help-0

[cold pumice]

Wait

[sterile hill]

Mhm plug in infinity

And then simplify

[cold pumice]

Why is this legal? I mean dividing each number inside the root

[vale wigeon]

$\sqrt{ab} = \sqrt{a} \sqrt{b}$

[ocean seal]

Ann

[sterile hill]

You’re force factoring^

[vale wigeon]

and since x is positive, 1/x = sqrt(1/x^2)

[cold pumice]

But you have a-b here

[cold pumice]

Yea from here it's easy I just wanna understand why it works

[sterile hill]

👍

[cold pumice]

@vale wigeon Why this is not what you get

[vale wigeon]

$\frac{1}{x} \sqrt{x^2 - 3x} = \sqrt{\frac{x^2-3x}{x^2}}$

[ocean seal]

Ann

[vale wigeon]

thats what im doing here

[sterile hill]

Force factor an x^2 for this

[cold pumice]

Oh wait

[sterile hill]

Square root of x^2 = |x|

[sterile hill]

Wait what is the limit here?

[cold pumice]

?

[cold pumice]

I very much prefer not getting absolute values in here

So the limit I get is -4 right?

[sterile hill]

What is limit approaching tho?

[vale wigeon]

x goes to +∞

[sterile hill]

It should be one here I believe

[sterile hill]

👍 ^

[cold pumice]

Ok wow thanks @vale wigeon !
A very neat trick, I will try to remember it

.close

[lone heart]

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[cold pumice]

.reopen

[lone heart]

✅

[cold pumice]

But wait, when calculating the same limit for -infinity, you can't do that, because 1/x will be negative, no?

[vale wigeon]

indeed, in that case 1/x will be **-**sqrt(1/x^2).

[cold pumice]

righttttt

Sorry my brain is not braining

Again tysm

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[timber cobalt]

I'm trying to find out the formula for the volume of a cone.

If we consider all the right triangles (with their heights being the height of the cone) constituting a cone, we might see that, just like it is with all the right triangles, they take up exactly half the area of the rectangle with the same height and base length.

So, obviously, the cone has to have half the volume of the cylinder with the same base radius and height. To make it more obvious, we might replace the missing half of the rectangle by the triangle located symmetrically to the considered triangle relative to the height of the cone. Then we will visually have exactly half of the cylinder.

But it's a known fact that the volume of a cone is exactly a third of the volume of a cylinder with the same height and base radius. What mistakes did I make while trying to get the formula myself?

[vale wigeon]

well intuitively

if you think about triangles all attached to the cone's axis, it is like they cover space "denser" the closer you get to the axis

like imagine them as made of paper or something

they're closer together near the center than further out

[timber cobalt]

Yeah, tho I am not sure it's applicable in math

[vale wigeon]

it's my attempt at explaining intuitively why the area ratio does not translate into the same volume ratio

[timber cobalt]

I understand your intention but am thinking over the explanation atm

[vale wigeon]

also for the record, i don't know of any good visual that'd illustrate how 3 cones come together into a cylinder.

[timber cobalt]

Why are you saying it?

Oh, I got it

[timber cobalt]

Are you sure it's the reason?

[vale wigeon]

reasonably so.

[timber cobalt]

Just sending a message so the bot doesn't think I'm away.

So however many triangles we take we won't have as many triangles / unit of space on the outskirts

So if we take the space at the center as the required planes density for counting space as volume, we won't be able to call anything farther from the center volume because it doesn't have sufficient planes density

Do you like my explanation?

Tho I think such a miscalculation should make volume even lower than 1/3. Why did it suddenly become 1/2?

I guess it's just because the method of calculation is inappropriate. For example we can take the outskirts as density standard. Also in most places there won't be space and just close triangles or the planes density will be too high depending on what density standard we choose.

I appreciate your help but I wouldn't be able to understand the answer without this page https://math.stackexchange.com/questions/4632714/why-is-the-volume-of-a-cone-not-half-the-volume-of-the-cylinder-of-which-it-is-a and if I didn't recently teach a person what area and volume are

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[void torrent]

Hey, can someone tell me how to use a surface integral to get the volume of a cone?

[lone heart]

@void torrent Has your question been resolved?

[lone heart]

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[feral blade]

in terms of the family of lines, (k-method), why is k undefined when m = 1/2

[vale wigeon]

"k-method"?

you may want to enlighten us on that.

[feral blade]

family equation using k-method straight lines, the q for the proof says, prove that the equation of any line through the point of intersection P of two lines l1 : a1x+b1y+c1=0 and l2: a2x+b2y+c2=0 has the form l : (a1x+b1y+c1)+k(a2x+b2y+c2)=0 where k is a constant

<@&286206848099549185>

[vale wigeon]

don't quite see where m=1/2 would pop up here honestly

[feral blade]

idk, it just says its undefined so i was just curious as to why

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[remote pelican]

I am now in

then how? hellpp

[unkempt vapor]

squares difference

[remote pelican]

how

[subtle birch]

1=1^2

a^2-b^2=(a+b)(a-b)

[unkempt vapor]

(20092007-1)(20092007+1)+(20092009-1)(20092009+1)

there are 2 20092008

[subtle birch]

well, don't just give out the solution

[unkempt vapor]

which cancel out

[remote pelican]

the 20092008 cancel out ?

[unkempt vapor]

yes

try it

[remote pelican]

wdym by this

i dont really understand

[unkempt vapor]

that's an equation

which works

[remote pelican]

ya what is a and b arent they both same?

[tardy tapir]

a and b are arbitary constants

[unkempt vapor]

a and b are just notations for numbers

[tardy tapir]

they can take any values

[unkempt vapor]

in this case 20092007 and 1

[remote pelican]

ohhhh!

I know alr

but

[unkempt vapor]

umm

what about the top part?

by canceling, i meant in both sides

top and bottom

[remote pelican]

then just left 20092006 + 20092010 ?

[unkempt vapor]

yes

but bottom part differs

[remote pelican]

but the answer say 1 over 2

why?

[unkempt vapor]

as i said

[tardy tapir]

the numerator isnt squared

[unkempt vapor]

oh

top

[tardy tapir]

correct it

[unkempt vapor]

yay

it is 20092008/(20092006+20092010)

[tardy tapir]

just add the denominator now

[remote pelican]

40184016

omg!

thank a lot

really thank you

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[clear comet]

Why isn't this true?

[unkempt vapor]

well

[vale wigeon]

hold on

[unkempt vapor]

expanding brackets

[vale wigeon]

it looks true unless you make some kind of typo

who is saying it isnt true

[echo socket]

,w is (sqrt(sqrt(41) - sqrt(5)) - sqrt(sqrt(41) + sqrt(5)))^2 = sqrt(41) - sqrt(5) - 2(sqrt(sqrt(41) - sqrt(5)) * sqrt(sqrt(41) + sqrt(5))) + sqrt(41) + sqrt(5)

[unkempt vapor]

i see no error

[sour dove]

Yeah same

[vale wigeon]

,calc (sqrt(sqrt(41) - sqrt(5)) - sqrt(sqrt(41) + sqrt(5)))^2 - (sqrt(41) - sqrt(5) - 2(sqrt(sqrt(41) - sqrt(5)) * sqrt(sqrt(41) + sqrt(5))) + sqrt(41) + sqrt(5))

[ocean seal]

Result:

-1.4432899320127e-15

[vale wigeon]

0

close enough

[unkempt vapor]

mb cause absolute value?

[clear comet]

1 sec

[vale wigeon]

@clear comet yeah, again, who is saying it is not true?

[unkempt vapor]

yeah, it looks correct

[clear comet]

wolframalpha

#❓how-to-get-help

[gleaming fjord]

oh

[vale wigeon]

please open your own channel.

[gleaming fjord]

mky bad

my bad

[past raven]

) How can you add a 5N and 4 N forces acting simultaneously on a body to produce a resultant of 7 N force?

[unkempt vapor]

sqrt((sqrt(41)-sqrt(5))*(sqrt(5)+sqrt(41))) = 6?

[unkempt vapor]

....

[clear comet]

#❓how-to-get-help

[vale wigeon]

please open your own channel. read #❓how-to-get-help for instructions.

[echo socket]

Those are the same though once you simplify

[vale wigeon]

WA is failing to simplify the second thing fully fsr.

[echo socket]

Again, it is nowhere implied that those two are different

[vale wigeon]

it is not actually claiming the two are different.

[unkempt vapor]

,w sqrt((sqrt(41)-sqrt(5))*(sqrt(5)+sqrt(41)))

[ocean seal]

Results provided by WolframAlpha

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Requested by tim_oder

[unkempt vapor]

they are same

[clear comet]

so wa just doesnt simplify it to the same thing

okay then

Thanks

.close

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[cloud scaffold]

how should i approach solving this?

[cloud scaffold]

i have no idea where to begin

( repost b/c of trolls in my help closed out the other one already )

<@&286206848099549185>

( time from previous help allots for ping sorry 🙇

[lone heart]

@cloud scaffold Has your question been resolved?

[exotic belfry]

just try some selected values. for example: use t = 0 for (a) to (f), calc the points (x,y,z) and look which graph possible matches.

[cloud scaffold]

ah you're right

ok thank you! @exotic belfry

.close

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[reef flame]

Question 3 how do you determine the order 1,2,3 like how do you know

<@&286206848099549185>

[lone heart]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[agile wind]

!status

[lone heart]

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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[reef flame]

I got the zeros: 3,-2 and 3/4 but idk how to indicate whether they are order 1,2 or 3

And idk how to determine how to solve algebraically if each function is even or odd.

[agile wind]

to check whether a fn is odd or even

u need to replace x by -x

in f(x)

[reef flame]

So how would I do that

[lone heart]

@reef flame Has your question been resolved?

[lone heart]

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[plush pivot]

Hello! I am looking to find a formula that can obtain the distance of a line originating from a point in space to any point on an XY plane below it. I will be provided the point of origin as well as the direction vector from that point of origin. The final result needs to be the length of the line segment that starts at Point P, travels in direction D, and ends at the XY plane some distance below the point.

This is a 3D equivalent of the more simple "SOH CAH TOA" style problems. I just don't know exactly how it might work in 3D

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[lone heart]

@plush pivot Has your question been resolved?

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[undone oar]

Hi i am stuck here with a expression unknown from formula (sorry if badly translated) i would like a simple explenation

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@undone oar Has your question been resolved?

[undone oar]

.close

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[sage hedge]

I did the first part but for some rason I kept messing up the second aprt

[tacit arch]

!show

[lone heart]

Show your work, and if possible, explain where you are stuck.

[sage hedge]

k one sec

[alpine sable]

bro, didn't you learn how to do this yesterday

$(b)$ is the exact same as $(a)$ just with a different $\epsilon$

[ocean seal]

Replaced by new brandon H

[sage hedge]

I think im messing up the math part

like the arithmatic

[alpine sable]

look at what you did with a

and change one number

[sage hedge]

is it 1.05647

I just used desmos and used the quadritic formula

[tacit arch]

...

can you show your entire work

[sage hedge]

😏

one sec

20 - 0.005 = 19.995

19.995 = x^2 - x

0 = x^2 - x - 19.995

[alpine sable]

btw

you wont have to use the quadraic formula if you just complete the square

Also learning how to use latex would be great

[sage hedge]

what is latex

[alpine sable]

$|f(x)- L| < \epsilon$

[ocean seal]

Replaced by new brandon H

[sage hedge]

how to complete square I haven't taken a math

in like 2 years

[alpine sable]

$ax^2+bx+c =0$

[ocean seal]

Replaced by new brandon H

[sage hedge]

oh

[alpine sable]

move the c to the other side

$$ax^2+bx=-c$$

[ocean seal]

Replaced by new brandon H

[alpine sable]

replace the left with $$(x-\frac{1}{2}b)^2$$

[ocean seal]

Replaced by new brandon H

[alpine sable]

add $\frac{1}{4}b^2$ to the right side

[ocean seal]

Replaced by new brandon H

[alpine sable]

so

$(x-\frac{1}{2}b)^2 = \frac{1}{4}b^2-c$

[ocean seal]

Replaced by new brandon H

[alpine sable]

then you can solve for x

$x-\frac{1}{2}b = \pm \sqrt{\frac{1}{4}b^2-c}$

[ocean seal]

Replaced by new brandon H

[alpine sable]

$x = \frac{1}{2}b \pm \sqrt{\frac{1}{4}b^2-c}$

[ocean seal]

Replaced by new brandon H

[alpine sable]

this only works if a is 1 tho

[sage hedge]

bruh

im lost in the sauce

I need to go to school I have to work on this later, thanks for the help tho.

[lone heart]

@sage hedge Has your question been resolved?

[lone heart]

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[indigo storm]

I wanna solve x

Tenth one

[alpine sable]

X?

Oh tenth

[indigo storm]

yup

[tacit arch]

use pythagoran theorem and write it interms of just cosine or sine

[charred summit]

Tenth ?

[indigo storm]

yup

10

[tacit arch]

$\cos^2(\th) + \sin^2(\th) =1$

[ocean seal]

riemann

[charred summit]

Oh

[tacit arch]

then factor or quadratic formula from there

[alpine sable]

Can you make it 2sin²x + sin²x + cos²x + 3 sinx = -1

Then sin²x + cos²x = 1

Then its a quadratic for sinx

[indigo storm]

this is what I did

what can i do next?

[tacit arch]

.

[indigo storm]

cannot factor

how to factor it?

[near apex]

Try checking if it's even factorisable.

[tacit arch]

,w factor 2x^2 + 3x + 2

[ocean seal]

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[tacit arch]

nope

so use quadratic formula

[near apex]

They probably need real values so no point.

[indigo storm]

will I be able to get general solutions

between 0 and 2pi

[near apex]

There aren't any solutions. At least not for real angles.

[indigo storm]

wait but that questions asks for solutions

let me send full

[near apex]

Well, then question is probably misprinted.

[indigo storm]

okkk

[indigo storm]

Can you tell me how to solve 1st one of 3rd question

[near apex]

Well, make everything into terms of cos(theta) and then solve the polynomial, i suppose.

[indigo storm]

did u try?

[near apex]

I am not certain if there is a simpler way.

[near apex]

Of course not. Lol

But i can see that it's going to a quartic. We know one root.

So that gives us cubic.

[indigo storm]

yes

[near apex]

After that, rational root theorem, i think.

[indigo storm]

idk rational root theorem

I should ask from the teacher

[near apex]

Do you know how to factorise a cubic?

[indigo storm]

yes

[near apex]

If yes, then you know what it is - you just don't know the name.

[indigo storm]

addition and difference between 2 cubes

I know that

[near apex]

Oh. You mean like, a^3 - b^3 ?

[indigo storm]

ye

[near apex]

No. Not those.

[indigo storm]

mmm then i really dont know

[near apex]

Oh wait. There is another simpler way.

[indigo storm]

telll

[near apex]

But i dunno if you know that formula.

[indigo storm]

just telll

[near apex]

familiar with this formula?

[indigo storm]

I know

C+D formule

RIght?

[near apex]

This one actually.

You can do directly using this.

[indigo storm]

yeah know those

[near apex]

Great then.

Do it.

[indigo storm]

ok

I'll try

Correct?

[near apex]

Why did you do this?

I told you to use this.

$\cos{(A)} - \cos{(B)} = -2\cdot \sin{\left(\frac{A+B}{2}\right)}\sin{\left(\frac{A-B}{2}\right)}$

[ocean seal]

Enemagneto

[indigo storm]

aaaaaaaaaaaaaaaaah

u mean I have to take cos 3theta left side

as minus cos 3A

[near apex]

Yes

[indigo storm]

cos 4A - cos 3A = 0

[near apex]

Yes

[indigo storm]

then formula

ahh

got it

thamks

this way?

well I substituted it

maybe it is correct

.close

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[ember ridge]

Hey! What's the logic here?

[lone heart]

@ember ridge Has your question been resolved?

[ember ridge]

<@&286206848099549185>

[foggy stirrup]

notice that in each column a dot is being removed from the shape in the images

[ember ridge]

What about the movement of the pattern in the first column?

What differentiates the 2nd and the 5th choice?

[lone heart]

@ember ridge Has your question been resolved?

[spiral umbra]

excuse me

I need help with my homework

[lone heart]

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[bleak geyser]

Send whole question pls

!status

[lone heart]

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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[wise jewel]

why is f(x) at x = 1 not differentiable

like how do yk if it is or not

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[olive urchin]

Can someone help me do this integral?

[olive urchin]

I think it is multiply by 1 and its probably obvious but I just cant think of what to do

[deep hornet]

Have a look at your integral rules, we have well defined patterns for handling fractions, constant factors and factors with subtractions

[lone heart]

@olive urchin Has your question been resolved?

[olive urchin]

Hmm

[olive urchin]

We do?

.close

[lone heart]

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[deep robin]

Is this integral improper?

[vague coral]

it is yes

[deep robin]

Bc ln(0) isn't defined right

[vague coral]

it is

[deep robin]

what

[vague coral]

wait ive read your sentence wrong

ln(0) isnt defined

[dim oasis]

ln(1) is defined, ln(0) is not

[deep robin]

oh ok lol

ye thats what i thought

Thanks guys 🙂

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[autumn crag]

Another problem i have that i feel i made a mistake somewhere

[jagged cobalt]

are you sure you havent been taught any other methods of differentiation? like the chain rule?

[autumn crag]

Nope havent heard of it 😦

It is the very basics i have been taught so far

[jagged cobalt]

i can give you a very quick rundown since it will generally save you a lot of time here

if you have some function like y=sqrt(7-74x)
then you can let u=7-74x
then y=sqrt(u)

from this follows that dy/dx=du/dx * dy/du
which will be dy/dx = -74 * 0.5 *1/sqrt(u)
=-37/(sqrt(7-74x)

[autumn crag]

but my teacher is having a quiz about this tomorrow and isnt allowing me to use different formulas outside of the ones i have been using

[jagged cobalt]

do you have a list of the formulae you can use?

[autumn crag]

yeah in my notes

[jagged cobalt]

may i have a gander

[autumn crag]

imma write them down since i dont have them both in one spot

[jagged cobalt]

damn

[jagged cobalt]

one key thing though is it has asked for an equation of a line

unless this line is just horizontal that cant be right

[autumn crag]

oh yeah cuz you use this equation after

[jagged cobalt]

thats fine and all, im saying youre missing the x in your answer

[autumn crag]

......

i feel so stupid right now lol

thank you

.close

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[left zealot]

how do I do this

[frozen plinth]

Plug in two x values get 2 y values , draw them and draw a line that goes through them

[left zealot]

wdym?

[frozen plinth]

Pick one x

Put it in your equation

And get a y

And that (x,y) is one point

Draw it in the coordinate system

Then do that again for a different x

Then you have 2 points in your coordinate system, draw a line through them

[left zealot]

how would I do that

I'm not understanding on what ur saying

[frozen plinth]

Well

Do you know what it means to plug in a value for x?

[lone heart]

@left zealot Has your question been resolved?

[left zealot]

<@&286206848099549185>

[forest cobalt]

can you create a table of points?

[left zealot]

how would I do that

[forest cobalt]

So you want to make a graph, right?

[left zealot]

yes

but how do I start

[forest cobalt]

Okay. You need the points on the graph first

(x,y)

x goes left and right, y goes up down

[wary stream]

You can literally do the same exact process you did here #help-36 message

You were shown this before

Or here #help-19 message

[left zealot]

this one is different tho

I can't really go down 7

as there's no space on the graph

[forest cobalt]

there is

[wary stream]

#help-36 message

And no it's not right, because -7/3 can either mean down 7, right 3 or up 7 left 3
This was mentioned too

Same logic with -7/5

[left zealot]

wait so if you can't go down, you have to go up?

and if you can't go up, you have to go down?

[forest cobalt]

instead of treating it as 7 down and 5 right, you can treat it as 7 up and 5 left

[left zealot]

okay but why can't I go right

[forest cobalt]

You can, but you said you don't have space, no?

So try the other way

[wary stream]

Because then you'd have a positive sloped line if you mean going up and then right

[forest cobalt]

yes exactly

now just draw a line between those two points

[left zealot]

so I just have to remember this?

[abstract crystal]

Are you still having concern with the function?

Yes

Locate the coordinates by inputting X values and get the Y

[left zealot]

alr ty

.close

[lone heart]

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[abstract crystal]

You will have to put the numbers you choose on independent variable(X) and solve it to get dependent variable(Y)

[wary stream]

.reopen

[lone heart]

✅

[abstract crystal]

then find the coordinates from the expression you just got (X,Y)

[wary stream]

Yes because $-\frac{a}{b} = \frac{-a}{b} = \frac{a}{-b}$ and slope is rise over run so if you did $\frac{-a}{b}$ that means down a units then right b units or if you did $\frac{a}{-b}$ that means up a units then left b units.
$\frac{a}{b} = \frac{-a}{-b}$ so that means if you did $\frac{a}{b}$ that means up a units then right b units and if you did $\frac{-a}{-b}$ that means down a units then left b units

[ocean seal]

dldh06

[wary stream]

.close

[lone heart]

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[languid bolt]

Okay so,

The general equation of a straight line is y = mx + c
How do I make x = h where h is a real number in that form?

[languid bolt]

like... for example: x = 4

or you can't?

[gray isle]

you don't

that form is only applicable for lines with defined slope

[languid bolt]

ohh okay

.close

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[earnest current]

$\lim_{x \to a} c = c$

[ocean seal]

texaspb

[earnest current]

hey how do you prove this using the formal definition? feel like this should be a nobrainer but I actually dunno how to write the proof in the most correct way possible

[lone heart]

@earnest current Has your question been resolved?

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[errant turret]

Hi I was wondering how I could find the domain of f(x) = 1 + ln(x^2+x-1)

[ionic jewel]

what is the acceptable inputs to ln()?

[errant turret]

anything other than x cannot equal 0

[ionic jewel]

that is not true

[ionic jewel]

The domain is x > 0

[errant turret]

i found two things:
x>1
x>0

when i tried to solve whats inside ln(x)

[ionic jewel]

well the trick you need to see is that ln(x) only accepts positive numbers

so x^2 + x - 1 has to be positive

and therefore you solve
x^2 + x -1 > 0

and you get the domain

[errant turret]

i couldn't factor this so I got x(x+1)>1

x>1 and x>0

[ionic jewel]

That is not a legal math step

the factoring is right, but whatever you did after is not

but anyways, have you heard of the quadratic equation?

[errant turret]

yes

[ionic jewel]

well then, you can use that to find the zeroes of the polynomial, then figure out which sections are positive

and thats the domain

[errant turret]

thanks, i have figured it out

.close

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[alpine sable]

need help with this, someone help asap

[lone heart]

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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[alpine sable]

i need someone to help

<@&286206848099549185>

[bleak geyser]

please refrain from pinging instantly and wait for 15 mins

[alpine sable]

alr

@crystal spade

ok i need help with it all

cause its confusing

can you help me

im confued

confused

[ocean seal]

quadrangle

[alpine sable]

0

to 1.5

where

[ocean seal]

quadrangle

[alpine sable]

i got

2

for the second one on the right side

but idk how to get it for the left side

[ocean seal]

quadrangle

quadrangle

[alpine sable]

so now what

@crystal spade

alg

i got all of the side ones btw

to the right

[ocean seal]

quadrangle

[alpine sable]

i have all of them besides this one now

how do i get that one

nvm got it

600

one last thing

@crystal spade

how can i find the domain and range of this

wait whats the range answer

cause theres 2 points at the same elevation

0

and 4

im unsure now

like

x<0?

@crystal spade

[ocean seal]

quadrangle

quadrangle

[alpine sable]

0≤4?

im so confused

now

what would the range be

@crystal spade

[covert wraith]

do you still need help ?

[lone heart]

@alpine sable Has your question been resolved?

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[alpine sable]

no

[lone heart]

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[potent oasis]

Can someone help me and see what I did wrong? We are supposed to use integration by parts but I can’t seem to figure this one out!

[sour dove]

yeah I'd do IPB. You conversion of cos^2(x) = (1 + cos(2x)/2 is correct though. Try setting u = x and dv = (1 + cos(2x)/2 dx and do IBP

[potent oasis]

Ill try that, how do you know which to use as U or as dV

[sour dove]

second sentence 🙂

[potent oasis]

Yes but how do you identify which would be better when you see the problem?

[sour dove]

just as a rule though if you see any power of x most of the time that should be your u

things like trig functions are cyclic, or e^x is it's own integral, so those are usually great candidates for dV

basically your job should be to find the easiest thing as a derivative and more importanly the easiest thing as the integral

plus, if we did u as a trig fucntion, and x as dV we'd get stuck in a loop as x continued to gain more powers and we'd just to between sin and cos over and over again with derivatives

[potent oasis]

Thats what would happen when i tried it

[sour dove]

yep pretty much

[sage hedge]

.close

[sour dove]

you can't close someone else's channel unless you're a helper

[sage hedge]

oh sorry

I JUST figured they were done cuz its been 12 mins

and im lost on this calc problem and wanted to use the channel

[sour dove]

nah I don't decide that for them, lol. Often people will go try stuff and come back or take a bathroom break or something

[sage hedge]

oh sorry didt know

[sour dove]

oh head to #❓how-to-get-help it'll show you how open an available channel

right now it's any of the these

[sage hedge]

I dont see any of them for some reason

[sour dove]

looks like you got it!

[lone heart]

@potent oasis Has your question been resolved?

[sage hedge]

ye

[potent oasis]

Ok wait I am still stuck!!! I cant figure out where I messed up

[sour dove]

Here's what I would do to solve this problem.

$\int x cos^2(x) dx$. Let $u = x$ and $dv = cos^2(x) = \frac{1}{2}(1 + cos(2x))$. Then $du = 1$ and $v = \frac{1}{2}(x + \frac{1}{2}sin(2x)\$ So then we get $uv - \int v du$, which in turn is this:
$\\ (x)(\frac{1}{2}(x + \frac{1}{2}sin(2x)) - \int \frac{1}{2}(x + \frac{1}{2}sin(2x) dx$. Once again we can integrate to get:
$\int \frac{1}{2}(x + \frac{1}{2}sin(2x) dx = \frac{1}{2}( \frac{1}{2}x^2 - \frac{1}{4}cos(2x)$. All together we get:
$\\ \frac{1}{2}(x)(x + \frac{1}{2}sin(2x) - \frac{1}{2}(\frac{1}{2}x^2 - \frac{1}{4}cos(2x)) + C$

[ocean seal]

MellowDramaLlama

[sour dove]

let me know if that makes sense

[potent oasis]

In the book it says the answer is different

It says to Partition the x axis, I am not sure what that means

[sour dove]

that's the same answer, just manipulated.

here's a breakdown of what they changed

$sin(2x) = 2sin(x)cos(x)$, so they converted $\\\frac{1}{2}sin(2x) = \frac{1}{2}(2sin(x)cos(x)) = sin(x)cos(x)$. Then we distribute the $\\\frac{1}{2}x$ and we get $\frac{1}{2}xsin(x)cos(x)$

[ocean seal]

MellowDramaLlama

[sour dove]

wait actually I think the book has the last portion wrong

[potent oasis]

really?

let me put it into my handy dandy calculator

[sour dove]

so the red line is xcos^2(x). h(x) is the proposed integral from the final answer

notice how h'(x), which should be the exact red line, is different?

that tells us the answer is incorrect

[potent oasis]

yeah shouldnt it be plus x^2 * 1/4

[sour dove]

x^2 * 1/4 is the same thing as x^2/4

[potent oasis]

yup, im saying in the book shouldnt it be added instead of subtracted?

wait im saying thr wrong thing

im confusing myslef lol

but shouldnt it be a plus sign in the book instead of a minus sign?

[sour dove]

ah my bad sorry all answers are correct here, they're all just vertically shifted

let me see if I can explain

so let's start with the asnwer I got

graphically we'll start with this

[potent oasis]

how did you get minus 1/4 cos(2x)

i got plus

waittttt

i got it

[sour dove]

ok cool

[potent oasis]

Ok so

The green is the calculator and the blue is the book

the black is what I got, which was wrong

[potent oasis]

So I am not sure what part I messed up

[ancient saddle]

Hello. I think the only mistake here is the last integral, sin(2x). You forgot the 1/4. Then the rest looks ok, don't forget the parentheses in (2x+sin(2x))

[lone heart]

@potent oasis Has your question been resolved?

[potent oasis]

So I did that and graphed it and it is different from the answer

[ancient saddle]

I graphed it with that correction and looks ok now 🤔

[potent oasis]

Hmm here did I correct it correctly?

[ancient saddle]

it's cos(2x)/8

[potent oasis]

Wait

Why over 8?

[ancient saddle]

you got cos(2x)/2, but you forgot 1/4

cos(2x)/2 * 1/4 = cos(2x)/8

[potent oasis]

No way

You’re right

This is aweosme! Thank you so much

I have been stumped on this for so long

[ancient saddle]

Ok no problem! It's normal to have some mistakes when doing one or more integrations by parts. Just practice it more, or use the DI method if it suits you (it's just an organized way of doing int by parts)

[potent oasis]

What is the DI method?!

[ancient saddle]

Look it up on youtube 👀 I think mistakes are less probables when you use that 🤔

[potent oasis]

That sounds so cool! I’ll definitely check that out

[lone heart]

@potent oasis Has your question been resolved?

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[alpine sable]

How would I calculate this?

[tardy tapir]

sub the limit.

thats not indeterminate

[alpine sable]

just sub pi into the sin(2x)+1?

[tardy tapir]

yes

[alpine sable]

what does that mean sorry

indeterminate

[tardy tapir]

there are indeterminate forms, in which you cant sub in the limits.

like sometimes 0/0

its not defined

[alpine sable]

ahhh i see

thx

so answer is 1

sin(360) is 0

+1

is 1

[tardy tapir]

yes, correct!

[alpine sable]

thanks

.close

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[fallow solar]

When I look at the irrational numbers in mathematics I think like where tf did these specific numbers come from like pi or e why is pi 3.14.... and not another number

[vapid shuttle]

pi refers to how much longer the circumference of a circle is compared to its diameter.

Formula for circumference of a circle given a diameter d is: C=pi*d

This is something that is true for all circles of all sizes. It wouldn’t be true for any other number than 3.14159265….

[fallen verge]

Just so happened to be that number

[vapid shuttle]

I’d search a video for, why is pi 3.14

And similarly for e. Probably those can explain it better than we can

[fallow solar]

Yup I know but why this number exactly

[vapid shuttle]

Are you asking how we calculate pi?

[fallow solar]

Nope

[fallen verge]

Why pi exactly

Idk, the universe decided

[vale wigeon]

"why is pi what it is"

[vapid shuttle]

It is that number exactly because that is the relation between the diameter of a circle and it’s circumference exactly

[fallen verge]

Then we get into the discussion of is math invented or discovered

[vale wigeon]

is kind of a philosophical question probably

[fallow solar]

That's what I'm talking about

[fallow solar]

Yup

Why did the universe "decide" these specific numbers

[vale wigeon]

there probably is not a "why"

[fallen verge]

It just is

[fallow solar]

Whyyyy

[fallen verge]

Thats like asking why is blue not green?

Thats just how it is

[zealous lichen]

well if you are not in Euclidean space pi might be different

if you need a reason then pi is 3.14 because we're in Euclidean space

[fallow solar]

If that's the case then who assigned these numbers , is that a proof for god's existence?

[zealous lichen]

no one assigned them

it's a consequence of the axioms of Euclidean space

[fallow solar]

Ok why this specific number

[tardy stag]

it would have different digits if we were using a different base

it's possible to arrive at a formula for π

well, an infinite series

if you really want to blow your biscuit look at the incomputable numbers

[fallow solar]

Well if you use a different base then you'll get different digits but the same question is there why those specific digits

[zealous lichen]

we measure the length of circumference and diameter

and their ratio is pi

[pale kestrel]

ask on a philosophy server

[fallow solar]

My honest reaction:🤯

[zealous lichen]

?

[pale kestrel]

Let me answer another way

The ratio has to be something

whether its 1.234567...

or 3.141...

in any case, youre still going to be asking the same question: why?

[fallow solar]

But this number is based on fundamental things it does not depend on units

[pale kestrel]

So rather than asking your original Q

you probably want to figure why ratio of lengths remains the same

when scaled

[fallow solar]

Nope

[pale kestrel]

not just a circle

The ratio of a square's perimeter to its side length is 4

no units also.

no matter how big the square is or how its rotated or translated

[zealous lichen]

I don't know why you are not satisfied

[tardy stag]

π is cool and all but what really gets me going is 0.669 or whatever it is

the bifurcation constant

[fallow solar]

Ok but that's est intuition because the square is made of 4 lines but why pi is 3.14...

[zealous lichen]

so you want to know how to calculate pi

[pale kestrel]

I told you the answer

the ratio has to be a constant

[zealous lichen]

bruh

[alpine sable]

_https://www.youtube.com/watch?v=d-o3eB9sfls&pp=ygUKd2hhdCBpcyBwaQ%3D%3D_

[pale kestrel]

because of this property

no matter how you scale the shape, this ratio of lengths is a constant

[tardy stag]

it's probably easy enough to show that the ratio must be irrational

[pale kestrel]

Since its a constant, it will be something

and no matter what that something is, youll be asking this question, no?

pi = 1.234567... why?
pi = 2.3947383746... why?

If you want a reason for why its specifically 3.141... though, you should have a look at a geometric proof perhaps

approximating pi with polygons, say.

That can give you a 'reason', but its up to you whether you want to accept it as intuition

[fallow solar]

You got a point

I'm stupid

.close

[lone heart]

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[hot trail]

I did term by term...

it did not cancel out nicely

[lone heart]

@hot trail Has your question been resolved?

[hot trail]

<@&286206848099549185>

[undone dune]

This sum does not converge. It goes to minus infinity.

[lone heart]

@hot trail Has your question been resolved?

[lone heart]

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[torn pagoda]

A bit stuck on assignment b 😅

[thorn veldt]

https://discord.com/channels/268882317391429632/1152141820101414942

Need help

[median oar]

!help

[lone heart]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[vale wigeon]

!noadvert

[lone heart]

Please do not advertise your help channel or thread in other parts of the server. There are many people who need help, and no one person can be prioritized over other people, so please patiently wait. Anyone who chooses to help you is a volunteer who is doing so out of their own kindness.

[torn pagoda]

@vale wigeon did my question get overruled? 😅

[vale wigeon]

no it didnt we are just redirecting the other person

[thorn veldt]

Sorry i dint mean to

[torn pagoda]

Thanks for reply! Got confused - No problem @thorn veldt

[hard mountain]

Is A supposed to be the bottom or the top of the main pipeline ABCD?

[torn pagoda]

Sry, A is the endpoint, in other words the lowest point

[hard mountain]

Ok

[torn pagoda]

I was trying to make a triangle and find the hypotenuse, But didnt Work out 😅

[hard mountain]

Why not calculate from B to E?

With the same method as you did from A to B

[torn pagoda]

I Think that was my first try But somehow i Got wrong result, might try again

Jesus Christ @hard mountain it did make sense. I just forgot to adjust for not going all the Way from A to D But stopping at B

Thanks!!

[hard mountain]

All good 👍

[lone heart]

@torn pagoda Has your question been resolved?

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[alpine sable]

hello
Who can explain to me what the backpack problem is?

[near apex]

Ah. That's same as knapsack problem.

Well, if you need an introduction, it's best if you watch a video or read up on internet.

If you have doubts about it, then you should delineate those.

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@alpine sable Has your question been resolved?

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[limpid abyss]

I don't get it on how to use L'hopitals rule here/ how to derive this

[subtle birch]

are you supposed to only use L'Hopital?

[vale wigeon]

are you required to do this with lhop?

[subtle birch]

Because rationalizing the denominator is a better and easier option

[limpid abyss]

No, Just trying to practice the l'hopital

[alpine sable]

hello
knapsack problem
who can explain to me?

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[limpid abyss]

.close

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[fast nest]

hi can someone help me

[near apex]

Where exactly are you stuck?

[fast nest]

so i’ve done

hold one

i have no idea how to do 4-6

[near apex]

Okay. So, 4 wants you to write an expression for finding all the ways to go from city 3 to city 2.

You also can't pass through any city twice.

[fast nest]

yes

[near apex]

Which means that you can't go through 3 and 2 in between. 3 will just be starting city and 2 will be end.

Now, how many possibilities do you have for middle stops?

[fast nest]

2…?

wait no

1..?

[near apex]

Which one is it?

Also tell me what are the choices?

[fast nest]

i literally don’t know i’m sorry

like this assignment has me genuinely so confused and lost

[alpine sable]

theres no reason to be sorry

[fast nest]

oh ok

[near apex]

Try to think at least.
You have to go from city 3 to city 2.
One possibility is that you could just directly go. Right?
Other possibilities are where you choose to take a roundabout i.e. first go to some other city, thence to some other, and so on... finally to city 2. In these cases, you go through multiple in between cities. You don't have many in-between cities to go to, though. Your question has only 4 cities. 2 of which are forbidden so that leaves you with limited options.

[fast nest]

3 to 1 to 4 to 2?

[near apex]

Yeah. That's one.

Any other possibilities?

[fast nest]

does it have to go through every city?

3 to 1 to 2

3 to 4 to 2

[near apex]

Yes. That works.

[near apex]

Good

[fast nest]

thags all of them

i think at least

[near apex]

Obviously not. Your question says that only condition is - Do not pass through any city twice.

[fast nest]

understood

[near apex]

There is one more.

[fast nest]

just 3 to 2?

[near apex]

We already saw that. That was the first one i mentioned.

[fast nest]

hmm

[near apex]

There's one more way where you pass through two cities in between.

[fast nest]

idk it’s not coming to me

[near apex]

How about you first go to 4 ?

[fast nest]

4 3 2 1

[near apex]

What? No

You are starting at 3.

[fast nest]

oh did you mean

3 to 4 and so on

[near apex]

Here, you are going 3 to 1 and then 4. How about you switch the order?

[near apex]

Yes. Tell me full path.

[fast nest]

3 4 2 1

[near apex]

No. You have to end up at 2. 😐

[fast nest]

ok well you can’t do 3 4 1 2

[near apex]

Why not?

[fast nest]

because look at the arrows

1 can only go to 4

not the other way around

[near apex]

Okay. Good.

But, you should still count it in your possible paths.

Matrix will take care of it automatically.

[fast nest]

oh

ok

[near apex]

When you calculate, that one term will be zero so that whole path would amount to zero.

[fast nest]

got it

[near apex]

Now, list all the paths together here. Fast.

So, we get to writing expression.

We have 5 paths in total.

[fast nest]

3 1 4 2

3 1 2

3 4 2

3 2

3 4 1 2

[near apex]

Nice

Now, for first path - what is the expression ?

[fast nest]

would it just be 3 + 1 + 4 + 2?

[near apex]

No.

[fast nest]

sorry

[near apex]

Look at what it means.

[near apex]

It's okay. I'll explain. Try to answer the questions.

[fast nest]

alright

[near apex]

So, First path basically says -
We go from 3 to 1, then from 1 to 4, then 4 to 2.

How many ways are there of going from 3 to 1 directly?

[fast nest]

1

[near apex]

Remember, you can answer that from matrix that you made.

Nice

Now, you have reached 1. We need to go to 4.

How many direct ways ?

[fast nest]

1

[near apex]

Nice

Now for 4 to 2 ?

[fast nest]

1

[near apex]

Yes. So overall, we multiply all three.

1 * 1 * 1

[fast nest]

1

[near apex]

So, there's only one way.

[fast nest]

yes

[near apex]

Do you understand why are we multiplying ?

[fast nest]

yes

[near apex]

Awesome.

So basically our expression for this just was:
$R_{31} \times R_{14} \times R_{42}$

[ocean seal]

Enemagneto

[fast nest]

ohhh

[near apex]

For path 3 1 4 2

[fast nest]

should i clarify the paths or just simply write the expressions?

like write the path : expression

[near apex]

Notice properly. We take 2 numbers consecutively, choose that term from matrix and just multiply them.

[near apex]

You can do this way as well - at the end you will have to do total.

Otherwise you can just write expression for all and put a plus sign in between expressions.

[near apex]

Did you get this? @fast nest

[fast nest]

yes

[near apex]

Nice. Write for other 4 paths. Show me.

Remember, write in terms of R

[fast nest]

alright

[near apex]

Don't write actual values.

You are supposed to write expression.

[near apex]

Write this way.

[fast nest]

yes got it

r31 * r12

is that it

or no

[near apex]

Yes. That's correct for path 3 1 2.

[fast nest]

i should be writing it like this on my paper?

r34 * r42

[near apex]

You can write it there as well but it seems like you are not used to writing in smaller fonts.

If your teacher will accept it on another paper then you can write there as well.

[fast nest]

i can write it smaller

but that’s how it’s supposed to be written correct

[near apex]

Yes. For better understanding, you can write path: Expression thing as well.

It's just i'm not certain what your teacher wants. Writing "Path : Expression" will sure make it easy for others to understand.

[fast nest]

well i could write it above?

ok i wrote the expression

[near apex]

Yeah. That's great.

Similarly do the rest.

I believe it's easy peasy at this point.

[fast nest]

well could you start me off for 5

i wanna make sure i have the hang of it

[near apex]

Well, try the part a. You will do it.

[fast nest]

i use the graph

[near apex]

List the possible cases. I'll check.

[near apex]

No. Just list possible paths. See. You have to start at 1 and end at 1.
In between, there can be only one city.

[fast nest]

1 2 1

1 3 1

1 4 1

[near apex]

Yes.

Now, just use the same formula to calculate.

This time you need to give values though. Not expression.

[fast nest]

is the first 6

[near apex]

Yes

[fast nest]

second is 2

[near apex]

yes