#precalculus

1 messages · Page 137 of 1

open apex
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No clue what to do after

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I'm sure you're supposed to find a conjuction

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Induction*

patent beacon
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You're pretty much there.
e^M = Σ Mⁿ/n!

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So, tell me, what's the top left value of e^M?

open apex
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2^n ?

patent beacon
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That's Mⁿ you're thinking of

open apex
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2

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e^2

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WOW

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POKKOSNJG

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ASF

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GADBNADFB GDFAB GN

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That's makes sense

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Ok

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Yikes

patent beacon
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The top left value of e^M is Σ 2ⁿ/n!

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Which is indeed e²

open apex
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Sorry it's sideways

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Basically you'd just multiply it in

patent beacon
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,rotate 90

open apex
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Uh

patent beacon
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Derp

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,rotate 180

obsidian monolithBOT
open apex
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Anyways

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You'd just mutliply that in

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And it would look extremely gross

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But it would end up coming as

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e^2 e^2 - 1 0 e

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So the answer would be B

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Mk

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Normally math team questions aren't that gross

patent beacon
#

The top right is
Σ 2ⁿ¯¹/n!

open apex
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Sorry if I took up your time

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Which ends up turning into

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e^2 - e

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Not -1

patent beacon
#

Oop, right

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Thx yeah that makes sense

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See? I barely did anything, you got this

open apex
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Yeah, normally I can do most of these by myself

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But the topic that I chose this year was matrices and vectors

patent beacon
#

It's np I'm happy to help

bright dock
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Aye I need some help

patent beacon
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Aye, aye

bright dock
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How do I condense #71

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Further

patent beacon
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ln(x) - 2ln(x² - 4)?

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Pull the constant up if that's the case

bright dock
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Ln(x)-Ln(x^2-4)^2, ----> = Ln(x/(x^2-4)^2)?

patent beacon
#

Yeah that's the right process

bright dock
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Well I fucked up there a little bit

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But thnx

dim charm
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how do you rotate a point around a point that is not the origin

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would you first translate the the center point to the origin and translate your point that needs to be rotated too

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and then rotate using rotation matrix

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and then translate back again?

patent beacon
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@dim charm
Translation needs an extra dimension if you want to do it with linear algebra

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You might have an easier time just using functions for something like that

swift glacier
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Why would csc x= .92 be impossible?

tawny nacelle
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csc x = 0.92

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sin x = 1/0.92

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,w 1/0.92

obsidian monolithBOT
tawny nacelle
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and as we know sin oscillates between -1 and 1

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1.086.... is clearly not between this range

swift glacier
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ooooh I see now

tawny nacelle
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great

swift glacier
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I didn't think about making it into sin

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Thank you!

tawny nacelle
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np

swift glacier
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Also, if sin is positive, csc must be positive as well, right?

tawny nacelle
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yeah

swift glacier
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cool cool

open apex
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Also queem

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Queen*

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That also applies to 1/sec if positive

open apex
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God, I hate word problems

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If somebody's able to help would appreciate it 👍🏻

open apex
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Come on man, the dog looks good and everything

swift galleon
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@open apex do you still need help with this?

open apex
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Yes

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I can't do world problems

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That's why I've placed only twice in competitions

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And this saturday I had a competition and didn't place by like 6 points

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Which made me pissy

swift galleon
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oof

open apex
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Out of 5 competitions

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I've only placed in 2

swift galleon
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I'm just suffering due to silly mistakes... :/

open apex
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Oh

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Can't agree

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Especially because the last pre-calc invitationals test I took was all graphing

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And I suck ass at graphing

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: )

swift galleon
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ok

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well the key thing to realize here

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is that this problem basically creates a region that is an ellipse

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with the foci at points P and Q

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so you just need to find the major and minor radius

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and then A = pi * a * b

open apex
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oof

swift galleon
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wait I'll send you a picture of two scenarios that let you find a and b

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@open apex

bright dock
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Ok idk where this belongs bit here goes

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But

swift galleon
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the bolded lines are the 30 yd rope

open apex
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Oh

bright dock
open apex
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Don't get me wrong, I don't have problems solving it

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I just have no fucking clue where to start on it

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Anyway, thank you, that makes sense

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Anyway

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This is algebra 2

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@bright dock

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But it's not hard to solve

bright dock
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Ok so the infos that we know are listed on top of the page and I'm trying to find the time to double amount

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Is my porcces correct?

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Process

open apex
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Mhm

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Pert

bright dock
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So I got 46 years

swift galleon
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the animation at the beginning of that page is pretty useful

bright dock
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Idk if 46 years is realistically correct

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Thats the issue lol

open apex
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Doesn't have to be realistic

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Sometimes math is just problems

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It doesn't have to be real

swift galleon
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^

open apex
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And btw @swift galleon

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Ty

bright dock
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Alrighty

swift galleon
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👌

bright dock
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Thanks alot

solar talon
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Can someone help me with vectors?

naive hawk
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What do you need help with about vectors?

open apex
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@solar talon

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What do you mean by "Help with vectors"

patent beacon
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What do you mean by "Help"

open apex
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uhhhh

pseudo flax
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This isn’t pertaining to math but it’s an extra credit opportunity for macroeconomics & I feel as though learning how to solve these can broaden my horizons if someone is willing to help?

📎 image0.png 📎 image1.png
ruby otter
fading token
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I think 3

ruby otter
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no 4 is correct

thick raptor
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"note x and X are different"

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Then what's x in the last image?

ruby otter
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small x is subset of capital x

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as u can see in first image

obsidian monolithBOT
thick raptor
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Oh you're using that notation

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f(A) = B, where B is set of outputs of f from inputs of A

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?

ruby otter
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ya

thick raptor
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Is the inverse well-defined?

ruby otter
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not given in the question

thick raptor
ruby otter
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its understood by U

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thats the question bro we cant question the question??

thick raptor
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Ofc we can

ruby otter
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where are u from??

thick raptor
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The internet

ruby otter
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country??

thick raptor
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dwai

ruby otter
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have u heard of JEE

thick raptor
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Nope

ruby otter
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thats why u r talking like that XD

thick raptor
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Assuming the inverse is well-defined, 3 and 4 both should have no problems

ruby otter
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ya the same

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i thought 3 AND 4 are correct

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but only 4 is correct

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thats my doubt if 4 is correct then how is 3 wrong??

thick raptor
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It isn't, from the information given and the assumption I made

ruby otter
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ya thats what im saying

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but 3 is wrong somehow and i dont know how its wrong

thick raptor
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If anything, 4 is the one that's wrong if f's inverse isn't well-defined

ruby otter
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check last question
28)

thick raptor
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(Can't be bothered)

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And answers provided can be wrong

ruby otter
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no the answer is correct it is given in many source

thick raptor
ruby otter
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and this is a JEE question

thick raptor
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Still means nothing to me tbh

ruby otter
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-_-

viscid thistle
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lol

ruby otter
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if u know the seriousness u wont talk like that

viscid thistle
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serious stuff can still be wrong

ruby otter
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it wont be when it can decide fate of 11 lakh students

thick raptor
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🤷 Just don't trust anything to be infallible tbh

ruby otter
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i have no choice coz im one in that 11 lakh 😰

thick raptor
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well

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🤷

viscid thistle
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i think the answer is (d) lol

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cuz the inverse goes Y -> X

ruby otter
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YA THE ANSWER IS D

viscid thistle
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you dont have to type in all-caps

ruby otter
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that came by accident

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then prove why c isnt the answer

viscid thistle
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because the output of the inverse has to be a subset of X not Y

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but the question is so vague

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who knows if the inverse is even well-defined

thick raptor
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if the inverse isn't well-defined i.e. f({-1, 1}) = {0} then you can get f^-1(f({1})) = {-1, 1} for example

viscid thistle
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also

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hello sa

thick raptor
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no

viscid thistle
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lol

ruby otter
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what just happened

thick raptor
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dwai

ruby otter
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what is dwai

thick raptor
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thonkeyes you're worrying about it already

ruby otter
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??

thick raptor
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precisely

ruby otter
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fine leave it have i another question

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f(x) = ax/ (x+1)

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find value of a such that f o f (x) = x

thick raptor
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well just plug it into itself

ruby otter
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its easy till that

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try to find the the final result

thick raptor
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?

ruby otter
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coz i got stuck

thick raptor
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$$f^2(x)=\frac{a\frac{ax}{x+1}}{\frac{ax}{x+1}+1}=\frac{a^2x}{(a+1)x+1}$$

obsidian monolithBOT
thick raptor
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set equal to x

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$$a^2x=(a+1)x^2+x$$

obsidian monolithBOT
ruby otter
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im sorry

thick raptor
ruby otter
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but fking skip all this

thick raptor
royal gull
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@thick raptor why squared f?

thick raptor
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?

ruby otter
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i know to simplyfy please skip to a point where u get quadratic

royal gull
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f o f =/= f^2(x) right?

ruby otter
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no

royal gull
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or is it jsut a ntoation

thick raptor
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that's function iteration notation

royal gull
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k

ruby otter
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its f( f(x))

thick raptor
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you're probably thinking of that whack trig notation

ruby otter
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composite function

royal gull
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yeyeye

thick raptor
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thonkzoom literally that trig notation is terrible tbh

vernal anchor
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Just set a=-1, and verify that it works

royal gull
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wai

ruby otter
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i know a =-1

thick raptor
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makes people think like sin^-1(x) means 1/sin(x)

royal gull
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isnt it ? xD

ruby otter
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but u cant set and verify here

vernal anchor
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a=1+x also works

thick raptor
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GWnanamiAWAUGERY not even consistent notation

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rEEEEEEEEEEEE

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no it not!

ruby otter
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i need to solve

thick raptor
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sin^-1 is arcsin (a.k.a. inverse function of sine)

royal gull
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ah ye

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well

thick raptor
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lmao no

royal gull
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so wahts the notation for 1/sinx

thick raptor
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a = 1+x 😂

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wew

royal gull
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I mean why not just type arcisn

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and leave 1/sin for the sin^-1

thick raptor
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normal notation says you should do [sin(x)]ⁿ

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not this whack sinⁿ(x)

royal gull
ruby otter
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but both are same rit??

royal gull
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well tehcnically, but naaah

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🎶

ruby otter
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sinⁿ(x) and [sin(x)]ⁿ are same if (n != -1)

viscid thistle
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🎶

vernal anchor
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what

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sin(sin(x)) is not the same as sin(x)sin(x)

viscid thistle
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f^(-1) (x) should be known as arcf

ruby otter
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u get
(a+1) x^2 + (1-a^2) x =0 right??

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@thick raptor

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arcf can also be af

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😂

thick raptor
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@ruby otter yes

ruby otter
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what should i do after that

royal gull
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lit af 😂

ruby otter
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im stuck therer

thick raptor
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And no to the other thing

viscid thistle
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factorise

ruby otter
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factorise what

thick raptor
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sin(sin(x)) is not the same as sin(x)sin(x)

ruby otter
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wut??

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(a+1) x^2 + (1-a^2) x =0
factorise with what variable??

thick raptor
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a+1 = 0

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1 - a^2 = 0

ruby otter
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a = +-1 or -1

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so a =-1

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ok we are asked to find a so why we form a quadratic with x??

thick raptor
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It's not a quadratic

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When a = -1 it's equal to 0

ruby otter
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it has x sq and x but not const termm

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so isnt it a quadratic??

royal gull
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but coefficients are 0 for a = -1

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0x^2 + 0x =0

ruby otter
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ya so??

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like we why put
a+1 = 0
1 - a^2 = 0

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we are we equating with zero??

royal gull
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wait post question again, what do you haver to do?

ruby otter
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f(x) = ax/ (x+1)
find value of a such that f o f (x) = x

royal gull
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you take everything on one side

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(a+1) x^2 + (1-a^2) x =0

ruby otter
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what u copying everything and pasting it agn??

thick raptor
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if the coefficient of x^2 is 0 it's not a quadratic

royal gull
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so I can see better and everyone else

thick raptor
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You don't call 0x^2 a quadratic

ruby otter
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ya so??

royal gull
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if you have a that makes you 0=0 that means its 0 no matter what x you put in

thick raptor
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So what you had was not a quadratic

ruby otter
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so u say what ever i put in x f(x) is not gonna change which is independent of a??

thick raptor
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No

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Whatever you put in for x, (a+1)x^2 + (1-a^2)x isn't going to change

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Since it has to equal 0

ruby otter
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@thick raptor u r soo high man
u r like what i said was wrong and say the same thing again

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-_-

thick raptor
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wtf no

ruby otter
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u said its not quadratic and then u say what we have is not a quadratic

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😂

thick raptor
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f(x) is definitely not the same as (a+1)x^2 + (1-a^2)x

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Yes

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It's NOT a quadratic!

ruby otter
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i meant f o f (x) is independent for whatever value u give in as x if a==-1

thick raptor
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IT'S NOT

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er

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Double negative >.<

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(fof)(x) = x

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It does depend on x

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And is NOT the same as (a+1)x^2+(1-a^2)x

ruby otter
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wait its kinda my fault coz i thought of smtng and typed smtng which have another meaning

thick raptor
ruby otter
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😂

pseudo flax
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@ruby otter Wait I’m confused how do I determine f & b & x & y?

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Unless the formula you posted wasn’t for my images?

ruby otter
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f & b & x & y??

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which formula u mean ?? @pseudo flax

azure lark
#

Can someone help me with these? I missed a day and I’m pretty lost

📎 image0.jpg
limber bone
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lets do C since its the hardest one

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first distrubute the power

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then collect same bases

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and minus them

sharp pagoda
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HEY OVER HERE

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how would i solve

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2sin(2θ) = √3?

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any help would be appreciated, been stuck on this question for too long and its making no sense to me

solemn tiger
#

Are you allowed a calculator?

sharp pagoda
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i believe so

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@solemn tiger

solemn tiger
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Your goal is to look for theta, right?

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How would you go about isolating it?

sharp pagoda
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but the answer is exact

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divide the 2 over to root 3

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sin(2theta) transforms into 2sinacosa

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2sinacosa= √3/2

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now im stuck

obsidian monolithBOT
sharp pagoda
#

u ?

solemn tiger
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Any variable

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Substitute it in for 2*theta

obsidian monolithBOT
sharp pagoda
#

never seen this before sorry

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substitute pi/3?

solemn tiger
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We're finding what angle measures yield a sine value of sqrt(3)/2 in [0, 2pi]

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That gives the solution for u, but u = 2*theta

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You can solve for the solutions that way.

sharp pagoda
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like the reference angle?

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you saying pi/3 and 4pi/3 ?

solemn tiger
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pi/3 is right

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4pi/3 yields a negative solution

sharp pagoda
#

... sorry wasnt thinking

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2pi/3

solemn tiger
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So for sin(u) = sqrt(3)/2

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We know that the solutions are pi/3 and 2pi/3

sharp pagoda
#

correct

solemn tiger
#

We also substituted u = 2*theta

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Now we can back substitute for the two theta solutions.

sharp pagoda
#

so your saying sin(pi/3) = sqrt(3)/2

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and sin(2pi/3) = sqrt(3)/2?

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my teacher has never taught us this substituting

obsidian monolithBOT
solemn tiger
#

Our intent is to solve for theta, however.

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We set u = 2*theta, and we have u. Solving for theta should be easy.

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All we're doing is seeing what angle(s) produce the correct solution

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Then dealing with the inside of the trig function.

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For example, if we have sin(whatever function) = 1

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Without looking at the function, we already know that it needs to = pi/2

sharp pagoda
#

anyone down to help solve this?

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@solemn tiger had to leave but i figured the question out

obsidian zenith
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@sharp pagoda

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yeah thats a tough one, were gonna need some professional help for this one

stark inlet
sharp pagoda
#

oh my

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someone ban him

tawny nacelle
solemn tiger
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@sharp pagoda Try getting the equation in terms of one type of trig function, then getting it equal to 0.

fossil sedge
#

Umuick question...

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What is derivative of 1000(30-t)^3?

fossil sedge
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Anyone? Know what I do to the function after I put it into product rule form?

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Sigh

fading token
#

You had better us the chain rule in this case

slow wharf
#

I have a line that's defines as the intersection of two planes

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$p... \ x+y-3z+6=0\x-y-z=0$

obsidian monolithBOT
slow wharf
#

That turns out to be

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$x=-3+2t\y=-3+t\z=t$

obsidian monolithBOT
slow wharf
#

Which is correct

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Now I have to find two dots that are a part of the line that have the same distance from the point

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$T(1,0,1)$

obsidian monolithBOT
slow wharf
#

The distance is $\sqrt{8}$

obsidian monolithBOT
slow wharf
#

The formula for distance is

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$\sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}$

obsidian monolithBOT
slow wharf
#

Which would be

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$\sqrt{(-3+2t-1)^2+(-3+t)^2+(t-1)^2}$

obsidian monolithBOT
slow wharf
#

Right?

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Yup

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It's correct

royal gull
#

why do you post a question if you can do it lol

vocal zenith
#

Would you be able to help me with a maths question?

slow wharf
#

@royal gull I was getting it wrong, then I write it here in latex and I find mistake.

unborn island
#

well anyway

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does 3cos^3(x) -9cos^2(x) +cos x-3 have any real zeroes

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when i solve it i get cos x = 3, which isn't possible

mortal gale
#

What are the other two roots?

#

@unborn island

vagrant horizon
#

@unborn island Do a substitution where t = cos(x) and plug the equation into some online solving tool. There should be 3 roots in total, check for all of them like you did with cos x = 3

idle dust
#

hi

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can someone explain to me

#

what complex number in exponential form is?

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i am already familiar with rectangular and polar form

serene heath
#

$re^{i\theta}=r\cos(\theta)+ir\sin(\theta)$

obsidian monolithBOT
serene heath
#

lhs is exponential form

slow wharf
#

I am having problems calcualting limit

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$\lim_{x\to\infty}\sqrt{x^2+1}-\sqrt{x^2-1}$

obsidian monolithBOT
slow wharf
#

I know I can write $x^2-1$ as $(x-1)(x+1)$

obsidian monolithBOT
spring thunder
#

that won't help you tho

#

just rationalize that whole stuff

slow wharf
#

Hmm

spark gyro
#

it's usually better to work from the outside in than the inside out

#

this looks like a $\sqrt\bullet - \sqrt\bullet$ type problem from the outside

obsidian monolithBOT
slow wharf
#

It wokred out

spring thunder
#

lul

#

think_left 🤔 think_right think_down

slow wharf
#

How about this one

#

$\lim_{x\to1}\frac{1}{1-x}-\frac{3}{1-x^2}$

obsidian monolithBOT
spark gyro
#

"how about"
what did he mean by this 🤔

slow wharf
#

._.

spark gyro
#

have you tried it

slow wharf
#

Yes

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I believe I am missing something

spark gyro
#

what have you tried? what do you notice?

slow wharf
#

When I put them as one fraction

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It turns into

#

$\lim_{x\to1}\frac{-x^3+3x-2}{(1-x)(1-x^3)}$

spark gyro
#

👀

spring thunder
#

wot

obsidian monolithBOT
spark gyro
#

how did that happen

slow wharf
#

No?

#

$\lim_{x\to1}\frac{1(1-x^3)-3(1-x)}{(1-x)(1-x^3)}$

obsidian monolithBOT
slow wharf
#

This is a step earlier

spark gyro
#

well first of all, it's a square, not a cube, isn't it?

slow wharf
#

Which one?

spark gyro
#

you wrote in x^2 initially

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and now everything else is being x^3

slow wharf
#

Mistake

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it's x^3

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$\lim_{x\to1}\frac{1}{1-x}-\frac{3}{1-x^3}$

obsidian monolithBOT
slow wharf
#

$\lim_{x\to1}\frac{1-x^3-3(1-x)}{(1-x)(1-x^3)}$

obsidian monolithBOT
spark gyro
#

anyway, regardless, now the question arises, what have you tried from here

slow wharf
#

$\lim_{x\to1}\frac{1-x^3-3+3x}{(1-x)(1-x^3)}$

obsidian monolithBOT
spark gyro
#

once you turn it into a fraction, what's next

slow wharf
#

Well, the idea comes that I should somehow write the top part that it consists of one of the bottom parts so I can delete them

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Or

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difference of cubes

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from

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$(1-x^3)$

obsidian monolithBOT
patent beacon
#

Great thing about polynomials.
If f(a) = 0, then (x - a) is a factor

#

So there is some way to write
1 - x³ - 3 + 3x = (x - 1)(quadratic)

slow wharf
#

Hmm

#

Is there a simpler way to do it

#

$-(x^3-3x+2)$

obsidian monolithBOT
slow wharf
#

$-(x(x^2-3)+2)$

obsidian monolithBOT
slow wharf
#

I remember a trick with it

#

Rewrite 3x as two numbers

patent beacon
#

There's always synthetic division if you can't find a trick

spark gyro
#

you're getting caught up in the weeds a bit, there's more problem afterwards to worry about

#

you already mentioned difference of cubes $1 - x^3 = (1-x)(1+x+x^2)$

obsidian monolithBOT
slow wharf
#

Yes

spark gyro
#

so i'll just tell you $\lim_{x\to1}\frac{1}{1-x}-\frac{3}{1-x^3} = \lim_{x\to1}\frac{1+x+x^2}{1-x^3}-\frac{3}{1-x^3}$ and let's move on

obsidian monolithBOT
spark gyro
#

there's still work to do

slow wharf
#

How did you do that

spark gyro
#

multiply top and bottom by 1+x+x^2 and used difference of cubes

#

anyway, even if you do that you still have a 0/0

patent beacon
#

Be careful not to split a limit over terms that don't exist

#

If you divide top out by x - 1 twice, you get
lim -(x + 2) / (x² + x + 1) = -1

#

If I made no mistake

spark gyro
#

||i was ready to just l'hopital it when it was quadratic/cubic||

#

:P

serene heath
#

hospitals PandaRee

spark gyro
#

if it's stupid and it works it ain't stupid PandaOhNo

slow wharf
#

Okay

#

I used an online tool to find the factor of

#

$x^3+3x-2$

obsidian monolithBOT
slow wharf
#

and I was able to solve the limit

#

But how do I find the factor on paper

#

It seems very complex

patent beacon
#

Do you know synthetic division? This allows you to pull an (x - r) out of any polynomial where r is a root

slow wharf
#

I've never heard about it...

patent beacon
#

It's very quick and easy once you know how, take a look on google

slow wharf
#

Okay

patent beacon
#

I used it to fully factor
x³ + 3x - 2 = (x - 1)²(x + 2)

idle dust
#

KAYNEX SENPAI

#

HIIIIIIII

#

ACTULLY I HAVE A QUESTION

#

HOW DO U GO FOR POLAR FORM TO EXPONENTIAL FORM

#

FOR COMPLEX NUMBERS

slow wharf
#

You have a formula for it

#

Or wait

idle dust
#

woot

slow wharf
#

That's from normal for to polar

#

But exponential form is just like polar form but written differently

#

$r(cos\phi+isin\phi) = re^{i\phi}$

obsidian monolithBOT
slow wharf
#

Right?

serene heath
#

right

idle dust
#

ok u just say its equal

#

but how??

#

why e^i

#

?????

#

i am very confuzzled

slow wharf
#

@idle dust It's just another form

idle dust
#

oK

hasty cairn
#

do i need to memorize the unit circle

patent beacon
#

Yes. It's not hard there's only like 3 potential outputs other than 0 and 1

green anvil
#

How do you right interval notation of x=2 ?

stark trench
#

how do I use (x-h)^2+(y=k)^2=r^2 formula

#

sorry for dumb question. I blame the canadian curriculum for not teaching me. I'm forced to learn this stuff for ap calc.

fringe stream
#

it represents the circle of radius r centered at (h, k)

#

what about it?

green anvil
#

(x^2+6) x (sqrt{8-x}) Can this be simplified ?

#

Help

#

,help

obsidian monolithBOT
#

A brief description and guide on how to use me was sent to your DMs! Please use ,list to see a list of all my commands, and ,help cmd to get detailed help on a command!

green anvil
#

,ask (x^2+6) x (sqrt{8-x})

#

Welp

#

That scared me more then helped

obsidian monolithBOT
outer wing
tawny nacelle
#

,w plot y = sqrt(|x|)

obsidian monolithBOT
tawny nacelle
#

@outer wing is that it ?

spark gyro
#

,w plot y = log(|x|+1)

obsidian monolithBOT
spark gyro
#

there are a lot of functions it could be

viscid thistle
#

Idk why I got the range wrong

#

can someone help?

#

Sorry If you cant see but I got the HA correct which was 0. But then it marks me wrong for the range which I put as
(-∞, 0)U(0,∞)

fading token
#

Can you please take a clearer screenshot of the questions?

fossil sedge
#

Pls a. I need to know how to have the sum of areas interms of only one variable

remote musk
#

i remember doing this exact question

#

alright let's call the lengths x & y

#

let's try writing down the information we know x + y = ?

#

area of the square of side x = ?

#

area of the circle of circumference y = ?

fossil sedge
#

And then we get two and find the total sum equation...and then derivative that and find the maximum or minimum right?

remote musk
#

exactly, when it's max/min problems always look to see if derivatives are involved

fossil sedge
#

Yeah ok I forgot X plus y equals 100

#

Aight thnx for reminding

hybrid pewter
serene heath
#

that is a separable differential equation

hybrid pewter
#

so i differentiate left and then right?

serene heath
#

you get all the x terms on one side and all the ys on the left

#

then integrate both sides

hybrid pewter
#

where does dy/dx go?

#

just stay on the left?

serene heath
#

yea

hybrid pewter
#

so the left, i would get lny (y)

serene heath
#

$\int \frac{1}{y} \frac{dy}{dx} dx=\int \frac{2-x^2}{x}dx$

hybrid pewter
#

?

obsidian monolithBOT
hybrid pewter
#

thank you

#

i'll try thata

#

that*

#

what do i get from differentiating dy/dx

#

i mean integrating***

#

do i not get y?

serene heath
#

whats the integral of 1/y?

hybrid pewter
#

ln y

tawny nacelle
#

|y|

#

$\int \frac{1}{y} \frac{dy}{\cancel{dx}} \cancel{dx}=\int \frac{2-x^2}{x}dx$

obsidian monolithBOT
tawny nacelle
hybrid pewter
#

oH

#

i can do that???

#

so i can cut it

#

ok thank u

tawny nacelle
#

mhmm

hybrid pewter
#

sorry

tawny nacelle
#

y sorry ?

hybrid pewter
#

cause wth am i even asking lmaooo

#

i havent learnt this before

#

ive done this but i didnt know separating it was a thing

tawny nacelle
#

i thonk @serene heath short circuited when i did this

obsidian monolithBOT
serene heath
tawny nacelle
hybrid pewter
#

is that how i go about it??

thick raptor
#

You're technically just using chain rule

hybrid pewter
#

so im wrong??

thick raptor
#

I meant for the differential thing

#

$$\frac{\mathrm d}{\mathrm dx}\ln|y|=\frac1y\frac{\mathrm dy}{\mathrm dx}$$

obsidian monolithBOT
thick raptor
#

Due to chain rule

#

I'm just telling you what's happening there

tawny nacelle
#

explain this stuff $ \frac{dy}{\cancel{dx}} \cancel{dx}$

obsidian monolithBOT
thick raptor
#

Ya that's what I'm saying

#

You're technically just using chain rule in reverse and writing it differently

#

$$\frac{\mathrm d}{\mathrm dx}f(y)=f'(y)\frac{\mathrm dy}{\mathrm dx}$$

obsidian monolithBOT
thick raptor
#

If you want to integrate the right side with respect to x, you basically need to do:

tawny nacelle
#

so $\int \frac{1}{y}\frac{\mathrm dy}{\mathrm dx}dy = \int \frac{\mathrm d}{\mathrm dx}\ln|y| dy$

#

?

thick raptor
#

$$f(y)+C=\int f'(y)~\mathrm dy$$

obsidian monolithBOT
thick raptor
#

Ya sure

tawny nacelle
#

but then wat ?

thick raptor
#

But the point is we're basically integrating w.r.t. y

tawny nacelle
#

i see

hybrid pewter
#

so i just add dx bc i'm integrating it??

#

like if i integrate 3x then i also add a dx at the back

#

so now when i integrate 3x dy/dx, i also add the dx

#

i'm just doing a levels like anything else is beyond me

tawny nacelle
#

shit bro

#

i see it now

thick raptor
#

$$\frac{\mathrm d}{\mathrm dx}f(y)=f'(y)\frac{\mathrm dy}{\mathrm dx}$$ $$f(y)+C=\int f'(y)\frac{\mathrm dy}{\mathrm dx}~\mathrm dx$$ $$f(y)+C=\int f'(y)~\mathrm dy$$

obsidian monolithBOT
thick raptor
#

These are all equivalent statements

#

You have dx on the end because you do the same thing to both sides: integrate with respect to x

hybrid pewter
#

can i just add a dx at the back of any equation with dy/dx

#

OH

#

ooOO

#

ok thank u

tawny nacelle
#

yw

thick raptor
#

:l you never do different things to each side of an equation

hybrid pewter
#

so whatever method i used was acceptable right?

thick raptor
#

Not sure where c = b-a came from

#

oic nvm

hybrid pewter
#

two diff unknowns so i made it into 1

thick raptor
#

Usually you write it all consecutively instead of separately

hybrid pewter
#

so i integrate the whole thing together??

thick raptor
#

$$\frac1y\frac{\mathrm dy}{\mathrm dx}=\frac2x-x$$ $$\int\frac1y\frac{\mathrm dy}{\mathrm dx}~\mathrm dx=\int\frac2x-x~\mathrm dx$$

obsidian monolithBOT
thick raptor
#

I'd write it like that

#

And maintain an equality throughout

#

And only have one +C after the integrals are done

hybrid pewter
#

ohhh alright i'll do that

#

thank u

thick raptor
#

no

hybrid pewter
#

what

tawny nacelle
#

u gotta say arrigatto

#

he's a man of culture

thick raptor
tawny nacelle
thick raptor
hybrid pewter
#

is it not pre

thick raptor
#

No?

#

This is definitely calculus

viscid thistle
#

why do all calculus books have an integral sign

#

like some people really think that that's what Calculus is all about...

empty mesa
elfin night
#

⬛

patent beacon
#

It's not a small part of it

elfin night
#

smol

slow wharf
#

Find the tangent on the parabola $y=x^2-7x+3$ parallel with the line $5x+y-3=0$

obsidian monolithBOT
slow wharf
#

Any idea?

serene heath
#

well whats the slope of the tangent

slow wharf
#

Hmm

#

Waiiit

#

Alright so the line is

#

$y=3-5x$

obsidian monolithBOT
slow wharf
#

And since the tangent is parallel to it, it's slope is -5

#

$y=k(x-x_0)+y_0$

obsidian monolithBOT
slow wharf
#

k is supposed to be -5

serene heath
#

yes

slow wharf
#

The only question is how I find the point

serene heath
#

well

#

u now the gradient

#

u can use the gradient function

#

to work out the point

slow wharf
#

gradient function?

#

Wait

spring thunder
#

you need to know for which value of x the slope of the tangent may be -5

#

ie take the deriv of x->x^2 -7x +3

slow wharf
#

I find the derivative of the parabolla and make it equal to -5

spring thunder
#

ya

slow wharf
#

AAAAhhhhh

#

Awesome

#

Thank you PandaHugg

serene heath
slow wharf
#

@serene heath The thank you was for you too catshrug

serene heath
slow wharf
#

Woot PandaOhNo

spring thunder
#

we're all becoming more and more like SA yk

slow wharf
#

WHat's that

worn forge
#

anyone not busy and able to help me on a project

elfin night
#

SA has infected the server PandaRee

#

and wat project @worn forge

worn forge
#

i have to record a video of information looking for a hand

slow wharf
#

What does this mean?

#

Is the sum a part of a?

spring thunder
#

yea

slow wharf
#

So, the limit of it would be like

#

$\lim \frac{1}{n^3}\infty$

obsidian monolithBOT
slow wharf
#

I mean the sum goes to infinity as n approaches infinity

spring thunder
#

yea

#

but you can express that sum in terms of n

slow wharf
#

hmm

spring thunder
#

$\sum_{k=1}^n k = \frac{n(n+1)}{2}$

obsidian monolithBOT
spring thunder
#

already saw this?

slow wharf
#

um...

#

no?

spring thunder
#

not familiar with sums at all ig

#

well that one is provable pretty easily

slow wharf
#

Um I am to a certain degree

#

I know how to calculate if they converge or diverge using cauchy and de'almbert theorem

#

And to what they converge to

#

This one obviously diverges to infinity though...

spring thunder
#

actually it doesn't

slow wharf
#

No?

spring thunder
#

that sum diverges to infty indeed

#

but 0 * infty is indeterminate my man

slow wharf
#

I was talking about the sum alone, without the sequence

#

Or without a

spring thunder
#

$$a_n = \frac{1}{n^3}\left(\sum_{k=1}^n k^2 + \sum_{k=1}^n k\right)$$

obsidian monolithBOT
spring thunder
#

right?

slow wharf
#

Yes

spring thunder
#

then if you use the fact that $$\sum_{k=1}^n k = \frac{n(n+1)}{2}$$ $$\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$$

obsidian monolithBOT
spring thunder
#

$$a_n = \frac{1}{n^3}\left(\frac{n(n+1)(2n+1)}{6}+ \frac{n(n+1)}{2}\right)$$ $$a_n = \frac{n(n+1)(2n+4)}{6n^3}$$

obsidian monolithBOT
slow wharf
#

It's 1/3

#

Oh wow

#

I see

#

But is that somekind rule that the sum of k as k goes from 1 to n is equal to the fraction?

spring thunder
#

let S = sum of k as k goes from 1 to n

slow wharf
#

I think I understand

spring thunder
#

S = 1 + 2 + ... + n

#

S = n + (n-1) + ... + 1

#

2S = (n+1)+(n+1)+....+(n+1) [n times]

slow wharf
#

I understand

#

Thank you 😁

spring thunder
thick raptor
#

Personally I find it more intuitive to say the mean value of 1, 2, 3, ..., n is (n+1)/2 so their sum is n*mean = n(n+1)/2

spring thunder
thick raptor
#

Also to compute those limits you only need decent bounds and telescoping

#

Well

#

That specific one is easy telescoping

#

$$n(n+1)=\frac{(n+1)(n+2)(n+3)}3-\frac{n(n+1)(n+2)}3$$

obsidian monolithBOT
viscid thistle
#

Someone please help me in the call...

#

I need a moderator in the call plz. To make sure I get the correct answer.

#

Plz come in call moderator

royal aspen
#

$ \frac{a}{b} \cdot \frac{c}{d} = \frac{ac}{bd} $

obsidian monolithBOT
viscid thistle
obsidian monolithBOT
viscid thistle
#

what's 12' ?

royal aspen
#

12 minutes

viscid thistle
#

wut?

#

epic maths

royal aspen
#

its some weird stuff with trig

#

$ \frac{7^{\degree}\pi}{180} + \frac{12'}{60} \cdot \frac{\pi}{180}$

viscid thistle
#

so it's a 60th of a degree?

royal aspen
#

yes

viscid thistle
#

the hec

royal aspen
#

its some trig angle stuff

viscid thistle
#

thats stupid

#

who uses minut GWchadThinkeyes

obsidian monolithBOT
royal aspen
#

nobody

viscid thistle
#

$$\frac{7 + \frac{12}{60}}(\frac{\pi}{180})$$

#

w0w nice tex

#

don't tell me there are also seconds that are annotated "

royal aspen
#

lmao

severe verge
#

@viscid thistle read #info

viscid thistle
severe verge
#

also, stop asking people to get in a voice call with you

#

also, we're not obligated to help you

viscid thistle
#

Kk

royal aspen
#

i am being a good samaritan

viscid thistle
#

$$(7 + \frac{12}{60})(\frac{\pi}{180})$$

obsidian monolithBOT
viscid thistle
#

wow epic tex

#

So... to combine these, would I have to put a common denominator?

#

@viscid thistle do I now find the common denominator?

#

60*180≠108

#

Hold on.

#

I think I did something wrong.

#

y

#

do this $$(7+\frac{12}{60})\frac{\pi}{180}$$

obsidian monolithBOT
viscid thistle
#

Kk

#

Thx

#

$$(7+\frac{1}{5})\frac{\pi}{180}$$

obsidian monolithBOT
viscid thistle
#

$$\frac{36}{5}\frac{\pi}{180}$$

obsidian monolithBOT
viscid thistle
#

$$\frac{\pi}{25}$$

#

maybe thats wrong hol up GWchadMEGATHINK

#

xD thx

obsidian monolithBOT
viscid thistle
#

😄

#

yeah

#

Would anyone like to help me find at least which formula to start out with? I think all I need is that spark to make me continue going.

#

For #2

#

have you done trigonometry?

#

I did. However I think I need that spark so I can almost solve it myself.

#

Ur question about my trigonometry level I think sort of offended me though...

#

sorry ma'am

#

you want to find that angle basically

#

Ah. That makes sense. Thanks.

viscid thistle
#

Would anyone like to be in a call?

severe verge
#

no

viscid thistle
#

:^)

#

@viscid thistle -_-

dim charm
#

this showed up on a precalculus test and i was wondering what the answer is

#

A mathematician confided
That it is simply one-sided
And you'll get quite a laugh
If you cut it in half,
For it stays in one piece when divided.

patent beacon
#

Probably a mobius strip @dim charm

dim charm
#

thanks @patent beacon

dim charm
#

how do you find the equation of the angle bisector of the angle formed between two lines

patent beacon
#

I'm guessing you have the equation of those lines?
y = mx + b

m = tanθ
Where θ is the angle between the line and the horizontal. Use that to find both line angles, and then average them.

#

@dim charm

pastel quest
#

If y=(a+(x)^(1/2)) / (a-(x)^(1/2)), then why does differentiating it directly and then differentiating it after rationalising the denominator yields different results?

plucky marlin
#

let me put that in tex because I cant fathom that lmao

#

sec

#

$y=\frac{(a+\sqrt{x})}{a-\sqrt{x}}$

obsidian monolithBOT
plucky marlin
#

is this your quation?

#

question*

pastel quest
#

Yes

plucky marlin
#

and you are asking

#

when you rationalise the denominator

#

you get a different result for the derivative?

pastel quest
#

Yes

plucky marlin
#

,wolf derivative $y=\frac{(a+\sqrt{x})}{a-\sqrt{x}}$

pastel quest
#

Does it happen or am I making any mistake?

obsidian monolithBOT
plucky marlin
#

is this what you get?

pastel quest
#

No

plucky marlin
#

it didnt take it correctly

pastel quest
#

Here a is just a variable

plucky marlin
#

let me help this guy in calculus and ill come back to you

#

I know

#

it just didnt take my input correctly

pastel quest
#

Ok

long pond
#

a is a variable?

#

not a constant?

pastel quest
#

Oh sorry

#

Constant

#

My bad

long pond
#

np

#

what do yuo get when you rationalize

pastel quest
#

[(a+ root x )^2]/(a^2-x)

long pond
#

mhm

obsidian monolithBOT
long pond
#

one sec

pastel quest
#

Can the bot here plot a graph?

long pond
#

nt very well tbh

#

use geogebra

pastel quest
#

K

#

Ok here is another question if you want to try...
A reservoir of square cross section has sides sloping at an angle of45 degree with the vertical.The side of the bottom is ‘p’ feet in length, and water flows in the reservoir at the rate of ‘c’ cubic feet per minute.Find an expression for the rate at which the surface of the water is rising at the instant it’s depth is ‘h’ feet.Calculate this rate when p=17, h= 4 and c =35.

south rivet
#

dont post in multiple channels

pastel quest
#

K

south rivet
#

and respect people when their question is still being answered

pastel quest
#

I wonder what’s the difference bw calculus and precalculus

south rivet
#

calculus is after orecalculus

long pond
#

calculus is the mathematics of change

south rivet
pastel quest
#

Means how do u identify ur question’s level?

south rivet
#

if it uses calc then put it in calc

#

or better yet, a question channel

long pond
#

precalculus is the mathematics you learn before you learn the mathematics of change

pastel quest
#

Ok..actually I am new to discord

#

@long pond did u got the earlier question.?

long pond
#

id have to see your work in order to see if you made a mistake

#

common sense would dictate that not chaning an expression would not cause its derivative to change

#

so it seem likely you made an error

pastel quest
#

Ok..probably I would have done a blunder

#

Thanks

long pond
#

if you like i can check your work, but theres not much else i can do except identify a mistake

open star
#

how do you prove the angle sum identities for angles greater than 180?

drifting compass
#

i'm assuming it's because of the restriction, but can someone elaborate

tawny nacelle
#

,w plot y = cos(x/2) from x = -pi/2 to x = pi/2

obsidian monolithBOT
tawny nacelle
#

yeah so cos(theta/2) is always positive on that interval

#

so y value always positive

#

thus only the positive version of sqrt

drifting compass
#

oh right, i need the - verision to plot as well?

tawny nacelle
#

why would u need the - version ?

drifting compass
#

i'm not understanding, why wouldn't i do x = sqrt(1-y^2) instead?

#

ah plugginm in numbers helped

#

but still not sure why x=sqrt(1-y^2) isn't valid answer

pastel quest
#

I’m the standard form Cartesian equations are of form y=... so, I am interpretating it would have been a standard notation.Also, when we need y as a function of x we have to write y=.. and vice versa if we need x as a function of y

arctic leaf
#

so terrible questions but

#

$\frac{\log_a\left(x\right)}{\log_a\left(y\right)}\cdot\frac{\log_b\left(y\right)}{\log_b\left(x\right)}$

obsidian monolithBOT
arctic leaf
#

When changing the bases, can they be the same or are they different

#

Otherwise can someone solve it :)

serene heath
#

solve what thonkzoom

arctic leaf
#

Simplifying that

#

Should have added that

#

My bad

wooden zodiac
#

you can change the log bases, by using Change-of-Base formula, then factorize the similar part

arctic leaf
#

Yea i know, but can you changed the base to the same value

#

for bot logs

#

*both

#

shouldn't be a problem?

wooden zodiac
#

you can change all log bases to base 10 (or base e)

arctic leaf
#

Oh ok

#

so answer is 1

wooden zodiac
#

yes, the answer is 1

arctic leaf
#

oh ok

#

Since the answers that were provided had

#

$\frac{\log_a\left(b\right)}{\log_b\left(a\right)}$

obsidian monolithBOT
wooden zodiac
#

once simplified, it is equal to that, or log_y x * log_x y

#

but I'm not sure how to get that, as it should be canceled out when you change base

arctic leaf
#

Yea..

rare zephyr
spring thunder
#

,rotate 90