help-27

[hard knot]

ty guys

:3

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[hard knot]

How can I find Y values of this

[late orbit]

This is direct calculation. There's no trick to this.

[hard knot]

im getting angles

for the range

tho

I got the calculation wrong but whatever I stuck with it

[magic pine]

this is best understood from the context of the unit circle

if you take the arctangent of sqrt(3/7) and -sqrt(3/7) you'll get two solutions

the problem is that the one for -sqrt(3/7) will be outside the desired range

since tangent has period pi, you could add pi to those solutions as many times as you want

and they would still give you the same tangent

[hard knot]

oh ok

Is it
possible to figure out the y values through the angles you got

[magic pine]

im confused by what youre asking

[hard knot]

Imagine they asked 0 <_ theta _< Pie

And you got the set of angles for that

right?

[magic pine]

there's no theta in the equation

[hard knot]

oh

I see

I understand

[magic pine]

the y is the angle

[hard knot]

so this

is not the same as this ^

[magic pine]

im not sure what youre asking

let's take a step back

and consider a simpler example

(\cos\theta = 1)

[woven radish]

maximo

[magic pine]

can you give me a solution to that

[hard knot]

0

[magic pine]

theta = 0

but yes

now consider

(\cos x = 1)

[woven radish]

maximo

[magic pine]

can you give me a solution to that

[hard knot]

(Is that y angle in radians?)

[hard knot]

theta = 0

[magic pine]

no

there's no theta there

[hard knot]

oh ok

[magic pine]

a solution would be x = 0

now how about (\cos y = 1)

[woven radish]

maximo

[magic pine]

can you give me a solution

[hard knot]

y= 0

[magic pine]

yes

notice how all of these equations

are equivalent

the letter you use for the variable

the symbol

is irrelevant

[hard knot]

if i may ask then

[magic pine]

they are all stating the same relationship

[hard knot]

Why is here the case

Where x is theta

[magic pine]

x is not theta

x is the argument of the trig functions

yes

[hard knot]

oh i think im following

[hard knot]

If they said 0 < x < 2Pi

as the range

would you have to convert all the angles into terms of pi?

[magic pine]

practically yes

in theory no

if the answer was x = 2pi

and you said

x = 360 degrees

your answer would be correct, since those are equivalent

[hard knot]

oh ok

[magic pine]

but it may be marked wrong

just use what they use

[hard knot]

Then last question

if i may

[hard knot]

The answers

are they here in terms of pi

or something else

[magic pine]

not in terms of pi

in radians

[hard knot]

oh right

[magic pine]

if they dont have a degree symbol, assume radians

[hard knot]

OH!

I understand

eg
60 degrees = 1/3 Pi = 0.333 rad ?

[magic pine]

60 degrees is pi/3 radians yes

[hard knot]

So here

its the same as y x Pi = degrees

[onyx rune]

tf its wrong

60 degrees = 1/3 pi radians

pi is just a number not a measure of an angle

[magic pine]

im not sure what youre trying to say

[hard knot]

If i wanted to convert it to degrees

[magic pine]

you multiply by 180 and divide by pi

[hard knot]

But thats radians no?

OH

no

yeah

I get it

can go both ways

Ok i got it

Thank you

That really helped!

Apprecaite it

:3

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[vital blaze]

How are they able to make (1+x)(1+x)^n = (1+x)^n +x(1+x)^n. I haven't doen these proofs in a while. I proved the rhs but couldn't with the left

[onyx rune]

they factored out (1+x)

(a+b) (c+d) =a(c+d) +b(c+d) see the pattern?

[vital blaze]

oh ok

So they factored out (1+x)^n and you are left with (1+x)^n + x(1+x)^n

makes sense thank yoy

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[prisma frost]

Can someone explain to me

[prisma frost]

Why -18 pi over 4 turned into -pie over 2 please

For 7

[stuck field]

Because the period of sine or cosine is 2pi. You can add/subtract 2pi until you get an angle you're more comfortable with. (preferably something in [-pi,pi])

And also! It's not pie.

[prisma frost]

Oh ok

So I want it to be in the range of of those too

Ok ty

[prisma frost]

Yea mb😭

Thanks for the help

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[restive river]

How do i solve this? I’ve tried the above and it doesn’t look right

Just want to find dy/dx

[cinder coral]

the curve looks like the folium of descartus if I'm not wrong

[solar goblet]

misread, my bad

[cinder coral]

first , find dy/dt

then find dx/dt

then divide them both

[restive river]

That’s what I’ve tried to do but they don’t look right

The dy/dt I’ve tried to figure out don’t look right

And I get different answers when trying product and quotient

[cinder coral]

omg

what have you done

(1+t)^-3???

[restive river]

Yh I wasn’t sure wtf I did there

[cinder coral]

that's wrong

$1+t^3$

[woven radish]

Victor Grignard

[restive river]

I tried searching it up but I couldn’t find whether that was right or wrong

[cinder coral]

this

and

$(1+t)^3$

[woven radish]

Victor Grignard

[cinder coral]

are different

[restive river]

I see

How would I write (1+t^3)^-1

[cinder coral]

why not just apply the u/v rule instead

[restive river]

Yh I tried that in the next one

But I did something wrong there too I think

[cinder coral]

its correct

don't solve the (1+t^3)^2 in the denominator

its going to cancel anyways after division

[restive river]

I got $1+2t^3+t^6$

[woven radish]

NotPlayingAgain0

[restive river]

As the expansion so I thought I did it wrong

Because I didn’t see how I could divide

[cinder coral]

the expansion is correct

[restive river]

Okay and how would I divide I by the top

[cinder coral]

don't do that

when you're going to

divide dy/dt and dx/dt

the denominators being same

will cancel out

[restive river]

Ahhh I see so do I leave the denominator expanded?

[cinder coral]

yes

unexpanded

[restive river]

Okay thanks I’ll try solve the rest

[cinder coral]

that would work too

[restive river]

Thanks found it 🙏

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[hard knot]

Uhhhh what can I do from here

to find an angle

[magic halo]

you can take out sin

and divide both sides by sin

to get 8 cos = 5

[hard knot]

so cos = 5/8?

oh

[magic halo]

yes, but that wont get you an angle

[hard knot]

I wasnt sure if you could do

5sin/8sin

[hard knot]

oh ok

[little totem]

i generally say this is not desirable

you run the risk of losing solutions

[magic halo]

so just keep it outside in brackets?

[little totem]

you cant divide both sides of an equation by c if c=0

yes

[magic halo]

¯_(ツ)_/¯

[candid lance]

well

[magic halo]

mb

[candid lance]

u can just get common factor

to get 2 different solutions

[hard knot]

How? you cant turn it into x^2 bc there is cos? or can u

[candid lance]

and u also get sintheta = 0 in one of them

[little totem]

its not "wrong", its just easy to forget and make a mistake

factor out sin:

$\sin{t} (8 \cos{t} - 5) = 0$

[woven radish]

CosmoVibe

[hard knot]

okay?

but what does that do

^

[little totem]

so now you have two things multiplied together

that get 0

what do you know about those two things

[hard knot]

5(5x+3)
x = 0 and x = -3/5

But idk with
5y (5x+3)

[little totem]

hold up, slow down

where are you getting those expressions from

[hard knot]

made them up

🙂

[little totem]

oh ok

so for this particular problem

what are the two things we multiply to get 0

[twilit comet]

wait i think there's a much simpler solution--

solution a is that sin theta = 0

[little totem]

thats what we are doing

[twilit comet]

solution b sin theta is not 0, then u just divide-

[little totem]

from first principles

[twilit comet]

so then use arccosine

[twilit comet]

...

ok

[hard knot]

Idk,

i just know
5(5x+3)
x = 0 and x = -3/5

[little totem]

did you see the first step here

do you see what i did

[hard knot]

Yeah

[little totem]

and do you see there are two expressions that multiply to get 0

[hard knot]

Sin x 8cos

= 0?

[little totem]

you have one of them right

the other one is wrong

you changed it

[hard knot]

um

sint x (-5) = 0

[little totem]

you don't see these two things are being multiplied?

[hard knot]

Yes? i do

[little totem]

that's the answer im looking for

[hard knot]

oh ok

I thought u meant like this

[little totem]

the two things being multiplied are sin t and 8cos t - 5

no i just meant the two expressions

these are the two things we multiply to get 0

suppose sin t = a

8cos t - 5 = b

ab = 0

[hard knot]

oh alright so thats how it works

[little totem]

yeah so

if ab = 0

what do you know about a and b

[hard knot]

a = 0

b = 5= 8cost

[little totem]

no you skipped steps

[hard knot]

oh otherway around

[little totem]

for now just think of a and b as numbers

if ab = 0, what does that tell you about a and b

imagine a and b are real numbers

[hard knot]

One of them is 0 or both

[little totem]

yes

a = 0 or b = 0

now we substitute

sin t = 0
or
8cos t - 5 = 0

[hard knot]

sint = 0 = a
cost = 5/8 = b

a = 1/2 Pi?

b = 0.89?

Or am i jumping

[little totem]

don't use a and b, you're confusing yourself

you're solving for t, not a and b

a and b was simply there to help you wrap your head around how we are splitting into two cases

[hard knot]

t = 1/2 pi and t = 0.89

[little totem]

sin pi/2 = 0?

check that

[hard knot]

= 1

[little totem]

yeah, we need sin t = 0

[hard knot]

oh

So i put 0 degrees

and not 0radian

[little totem]

i mean, that doesn't change anything

0 deg = 0 radian

it's 0

[hard knot]

Idk, i put in the calc Sin^-1 (0) when its in radian form

but yeah i got t= 0

[little totem]

you really should not need to punch inv sin 0 into a calculator

imagine someone had to punch 5+5 into a calculator

could you? absolutely

but if someone did that you'd be like "wow they don't know how to add"

you should have a very solid understanding of what inv sin 0 is

[hard knot]

I need to get a feel of the values a bit more, i dont use circle theorem

Thats probably why

[little totem]

yeah if you want we can work on that later

but for now, sure

t = 0

[hard knot]

Yes

and t= 38.7 C

[little totem]

im guessing that's degrees

[hard knot]

Yes

[little totem]

sure

but another question

the problem probably states that it wants all solutions of a certain range

[hard knot]

Its for a triangle

[little totem]

what is that range? or does it just want all solutions

ohhh

[hard knot]

So less than 90

[little totem]

ok then you're good, those are both solutions

[hard knot]

I had to do one of these things

[little totem]

ok just checking

because you always should check the desired range of the problem you're solving

[hard knot]

Learnt you can split it

[little totem]

now do you want to go over the circle thing really quick

[hard knot]

But apparently

0 is not a solution

[little totem]

it makes sin / cos super easy to understand

i mean, does it make a degen triangle?

if it makes a degen triangle you generally toss the solution

[hard knot]

No that makes sense

yeah

[hard knot]

Ik ik but, maybe in a month lol :3 thank you for the offer tho!

Everyone is trying to make me swallow circle theorem

but i survived

so

Ill see

Maybe another date

[little totem]

idk what "circle theorem" is but

the unit circle is how we define sin and cos

so if you don't have a good grasp of that you might struggle with trig

[hard knot]

unit circle*

[little totem]

you're just feeling away at things rather than understanding why it is

yeah I don't use that

[hard knot]

oh?

THEN HIT ME

[little totem]

so like i said

the unit circle is how we define sin and cos

let me give the definition of sin and cos now

so obv, the input of a trig function is an angle measure

so imagine that theta is your input

you turn theta angle measure

then draw the ray until it hits the unit circle

the point where it hits the unit circle can now be written as (x,y)

and very simply

x = cos theta
y = sin theta

that's it

[hard knot]

🤨

Yeah you use that in mechanics

If I remember correctly

[little totem]

probably

[hard knot]

Horizontal and vertical

[little totem]

so now watch how easy it is

to solve something like sin t = 0

as you know from the definition we just gave

sin t corresponds to the y coordinate of where the ray hits the unit circle

so sin t = 0 is saying the y value is 0

you have t = 0 and t = pi

that's it

the red line is y = 0

[hard knot]

Because it can’t be negative

[little totem]

what can't be negative?

[hard knot]

Oh sorry

I jumped

You can’t have a negative X value

Sin value

And the range is 0 to Pi

[little totem]

you totally can

both sin and cos range from -1 to 1

for instance, suppose you wanted to solve

sin t = -1/2

draw y = -1/2

and mark where it hits the unit circle

those are your solutions

in this case -pi/6 and 7pi/6

easy

[hard knot]

Oh

How does that change with cos and tan

[little totem]

cos t is the x value

so suppose you wanted to solve like

cos t = 1/2

draw x = 1/2

and then mark

your solutions are pi/3 and -pi/3

[hard knot]

Oh ok

And if it was in a specific range

You would just +- 2Pi

Right these values?

[little totem]

exactly

[hard knot]

Oh ok

[little totem]

you can always add or subtract as many 2pis as you want

so most people do what is called finding the principal solutions

[hard knot]

Is tan possible on that as well

Because it’s weirdly shaped not circular

[little totem]

the solutions between 0 and 2pi

and then just add a +2pi*n

where n is any integer

tan is also really really easy

you know that tan = sin/cos right?

[hard knot]

Yeah

[little totem]

so this is one way of getting to the point (x,y), yeah?

[hard knot]

Righttt

Yeah

[little totem]

y = sin t and x = cos t as we have established

[hard knot]

But that’s only one solution?

Or all possible solutions?
Like in the other you always had 2 angles

[little totem]

so tan = sin/cos = y/x

one solution, im showing you the visualization

what the concept is

does y/x ring a bell for you?

[hard knot]

Gradient

Oh right

[little totem]

idk if that's the same concept, but here in the US we more commonly refer to it as slope

[hard knot]

Y= 3/5x
Find tan0
Tan0= 2/5

[little totem]

so the tan function

converts angle to slope

[hard knot]

If you had
y= 3/4x + 7

Could you find the tan0 value?

Is it still tan0 = 3/4

[little totem]

not sure what y = 3/4 x + 7 has to do with anything

tan 0 = 0 always

because if you are at an angle of 0, the slope is always 0

[hard knot]

Tan theta sry*

[hard knot]

Cause y/x is gradient

So I just gave an equation of the line

To check if you can mix it with that

Or too jumpy

[little totem]

too jumpy

so let's do this

[hard knot]

[little totem]

suppose I want to solve

tan t = 3/4

it turns out this doesn't have a nice answer, but you can ballpark estimate it with a solid conceptual understanding

draw the line passing through the origin with a slope of 3/4

the solution on the right is approx like 35-40 deg?

the solution on the left is gonna be like

180 + whatever that 35-40 is

[hard knot]

something like that

[little totem]

yeah something like that

it's gonna be higher than 30 though because

at an angle of 30 the slope is 1/2

but that's the idea

[hard knot]

oh ok

[little totem]

tan converts angle to slope

[hard knot]

how do i get the correct angles?

[little totem]

inv tan converts slope to angle

[hard knot]

okok

[little totem]

well if it's a nice 30-60-90 ratio or something like that

then you just work it out geometrically

[hard knot]

Oh right

yeah makes sense

[little totem]

but if it's not, then it's probably really ugly

[hard knot]

They always give you a nice Pi

[little totem]

you would be able to get a good estimate from this conceptual understanding

but a more precise answer you probably want to grab a calculator

[hard knot]

Do you know if there are any circle questions

Where you have to use this?

[little totem]

but it's important to be able to see trig functions in this way or you might not really understand where all of these different solutions are coming from

basically almost any trig question

like

sin t = 0

a lot of students will just punch inv sin 0 into the calc

get 0

and miss the pi solution

the inv trig functions can only output one principal value, but it misses all of the other valid solutions

by thinking about trig functions in this way, you know how to think about them in a way that you won't miss them

rather than just memorizing obscure rules about trig

everything you learn about trig in school can be derived from first principles using this visualization

[hard knot]

this stuff right?

I saw this a while ago

[little totem]

oh that's a little abstracted away

i thought you were starting a unit on trig

[hard knot]

no i am

[little totem]

this is like second year calc

[hard knot]

its just when i searched up all trig formulas

this came up

[little totem]

yeah these?

[hard knot]

and Ik most of it

[little totem]

these are all derived from first principles

[hard knot]

except for reduction

[little totem]

reduction you won't need until calc, since it involves integrals

if you don't know what integrals are don't worry about it

[hard knot]

oh ok

good news for me

I think

[little totem]

oh

wait

you linked like

two different "reduction" formulas lmao

[little totem]

this is like calc, you can ignore this

[little totem]

the reduction formulas here you do actually need to know

[hard knot]

Oh idk

Ohhh

Ok

[little totem]

these are important

i didn't see that they were called reduction formulas

so yeah if you understand what i just explained

you actually don't need to memorize any of the reduction formulas

they take like a few seconds to visualize in your head once you get used to it

so take for example

the first one

sin(-t) = -sin t

[hard knot]

Its similar to transformations

no?

[little totem]

yeah

[hard knot]

If not the exact same

[little totem]

it's all transformations

it's saying

suppose i have an angle t

that gives me a particular x, y, and the y value is the sin value we want

sin t = y

if i flip the angle to be negative

that's the same thing as flipping y to be negative

that's it

why would anyone ever memorize the reduction formulas

[hard knot]

I think im following

[hard knot]

You just need to memorize transformations

which links to unit circle???

kinda

no

f(x)

no

Well yeah no

Is this transformations?

bc thats the next topic :3

[little totem]

this is a bit more advanced

[hard knot]

great

[little totem]

so like

[little totem]

imagine the top row here

pythagorean and reduction

• the reciprocal and quotient identities

those are like "level 1"

with only the circle thing i showed you

all of these identities can be understood from first principles

[hard knot]

yeah

[little totem]

law of cosines and law of sines (not shown on the sheet) are like level 2

once you have a basic understanding of trig functions, you can apply a little bit of geometry to derive those as well

and the two of them together allows you to solve all triangles

[hard knot]

oh wow

theres more

very disapointed

[little totem]

everything in the middle, starting with the angle addition formulas

that's like level 3

if you want to understand them from first principles, you need to take a detour to understand complex numbers

but once you have the angle addition formulas, everything else can be derived fairly easily

[hard knot]

good news

[little totem]

they probably either

threw the angle addition formulas at you

and you just memorize them

[hard knot]

Yeah i do, and also the double angle formulas
Acc no, addition formulas are given double angle is memorize

[little totem]

or they did some geometry proof of them, which technically works, but i don't find them elegant or particularly helpful conceptually

double angle formulas are pretty easy as long as the angle addition formulas are memorized

so worst case scenario you memorize the angle addition formulas

and then derive everything else

[hard knot]

Oh actually?

Oh its just

[little totem]

yeah a = b

[hard knot]

cos(A+B)

yeah

i do B =A

[little totem]

same thing

[hard knot]

yeah

oh ok

maybe ill learn how to prove them

quickly

before i go next topic

Thats good

!

[little totem]

i think all trig students should try to practice the derivation of the trig formulas

because it's good practice and you feel that they are justified while getting familiar with them

and that experience will help you when you get to calc stuff

anyways i hope that helped

trig functions should not be difficult

they're very simple and very visual

[hard knot]

No that definetly helped

Im drawing a circle now

As practice

and do some questions again

Thank you, very much haha

Was really nice of you

[little totem]

np

[hard knot]

Shukran

emerati? or just emiritus

nvm emiritus is a word

lol

thanks xD

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[random shoal]

Hey guys, what'd I do wrong?

[random shoal]

kinda need this rn, my test's tmr

[restive river]

no real roots
downward opening parabola

you did that ?

[random shoal]

parabola?

lemme google

what i did was to identify a, b and c, then substitute in the values for b^2-4ac

then i drew that graph to find the range of values

yo wait i just realizedwhat the problem is

4th to 5th line lol

aight bye. imma ace this test

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[devout snow]

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[marble vine]

Hello

[marble vine]

We take three distinct points at random from the points on the grid opposite.
● ● ● ●
● ● ● ●
● ● ● ●
How much different ways of obtaining three aligned points. And what's it probability
(If you find 14,it's false).

can someone help me how to find this

[devout snow]

@marble vine Has your question been resolved?

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[marble vine]

nvm i find 20

.close

.reopen

[devout snow]

✅

[marble vine]

.close

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[hot epoch]

Given a system of 3 linear equations (Ax +By +Cz +D=O, A1.., A2.., A3..) in 3D-space, what are the solutionspaces and the conditions for the coefficients?

I did this for two planes first, then did this one but was told to find a systematic method for it and use pivots, and idk how that works here

[frank zinc]

A pair of dice is rolled, what is the probability that a number is divisible by 2 and 3

am I correct? Not sure

[pastel pasture]

Cramer's rule

[pastel pasture]

open your own channel

[frank zinc]

wait this is taken

my bad

[pastel pasture]

yeah

[hot epoch]

But I’m supposed to deduce the different possible solutionspaces and give conditionals with a general case

[devout snow]

@hot epoch Has your question been resolved?

[devout snow]

@hot epoch Has your question been resolved?

[devout snow]

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[frail bloom]

help

[frail bloom]

why is it not 1

and if it is 1 what mistakr did he make

[twin thunder]

math problems like this create wars

long ones

sed that math like this can't be subjective

[restive river]

the problem could be interpreted as either of the following: [
\p{6\divsymbol2}\p{1+2} \tss{or} 6\divsymbol\p{2\p{1+2}}
]
so either answers can be interpreted to be correct

[restive river]

So the moral of the story is to use either parentheses or fractions to be less ambiguous

[frail bloom]

😓

im insatisfied

i wanted my answer to be right

[twin thunder]

sometimes, math is like that

[frail bloom]

math l

well thanks

.close

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[thorn umbra]

the plus c is thrwoing me off

[robust cradle]

so what is the problem here?

[thorn umbra]

i can factoerisde

but how do i get the plus c

[thorn umbra]

would c be zero

[robust cradle]

yes

-b+c = -4

[thorn umbra]

lol i got baited

i thought c would be a number so i didnt attempt

[robust cradle]

0 is a number

lul

[thorn umbra]

yhbut like u know

anywayssssss

.close

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[split ridge]

Could anyone please help me solve this question about regular expressions? I hope you can help me, as this question is more about theoretical computer science.

[supple knot]

#old-network for CS server

[split ridge]

This is my question. This is my answer: L((a + b)(abb(a+b)+λ)) This is my mistake: Your regular expression generates the word abbabb while it should not.

[split ridge]

Thank you, but the server is not as active as this one and I would really like to figure this out asap. So if anyone knows, please let me know!

[supple knot]

Still in the wrong place

[devout snow]

@split ridge Has your question been resolved?

[split ridge]

.close

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[split ridge]

.close

[urban vector]

Hello! Sorry to be a bother but I have a few problems on my practice exam that I have no idea how to solve because we weren't taught, and I would like to at least familarize myself with the concepts before my exam.

Here is the first problem:

[urban vector]

I understand that we have to use trig substitution here purely because I had to look it up, but I don't know how this will equate to cot^2x, and its mildly concerning given we just learned how to do trig sub using trig indentities this friday and class was cancelled today

[scarlet sequoia]

Emm

Let x = 3sin(u)

Have you considered that?

[urban vector]

i have, i just don't understand why this is a trig sub problem in the first place, if that makes sense

also, do I have to open a new channel for each problem? I don't mind but would like to know

[scarlet sequoia]

it's sort of an intuition but you can see that it leads into Pythagorean Identity

which is just nice

in a way to solve that

and we get rid of the square root which is problematic here

[urban vector]

uh

oops hold on

that also didnt work

so i have:

[supple knot]

du is not dx

[urban vector]

yes latex to image is being wonky

[supple knot]

find du in terms of x and dx

[urban vector]

yes yes

i know du is this

however

i dont know how we get 3cos^2(u)

in terms of what we have in the equation

oh wait a minute

[scarlet sequoia]

9 - 9sin^2(u) = 9(1-sin^2(u)) = 9cos^2u ---> 3cos(u)

(after taking square root)

[urban vector]

ok ok so that makes sense

[scarlet sequoia]

it should be dx = 3cos(u) du, right?

[urban vector]

yes im having trouble typing in latex apologies

[scarlet sequoia]

no problem, it's fine

[urban vector]

so that basically leaves us with

uhhh

[scarlet sequoia]

it gives

[urban vector]

right? or am I blanking here

[scarlet sequoia]

3cos(u)/9sin^2(u) * 3cos(u)

[urban vector]

oh wait

so its just 1/9sin^2(u)?

[scarlet sequoia]

no, how?

[urban vector]

wouldn;t the 3cos(u) cancel

because its on both the top and the bottom?

[scarlet sequoia]

it's not on the top and the bottom

it's twice at the top

first one comes from sqrt(9-x^2) --> 3cos(u)

and secone one comes from dx = 3cos(u) du

[urban vector]

oh you wrote this as cos3(u) on the bottom sorry

i got confused

ok so now the cot^2 makes sense

[scarlet sequoia]

Ah, brackets would be great haha

Yep, can you proceed?

[urban vector]

LMAO

ok so

integral of cot^2 is -x - cotx - c

but we are in thr form of u

[scarlet sequoia]

but you know that x = 3sin(u)

then what's u

[urban vector]

u isssss

arcsin(x/3) ?

[scarlet sequoia]

perfect

[urban vector]

so the answer is :

and i do not believe I need to simplify there

ok sweet thats great

[scarlet sequoia]

I think it can be as it is, but 2nd one is sqrt(9-x^2)/x

[urban vector]

thats easy to do once you knoq hoq LMAO

[urban vector]

huh

[scarlet sequoia]

(it can be obtain from the substitution)

[urban vector]

how in the hell do we get to tere LMAO

[urban vector]

yeahhhhh im leaving as is

would you be able to help me with anymore or should I open a second ticket/

?

[scarlet sequoia]

Well, you can post it here

[urban vector]

gotcha

one was stumping me but the calculator online says its integration by partial fractions so I'll watch a video on that instead

this one I think I have to use u-sub for but im unsure

I set u = 2x + 1 and du = 2dx

so the integral would come out to

2 integral from -3 to -1 of f(u)du

[scarlet sequoia]

x = 3sin(u) ---> sin(u) = x/3
x^2 = 9(1-cos^2(u)) ---> cos^2(u) = 1 - x^2/9
cos(u) = sqrt(9-x^2)/3
cot(u) = cot(arcsin(x/3)) = sqrt(9-x^2)/3 * 3/x = sqrt(9-x^2)/x

[urban vector]

do I need to redo the bounds for the integral for the graph? they would come out to -5 as lower bound and -1 as the upper bound

[scarlet sequoia]

why 2?

there is also 4 there

[urban vector]

i pulled the 4 out from f(2x+1)

[scarlet sequoia]

ok

[urban vector]

and then du = 2dx, do 1/2 du=dx

so i pulled the 1/2 out as well

[scarlet sequoia]

okay, fine

[urban vector]

then do I rearrange the bounds?

[scarlet sequoia]

yes, when dealing with definite integrals and u-sub we are supposed to change bounds

[scarlet sequoia]

I agree

[urban vector]

ok

so we have the new integral and we know the original integral is = 16

is the area of the new integral just 8?

[scarlet sequoia]

the original is

[urban vector]

the original is = 16

[woven radish]

Modus

[urban vector]

mhm the same as the original integral

except we have * 2

oh

32 not 8

i was dividing

[scarlet sequoia]

yeah

because we have 2 * this

[urban vector]

ok sweet i actually did that one

[scarlet sequoia]

and the variable is just a variable, u or x, doesn't matter, integral is same

[urban vector]

yeah

final one (for now I believe):

is this integration by parts?

[scarlet sequoia]

We can try

Seems so

[urban vector]

ok so u = x, dv = f''(x), v = f'(x), du = 1

so uv - int(vdu)

xf'(x) - int from 0-2 (f'(x)dx

so the integral is equal to :]

just 0

because f(2)=f(0)

so we just end up with x*f'(x)

what do we plug in for x here/

?*

[scarlet sequoia]

same

2 then 0

so it's

2 * f'(2) - 0 * f'(0)

[urban vector]

so -2 is the answer

[scarlet sequoia]

yes

[urban vector]

thank you for the help! i appreciate it its been very helpful

.close

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[devout snow]

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[tawdry wave]

Hey I wanna understand the logic behind this
(I just used a calculator to try the choices)

[dawn iris]

notice $\cos(x^{\circ}) = \sin((90 - x)^{\circ})$

[woven radish]

AlphaNull

[tawdry wave]

After that?

[dawn iris]

start writing out terms from the beginning and the end of the series

you will see a lot of cancellation

[tawdry wave]

Ah

[tawdry wave]

Ty

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[pulsar crater]

if i want to make a position time graph out of a velocity time graph that looks like this

[pulsar crater]

and im finding the rough displacement with the area method

would i do this

red area plus green area

to find rough displacement

[acoustic leaf]

yes, although area under the x-axis should count as negative for displacement

[pulsar crater]

ty

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[high notch]

Help me

[bright phoenix]

just do exactly what the paper says

[high notch]

I don’t get what to do

Idk how to find the slope

[bright phoenix]

Slope is (how far you went up) / (how far you went over)

If the line is like / it’s positive

and if it’s like \ it’s negative

[high notch]

So the one on the paper is positive basically

[bright phoenix]

Yes

[high notch]

and to find the slope I go up by the units??

[bright phoenix]

Count how many units you go up (or down) from the start point

And divide by how many units you went left or right

[high notch]

The start point of the triangle right?

[bright phoenix]

Start point of either Brandon or Madison

They each start in different places

[high notch]

so -3 , -3 and 0 , -1

[bright phoenix]

Yes

[high notch]

And go up

Okay thank you

[bright phoenix]

Then over

[high notch]

So up then over, then add them together?

[bright phoenix]

Divide them

Not add

Up/over

=slope

[high notch]

So I would start by -3 and go up

Then count the units goikg over and divide them

[bright phoenix]

Yes

It also tells you

To go up six

Over 9

So 6/9 = slope

Or 2/3

[high notch]

So 2 and 3 and 6 and 9

And that’s out slope

Our*

[bright phoenix]

2/3 by Brandon and 6/9 by Madison

[high notch]

Okay thank you

I also have 2 other pages if that’s fine

[bright phoenix]

No it’s not

You can do this

Just think for yourself

[high notch]

Okay I’ll try

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[torpid locust]

Could someone help me with this question? I had to translate it from French to English so please forgive me if the translation is choppy

[sand thorn]

Ok @torpid locust so let's do part (a)

[torpid locust]

Alright

[sand thorn]

do you know what a secant line is

[torpid locust]

The lign that passes by 2 points?

[sand thorn]

yeah

do you understand that if I have two points in space, there is exactly one line which crosses through both of them?

[torpid locust]

yes

[sand thorn]

do you know how to represent a line as an equation with variables and coordinates

[torpid locust]

delta y / delta x ?

[sand thorn]

I mean sure? what is that

what's x and y, and what is delta

[torpid locust]

i mean to find the secant line

[sand thorn]

yes

[torpid locust]

y2 - y1 / x2 - x1

[sand thorn]

yeah

well what are y2 and y1 and x2 and x1

[torpid locust]

y2 is f(2), y1 is f(1), x2 is 2 and x1 is 1

so basically 2 - 1 / 2 - 1

[sand thorn]

sure yeah

[sand thorn]

okay and what does this quantity represent?

[torpid locust]

the secant line?

[sand thorn]

well no not really

did you just memorize that or something lol

[torpid locust]

random guess lol

[sand thorn]

okay

do you understand how coordinate space works? if I have a piece of paper, and I pencil in a dot somewhere, how do I fully describe where the dot is?

[torpid locust]

Well you need to know where the penciled dot is located no?

[sand thorn]

What do you mean

No, I took a blank piece of paper, drew a dot in some random place, and I want to communicate to you exactly where it is

Suppose I have access to a ruler...

[torpid locust]

Well you’d want to use that ruler to mesure the distance of the dot

[sand thorn]

Distance to what?

[torpid locust]

From a y and x axis

[sand thorn]

Okay cool

be more specific

My paper is square 10 inches by 10 inches

(let's say)

how are you describing where the dot is

[torpid locust]

okay so

use the ruler to measure it out horizontally and get let’s say 6 inches

That would be your x

[sand thorn]

great, okay

well wait hold on, measure it out from where?

[torpid locust]

Well from both ends of the paper

[sand thorn]

like, put one end of the ruler on the bottom-left corner, and lay it out parallel to the horizontal edge?

and then wherever the dot lines up with the ruler marking, that's the x coordinate?

(I know I'm being pedantic lol, this is just for the sake of understanding)

[torpid locust]

No I mean from both ends of the paper

So for example

for the left end, I get 7 inches

and from the right end, I get 1 inch

And so 7-1 inches would be x axis

[sand thorn]

like this, you're saying

from the left edgwe, you get 7 inches

[sand thorn]

right, right

okay cool

well wouldn't you say the right-edge information is superfluous?

[torpid locust]

It would sure

Cause I felt I would be missing the other end

[sand thorn]

oh, sure

okay yeah

anyway okay so you have 7 from the left edge, you're calling this the x-coordinate

how about the vertical measurement

[torpid locust]

Sorry that I drew over your line lol

[sand thorn]

no ur good

yeah that's right

Okay so now we have a system for completely describing the location of points on this paper

Now completely describe this line

@torpid locust

[torpid locust]

Well I’m guessing that this line takes up the distance of the paper from one edge to another

[sand thorn]

Yes it goes from edge to edge

what's special about the points on this line?

can you describe the line not in terms of what it looks like visually, but what pattern the x- or y- coordinates must obey?

(there is a pattern)

[torpid locust]

Well the x and y coordinate should be passed by the line

[sand thorn]

what? no I'm saying, write a statement like:

Every point on the line with x-coordinate "A inches" and y-coordinate "B inches" satisfies ...

... A = B.

do you understand what I just wrote lol

that's the pattern which these points exhibit

every point on the line looks like (1,1) or (5,5) or whatever