work

is better if i draw it

[fiery marten]

Oh I misread it, I thought you asked me to draw something

[fiery marten]

Check the quadratic formula, here a=3λ0, not just λ0

Also I think there's no need to write the general formula because in almost every case (like λ1>0, λ2>0 or λ1>0, λ2=0 or λ1=0, λ2>0, ...) the first two equations from the first image simplify

[spring oasis]

what now

[fiery marten]

#help-27 message

[fiery marten]

Case work

[spring oasis]

what does this mean in human

[fiery marten]

λ1>0, λ2>0
λ1>0, λ2=0
λ1=0, λ2>0
λ1=0, λ2=0

[spring oasis]

12 cases

[fiery marten]

It means that equations 1 and 2 from the system look much simpler when you consider concrete cases

[spring oasis]

so just this?

or do i erase more

[fiery marten]

For example, when λ1=λ2=0, the first two equations reduce to 3λ0 x^2 = 0 and -2λ0 y = 0

[fiery marten]

Yes, this is a good start

[spring oasis]

or do i erase more

[spring oasis]

what do you mean

[fiery marten]

Right, we also have λ0 to deal with. However, these are the main 4 cases.

[fiery marten]

These are 4 main cases. We completely solved the first one

[spring oasis]

can we start with lambda0 is 0

[fiery marten]

The second one we started but we haven't finished it yet

[fiery marten]

Okay

[spring oasis]

0 is the easiest

[fiery marten]

I assumed that it would be faster to label the cases this way, but okay

[fiery marten]

Okay, go on

[spring oasis]

ok

[fiery marten]

Okay, go on

[fiery marten]

Yes

[spring oasis]

wait

[fiery marten]

yep that works

[spring oasis]

ok

[spring oasis]

let me do 2)

[spring oasis]

AHAHAH math is MATHING

i finished lambda0 = 0 @fiery marten

[fiery marten]

nice

Now we have +-1

[spring oasis]

now what

[spring oasis]

yeah because lambda0=0 is garbage

[fiery marten]

mhm

[spring oasis]

can lambda1 be negative?

[fiery marten]

no

[spring oasis]

how do i even solve this shit

[fiery marten]

we've actually already done that today

well, you have y=sqrt(2)x

And then solve third as a quadratic in x after subbing y

[spring oasis]

ok

[devout snow]

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[feral hare]

Unsure how to find b_n. Do I use the same formula as I used for a_n?

[feral hare]

here is my work so far

[devout snow]

@feral hare Has your question been resolved?

[supple knot]

Like i said before, b_n has its own formula similar to a_n

[supple knot]

8 and 9 here

[feral hare]

So it's practically the same formula

You just replace n with an m

I'm still confused on what to do wit that

[feral hare]

Sorry, 8 and 9 here what?

[supple knot]

https://mathworld.wolfram.com/FourierSeries.html

My bad forgot to paste

[supple knot]

I don't see how you did all this with a_n but can't for b_n

[feral hare]

oh ok, yeah

So bsaically i was just given the formula for a_n

and the reason im not doing it for b_n is because i thought it would just turn out to be 0 again, just like a_n

thank u for this website, it's helpful

ill keep trying

do u see what im saying though

itll just turn out to be 0 again

ok, i have a better question to ask

This part, where it asks me a_n = 0 for all values of n BUT

That confused the hell out of me

and i lowkey just copied the example of a previous problem for that and got it right

i do sort of understand it though

but i just dont

should i just replicate this last step for b_n?

I really don't know how to tbh, I did try here

how do i solve for b_n here?

[supple knot]

You plug f into here

Read example 2 here then try your problem again
https://tutorial.math.lamar.edu/classes/de/FourierSeries.aspx

[feral hare]

ok gotcha

wait im a dumbass

SORRY

i didnt see it's cos and sin 💀

thank you

Sorry, did I do a_n wrong?

[supple knot]

Oh yes cos and sin are different functions

[feral hare]

Since I used sine, not cosine

I still got the right answer but

[feral hare]

LOL

[upper schooner]

Do you have to do that

[lapis quarry]

[devout snow]

@feral hare Has your question been resolved?

[devout snow]

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[nova marsh]

(1994 AIME 14, Reflective Geometry)

https://artofproblemsolving.com/wiki/index.php?title=1994_AIME_Problems/Problem_14

Can somebody explain the inequality? I get the k*b part but I’m not sure how you get 180 - 2a and what it is supposed to represent

[devout snow]

@nova marsh Has your question been resolved?

[lunar harbor]

(from the official solution packet)

@nova marsh

[devout snow]

@nova marsh Has your question been resolved?

[nova marsh]

Oh shoot that makes more sense

[devout snow]

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[ocean gale]

I think I have a solution to the problem but I don’t know if it is correct

So I’ve assumed that the longest AB can be is if AB is straight across and I assumed this because I drew a picture of it and measured the lengths of various possible AB and straight across was the longest

I don’t know how to prove this tho

And I also am not sure my assumption is correct

Nvm

.close

[devout snow]

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[devout snow]

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[native shadow]

Why is the number for minutes squared (3600) squared again

[native shadow]

I thought you'd be looking for seconds squared but seconds squared isn't 3600^2

[hardy trail]

Hi! How do you convert minutes into seconds?

[native shadow]

to 60 seconds, and squaring it will get us to 3600

[hardy trail]

Yep! Sooo that's it :)

1 minute is 60 seconds, so 1 minute^2 is 60^2 seconds^2.

[warm karma]

I don't see where it says 3600^2 ?

[warm karma]

that's 3600 (s^2), not (3600s)^2

powers bind tighter than multiplication

,tex $ a b^2 \neq (ab)^2 $

[woven radish]

CyclicTree

[native shadow]

so the unit is being squared but not the number I'm a bit confused

[warm karma]

well, the number is also squared, but it's 60^2

[native shadow]

so it's not being squared again

[warm karma]

no

units behave like indeterminates (variables) generally. Except dimensionless units. Those are weird (e.g. sin taylor series only makes sense if rad = 1 ...)

[native shadow]

the thing I'm confused about is the unit is looking for second squared but instead it's a minute squared

[warm karma]

well you know that 1 min = 60 sec

so you can solve for min^2 = (60 sec / 1)^2 = 3600 (sec^2)

and now you can just replace min^2 in the answer with this expression

to get $2160 \frac{\text{degrees}}{3600 \text{s}^2}$

[woven radish]

CyclicTree

[native shadow]

.close

[devout snow]

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[orchid umbra]

am i tripping or did he draw it wrong, i thought (-2,-120) was in third quadrant

[alpine python]

oh, i guess because the radius is negative, it flips it to the first quadrant

[woven vale]

exactly - not to mention, the negative angle means you go clockwise from (-2,0)

[vital edge]

...what the fuck does it mean to have negative radius

[alpine python]

positive radius means go away from the origin
if you go in the opposite direction, you'll cross the origin

[vital edge]

Weird, all formulations of polar coords that I've seen treat radius as distance from origin

[orchid umbra]

i dont get it

whats the point of - radius then

[alpine python]

fair enough, you don't need negative values

you can cover the entire plane using only nonnegative radius

[vital edge]

Is this how they're doing it for you in school or is this some online resource

If it's the former then unfortunately you have to go along with it

[alpine python]

a looped limaçon uses negative radius

this fellow

[granite venture]

same

i havent seen r < 0 except in equations like above

[orchid umbra]

i saw it from youtube cuz my prof didnt teach us

[orchid umbra]

oh

[orchid umbra]

oh really

[alpine python]

yeah, the negative values are redundant in a sense, but they can be used for equations like the one i posted above

[granite venture]

yeah, r ≥ 0 and theta from 0 to 2pi does it

seems like if you limit theta to be 0 to pi then you need neg r values

[orchid umbra]

oh i see

てか日本人ですか

[alpine python]

heheh

[orchid umbra]

wait can yall check my work cuz like

i use - radius 😭

[alpine python]

nah i just studied japanese

[orchid umbra]

日本語読める?

[exotic vessel]

Look

[alpine python]

うん、読める

[orchid umbra]

ｊｌｐｔ知ってるの

[alpine python]

ye

you took it?

[orchid umbra]

nah

im too bad

wbu

[alpine python]

i took it a while ago

[orchid umbra]

woah

n what

[alpine python]

N1

[orchid umbra]

ええー

すごい

how long have you been learning for

[alpine python]

i think i studied for around 5 years before taking the test

i don't study anymore

oh you're in the refold server

[orchid umbra]

wow

impressive

[alpine python]

how do you like to study?

[orchid umbra]

yup

[orchid umbra]

idk tbh like i dont rlly study nowadays i used anki for vocab

and just kept calling natives on an app

to practice speaking

wbu

how do you get to n1 lmaooo

[alpine python]

watching anime and reading manga and books

i did vocab similar to anki too

[orchid umbra]

o

[alpine python]

we are so off topic, but it's your channel so i guess it's fine

[orchid umbra]

how do you learn grammar tho cuz grammar is so hard

[orchid umbra]

sorry

[alpine python]

😂

i don't mind

[orchid umbra]

lmaoo

ok but can u check my graph first

(to stay on topic ofc)

[alpine python]

yeah -13pi/7 doesn't count right?

oh you changed the range to -2pi < theta < 2pi

i guess it counts then

it seems right

[orchid umbra]

thanks

[exotic vessel]

Wait im so confused

For the highlighted point

How do I know if it’s the red dot in quadrant 2 or if it’s the green dot in quandrant 4?

<@&286206848099549185>

[alpine python]

isn't it in quadrant 1?

[orchid umbra]

huh?

[alpine python]

i think my way of thinking is different so i'm not sure if i should explain

[exotic vessel]

😭

I’m so confused now

[alpine python]

because the image you posted here...

[alpine python]

seems to match what this person said

[orchid umbra]

oh uhh

can someone else help

bruh i dont understand

[alpine python]

5pi/4 is third quadrant
since it's negative radius, flip it to the first quadrant

[orchid umbra]

bruh.

[alpine python]

cause it's diagonally opposite

kitty-corner

[orchid umbra]

why

why cant it be in quad 3

[alpine python]

because the radius is negative

[orchid umbra]

huh

[orchid umbra]

?

i dont undertsand brh

[alpine python]

sorry

[orchid umbra]

whats wrong w radius being negative

no

i should be sorry

😭

for not being able to understand

[alpine python]

nothing's wrong with the radius being negative