help-23

[lime nexus]

the empty middle bit would be A’ n B’

the intersection shares a B?

[timid scroll]

Do you guys memorize this

or do it step by step?

[lime nexus]

the coloured intersection basically tells us the part where it has BOTH A AND B

[vagrant ice]

I would just draw out B' and A on two separate Venn diagrams

then see which areas are in common

[glacial meadow]

I usually work it out by drawing the diagrams

What Kristy said is a pretty good way to think of it

[vagrant ice]

don't forget the space outside the circles, cause everything in the rectangle is the universal set

[glacial meadow]

A' is the “not A“ element

And successively

[lime nexus]

i would normally try to understand what it’s asking us to do, then colour accordingly!

[vagrant ice]

yeah so if you do that, their intersection is A \ B, A not B

[timid scroll]

right...

[lime nexus]

when it asks for A n B’ it basically asking what section has A ONLY and NO B, there’s 3 sections in totoal
the left middle and right
both the middle and right has B included in it, thus y it ISNT coloured, and the left side ONLY has A which is y it’s coloured

[timid scroll]

But sometimes NO B

means color outside

and sometimes it means not

[glacial meadow]

Depends on if it's asking to intersect something

[timid scroll]

right

[glacial meadow]

Anything outside the circles is neither of the things happening

[lime nexus]

colour outside means NO A AND B

[glacial meadow]

There it's asking ya for A not to happen and B to happen

Outside the circles, B does NOT happen, thus why it's not colored

[timid scroll]

why is NO A and B color outside, and NO A or B (outside isnt colored)

[glacial meadow]

Wdym

[timid scroll]

Before anyone types

This is my logic if I may

NO A and B

Step 1, NO A

Step 2 what do I do with 'and' 'B'

[vagrant ice]

colour in B with another colour

where there are two colours is the intersection

[glacial meadow]

Decolor any region in which B does not happen

[vagrant ice]

this is a neater way of doing that, yeah

[glacial meadow]

Intersection(or 'and’) requires both conditions to be fulfilled

[timid scroll]

Ok so step 2

[glacial meadow]

Simultaneously

[timid scroll]

what do I do step 3 'and'

[glacial meadow]

You've decolored where B happens

[timid scroll]

hmm

ok

Step 1, NO A

Step 2 Decolor region B doesnt happen

?

[glacial meadow]

Yup

[timid scroll]

Why decolor?

[glacial meadow]

Think of it like this: you're looking for a region where 2 conditions are satisfied:
1: A does not happen
2: B does happen

Fulfill one condition in each step, and consider what remains

[timid scroll]

Ok

now we got this

[glacial meadow]

There we go

That's what you're looking for

[timid scroll]

and the 'and' ?

[glacial meadow]

You've fulfilled both conditions

That's all the “and“ means

[timid scroll]

oh

[glacial meadow]

For both conditions to be satisified

In that region, B happens and A does not

[timid scroll]

OHH!!!

Thank you

mind if we do this quickly

[glacial meadow]

Well, I've gotta do something, try to work it out and see if someone else can check it out

[timid scroll]

no worries

ty :>>>>>

very much

Have fun

[lean otter]

what

do you need help

with

[vagrant ice]

cool, so first colour in B' as you did

now colour in A

if you combine the two regions, that is the union

[vagrant ice]

yep, that's it

[timid scroll]

Oh

well

suprisingly easier?

than i thought

[timid scroll]

lol

[vagrant ice]

mhm

it's just knowing what they mean by the notation

and following the procedure

[timid scroll]

yeah haha

[vagrant ice]

ye

[timid scroll]

tyty

Then

:3

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[timid scroll]

https://tenor.com/view/cat-meow-fat-augh-gif-24948731

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[steep magnet]

I dont know much about similar triangles

It just means the angles are the same right?

The triangles dont look like they have the same angles 👀

Is COA supposed to be the same as BCA?

[mossy lotus]

yes

[drifting knot]

Ok

[steep magnet]

Im not really sure what the first step is still...

[mossy lotus]

what do you know about the sides of triangles in any pair of similar triangles?

[steep magnet]

Theyre proportional somehow?

[mossy lotus]

Yes

use that here

[steep magnet]

Oh I just noticed that the length of CA is 3+sqrt(3)

Wait no

[mossy lotus]

why did you claim this?

i mean i see you said no, but why not?

[steep magnet]

Because the length of OA is 3+sqrt(3) so the line that is equivalent to that in the similar triangle is CA, so I thought that that meant it had the same length, but I then remembered theyre proportional not the same

[mossy lotus]

Right. And what are the proportions between the corresponding sides?

[steep magnet]

Im trying to figure it out

We dont know C so idk how we can get any of them

we dont have any pair from the similar triangles

[mossy lotus]

Ok, first tell me which pairs of sides are supposed to be in proportion

[mossy lotus]

we do

[steep magnet]

BC and OC, BA and CA, CA and OA

[mossy lotus]

And, you know their lengths?

at least for some sides, without the vertex C?

[steep magnet]

I can figure out the lengths of only 1 from each pair

We need both to figure out the proportions, right?

Cause every pair has something with C

[mossy lotus]

Yes, but there is also a fact that the ratio of any pair is same

[steep magnet]

Yes but we need a pair to get the ratio, no?

[mossy lotus]

Just write it out like this (ofc the names here are different coz i cropped an image off of the internet)

[steep magnet]

Does it matter which one goes on top? Big or little I mean

[mossy lotus]

Nope, the ratio k just becomes 1/k

[steep magnet]

BC/CO = BA/CA = CA/OA

[mossy lotus]

so about the last two, can you rearrange the terms so you have the known values on one side and unknown values on the other?

[steep magnet]

Ohhhhhhhh

I see

so (BA)(OA) = CA^2

[mossy lotus]

Yes. So you have CA now

[steep magnet]

Ok 1 moment I will get the numbers

[mossy lotus]

No no, let it be. You can put int the numbers in the end

[steep magnet]

Ok yeah good idea

[mossy lotus]

So, with CA, you can find location of C right?

It lies on OB, and is AC distance away from A

then you can use the formula for sin to get the exact angle

[steep magnet]

Sry this is taking me a min to figure out this is so confusing 😵‍💫

[steep magnet]

Ok sry im not sure what to do still 😭

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[edgy vine]

can someone help me sketch |x+y|-|x-y|=0

[edgy vine]

im used to absolute values with just x in them

but dont know how to deal with this

[sacred sentinel]

you can rewrite it as |x+y|=|x-y|

[edgy vine]

sorry suposed to be >0

[sacred sentinel]

then cancel the absolute value signs but with plus or minus

and you can graph the two functions seperately

i think

oh

yeah

[faint seal]

$|x|=|y|$ and $x^2=y^2$ are equivalent equations (as long as we're talking about real numbers only)

[flat frigate]

kheerii

[edgy vine]

|x+y|-|x-y|>0

ok, how do i use that

@faint seal

[sacred sentinel]

i think you should just do it piecewise so check whether x is positive and y is positive, x pos y neg, x neg y pos and x neg y neg

and then for all check whether the inequality holds

[edgy vine]

ok

i did that originally but it got kinda messy

[sacred sentinel]

can you show me what you've done?

or are you trying again rn?

[edgy vine]

ill try again

goos

good so far?

[sacred sentinel]

i was thinking we just say x is positive and y is positive, then check whether |x+y|>|x-y|. well, because x and y are both positive, |x+y| will be just x+y. if x>y then |x-y| will be x-y will always be smaller than x+y right?

now check for x<y, but both still positive

[edgy vine]

ah i see

[sacred sentinel]

i don't know if this is the fastest method but i think it works

[edgy vine]

ok so say x is posive and y is positve

but why are we only considering this case

[sacred sentinel]

we will do the other cases later

[edgy vine]

is that not what ive already done

lke written out the cases

theres 4

can i do it the way ive written out because thats just the way im used to doing them

[sacred sentinel]

uh

i don't think so

maybe im just dumb

i think what you wrote is just x+y>0, that doesn't mean x and y are positive

right?

[edgy vine]

well i have x<-y

and x>-y

ive considered when the absolute values are negative and positive

[lime dust]

x and y can never have different signs

[edgy vine]

huh

[sacred sentinel]

why?

[lime dust]

Do you agree that |a| = sqrt(a^2)?

[edgy vine]

what

yeah

ok

[lime dust]

Ok so we have sqrt((x+y)^2) in one side

Right?

[edgy vine]

ok

[lime dust]

If we expand it we get sqrt(x^2+2xy+y^2)

[edgy vine]

but also the thing inside the absolute value can be negative

[lime dust]

Lets start with opposite signs and sale value

If this happens then sqrt would be 0

[edgy vine]

sale value?

[lime dust]

But we have > not >=

Same

[edgy vine]

hm?

[lime dust]

When |x| = |y| if x and y have different signs

We would have

[edgy vine]

what do u mean lets start with same value opposite signs

[lime dust]

x^2 -2xy + y^2

Which is 0

This specific u understand?

[edgy vine]

im a bit confused,

where did we start from

[lime dust]

Ok step by step dont worry

[lime dust]

Lets start from here

[edgy vine]

ok ok

[lime dust]

Until here everything is ok?

Lets take x positive and y negative

[edgy vine]

i gonna state |a|=(sqrta)^2

[lime dust]

No that is false

[edgy vine]

u said that was right

[lime dust]

|a| = sqrt(a^2)

[edgy vine]

ohhh

[lime dust]

Is not the same

[edgy vine]

i see i see

ok so after stating that

what should i write down

[lime dust]

We replace a with one of ur absolute values

|x+y| for example

[edgy vine]

ok lets do that

[lime dust]

So you get this

[edgy vine]

|x+y|=sqrt(x+y)^2

[lime dust]

Ok now expand (x+y)^2

[edgy vine]

cool

ok

[lime dust]

U get this

Right?

[edgy vine]

uh huh

[lime dust]

Do u have difficulties on that step?

[edgy vine]

no ive got

[lime dust]

Ok now we start analyzing from here

[edgy vine]

ok

[lime dust]

I said x and y cant have different signs and now i will prove it

Lets suppose they have different signs

[edgy vine]

ok

[lime dust]

x is positive and y negative ok?

Now lets say they are “equal” in size

For example one is 10 and the other is -10

In other words

|x| = |y|

This u follow?

[edgy vine]

ok

got it

[lime dust]

Now, i am gonna replace that in what we had

Lets use the 10 and -10 to visually see

[edgy vine]

ok

[lime dust]

10^2 +2(10)(-10) + (-10)^2

[edgy vine]

ok got it

[lime dust]

100 -200 + 100 =0

[edgy vine]

ok

[lime dust]

So if they are both equally big with different signs

The result is 0

And sqrt(0) = 0

[edgy vine]

i see

[lime dust]

Now in the other side of the inequality

We would have x^2-2xy+y^2

Because it was |x-y|

If we replace here with same thing

[edgy vine]

ok

[lime dust]

We would have 100-2(-100)+100

400

0>400?

False

[edgy vine]

ahhhhhh

[lime dust]

Now lets say hey are different in size

Lets say x>y and x is positive and y negative

So that would mean x^2+2xy+y^2 is negative

[edgy vine]

ok

[lime dust]

But sqrt of a neg number is undefined

Over reals

[edgy vine]

wait let me write this down

why would it be negative

hold on

if x>y

x^2 >y^2

[lime dust]

Yes

Wait i made a mistake

[edgy vine]

this would be positive

[lime dust]

You are right i was thinking in other thing

[edgy vine]

should we put x<y

[lime dust]

Sorry

[edgy vine]

and y negative

[lime dust]

No, we check the same scenario

[edgy vine]

nws

[lime dust]

Look

x and y different sign correct?

Idk why didnt explain like this before

[edgy vine]

ok

[lime dust]

Focus on 2xy and -2xy in the other side

What sign is 2xy

If they have diff signs

2(positive)(negative)

[edgy vine]

y is negative right and x is positve

[lime dust]

Yes

It is negative right?

[edgy vine]

ok yeah

[lime dust]

And -2xy?

[edgy vine]

positve

they switch

[lime dust]

So we have

Sqrt(x^2+2xy+y^2) > sqrt(x^2-2xy+y^2)

When x and y have different signs

2xy is negative and -2xy is positive

Due to this

[edgy vine]

yeah

[lime dust]

Left side is actually no >

But the opposite

[edgy vine]

yeah

[lime dust]

Because the other stuff inside sqrt is the same

Thats is why x and y cant have diff signs cause

Then the inequality is false

[edgy vine]

yeah

[lime dust]

So after proving this you prove both same sign

Both negative or both positive

Both positive is esasy

[lime dust]

Focusing on 2xy and -2xy

You can realize is true

And both negative u can see that is also true

Can u follow everything?

[edgy vine]

yeah

[lime dust]

So those are the solutions. x and y > 0 or x and y < 0

[edgy vine]

ahhhhh cool

thats easy enough to graph

thanks sm

useful identity

[lime dust]

It is a definition

|x| = sqrt(x^2)

[edgy vine]

yeah just never thought of it before

thanks tho

\good to know

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[steep magnet]

The intersections are at x = 0 and x = 2

Then we get the integral with limits 2 and 0 of g(x)^2 - h(x)^2

integral of (sqrt(8x))^2 - (x^2)^2

integral of 8x - x^4

4x^2 - (x^5)/5

Then sub in 2 and 0

(4(2)^2 - ((2)^5)/5) - (4(0)^2 - ((0)^5)/5)

The answer they wanted was 8/3

What did I do wrong?

Omg its area not solid of revolution I just saw 🤦‍♂️

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[magic junco]

Damn

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[proper swift]

Trying to resolve this formula...

[proper swift]

L=800, B=106, A=5

https://www.wolframalpha.com/input?i2d=true&i=Sqrt[Square[800]%2BSquare[106]]*SIN\(40)\(40)Divide[ArcTan[106]%2C800]\(41)%2B5\(41) is getting -773, but i know thats not correct 😦

[burnt notch]

Is A = 5° or 5 radians?

[proper swift]

5 degrees

[safe radish]

@proper swift Has your question been resolved?

[proper swift]

I think I got it, 5 should be in radians not degrees...

I'm getting 173.22

[burnt notch]

I get 175.32...

https://www.wolframalpha.com/input?i2d=true&i=Sqrt[Square[800]%2BSquare[106]]*SIN\(40)\(40)ArcTan[Divide[106%2C800]]\(41)%2B5Divide[π%2C180]\(41)

[proper swift]

great, thank you~

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[proper swift]

yes

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[frozen wren]

Can someone help me with Geometry? Its area addition and subtraction, mostly circle shapes I'm not sure how it works

[safe radish]

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

[frozen wren]

its full of equations like this

[empty gyro]

,rcw

[empty gyro]

Ask your specific questions and we will try to help

[frozen wren]

i'm sorry i don't know how to be more specific that's the whole question that's how they all are

[empty gyro]

Okay well, you're given the angle of the triangle, which is also the angle of the circular sector. So you can use the sector area formula to get the area of the sector since you have the radius and angle. And once you find h, you can find the area of the triangle inside of that sector. Then the area of the shaded region is just the difference of the two areas you just found.

[frozen wren]

whats the circular sector

and how do i find h

[flat frigate]

Randel_

Randel_

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[stark yew]

Is this right

[stark yew]

Or is it second and fourth

[prime locust]

what’s a cosine “ratio”?

[rancid sand]

every trigonometry functions are ratio of length

e.g. cos (x) = a/b

[soft matrix]

Ya so like what's cosine ratio ratio of the function itself

[rancid sand]

What do you consider when you do that question?

What makes u think its second, third or fourth

[stark yew]

The left quadrants are where x is negative

So it’s q2 and q3

[lone leaf]

Yea it’s right

[rancid sand]

gd

[stark yew]

The only quadrant that shares both qualities is q3

[lone leaf]

My teacher actually once taught me a way to memorize that

From quad 1-4 it’s just “add sugar to coffee” which stands for all, sin, tan, and cos

[rancid sand]

think, rather than mermorize

[lone leaf]

Those are the positive values

[lone leaf]

Not wrong

[soft matrix]

Memorize?
Just draw a unit circle and a right triangle and derive it

[rancid sand]

ye

[lone leaf]

But sometimes if you have to waste time and think for something that can be done quick it’s not efficient

If you know the theory behind it it’s good

[soft matrix]

I already have alot else to memorize lol

[lone leaf]

Aight

[soft matrix]

Btw it's correct cos is +ve in first and 4 rt quadrant

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[glacial cosmos]

how can you easily factor to solve the quadratic 16x^2 - 48x +27 = 0 without a calculator? i can do the sum product pattern in my head for this one

[quick crater]

i dont think you can do better unfortunately

[glacial cosmos]

can't*

[quick crater]

ow

do by hand

[glacial cosmos]

alr thanks

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[obsidian oracle]

16x^2 = (4x)^2. I think you can easily solve for 4x if you look closely

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[ancient peak]

hi, my school has given me some homework on compound interest and i'm stumped on this question: "If I invested $500 in an account that pays 0.6% a month compound interest, how much would my investment be worth in a year's time?"

I have done a lot of practice questions and have gotten them all right but this one says i'm wrong and i would like to understand why. It says the answer is 537.21 but i am sooo confused how they got that and they don't even show working out or anything...

if the formula is P((1+(r/n))^nt), then wouldn't that give me:

500 * ((1 + (0.006 / 12))^12) = 503.00

Is there something im doing wrong or is the question broken? how did they get 537.21?

[maiden marlin]

I figured a way they could’ve gotten 537.21:
(1.006^12*500)

You divided 0.006 by 12, but 0.006 is per month already.

[ancient peak]

Oh... they haven't taught us that... i didn't know

Actually it makes sense now since 0.6 / 12 is 0.6% a month I think i just got thrown off by the wording

ty ❤️

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[quick crater]

its the same as your formula

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[fickle rapids]

phelp how to start e

[stoic dune]

2^(n-2) = (1/2) 2^(n-1)

So that's really u - u/2 for u = 2^(n-1)

[fickle rapids]

how do we prove this? which side do we use rhs/lhs?

i think i ate

but... i think we need to have the first line at the end

eii D:

<@&286206848099549185>

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[old basin]

Please explain to me the concept of linear transformation of matirces

[old basin]

i have watched videos, read explanations, and looked at the AP Pre-calc book, and I still dont understand what that means

Like waht does this mean???

A linear transformation is a function that maps an input vector to an output vector such that each component of the output vector is the sum of constant multiples of the input vector components.

[royal kiln]

The new vector is a linear combination of the n axis coefficients of a vector in R^n

In R^3, that would be a linear a combination of the x, y, and z components

[old basin]

what is r

R

[royal kiln]

The real numbers

But you know

We don't have to stick to real numbers necessarily

We just usually do

[old basin]

what is the n axis

[royal kiln]

There are n axes

R^3 is the 3D world

There are n=3 axes, x y and z

[old basin]

3^3 = 9???

since 3 is a real number

[royal kiln]

R doesn't get substituted

R stands for the real numbers

It's notational

[old basin]

so its power means how many axes we have?

[royal kiln]

Yeah, that's a way of thinking of it

[old basin]

A linear transformation is a function that maps an input vector to an output vector such that each component of the output vector is the sum of constant multiples of the input vector components.

what is the constanst part mean

[royal kiln]

Linear algebra is a very dry field (imo) and it has been horribly abstracted beyond 3d space

[old basin]

a linear transformation itself is a function???

[royal kiln]

Just numeric multipliers. "Non-variable"

The multipliers won't be unknowns or functions

[royal kiln]

Yes

[old basin]

so actual numbers like 1,2,3...

[royal kiln]

Yeah, or-pi/13

[old basin]

can you maybe give me an example of a question so maybe i understand better

and explain what each part means when you type the question?

[royal kiln]

Um...

"Find a linear transformation matrix that maps [1,2,3]^T onto [2,4,6]^T."

Well hang on

That's too simple

[old basin]

what is T?

is that the axes to the power thing?

[royal kiln]

"Find a linear transformation matrix that maps [1,2,3]^T onto [2,6,12]^T."

[royal kiln]

That means "Transpose"

I printed the vector as a row, so the T means it's a column vector

The transformation matrix is

[old basin]

The [1,2,3] matrices can be of any size right??

like 2x2, 3x5, etc.

[royal kiln]

$\begin{bmatrix}
2 & 0& 0 \
0 & 3 & 0 \
0 & 0 & 4
\end{bmatrix}$

[flat frigate]

Melvin Eugene Punymier

[royal kiln]

$\begin{bmatrix}
1 \
2 \
3
\end{bmatrix}

\begin{bmatrix}
2 & 0& 0 \
0 & 3 & 0 \
0 & 0 & 4
\end{bmatrix}

=

\begin{bmatrix}
2 \
6 \
12
\end{bmatrix}$

[old basin]

hey man you don't have to work this hard to tell me

I think i get what youre trying to type

you dont have to type the rest

but can you explan why all the zeros are for

[flat frigate]

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

[royal kiln]

I'm just trying to remember how to write this kind of thing up

[royal kiln]

Try multiplying the first column vector by the transformation matrix and see

[old basin]

oh i see

alr man, i think this should be enough for now

thanks for your help

👍

[royal kiln]

Np

[old basin]

.close

[safe radish]

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[royal kiln]

.reopen

[safe radish]

✅

[royal kiln]

@old basin

What I showed you was very simple though

I suggest you read this

https://mathinsight.org/matrices_linear_transformations

[old basin]

will do, thank you

[safe radish]

@old basin Has your question been resolved?

[safe radish]

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[vivid trellis]

is this right or am i completely wrong

[junior smelt]

I don't think that's the wording they're looking for tbh

[vivid trellis]

what changes can i make?

[junior smelt]

They're more looking for a "name" of the type of relationship, for example, "linear" is an example, you could say variables (though, spoiler, not here!) are linearly related

Do you know any other types of relationship?

[severe pond]

does N increase by a factor of 1.7 for each increase in t?

[vivid trellis]

yeah exponential and power law, i think i got itt from here thank youu so much

[vivid trellis]

sorry i don't get what u mean??

[junior smelt]

(yea also I was gonna point that out too)

[severe pond]

i was just pointing out that the left side isn’t just N, it’s ln(N)

[junior smelt]

(rewriting N explicitly will answer the question here for you, and also help you see answer the other question about the factor of increase)

[vivid trellis]

.closed

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[uncut prairie]

small question
I want a restrict a function's x and y values to a circlular boundary
say f(x)=mx and I have a circle y2 + x2 = 36
how do I restrict f(x) so that it doesnt extend outside the 6 unit radius

[median vigil]

if you want to find the appropriate restricted domain, you can solve for where the curves intersect

[uncut prairie]

oh cool thanks

[uncut prairie]

idk though ill do smth rq

[uncut prairie]

https://www.desmos.com/calculator/t9nnws9mfm here

ill be toggling the circle off eventually but I want to restrict the line to the circle

(and i dont know how to calclate intersection points ;-;)

[median vigil]

if you want to do it in desmos you can just add {x^2 + y^2 ≤ 36}

[uncut prairie]

didnt seem to work

[median vigil]

it works on my side. make sure you enter the same circle equation you want to restrict it by and put it in the same box as the line equation

[uncut prairie]

wait

[uncut prairie]

fixed it

thanks

.close

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[flat igloo]

If P(x) is an odd function
Then can it be said that P(-x)= -P(x) ?

[drifting knot]

Yes

[flat igloo]

Please explain

[drifting knot]

See

,w plot sinx

[flat frigate]

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[drifting knot]

Can u see

That it is symmetric bout the origin

[flat igloo]

Yes

[drifting knot]

This is the only fact

[spice furnace]

that is the definition of an odd function

[drifting knot]

Odd functions are symmetric

Bout the origin

[flat igloo]

But not even functions?

[drifting knot]

See this example my guy

,w plot cosx

[flat frigate]

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[drifting knot]

Can u say is this symmetric bout origin?

Yes or no?

[flat igloo]

Looks symmetric

[drifting knot]

Ok we go into more basic things

Do u know what symmetry bout origin means

[flat igloo]

No

[drifting knot]

origin is a point

[flat igloo]

Ik that

[drifting knot]

Right

Good

Now see symmetry means

Wait imma draw

Points in 1st and 3rd quadrant

Are equidistant

From the origin

[flat igloo]

Yes

[drifting knot]

And points in 2nd and 4th quadrant are equidistant from the origin

This is what you can say is symmetry

Bout the origin

[flat igloo]

But cos x also looks symmetrical on both sides of origin

[drifting knot]

Ok take a point

In 1st and 3rd quad

And show that they are equidistant

,w plot cosx

[flat frigate]

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[flat igloo]

Pi

[drifting knot]

Does it lie on the graph?

(-π,0) and(π,0) don't lie on the graph

Symmetry bout origin means

[flat igloo]

Ok

[drifting knot]

That graph in

1st and 3rd quadrant is similar

And graph in 2nd and 4th quadrant is similar

This comes from the equidistant fact

[flat igloo]

Oh understand it now

[drifting knot]

Like see this

,w plot sinx

[flat frigate]

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[flat igloo]

What if the equation has a constant term still can it be symmetric?

[drifting knot]

If the equation has a constant term

Like

f(x)=sinx+c

Now if u put -x in it

We get f(-x)=c-sinx

Now check yourself

f(x)=f(-x)?

,w plot sinx+2

[flat frigate]

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[drifting knot]

See

It's not symmetric bout origin

It is symmetric bout (0,2)

But we consider the definition w.r.t origin

[flat igloo]

Ok i understand it now

Thanks

[drifting knot]

Welcome

Fun fact

for even functions

f(x)=f(-x)

[flat igloo]

Even functions?

[drifting knot]

Even functions are those functions

[flat igloo]

Does this include the constant terms also?

[drifting knot]

Which are symmetric bout an acis

Axis*

[flat igloo]

Like in quadratoc functions

[drifting knot]

No

x^2 is symmetric

[flat igloo]

x^0 c is not an even term?

[drifting knot]

oh you wanna say constant functions?

Like f(x)=c

See

[flat igloo]

Are they considered odd or even?

[drifting knot]

They are not even nor odd

0 is the only function which is odd and even

Cz it has symmetry bout origin

,w plot 0

[flat frigate]

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[drifting knot]

Symmetry bout origin

[flat igloo]

Oh thanks

Now here’s the actual reason I came to ask here

[drifting knot]

Okik

[flat igloo]

Can you see Q 15 ?

[drifting knot]

Yes

Ask

[flat igloo]

Yeah it has to be solved

[drifting knot]

See

[flat igloo]

So if you can then please do

[drifting knot]

!noans

[safe radish]

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

[flat igloo]

Ik know the answer

[drifting knot]

I can help u learn

[drifting knot]

!nosols

[safe radish]

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

[drifting knot]

We will understand it from basic

f(x)=x^3 is this odd function?

[flat igloo]

Well fx can have any term of odd degree

[drifting knot]

Pls answer what I asked it will help u

Finally

[flat igloo]

x^9+x^7…..

[flat igloo]

Yes

[drifting knot]

is f(x)=x^5+x^3+x too a odd function?

[flat igloo]

Yea

[drifting knot]

is f(x)= sum of all odd powers of x is also a odd function?

[flat igloo]

Yes

[drifting knot]

So now can u tell is P(x) a odd or even function?

[flat igloo]

What if it has coefficients?

Even coefficients?

Will it still be odd?

[drifting knot]

Let's me give u a examplw

f(x)=2x

f(-x)=-2x

What u think now?

[flat igloo]

Odd

[drifting knot]

Good

[drifting knot]

Now can u answer this

[flat igloo]

Right

[drifting knot]

Good

Now can u tell me will P(x) be a odd or even function

[flat igloo]

Odd

[drifting knot]

Good

now one more

Thing to ask

Don't u think

P(6)=P(-6)?

yes/no

[flat igloo]

Yes I did that and solved the answer

[drifting knot]

Good

[flat igloo]

I’m thinking if there’s another way to solve this

[drifting knot]

So what doubt u have ?

[flat igloo]

Without using this concept

[drifting knot]

Oh

[drifting knot]

This

[flat igloo]

I’m thinking using only factor and remainder theorem

[drifting knot]

Yes

See the other method is

That you find P(x) first

[flat igloo]

Can’t we just assume some random odd function as px

[drifting knot]

See assuming it

May lead to diff solutions

[flat igloo]

Let’s say it is ax^3 and box

[drifting knot]

Don't u think

U will get diff and?

Ans*

Like if I assume as

ax^7+bx^5

Is this not odd too?

And what about individual terms

Like ax^3

[flat igloo]

Maybe not

The answer should be a general answer independent of degree etc.

[drifting knot]

Ik

[flat igloo]

Let’s say we give px by x^2-9

[drifting knot]

That's why they have used remainder factor theorem

[flat igloo]

Divide**

[drifting knot]

Cz it does not relate with P(x)

[flat igloo]

Let the quotient be Q

[drifting knot]

Only the definition of P(x) in terms of (x^2-9)

[flat igloo]

And denominator be ax+b

[drifting knot]

See imma prove this see

[flat igloo]

Px= Q (x^2-9 )+ ax+b

[drifting knot]

,w (ax^3+bx^2+cx)/(x^2-9)

[drifting knot]

My guy

Is this not the only thing u do by remainder factor theorem?

This is the only way my guy

[drifting knot]

Put 9 in this

What u get

P(9)=9a+b

Right?

[flat igloo]

Yes

[drifting knot]

Put -9 in this

What u get

P(-9)=b-9a

But we know P(x) is odd

[flat igloo]

-a9+b

[drifting knot]

So P(-9)=-P(9)

So we get

9a+b=P(9)
b-9a=-P(9)

Now add them

2b=0

[flat igloo]

Yeah

Wow so that was the only way

[drifting knot]

Yes

You are using the remainder factor only my guy

P(x)=Q(x)(x^2-9)+R(x)

This only comes from the definition of remainder factor theorem

Then u used the property of odd function

[flat igloo]

You’re a genius!

Thank you

[drifting knot]

Welcome

Thanks for the compliment

[flat igloo]

Did you learn all this in school?

[drifting knot]

No problem

[drifting knot]

Yea

But much by myself

Books

[flat igloo]

Great

[drifting knot]

Good

[flat igloo]

My school also doesn’t go over more advanced concepts

So self study it is

[drifting knot]

But yes

Remainder factor theorem is goated

[flat igloo]

Def

[drifting knot]

If you have another doubt u can ask

Or if your query is resolved

!done

[safe radish]

If you are done with this channel, please mark your problem as solved by typing .close

[drifting knot]

You can go to topic wise discussion channels

[flat igloo]

Well I still got 40 more such questions 😅 so will definitely be here for a long time

[drifting knot]

Like #calculus

[flat igloo]

For now we good

[drifting knot]

Ok

[flat igloo]

Thanks

.close

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[drifting knot]

Welcome

[safe radish]

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[safe parrot]

b) Determine a approximate value for f(B) if B is of distance 0.01 from A in the direction where f is growing the fastest

[safe parrot]

Does someone know how i determine first of all "the direction where f is growing the fastest" ?

[mellow cypress]

Are you familiar with the gradient operator?

[safe parrot]

Yes. I determined the gradient for the given function f in (a)

$\frac{\partial f}{\partial x} = cos(x+yz)(x^2+y^2+z^2)^{1/2} + sin(x+yz)\frac{(x^2+y^2+z^2)^{-1/2}}{2}2x \$
$\frac{\partial f}{\partial y} = cos(x+yz)z(x^2+y^2+z^2)^{1/2} + sin(x+yz)\frac{(x^2+y^2+z^2)^{-1/2}}{2}2y \$
$\frac{\partial f}{\partial z} = cos(x+yz)y(x^2+y^2+z^2)^{1/2} + sin(x+yz)\frac{(x^2+y^2+z^2)^{-1/2}}{2}2z$

[flat frigate]

Merineth

[mellow cypress]

But do you know what the gradient tells you about the function?

[safe parrot]

I'm not 100% sure but i'll try and explain as much as i can.
So what i know is that when we derive a function in 2d space for example f(x) we get a function which describes the increase/decrease at a given x value of a function.

I would imagine it works somewhat in the same sense with gradients where we derive wrt x,y and z and these new functions describe the increase/decrease of the function at a given point?

[upbeat ridge]

Wait

[mellow cypress]

More specifically, it tells you the direction that f is increasing fastest

Which is exactly what you need

[safe parrot]

Perfect, so i have to utilize the gradient somehow to find the direction where it increases the fastest?

In 2D space we determine the fastest growing point by doing f'(x) = 0. Would this method work for gradients?

[mellow cypress]

The gradient is the direction

[upbeat ridge]

Gradient tells you the steepest slope

[upbeat ridge]

wdym?

We don't though

[mellow cypress]

Nabla f = ( del f / del x, del f / del y, del f / del z)

[safe parrot]

?

[upbeat ridge]

This is a matrix

Nabla is the name for the gradient symbol

[safe parrot]

$\grad f = [\frac{\partial f}{\partial x},\frac{\partial f}{\partial y},\frac{\partial f}{\partial z}]$

?

[upbeat ridge]

No

$del$

[flat frigate]

Chaewon

[mellow cypress]

What do you mean no

[flat frigate]

Merineth

[upbeat ridge]

Yes

Compute them one by one

[mellow cypress]

This is what I wrote???

[safe parrot]

I've already computed the gradient for my given function

I'm just not entirely convinced i know what a gradient is anymore

[upbeat ridge]

Yeah but they didn't understand it

I was just explaining

[safe parrot]

Do we have a formal definition or explanation on what a gradient is?

[upbeat ridge]

Umm

It's the same as normal derivatives

But in three directions or more

[mellow cypress]

It’s a vector that points in the direction of greatest change. 0 vector when evaluated at a local extrema

[upbeat ridge]

Gradient is a combination of derivatives in each axis that results the steepest slope in a function

[mellow cypress]

What you have computed is the gradient asa vector field of f. So you want just a single direction which means you need to evaluate it at a point

What point do you think you should evaluate it at?

[safe parrot]

at A

I have a question tho about derivatives

[mellow cypress]

I’d agree

[safe parrot]

When i derive f(x) which results in f'(x). The new function f'(x) is a function that describes the functions increase/decrease at each point, right?

[mellow cypress]

f’ describes the rate of change, yes

[upbeat ridge]

Altho gradient is a vector not a function

[safe parrot]

Right okay

So when i find the gradient. It works the same way as f'(x)? The rate of change at a given point wrt x y and z?

Which results in a direction?

[mellow cypress]

https://youtu.be/TEB2z7ZlRAw?si=C1JoPDlK_dRe5RMf

Give this a watch

[safe parrot]

[mellow cypress]

?

[safe parrot]

@glacial cairn Hey Nel Are you online? :>

[upbeat ridge]

This video explains it well

[safe parrot]

"A gradient is a vector that points in the direction of the steepest increase of a function."

This is what i was looking for

[mellow cypress]

.

[safe parrot]

🙂 👍

[upbeat ridge]

Um

Hope we've helped

[safe parrot]

mhm, thanks

.close

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[velvet lark]

Hello im not quite sure why in this solution b100 is written as b^100-2 x b2 instead of b^100-1 x b1 which gives a completley diffrent answer

[velvet lark]

here is the orignial question

[main mural]

you mean q^(100-2) * b2 right?

the reason is that b2, b3, ... form a geometric sequence

doesn't include b1

[velvet lark]

ah I see

so it should work if it is b3 right?

yep nvm

i got it

thx

.close

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[brazen parrot]

did they get the answer

[brazen parrot]

by using determinant and then solving for k?

[mossy lotus]

step 2 is the determinant

Which I assume they solved for det(A) = 0 to get the k values

[brazen parrot]

okkk thx

:D

.close

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[ashen tinsel]

can anyone help on how answer c and d was solved?

[ashen tinsel]

or any yt video explaining it would also be very help full

[safe radish]

@ashen tinsel Has your question been resolved?

[bold sandal]

The exponent outside of the brackets can be distributed to all terms within the brackets

And the identity which is used is
(x^a)^b = x^(a*b)

Does this help you?

[ashen tinsel]

thank you

[safe radish]

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[tiny sand]

Hi, why are they capitalizing the function here? is F'(x) just the derivative of f(x)?

[tiny sand]

or does it mean something else

[tired whale]

derivative of integral should be the function itself ..??

Seems to be some kind of misprint

Or I must be missing something

[tiny sand]

wait

so

normally F(x) signifies the integral of the function?

[tired whale]

It’s not a convention or a standard notation of any sort I believe

But a lot of texts use it to denote the anti derivative

[tiny sand]

gotcha, ty

and you're right it was a misprint

.close

[safe radish]

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[wet plume]

how do i do this

i feel like the antiderivative is wrong

or wtv

idk

[severe pond]

what is this

[wet plume]

integral

the bottom is my work

[tired whale]

Well you missed the coefficients

[wet plume]

oh yeah

thx

[tired whale]

And also

You wouldn’t write it that way

You’re putting in the limits

What you’ve written is integral of the integral

[severe pond]

i’m confused

[wet plume]

how would i write it ?

oh

[severe pond]

you rewrote the problem?

but

[wet plume]

idk im new to this

[severe pond]

wrong

[wet plume]

idk what im doing

[severe pond]

it’s ok

if you found the anti derivative which is wrong

you don’t need to write the integral again

[wet plume]

oh ok

[wet plume]

ohh you write it in brackets?

what is dn

[golden verge]

dx

[wet plume]

what is dx 😰

[severe pond]

oh no

[golden verge]

How did u integrate the function without knowing what is dx

Bro what are u doing genuinely

[wet plume]

idk the course im doing is weird

[severe pond]

damn xander calm down

[devout shale]

@golden verge can u teach me rizz

[golden verge]

Oh

Go first learn what integration is

[wet plume]

okay

[golden verge]

The definitions and shit

[devout shale]

facts preach ‼️

[golden verge]

Then touch these problems

[wet plume]

alr thanks

ill tocuh the problems when im familiar with everything

.close

[safe radish]

Channel closed

Closed by @wet plume

Use .reopen if this was a mistake.

[devout shale]

ya dont touch them too early

[golden verge]

Yeah

[severe pond]

pause

[wet plume]

ill touch them a lot later

[valid rock]

so I shouldnt take ap calculus if last year I took geo

[devout shale]

idk how often did u touch the problems

[severe pond]

nah you should bro

💯

[devout shale]

I say fuck it bro do it

[valid rock]

never seen a ap calc problem

[devout shale]

I can tell you get this

[valid rock]

thanks man

will take 5 aps

[severe pond]

take 7

why not

[devout shale]

5 is nothing