help-23

[indigo sierra]

ik the 10*11 comes from 10 spots then 11 spots

but why the division by 2

ik 2 letters

but why does that make sense

[left gyro]

why did you divide by 2! in 2.6.38?

[indigo sierra]

because it repeats so doesnt matter which one goes where

[left gyro]

would that be different here?

[indigo sierra]

oh is that 2! seeing it like that makes more sense in my head lol

[left gyro]

I did confirm it was 11C2

its the same 2!

[indigo sierra]

indeed, but i got lost on (11c2)

[left gyro]

if you want a more literal way to see 11C2

[indigo sierra]

cause wouldnt that account for lesser spots

like 5 spots for example at the start

11c2 is a bigger number

[left gyro]

if you want a more literal way to see 11C2,
L L I L O S M G N is 9 letters, so adding in 2 more letters would make the full 11 letters
so so choosing 11C2 would mean to choose 2 out of the 11 letters to be U, then filling in the remaining 9 letters with L L I L O S M G N

[left gyro]

if you look closely, the calculation is 4 * 3 * 2 * 1 * 8 * 9 * 10 * 11 / 2
that / 2 can be placed anywhere: 4 * 3 * 2 * 1 * 8 * 9 / 2 * 10 * 11 also works

this would correspond to placing the consnants, then placing the Us first, then the I and the O

4 * 3 * 2 * 1 * 8 * 9 * 10 / 2 * 11 would placing the I, then placing the Us, then placing the O

[indigo sierra]

where would you put (11c2) in that formula

just so i can wrap my brain around what its doing exactly

[left gyro]

???????

[indigo sierra]

i understand the placement math now

but like is it 4! * (11c2)

[left gyro]

no

[left gyro]

didnt you just say you wanted to not see 11C2 because the number wouldnt be there if the vowels were placed in a different order?

11C2 only appears if we place the two Us at the end

11 letter word C 2 Us

[left gyro]

for this for example,

youd have:
4! ways to place the consonants
then 9C2 ways to place the Us
then 10 ways to place the O
then 11 ways to place the I

[indigo sierra]

yeaaaa ok so i had a flawed understanding of what formula 11c2 led to

i thought a lower "k" lead to more numbers being multiplied in the numeral

which is why i was getting tripped up

[left gyro]

you can see depending on the calculation that it could be:
4 * 3 * 2 * 1 * 8 * 9 * 10 * 11 / 2 (I O UU)
4 * 3 * 2 * 1 * 8 * 9 * 10 / 2 * 11 (I UU O)
4 * 3 * 2 * 1 * 8 * 9 / 2 * 10 * 11 (UU I O)

all of these are the same,

[indigo sierra]

so the formula with 11c2 would be 4! * 8 slots * 9 slots * (11c2)

[left gyro]

and the C that appears would be 11C2, 10C2, or 9C2

[left gyro]

yes

[left gyro]

9C2 < 10C2 < 11C2 you can view is balanced by the rest of the numbers being multiplied with it

9C2 * 10 * 11 = 8 * 10C2 * 11 = 8 * 9 * 11C2

again this is both from that theyre calculating the same numbers, and that we expect the number of ways to be the same regardless of which order we place the letters

part of what makes this placing idea so useful

[indigo sierra]

what if you placed the U's first

nvm you already showed that

lol

[left gyro]

twice

[indigo sierra]

yea ok its churning

[left gyro]

this is a good sign, you know

[indigo sierra]

you ever hit a topic you just cant wrap your head around

like you do good on everything

but theres one thing your brain just wont make the connections with

thats me rn

[left gyro]

it means youre instinctively asking questions like sparks that come out regardless of what youre consciously aware of

that means its not just "do I think I understand this," its not a conscious decision

at a point youll know you have a handle on this

if you keep thinking on this process, it might make enough sense for you to try using it

keep in mind this kind of calculation is easy to do for relatively simple problems like this one

it wont pan out for other problems if they dont allow it

[indigo sierra]

makes snese thank you, i think its making sense, as for the other problem i sent, I saw that the solution was to pair the sy together

my question is does this intuitively come to you

like you see this problem and your like yea that y needs to be with that s as one character

[left gyro]

intuitively I get used to the methods I can use which I trust

then when I get used to the methods, that intuition really means as a shortcut to speed up the method

you know the method wont change if I place the Ls first

so I place the Ls first, which inevitably means to place the 3 Ls as one thing

similarly with the S Y

[indigo sierra]

ah true i did mentally place the L's together initially

[left gyro]

you can also view that when rearranging the letters in 2.6.32, the SY are stuck together

the idea of permutations can rearrange the letters into any order
since the SY are stuck, that indicates to treat SY as one letter

either way, its the method that hints that you can do this

you can find intuition in the methods you use, as for the bare problem itself it might not exist

only in the sense of "these methods should work well on this problem, theres nothing else that stops it"

theres many problems out there in math where the only way you solve them is to do them and see what happens

[indigo sierra]

Yea, makes sense,

ima try this one

2.6.47. In how many ways can the letters of the word
BROBDINGNAGIAN
be arranged without changing the order of the vowels?

Similar concept itd seem, lets see what i can do

[left gyro]

gl

[indigo sierra]

So, what I did was I wrote out the vowels, as in
OIAIA
then i wrote out all the consonants and how many dups of each, counted the intial slots between the vowels which was 6, placed a consonant, and incremented, for the duplicates like G since theres 2 i incremented twice and divided by 2!, for N i did 3! and incremented 3 times,
the math came out to be
6 * 7 * 8 * 9 * 10 * 11 * 12 * 13 * 14 /(3!3!2!)
so thats the amount of ways those leters can be dropped between those vowels, which should be the answer?

[left gyro]

it should be 2!2!3! (you miscounted the duplicates), but other than that this is correct

you can also view this in terms of rearranging the letters

[indigo sierra]

bruh lol ofc i miscounted

[left gyro]

there are 14 letters, you are looking to place 14 - 5 = 9 of them
so 14 * 13 * 12 * 11 * 10 * 9 * 8 * 7 * 6 ways to place them (9 numbers multiplied together, starting from 14)
then as before divide by 3! 2! 2! for duplicates

in other words (14P9) / (3! 2! 2!)

[indigo sierra]

how does that keep the vowels intact tho

i guess since your not selecting to place them

thats my biggest qualm with these probs is having things that come out to an answer, by not refrencing it at all

[indigo sierra]

like that keeps the vowels in place, but vowels werent refrenced at all so it just confuses me

there couldve been anything between them cons

[left gyro]

yes?

doesnt mean the order of that anything needs to be mixed

if you wanted to also rearrange the vowels, you would need a 5!

[indigo sierra]

yea it makes sense

[left gyro]

5! / (2! 2!) more specifically

[indigo sierra]

like not in an understanding way, as im still working on that, but i see why

[left gyro]

you can view that theres only one allowed order of the vowels, so instead of a 5! / (2! 2!) theres a 1

also, if the order didnt matter, the math in all would be
1 * 2 * 3 * 4 * 5 * 6 * 7 * 8 * 9 * 10 * 11 * 12 * 13 * 14 / (3!2!2!2!2!)
which is the same as ordering the letters in the whole word

14! / (3! 2! 2! 2! 2!)

[indigo sierra]

yea

alright

getting somewhere

last question for now

i think i understand it

just need a confirmation

so it has to go left 5 times, and has to go right 5 times

theres 10! total combinations it can choose to get down

so you do 10!/5!5!

[left gyro]

thats correct

another way you can view it is that there are 5 "Lefts" and 5 "Rights" strung together into one phrase

so there are 10C5 to place the Lefts

from which the Rights fill in the remaining spaces

10C5 = 10! / (5! 5!)

[indigo sierra]

i thought i had it and lost why im dividing by 5! lol

[left gyro]

you divided by 5! twice, one for the Lefts, one for the Rights

[indigo sierra]

but why does the division coorelate to how many its gonna take to get there

[left gyro]

the division means the Lefts or the Rights can be freely swapped with each other

[indigo sierra]

ah yes cause its just removing the duplicates

[left gyro]

had the Lefts each had a specific color, and you needed how many different strings of colors you coul dmake, itd just be 10!

yep

notice here theres again two different ways to see the 5! 5!

[left gyro]

one is from 5! ways to rearrange the lefts and 5! ways to rearrange the rights

[left gyro]

another is that its just the denominator of the 10C5 that I used for the lefts

10! / (5! 5!) (your way) = 10C5 * 1 (my way)

[indigo sierra]

yea cause you place the lefts and the rights just fill in regardless

[left gyro]

1 being from that once the lefts are placed, theres only one way to fill in the rights, which is to place them in

yep

very good

[indigo sierra]

shi after a 2 months taking this class feel like shits actually clicking

i say this until i see another word problem and im lost again

thank u mtt if i need more help would you be fine with a ping, could offer some mild amounts of $$ aswell

[left gyro]

lol np

[indigo sierra]

.close

[safe radish]

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Use .reopen if this was a mistake.

[indigo sierra]

.reopen

[safe radish]

✅ Original question: #help-23 message

[indigo sierra]

i lied im back bro 2.7.2. An urn contains six chips, numbered 1 through 6.
Two are chosen at random and their numbers are added
together. What is the probability that the resulting sum is
equal to 5?

i dont even know where to begin choosing a situation where they are equal to 5

ik we are choosing 2 coins of the 6 thats an easy (6c2)

and for the probability itll be X/(6C2)

[velvet nebula]

You should be able to figure out which cases are by hand

You can count them on one hand

[indigo sierra]

I could probably do it manually but im wondering what the math would look like for it

[velvet nebula]

Just do it manually

[indigo sierra]

whats the proper method, cause ik 1+4, 2+3, and vice versa so 4/(6C2) ?

[velvet nebula]

You double counted

[indigo sierra]

i thought 1+4 would be different than 4+1 in terms of counting probabilities

[velvet nebula]

When you did 6C2, did it count 1+4 and 4+1 different?

[indigo sierra]

no

so makes sense

yea 2/(6C2)

[velvet nebula]

Yeah, you should make sure your numerator and denominator counts in the same way

[indigo sierra]

would it be possible to count that if it was bigger tho

or was this intended to be hand counted

so doing combinatoric probability seems to be mainly counting a total

and counting favorable outcomes

[velvet nebula]

There may be ways, but you'll probably learn them in due course

[velvet nebula]

This is probably intended to be hand counted

[indigo sierra]

makes sense

so for this one

theres (20C2) total, now we need to count favorable outcomes, but thats alot to count

so what i did is, counted the chips that dont differ by more than 2, i hand counted and theres 19 that differ by 1, so im assuming that also 19 that differ by 2 and 20 that differ by 0, so 58 total that dont differ by more than 2?

so 1-(58/20C2)

well i guess itd be 18 that differ by 2, idk the math but

so itd be 1-(57/20C2)

ah wait

i assumed replacement and that there could be duplicates

so id have to remove the 0 case

so 1-(37/20C2)

[safe radish]

@indigo sierra Has your question been resolved?

[last wren]

that sounds right to me

your most recent answer

[safe radish]

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[north escarp]

Hi. I am in a Calc 1 course, doing some differentiation problems.
Right now, I'm on one that looks like this:

e^(x+1) + 1
Find the derivative.

I know that the derivative of e^x equals itself.
And, I think that the +1 constant would equal zero.
But I don't know what to do with the +1 power.
I have tried entering, simply e^x into the answer box, but that didn't work.
I haven't encountered anything like this yet, and there's nothing that looks like it in the textbook.

[timid escarp]

this requires the chain rule, I think.

but this chain rule would be trivial. have you heard of, or are you allowed to, use the chain rule?

[exotic tundra]

Try e^(x+1) as answer

[austere goblet]

!nosols, might wanna explain how to get there?

[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.

[exotic tundra]

Yes

Well its just they alr had the right idea

That the deriv of e^x equals ifself

They know that the deriv of any constant is 0

[north escarp]

we are allowed.

[timid escarp]

would be a nice opportunity to introduce the chain rule for more complicated problems, but I digress.

[north escarp]

that did work.
though i am also here for the explanation.

[exotic tundra]

Yes

Ok

So do solve with the chain rule

And it should clarify any confusion

[north escarp]

ty

[safe radish]

@north escarp Has your question been resolved?

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[safe radish]

@woven mica Has your question been resolved?

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

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[cloud ermine]

i need help on understanding this one!! i've watched a lot of videos and i really dont get it can someone explain very simply..

[glass egret]

hi

[median vigil]

,rccw

[cloud ermine]

hello

[glass egret]

what

[cloud ermine]

wait..

what

[glass egret]

lol same time

[stoic saddle]

get your own help channel

[cloud ermine]

LOL

[austere goblet]

too late, sorry lol

[median vigil]

sleepy got the channel slightly before you, please open a new one

[glass egret]

mf i tried to

[cloud ermine]

can someone help me with mine.. i just need it to be explained in more dumb friendly terms 😭

[median vigil]

which one in particular?

[cloud ermine]

the whole thing, the whole process is kinda confusing for me i've tried studying the solutions then trying it for myself with more advanced problems and i still get it wrong

is there any other way to like simplify this the book isnt helping at all and theres not much vids tackling abt this subj online

[stoic saddle]

how about you suggest a particular problem so that we can help you better

[cloud ermine]

ok im kinda confused when and what formula to use when finding the derivatives

[stoic saddle]

ok lets start small

how would you find the third derivative of x^5

first, tell me what you know by third derivative

[cloud ermine]

i dont know.. do u differentiate it for like 3 times

[stoic saddle]

yes

third derivative means, that you differentiate the function thrice

ok, whats the first derivative of x^5

[cloud ermine]

5x?

[stoic saddle]

5x^4 right?

[cloud ermine]

5x^4 i mean

[stoic saddle]

nice

[cloud ermine]

cuz n-1 or something

[stoic saddle]

yes, take the power to the front then subtract one

so whats the second derivative

[lean otter]

(x^n)' = nx^{n-1}

{nx^(n-1)}' = n * (n-1) * x^(n-2)

[cloud ermine]

what formula should i use to derive

5x^4?

[stoic saddle]

ok how would you find the derivative of x^4

[glass egret]

4x^3

[stoic saddle]

bruh i didnt ask you

[glass egret]

i thought you were in my channel

[cloud ermine]

is it 20

20^4

[stoic saddle]

no

ok before repeated derivatives, you need to know how normal derivatives work

to find the derivative of 5x^4

[cloud ermine]

ok but how

[glass egret]

nx^n-1

[stoic saddle]

$\frac{d}{dx} 5 x^4 = 5 \frac{d}{dx} x^4 = 5 \cdot 4 x^3$

[flat frigate]

CherryMan

[stoic saddle]

we can take constants out right?

[cloud ermine]

yes

[stoic saddle]

is there anything that doesnt make sense here

[cloud ermine]

well.. all of them actually

so like do u just disregard whats in the middle

[stoic saddle]

right

so if we are finding the derivative of something

that looks like k f(x)

where k is any constant like 2,3 or 5 in our case

and f(x) is any function like x^4 in our case

then d/dx kf(x) = k d/dx f(x)

which means we can take constants out

[cloud ermine]

and then u multiply them?

[stoic saddle]

multiply what?

[cloud ermine]

the 5 and 4x^3?

[wicked root]

yes

[stoic saddle]

yes

so now we have 20x^3

how would you find the next derivative?

[cloud ermine]

is it 60x^2

[stoic saddle]

yes

[cloud ermine]

i just dis the same thing above

[stoic saddle]

so we get that the third derivative of x^5 is 60x^2

[cloud ermine]

yes

[stoic saddle]

say i have something more general

like x^n (where n is some natural number)

how would you find the second derivative of x^n

[cloud ermine]

i only get nx

but what abt the exponent what should i do with that

[stoic saddle]

tell me how you would find the first derivative

[cloud ermine]

well i just went straight to multiplying both x^n

and thats why i got nx

^n-1 but idk if i should include that

[stoic saddle]

the power rule says we bring the power to the front then subtract one from the power

so the first derivative is nx^(n-1)

following the power rule again

[cloud ermine]

ohhh

[stoic saddle]

what do we get?

[cloud ermine]

wait what do u mean

[stoic saddle]

whats the derivative of nx^(n-1) ? (remember i asked for second derivative)

[cloud ermine]

ok wait

ok i dont get it

[stoic saddle]

what do you not get

[cloud ermine]

bc i only got n^n-1

i multiplied n and x^n-1

but what about the exponent?

the n-1?

[stoic saddle]

lets look at it

nx^n-1

we have too parts

[cloud ermine]

ok sorry

[stoic saddle]

*two

[cloud ermine]

yes

[stoic saddle]

the constant n

and the x^(n-1)

the constant, as i said earlier

can be left alone

that is, d/dx nx^n-1 = n times d/dx x^n-1

as for the x^n-1

lets apply the power rule

power rule says "bring the power to the front and then subtract one"

so we bring n-1 to the front

[cloud ermine]

so its n-1 x?

[stoic saddle]

and subtract one from n-1 to get n-2

so it will be n-1 x^n-2

and bringing the constant from earlier

the total derivative is $n(n-1)x^{n-2}$

[flat frigate]

CherryMan

[stoic saddle]

does this make sense

[cloud ermine]

yes it makes sense

[stoic saddle]

ok, just to test your understanding

how would we take the next derivative

(the third derivative)

[cloud ermine]

wait ill try

is it n(n-3)x^n-4

or n(n-2)x^n-4?

[stoic saddle]

explain pls

explain your thought process

[cloud ermine]

what i did is i applied the power rule and u said that we bring it to the front and subtract one

[stoic saddle]

yes

[cloud ermine]

and i did that so i got n-3

and the exponent changes?

[stoic saddle]

my mistake

i understand your confusion

[cloud ermine]

so i assumed its n-4

[stoic saddle]

the complete rule is that: "bring the power to the front and subtract one from the power of x"

like if we have x^5

we bring 5 to the front

and subtract one from the power of x to get x^4

so all in all we get 5x^4

[cloud ermine]

ok i got it

so my answer is wrong?

[stoic saddle]

so know what do you think the derivatigve o this is

[stoic saddle]

you shoukld try once more

[cloud ermine]

ok wait

is it n (n-1)x^n-3

[stoic saddle]

close!

n(n-1)(n-2)x^n-3

but why?

[cloud ermine]

why is there n-1 and n-2

[stoic saddle]

so

derivative of n(n-1)x^n-2

we have two parts

the constant n(n-1)

and x^n-2

as i said earlier we can take the constant out

as for x^n-2

we will have n-2 x ^n-3

because i took the power to the front and subtracted one from the power of x

[cloud ermine]

ohhh

[stoic saddle]

putting these two together i have n(n-1)(n-2)x^n-3

[cloud ermine]

ok i got it now that was so very confusing but it makes sense

[stoic saddle]

how would i take the next derivative?

(just to test you one last time)

[cloud ermine]

ok wait

n(n-1)(n-2)(n-3)x^n-4?

[stoic saddle]

yes!!

[cloud ermine]

i think i kinda get it now when deriving

do i have to apply power rule in every problem?

[stoic saddle]

only problems where it is applicable

which means derivative of things that look like a * x^b

now coming back

[stoic saddle]

what is the derivative of ln x

[cloud ermine]

do i use the same process?

like when i did with the previous one

[stoic saddle]

like the power rule?

[cloud ermine]

yeah how do i derive it if its judt ln(x) is there a diff formula for that

[stoic saddle]

firstly

ln x does not look like a x^b

so we cant apply the power rule

secondly, the derivative is a little involved

do you know the derivative of e^x

[cloud ermine]

all i know is that u get the same thing when deriving e^x?

[stoic saddle]

yes

derivative of e^x is itself

[cloud ermine]

like ut goes back but idk much in depth abt that topic

[stoic saddle]

now if you wish i could go on a tangent and prove that the derivative of ln x is 1/x

[cloud ermine]

ok pls do explain

[stoic saddle]

right

so we have $y = \ln x$

[flat frigate]

CherryMan

[stoic saddle]

by the definition this means that $e^y = x$

[flat frigate]

CherryMan

[stoic saddle]

now we can take the derivative with respect to y of both sides

so $\frac{d}{dy} e^y = \frac{d}{dy} x$

[flat frigate]

CherryMan

[stoic saddle]

now the derivative of e^y with respect to y is just e^y

so $e^y = \frac{dx}{dy}$

[flat frigate]

CherryMan

[stoic saddle]

are you following so far?

[cloud ermine]

yes

[stoic saddle]

sure

now recall that e^y =x

so $x = \frac{dx}{dy}$

[flat frigate]

CherryMan

[stoic saddle]

which is equivalent to saying $\frac{dy}{dx} = \frac{1}{x}$

[flat frigate]

CherryMan

[stoic saddle]

so the derivative of ln x is 1/x

pls tell me if any part feels contrived or doesnt make sense

[verbal vector]

wouldnt u differentiate with respect to x and use the chain rule?

y'e^y=1

y'=1/e^y

[stoic saddle]

sure, both ways are valid

[verbal vector]

e^y was x

so y'=1/x

can u explain d/dy(x)=dx/dy?

i hate this notation

[cloud ermine]

i get it but its confusing since it does go back and forth kinda

[stoic saddle]

derivative of x with respect to y is dx/dy

yeah i guess its kind of abuse of notation

[verbal vector]

this notation is trash

[stoic saddle]

right so the derivative of ln x is 1/x, then whats the second derivative

[verbal vector]

do ' or if its a partial derivative use a subscript

[cloud ermine]

1x^2?

1/x^2

[stoic saddle]

maybe its easier if i write it as x^-1

[cloud ermine]

so its negative

-1/x^2?

[stoic saddle]

yes

correct

and the next one?

[cloud ermine]

-2/x^3?

or sctually without -

[stoic saddle]

yes

2/x^3

[stoic saddle]

now do you know why the formula here works

[cloud ermine]

yes i suppose it does make sense now

[stoic saddle]

how would you do this question now

[cloud ermine]

is it necessary to derive it for the fourth time?

bc in the book they did so im not sure if its always required to do so

[stoic saddle]

meaning?

[safe radish]

@cloud ermine Has your question been resolved?

[safe radish]

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[swift sun]

Is my proof alright

?

[cedar widget]

,rotate

[swift sun]

To prove
1- Two adjacent sequences converge

2- Converge to the same limit let it l

[safe radish]

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[elder cypress]

I often come across second order differential equations like this one when solving physics problems. Second order differential equations are not a part of our syllabus, but we are told to create an analogy to harmonic motion (x'' = -ω²x) to solve these equations.

I find this method inconvenient. Is there a simpler way to solve these equations

[elder cypress]

In general, Im talking about equations of the form-

$\frac{d^2x}{dt^2} = a(x - x_0)$

[flat frigate]

Dhruv

[primal blaze]

i dont get how it is inconvenient, can u explain how u have been doing it?

[elder cypress]

If we take the equation in i as an example,

I would've made an analogy to a harmonic motion where current corresponds to displacement
di/dt <-> velocity
1 <-> mean position
100 = ω²
and so on

And then I would apply formulae related to harmonic motion

[primal blaze]

ohh u only need to know the general solution.
For x'' = -w^2 x it is x = Asin(wt) + Bcos(wt)

so for the form u sent, just take (x-x0) as some y and w = root(-a), it will be converted into this form and u get the solution by putting those in

[elder cypress]

Thank you

.close

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[midnight fern]

I need help with these Integrals

[low echo]

Something I actually know how to do in this server?

[midnight fern]

I tried replacing the denominator with t then doing the whole dx =dt/2x + 2 but i dont think that works

Or i dont know how to make it work

[zealous ingot]

first one?

[midnight fern]

Yes!

[zealous ingot]

have you heard of partial fractions?

[midnight fern]

I dont think so.

[upper mulch]

Right path, now think what you can do with the numerator 3x+5 so that you can make it to two fractions.

[low echo]

I'll try

[midnight fern]

Mm Should I make it the 2x + 2 appear in there? Something like 3/2(2x + 2) + 2?

[zealous ingot]

$\frac{x+c}{(x+a)(a+b)} = \frac{A}{x+a} + \frac{B}{x+b}$

[low echo]

Which part are you talking about

[flat frigate]

Donkey

[low echo]

How'd you talk like that

[upper mulch]

Exactly, you're trying to get 2x+2 in numerator from 3x+5.

[zealous ingot]

Texit

[midnight fern]

I’ve never seen that tbh so I dont think that’s how my teachers want me to solve it

[low echo]

How

[upper mulch]

You don't need that.

[zealous ingot]

#latex-help message

[low echo]

brb

[midnight fern]

So then I could use the ln rule and make Integral 3/2 * 2x + 2 /x^2+2x-3 into 3/2 * ln (x^2+2x-3)? But then how do I solve 2/(x^2 + 2x - 3)?

[upper mulch]

No, you want to like make 2x to x+x (2x = x + x).

[midnight fern]

Now I am a bit lost

[upper mulch]

Say 3x, how would you decompose it be a sum of 2x and something else?

||3x=2x+x||

[midnight fern]

Ohhhh ok got it and then I do the same with 5? As in 3 + 2?

[upper mulch]

Now it's regular integral $$\int \frac{2x+2}{x^2+2x-3}+\frac{x+3}{x^2+2x-3}dx$$

[flat frigate]

Good

[midnight fern]

OHHHH got it! Tysm!

[brittle estuary]

hlo guys

[midnight fern]

I am now gonna keep doing problems, should I close the channel and just reopen it when I need help again?

[zealous ingot]

yeah

[midnight fern]

.close

[safe radish]

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[brittle estuary]

hlo

[safe radish]

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[hoary seal]

question regarding AP/GP

do we only need to calculae the difference/ratio of two terms to conclude its a GP/AP?

[hoary seal]

or certain ammount of terms

[lean otter]

two is enough

[hoary seal]

oh ok

[lean otter]

but

if you want to double check and be sure u can do with 2 pairs

like a2-a1 and a4-a3

for ap

if a2-a1=a4-a3 then its ratio

[thin bridge]

you'd need every single one
or deal with general terms to conclude whether you have a GP or AP

but if you're told for certain what type you have
you can use consecutive terms to determine the ratio or difference

1,2,3,6, 13 , 42
you can't just say that there's a difference of 1 between t_1 and t_2
same with t_2 and t_3
and say that this is an AP

[hoary seal]

wdym?

[thin bridge]

similar issue when considering if something is in GP

[hoary seal]

I think im confused when u said what type I have

[thin bridge]

e.g. if you are told that 2,3,4 are consecutive terms in an AP
and you can say the common difference of that is 1

but if you're given a sequence, you'd need to determine every ratio or difference (or until you find something different)
to determine whether the sequence is in AP or GP

e.g. if you're given the sequence (and told nothing else about it)
1,2,3,4,5,6,13,42
to determine if its in AP you'd want all the differences (or until you spot something different)
1,1,1,1,1,7,29

[hoary seal]

ok so we dont only find the d/r of the first two terms

we need to do it for the whole sequence given

[thin bridge]

in other words,
given a random sequence
you can't just randomly take two consecutive terms and say stuff
that this is an AP with difference of 2
and this is a GP with ratio of 4

[hoary seal]

what if theres mulitple terms

we wouuld need to do it for all?

[thin bridge]

yes

[hoary seal]

what about when the quesiton tells us if its a GP/AP

[thin bridge]

but if you're told for certain what type you have (AP or GP)
you can use consecutive terms to determine the ratio or difference

[unique hawk]

Can some one alive and explain it in easy ways these question !

[thin bridge]

!occupied

[safe radish]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[hoary seal]

ok so just use 2 terms if they tell us if its a GP/AP

to find d/r

[thin bridge]

consecutive terms yes

or adjust the calculations accordingly

[hoary seal]

I see

my other question is for the sum, how can we tell if its a GP/AP sum

[thin bridge]

consider writing out the firsts few terms
but don't expand/simplify

and it should be clearer

e.g.
$$2^1, 2^2, 2^3, \dots$$

[flat frigate]

ραμOmeganato5

[thin bridge]

what is happening when the power is increasing by 1

[hoary seal]

its exponentially increasing?

[thin bridge]

yes

[hoary seal]

so that is

GP

[thin bridge]

yes

the terms will be in GP

[hoary seal]

what about for more complicated ones

do they all follow this same rule?

[thin bridge]

the ones in Q2?

try the same approach

write out the terms and calculate the
difference / ratios
if you feel the need

[hoary seal]

once we know if its GP/AP, we dont need to calculate d/r for each terms right?

just the consecutive terms

[thin bridge]

yes

[hoary seal]

also for this one, shoudn't it be + between the terms

since its sigma notation

not a sequence

[hoary seal]

also I think we could tell if its GP by observation

[safe radish]

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[dire trout]

Good day, I'm trying to do number 1 and I know, the method-ish.

[dire trout]

But I cannot figure out how to impliment it properly.

[last wren]

to find a particular solution to a second order ODE like this one, the easiest way by far is to assume y looks like the right hand side and its derivatives

in this case, since the RHS is just sin(3x), y should look like Asin(3x) + Bcos(3x)

then you plug that in for y and solve for A and B. that's your particular solution

[dire trout]

I'm a bit lost. Is this the same method the slides are showing me?

Okay so it is. We assume RHS is just sin(3x). Wouldn't it still be weird since we don't have y'?

gonna put it on paper to see it first. One sec.

[lone void]

isnt the first one a linear differential equation

[dire trout]

2nd order differential non-homogenous equation.

[lone void]

oh alright sorry then thought i could help out

[dire trout]

I'm lost on what to do here.

[last wren]

look in the method of undetermined coefficients at the bottom here.

your b is 0, your c is 9, and g(x) = sin(3x)

so since g(x) involves sin, your guess (the particular solution) should involve sin and cos. in other words,

y_p = Asin(3x) + Bcos(3x)

[dire trout]

Ok so y'p = Asin(3x) + Bcos(3x) and y"p = Uhhh.. -Acos(3x) + Bsin(3x)

Wait no.

y'p = Acos (3x) + Bsin(3x)

y''p = -Asin(3x) - Bcos(3x)

[last wren]

you have the right idea

y_p = Asin(3x) + Bcos(3x)
y_p' = Acos(3x)(3) - Bsin(3x)(3)

then try finding y_p''

(you forgot the chain rule)

[dire trout]

So since y'p = 0

Then 3Acos(3x) - 3Bsin(3x) = 0?

[dire trout]

Would the b=0 affect it? Or just look at y'p and y''p?

[last wren]

b doesn't have anything to do with this at all. the only reason I mentioned it is so that you would realize this ODE is the correct form to use undetermined coefficients

notice that in the ODE, y' doesn't appear at all. what you need to do is find yp'' and plug that and yp into the ODE

then solve for A and B

[dire trout]

So,,, uhhh... -9Asin(3x)+9Bcos(3x) + Asin(3x) + Bcos(3x) = sin3x

So... -8Asin(3x) + 10B cos(3x) = sin(3x)

[last wren]

super close, should be -9Bcos(3x) instead of positive

[dire trout]

Oh.... So.... -8ASin(3x) - 8Bcos(3x) = sin3x?

[last wren]

yep! do you know where to go from there

[dire trout]

Uhhh.....-8A - 8B = sin3x?

[last wren]

not quite but worth a guess, what you need to do now is find A and B that make this equation true

for example, the right hand side has no cosines, but the left does. what does that tell you about B?

what does it have to be

[dire trout]

0?

[last wren]

exactly

[dire trout]

B=0 and A = to -1/8?

[last wren]

correct

so plug those back into your formula for yp and you're done

[dire trout]

Thus the solution yp = -1/8sin(3x)

[last wren]

there you go

[dire trout]

Huh... thanks. Now to read the hand out for number 2.

[last wren]

np, variation of parameters is a bit gross so have fun lol

[dire trout]

That does not give me hope. xD Thanks again.

.close

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[burnt pewter]

I'm trying to solve a volumes of revolution problem. I have to find the volume of the region which is enclosed by the functions y = 2x^(1/2) and y = x, when y = 2x^(1/2) is rotated about the line y = x.
I tried to do this by finding a function that would tell me the radius at a certain value of x, and then using the integration formula for the volume of revolution. On my graph, the green function represents the distance between A and B (the radius) and it takes the x value of A, as A moves along the red curve. This green function should be my radius function and when I use it with the integration formula for the volume of revolution I get an answer of (16/15)pi, however the answer is (16/15)(root2)pi. Where did I go wrong?

[safe radish]

@burnt pewter Has your question been resolved?

[safe radish]

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[thorny isle]

since x-->0, i inserted x as 0, so on bottom, we have 2x^2, which is 2 * 0 * 0, which means divide by 0, you can't divide by 0, so the answer should be does not exist (or that's what i thought_

[timid escarp]

not so fast. you can, and probably should, try to manipulate the expression given. maybe try dividing by something with x in it?

[open wedge]

Probably you should simplify first

[thorny isle]

oooh, divide by x^2

TY

2/4, = 1/2

so answer is 1/2, right?

[empty gyro]

Evaluating limits and simply plugging in the number are two different things

[timid escarp]

2/4?

[open wedge]

No

[timid escarp]

what do you get after dividing by x^2?

[thorny isle]

ooh, that's x^3 on top, not another x^2

this is y i need more sleep

3/2

[timid escarp]

and that would be right.

[thorny isle]

Thank y'all 1 million

[timid escarp]

also, quick reminder about limits.

[thorny isle]

yeah?

[timid escarp]

with limits, we don't care about what happens at exactly that point.
we only care about what happens around that point, a concept known as the deleted/punctured neighbourhood of that point (you don't have to remember the term, but the concept would be enough).

[thorny isle]

so we care about the numbers really close to x=0, but not quite?

[timid escarp]

yes!

so if substitution doesn't work, sometimes it may help to look at what happens to the expression as we let x tend to the limit.

(from both sides, btw.)

[thorny isle]

thankns for the help!

[timid escarp]

glad to help!

[thorny isle]

.close

[safe radish]

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[gray verge]

I found an exercise which goal was to fidn to the derivative of verious functions, the first one was easy, but what do I do about this: $2x * (x-1)^{-1}$?

[gray verge]

opps

[flat frigate]

ℕ∈ℝD ALERT: Gonçalo Gonçalves

[gray verge]

there

I've tried to do it

[lone void]

quotient rule

[gray verge]

yeah, but in teory it should work with the power rule, and it was wrong

it gave me

oh no wait

too many exercises, yeah I did use the product rule on that one

but it still gae wrong I believe

at least it's highlighted

let me double check

yup

no

it's wrong

anywh

I'll show the calculations

if you cannot read say

(maybe it's a simplication issue)

[supple radish]

man

unreadable

atleast for me

[gray verge]

ok

a sec

I can read this

I'll type it out

[supple radish]

ill try to see wait

[gray verge]

zoom in

[supple radish]

maybe u did a calculation error?

[gray verge]

possible

but

I checked for that

let me see

[supple radish]

i dont see one tho

[gray verge]

I don't think so

what it gave me was

$$\frac''$$

[flat frigate]

ℕ∈ℝD ALERT: Gonçalo Gonçalves
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[supple radish]

whats the answer

in the exercise u doing

[gray verge]

give me a second, I can-t multitakes

[supple radish]

lmao yea no issue

[gray verge]

$$\frac{2}{x-1}+\frac{2x}{- x^2 + 2x - 1}$$

wait

[supple radish]

hey maybe u are missing a minus

in the first step of calculatoin

[gray verge]

I am trying to type but my keyboard isn't helping

[supple radish]

where u applied product rule

check it

[gray verge]

there we go

now¡

I'll do all that

I HATE my keyvoard

[supple radish]

oh

u kept it below the minus

nvm

[gray verge]

let me type it...

[supple radish]

i think

it will +2x

u wrote +2x too in the original photo which is correct

[gray verge]

oh wait

[supple radish]

okk

[flat frigate]

ℕ∈ℝD ALERT: Gonçalo Gonçalves

[supple radish]

yea

[gray verge]

I do really hate my kybord believ me bro

[supple radish]

lol

[gray verge]

for the right answer

maybe you're right

it's a calculation error

[supple radish]

whats the answer

man if uare gonna troll go away

<@&268886789983436800>

ignore these guys

[gray verge]

I am, I got worse in my neighbourhood

$$- \frac{2}{x^2 - 2x + 1}$$

[flat frigate]

ℕ∈ℝD ALERT: Gonçalo Gonçalves

[gray verge]

this is the result

the right one

[empty gyro]

<@&268886789983436800> spam here too

[supple radish]

alright

hm

[gray verge]

you're right

probably a sign error

or something

calculation error

but

half of the thing just disapears, why?

[supple radish]

i believe u are getting the same answer...

[gray verge]

oh, its not simplified is that it?

[supple radish]

yeah..

take lcm

u will see it

[gray verge]

how do I simplify it?

[supple radish]

just take lcm

[gray verge]

lcm??

[supple radish]

least common multiple

like how u

simply polynomial fractions

have u doen it before?

[gray verge]

not sure

[supple radish]

ill show u

u will recall

[gray verge]

by least common multiple give an example

[supple radish]

like

1/2 + 1/3

u make it

3/6 + 2/6

[gray verge]

oh

[supple radish]

making the denominators same

[gray verge]

gotcha

[supple radish]

then adding

making like fractions

thats done by taking lcm

[gray verge]

I didn't know that was the name

[supple radish]

no issue

[gray verge]

but, how do I find it

[supple radish]

see

one fraction has (x-1)

as a factor

and the other has -(x-1)^2 as a factor

in denominator

u do the same thing just like u do with numbers

multiply numerator and denominator of first fraction with the same polynomial

so that u have same denominator both sides

*both fractions

do u want me to show it to u?

[gray verge]

gotcha, ke me try

[supple radish]

alrirght

[gray verge]

nono

I'll do it

[supple radish]

yes nice

[gray verge]

alright, let me check

done

so

it's correct

that was it

thanks

@supple radish

I'll practice with the other exercises I missed

[supple radish]

alright

np

[gray verge]

.close

[safe radish]

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[safe radish]

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[real arch]

Could someone give me some tips solving this trig. I need to express x and y by trig functions though I am confusing myself on which needs beta or theta

X and Y are perpendicular to line h

[safe radish]

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as this action is irreversible, and this channel may abruptly lock.

[real arch]

.open

Shoot

.close

hm

[median vigil]

@real arch open a new channel and don't delete your first message there

[safe radish]

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[split elm]

In a circle with center O, EA and EB are secants, and CD is a chord. If AB=AE and BE=12, find the length of the chord CD.

[flint vessel]

6??

it's a little hard work but i mainly used the sin law , congruence and asp of the triangle

[split elm]

can you explain bit more please

[flint vessel]

hmm first join A with D

[split elm]

okay

[flint vessel]

angle ADB would be 90

[split elm]

oh that makes sense

[flint vessel]

and AEB and ABE are given equal

[split elm]

yeah

[flint vessel]

use the RHS congruence criteria to get ED = DB = BE/2 = 6

[flint vessel]

good till here ??

[split elm]

yeah wait a sec please

oh i understand it now

[split elm]

ED=DB cause one same angle and one same side

[flint vessel]

and one right angle

angle ADB and ADE since EB is a straight line

so if one is 90 the other should be too

[split elm]

yeah thats what i meant

[flint vessel]

ok now keep ts saved

[split elm]

im drawing this rn dont worry

[flint vessel]

and draw a new figure with B and C joined and A and D joined

done ?

[split elm]

yea

[flint vessel]

now ACB should also be 90

[split elm]

yes

[flint vessel]

take the meeting point as M

the diagnols

[split elm]

done

[flint vessel]

now try to find angles MCD and CDM

note that opposite angles of a cyclic quadrilateral are supplementary

[split elm]

okay

[flint vessel]

what did you get

[split elm]

i only get that they are same angke

angle

[flint vessel]

not rly they shoudnt be same

[split elm]

oh...

[flint vessel]

okay lets do it step wise

[split elm]

okay

[flint vessel]

lets calculate MCD first

[flint vessel]

whats ACD by this method

[split elm]

sorry to ask but i dont know what these two words mean

im not good at english

[flint vessel]

the sum of opposite angles of a quadrilateral having all it's vertices on a circle

is 180

quadrilateral is a 4 sided figure

[split elm]

i understand

[flint vessel]

so taking angle ABE as some variable lets say @ find angle ACD

[flint vessel]

what did you get

[split elm]

180 - @

no

90-@ ?

[flint vessel]

yes

[flint vessel]

nope

[split elm]

oh you said ACD

[flint vessel]

yes

[split elm]

MCD is 90 - @

[flint vessel]

good

now look at triangle AEB

[split elm]

okay

[flint vessel]

can you find EAB

knowing AE=AB

in terms of @

[split elm]

180-2@

?

[flint vessel]

yes

now whats CDB

[split elm]

hmm

if CAB +CDB =180

[flint vessel]

yes

[split elm]

CDB=180-CAB

[flint vessel]

yes

so what do you get

[split elm]

2@

[flint vessel]

yes

now find CDA

[split elm]

90-2@

[flint vessel]

nope

[split elm]

oh wait

2@-90

[flint vessel]

yes

now look at triangle CMD

we have two angles whats the third

[split elm]

180- other two

[flint vessel]

yes what do you get

[split elm]

180-@

[flint vessel]

good

now find BMD

after that we just have some trignomatry left

[split elm]

umm

[flint vessel]

mb did a typo there

[split elm]

oh

its okay

[split elm]

soooo

how do i continue

[flint vessel]

find BMD

[split elm]

@

[flint vessel]

good

now how much do you know abt trignomatry

[split elm]

i am at 11th grade

so i know a little but im not that good

[flint vessel]

ohh wait

no need for that

[split elm]

oh

really

[flint vessel]

check ts

the circle angles are the same

so the sides opposite to them should be equal

[split elm]

OOOH

[flint vessel]

so CD=BD=(1/2)BE = 6

[split elm]

i understand

omg

[flint vessel]

me as well

[split elm]

damn

[flint vessel]

ye thats why i love maths

[split elm]

im dumb af

[split elm]

i love math too but sometimes its just hard

[flint vessel]

i just saw that my orignal method by sine law was wrong btw

so god saved you here

[split elm]

thank god then

[flint vessel]

you fine geomatry just confusing

[split elm]

btw can i ask you a question

[flint vessel]

yes

[split elm]

its kinda personal tho

where are you from

[flint vessel]

Asia

India specificaly

[split elm]

and is 11grade last grade in your country

[split elm]

same

[flint vessel]

no

[split elm]

in Azerbaijan 11th grade is schools last year

[flint vessel]

I am pretty sure we should do ts in our dms now ts not allowed in help chanels

[split elm]

oh mb

[flint vessel]

.close

do that

[split elm]

yea ik

.close

[safe radish]

Channel closed

Closed by @split elm

Use .reopen if this was a mistake.

[flint vessel]

nice

[safe radish]

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[fair goblet]

bro how do i evaluate this

[fair goblet]

is there an easy way

to do this

brb

[fathom jewel]

maybe you can factor it or complete the square

,w factor 25cos^2(t)-30cos(t)cos(5t)+5cos^2(5t)

[flat frigate]

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[fathom jewel]

bruh

i would try some shenanigans with trig identities and formulas

experiment

[brisk shard]

integral of middle term will be 0 after you apply defactorization formula

[quasi bison]

factorizing is gonna do no good

however

[fathom jewel]

i realized that soon enough

[quasi bison]

you can pull these things apart

[brisk shard]

integral of all cosine terms will be 0 infact

[quasi bison]

$25 \int_0^{2\pi} \cos^2(t)\dd{t} - 30 \int_0^{2\pi} \cos(t)\cos(5t) \dd{t} + 5 \int_0^{2\pi} \cos^2(5t) \dd{t}$

[flat frigate]

Ann

[quasi bison]

and then you can apply half angle formula on the first and third integrals as well as prod->sum formula on the second

in any case $\int_0^{2\pi} \cos(nx)\dd{x} = 0$ for $n \in \bZ \setminus {0}$

[flat frigate]

Ann

[brisk shard]

thats what i said

[fair goblet]

Oh ok

I will try

[quasi bison]

you said it in a way that can (and prob will) be misinterpreted @brisk shard

listening to you it sounds like the integral of cos^2(t) should be 0 as well for some reason

which it very obviously is not

[brisk shard]

mhmm my bad

[safe radish]

@fair goblet Has your question been resolved?

[safe radish]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[safe radish]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.