precalculus

[hard hornet]

also, its worth noting that another reason why we multiply by conjugate

is to cancel out the middle terms

take a+b for example

if we did (a+b)(a+b), we would get a^2 + 2ab + b^2

but if we did (a+b)(a-b), we would get a^2 - b^2

notice how it comes out a lot cleaner that way (in both the trig case and the squareroot case)

[tropic crown]

yes i do

[hard hornet]

its a strategy we often use

[tropic crown]

thanks for the quick lesson, appreciate

the help

[hard hornet]

but it doesnt work all the time

ye ye anytime

[grave furnace]

How do you know if something is an inverse variation function

[fluid scarab]

hi

[hard hornet]

hi

[hybrid charm]

👋😼

[honest igloo]

I need help with the question: A certain type of bacteria, given a favorable growth medium, doubles in population every 8 hours. Suppose one single bacterium was put in a container with suitable environment. Write a function that gives the population of bacteria, B(t), after t hours.

[rocky bison]

This is an exponential growth

Are you familiar with this type of thing?

[honest igloo]

Correct

Yes

[rocky bison]

Okei

So you know it's going to be somewhere along the line B(t)=2^t

[honest igloo]

I have an idea of the correct answer but Im not sure if its correct

[rocky bison]

All you need to do is pad the time a bit

Now t is in hours

So just divide by 8

because 8 hours

,$ B(t)=2^\frac{t}8

[obsidian monolith]

Pseudo:

[rocky bison]

So one bacteria at t=0

[honest igloo]

Oh nice

[rocky bison]

Which we know is true

Two at t=8

Yep

Then another eight hours

Four at t=16

[honest igloo]

I got this but I put a B in front of 2 to represent the initial amount of bacteria

Is that fine

[rocky bison]

No

It specifically says on single bacteria is placed in a container

[honest igloo]

ahhh

ok got it thanks!

[rocky bison]

npnp

[honest igloo]

Oh can you help me with part 2?

It gets a bit tricky as it deals with logarithms

[rocky bison]

just post

[honest igloo]

It says: How long will it take for the population to reach 1000?

[rocky bison]

Ok

So in this case we know B(t) = 1000

and we gotta solve for t

[honest igloo]

yes

[rocky bison]

,$ B(t)=2^\frac{t}8\B(t)=1000\2^\frac{t}8=1000

[obsidian monolith]

Pseudo:

[rocky bison]

Have you ever used logs before?

[honest igloo]

yes I have that so far

and yes

[rocky bison]

Ok

So the general idea is

,$ a^b=c\iff\log_a\left(c\right)=b

[obsidian monolith]

Pseudo:

[rocky bison]

So if we apply this idea and take log_2 of each side

[honest igloo]

why log 2?

[rocky bison]

You see how c=a^b

[honest igloo]

yes

[rocky bison]

If we sub that into the equation on the left

,$ \log_a\left(a^b\right)=b

[obsidian monolith]

Pseudo:

[rocky bison]

Notice how it cancels down to just b

[honest igloo]

ahh yes thats convenient

[rocky bison]

Yep

And we have 2^t

So we take base 2

To isolate the t

,$ \log_2\left(2^\frac{t}8\right)=\log_2\left(1000\right)

[obsidian monolith]

Pseudo:

[rocky bison]

Now that log on the left will cancel to just t/8

[honest igloo]

nice, I think I get it

[rocky bison]

,$ \frac{t}8=\log_2\left(1000\right)

[obsidian monolith]

Pseudo:

[rocky bison]

And I'm sure you're capable of solving that ::)

[honest igloo]

Yes indeed thanks for the help!

[rocky bison]

npnp

[honest igloo]

In regards to this same question if I were to ask how many bacteria in a day would I do: 2^3/8?

since 3 * 8 is 24

[viscid thistle]

Anyone online?

[torn swift]

plenty, yes lol

[viscid thistle]

I have a math question, im not sure if im vomplicating things

complicating,

[torn swift]

[viscid thistle]

How would you expand

(2x^2 -xy -3y^2)(3x^2y^2 -3xy^2)

[torn swift]

$(2x^{2}-xy-3y^{2})(3x^{2}y^{2}-3xy^{2})$

[obsidian monolith]

⚡Amphy⚡:

[viscid thistle]

yes that

[torn swift]

it says you have to multiply it out?

[viscid thistle]

yes, but im confused of how many terms i should end up with

[torn swift]

there appears to be a chance that a couple will cancel in the end

did you already do the multiplication?

or you stuck on that?

[viscid thistle]

Im going to complete it and ill get back to you

[torn swift]

ok then, do that first then we can check

typically I like to put the expression with the least amount of terms on the left

$(3x^{2}y^{2}-3xy^{2})(2x^{2}-xy-3y^{2})$

[obsidian monolith]

⚡Amphy⚡:

[torn swift]

for me, that's just easier to do

[viscid thistle]

oh ok, thanks

I got...
$(6x^{4}y^{2}-6x^{3}y^{2}-3x^{3}y^{3}-9x^{2}+9xy^{4})$

[obsidian monolith]

Uniyou:

[viscid thistle]

sorry the second last term should be

[torn swift]

just edit the message

[viscid thistle]

9x^2y^4

[torn swift]

then the bot will re render it for you

[viscid thistle]

ok

I got...
$(6x^{4}y^{2}-6x^{3}y^{2}-3x^{3}y^{3}-9x^{2}y^{4}+9xy^{4})$

[obsidian monolith]

Uniyou:

[torn swift]

this one has six terms

so you are missing one

[viscid thistle]

hmmm

[torn swift]

all other terms were correct

you just missed one somehow

[viscid thistle]

i see the problem, thanks

[torn swift]

like I said, plenty of x's y's and numbers going around so you have to be careful

because you can easily mess up

[viscid thistle]

$(6x^{4}y^{2}-6x^{3}y^{2}-3x^{3}y^{3}-9x^{2}y^{4}+9xy^{4}+3x^{2}y^{3})$

[obsidian monolith]

Uniyou:

[torn swift]

that is correct

[viscid thistle]

thank you so much!

[torn swift]

you can then factor out an xy^2 term to simplify

[viscid thistle]

Right!

[torn swift]

it's xy^2 that is common to all

[viscid thistle]

cant you also take out a 3?

[torn swift]

that too lol

I was focusing on the variables, but there is a common 3 as well

[viscid thistle]

Few, I thought I put something funny

thanks

[keen heath]

So what’s the content of pre calculus, just to know. I’d like to get into calculus when I get everything in pre calculus

[torn swift]

topics usually covered are large amounts of trig, basics of series and summations, complex numbers, lots of other things. The book my highschool used covered a broad range

but the main focus in precalc is the trig

[keen heath]

Well that’s where I’m at I guess lol

So basically it’s finding values out of angles right

I’ve seen some formulas

[torn swift]

you really need to get the basics of trig down, as trig will be important in calc

usually you don't use anything but the special angles in calc

[keen heath]

Special angles ?

[torn swift]

I haven't ever had to use a sum and difference formula

I mean like, 30,60,45,90 degrees

[keen heath]

Oh

[torn swift]

has a calc question ever asked me to use like 38 degrees? no

[keen heath]

I forgot about what 30 degrees does

I think it’s like the opposite side is 1/2 the length or something

But yeah so what about them

[torn swift]

you just need to remember those values

[keen heath]

K then

Except learning formulas, what’s there to understand in trigonometry

[torn swift]

the common trig identities will be very useful as well, there's that whole thing in pre calc where you have to prove identities of large expressions, but you won't encounter a prove sec^5(x)cos(10x) = blah blah blah

[keen heath]

What are cos and sin made out of

[torn swift]

what do you mean by that?

[keen heath]

How did we get these wave boys

[torn swift]

you mean how did we end up with the graphs of sine and cosine?

[keen heath]

Yes

I think it’s degrees or something, since they are linked with Pi and stuff

Like at a certain degree it will give a specific value

Since Pi is 180 degrees

[torn swift]

The first thing you ever learn about sine and cosine is that:\
\
$\sin(x)=\frac{\text{opp}}{\text{hyp}}$ and $\cos(x)=\frac{\text{adj}}{\text{hyp}}$

[obsidian monolith]

⚡Amphy⚡:

[keen heath]

Opposite and adjacent sides

[torn swift]

later in pre calc you learn how these are related to the unit circle

[keen heath]

Ok

Hyp is hypotenuse right

[torn swift]

yes

[keen heath]

And we are talking opposite sides or angles

Probably side length lol

[torn swift]

side lengths lol the only thing that is an angle is x

x is referred to as the "argument" of the function

[keen heath]

Oh ok well that’s easy now to find the values based of angles

Lol

[torn swift]

that's not something you really have to worry about

just know that's what it is referring to when you see that

[keen heath]

Ok then

So they go from 1 to 0

No 1 to -1

Sin and cos

[torn swift]

have you started pre calc yet?

[keen heath]

Well idk what’s pre calc, I’m doing math at school

Example ?

Maybe give an exercise or something

[torn swift]

sure

[keen heath]

I guess you are solving it before presenting it

[torn swift]

what?

[keen heath]

Well I said “maybe give an exercice or something”

[torn swift]

oh I thought you were going to give it lol

[keen heath]

And you said sure

Oh lol

Maybe I’m not clear, English isn’t my first language lol

[torn swift]

what kind of trig problem do you want to see then?

I can show difficult ones if you want

[keen heath]

Well you asked if I did pre calc before, just a general pre calc exercise

Anything you want

[torn swift]

This encapsulates the thing you learn in pre calc in terms of trig

here's a close look at them

[keen heath]

That’s not overwhelming but still a lot of stuff

[keen heath]

Are logs in pre calc ?

[torn swift]

I believe so

you like logs?

[keen heath]

I mean I know how to use them

Also the 0 with a bar

Is just angle right

I forgot it’s name

[spring thunder]

it's theta

[keen heath]

Oh yeah that

[slender river]

psi rotated

[torn swift]

Give an upper and lower estimate on the value of $\log_{10}(36.82)$

[keen heath]

I mean I know how to use them in algebra

[obsidian monolith]

⚡Amphy⚡:

[keen heath]

I don’t rly know how to do estimates with it

Didn’t know they had a upper and lower value or something

Thought it was fixed

[torn swift]

it's between 1 and 2 if you are doing a gross estimation

the value of log(36.82) is certainly fixed but sometimes you just want to know what values if could be

it's just a bit smaller than the 1+log(4)

and certainly larger than 1+log(3)

[keen heath]

Oh so that’s what you mean

Well I know it takes a long time after like 1.5 to go up

Like with log(x)

[torn swift]

there are certainly interesting things you can go with logs though

you can take the nth root of anything using log10

[keen heath]

Is there a max value of log

It seems to always go up

[torn swift]

infinity I suppose

[keen heath]

But very slowly

[echo plaza]

no max no min

[torn swift]

you can take the nth root using log base anything actually, 10 is most convenient though

[keen heath]

Hey give me log(-1)

(:

[echo plaza]

complex logarithms have branches

so it depends which one you pick

[static prairie]

the one which is continuous :kappa:

[echo plaza]

[keen heath]

Oh crap I gtg go eat

Brb

[tropic crown]

when A is to the power of lets say 3

it looks like AAA

but lets say if A is to the power of -3

what would it look like

?

[torn swift]

$A^{-3} = \frac{1}{A}\frac{1}{A}\frac{1}{A}$

[obsidian monolith]

⚡Amphy⚡:

[keen heath]

K I’m back

Why do you put this equation so weirdly

[tropic crown]

T_T, ty, idk why i didnt realize that

[keen heath]

Lol

[torn swift]

what is A^n though? it's AAAAAAAAAAAAA... lol

[keen heath]

Oh this vid

[torn swift]

I hate these kinds of problems in precalc

[echo plaza]

what triggers me is when people start with the identity in order to prove it

also that's technically wrong because the left isn't defined for theta=pi/2

[keen heath]

Verify Euler identity

[torn swift]

Taylor Expand, done

[keen heath]

See, it’s in the cos and sin and tan subject

[fallow temple]

Is this correct?

[short sorrel]

not the way i interpreted the question, no

only two people are allowed to sit in front, which gives you only 2 options for who can sit in front

if we then remove that person from the pool, we have 5! ways to order the remaining people

(im assuming the toboggan fits 6 people)

this would give us 2*5!

2 options for the leader * 5! options for the remaining people

which is 240

fyi its good practice to sort of "gut check" your answers in combinatorics, since the answers are often tricky to get an intuition for: for this problem, clearly there should be less possibilities than if there was no rule on who sits first

which would be 6!

so our answer should be less than 6!

your answer is much more than 6!, which doesnt make much sense

as adding more restrictions should remove options, not add them

[fallow temple]

that makes a lot more sense thank you

[viscid thistle]

Im doing a question and got to this point in the question and am now stuck if someone can help me.

$[(x^{2}-x) -6][(x^{2}-x) -2]$

Inside the brackets would i multiply the -6 through or add it to the equation?

[obsidian monolith]

Uniyou:

[short sorrel]

you're subtracting 6 from (x^2 - x)

[short sorrel]

so that bit can be rewritten as just [x^2 - x - 6]

[keen heath]

Kind of weird, didn’t know that relation

[short sorrel]

well, yeah, log_10(10^0) = 0, log_10(10^-1) = -1, log_10(10^-2) = -2, etc

[keen heath]

I think you are supposed to multiply the -6 with (x^2-x), same for -2 and just multiply the results

Wait nvm

[short sorrel]

$[x^2 - x - 6][x^2 - x - 2]$

[obsidian monolith]

Namington:

[short sorrel]

we can rewrite that as just this

[keen heath]

Yeah

[viscid thistle]

so it would just form a quadratic?

[keen heath]

It’s not the same if the -6 is before or after

[short sorrel]

well, a product of 2 quadratics

[keen heath]

Quadratic ?

I mean if you can put it into quadratic form

[short sorrel]

which wouldnt be a quadratic, itd be a quartic

[keen heath]

There is a quartic formula

[short sorrel]

but each individual factor is a quadratic, sure

in that form

[keen heath]

It’s so big it uses multiple letters to not make it unreadable

Btw

[viscid thistle]

the -6 is after so you would just add it to the brackets? to make x^2-x-6

[short sorrel]

yes

its like if we have

(1 + 2) + 3

[viscid thistle]

thanks

[short sorrel]

thats the same as just

1 + 2 + 3

[viscid thistle]

Ahh thankd

thanks*

[keen heath]

What are we tryin to do

If you want quadratic you can change it into y(y+4)

But that doesn’t do much

[short sorrel]

?

[keen heath]

Idk you people saying things about quadratic

I mean there isn’t an equality so I was kind of lost

[viscid thistle]

@keen heath sorry I was just connecting terms, my teacher really wants us to realise when there is a quadratic even if we dont need to solve

[keen heath]

I mean if your equation was equal to 0 I had a way to do it

Replace the -6 one by y and the -2 by y+4

y(y+4)

y^2+4y=0

Quadratic gives us either 4 or 0

Then you do : x^2-x-6=4 or 0

You solve for x and boom, it’s solved

I think I just decomposed a quartic to turn it into a quadratic, since there are 4 possible results

[viscid thistle]

@keen heath Ok I see thanks

[tropic crown]

is there a name for this method?

for the red circle part

[torn swift]

making a ratio?

you can arrive at the same ratio by solving each equation for C then equating that

$Ca^{3}=24$, $Ca=6 \to C=\frac{24}{a^{3}}$, $C=\frac{6}{a}$\
\
$\frac{24}{a^{3}}=\frac{6}{a} \to 6a^{3}=24a \to a^{2}=4$

[obsidian monolith]

⚡Amphy⚡:

[nimble ridge]

Hey would any of you mind checking my work?

[magic abyss]

The graph of y = h (x) is a transformation of the graph of y = g (x).

Write a formula for the function h in terms of the function g.

h(x) = ?

I need help w/ this guys

The teacher put this homework but didn't explain anything...

[nimble ridge]

Yikes

[torn swift]

well you can do that based on how the vertex shifted

[magic abyss]

Counting the distance bt the points?

[torn swift]

quite literally yes lol

it was shifted how many to the left and how many up?

[magic abyss]

It doesn't say anything about that

[nimble ridge]

(Can I get help after him?)

[torn swift]

what is the vertex of g(x)?

what point is it at?

[magic abyss]

I don't remember what a vertex is

The topic is transfer functions btw

I'm practically new to precalculus

[torn swift]

that doesn't matter lol. We just have to write a transformation in terms of g(x). Easiest way to do that is see how the points moved

what is the easiest point to read off the graph of g(x)?

it's that point all the way at the top right?

[magic abyss]

Well, in the y = h(x)counting from -5 I can see it's 11

[torn swift]

you are correct it shifted 5 units to the left, so it went -5 on the x axis, but 11 has nothing to do with this

[magic abyss]

I thought it was 11 units to the top

[torn swift]

no...let's not do that

[magic abyss]

Alright

[torn swift]

the easiest point to read on the graph of g(x) is the point at the very top correct?

[magic abyss]

Yes

(0,6)

[torn swift]

yes, and that same point at the very top of h(x) is also pretty easy to read as well

what is it?

[magic abyss]

(-5,7)

[torn swift]

good, so we have two points on the same location of the graph, both these points are at the top of their respective graphs
we can then use those to find the shift that created h(x)

how much did the graph shift in the x direction?

[magic abyss]

-5?

[torn swift]

be sure of your answer

but counting squares alone, it is -5, nothing to be doubtful about

[magic abyss]

Alright

[torn swift]

now how much did it shift up in the y direction?

[magic abyss]

1

[torn swift]

good

now we are ready to write down the transformation

so we start with g(x), but we can see that the x inputs have been shifted by -5

[magic abyss]

Oh

[torn swift]

so we have to show that in g(x)

[magic abyss]

So g(-5)?

[torn swift]

almost

that only describes one point

you need to describe x amount of points

so it would really be your input (x) -5

so it's g(x-5)

[magic abyss]

Oh

So it's x cuz it's in the x line right?

if it was at the top would be y

[torn swift]

what do you mean by "at the top"

[magic abyss]

Like, if the graph shifts in y

[torn swift]

we still have to account for that so we will get to that actually

we wrote g(x-5) because we have to describe how all the x inputs have been modified

[magic abyss]

Got it

[torn swift]

x-5 only shifts the x coordinates

[magic abyss]

So now we have to write the 1 in the y coordinates right?

[torn swift]

so we still have to add onto g(x-5) to also shift the y

correct, exactly

how can we do that?

let me re word that

the output,y, is equal to the result of the input, which is g evaluated at x

so the y coordinate is given by y=g(x-5), but we have to add on 1 to this so it can shift all y values up by one

[magic abyss]

And we can do that by putting that function inside the other function?

[torn swift]

wait give me a second, I think it's actually g(x+5), and I'll explain just why in a second

[magic abyss]

Alright

[torn swift]

I knew something seemed off lol

let me check myself then get back to you

[magic abyss]

Ok

[torn swift]

yes, it actually is x+5

now we have to see why that is

recall that h(x) is the shift from g(x)

[magic abyss]

Yes

[torn swift]

starting from h(x) we need to find out how to get back to g(x)

if g(x) has an x coordinate of 0, and h(x) has an x coordinate of -5 the way to get back to g(x) is to do what?

[magic abyss]

By moving 5

[torn swift]

to the right to be precise, so add 5

[magic abyss]

But what happened to the 1 in the y coordinates then

[torn swift]

we did it backwards lol

for x that is

you know that is interesting because now I made it more confusing
I didn't know if you would understand if I wrote
h(x)=g(x-h)+k

[magic abyss]

Ok so basically what I'm understanding is that the value of the Y coordinates depends on the value of the X ones

Is that correct?

[torn swift]

yes, that is correct

but I also wrote down for you the general formula for a transformation

h(x)=g(x-h)+k
h is the horizontal shift (difference in x)
k is the vertical shift (difference in y)

[magic abyss]

Alright

[torn swift]

does that happen to make sense for you?

[magic abyss]

Yes it's like when I see conic functions

[torn swift]

ok well using this general formula for a transformation it's pretty easy now

h=-5, like you said
k=1

[magic abyss]

We just have to put values

Alright

[torn swift]

so then h(x)=g(x-(-5))+1

[magic abyss]

I get your point

[torn swift]

so then h(x)=g(x+5)+1

[magic abyss]

Nice

Tysm

I'm gonna keep practicing the topic by myself

[torn swift]

it is a bit weird that adding will shift left

and subtracting will shift right

it goes against the general notions of left being negative and right being positive

[magic abyss]

Lol

[magic abyss]

So did we just make a new law or what

[torn swift]

I was always a bit weary with transformations, and definitely look up more things if it's confusing because it really is at first

we didn't make a new law

lol

[magic abyss]

Alright I would keep doing exercises of this topic

Thanks for your time

Really appreciate it

[torn swift]

I've always just remembered that general form and wrote transformations that way

[ruby otter]

$\lim_{x \to 0} \frac{x\left(2a\cos\left(x\right)+1\right)-b\sin\left(x\right)}{x^3} = 2$

[obsidian monolith]

Radical Ninja:

[ruby otter]

how to find a , b

i did L'hospital

[willow bear]

l'hôpital 🤢

[ruby otter]

then what shall we do?

[willow bear]

let $f(x) = x(2a \cos(x) + 1) - b \sin(x)$

[obsidian monolith]

Ann:

[willow bear]

you want $f(0) = f'(0) = f''(0) = 0$ and $f'''(0) = 2 \cdot 3!$

[obsidian monolith]

Ann:

[willow bear]

if any of f(0), f'(0) or f''(0) were not zero then the limit would either be infinite or wouldn't exist

[ruby otter]

why and how f'''(0) = 6?

[willow bear]

did i say f'''(0) = 6

[ruby otter]

ohh thats a factorial there

[willow bear]

...i mean ok i guess one could like

[ruby otter]

but any how its gonna give you the same result as l hospital

[willow bear]

LH that three times

if you want

i just hate LH because most of the time it can be replaced by something that's harder to misapply

[ruby otter]

but if you do that you would get
b- 6a = 12

after that?

[willow bear]

that's not the only thing that needs to hold

you also need $f(0) = f'(0) = f''(0) = 0$

[obsidian monolith]

Ann:

[ruby otter]

can i take any one and equate to zero to get the 2nd eq?

[willow bear]

no, you need all three.

two of them are going to degenerate into 0=0 but you need all three.

[ruby otter]

also how did you know to stop exactly at 3rd derivative bc of the Dr?

[willow bear]

the higher derivatives are irrelevant

you could see it as like

LH no longer applying after doing it three times

[ruby otter]

thats what im asking so you need to do trial and error to know that you have to stop at 3rd derivative ??

[willow bear]

i mean if you want i could tell you that i knew the function was smooth and so had a taylor series of which i only cared about the terms up to and including x^3

[ruby otter]

[dapper forge]

Anyone can help with a few problems

[fading token]

[dapper forge]

http://prntscr.com/no0lxs @fading token

[hard hornet]

i recommend drawing the triangles in each quadrant

and then looking at the x and y component

and then taking the ratio, that should help

[torn swift]

Recall how x,y coordinates are related the cos(x) and sin(x) then you can use the signs of the numbers on each quadrant to help you figure it out

[dapper forge]

Whats the exact name for this http://prntscr.com/no1aze

[serene heath]

theta

[pure pine]

Hey everyone. First time poster here. Not usually stuck on PreCalc, but I feel a little brainless here.

(cos - cot) / (1 - sin) = - cot. I get to (cos - (cos/sin)) / (1 - sin) but am stuck on how to flip it to ((cos) (sin - 1)) / (1 - sin) (sin).

[hard hornet]

wat r u trying to do?

[pure pine]

Oh, my bad. Proving (cos - cot) / (1 - sin) can be turned into - cot

[hard hornet]

gimme a sec

o ok

so the tricks for things like this

is you want to work on one side

and leave the other side alone

which side would you choose to work on

o actually u had the solution above

ur question was how to flip it

recall

that

(sin - 1) = - (1 - sin)

you can factor out a negative

hope that helps

[pure pine]

Thanks! I will give it a try.

[hard hornet]

when you factor out the negative, and cancel out something, you should get your solution

[rocky bison]

,$ \frac{\cos\theta-\frac{\cos\theta}{\sin\theta}}{1-\sin\theta}

[obsidian monolith]

Pseudo:

[rocky bison]

If you've got to there it's p simple to get it

Multiply by sin/sin

,$ \frac{\cos\theta\sin\theta-\cos\theta}{\left(1-\sin\theta\right)\sin\theta}

[obsidian monolith]

Pseudo:

[rocky bison]

Factor cos on the top

,$ \frac{\cos\theta\left(\sin\theta-1\right)}{\left(1-\sin\theta\right)\sin\theta}

[obsidian monolith]

Pseudo:

[rocky bison]

Now

,$ \sin\theta-1=-1\cdot\left(1-\sin\theta\right)

[obsidian monolith]

Pseudo:

[rocky bison]

Or

,$ \sin\theta-1=-\left(1-\sin\theta\right)

[obsidian monolith]

Pseudo:

[rocky bison]

So make this substitution in the numerator of your fraction

,$ \frac{-\cos\theta\cancel{\left(1-\sin\theta\right)}}{\cancel{\left(1-\sin\theta\right)}\sin\theta}

[obsidian monolith]

Pseudo:

[rocky bison]

,$ -\frac{\cos\theta}{\sin\theta}

[obsidian monolith]

Pseudo:

[rocky bison]

@pure pine

[pure pine]

@rocky bison Thank you ever so much. That helped me a lot.

[rapid edge]

what exactly is precalculus? just algebra + arithmetic?

[torn swift]

lots of trig too

[rocky bison]

Integration/Differentiation not included too?

[royal gull]

Not in precalc I believe

[rocky bison]

Rip

[hard hornet]

stuff u shud know before calculus

basically

[grizzled orchid]

also limits and vectors

and sometimes matrices

[hybrid charm]

Every single thing you mentioned is difficult

[grizzled orchid]

eh

relative to further stuff it isn't

but i guess that can retroactively apply to lots of math

[hard hornet]

limits is bordering pre calc and calculus, i'd say thats more calc, unless u have like some advanced pre calc classes

vectors is probably towards the end of calculus

i'd say for pre calc, learn conics, trig, polar coordinates, a bit of matrices, and thats about it

[grizzled orchid]

rly? the precalc class at my school covers limits and vectors

[hard hornet]

vectors is okay i guess, super basic vectors though

[grizzled orchid]

we do conics in alg II

basic vectors yah

[hard hornet]

sounds about right

curriculum differs from school to school

[grizzled orchid]

yeah

also i've only seen the honors precalc curriculum so it might not be covered in the non honors class

lol the #discussion flamewar about cats was legendary

[hard hornet]

but tbh, i'd say the most important things in preparation for calculus would be trigonometry, polar coordinates, and like lots of algebra, thats about it

u dont need to learn vectors, matrices, or limits in calculus

you do that later

[grizzled orchid]

yeah pretty much

precalc is a weird class b/c it seems to cover a lot of things very briefly and then you come back to them later in calc itself

[hard hornet]

its weird because it differs from school to school

there is no like

[grizzled orchid]

tbh the best thing precalc can do for you is familiarizing you with derivatives and thinking about calc in the last 2 months

[hard hornet]

solid pre calc curriculum kind of thing

[grizzled orchid]

ye

[hard hornet]

its just an extension of other maths that are important for calculus, problem solving techniques n stoof like that

[grizzled orchid]

yeah

for calc tbh building fundamentals and reasoning/logic skills is more useful than learning a bunch of vaguely related topics

[hard hornet]

but honestly, u dont have to stress about it @rapid edge if you're taking it soon

[grizzled orchid]

yah! the class will cover everything you need

and if you are prepared going into it it isn't that big a jump from alg II

[tropic crown]

im having a little trouble understanding this

from my thought process, sin0 = 1

so ln1 = 0

cause e^0 = 1

[hard hornet]

sin(0) is 0

0 degrees, or 0 radians doesnt matter, sin(0) is 0

[grizzled orchid]

yeah

that's probably where the confusion is coming from

[hard hornet]

but i think the more important question is recognizing the limit coming from the right hand side vs a limit coming from both hand sides, hopefully you understood that

hope it helps

[grizzled orchid]

$\lim_{x \to 0^+} \ln (\sin x)$

[obsidian monolith]

hegel:

[grizzled orchid]

$\lim_{x \to 0^+} \ln x = -\infty$

[obsidian monolith]

hegel:

[grizzled orchid]

from there you should be able to see the steps taken in the equation you posted

[tropic crown]

thanks for the clarification

[grizzled orchid]

yah i think you just messed up with sin(0) lol

[tropic crown]

lol yup

[sour token]

Hey, having an issue with this question. Hoping someone can help me out

Write a polynomial function of minimum degree with real coefficients whose zeros include those listed. Write the polynomial in standard form.

8, -14, and 3 + 9i

[grizzled orchid]

this isn't where questions go lol

[sour token]

rip

sorry

[grizzled orchid]

try one of the open questions channels

lol it's fine

just got to keep it organized cause there's so many ppl

[serene heath]

wdym @grizzled orchid

this channel.is meant for questions

[grizzled orchid]

example questions go in one of the like. 10 question channels don't they

[serene heath]

they can go here as well

it's just the question channels are more.general

[grizzled orchid]

then why do we have those channels too

[serene heath]

and these are area specific

[grizzled orchid]

i confuse, other honorables tell me not to post qs here

[serene heath]

who

[grizzled orchid]

like literally everyone anytime anyone posts a question

[serene heath]

people been asking and answering qs here for ages

unless they changed it

[grizzled orchid]

w a t

[serene heath]

recent

I think ur mistaking it with #math-discussion

[grizzled orchid]

idk the system is kind of confusing

[sleek pawn]

I have asked questions here

[grizzled orchid]

idk maybe my IQ is too low

[hard hornet]

@sour token still need help?

[sour token]

yea

[hard hornet]

alright

i can help

so you have solutions 8, -14, 3+9i right?

and you have to write a polynomial

is that right?

[sour token]

yea

[hard hornet]

alright cool

lets start

first things first, there is something yo uwant to see off the bat

and that is, they want REAL coefficients polynomial

what does that imply?

if you didnt learn this in class, it might take some time to explain

[sour token]

Not sure I know what you mean

[hard hornet]

ok lets simplify it a bit

what will the highest degree of this polynomial be?

[sour token]

I understand that those are 4 roots including the opposite of 3+9i and 3-9i

[hard hornet]

ah ok

there u go

perfect

[sour token]

4th degree

[hard hornet]

that was where i was getting at

you know conjugate, and if you need a polynomial of real degree, that your imaginary solutions must have its conjugate with it

ok you got that part

that was actually the hard part imo

now, here is where the computations comes in

[sour token]

the computations have been throwing me off b/c I have done it by hand and calc and cant seem to get any of the answer choices given

[hard hornet]

if you have a polynomial with solution 3+9i, 3-9i, 8, -14, what would your equation look like

*hint: if you had a polynomial in which the solutions were 1, and -1, your equation will be (x+1)(x-1) = 0

oh the calculations are throwing you off?

ok so how would the polynomial look factored out like above

[sour token]

(x-8)(x+14)(x+(3-9i))(x+(3+9i))

[hard hornet]

you might want to be a bit careful with your imaginary

try again

the (x-8) and (x+14) is correct

the last two terms is a bit off

remember, if you plug in (3+9i) and (3-9i), it should give you 0

[sour token]

not sure

[hard hornet]

how did you do the 8 and 14 part?

if 8 is a solution, that mean (x-8) is 0 right?

do the same thing with the imaginary

[sour token]

what sets it equal to 0

[hard hornet]

its a polynomial with those solutions

[sour token]

(x-8)(x+14)(x+(3-9i))(x+(3-9i))

[hard hornet]

mmmm still a bit off

if i were to plug in 3+9i and 3-9i, i would NOT get 0

[sour token]

(x-8)(x+14)(x+(3-9i))(x-(3+9i))

[hard hornet]

you're very close

ok you got the (x-(3+9i))

the (x+(3-9i)) is still a bit off

hint: your sign is a bit off

another hint: treat 3+9i and 3-9i as a whole entity

its easier than you think, i think you might be over thinking a tad bit

[sour token]

im a bit confused

[hard hornet]

ok lets go back to the basics

lets say

i have solutions of 1 and -1

what would my polynomial be?

[sour token]

(x-1)(x+1)

[hard hornet]

correct

now

lets say i have solutions (1+i) and (1-i), what would my polynomial look like?

its the exact sam ething

trust me

you're overthinking

[sour token]

(x+(1+i))(x-(1-i)

[hard hornet]

where are you getting the + from?

in the (x+(1+i))

the x +

[sour token]

is it both minus

[hard hornet]

where u get the +

yes

[sour token]

oh

(x-8)(x+14)(x-(3-9i))(x-(3+9i))

[hard hornet]

Does it make sense?

[sour token]

seems to since the 3-9i and 3+9i are their own entity so the x - is how they reach 0

[hard hornet]

EXACTLY

[sour token]

ok

I can do the question now, THANK YOU SO MUCH

been staring at this for an hour

[hard hornet]

yea np

[tropic crown]

is f(x) = 5-sin(x)

1 to 1?

i know if they restrict it, it can be, but it doesnt give context that they did

[hard hornet]

what does it mean to be 1 to 1

ok yea it kinda depends on ur domains

hmmmm

[tropic crown]

1 to 1 is if a unique x value produces a unique y value in a function

[serene heath]

not unless u restrict

[hard hornet]

wait shit i forgot, is 1 to 1 bijection

or surjection

[serene heath]

its injection

[hard hornet]

o boi

welp find a counter example

if u can find a counter example, ez klap

awww i wanted him to answer that

[serene heath]

my b I was planning to say that anyway

[hard hornet]

ah its k

but u get it?

@tropic crown

[tropic crown]

yah i got it, thanks

[hard hornet]

cool

[idle dust]

hey

i have a question thats kind of stupid

is it 5e^5y or is it e^5y

[willow bear]

neither

it is in fact $5e^{10y}$

[obsidian monolith]

Ann:

[willow bear]

$\dv{x} [e^{5y}] = 5e^{5y} \dv{y}{x} = 5e^{5y} \cdot e^{5y}$

[obsidian monolith]

Ann:

[viscid thistle]

$[(2p-1)(p+1)]/ (6p-3)$
How would I simplify this?

[obsidian monolith]

Uniyou:

[torn swift]

can you factor anything out from 6p-3?

[earnest wedge]

can anyone help me understand parametric equations?

[torn swift]

you have to post an example lol

[hard hornet]

solve as a function of x and y, and then set them equal to teach other

t = f(x) = g(y)

[earnest wedge]

ok, i don't understand like the last two examples

[hard hornet]

ah ok

heres a hint

polar/conics

if you still dont get it, i can show u a bit

[earnest wedge]

yeah, could you like show how to approach those kind of examples?

[hard hornet]

heres something that might help

lets say you have an equation of a circle

[earnest wedge]

yes

[hard hornet]

x^2 + y^2 = 1

[earnest wedge]

ya

[hard hornet]

how would you parametrize it with time

theres a neat trick that utilizes a trig pythag identity

[earnest wedge]

um

[hard hornet]

take some time to think about it

there is a trig idnetity used

when parametrized with t

will give you the same exact equation, but with t instead

[earnest wedge]

all i know is sin^2 + cos^2 = 1

[hard hornet]

correct 😄

now

how would you parametrize that circle with time

[earnest wedge]

that make no sense tho

you add the two terms together

[hard hornet]

what doestn make sense

sin^2 (t) + cos^2 (t) = 1

[earnest wedge]

yes

[hard hornet]

that sure is true for any value of t

[earnest wedge]

ok

yeah, but how would you be able to get the t by themselves with the trig functions

[hard hornet]

sin^2(t) + cos^2(t) = 1
x^2 + y^2 = 1

you see a pattern?

[earnest wedge]

yes

[hard hornet]

what is x euqla to, and what is y equal to?

[earnest wedge]

x is cos

y is sin

[hard hornet]

perfect

sry i was doing something in the meantime

@earnest wedge

so you would have an equation

x = sin(t) and y = cos(t)

which is the parametrization for a circle of radius 1

does that make sense?

[earnest wedge]

yeah

[hard hornet]

notice how u can abuse the pythagorean trig rule

wait mb

x = cos(t) and y = sin(t)

[earnest wedge]

oh

[hard hornet]

x^2 = cos^2(t), y^2 = sin^2(t)
that means

cos^2(t) + sin^2(t) = x^2 + y^2 = 1

tada

now

do the same thing

sin^2(t) + cos^2(t) = 1

to solve c and d

lmk how it goes

[earnest wedge]

uh ok

do you know how i would do part c and d then?

[hard hornet]

use the pythag identity

x = 1 - sin(t)
y = 1 - cos(t)

question c for example

how would u use the pythag identity to ur advantage here @earnest wedge

[earnest wedge]

um

[hard hornet]

remember that sin^2(t) + cos^2(t) = 1 is ALWAYS true

no matter what

[earnest wedge]

yes

[hard hornet]

how would u use that

to ur advantage

[earnest wedge]

square the terms

[hard hornet]

x = 1- sin(t)
y = 1 - cos(t)

mmmm

i wouldnt square them

actually

how would u square them

show me

[earnest wedge]

x^2 = (1-sint(t))^2

[hard hornet]

mmmm yea

no

try something else

if you do that

you will get something nasty

[grizzled orchid]

yah that would be very gross trig

[earnest wedge]

x+sin(t) = 1

[hard hornet]

close....

hint: ||isolate sin by itself||

[earnest wedge]

sin(t) = 1-x

[hard hornet]

correct

[grizzled orchid]

[hard hornet]

now whats next?

[earnest wedge]

square terms

[hard hornet]

mhm

[earnest wedge]

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

[hard hornet]

mmmmm dont distribute for now

leave it as it is

[earnest wedge]

ok

[hard hornet]

sin^2(t) = (1-x)^2 is just fine

now what would you do for the y-equation

[earnest wedge]

i did the y equation

[grizzled orchid]

awww the cool part was about to happen

[hard hornet]

what did u get

punch it out here

[earnest wedge]

wasn't that sin

.-.

oops

[hard hornet]

oh shit mb

oops

[earnest wedge]

LOL

[hard hornet]

ok ye

[grizzled orchid]

hahahahaha

[hard hornet]

so u shud get

wait thats hella weird

ok whatev

u shud get

[earnest wedge]

cos(t) = (1-x)^2

[hard hornet]

cos^2(t) = (1-x)^2
sin^2(t) = (1-y)^2

right?

[earnest wedge]

yeah

[hard hornet]

now

final step

what do u do

[earnest wedge]

add the terms

[hard hornet]

add both of em

yup

cos^2(t) + sin^2(t) = (1-x)^2 + (1-y)^2

is equal to 1

[earnest wedge]

(1-x)^2 + (1-y)^2 = 1

[hard hornet]

ding ding ding

u got it

[earnest wedge]

do i need to distribute

[hard hornet]

i wouldnt

i think that looks fine

[earnest wedge]

ok

[hexed dagger]

omg i remember learning that earlier in the year

[hard hornet]

now

for part d

do the same thing

BUT

what i want u to do

is ignore the 2t for now, think of 2t as like theta

[earnest wedge]

ok, so its just the angle

[hard hornet]

i can explain later what t and 2t means after u finish

well if u want

[earnest wedge]

it means 2 times the period

[hard hornet]

yup

[grizzled orchid]

what a spooky circle

[hard hornet]

for example in a circle

t means one revolution, 2t means twice as fast, 2 revolutions

basically

[earnest wedge]

k

[hard hornet]

in the context of circles that is

[grizzled orchid]

,w graph (1-x)^2 + (1-y)^2 = 1

[obsidian monolith]

Display results online and refine query

Requested by hegel#5301

[grizzled orchid]

[earnest wedge]

(1,1)

[hard hornet]

are you required to describe the graph you reparametrized into cartesian coordinates?

[earnest wedge]

https://cdn.discordapp.com/attachments/363224154469826562/578427653611323404/unknown.png

[hard hornet]

nvm

[earnest wedge]

wut

part d

[hard hornet]

ok part d

[earnest wedge]

don't i just divide by 2

then square

[hard hornet]

yup

LOL u got it

[earnest wedge]

ok, but what about 2t

[hard hornet]

sin^2(t) + cos^2(t) = 1

sin^2(2t) + cos^2(2t) = 1

remember

that the trig pythag identity

always works

[earnest wedge]

ok

[hard hornet]

as long as the angles in sin and cos are the same

trig pythag identity always holds

[earnest wedge]

ok

x^2/4 + y^2/4 = 1

[hard hornet]

and what does that graph

[earnest wedge]

a circle

[hard hornet]

awesome

now

the thing to note here is that

a parametric graph and a cartesian graph are NOT the same

they give you the same exact graph

but parametrizing with t

essentially gives you some sort of "path", compared to carteisan, whch only gives you a bunch of points

lets say

you have two equations of a circle, one that is carteisna, and one that is paratrized with a third variable t

x^2 + y^2 = 1
AND
x = cos(theta)
y = sin(theta)

both of these

are the same exact graph

BUT

x^2 + y^2 = 1 only gives you a bunch of points, which gives you a circle

on the other hand

x = cos(theta) and y = sin(theta) gives you a bunch of points, but the thing that makes parametric cool, is that its not just a bunch of points. its a bunch of points that makes a line that makes a direction

in this case

the point starts at (1,0)

and goes counterclockwise

does that make sense?

o god i think i scared him

yup i think i scared him

rip

[hard hornet]

need help?

[viscid thistle]

Really confused

[hard hornet]

have you gone over inverse trig stuff

what it means

when you take inverse

[viscid thistle]

Yeah I guess

[hard hornet]

and have you learned the 4 quadrant thing

quadrant 1, what happens, quadrant 2, whath appens, quadrant 3, etc etc etc

[viscid thistle]

Lol Im supposed to be learning this so idk

[hard hornet]

and have yo ulearned that inverse trig has some sort of domain

oh

uhh

ok lets find some starting point

u know ur range for inverse trig functions?

[viscid thistle]

The range for regular trig?

Oh

inverse

[hard hornet]

not regular trig

inverse trig

inverse trig has a range

[viscid thistle]

Sure

[hard hornet]

in order to be a defined ufnction

do u know it or naw

[viscid thistle]

It has a restricted domain since we are using the unit circle

[hard hornet]

ok just for test

what is the range for inverse cos

[viscid thistle]

The range for it should be (-infinite, infinite)

[hard hornet]

uhhhhh no

[viscid thistle]

Oh is that sin

[hard hornet]

o boi

sin not even close

uhhh

[viscid thistle]

Oh wait

[hard hornet]

u got is

dis

i believe

[viscid thistle]

I Have sin graphed in my notes

[hard hornet]

not sin, cos, we talking about inverse trig function

inverse sin, inverse cos

[viscid thistle]

but should I just memorize the elementary functions of cos and sin then

[hard hornet]

they have some sort of range in order for them to be defined as functions

[viscid thistle]

[-pi/2, pi/2]

for sin

[hard hornet]

😄

not sin, inverse sin

[native sequoia]

arcsin

[hard hornet]

yup thats right

and how about cos

i mean

arccos

[earnest wedge]

i think i lost track of time

[hard hornet]

read my brick of text above

and pm me if u still have questions

[earnest wedge]

parametric usually are curves

cartesian usually are points

[viscid thistle]

arc cos is [0,pi]

[hard hornet]

correct

[viscid thistle]

Wait