precalculus

[radiant ibex]

Wut.gif

[craggy dune]

look at the term you have a sum left

now you want to get rid of that and you do that using a closed formula

thats why you prove it, then you can use it and you have the answer to the orignal question

[radiant ibex]

ok

thank you

Im not 100% on the cloased formula, but I'll try to figure it out

[craggy dune]

if you need help just ask here it's the standard introduction to induction

[radiant ibex]

@craggy dune would you mind walking me through this?

[craggy dune]

+6

ya i can do that

okay with induction you always start with the base case

in this case that is n = 1

[radiant ibex]

right

[craggy dune]

so you show that the formula is correct in the case of n = 1

that should be simple enough right?

[radiant ibex]

yea

[craggy dune]

okay

now you need to prove it for the case (n + 1)

$\sum_{k=1}^{n+1}k$

[obsidian monolith]

deekaan:

[craggy dune]

do you know how to get the upper bound back to n?

[radiant ibex]

not a clue, do you multiply it to k?

[craggy dune]

no

$\sum_{k=1}^{2}k$

[obsidian monolith]

deekaan:

[craggy dune]

you know what this sums equals to?

[radiant ibex]

3?

[craggy dune]

yes

[radiant ibex]

OK

[craggy dune]

how do you get there?

[radiant ibex]

1 + 2

[craggy dune]

okay

$\sum_{k=1}^{1}k + 2$

[obsidian monolith]

deekaan:

[craggy dune]

right?

[radiant ibex]

yeah that would be the same

[craggy dune]

okay

$\sum_{k=1}^{10}k$ say we want that sum to have the upper bound 9

[obsidian monolith]

deekaan:

[craggy dune]

how would we do that?

[radiant ibex]

ok, Im not sure how to do this in my head

but give me a sec

[craggy dune]

don't overthink it

look at the previous example

how did we isolate the 2?

[radiant ibex]

by adding it

[craggy dune]

exactly

and how do we isolate the 10?

[radiant ibex]

we could add it

[craggy dune]

exactly

[radiant ibex]

and that would make the upper 1

right

[craggy dune]

not 1

9

[radiant ibex]

ooooooo

ok

[craggy dune]

we just shaved the 10 of

$\sum_{k=1}^{n+1}k$

[obsidian monolith]

deekaan:

[craggy dune]

okay how do we get to the upper bound n?

[radiant ibex]

add 1

[craggy dune]

no

the upper bound is (n + 1)

1 2 3 .... n (n +1)

a sum is just shorthand for adding the same thing again and again

so $\sum_{k=1}^{5}k$

[obsidian monolith]

deekaan:

[craggy dune]

means we have 1 + 2 + 3 + 4 + 5

[radiant ibex]

yea

[craggy dune]

and we can just shorten a sum like so

$\sum_{k=1}^{3}k + 4 + 5$

[obsidian monolith]

deekaan:

[craggy dune]

this way we can move the upper bound

we have to calculate 1 + 2 + 3 + ... + n + (n + 1)

or $\sum_{k=1}^{n+1}k$

[obsidian monolith]

deekaan:

[craggy dune]

so how do we isolate the (n+1)

don't overthink it we're doing simple addition

n is just a placeholder for an integer

so if n = 9, (n + 1) would be (9 + 1) = 10

[radiant ibex]

so i add 2?

Im so flippin confused

[craggy dune]

ya

do you have pen and paper?

[radiant ibex]

yea

[craggy dune]

okay

try to do some calculation with sums

the issue right now is that either the n is too abstract for you or that you don't understand sums

$\sum_{k=1}^{n+1}k$

[obsidian monolith]

deekaan:

[craggy dune]

take this sum and set n to 1, 2, 3, 4 and 5

and try to reduce the upper bound

use the rule before

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

[radiant ibex]

im gonna take a break

[obsidian monolith]

deekaan:

[craggy dune]

or this way

[radiant ibex]

and look it over

thank you for the help tho

[craggy dune]

no worries

[fleet yew]

@radiant ibex hi

Is your name supposed to just be a bunch of boxes

Or is it unicode i cant read

[thorn mountain]

looks like boxes to my computer too

[hard mulch]

I keep getting this wron

T_T

I tried doing -1t +28(5) for a

but I dont think thats how its done

[gritty sandal]

u sub in t=5

U missed the quadratic at the front

so it would be $-(5)^2 + 28(5)$

[obsidian monolith]

Parking Ticket:

[gritty sandal]

or 115 dollars

[full garden]

hey guys does it make sense to ask for the acute angle of -53?

because the question says what's the acute angle of -53

and like i think -53 is already acute

[gritty sandal]

could u send a picture of the question

probably go to #geometry-and-trigonometry

[full garden]

I'll send u a screenshot because it's from online class

[gritty sandal]

you should probably go to geometry and trig

[full garden]

sorry

[gritty sandal]

nah dw

[violet nebula]

I am lost on finding a second solution to a problem, I need to find 2 alternate solutions to this on top of my first find which was done with a summation of 4(n-1)+1 with n=1 below and an x on top to represent the layers. can anyone guide me to another way of solving it?

[willow bear]

you can view it as a central column and four triangles

[violet nebula]

would that end up solving for area then multiplying by 4? or would it be another way. I dont see how i would go from area, and translate it to a number of cubes with the diagonal

[slim wigeon]

Is counting an option?

[violet nebula]

i would assume, but i might get docked points for not having an original solution

[slim wigeon]

Well

[violet nebula]

this project is about describing the process

[slim wigeon]

Yeah

hmmm

[violet nebula]

ah, I understand the triangle method now

thank you Ann

i worked it out a bit and that makes a lot of sense

[fleet yew]

$6+4(\sum\limits_{n=1}^5 n)$

[obsidian monolith]

AMD:

[fleet yew]

,w $6+4(\sum\limits_{n=1}^5 n)$

[obsidian monolith]

Display results online and refine query

Requested by AMD#4434

[willow bear]

amd you're late

[fleet yew]

Lmao

Texit be like that

[obsidian monolith]

AMD:

[fleet yew]

Why

[maiden pebble]

Oh, I did it as the sum of numbers from 1 to 11

$ (\sum\limits_{n=1}^{11} n)$

no way I got it to work

[exotic grove]

$ (\sum\limits_{n=1}^{11} n)$

[obsidian monolith]

MeepMoop:

[maiden pebble]

Thanks

[exotic grove]

Judicious use of {these}

[maiden pebble]

mhm

[obsidian monolith]

Bridge:

[viscid thistle]

How would I simplify this

[tawny nacelle]

factor out cos(x)

something magical will happen

[harsh smelt]

you ll get tan btw

[tawny nacelle]

tan^2(x)

also spoilers

[harsh smelt]

SPOILEEERS

[tawny nacelle]

[viscid thistle]

How

Please walk through it

Please help

[tawny nacelle]

have you factored cos(x) out ?

[viscid thistle]

How do I

[tawny nacelle]

[viscid thistle]

I’m dumb

[tawny nacelle]

no no we'll get through this

theres still hope

lets factor something nicer first

like say we have the expression 6 + 10x

now can you factor out 2 from both ?

[viscid thistle]

2(3+5x)

[tawny nacelle]

nice 🎊

so when you factored out 2, you just divided both of the terms by 2

we kinda do the same thing for your expression

factor out cos(x)

and as there is cos(x) in both of the terms, they will cancel out to 1 when you divide by cos(x)

so now can you factor out cos(x)

[viscid thistle]

So I don’t make it 1/sec?

Sorry I’m mega stupid

@tawny nacelle

[tawny nacelle]

ok think like this

[viscid thistle]

Also is sin^2x/csc^2x just 1?

[tawny nacelle]

$\frac{\cos(x)}{\cos(x)}(\sec^2(x)\cos(x) - \cos(x))$

[obsidian monolith]

soαρ:

[tawny nacelle]

do you agree that this is the same thing as you had before

[harsh smelt]

ab+a=a(b+1)

it is distributive law

[tawny nacelle]

nice

[viscid thistle]

Yes

[tawny nacelle]

ok so now we move the 1/cos(x) inside the parentheses and distribute it to both of the terms

[viscid thistle]

What 1/cos

[tawny nacelle]

$\cos(x)(\frac{\sec^2(x)\cos(x)}{\cos(x)} - \frac{\cos(x)}{\cos(x)})$

[obsidian monolith]

soαρ:

[tawny nacelle]

you cool with this

[viscid thistle]

Yee

[tawny nacelle]

well gg man now you just cancel out the cos(x)

acutallly i might have fucked it bad

this was prolly more confusing

[harsh smelt]

why on earth did you do that, i mean it is correct but what is the sense

[tawny nacelle]

idk man i thought it would make sense to him

[harsh smelt]

by distributivity (sec^2cos-cos)=cos(sec^2-1)

[viscid thistle]

Also is sin^2x/csc^2x just 1?

[tawny nacelle]

but now i think using distributivity wouldve been much much beter

no it isnt

sin^2(x) * csc^2(x) = 1

[viscid thistle]

So what is it

[harsh smelt]

by distributivity (sec^2cos-cos)=cos(sec^2-1)
@viscid thistle also, since tan^2+1=sec^2 you have cos(tan^2)

[tawny nacelle]

mfw cos(tan^2(x))

[harsh smelt]

since tan^2=sin^2/cos^2 youhave cos(sin^2/cos^2)=sintan

that's all in fact

where from you've taken csc?

[ember wave]

Does anyone know how to do this problem? Normally, to find tangent lines; I thought you need coordinates.

[uncut mulch]

solve y' = 0
there are infinitely many solutions, but you only need 2

alternatively, look at the graph as suggested

these types of lines have a slope of 0: ||h|||| ||||l||

[ember wave]

but for a slope of 0 for sin x wouldn't it just be a horizontal line?

[undone pawn]

it would

[ember wave]

so just two random x values? or am I still missing something?

[undone pawn]

not two random x values..

at which points of sinx, are the tangents perfectly horizontal???

,w plot sinx from -15 to 15

[obsidian monolith]

Display results online and refine query

Requested by FlynnXD#2733

[ember wave]

So would x=3 be one of them?

-3,3,6,9

[undone pawn]

no..

where are you getting these numbers from

[viscid thistle]

look at the x coordinates of the peak point

[ember wave]

They touch the x axis at the horizontal line of sin x w sloe of 0

[undone pawn]

but that's not where slope is 0

what the heck

do you know what the slope is?

and specifically, what a tangent is?

the tangent to a curve

[viscid thistle]

how do i start this problem?

<@&286206848099549185>

[viscid thistle]

@viscid thistle So what is troubling you

well where do you start?

first compose the function

with taking proper conditions

but the conditions, some of them are numbers and the others are in radians soooo

So take pi is like 3.14

oh that's it? i'll try again ig

Yess

Just take intersection of all the conditions

I mean if

for example if you take f(x) = [x] as function then -2 <= x<= -1 right

Ria u here

@viscid thistle

yea dw i think i got it

thx for the help

Okky

[wide lynx]

How would I factor 6x^3+17x^2+11x+2

[uncut mulch]

apply rational root theorem, to test for factors.

[wide lynx]

didnt work?

[thorn mountain]

You tried all 16 option?

Cause I’ve tried 2 of them in my head and pretty sure I found one

[wide lynx]

i must be doing something wrong lol

[past meadow]

ye there are some simple rational roots for this one

[wide lynx]

i dont get remainder of 0 on any of my synthetic division

assuming you do synthetic division

[past meadow]

have you tried (x+2) as a factor

[wide lynx]

i just followed this video https://www.youtube.com/watch?v=4XytYH35AP0

[past meadow]

also, you shouldnt have to divide to know if you're gonna get a remainder of 0

[wide lynx]

I followed that and don't get past synthetic division

[past meadow]

he doesnt test if theyre factors before doing the division

idk why he does that

anyway use factor theorem to test if the roots are factors or whatever before you actually divide

[wide lynx]

ill just watch a different video

[past meadow]

what do you think the possible rational roots are? test those by substitutuing them back into the polynomial, and if theyre zero, you have yourself a factor

[wide lynx]

you find possible rational roots by doing p/q

like the constant/coefficient of first term

[past meadow]

yeah, so which ones have you worked out?

[wide lynx]

1,(1/2),(1/3),(1/6),2,(2/3)

[past meadow]

+-

[wide lynx]

yes

can i use negative roots? like for the synthetic division

i didnt know that was allowed

[past meadow]

im 99 percent sure you can

although ive always just done polynomial long division

so have you tried the negative ones?

to see if there are any roots there?

(spoiler: all the roots are negative)

[wide lynx]

no lol

ill do that rn ty

[past meadow]

so you found any factors yet?

[wide lynx]

yeah i did it

sorry for late response i wasnt checking discord

[past meadow]

no worries

so you got the problem sorted now?

[wide lynx]

yeah

but

in the workbook it had an x+2

but i got x-2

i saw this rule on yt about always having it be x-n but that would make my other answers incorrect

[thorn mountain]

@past meadow synthetic division is just fine with negative roots. It even works with complex roots if you do if sequentially for both roots in the conjugate pair... as long as the factor can be written as x-c (and c can be negative or complex), synthetic division can be used.

[past meadow]

i think you meant to ping the person who i was helping, but ive just always found polynomial long division makes more sense personally and i was never taught synthetic division

so i didnt want to give any certainties about what you could do with it

[thorn mountain]

EH, you expressed a small amount of doubt over whether it was applicable

[past meadow]

well thanks for the info

[thorn mountain]

So I was gonna clear that up for you if someone else in the future asked

[past meadow]

thanks then. that makes sense

[remote bison]

@past meadow

[past meadow]

sorry if I'm being thick but how is that relevant to what i said?

[remote bison]

@past meadow

[past meadow]

ok then

[remote bison]

sneaky boi

[past meadow]

oh i see

[remote bison]

can someone explain hyperbola equations to me please

please dm me

[viscid thistle]

Do sequences belong under precalc?

[thorn mountain]

Kinda depends...but the questions channels could always be used

[fleet yew]

Yes they do

I would even say algebra maybe

[scarlet mirage]

I don’t know what to do next

[patent beacon]

@scarlet mirage
Good start! Let's solve that first one. When is cos(x) = -2?

[still meadow]

cosx = -2 and cosx = -1/2 is correct.

[scarlet mirage]

Is this correct

[surreal pivot]

Anyone know this

[remote bison]

bruh

the mode is basically the most frequent term

which is between 20 and 25

@surreal pivot

for the string box question, the minimum is 1, median is 3, maximum is 5

q1 and q2 are 1.5 and 4.5 respectively

[surreal pivot]

Thanks bro

@remote bison but the dot graph is hard this my first time doing this ever

[remote bison]

@surreal pivot wait please, I'm in online class rn

btw you sure this is precalc?

[undone pawn]

technically prob stat

[remote bison]

doesn't precalc stats include combinatorics and stuff?

[undone pawn]

ig

[silver matrix]

Ping me when answering, thanks.

[maiden pebble]

@silver matrix I got F

https://i.imgur.com/soyn3Oi.png

here is my work (good luck on reading the handwriting)

[willow bear]

4x < 4

x < ... what

[maiden pebble]

OOH

I erased it and never wrote i back

pretty sure that dividing 4/4 wouldnt have been too hard though Ann

[willow bear]

don't take me for someone who can't do arithmetic

i was specifically questioning your omission of the 1

[maiden pebble]

just making a joke

[silver matrix]

Well it's not f

It's B.

Use Wolfram for checking

[remote oyster]

A club consisting of 6 juniors and 8 seniors is to be formed from a group of 13 juniors and 16 seniors. How many different clubs can be formed from the group?

[warped crypt]

What have you tried @remote oyster

[robust edge]

yeah try researching stuff about ncr and npr

[remote oyster]

i subtracted the possible seniors and juniors from the groups u choose

and i got & and 8

sorry 7 and 8

and i tried just adding 7! and 8!

unless youre supposed to multiply 6 and 8 and 13 and 16 and make no distinction between seniors and juniors

[warped crypt]

No

[remote oyster]

?

[warped crypt]

What you're trying to do is find the total combinations you can have of 6j and 8s

From a total of 13j and 16s

[remote oyster]

i thought you make one "object" out of the 6 juniors and so youre left with 7 possible junior combinations

or 7!

[warped crypt]

What is the number of combinations of 6 juniors when you have 13 juniors to pick from

[remote oyster]

13!/7!

ohhhhhhh

bruh

wait is that right

[warped crypt]

No

[sour eagle]

13!/(7!*6!)

[remote oyster]

i thought u subtract it

oh i also tried what dan said

[warped crypt]

That would be right if the order to juniors mattered

[remote oyster]

yeah i did the combination equation too

[warped crypt]

But if order doesn't matter it is the combination, not the permutation

[remote oyster]

but my calculator gave me a huge number i assumed it was wrong

wait let me try it again

[warped crypt]

You should get 13C6 = 1716

[remote oyster]

and then u would do the same combination equation for the seniors right

[warped crypt]

Yes

Then add them

[remote oyster]

ok thank you

[uncut mulch]

shouldn't it be multiply

[sour eagle]

Yea i thought the same.

[uncut mulch]

considering the simplest case:
selecting 1j,1s from 1j, 1s
there would only be 1 'group'

[sour eagle]

I mean they are not separated events i guess?

[uncut mulch]

not 1C1+1C1 = 2

[sour eagle]

😔

[warped crypt]

Ah yes you're right

Not sure what I was thinking

[inner karma]

why is there a precalculus channel

precalculus = everything before calculus [+ some calculusy stuff]

[willow bear]

not quite

in america there's a class specifically called precalculus which deals with trig, exponentials and logs with a hearty helping of algebra

[daring yarrow]

what would this be expressed as a limit?

[stuck lark]

recall the definition of a limit and what each part of the definition means, then carefully reread the given statement with the definition in mind and understand what each part of the statement says about the whatever limit the statement is talking about

[daring yarrow]

the right limit (of f(x)) as x aproaches 1 = -infinity

?

@stuck lark

[stuck lark]

that's it @daring yarrow

[daring yarrow]

thank you!

[stuck lark]

no prob

[gloomy mortar]

I really need help on my online PreCalc homework. I’ve looked over our past work and videos and stuff and I still can’t figure them out. It’s completely different than our other problems and I don’t even understand how to do them. My teacher didn’t explain it and it’s super frustrating

1. A train is travelling at the rate of 10 mph on a curve of radius 3000 feet. Through what angle will the train turn in one minute? Round to the nearest whole number of degrees.

2. Two pulleys, one 6 in. and the other 24 in. in diameter, are connected by a belt. The larger pulley revolves at a rate of 60 rpm.
   a. Find the speed of the belt in feet per minute.
   b. Determine the revolutions per minute of the smaller pulley.

   3. Calculate the angular speed in radians per minute of a Ferris wheel 250 feet in diameter that takes 45 seconds to rotate once. Express the answer in terms of π.

3. If you sat on the rim of the Ferris wheel in problem 3, what would your speed be in feet per minute?

[willow bear]

am i right to assume you need help on all of these

@gloomy mortar

[gloomy mortar]

Yes

[willow bear]

ok

let's start with number 1

how far did you get for number 1

also, will i have to ping you for every single reply? because i'd really like to not have to do that.

@gloomy mortar

[gloomy mortar]

Sorry I’m doing another assignment at the same time I’m trying to keep track of what you’re saying too it looks like I’m gonna have to turn on notifications for this channel since it didn’t do that automatically

[willow bear]

maybe focus on one thing at a time

[gloomy mortar]

I got to 52800ft/60min = 6000xπ/1min but I was like “Wait am I even doing this right?”

[willow bear]

sounds like you're not entirely sure what you're doing

52800 ft (which is 10 miles, converted to feet) is the distance your train travels in an hour.

you'll want to ask yourself the following question first: how far does the train travel in a minute?

[gloomy mortar]

I meant 52800/60min sorry I’m not thinking well rn

I corrected it in the message

[willow bear]

do one thing at a time

how far does the train travel in a minute?

[gloomy mortar]

52800•60?

[willow bear]

really?

so if the train moves at the same speed, it travels sixty times further in a minute than it does in an hour?

[gloomy mortar]

Ah I feel stupid. 52800/60

[willow bear]

that's more like it

so now, what's the angle in radians that makes an arc of (52800/60) feet on a circle of radius 3000 feet?

this is taking you a mighty while to answer

if you're trying to jump ahead, please don't

[gloomy mortar]

I’m not my brain is just really slow because I’m not having any motivation and pretty much having to force myself to do this

Uhhh well I know 52800/60=880. Arc length=central angle•radius.
880=theta•3000.
880/3000=22/75 but is that the radians? Does the radians have to have pi?

[willow bear]

you're kinda overthinking it already.

also, 52800/60 is not 880.

[gloomy mortar]

That’s my speciality 🙈

[willow bear]

,calc 52800/60

[obsidian monolith]

Result:

880

[willow bear]

oh it is.

my bad.

somehow i was under the impression that 5280 was not perfectly divisible by 3.

aight whatever

angle = arclength/radius, and yes in your case it works out to 22/75 radians, and no a radian measure need not contain pi in it

great, so now all that's left is to convert 22/75 radians into degrees

can you do that?

[gloomy mortar]

Uhhhh without pi in there equalling 180° I have absolutely no idea how to convert that to degrees

[willow bear]

1 radian = (180/π) degrees, don't overthink it

if you can convert between miles and feet you can convert between radians and degrees

[gloomy mortar]

Like if it were 3π/2 I know that’s 270°

Oh

[willow bear]

it really is just $\frac{22}{75} \times \frac{180}{\pi}$ degrees

[obsidian monolith]

Ann:

[gloomy mortar]

Yeah I was calculating that

17°?

[willow bear]

,calc 22*180/(75 * pi)

[obsidian monolith]

Result:

16.806761990504

[willow bear]

yup

[gloomy mortar]

Let me write those down for this problem then we can move on

Okay

[willow bear]

ok aight so

problem 2

2) Two pulleys, one 6 in. and the other 24 in. in diameter, are connected by a belt. The larger pulley revolves at a rate of 60 rpm.
 a. Find the speed of the belt in feet per minute.
  b. Determine the revolutions per minute of the smaller pulley.

as before i'd like you to show me your attempt

[gloomy mortar]

Sorry had to eat breakfast. I don’t really know where to start with this one

It’s all running together in my head and confusing me a lot. I think I’m overthinking again, I tend to do that a lot

[willow bear]

you might start by finding the speed at which points on the edge of the first pulley rotate

[gloomy mortar]

What

Sorry my brain isn’t really functioning very well today I’m super confused already

[willow bear]

er

sorry, bad wording

you might start by finding the speed at which points on the edge of the first pulley move

[fleet yew]

@willow bear you can add the digits to see if its divisible by 3

[vague aurora]

What would be the best way to learn precalculus over the summer

?

[spare walrus]

What resources do you have?

[vague aurora]

I got the school workbook for the course, but it’s mostly practice problems without the actual learning

I was thinking khan academy?

[north crescent]

Khan academy is helpful yes

[remote bison]

can someone please help me with hyperbola equations from conic sections in my dms

god bless if you do

[maiden pebble]

When taking the derivative of something with multiple variables, such as (x^4)(y^3) do I 1. take the derivative of both? 2. Take the deriv of the one with the largest exponent? 3. do something else

[uncut mulch]

apply product and chain rule

[maiden pebble]

yes duh

thanks

wait, and the chain rule?

but its not a function

$ x^{2}y$

[obsidian monolith]

Bridge:

[maiden pebble]

just to clarify, the derivative of that would be 2xy+x^2?

[smoky pagoda]

Partial derivative?

If it's simple derivative in terms of x

Then apply product rule to get dy/dx * x^2 + y*2x

[maiden pebble]

So I was right?

[smoky pagoda]

Not quite

x^2 needs to be multiplied by dy/dx

[maiden pebble]

https://i.imgur.com/6q5JtBl.png

[smoky pagoda]

Yes

[maiden pebble]

the product rule has no dy/dx in it

[smoky pagoda]

Here f is x^2

And g is y

[maiden pebble]

yep

[smoky pagoda]

dy/dx is simply y'(x)

Because we dont know the derivative of y with respect to x

So we simply mark it as "derivative of y with respect to x"

[maiden pebble]

agreed

[smoky pagoda]

Which is why you multiply x^2 with y'(x)

[maiden pebble]

Why do you keep saying y'(x)

[smoky pagoda]

It's the derivative of y with respect to x

Like g' in your product rule definition

[maiden pebble]

oh so its the fg' portion

[smoky pagoda]

No, the f'g portion is 2xy

[maiden pebble]

yea I corrected

[smoky pagoda]

The fg' portion is x^2 * y'(x)

Ah okay then yes

[maiden pebble]

Ok, awesome

thanks

[smoky pagoda]

Np

[maiden pebble]

@smoky pagoda would the derivative of 6xy be 6y or 6y+(6x * dy/dx)

[smoky pagoda]

the second option

[maiden pebble]

LETS GO

[smoky pagoda]

easier to do if you consider it to be 6 * the derivative of xy

so you get 6 * (y + x * dy/dx)

[maiden pebble]

Ahhh, smart smart

if there is a y on both sides of my equation then the dy/dx would cancel out right

[smoky pagoda]

wdym

[maiden pebble]

https://i.imgur.com/wYNGPoZ.png

on the left I have: 3x^2 + 3y^2 * dy/dx

on the right I have 6y + 6x * dy/dx

[smoky pagoda]

oh so you differentiate both sides first

[maiden pebble]

sorry, i Dont know what that means

[smoky pagoda]

to get 3x^2 + 3y^2 * (dy/dx) = 6(y+x*dy/dx)

and then you isolate dy/dx

so that it gives you the solution

id recommend looking at resources on implicit differentiation

because that's hwat you're doing

[maiden pebble]

it just so happens that thats the name of my worksheet

[smoky pagoda]

haha

yeah get a good understanding on it before you keep going

[maiden pebble]

Would the derivative of sin(xy) be sin(xy)' + sin'(xy)

[uncut mulch]

no

[willow bear]

sin(xy) is not the product of sin with xy

[uncut mulch]

lagrange notation is also horrible here

[maiden pebble]

?

https://i.imgur.com/bYBkptM.png how would I find the derivative of this

[pale bison]

with chain rule

[uncut mulch]

^

[maiden pebble]

well thats what I tried but you said it wasnt right

[uncut mulch]

show your work

[maiden pebble]

oh I did the product rule

[uncut mulch]

it involves combinations of both

[maiden pebble]

not sure how to do both, but if I just did the chain rule I would get cos(xy)(xy'+y)

[uncut mulch]

= 0

[maiden pebble]

yes

[uncut mulch]

and that's fine.

[maiden pebble]

ok cool

[uncut mulch]

rearrange and isolate y' if required

[gloomy mortar]

Hey sorry I disappeared yesterday. Just needed a break. Here are the problems I still need help with:

2. Two pulleys, one 6 in. and the other 24 in. in diameter, are connected by a belt. The larger pulley revolves at a rate of 60 rpm.
   a. Find the speed of the belt in feet per minute.
   b. Determine the revolutions per minute of the smaller pulley.

   3. Calculate the angular speed in radians per minute of a Ferris wheel 250 feet in diameter that takes 45 seconds to rotate once. Express the answer in terms of π.

3. If you sat on the rim of the Ferris wheel in problem 3, what would your speed be in feet per minute?

[gloomy mortar]

Also what am I doing wrong here? I tried 36 as the arc length and I also tried it as the radius and I am still apparently not getting the “correct” answer of 26.4cm

[lilac pier]

It's not you, it's the answer that's wrong. They took 180 degrees as 2pi

you can see the difference b/w your and their answer is a factor of 2

@gloomy mortar

[gloomy mortar]

Ooooohhhhhh good point

[maiden pebble]

https://i.imgur.com/cs33nLK.png

$ \frac {1}{a-x}{a+x}$

[undone pawn]

what do you want to do with it

[drowsy quail]

what is the q?

[maiden pebble]

Trying to find the derivative of that picture 6.)

[drowsy quail]

ahh ok

[undone pawn]

$\ln\bigg ( \frac{x+a} {a-x} \bigg) $

[obsidian monolith]

FlynnXD:

[maiden pebble]

well i was trying to type my answer but thanks

[undone pawn]

oh ok

[drowsy quail]

do u want to know the derivative?

it s

[undone pawn]

$\frac{1}{(a+x)(a-x)}$

[obsidian monolith]

FlynnXD:

[maiden pebble]

let me get a pic of my work

[undone pawn]

that's what you were typing?

[maiden pebble]

il just use mathway or something

makes typing stuff 200 times easier

[undone pawn]

aight

[maiden pebble]

https://i.imgur.com/aeDFfR5.png

oh i never used the quotient rule

[undone pawn]

wtf 0 2?

dude wait, just split the log before differentiating

$\ln(\frac{a} {b}) =\ln(a)-\ln(b)$

[obsidian monolith]

FlynnXD:

[maiden pebble]

didn't know that, looks helpful

[willow bear]

you're studying calculus buit don't know the most basic of log laws?

[maiden pebble]

guess not

im in precalc if that defends me at all

[willow bear]

no

[maiden pebble]

Flynn I got:

$\frac{1}{a-x}(0)-\frac{1}{a+x}(2)$

[obsidian monolith]

Bridge:

[drowsy quail]

@maiden pebble i would recommend learning log laws and trig identites before calc stuff

that'll be a huge help

[maiden pebble]

ok, il review them between homeworks

[drowsy quail]

good idea

[maiden pebble]

I still need help with this problem

[drowsy quail]

ok

@maiden pebble what do you need help with

[maiden pebble]

turning ln(a/b) brought be to a dead end, so I went back to my first method of 1/a-x/a+x

[drowsy quail]

oof dont do that

[maiden pebble]

Thats the left side of my work, now im doing the quotient rule on the right side

😒

[drowsy quail]

${ln(a/b)} = ln(a)-ln(b)$

[obsidian monolith]

MisterLJGibson:

[drowsy quail]

split it up like that

and then differentiate if you're still using l'Hopital's Rule

[maiden pebble]

idk what that is

[drowsy quail]

...

so you're being asked to differentiate f(x) = ln((a-b)/(a+b)) correct?

[maiden pebble]

yes

[drowsy quail]

do you still need to know the answer?

[maiden pebble]

Im not sure how to find the deriv of a+or-x

[drowsy quail]

ok well we'll start from the beginning

$simplify ln((a-x)/(a+x)) first $

[obsidian monolith]

MisterLJGibson:

[maiden pebble]

$ ln(a-x)-ln(a+x)$

[obsidian monolith]

Bridge:

[drowsy quail]

yes

[maiden pebble]

$\frac{1}{a-x}(a-x)'-\frac{1}{a+x}(a+x)'$

let me edit that

[drowsy quail]

no

[maiden pebble]

ok...

that was a big jump

[drowsy quail]

differentiate both of the natural logs

[maiden pebble]

(a-x)'/(a-x) - (a+x)'/(a+x)

[drowsy quail]

no

[maiden pebble]

oh yes

[drowsy quail]

differentiate after you've used the log law

[maiden pebble]

(a-x)'/(a-x) - (a+x)'/(a+x)

[drowsy quail]

no

[maiden pebble]

what!?!?

[drowsy quail]

wait

what does (a-x)' equal?

[maiden pebble]

Derivative of (a-x)

[drowsy quail]

yes but what is the derivitive

[maiden pebble]

0

[drowsy quail]

no

that's all wrong

d/dx ln(x) = 1/x

so d/dx ln(x-a) = 1/x-a

[maiden pebble]

a-x*

but isnt that what I did

[drowsy quail]

no

you said the derivative = 0

which is wrong

[maiden pebble]

ok, how do I solve for that

[drowsy quail]

well d/dx (a-x) = -1

a is a constant and is discarded

[maiden pebble]

oh, that would hlep

[drowsy quail]

and d/dx (-x) = -1

[maiden pebble]

ok good, that makes sense

[drowsy quail]

you got it?

[maiden pebble]

Il keep going through the problem

[drowsy quail]

bc remember you're differentiating with respect to x

so any non-x variable when differentiating goes

[maiden pebble]

REALLY

That helps so much

[drowsy quail]

yeah hahaha

were you differentiating a and x?

and then got 1-1?

[maiden pebble]

yes

[drowsy quail]

ahh

i see

did you not think that that would give you ln(0)?

which is undefined

[maiden pebble]

(6xy)' wrt x = 6y right?

[drowsy quail]

ye

[maiden pebble]

So why doesnt the y go?

[drowsy quail]

bc it is being multiplied by x

d/dx (a x X) = a

[maiden pebble]

oh ok

[drowsy quail]

small x means times btw

[maiden pebble]

yep that amkes sense

[drowsy quail]

does that help?

[maiden pebble]

yep

LETS GO I GOT IT

🔥

[drowsy quail]

yay! well done

[gloomy mortar]

I still need help with these problems:

2. Two pulleys, one 6 in. and the other 24 in. in diameter, are connected by a belt. The larger pulley revolves at a rate of 60 rpm.
   a. Find the speed of the belt in feet per minute.
   b. Determine the revolutions per minute of the smaller pulley.

   3. Calculate the angular speed in radians per minute of a Ferris wheel 250 feet in diameter that takes 45 seconds to rotate once. Express the answer in terms of π.

3. If you sat on the rim of the Ferris wheel in problem 3, what would your speed be in feet per minute?

[drowsy quail]

this is more physics isn't it?

[gloomy mortar]

Precalc

Uhhhh trig I think this part is

[drowsy quail]

rlly?

[gloomy mortar]

Yeah

[drowsy quail]

im afraid im from the uk

we do this in physics hahaha

[gloomy mortar]

Oh haha

Think you can still help?

[acoustic harbor]

nah thats physics

but u can use it in precalc

[viscid thistle]

@gloomy mortar This is trig?

[gloomy mortar]

Our teacher said we were doing trig. Unless we switched to physics and I didn’t realize

[viscid thistle]

well $s=\frac{d}{t}$

[obsidian monolith]

Slavoj:

[tepid basin]

Can anyone help me with this? Find the exact equation for the end behavior line of y=2x^2+4x+1 / X+2

[heady harness]

@tepid basin do you still need help with that problem?

[viscid thistle]

**If anybody is having trouble understanding pre-calculus, I can help you out. DM me! 🙂 **

[echo knot]

There are more than 2000 people here

[sinful parcel]

o.O

[queen cave]

okay, this might seem like a stupid question but i really just need to ask it: is dy/dx * dx/dy = 1 ??

[smoky pagoda]

you shouldnt treat leibniz notation as regular fractions

[queen cave]

so yes?

[north crescent]

No

[queen cave]

thanks

[smoky pagoda]

well no it is true

you just shouldnt come to that conclusion from the fractions

[north crescent]

Not always tho

[smoky pagoda]

only when they arent equal to 0

[north crescent]

Well that case does disprove it tho

[queen cave]

so would it be alright to say this is equivalent to R + dt/dQ * Q/C ?

[smoky pagoda]

yeah of course

but it you put the constraint that they arent equal to 0 it's true

im not saying ofc to your picture @queen cave

and again, you must be careful with constraints

generally speaking, I dont think dy/dx * dx/dy = 1 is applied a lot for conversions

[queen cave]

well here's the whole question

tbh im kinda stuck

[smoky pagoda]

is C the capacitance?

[queen cave]

i believe so

[smoky pagoda]

im guessing by knowing the subject you would know how one of these variables changes with time, but I dont know enough about circuitry to be able to tell you

because otherwise you cant know what the rate of change of Q with time is

[queen cave]

yeah this is from maths, so i know as much about circuitry as you :/

[north crescent]

U can find charge by isolating dq/dt and integrating both sides ig

[smoky pagoda]

im pretty sure the equation isnt separable

but i have very little experience with differential equations

[queen cave]

Does this make sense??

<@&286206848099549185>

[stuck lark]

@smoky pagoda this IS separable

@queen cave didn't expect DEs in precalc. this is fine so far

[smoky pagoda]

oh yeah I see now

big dumb

[queen cave]

@stuck lark sorry am i in the wrong channel? im new here ;))

[paper pecan]

#multivariable-calculus

That's the channel you're looking for

[stuck lark]

too much of a bother to move, just know for future reference there's a more appropriate channel for this

anyway keep going, isolate Q

[north crescent]

O it’s multi no wonder I was so confused lmao

[queen cave]

Alright so I got this, and below it are the variables given in the question...

But this is the answer the textbook gives (q16)

[stuck lark]

you forgot this is an initial value problem, infer an initial condition from the text @queen cave

[remote bison]

can someone please explain how to solve this in my dms

[tawny nacelle]

y it gotta b in dms

[remote bison]

bc the chat might get flooded with something else and I am busy

@tawny nacelle

<@&286206848099549185>

can someone please explain how to solve this problem (https://discordapp.com/channels/268882317391429632/363224154469826562/715180324342202429) in my dms

[craggy dune]

yeah soaps you pleb, he doesn't want to actually have to pay attention while you solve his problems for him @tawny nacelle

[viscid thistle]

@remote bison we don’t do your problems for you m8. You can get help, but we will not do the work for you.

[tawny nacelle]

[remote bison]

I don't want you to solve it for me, I asked for an explanation for how to solve it, sherlock (I didn't mean to hurt you with 'sherlock' there)

[uncut mulch]

the information indicates that the parabola has the parametric equations
x = bt^2 and y = 2bt

differentiate wrt t, and apply chain rule then sub in t = t_1 to get the slope of the tangent at (b(t_1)^2,2bt_1)
then get the slope of the normal from that

then apply point-slope formula and use the fact that (b(t_2)^2,2bt_2) also lies on that line
alternatively, the slope between those 2 points is the same as the normal
(different way to view it but the calculations are essentially the same)

then rearrange, solve for t_2

@remote bison

[viscid thistle]

?membercount

?purge

[vivid lake]

x^2+y^2+6x-24=8y

already asked about it before, but i didn't see the response until later and it was too late for me to ask a question about some stuff I didnt understand.

Anyone know how to convert the equation into a standard circle equation?

[past meadow]

complete the square

or whatever its called

[vivid lake]

Oh I think I remember what that is

[past meadow]

look to get something of the form x^2+2ax+a^2 and y^2+2by+b^2

[vivid lake]

Let me look that up rq

[past meadow]

i would suggest starting by moving the constant to one side, the x's and y's to the other, grouping the x's and y's and looking to complete the square for the x terms and the y terms

i gtg but ill recommend this video if you still have any trouble after rereading on completing the square:https://www.youtube.com/watch?v=Fx4rUC1EZPI

[vivid lake]

Alright, thanks man

[viscid thistle]

ruh

[remote bison]

sneak 100

[viscid thistle]

Ending precalc this quarter with a 65

Dear trig identities,

Fuck you.

Unsencerely,

Me

[fossil raptor]

F

[viscid thistle]

I mean I can still fix it if I do well on the last test

[fossil raptor]

try your best

[mellow sapphire]

Good luck bud

[wicked jacinth]

anyone able to help me with forming rational functions in the format of y=ax + b + f(x)/g(x) where f and g are polynomials? basically they give me holes and slant asymptotes. I'm really struggling to figure out this concept

here is an example i found from my text book

[fleet yew]

@wicked jacinth a hole is a 0/0 limit

[wicked jacinth]

kk

i think i got it 😄

[small kraken]

Hi! Does anyone know how to prove 4ii)?

[willow bear]

make a diagram

[mellow sapphire]

How is this precalc lol

[small kraken]

O i got it tqtq :>

We dont go by calculus and pre calculus so i dont really know which this topic is under 😦

[mellow sapphire]

How'd you do it

[willow bear]

if it involves derivatives it's calculus

[small kraken]

Implicit differeciation

[mellow sapphire]

Yeah but that's aids

Wow

This sucks

[small kraken]

:’>

[mellow sapphire]

Actually it's not that bad

I just suck at calc now

[viscid thistle]

can someone please check if i did those 4 right?

and if i could add anything else to this response

thanks in advance

ping if u answer ughnh mom told me to take shower now

be back in couple mins

[sour eagle]

(fg)(x) = f(x)g(x) = (4x+6)2x^2=8x^3+12x^2
(f/g)(x)=f(x)/g(x)=(3x+2)/(3x-5), x =/= 5/3
(f+g)(x)=f(x)+g(x)=2x+6+4x^2=4x^2+2x+6
f(g(x))=f(8x-13)=sqrt(8x-13+9)=sqrt(8x-4)=2sqrt(2x-1)
I dont think you need to add anything else to the last problem.

[viscid thistle]

thank you so much @sour eagle i saw u typing so i didn't go shower yet but when i am back i will ask more..

[sour eagle]

No problem

[viscid thistle]

idk what they mean in the 1st question

I know #2 is right

#4 i know is right

for #3 i know we have to use transformations

if y=f(x)+k, then the y value of any given point will be f(x) + k

[uncut mulch]

if you understand 2 and 4, you should be able to do 3

i mean this whole topic is transformations

and that reflection property should also have been covered in your notes

[viscid thistle]

what about #1 idk how to do that

[uncut mulch]

do you know what reflections are?

[viscid thistle]

for #3 i would say it's 2nd option am i right/

[uncut mulch]

yes

[viscid thistle]

thanks and could u explain to me for #1 a similar question or at least explain to me what it's asking i am confused

[uncut mulch]

do you know what reflections are?

also look up a summary of graph transformations

[viscid thistle]

I know reflection about y axis is f(x) = (x)^2

x axis is f(x) = -(x)^2

if we are given f(x) = x^2

[uncut mulch]

y = -f(x) is a reflection of y=f(x) about the x-axis

and doesn't only apply to x^2

[viscid thistle]

for #1 i think it's 3rd option right?

https://www.desmos.com/calculator/u7omjasicq

it goes across x axis so 3rd option right?

It doesn't "go across the x axis" if you mean its a reflection respect to x axis, then yes

the question is above

https://discordapp.com/channels/268882317391429632/363224154469826562/716014174156750850

Oh wait

#1

Well thats weirdly worded

yea

But yeah it is the 3rd option

thanks

Nw

wanna help me with more questions?

the next thing is 22 questions

i can do some but i like checking if i do them right

Jeez i wont go through 22 questions wtf

I have to sleep almost now so you'll wait for another one else

oh ok no worries

[viscid thistle]

is the 1st one right and for 2nd one i dont get it cause none answer choices match

<@&286206848099549185>

is it that i can ping helpers if i stated i was going to have questions soon or it starts the 10 minutes when i posted the new question?

/questions

on the 1st question i forgot how to tell if there is no point like -1
but the lines never decrease so i guess it's 3rd option?

is #6 right as well?

idk if someone would be nice to help me finish these questions

i will try tho

[hexed ermine]

Right the rate of increase increases near x=0 and it dies down on both ends

So it's always increasing never decreasing

Also no 6 isn't right

There isnt a minimum at x=0

By that, that means that the vertex is there which it isnt

[viscid thistle]

yea i finished those already and fixed the wrong ones

sm1 else is helping in dms

[hexed ermine]

Ok

[viscid thistle]

oh they went busy again so onto next questions

is (f + g)(x) the sum of two functions?

[hexed ermine]

Yes

[viscid thistle]

k got it now

done for today

[keen sigil]

could someone help me with this?

[willow bear]

@keen sigil do you sitll need help with this

[keen sigil]

Oh I figured it out, thanks though

[smoky needle]

Can someone help me solve this?

[patent beacon]

sinθ = -12/13
cos²θ + sin²θ = 1

[smoky needle]

Ohhh I didn’t know we had to use the Pythagorean trig identities

Thank you!

I don’t think I’m getting the answer when I solve it

I got this but I am not sure if it’s pos or meg

[harsh smelt]

$\sqrt{(-x)^2} = |x|$

[obsidian monolith]

Commander Vimes:

[willow bear]

I am not sure if it’s pos or meg

read the hint

you're told

-π/2 ≤ θ ≤ π/2

[smoky needle]

Yes, I considered those as well, but I’m having a hard time knowing which quadrant it would be in

I’m not too familiar with -pi/2

[harsh smelt]

"-" means clockwise direction

[smoky needle]

So it is between Quadrant 2 & 3 right?

[willow bear]

no

quite the opposite, in fact

[harsh smelt]

where are quadrants 2 and 3?

where are all quadrants?

[smoky needle]

Oh so Quadrant 1 & 4?

Sorry can you explain what you mean by where all quadrants are

[willow bear]

this is the arc from -pi/2 to pi/2

[smoky needle]

Ohhh

[willow bear]

as you can see

cos(θ) ≥ 0 on this arc

[smoky needle]

I see

[viscid thistle]

if someone can please check if i did some questions right thanks in advance

[willow bear]

ok

post your questions and your answers

[pale bison]

what in the fuck

[willow bear]

woah wow

this is a Lot

uh

wait

is this a timed test or is this just an assignment

[viscid thistle]

a assignment

like homework

[willow bear]

ok

aight well

good news, you only messed up on one question out of the 19 you submitted

and that's question 18, which should've been Reflect the graph of f across the x-axis

everything else is correct

[viscid thistle]

thanks for checking it but can u explain #18 why

[willow bear]

i can show you the graphs if you want

[viscid thistle]

yes please

[willow bear]

red is y = x^4, blue is y = -x^4

[viscid thistle]

oh yeah it intersects close to origin

[willow bear]

not really the point

my point is if you were to reflect y = x^4 across the y-axis you'd get y = (-x)^4

which is just y = x^4 again

[viscid thistle]

oh ok i got it thanks @willow bear

[full garden]

hey guys can someone please help me with this question

I really don't know how to start it

if someone can please help me it will be greatly appreciated

[pale bison]

a function means well defined,
i.e. 1 input does not map to more than 1 output

[full garden]

Yes

[pale bison]

this is the same as the vertical line test

[full garden]

Yes sir

[pale bison]

so what can you conclude from this?

[full garden]

2x+1 is the y intercept 😅 ?

[willow bear]

no

[pale bison]

the order pair are of the form
(input, output)

[full garden]

ohhh

y = 2x+1

[pale bison]

yup yup

now do the same for the other two ordered pair

and precede from there

[full garden]

bro but like

[pale bison]

[full garden]

1 produces both 3x and y+2

does that mean I have to put them equal to each other 😅 ?

[pale bison]

ok maybe this will help:

this means
f(0) = 2x + 1
f(0) = y
f(1) = 3x
f(1) = y + 2

so since a function means same input must go to the same output, you have
f(0) = f(0) and f(1) = f(1) because 0 = 0 and 1 = 1

[willow bear]

does that mean I have to put them equal to each other 😅 ?
precisely.

[full garden]

okay I'll try rn

i got y = x -1 but idk if that's correct 😅

for the final equation

[pale bison]

can you show us all your work

since you're suppose to get some precise value for x and y

[full garden]

yes yes

f(0) = y so where it says f(1) = y +2, I subbed f(0)=2x+1 in y+2, and got f(1) = 2x+1+2

and u showed me that f(1)=3x

so i put 3x = 2x+1+2

and then got y=x-1

so is that correct 😅 ?

[pale bison]

so i put 3x = 2x+1+2
solve for x

this will get you y

[full garden]

oh

i got -2

im so confused bro im sorry

but why will it give me y

like is the old equation wrong ?

I got x = 1

and when I subtituted it i got -2

please bro ik its too much to ask but please clarify it for me