precalculus

[lime bolt]

As radius is constant the width is determined by length

Due to Pythagorean

[wide lynx]

like this?

[lime bolt]

Only consider one half as the other is symmetric

Write the width in terms of the length and the double lengths and width = perimeter

[wide lynx]

how would i write width in terms of length

[lime bolt]

Due to Pythagorean
This

[blissful ridge]

Apply Pythagorean theorem

[wide lynx]

ho i see

[blissful ridge]

You'll get a relationship between L and W

[wide lynx]

this is what you mean right

not some fancy thing im missing

[lime bolt]

That is diameter not radius

Also it is half width

[blissful ridge]

Why use half width?

[wide lynx]

why is it half width

yeah

[blissful ridge]

Just use full W and L

[lime bolt]

Oh actually it depends on what triangle u use

[blissful ridge]

It's a rectangle anyway

[lime bolt]

This also works, my idea was different but it is ok

[blissful ridge]

The diameter is the hypotenuse

[wide lynx]

The diameter is the hypotenuse

[blissful ridge]

Now you need to separate W in terms of L

Rearrange and do that

[lime bolt]

Then do 2(w+l) and finish

You need to square root first

[blissful ridge]

First bring W on LHS

You need answers in terms of L

[lime bolt]

Yea other way around

[wide lynx]

oh

[blissful ridge]

It doesn't matter what L or W is but you've specified it that way

[wide lynx]

so like this?

[lime bolt]

Then do 2(w+l) and finish
yes now this

But RHS instead of w

[wide lynx]

thanks

ive never seen a problem like that before

[blissful ridge]

Well, now you have.........

[wide ocean]

what do they mean by 3rd term?
like (4b-5) (4b-5) (4b-5)

[sand void]

doesn't really help with their question

lol

i think they might be looking for B?

[wide ocean]

ok thanks @viscid thistle @sand void

let me check

no

[sand void]

yeah

15 * (4b)^4 * 5^2

is what they want

[wide ocean]

ok

[viscid thistle]

look

the first term is $a_1b^6$

[obsidian monolith]

polynomial:

[viscid thistle]

the second term is $a_2 b^5$

[obsidian monolith]

polynomial:

[viscid thistle]

the third term is $a_3 b^4$

[obsidian monolith]

polynomial:

[viscid thistle]

i guess that's how they did it

well w/e

[sand void]

yeah

[viscid thistle]

i wanted to say B

just because it's the only answer that is a "term"

[sand void]

@wide ocean you're just expanding (A+B)^n as normal, then getting coefficients from pascal's triangle. Does that make sense?

[viscid thistle]

i hate badly worded questions

[sand void]

i mean i think at this level "third term" as "third highest power of A in the expansion" is probably ok?

idk

[viscid thistle]

it's not hard to write find the coefficient of $b^4$ in the expansion of ...

[obsidian monolith]

polynomial:

[viscid thistle]

since that's basically what they actually wanted

[sand void]

it's probably a product of the entire curricula vocab

not this particular question

[opaque olive]

https://gyazo.com/3241997c9b08c703b06e6f5c884959fb

Can i get a hand with the second part of this question

I dont know how to calculate the area of RST or OPQ (hints only please)

[opaque olive]

do i just need to use cosine and area of a triangle rule?

[blissful ridge]

You'll have 3 points

Use the formula for area of triangle in coordinate geometry

That matrix one

Or just google it

[opaque olive]

i cant use matrix formula here

[blissful ridge]

Why?

[opaque olive]

cuz the syllabus for this test doesnt include matrix

algebra

[blissful ridge]

Okay then use the formula

1/2 {x 1 (y 2 − y 3 ) + x 2(y 3 − y1 ) + x 3 (y 1 − y2 )} sq.units.

This one

[opaque olive]

hm

wont half base * height work here

[blissful ridge]

Why do you want to increase your workload

After finding the coordinates you'll have to find the altitude and that will be more unnecessary calculation

[earnest barn]

What is the average of numbers between 0 and 1000 inclusive which are not divisible by 2 or 5?

My response:

Not divisible by 2 or 5 implies they end in 1, 3, 7, or 9, so that's 4/10 * 1000 = 400 numbers.

For the average I need to sum the numbers with these endings. Ie, the numbers ending in 1 can be described by

$$1 + \sum_{n = 0}^{99} 10n$$

So extending that logic,

$$\frac{ 1 + 3 + 7 + 9 + 4\sum_{n = 0}^{99} 10n}{400} = 495.05$$

[obsidian monolith]

Daniel Cann:

[earnest barn]

My response was incorrect

The answer is 500

Why was my average incorrect?

(ie, why does that not work)

Ohhh

I know why

I need to write 100 ( 1 + 3 + 7 + 9)

lol

no need for help

Stupid idiot thing to do

[keen parrot]

wouldn’t that be the same as the average of numbers that ARE divisible by 2&5?

so i’d say 500 straight away

[opaque olive]

would the area of the triangle be $\frac{1}{2}aq^22aq$ or $-\frac{1}{2}aq^22aq$

[obsidian monolith]

Yes:

[upper kelp]

We apply the formula for the area, and use the absolute values in order to get the lengths: $\$
$S = \frac{|aq^2| \cdot |2aq|}{2}$ $\$
We know that $aq^2 > 0$ because the point is to the right of the Y-axis, and we know that $2aq < 0$ because the point is beneath the X-axis. This means that $|aq^2| = aq^2$ and $2aq = -2aq$, and therefore: $\$
$S = \frac{aq^2 \cdot (-2aq)}{2} = - a^2 \cdot q^3$.

[obsidian monolith]

RoiKadmon:

[opaque olive]

ok at 2aq

youve said that 2aq = -2aq

from the question we were told that q < 0

[upper kelp]

Sorry, I forgot the absolute value. It should have been |2aq| = -2aq.
And yes, the reason for that is because q > 0. Since we know that a > 0, then 2aq must necessarily be negative, and therefore its absolute value is its additive inverse.

[opaque olive]

ok so the height of the triangle is -2aq

?

[upper kelp]

Yes, because the height has to be positive. (since it symbolizes length, and lengths are never negative)

[opaque olive]

sorry i didnt understand lol

what do you mean by |2aq| = -2aq

i understand that 2aq has a negative value

[upper kelp]

Sorry.
In the formula for an area of a triangle, we have the length of the base, which must be a positive number, and a height. The height is the distance between the opposite vertex, and that distance has to be positive.

In order to get that distance, we need find the difference between the Y-values of the different points, and in order to calculate the difference of numbers a and b, we find the absolute value of a - b, meaning |a - b|.

In this case, the Y values are 0 and 2aq. We would like to find the difference between them, and that difference is the height. In that case, we calculate:
|2aq - 0| = |2aq|.
Now, we must determine whether 2aq is positive or negative to know whether |2aq| = 2aq or |2aq| = -2aq. In this case, since it's negative, we have |2aq| = -2aq, and therefore the difference, which is the height, is |2aq| = -2aq.

[opaque olive]

ok

but q is already a negative right

[upper kelp]

Yes.
Because q is negative, then since we know that a is positive, then 2aq is a negative number, and that's why |2aq| = -2aq and not |2aq| = 2aq.

[opaque olive]

that makes sense

but when we go to calculate the area

we do 0.5 x aq^2 x 2aq ?

since we use absolute values?

[upper kelp]

In this case, no.
We need the lengths (which you find with absolute value) in the formula, not just the terms as they are.

Think of it this way:
Suppose you were to find that the Y-value of one point is 0 and the value of the other point is -3. In this case, you'd calculate -3 - 0 = -3, and then you'd plug it into the formula, and get a negative area, which doesn't make sense.
What you must do, instead, is use an absolute value: |-3 - 0| = |-3| = 3, and then you'll get a positive area.

In the same way, you need absolute values in the formula.

[opaque olive]

ok

so as 2aq is a negative

we need a positive to calculate area right

so thus we need -2aq

?

-2aq is a positive number

[upper kelp]

Exactly. You use:
0.5 |aq²| |2aq|, which comes up to:
0.5 (aq²) (-2aq) and this would be the area.

[opaque olive]

great i understand!

thank you!

[severe flare]

@daring gate Go to bed

[solemn tusk]

Hey anyone online?

I wanted to ask an question

[harsh smelt]

#❓how-to-get-help

[willow bear]

@solemn tusk no clearly nobody is online in a server of tens of thousands of people /s

[viscid thistle]

5k people online

[solemn tusk]

Why can't we use the derivative formula of e^f(x) in e^logx

[past meadow]

you can, but it would be pointless

[blissful ridge]

You can

[past meadow]

since e^logx=x

(assuming you mean log_e)

[solemn tusk]

Yeah but if we can then why are there two different answers

[past meadow]

show me your 2 different answers

is one of them e^logx/x by chance?

[blissful ridge]

You've made some error

[echo wagon]

What's the formula you are using?

[solemn tusk]

If we use the formula of e^f(x) then the ans will be e^logx × 1/x

[echo wagon]

And what is e^log x?

[solemn tusk]

X

[blissful ridge]

Simplify it

[echo wagon]

So x/x

[solemn tusk]

Ah got it

Thanks guys

[echo wagon]

Np

[solemn tusk]

In the number 7
Why couldn't we keep x= cos (theta)

[echo wagon]

What is the question?

You could make x=cos θ, and you would get the exact same thing

I think

Wait

Yes, looks like you also get $sin^{-1}(sin 2 \theta) $. Why they chose sin, would either depend on the question or you could do either

[obsidian monolith]

Lunasong:

[solemn tusk]

No the answers aren't the same

[echo wagon]

@solemn tusk?

Whichever one you choose, the sqrt gives you the other one. So how is it different?

[winged shell]

what are some of the main concepts to know from alg 2 when going to pre calc

[uncut mulch]

depends on what your being taught in Algebra 2

depending on structure, certain topics can fall into both categories

general algebraic manipulations, functions/polynomials, trig, logs/exponents

[pseudo cargo]

Hey, I got a question about rolle's theorem

So imagine there's a person travelling in a circle path. He starts at a point and ends in the same point, with constant velocity

When sketching the graph, its a parabola

[blissful ridge]

Constant speed*

[pseudo cargo]

Yeah, constant speed srry

Something like this

[blissful ridge]

Anyway what's your question?

[echo wagon]

What are you graphing there?

[pseudo cargo]

My question is, since rolle's theorem says there's a point c where f'(c)=0, that would mean there's a point where velocity is 0

What are you graphing there?
@echo wagon Position-time

[echo wagon]

Clearly the speed is not constant

Since it's not a straight line

[pseudo cargo]

How would the graph be if it has constant speed and ends up in the same point?

[echo wagon]

A spike

Which is not differentiable

So Rolles theorem doesn't apply

[pseudo cargo]

Oh

[echo wagon]

Make sense?

[pseudo cargo]

Yep, ty

[echo wagon]

Np

[kindred garnet]

Anyone know how to graph name?

[fossil crown]

From

#❓how-to-get-help

Rule 1

The help channels are solely for help with math, so feel free to post your question. Asking whether you can ask a question or if anyone knows about some specific topic is unnecessary, so please try to avoid questions of that nature.

[kindred garnet]

Lol

[echo wagon]

Huh?

How to graph name? I cannot compute this question

[lime bolt]

u mean write a graph of your name ?

seems pretty complex

[blissful ridge]

You can do it if you include multiple fucntion

I mean we have enough graphs to accommodate all the possible shapes

[echo wagon]

If you approximate the letters with straight lines, you can just plot a bunch of straight lines

If you want to be fancy with curves, it's possible

[blissful ridge]

You just have to define them in such a way

[vivid glacier]

when I master pre-calculus I'll be ready to learn calculus 1?

[blissful ridge]

Master is a strong word

[echo wagon]

Presumably yes, unless you have other more basic prereqs like algebra or trig you don't understand 100%

[acoustic harbor]

yea you don’t even need to master pre calc

some knowledge is good enough

[willow bear]

what you do need to master is algebra

so that you don't pull an alge-BRUH

[acoustic harbor]

yeah most of my mistakes come from algebra

like adding incorrectly you feel

[echo wagon]

Because you don't know how to add, or because you have a brain fart and think 12 + 7 is 20 for a second?

[acoustic harbor]

more like 13+19=42

[uncut mulch]

there's crap atm about 2+2=5

[lime bolt]

is 91 prime

[vivid glacier]

oh ok thank you

[blissful ridge]

No

[acoustic harbor]

no

[lime bolt]

ik

[vivid glacier]

I'm learning pre-calculus

[lime bolt]

just testing your arithemetic

[acoustic harbor]

y do they call it precalculus

most of the shit i learned in precalc isn’t even used in calculus

[echo wagon]

Lol, I still can't find a divisor for 91,help

[blissful ridge]

In my country they don't

[acoustic harbor]

7 13

[blissful ridge]

13 times 7

[lime bolt]

70 + 21

[echo wagon]

Uh

Jk?

[jade heron]

Un
M

[echo wagon]

Pear!

Are you a Mr or Ms Pear?

[lime bolt]

c'est d'ane ou de l'ane?

is that proper french idk

[echo wagon]

Seems fine

[viscid thistle]

that doesnt mean anything lo

lol

C'est un âne

C'est l'âne

[lime bolt]

woah l'ane means donkey

i didnt know this

that doesnt mean anything lo
@viscid thistle isnt du masculine and de la feminie

[echo wagon]

I mean the sentence construction seemed fine? Wouldn't it be: Is it ___ or ___?

[viscid thistle]

No

That's un or une

for masculine or feminine singular

You could phrase your question like this: "C'est un âne ou une âne?"

And the answer would be "C'est un âne"

[lime bolt]

j'ai oblie beaucoup

[echo wagon]

Thanks for the French lesson

[viscid thistle]

You guys have choice between french and spanish ?

Or only spanish

in high school

[lime bolt]

my choice was french, spanish or german

[sudden mauve]

I am a spanish native 🙂

[lime bolt]

hola

[jade heron]

Are you a Mr or Ms Pear?
@echo wagon hello u can use mister =)

[lime bolt]

bien

[proven marten]

@echo wagon hello u can use mister =)
@jade heron What can I use?

[lime bolt]

you are quoting his answer?

[echo wagon]

Merci beaucoup, messieur pear

[blissful ridge]

Didn't you had a test JC?

[proven marten]

No?

Im on summer break rn

[lime bolt]

why would he have a test now

[echo wagon]

He wished someone luck with their test

[blissful ridge]

Oh

I didn't read tbh

[proven marten]

Oh no worries

[jade heron]

Yes yes u may all use he him or they them for me

[proven marten]

Mr Pear, do you play chess?

[jade heron]

Not in a long time I'm pretty garbage now

[lime bolt]

can i call you donkey

[proven marten]

Theory: All mathematicians have played chess once in their lifetime

[lime bolt]

i play chess sometimes

[blissful ridge]

Prove it using induct

Induction*

[proven marten]

kane vs Luna 👀

[lime bolt]

ez win

[echo wagon]

No why?

[blissful ridge]

I'll spectate

[jade heron]

I am donkey yes

[blissful ridge]

Go you two

[echo wagon]

Probably an easy win. I'm really not good, lol

[proven marten]

I shall spectate too

[jade heron]

Pretty sure thats the horse emoji my friends

[lime bolt]

yea well it looks a bit like one

[jade heron]

Valid

Le Cheval

🍎

[proven marten]

So, will there be a game?

[echo wagon]

Up to kane

[lime bolt]

🆗

[echo wagon]

Lichess?

[lime bolt]

fine

[blissful ridge]

How do I get that chess role?

[proven marten]

Spoiler: Kane is like 2300

[echo wagon]

Don't laugh at how bad I am, lol

[lime bolt]

ill make a new account to hide my rating

[echo wagon]

Kane is a super GM

I also made a new account to play with people here, lol, but only because I don't want to dox myself

[proven marten]

send link

[blissful ridge]

Start the god damn game

[echo wagon]

Time control preference? @lime bolt

I will send link now, he is still making an account anyway

[blissful ridge]

How do I get that chess role?
@blissful ridge

[lime bolt]

maybe 5mins per person

[proven marten]

@blissful ridge
@blissful ridge Talk to Gomez I think

[echo wagon]

lichess.org/ilpheJgh

[sudden mauve]

off topic..

use #chess-go-shogi

[proven marten]

No

Its fine

[echo wagon]

I retyped that, so possibly it is lichess.org/iIpheJgh

[proven marten]

to talk here unless theres a question

Same thing

[echo wagon]

No, the one is iLphe the other is iiphe

[proven marten]

Evans gambit

I see a not beginner probably

[blissful ridge]

How can I watch this game

[proven marten]

Tip: Bd6 locks the pawn so that you cant move it so its usually not a good move

Copy paste the link

lichess.org/ilpheJgh
@echo wagon this

[blissful ridge]

Where?

I have the app on android

[proven marten]

The link

I want to play kane now

[echo wagon]

Well, I was only one pawn down since he accepted my take back on that shit blunder

I want to spectate

[lime bolt]

ok in ready if u want

[proven marten]

https://lichess.org/LyUEa5Qg

he knows theory

[proven marten]

I make stupid mistakes

Ive done this with Ramonov as well

[echo wagon]

You were up for a while though, so that's good

[lime bolt]

what rating is ramanov

[viscid thistle]

At least 7

[echo wagon]

And at most 3000

Now we just need to decrease the range

[viscid thistle]

Why not more than 3k

[lime bolt]

ok so we have two a lower and upper bound

[viscid thistle]

He could be a super gm

[lime bolt]

Why not more than 3k
that is true maybe he is better than carlsen

[viscid thistle]

He is very active in the server

[lime bolt]

well at minimum we know he is between 1 and 5000 inclusive

[viscid thistle]

So I'd say <2000

[echo wagon]

Lol, there are only two people with a rating over 3000 on lichess

Ramonov ain't one of them

[viscid thistle]

:(

[lime bolt]

[silver matrix]

I know I should apply the exponential formula and forget this formula but they ask me explictly to use this particular formula (sorry for the redundancy)

[blissful ridge]

First of all, how can a fucntion of x contain no x

That u should be x

[viscid thistle]

Lmao true

Wtf

[silver matrix]

yeah I suppose is an error in the photo where u might be x

[uncut mulch]

um...wtf is that...

[silver matrix]

it's from google

[blissful ridge]

Anyway coming to your question, just write kth root of x as
x^(1/k)

Then just apply the power rule

[silver matrix]

read the message please

[blissful ridge]

Can you show me your work

I think you are missing the chain rule

[uncut mulch]

the image is incomplete

u is supposed to be the inner function

[silver matrix]

my work is basically (as I have to use this rooth formula)
[ f'(x) = \frac{2x - 2}{2 · \sqrt{}} ]
and then I don't continue as I can't continue as I have the doubt

[obsidian monolith]

Pa_u_los:

[silver matrix]

u is supposed to be the inner function
wdym? I don't think the formula is icomplete

anyways

i was just looking for derivatives exercises for practising

and I got a page where they explicitly asked me for using the root formula derivaties

and I looked at google and that's what i got

[blissful ridge]

Please from next time onwards just send a pic or screenshot

[silver matrix]

?

that's what i did...

[uncut mulch]

the should've mentioned what u was

[silver matrix]

yeah I suppose is an error in the photo where u might be x

[uncut mulch]

no there isn't an error

u is a function of x

[silver matrix]

i don't get you

[uncut mulch]

for your question, if you were to apply that formula,
u = x^2 - 2x + 3

[silver matrix]

alright...

so then

for u i have to subtract the exponents 1?

what about the constant?

[uncut mulch]

i'd actually advise against applying that formula

and would recommend going through the steps of power/chain rule

[silver matrix]

I haven't seen yet properly chain rule

i was thinking just applying the power rule

with 1/2

so you think this page is worthless to continue?

cuz they ask me to do it like that explicitly

[uncut mulch]

i mean if you're "told" you could just apply it

[silver matrix]

yeah but my question is how do I subtract the one to the constant

or to the x

[uncut mulch]

your k here is 2

you subtract 1 from that

[silver matrix]

so it remains the same?

[ (x^2 - 2x + 3)^{2-1} ]

[obsidian monolith]

Pa_u_los:

[uncut mulch]

yes, that would be the component under the radical in the dominator

[silver matrix]

alright thanks

[uncut mulch]

which simplifies to just x^2-2x+3

[blissful ridge]

Wasn't there sqaure root not the square?

[uncut mulch]

the above is only part of the expression

[blissful ridge]

Oh, okay

[fleet yew]

2-1=1

[opaque olive]

could someone find the gradient between these points in the simplest form for me please $(q^2,2q)~and~(p^2+q^2+1,p+q)$

[obsidian monolith]

Yes:

[opaque olive]

its supposed to be -q i think

i cant seem to get it

[viscid thistle]

why would it be -q?

it doesn't seem like this should be invariant to p

[fleet yew]

let those two points be

(a,b) and (c,d)

then just find

[opaque olive]

i know how to do it

[fleet yew]

$\frac{d-b}{c-a}$

[obsidian monolith]

AMD:

[opaque olive]

i just cant seem to simplify it

[fleet yew]

what's d-b

or what are you getting

[opaque olive]

i need to show that 2 lines are perpendicular

one sec lemme show question

https://gyazo.com/fdcd4bee723867dc81a53f5280c79828

on the last part, i need to show PSQR is a rectangle

ive proven that 3 intersections that cause the rectangle to be formed are perpendicular

just need to show that grad of QS and grad of PS multiple for -1

[plain grail]

ive proven that 3 intersections that cause the rectangle to be formed are perpendicular
it's actually enough to show that this is a rectangle

[opaque olive]

i know

but

its possible to show the last angle is also 90

i just cant seem to simplify it?

[fleet yew]

i mean what are you getting when you apply the slope formula

in the form (d-b)/(c-a)

[opaque olive]

oh ok

$\frac{2q-(q+p}{q^2-(p^2+q^2+1}$

[obsidian monolith]

Yes:

[opaque olive]

and then $\frac{q-p}{-(p^2+1)}$

[obsidian monolith]

Yes:

[opaque olive]

anything im missing to see?

its late here

[fleet yew]

find the other slope now

[opaque olive]

tried that

[fleet yew]

and see if they multiply to -1

[opaque olive]

it doesnt

the other slop would be $\frac{p-q}{-(q^2+1)}$

[obsidian monolith]

Yes:

[opaque olive]

btw we also know that pq = -1

[viscid thistle]

I think that does it

[fleet yew]

multiply them together and simplify

[viscid thistle]

set p=-1/q and simplify

[opaque olive]

ok

[viscid thistle]

in $\frac{q-p}{-(p^2+1)}$

[obsidian monolith]

Botn:

[opaque olive]

youre right

it does = -q

the other equals -p

and then pq = -1 and thus perpendicular

qed

lol

i thought i wouldnt have worked so i didnt try it

[viscid thistle]

btw we also know that pq = -1
this is what we needed lol

[opaque olive]

lol sorry

its nearly 4am here lol

[viscid thistle]

all good

[reef jasper]

is this correct ?

[elfin dirge]

Just checked it on desmos and that gives the outer loop

[reef jasper]

how to find inner loop then

the only answer is by setting it to 0

[elfin dirge]

-π/3 , π/3 is my solution

[harsh smelt]

@reef jasper consider which value of theta provides upper part of inner loo

and whichvalues upper

i mean what is r(0)?

[reef jasper]

it’s 1/2?

[harsh smelt]

why?

what is cos(0)?

[reef jasper]

pi/2 and 3pi/2

[harsh smelt]

what

,w cos(0) == pi/2

[obsidian monolith]

Results provided by WolframAlpha

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Requested by Commander Vimes#7532

[harsh smelt]

@reef jasper what do u know about cos?

[reef jasper]

oh it’s 1

[harsh smelt]

yes, and 4-8cos(0)?

[reef jasper]

-4

[harsh smelt]

so at the begining it is on loop

on here

where will it go as theta increases?

[reef jasper]

to the right?

[harsh smelt]

to the right?

how

what is direction of increasing theta given that -4 is initial r?

where theta increases to?

[reef jasper]

oh sorry it increases towards -3

like on a unit circle

[harsh smelt]

yes, but i mean in which direction it will go

up

or down?

[reef jasper]

down

[harsh smelt]

so it will describe lower branch of inner loop

and finish it at 0

when r = 0?

[reef jasper]

oh ok so it would go from -4 to 0

[harsh smelt]

yes but for which values of theta?

[reef jasper]

pi to 0?

[harsh smelt]

no

when r = 0?

r = -4 at theta = 0, but for which theta r = 0?

[reef jasper]

pi/3 and 5pi/3

[harsh smelt]

nice

and to finish it

what is cos(2pi)?

[reef jasper]

yeah that was my original answer

it’s 1

[obsidian monolith]

Commander Vimes:

[harsh smelt]

which can be written as

-π/3 , π/3 is my solution

[reef jasper]

oohhhh ok i see now

i just had to go backwards a little

i should’ve put it on desmos

tnx gn

[harsh smelt]

yw

[winter marten]

Just wanted to double check this looks fine or something wrong?

[elfin dirge]

Is this the one you started with?

Or did you get there from other steps

[winter marten]

Started

[elfin dirge]

Alright then everything looks fine to me

[winter marten]

Thanks

Wolfram was confusing then i realised it have given answer as dx/dy not dy/dx

[elfin dirge]

Oh your notation on the first couple lines was weird actually

You put sinxy(xy²) instead of xysin(xy²)

[willow bear]

first line

derivative of product is not the product of derivatives, AND the derivative of ln^2(sin(x)) is not 2ln(sin(x))

[echo wagon]

Look up the product rule and the chain rule for derivatives, if you don't know them

[silver matrix]

is this correct?

[blissful ridge]

How did you distribute log in there?

[silver matrix]

what do you mean?

[blissful ridge]

You took natural log in second step, right?

[silver matrix]

I've used the log rule $\ln A^B = B\ln A$

yes I did

[obsidian monolith]

Pa_u_los:

[blissful ridge]

Or is it something esle?

[silver matrix]

no, that's right

[blissful ridge]

The second to third step

[silver matrix]

it's the only way I know how to get rid off the x on the exponent

[blissful ridge]

How are you distributing log

[silver matrix]

it's the same

[blissful ridge]

No, it's not the same

Log(a-b) does not equal to log(a)-log(b)

[silver matrix]

I mean I haven't used that log property

just distributed all the ln(arguments )

with the property I said

I don't know if that's correct

tho

[blissful ridge]

You cannot do that

[echo wagon]

You cannot distribute log over addition

[silver matrix]

alright.. :'(

[blissful ridge]

Though, that panda emoji is cute

[echo wagon]

Remember log(ab) = log(a) + log(b). So clearly log(a+b) is not the same thing too.

[silver matrix]

Then I don't know how to do it

[echo wagon]

What's the question?

[blissful ridge]

Just diffrentiate without log

[silver matrix]

[ f(x) = e^{x+1} - 3e^x + 2^{e^{3}} ]

[obsidian monolith]

Pa_u_los:

[echo wagon]

To differentiate?

[silver matrix]

yes

I just wanted to get rid of the x on the exponent

[echo wagon]

Look up the product rule and the chain rule for derivatives, if you don't know them

Did you do this?

[silver matrix]

yes I did

[echo wagon]

What is the chain rule?

[silver matrix]

it's like an onion, you are differentiating from the biggest function into the more smaller pieces/arguments

[echo wagon]

Lol

Sure, that's good enough for now

What is the derivative of e^x?

[blissful ridge]

I like that terminology

[silver matrix]

e^x again?

[echo wagon]

What is the derivative of x+1?

[silver matrix]

1

[harsh smelt]

What is the chain rule?
@echo wagon rule by which we exploit working people!

[echo wagon]

Now apply the chain rule to e^(x+1)

[harsh smelt]

⛓️ 📐

[silver matrix]

then it's e^x?

e^x · 1

[echo wagon]

You are not using the chain rule correctly

You take the derivative of the outer function in terms of the inner one

So it should be e^(x+1) still

[silver matrix]

mmm, i don't get it

could you recommend me a particular resources for reviewing this topic?

[echo wagon]

Not really tbh, but literally just googling chain rule and watching videos on it should make it clear and have a lot of examples

I'm sure Khan Academy has a video on it

[silver matrix]

alright, let's review it one more time..

it's because

for example i

saw a video

on this function

Derive
[ e^{x^{x}} ]

[obsidian monolith]

Pa_u_los:

[silver matrix]

and the result is

[ y' = e^{x^{x}} \cdot x^{x^{x}} \cdot \left(x^x (\ln x +1) \ln x + \frac{x^x}{x} \right) ]

[obsidian monolith]

Pa_u_los:

[silver matrix]

and he was applying that log property

for getting rid of the x on the exponent

[frozen needle]

that third factor is suspiciously long

[blissful ridge]

In this question you can take log to simplify

But your original question is completely different

[fleet yew]

x^x = u

So just use chain rule

[pseudo cargo]

Hi, I dont know how to solve a limit, can anyone help me?

[fossil crown]

From

#❓how-to-get-help

Rule 1

The help channels are solely for help with math, so feel free to post your question. Asking whether you can ask a question or if anyone knows about some specific topic is unnecessary, so please try to avoid questions of that nature.

[pseudo cargo]

$\lim_{x\to 0}\frac{f(2+3x)+f(2+5x)}{x}$

[obsidian monolith]

Lobo:

[pseudo cargo]

The exercise tells me f(2)=0 and f'(2)=7

I used L'Hopital because its a 0/0 indetermination

And I got $\lim_{x\to 0}(f'(2+3x)+f'(2+5x))$

[obsidian monolith]

Lobo:

[blissful ridge]

Use chain rule

[pseudo cargo]

I then used the fact that f'(2)=7 to get 7+7=14

But the exercise says the answer is 56

[blissful ridge]

You have to use chain rule, because there is function inside a function

[pseudo cargo]

Oh, true

$\lim_{x\to 0}3f'(2+3x)+\lim_{x\to 0}5f'(2+5x)$

[obsidian monolith]

Lobo:

[pseudo cargo]

Like this?

[willow bear]

excessive

[pseudo cargo]

Excessive?

[willow bear]

you can just write your thing as $$\lim_{x \to 0} \paren{3 \times \frac{f(2+3x) - f(2)}{3x} + 5 \times \frac{f(2+5x) - f(2)}{5x}}$$ yknow

[blissful ridge]

You don't really need to distribute limits

[willow bear]

with some clever additions of zero and multiplications by 1

[obsidian monolith]

Ann:

[pseudo cargo]

I think those fractions dont solve it

Couldnt I do $$\lim{x\to 0}3f'(2+3x)+\lim{x\to 0}5f'(2+5x) = 3f'(2)+5f'(2)$$?

[obsidian monolith]

Lobo:

[pseudo cargo]

I dont know why the limits are showing that way but isnt this well done?

[blissful ridge]

Yeah, you've solved it right,
Ann was showing you an another way to solve it just by some manipulation

[pseudo cargo]

Oh, nice

Ty @blissful ridge @willow bear

[willow bear]

underscores

[viscid thistle]

may i get some help with this?

[compact crystal]

I should know about this, but it has been a long time ago

[viscid thistle]

i got P = (0 & 1; 1 & 0)

[proven marten]

Play the Lunasong and summon @echo wagon

there you go

he needs help

[echo wagon]

What if I can't help? Why you embarrass me like this?

[proven marten]

Oh

[viscid thistle]

its better than what i can do

[proven marten]

Up until now I thought you were some omni knowledgable math person

I still think you

are

Wait this is AOPS

[viscid thistle]

yea 😦

[echo wagon]

What is AOPS?

And how did you get your answers?

[viscid thistle]

art of problem solving

and so i made myself a 2 by 2 matrix to try and get the projection

but that was very flawed

[echo wagon]

Do you know what projection is?

Because your answers are not projections onto u

[viscid thistle]

😐

do you have a mic?

[echo wagon]

It's not a rhetorical question

And no I don't

[viscid thistle]

oh okay

and im aware

projections are kind of like shadows

right?

[echo wagon]

I don't like analogies, they are vague and imprecise

[lime bolt]

analogies can be good

[viscid thistle]

sometimes

[echo wagon]

If you project a vector v onto a vector u, it is the same as the orthogonal projection of the vector onto a line parallel to u

[viscid thistle]

so its the same thing?

[lime bolt]

do u know about projective geometry

[viscid thistle]

no

no i do not

[echo wagon]

There you have a vector u and a vector v you want to project ORTHOGONALLY onto u

So find the line that goes from the endpoint of v onto the line through u perpendicularly

[lime bolt]

projecting doesnt have to be orthogonal

[echo wagon]

This is orthogonal projection though

[lime bolt]

i guess u can assume it if the context is clear like in vectors

[echo wagon]

I've never seen vector projection not refer to orthogonal projection

Anyway

@viscid thistle Do you understand my picture?

[proven marten]

kane how far have you gotten in olympiads?

[viscid thistle]

YES

[blissful ridge]

In linear algebra and functional analysis, a projection is a linear transformation from a vector space to itself such that. That is, whenever is applied twice to any value, it gives the same result as if it were applied once. It leaves its image unchanged. Wikipedia

[lime bolt]

well ive only started doing it in march but i have improved a lot

next year i think i can do well

[viscid thistle]

so i got the result

[blissful ridge]

I Googled it for you guys

[proven marten]

Damn I thought you wrote that

[echo wagon]

OKAY, NOW DRAW YOUR U AND THEN PROJECT THE VECTORS ONTO IT

[proven marten]

and I was about to clap my hands

Luna calm down :(

[viscid thistle]

(2 & -1; 0 &0)

[echo wagon]

I am replying in all caps because he did, lol. I'm always calm

And was it right?

I didn't do it

[viscid thistle]

;-;-;-;-;-;-;-;—;

it was correct

[lime bolt]

projective geometry is interesting stuff

[echo wagon]

@viscid thistle don't write like that, it's horrific. But yay, glad I could help

[viscid thistle]

[lime bolt]

the proper notation is (x,y;z,w)

[viscid thistle]

@echo wagon

i have another question

i cant figure out either of these

@fringe stream

do you think you could help?

[echo wagon]

You need to project the unit vectors onto u to find P

[viscid thistle]

im not sure how to do that

[echo wagon]

The basis vectors

Sorry, busy, will help in a bit

(1,0) and (0,1)

[viscid thistle]

right i got that

[echo wagon]

Try to project them like you did the others

[viscid thistle]

ok

[echo wagon]

@viscid thistle And?

[viscid thistle]

im lost

@echo wagon

completly

[echo wagon]

How did you get the first one right then?

[viscid thistle]

it was just looking at the projections

right now im lost entirely

[echo wagon]

@viscid thistle I'm sorry, but I absolutely have to sleep now. Maybe someone else can help you, but if not: draw a picture

Draw the vector u starting from the origin

[viscid thistle]

I FIGURED IT OUTTTT

[echo wagon]

Okay, great, good job!

[viscid thistle]

ty;

[echo wagon]

Good night

[opaque olive]

https://gyazo.com/0861d4c2100bdc0e13097a45ce9cc51a

ive shown the first part. I set k = 1 and i get $2xy\frac{dy}{dx} = y^2 + x^2\frac{df}{dx}$

[obsidian monolith]

Yes:

[opaque olive]

i dont see what to do next

[patent beacon]

What if df/dx = 1 - 1/x²?
@opaque olive

[opaque olive]

@patent beacon am i supposed to use a substitution?

[patent beacon]

That's not really a substitution haha. f is any function

By choosing one, you get the form you want

@opaque olive

Of course, the function you'd be choosing is f(x) = x + 1/x

[opaque olive]

ah makes sense

thank you

[dusky chasm]

I don’t understand how they substitute the first eruption for that?

Trying to solve this problem

[rare lava]

is just a variable change

it can only be on the range -1,1 though

[viscid thistle]

how can I write $1 \leq b \leq a \leq4$ normal? Like $1 \leq b \leq a$

[obsidian monolith]

Alphonse [Maxwell]:

[viscid thistle]

Wym normal?

I do not know what it means

I only know those ranges if it is written like

$1 \leq b \leq a$ this makes sense to me

[obsidian monolith]

Alphonse [Maxwell]:

[viscid thistle]

I do not know what it means 😦

So it makes sense to you but you don't know what it means???

What

[willow bear]

@viscid thistle it means 1 <= a and a <= b and b <= 4

or alternatively you could look at it like "a and b are both between 1 and 4 inclusive, and additionally a <= b"

[viscid thistle]

ah okay that I understand

but why do ppl write it so difficult . Why not give two ranges

[willow bear]

cause it's shorter, and the transitive law of inequality is something you are kind of supposed to be comfortable with

[viscid thistle]

hm. Then I will brush up my skills about that

thank you for helping 🙂

[obsidian monolith]

Pa_u_los:

[willow bear]

$(-3^{e^x})' = -e^{e^x \log(3)}$ is false

[obsidian monolith]

Ann:

[willow bear]

$-3^{e^x}$ itself equals $-e^{e^x \log(3)}$

[obsidian monolith]

Ann:

[willow bear]

$e^{\log(3) \cdot e^x}$ is the composition of three functions, in this order from left to right: the exponential function, multiplication by $\log(3)$, and the exponential function again

[obsidian monolith]

Ann:

[willow bear]

@silver matrix

[silver matrix]

Alright let's try. Thanks.

And that would be $e^{x^{2}} \cdot \log 3$?

[obsidian monolith]

Pa_u_los:

[willow bear]

no

[silver matrix]

The procedure I've followed has be the following:
\begin{equation*}
\left(\minus e^{e^{x} \log(3)}\right)' = \minus e^{e^{x} \log(3)} \cdot e^x \cdot \log (3) + e^x \cdot \frac 1 3 \cdot 0 \cdot e^x
\end{equation*}

[obsidian monolith]

Pa_u_los:

[willow bear]

this is excessive

log(3) is a constant

your other term is correct though

[silver matrix]

i.. i don't understand

then it's the same original function

because now that I see that 0 is going to get rid off the e^{x^2}

$\minus e^{e^{x} \log(3)} \cdot e^x \cdot \log (3) $

[obsidian monolith]

Pa_u_los:

[silver matrix]

that is the solution?

[willow bear]

this is the derivative of -e^(e^x * log(3)) yes

[silver matrix]

ok tyvm

[obsidian monolith]

Pa_u_los:

[silver matrix]

Is the solution x?

[viscid thistle]

Try to get something like this $f(x) =e^{g(x)} $

[obsidian monolith]

HoboSas:

[silver matrix]

How do I do that?

I think that I have to apply limits?

I don't remember very well how to do that

[viscid thistle]

Write 2^{x/2} in terms of e

[silver matrix]

that would mean e^{2^{x/2}}?

I don't understand that btw

[viscid thistle]

Nope

You missed some ln

[silver matrix]

e^{ln(2^{x/2})}?

[viscid thistle]

Yes

Use log rules now

[silver matrix]

e^x/2 · ln(2)?

And I guess this is still f(x), no?

[viscid thistle]

Nice

And I guess this is still f(x), no?
You forgot -

[silver matrix]

where?

[viscid thistle]

But yes

[silver matrix]

this is -f(x)?

what?!

[viscid thistle]

F(x)=-e^....

[silver matrix]

🤔 I don't understand all this, where could I search for this information?

[viscid thistle]

You did this all alone

[silver matrix]

I'm seeing a lot of courses on derivates but they talk about other stuff always (chain rule, higher order derivaties, implicit ones, exponential, log, etc.) but not about this type where x is the exponent and I have to apply all this e thingy

Yeah

[obsidian monolith]

HoboSas:

[viscid thistle]

The thing is that you know the derivative of e^x

[silver matrix]

well although I do, I don't understand it

[viscid thistle]

if you have e^f(x), the derivative is f'(x)*e^f(x)

[silver matrix]

I don't even understand the proper relation between e and the natural log, etc. But I have so many questions that is better just to focus on grinding the exercises, if not I'm not able to continue

[viscid thistle]

ln(x) is the inverse function of e^x

$x=e^{\ln(x)}=\ln(e^{x})$

[obsidian monolith]

HoboSas:

[silver matrix]

alright, so coming back to
$\frac{-1}{-e^{\frac{x}{2}\ln(2)}}$

[obsidian monolith]

Pa_u_los:

[silver matrix]

Now I should be able to derive?

[viscid thistle]

Too many -

[silver matrix]

as this would still be f(x)

[viscid thistle]

And try to move e to the numerator

[silver matrix]

alright so could I do $1 \cdot e^{\frac{-x}{2} \ln(2)}$ ?

[viscid thistle]

Still missing a minus

[obsidian monolith]

Pa_u_los:

[viscid thistle]

Why is there a 1?

But yea this is good

[silver matrix]

because it was on the denominator

and a fraction

when passing to the numerator is

x^-1

alright, so let's derive then... Let's see if it goes well

thanks

[viscid thistle]

$- e^{-\frac{x}{2} \ln(2)}$ ?

[silver matrix]

why the minus?

[obsidian monolith]

HoboSas:

[silver matrix]

as in the numerator it was -1

and in the denominator it was -e^...

[viscid thistle]

Look at the original question

[silver matrix]

then the minus cancel no?

[viscid thistle]

and in the denominator it was -e^...
Nope

I was talking about the whole function

[silver matrix]

oh alright

[viscid thistle]

Remember the chain rule

[silver matrix]

Yeah that was what I was going to tell you; i mean first derive e^..., then the argument (product differentiation) and then should I derive the x/2?

that would be 1/2?

cuz then the log (as ann said) would be a constant, meaning that is 0

[viscid thistle]

What's the derivative of $-\frac{x}{2} \ln(2)}$

[obsidian monolith]

HoboSas:
Compile Error! Click the reaction for details. (You may edit your message)

[silver matrix]

Because I'm thinking that $\minus \frac x 2$ has to be $\minus \frac 1 2$

[obsidian monolith]

Pa_u_los:

[silver matrix]

The derivative is $\left(\frac{-x}{2}\right)' \cdot \ln(2)$

[obsidian monolith]

Pa_u_los:

[silver matrix]

and that has to be -1/2

bc if not it would -1/0 which would be an indetermination

[blissful ridge]

What's the final answer?

[silver matrix]

the final answer would be
[ f(x)' = \minus e^{\frac{-x}{2}\ln(2)} \cdot \left(\frac{-x}{2}\right)' \cdot \ln(2) ]

[obsidian monolith]

Pa_u_los:

[silver matrix]

and I don't know if i have to also derive then each argument separately

I don't think so

i think the chain rule stops there

[blissful ridge]

Why is there a prime sign on (-x/2) term?

[silver matrix]

bc I don't know the derivative of that

I think it's -1/2

but I'm not sure

[blissful ridge]

Yes, it's -1/2

[willow bear]

$-\frac{x}{2}$ is just $(-\frac{1}{2})x$ you know

[obsidian monolith]

Ann:

[silver matrix]

well then substitute and that would the answer

[blissful ridge]

First of all show me the original question

[silver matrix]

yeah better to do that

[viscid thistle]

.. You have the answer

[silver matrix]

ok thank you very much

[obsidian monolith]

L'Âne 🍐:

[silver matrix]

👍

[jade heron]

=D

[silver matrix]

Is it correct?

[jade heron]

no

[silver matrix]

well now that I see mathematicians don't like negative exponents

alright :'(

I can show the process to see more in detail where the error

Here we go

[blissful ridge]

I'm not going to calculate that but by just intuition, I can say that there has to be two terms

[obsidian monolith]

L'Âne 🍐:

[jade heron]

Ugly but what I got

[obsidian monolith]

L'Âne 🍐:

[reef jasper]

is my picture correct? i don’t know what the 23 cm means

[rare lava]

the 23 cm is correct

[reef jasper]

ok what am i even looking for

linear or angular speed?

[echo wagon]

What does the question say?

[rare lava]

you are looking for a function for the position of a point in the edge of the wheel

the point starts at the bottom

[echo wagon]

Find an equation in t which models the height above the ground of the point currently at the bottom of the wheel

[reef jasper]

yeah ik but like how do i start

[rare lava]

circular coordinates

[reef jasper]

huh

[rare lava]

put the axis y in terms of the angle of the wheel

sometthing something sin()

[reef jasper]

still can’t do it

[reef jasper]

for half life problems do u have to find the original amount

like this one

nvm i got it tnx guys

[slender river]

yw

[narrow peak]

hi i need help

Express $\frac{x^2-2x-6}{x(x^2-x-6)}$ as a sum of 3 partial fractions.

[obsidian monolith]

Hmm:

[narrow peak]

what does "sum of 3 partial fractions" mean

[harsh smelt]

as sum of simplest fractions

[narrow peak]

im assuming i just do it normally

right

[harsh smelt]

what are factors of x(x^2-x-6)?

[narrow peak]

x-3 and x+2

i did it normally is that correct

[harsh smelt]

so den is x(x-3)(x+2) am i right?

[narrow peak]

$\frac{x^2-2x-6}{x(x^2-x-6)}=\frac{A}{x}+\frac{B}{x-3}+\frac{C}{x+2}$

i did this

[obsidian monolith]

Hmm:

Commander Vimes:

[harsh smelt]

oh ye

[narrow peak]

yep guess im right

[harsh smelt]

that what they want

order does not matter

so yep

[narrow peak]

was just a bit confused about the sum part

[harsh smelt]

this method is called partial fraction decomposition

[narrow peak]

ik

usually they just ask for "express xxx in partial fractions"

[harsh smelt]

xxx

[narrow peak]

;)

[harsh smelt]

18+ partial fractions?

[narrow peak]

what

[harsh smelt]

nothing

[fleet yew]

I honestly dont get the point of partial fraction

Specifically in a precalc class

Because its entirely pointless until you actually have to do integrals

But then again i also think that precalc class itself is kind of pointless

99% filler and recap

Do not recommend

[blissful ridge]

What actually do they cover in pre calc

There is no such distinction in my country, so I'm curious

[lime bolt]

partial fractions are helpful for evaluating series

[echo wagon]

Isn't the whole point of precalc to teach you the noncalculus things you need to know before calculus? Seems like the perfect place for partial fractions

[lime bolt]

also precalculus isnt just preparation for calculus it has some things that are useful for their own right

like trigononemetry for example

[blissful ridge]

Is analytical Geometry (Coordinate geometry) precalculus or post calculus

[echo wagon]

Sure, but the trig is part of what you need to know before calculus. I never claimed you can't use anything you learn outside of calculus

[lime bolt]

anayltic is both(after and before)

[viscid thistle]

I'm actually not sure which part of math this fits into, so i'm going to put it here. it's a challenge problem that I was given, so if u wan't some clarification on the way I wrote it down just tell me

[blissful ridge]

Is that implies or equality?

And what is tx

[viscid thistle]

tx i the number of nths roots

and the x is just the number of terms

it's a sequence

or at least that's how I interpreted it

[blissful ridge]

Does that nested roots equal to x?

If yes,
Then raise the whole equation to the power of n

Divide by a^n on the both sides and then see, if something clicks in your head

[viscid thistle]

no, it's just n terms

it's like each term multiplies the last by another nth root, and fits under each of the last terms

[blissful ridge]

It looks like roots are nested infinite times

[viscid thistle]

yea, but the challenge is to find it for a specific amount of terms

I solved for infinity alreaday