precalculus

[vale urchin]

when you mix the two

x(2) y(7)

[winter quartz]

Our point should be (2,7)

But we know every second we move down 4 and right 1

[vale urchin]

yes

so next point is 3, 3

[winter quartz]

You got it

[vale urchin]

Good wow okay

[winter quartz]

And now this pattern is the same, so we have a line

[vale urchin]

and that line is slowly curving to the right and rapidly moving down

[winter quartz]

So it will look like that

[vale urchin]

but it never stops?

it doesnt curve?

[winter quartz]

No, the pattern is the same through out

This is a linear function

[vale urchin]

oh just linear okay

cool then

[winter quartz]

It never stops because we are letting $0\leq t< \infty$

[vale urchin]

oh okay cool good

[obsidian monolith]

dackid

[winter quartz]

There we go

[vale urchin]

yep okay infinite

and you did a triangle

so it cant change into a different triangle

[winter quartz]

The triangle was more of a visual aid

[vale urchin]

cant curve a triangle

[winter quartz]

It let me know how much we were moving up and down

[vale urchin]

it was greatly needed

haha

moving down 4

[winter quartz]

But the red line is our actual graph

[vale urchin]

okay so whats the big deal about parametric equations if theyre just linear?

[winter quartz]

No no no my friend

this parametric equation was linear

[vale urchin]

oh

oh dear okay

[winter quartz]

Let's try a different one. Let's see what you can gather from it.

[vale urchin]

also why is subscript 0 a thing

[winter quartz]

$x(t)=t^2$
$y(t)=t^3$

[obsidian monolith]

dackid

[winter quartz]

Try to graph that with just the tools we have

[vale urchin]

okay so sqrt(x)=t

y=cubroot(x)

[winter quartz]

We'll let $0\leq t\leq 2$

[obsidian monolith]

dackid

[vale urchin]

so I draw y = cuberoot(x)

[winter quartz]

False. Ignore this elimination process for now. Do what we did with our line. Develop the intuition

[vale urchin]

okay I'll try

so goes up by squareroot

so if t is 1

then x is x^4

[winter quartz]

No no. It goes up by t^3

[vale urchin]

I mean to right sorry

starting with x

[winter quartz]

For $x(t)=t^2, x(1)=4$!?

[obsidian monolith]

dackid

[winter quartz]

I call lies

[vale urchin]

no its ^4?

[winter quartz]

Well, technically right, but just because x(1)=1

[vale urchin]

but if was 2

2^6?

[winter quartz]

-_-

[vale urchin]

or 2^2

[winter quartz]

If $x(t)=t^2$ what is x equal to when t=1?

[obsidian monolith]

dackid

[vale urchin]

just 1^2

[winter quartz]

Which is?

[vale urchin]

1

[winter quartz]

Yes

[vale urchin]

so 2^2 is 4

so when t = 2

that is

2^2 4

[winter quartz]

For x(t), yes

[vale urchin]

so y(t) = 1^3 is also 1

but 2 is 2^3

9

[winter quartz]

Let's actually do from $-1\leq t\leq 2$

[obsidian monolith]

dackid

[winter quartz]

3 is gonna be massive

[vale urchin]

oh okay

3 ^3

[winter quartz]

Ooo, this is actually really interesting

[vale urchin]

27

[winter quartz]

Let's actually do $-2\leq t\leq 2$

[obsidian monolith]

dackid

[vale urchin]

okay is that better?

[winter quartz]

There is something really interesting going on in this interval

[vale urchin]

its going up and to the right a lot

[winter quartz]

Like I said, focus our t values on the interval above now

[vale urchin]

can be -2 or 2

[winter quartz]

t can be between -2 and 2

[vale urchin]

but nothing lower or higher

[winter quartz]

So check (x(t),y(t)) when t=-2,-1,0,1,2

That will give you a good idea fo what will happen

[vale urchin]

it will be the same as when -2 and 2 it is the same

because squaring removes negative

[winter quartz]

Ah, x will. But what about y?

[vale urchin]

y no

y is negative

sometimes

every other

it is

[winter quartz]

Your graph will end up looking something like this

[vale urchin]

wow thats a new one

so it repels

and bends away

[winter quartz]

I like that description. Pretty much 😁

But something that may seem strange is it is symmetric across the x-axis

[vale urchin]

whats that

[winter quartz]

Do you know what the x-axis is?

[vale urchin]

yes haha

the horizontal

[winter quartz]

So if you look at the horizontal line. The top looks like it is a mirrored version of the bottom

[vale urchin]

oh thats what you meant

my terminology is bad

yes mirror makes sense

[winter quartz]

But one of the first things you learn in algebra 1 and 2 is the vertical line test

If you draw a vertical line anywhere, and two points on the graph are on the line, then it's not a function

[vale urchin]

but center can be 0,0 though?

[winter quartz]

I mean, that's not really relevant for the vertical line test

The problem is two points on the graph hit the red line

[vale urchin]

yes

[winter quartz]

So, technically... this isn't a function

[vale urchin]

you tricked me all along!

[winter quartz]

But it is a parametric function

[vale urchin]

😦

[winter quartz]

Parametric functions allow you to represent any curve as a collection of functions over time. So any curve you draw on a graph, function or not, can be parametrized!

[vale urchin]

so no more graphs then?

[winter quartz]

No one said that

[vale urchin]

Okay now im shook

should I put the question now?

[winter quartz]

Sure

[vale urchin]

should I start by drawing and labeling the diagram below

[winter quartz]

Well, A is given to be 40°

[vale urchin]

So theta is given

[winter quartz]

Yep, and so is v

[vale urchin]

why do they give meters when we have feet for answer?

[winter quartz]

The building is 63 ft high, so y_0=63 and x_0=0

Everything needed is in ft

[vale urchin]

we have speed

32ft/s

[winter quartz]

No no, gravity is $32 \frac{ft}{s^2}$

[obsidian monolith]

dackid

[winter quartz]

And speed is $14\frac{ft}{s}$

[obsidian monolith]

dackid

[vale urchin]

oh

unfortunate

y looks harder so lets start with x?

[winter quartz]

Read the problem. Everything we need is given to us

[vale urchin]

14 is initial speed

which is v

I get 10.7246222t + x_0

for x(t)

[winter quartz]

X_0=0

[vale urchin]

10.7246222t just this yes

[winter quartz]

If that's what it comes out to, yeah

[vale urchin]

okay good good

now y

is tough

but g is gravity?

t^2 we need to get rid of the square somehow

[winter quartz]

No we don't

This is a function of t

[vale urchin]

-1/2gt^2 + 8.999026536t + 63

[winter quartz]

Also, just gonna say it right now, never plug numbers in until you are at the final stage

[vale urchin]

oh

[winter quartz]

And I can guarantee you 14sin(40) os perfectly acceptable for this program

[vale urchin]

okay I do that then

[winter quartz]

It is also more accurate than your approximation, as yours is just an approximation

[vale urchin]

oh

-1/2gt^2 + 14sin(40)t + 63

[winter quartz]

And g is a number

After that, yes. That's all you need

[vale urchin]

g us 32

is

so -16

[winter quartz]

Yep

[vale urchin]

-16t^2 + 14sin(40)t + 63

[winter quartz]

$y(t)=-16t^2+14\sin(40°)t+63$

[obsidian monolith]

dackid

[vale urchin]

and no approximation

[winter quartz]

Nope. This answer is as exact as it gets

[vale urchin]

wow thats fire

[winter quartz]

So now we now how high the canteloupe will be for any time t, and how far away it will be for any time t.

[vale urchin]

So that has real world connection

well thanks for the carry!

[winter quartz]

You bet. Parametrized functions are simply a way to measure things with respect to time.

[rose vapor]

$f_y$ where f(x,y) means what exactly? $df(x,y) \over dy$?

[obsidian monolith]

Daemon

[winter quartz]

Nothing actually
Now $\frac{\partial{f(x, y)}}{\partial{y}}$ does mean something

[obsidian monolith]

dackid

[rose vapor]

aaaaah

damn it. I need to get better at latex

[winter quartz]

No worries. I just wanted to make that clarification

[rose vapor]

$\frac{d}{dx} [f(x,y)\frac {y'}{\sqrt{1+y'^{2}}}]$

[winter quartz]

Essentially what you're doing is looking at a slice of the surface in the y-z plane and seeing what the derivative of f is with respect to y.

[obsidian monolith]

Daemon

[winter quartz]

So say we have this amazingly drawn surface by yours truly

[rose vapor]

haha. I think I got it :) So $f_y$ is taking a slice of f(x,y) and find it's tangent on the y-axis?

[obsidian monolith]

Daemon

[winter quartz]

On the y-z axis

[rose vapor]

the above derivative is more calculus though right? :S

oooh. right. you're doing it on xyz axis!

[rose vapor]

or that :P

[winter quartz]

So when you are taking a slice of the graph, you are looking at the y-z plane where x is a fixed number

[rose vapor]

cooool :D

[winter quartz]

Does that idea make sense?

[rose vapor]

yeap

it's hilarious that I'm doing an MSc on applied math and I'm excited when I understand these things XD

[winter quartz]

Here is a quick example
Let $f(x, y)=xy$. Then $\frac{\partial f}{\partial y}=x$, since x is considered a constant

[obsidian monolith]

dackid

[rose vapor]

and $\frac{\partial{f}}{\partial{x}} = y$

[winter quartz]

Almost had it 😆

But yes

[obsidian monolith]

Daemon

[rose vapor]

ha!

[winter quartz]

There ya go

Also, some shorthand notation for that same thing is $f_y$ if it's a y partial and $f_x$ if it's an x partial.

[obsidian monolith]

dackid

[winter quartz]

Quick question, why are you asking this in #precalculus 😆

[rose vapor]

I thought it was precalculus :S

[winter quartz]

Oh goodness no. This is multivariable calculus

[rose vapor]

And this is? $\frac{d}{dx} [f(x,y)\frac {y'}{\sqrt{1+y'^{2}}}]$

[obsidian monolith]

Daemon

[rose vapor]

for #multivariable-calculus ?

[winter quartz]

Again, you are taking a partial here.

[rose vapor]

wait...

It's part of the euler-lagrange equation. It's written as d/dx. No θ/θx :\

Gahhh. I'm calling it a night... WIll think again on it tomorrow.Many thanks for the help dackid :))

[winter quartz]

Okay. Ask these questions in #calculus next time. This is definitely not pre calc.

[rose vapor]

aye aye o7

[vale urchin]

dack my guy I did the hard part of this problem but now they're asking for distance

[vale urchin]

<@&286206848099549185>

[lilac storm]

so i have solved this question but i just want to make sure that i did it right

so first i started off by listing out all the things that are given which are, A=58750, r=0.05 and t=4

[vale urchin]

?

[lilac storm]

then i used the A=Pe^rt formula and plugged everything in and rearranged the eq accordingly to find P and the answer i get is 48100.43 or simply 48100, i just want to ask that did i do everything correctly?

[vale urchin]

Looks good to me

[lilac storm]

so the formula is correct and just want to make sure just in case that 58750 is the A value (the future value) right?

[vale urchin]

The formula is correct but did you plug in the right numbers?

58750 is the A value yes

thats your start

[lilac storm]

ok, thank you so much for your help

[vale urchin]

np

[vale urchin]

Hey dack you understand how I can get the distance?

[winter quartz]

Of course. Solve for when y(t)=0 and plug that t value in to x(t).
Also, stop writing the approximations!

[vale urchin]

Oh perfect so it'd be just 86. Also, I have to write the approximations! They say its wrong when I dont its weird ik!

then 18.70283481(86)

[winter quartz]

No, it is not 86

[vale urchin]

😮

well

okay then

[winter quartz]

I didn't say when solve y(t) when t=0, I said solve for t when y(t)=0

[vale urchin]

ahhhh

its that diagram thing you should me again

[winter quartz]

Then you need to tell the program to stop being approximately wrong! I'm sure it'll listen to you 😆

[vale urchin]

I wish I could haha

[winter quartz]

Also, for approximations, you probably only need to round to the first two decimals

[vale urchin]

so y(t) equals 0 it will be -4.9 + 18.06.... + 86

[vale urchin]

Oh true it says 4 for this one but I never round for some reason

[winter quartz]

$h(t)=0=-4.9t^2+18.06t+85$
Solve for t

[obsidian monolith]

dackid

[winter quartz]

Ah, do 4 then.

[vale urchin]

oh so first sqreroot both sides?

get rid of ^2?

[winter quartz]

Well, this is a quadratic equation

[vale urchin]

oh so factor

[winter quartz]

Time for the Quadratic Formula!!!!

[vale urchin]

sadness

I get two answers

[winter quartz]

And one of them should be negative

[vale urchin]

x = 6.39747242, -2.71153005

one is ! lets gfoooo

[winter quartz]

A negative time does not have any meaning

[vale urchin]

so pick the positive

[winter quartz]

So you only care when t is positive

Correct

[vale urchin]

yes!

[winter quartz]

Now plug in that t in to x(t)

[vale urchin]

oh okay will do

[winter quartz]

After all, you're trying to show how far it traveled when it hit the ground. Well, it hit the ground at the time when y(t)=0

[vale urchin]

119.650869873

meter

[winter quartz]

I mean sure. I'm not plugging it in

[vale urchin]

Okay thanks for the carry

[vale urchin]

Yo my guy congrats on the promotion, well deserved.

[winter quartz]

I asked for it 😆

[vale urchin]

and you deserve it haha

[winter quartz]

Thanks, that's very kind of you :)

[pastel carbon]

,w x^4-2x^2-24=0

[obsidian monolith]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by Suikae#5524

[golden needle]

Can someone help me with my hw?

[wispy aurora]

Yes, but post it

[golden needle]

@wispy aurora

[wispy aurora]

Why ping?

[golden needle]

You said to post it earlier

[wispy aurora]

Hm well ask someone else ig, rip

Im tired

[golden needle]

Ok

[faint nest]

How do u do the second

Ignore my graph idk what i’m doing lol

[thorn moat]

x=log7^8-5/2, what does it mean to have log present after you have solved for the unknown? im having trouble understanding what a log truly is. i know that its used to find the unknown exponent but why keep log present once we have solved for x?

[stable yacht]

You know that from a half rotation, the height of the stone goes from 3 (the maximum height) to 0 (the minimum height). You need to figure out at which height the stone starts to model your graph. You know the span of one period since 0.4 seconds is one revolution.

I’d fix the graph first and see if you can determine what equation to use after that.

[wispy aurora]

How would you get rid of the log?

[bold lintel]

I'm sure this is pretty simple and I'm just missing something here but everything I've thought of to solve for t would just contradict eachother when getting ln to the other side

[jolly raven]

they isolated t by taking just the 4 and the ln

because its like 4 times t times ln

so they just divided 4 and ln to leave the t out

[blissful hatch]

hi guys anyone know how to do question b?

[uncut mulch]

quotient + chain rule

[blissful hatch]

that works for b too?

[sick steppe]

yes, cause f is a quotient of functions

[blissful hatch]

but how do i get the g function? it's not 2gx right?

[sick steppe]

you're given enough information about g

[uncut mulch]

differentiate g(x^2) using chain rule

[blissful hatch]

sorry i just dont understand it huhu

okayy i got it thanksssssss

[sick steppe]

Do what Ramonov said

[still quarry]

anyone know where i went wrong here

[sick steppe]

f is increasing on (1,inf) @still quarry

[still quarry]

i see

thanks

[thorn moat]

i suppose i would do the opposite steps that i took to get to the log

[shrewd storm]

also that is increasing from 1 to inf

in its derivative

[tender lotus]

what can't x be?

not sure how to find that

[sick steppe]

can you divide by 0?

[tender lotus]

obviously not lol but i tried 0 and its wrong

[sick steppe]

well when is x+5 = 0?

[tender lotus]

-5, there we go, now it makes sense thanks

[thorn moat]

How can i prove this to be true long hand with out a calculator? i am not sure i truly understand logarithms.

[lethal oracle]

@thorn moat look up logarithm rules

[wooden garnet]

Any ideas on how to do this

[sterile wedge]

Can someone please help me with the 2 I have circled? I don't know how to transform them based off of D and C and I also don't understand how to know where to put the line of the graph through

[tranquil bronze]

You are aware of this, right?

[tranquil bronze]

For example, question 37.
If you let your x --> +-infty. What are you left with? 🤔

[sterile wedge]

I understand how to get those but I dont know how to graph it

[tranquil bronze]

I was going to write a lot here, but there are so many videos on YouTube. I will refer you to one. I don't know if I can post the link here, so, check your pm.

[lost mesa]

do lim x-> 0- and 0+ exist

[sick steppe]

yes they exist

[lost mesa]

and it's 0 right?

[sick steppe]

yes they both equal 0

[lost mesa]

kk

so therefore lim x->0 f(x) = 0 is true

[sick steppe]

yes

[lost mesa]

kk ty

[sick steppe]

but f(x) is not continuous at x=0

[lost mesa]

right

[arctic walrus]

https://puu.sh/Howsj/0431b59912.png

How do you simplify that?!

How did they go from left to right

[rustic glacier]

I'm not really sure but I tried it and I just multiplied both sides by 4 and got that it was equal

so probably factor out 1/4

[arctic walrus]

https://puu.sh/HowyU/29d450b462.png

0oh no

there's no both sides sorry

They simplified from left to right

https://puu.sh/HowzA/694d72b719.png

[rustic glacier]

also in polar equations of conics is p directed distance or absolute value

[arctic walrus]

Original problem for context/.

[rustic glacier]

i'm getting the same answer as them but (sqrt 3 -1) in parentheses

I just factored out 1/4 then -sqrt 2

[arctic walrus]

i dont get it at all

im sorry :/

could you show your work?

[rustic glacier]

@arctic walrus

top part I just forgot to factor out negatives properly

[arctic walrus]

wow dude thank you so much. So odd that that isn't taught in my book or by the teacher yet is expected.

https://puu.sh/HowVW/4e3a725dc4.png

could you also show me this one?

[rustic glacier]

yeah that’s taught in algebra but expanded on in precalc

uhh try it first by multiplying by 3/3

step 2 to step 3 is just substituting tangent values

and step 3 to 4 is expanding and multiplying by 3/3 which is 1 so it doesn’t change the value

[rustic glacier]

last step to final answer is multiplying by (3 - sqrt 3) divided by itself

[arctic walrus]

that gives a crazy answer though

oh shit okay

[jovial salmon]

i just need help with finding the start and end pt, for some reason I get the same number for both

[dapper night]

Does anyone know how to solve this

[sick steppe]

go through w/ first principles?

[dapper night]

what’s that

I just got in calculus with no context and they randomly give this hw to me

[sick steppe]

the limit in the question

[dapper night]

Without telling me steps

Watched videos but still stuck

[sick steppe]

$f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$

[obsidian monolith]

moshill1

[dapper night]

👀

[sick steppe]

so what's f(x+h)?

[dapper night]

If you want me to be completely honest, I have no idea

They put me from algebra one, to calculus

[sick steppe]

ok so plug in x+h in for x and that's f(x+h)

[dapper night]

Cause they said I had talent or something

[sick steppe]

$\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to 0}\frac{3(x+h)^2-5(x+h)-3x^2+5x}{h} \ =\lim_{h\to 0}\frac{3x^2+6xh+3h^2-5x-5h-3x^2+5x}{h}$

[obsidian monolith]

moshill1

[dapper night]

Do you cancel out 3x^2, -5x, -3x^2?

[sick steppe]

after expanding the brackets stuff will cancel, yes

[dapper night]

This is what I got

But I’m now stuck

[sick steppe]

didnt distribute properly

also forgetting the limit in every line

[dapper night]

😞

Would you mind taking me step by step?

[past meadow]

lost some 3's while distributing

[sick steppe]

almost like that's what I was referring to

$\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to 0}\frac{3(x+h)^2-5(x+h)-3x^2+5x}{h} \ =\lim_{h\to 0}\frac{3x^2+6xh+3h^2-5x-5h-3x^2+5x}{h} \ =\lim_{h\to 0}\frac{6xh+3h^2-5h}{h}$

[dapper night]

Hmmm

[obsidian monolith]

moshill1

[dapper night]

👀

[winter quartz]

Almost there :)

[dapper night]

I’m so stuck 😣

[winter quartz]

Well, factor out an h from the top

So $\lim_{h\to 0} \frac{h(6x+3h-5)}{h}$

[obsidian monolith]

dackid

[dapper night]

Do you cancel h now?

[winter quartz]

I don't see why not

[dapper night]

6x + 3h - 5

?

[past meadow]

yeah, and as h approaches 0...

[dapper night]

6x-5?

[winter quartz]

Bingo!

[dapper night]

Thank y’all!

[woven ravine]

hello,new here. i am really bad at these type of Qs where it asks of me to prove something is something, it would be really appreciated if someone could help me out a bit with the process

[woven ravine]

<@&286206848099549185> apologies for the ping. This type of question really confuses me a lot

[grave tartan]

Basically

It’s just a little analysis

When something is symmetrical along y=x

[woven ravine]

isn't it a straight line?

[grave tartan]

You gotta ask Well how do we go about proving that

Y=x is straight but also

[woven ravine]

english not my 1st language,so I do apologize for my crappy eng lol

[grave tartan]

What do we know is associated with y=x

No worries

Lmk if you want a hint

[woven ravine]

tbh

I don't understand this question almost at all

[grave tartan]

Yeah so we know that inverses are reflected along y=x

[woven ravine]

all i could kind of guess was since y=x symmetry, it would be a straight line

yeah

[grave tartan]

Well pull up desmos rq

Plot y=x and y = 1/x

Lmk when you have

[woven ravine]

y=x is just a tilted line going straight through at a 45 degree

[grave tartan]

Yeah but also

[woven ravine]

and the y=1/x is an even function

[grave tartan]

If you flip the graph over the line y=x

You get the same exact graph

That’s what symmetrical means

[woven ravine]

oh so it's a reciprocoal?

[grave tartan]

One side is identical to the other

It’s an inverse

A special property of 1/x is that it is its own inverse

[woven ravine]

is y=x

[grave tartan]

Basically the question is asking you to prove that expression is it’s own inverse

Do you know how to find inverses?

[woven ravine]

isn't it like

you get the function

then switch the y and x values

to solve for f-1(x)

I had to serve a mandatory military service and everything about math is jumbled into a mess in my head...

[grave tartan]

Nah bro you’re good

That’s exactly it

You flip the x and y and solve for y again

Basically a neat thing about the function it gave is that it is it’s own inverse and the question is just asking you to prove that fact

[winter quartz]

Interestingly, that function is significantly linked to a 2x 2 matrix's determinant

Which I dunno, I find super interesting

[woven ravine]

I really appreciate the help guys, but I'd like to be a bit selfish and ask for just the first step in as to how to begin

so what I got down so far

[winter quartz]

Start here: $x=\frac{ay+b}{cy-a}$

And now try to solve for y

[obsidian monolith]

dackid

[woven ravine]

oh god

my pea sized brain

[winter quartz]

[woven ravine]

i know we have to single out y

but i am not sure what number to multiply both sides by,in order to single out the fraction

[winter quartz]

Well, the number in the denominator

$\frac{a}{b}\cdot b=a\cdot\frac{b}{b}=a$

[woven ravine]

so then x(cy-a)=ay+b

[obsidian monolith]

dackid

[winter quartz]

Quick algebraic reason as to why we do that

[winter quartz]

Correct

[woven ravine]

then...divide by a...?

[winter quartz]

So the trick you wanna do is distribute x on the LHS

Then see of you can get terms with y in them on their own side. That is:

$cy+ky=\dots$ where c and k are some random constants.

[obsidian monolith]

dackid

[winter quartz]

But of course, they are very specific in this problem

[woven ravine]

So now that we got (cy-a)x=ay+b

We can inverse the x and y to get the value

This is how I interpreted it based on you fellas” hints

[winter quartz]

You got it!

Also keep in mind, this is the inverse, so $f^{-1}(x)=\frac{ax+b}{cx-a}$

[obsidian monolith]

dackid

[winter quartz]

But since $f(x)=f^{-1}(x)$, f is symmetric across y=x

[obsidian monolith]

dackid

[woven ravine]

aha,i see thanks bro!

[sterile wedge]

Can someone help me with 37 please

[main cedar]

give the inverse sine function some arbitrary value y. so sin^-1 (x-1) = y

build a triangle where one of the angles is 'y' and then find the sides of the triangle

do you need further explanation?

[sterile wedge]

sorry I was away from my computer for a bit. I don't understand how to find the sides of the triangle from that equation. Also can I cancel out the sin^-1 with sin so it is sin(x-1)=y

[main cedar]

sin^-1 is an inverse function. so if sin^-1 (x) = y, then sin y = x

[sterile wedge]

so with sin(y)=x I make the x x/1 so then my b side ill be x and the hypotenuse will be 1?

[main cedar]

ayy correct

awesome

[sterile wedge]

and then from there I get all the sides and then get the tan of the triangle?

[main cedar]

yes

keep in mind that you have (1-x) here, not (x)

[sterile wedge]

I got the right answer! You are a legend thanks for the help!

[main cedar]

awesome. np!

[sterile wedge]

can I ask you another question real quick? Its okay if not

[main cedar]

go ahead

[sterile wedge]

from the picture I posted earlier for number 41 I messed up on the x+1/2 but I don't know how to correctly do it. I know that I need to combine them somehow so I can get my 2 sides but I got x+2 which is wrong

[main cedar]

type in your steps

so i can see where you made a mistake

[sterile wedge]

I did (x+1/2)/1 and then multiplied the 1/2 and denominator by 1/1 and got x+2

[main cedar]

wait what

so your opposite side is x + 1/2 and your hypotenuse is 1

(idk what your step did, but x+ 1/2 would simplify into (2x+1)/2)

[sterile wedge]

wait I can just leave it as (x+1/2)/1?

[main cedar]

yes

[sterile wedge]

and ohh I was way off with the simplifying

[main cedar]

its a ratio so it doesnt really matter

[sterile wedge]

for some reason I thought I had to combine them

sorry for the bad question thank you for helping!

[main cedar]

again, no problem lol

[gilded oasis]

Yo what topics are under precalculus?

[main cedar]

read the channel info

[gilded oasis]

you mean #info or am I missing something

oh nvm found it

mobile sucks 🗿

[strange plinth]

Prove that if a^2+bc≠ 0, then the graph of f(x) = ax+b/cx-a is symmetric about the line y=x. Can you help me with this problem please?

[native creek]

Please show me how ro do thia

This

[swift mountain]

assuming that's x+4 on the right hand side, collect the x terms

[trim hemlock]

is this a test?

[main cedar]

everybody hates i. try rationalizing the denominator

[trim hemlock]

and wrong category

[arctic walrus]

It's review for a test

[trim hemlock]

the paypal part makes this skeptical

[arctic walrus]

The teacher didn't upload his lectures for that week

I mean, I have the textbook. But my friend comes over in an hour or less. I'd much rather buy someone a coffee to get it done tonight and not worry about it. But I can just go through my textbook as well

[trim hemlock]

welp, just like el chupacabra suggested

divides them and get rid of the i in the denom

try to use conjugates in the process as well

[arctic walrus]

God, I have no idea what that even means. Guess it's textbook time

[main cedar]

@arctic walrus im assuming you know how to rationalize complex fractions

[arctic walrus]

no i dont XD

[main cedar]

do you know what i^2 is

[arctic walrus]

It's fine, i'll learn the old fashioned way

Nope

[main cedar]

bruh

do you know what i

is

[arctic walrus]

I don't want to waste anybody's time to teach me an hour for a cup of coffee

An imaginary number or a vector

[main cedar]

i dont need a cup of coffee. im just tryna help

what is the definition of i

[arctic walrus]

Im so sorry, hold on, I did not expect such a quick reply. I am reviewing a previous section

I do have a question though

How do i distribute the second part correctly?

is it -9i - 12j?

[main cedar]

yes it is

ayo

[arctic walrus]

WTF

There's a distributive error here, is there not?

The 2j was distributed with a negative, and re-distritubed again..

[main cedar]

nah i dont see anything wrong

the - was only distributed once

[arctic walrus]

OH WAIT. It makes sense.... They distributed it to the second part first, then the first part last

[uncut mulch]

just seems there's a stray )

[main cedar]

the () doesnt include the 4j anymore after the first distribution

[arctic walrus]

That's what was confusing.....

[main cedar]

theres an extra ) for no reason

[arctic walrus]

Thank you you guys are great lol

[main cedar]

complex time

[arctic walrus]

not yet

it's directional angle time D:

i got it tho. hold up 🙂

[main cedar]

post this shit in the right section bruh

this isnt precalc

#geometry-and-trigonometry would be better imo

[arctic walrus]

Oh, sorry. I'm in PRECALC 2 right now.

So confused

[harsh cipher]

Hi

can someone help me with a question

[kind horizon]

Hi guys! this week i started learning complex numbers, but it's a huge confusion for me. So i don't understand the part why negative roots exist since we learned that we don't have a solution for the quadratic equation when the answer is negative. Can someone pls explain me a bit? Thankyou!

[sick steppe]

we just define sqrt(-1) as i

and fundamental theorem of algebra says there are always 2 roots to a quadratic

Your teacher likely said no R roots exist if b^2-4ac<0

[kind horizon]

exactly

[sick steppe]

yeah, that's not "there are no roots period"

No roots vs no R roots are different

[kind horizon]

and how come i^2=-1

[copper musk]

by definition

[kind horizon]

can i find the def on wiki or ...

[copper musk]

yes you can

[winter quartz]

You'd be surprised just how expansive math gets by letting i^2=-1

It seems so simple, but complex numbers is like a whole new world in comparison to the reals.

[wispy aurora]

A WHOLE NEW WORLD

[hallow thunder]

they are unreal

[winter quartz]

Welp, someone had to do it 😆

[timber ice]

Hey guys, i have a calculus question about a integral rule and i wonder is it ok to ask here or should i wait for a question channel?

i just need the name of the rule :3

[proud raven]

?

[timber ice]

what is this rule called?

[proud raven]

its unclear what the rule is but are you thinking of u sub

[timber ice]

basically you use it when you want to integral a rational function

[proud raven]

quotient rule?

[timber ice]

all i get is partiell integral

let me look it up one sec

yuuup

that its

thank you!

I knew how to use it before but i totally forget about it for some reason lol

[proud raven]

you should

its a bastardized version of chain rule

[timber ice]

i just can't seem to get it to work

i knew how to do it before but for some reason it does not make any sense to me

[proud raven]

rederive it

[timber ice]

and all i see on youtube is partiel integration and u-substitution which is something that i don't need for this course

[proud raven]

$\int \frac{ f(x) }{ g(x) } \dd x $ ?

[timber ice]

$\int \frac{ f(x) }{ g(x) } \dd x$

[obsidian monolith]

steppaz

[proud raven]

fugc

[timber ice]

hehe

but like

look at this

[proud raven]

$\int f(x) \cdot g(x) ^{-1} \dd x $

this is what i meant

idk whats up w texit

[timber ice]

if you look at b)

where does 1/2 come from?

[proud raven]

u sub

[timber ice]

i remember something about moving out the denum outside of the integral but that doesn't seem to work and stuff would cancel out

[proud raven]

$2 \cdot \frac{1}{2} = 1$

[obsidian monolith]

jan Niku

[proud raven]

youre only looking at the 1/2

consider the 2 as well

i guess really theyre also applying the constant multiple rule

$\int x = \int \frac{1}{2} \cdot 2 \cdot x = \frac{1}{2} \int 2 \cdot x$

[obsidian monolith]

jan Niku

[proud raven]

which you want there because 2x is the derivative of x^2+1

alternatively, you could do u sub first and get to a similar answer

[timber ice]

yeah i'm with you with the 2x is integral of x^2+1

[proud raven]

which is probably better since the approach yo showed requires you to have some foresight

well

[timber ice]

this is my problem

[proud raven]

x^2 + C is the integral of 2x

[timber ice]

the course contains derivatives and integrals but no u sub

or partiell integration

[proud raven]

u sub is necessary

well maybe not

[timber ice]

tell that to my teacher haha

[proud raven]

but its a good idea i think

,w partiell integration

[obsidian monolith]

Wolfram Alpha doesn't understand your query!
Perhaps try rephrasing your question?
Click here to refine your query online

Requested by jan Niku#2119

[timber ice]

there is literally 5 or 6 pages about the substitution method in my book

[proud raven]

oh

parts

i had no idea this had a better name

[timber ice]

😄

[proud raven]

parts is a really dumb name

eh just watch some videos

[timber ice]

i think it sounds cool

Partiell integration

[proud raven]

u sub is really just algebra disguised as calculus

[timber ice]

integration sounds cool in general

alrighty lets see if i can find some videos of it

[proud raven]

khan academy maybe or

theres no shortage of calculus tutorials

[timber ice]

Yes, thank you

[proud raven]

[timber ice]

AH

found it

it's called "Reciprocal rule" or just "Reciprocal"

there is a perfect 2 minute video that explains exactly how it works

[winter quartz]

This isn't quite a rule, it is just a lucky u-substition

[sick steppe]

yeah it's just u-sub

[pliant locust]

That is not the reciprocal rule. That is just a standard integral recall that in highschool we were taught that the derivative of ln(x) = 1/x. This integral is just a generalized version which you get through u-substitution.

Reciprocal rule is a rule for differentiation not integration.
Pic related is kvotregeln.

[pliant locust]

^

[rustic glacier]

could someone please help w the polar equation of a parabola with its focus at the pole and vertex at (1, -pi/2)?

I don't really understand why p is 10 instead of 5 and how the directrix is to the left of the pole

[timber ice]

Then how would you take the derivative of a rational function without using the u-substitution? Because in my book we have been taught that rule. Where you basically derivative the function and you put it in the denum and you re-write it as "ln I f(x) I + C". But that don't seem to work for me as i am missing one step and i am not sure what it is lol

[pliant locust]

I was just saying that
$\int{\frac{dx}{x}} = \log{\abs{x}} + C \
\int{\frac{f'(x)}{f(x)}} = \log{\abs{f(x)}} + C$
Isn't called the reciprocal rule, reciprocal rule is for differentiation. This integral is just a standard integral, an identity that pops up when solving some integrals which you can reuse. There are a list of standard primitives you have to your use when solving integrals. I assume your book has a proof of the integral somewhere.

[obsidian monolith]

rts

[pliant locust]

For example, this integral
$$\int{\frac{e^{x}dx}{1+e^{x} } }= \log{\abs{1+e^{x}} } + C$$
I identified that the numerator is the derivative of the denominator so i applied the standard integral $$\int{\frac{dx}{x}} = \log{\abs{x}} + C$$ to solve it

[obsidian monolith]

rts

[timber ice]

Oh

It says that any rational function has the integral of f'(x)/f(x)= ln I f(x) I + C

[pliant locust]

Yeah no biggies, you can say it is rule i just didnt want you to confuse it with the reciprocal rule in the youtube video you linked which was him just differentiating a rational function

[dark sky]

how would you graph overall shape of this by hand

[viscid thistle]

What's 9+10

[odd rivet]

Find their asymptotes x intercepts y intercepts

[dark sky]

ok i did that

but what do i do after

[odd rivet]

So u can plot a rough graph using that info

And see as x tends to -inf and +inf

If f(x) tends to 0 or inf or whatever

[pliant locust]

Did you factorize the denominator and then check for vertical asymptotes for those x where the denominator equals zero?

You find the intercept points, find the stationary points by having the derivative, then solve the limit as x approaches infinity and negative infinity

lastly

check for vertical asymptotes by checking what happens to a function as it approache a singularity. (a singularity in this case is divison by zero). You have singularities at x = -8 and x=3, Solve the limit of the function when x approaches 3 and when x approaches -8 both from left and from right.

If it equals infinity you have a vertical asymptote.

If the limit exists from right but it equals infinity from left then the function approaches an open endpoint from right but it will be a vertical asymptote from left.

[dark sky]

Ok thanks

What’s the end behavior of this function?

Is it as x approaches -infinitely, y approaches positive infinity and as x approaches infinity, y approaches negative infinity?

[uncut mulch]

technically true, however it seems like an oblique asymptote would best represent what's happening

[rancid crest]

can uh

someone help me with this

i know you can factor it

cosx(sinx + 1)

set that to 0

cosx = 0

sinx = -1

[hallow thunder]

you can do inverse sin/cos here

then consult the graph to find the other values between 0 and 360 degrees

or unit circle

[rancid crest]

wdym

how would i do that

[hallow thunder]

here is how i would do it for cosx

[pliant locust]

Looks, good, so for what x is cos 0?

[rancid crest]

3pi/2

and pi/2

90 or 270

[pliant locust]

So what is the problem? 😄

[rancid crest]

uh

um

is it just 270

[tall cloud]

Can I get help with this question please?

[pliant locust]

Remember that trig functions are periodic so when they ask for all solutions you add with 2pi right

[rancid crest]

right

but only 270 meets both of requirements so thats where im sorta confused

[pliant locust]

I see

[rancid crest]

ye

[pliant locust]

Well if x=3pi/2 implies sin(x)=-1
Then i can say sin(x)=-1 for x=3pi/2 + 2*n*pi
where n is an integer since sin is periodic.

[rancid crest]

hm

[pliant locust]

If i rotate 360 degrees i end up in same place

2pi = 360 degrees

[rancid crest]

as you go through the unit circle

90 would technically still work

?

[pliant locust]

yeah you get zero whenever cos is zero aswell

[rancid crest]

alr so imma put 90 n' 270

[pliant locust]

remember to include all solutions since it is periodic

[rancid crest]

it says 0 < theta < 360

[pliant locust]

Well then should be those two then

[rancid crest]

oki

[crimson cedar]

Yo, if you have dy/dx of two graphs, how do you use that to find where the gradient is the same?

[pliant locust]

you mean slope?

if f and g are two function, just solve for x

where
f'(x) = g'(x)

[rancid crest]

what does simplest radical form with a rational denominator mean..

my answer is sqrt30/10

[pliant locust]

what is the question

[crimson cedar]

Part B. Do I just do 1=-2x and then solve for x?

[pliant locust]

ye

[crimson cedar]

does x = -1/2 seem right for that? idk if thats right at all lol

[pliant locust]

Well the derivative of a function at point x=c will give you the slope of the line that tangent the graph at x=c

The function x is its own tangent, recall the equation of a linear function being kx+m where k represent the slope

and k = delta y / delta x

[crimson cedar]

-1/2 is the slope for the tangent cause x is the slope of that norm right?

[pliant locust]

yeah

[rancid crest]

sooo

wa

[pliant locust]

sorry man, i am not sure what they mean by rational denominator

I agree with your answer but then again im not sure

[main cedar]

do you know double angle formulae?

[rancid crest]

yea

i did it

so

we good

[main cedar]

oh ok then

[lilac storm]

ln(1/2)/-27 can anyone simply just help me on how do you simplify this?

i know that the answer is ln(2)/27 but i just need a little help on how do you get from ln(1/2)/-27 to ln(2)/27

[uncut mulch]

apply power rule to ln(1/2)

[lilac storm]

so how do i do that?

so what i have been trying to do is taking the reciprocal of the denominator meaning i did ln(1/2) times -1/27 which gives me ln(-1)/54, but this isn't the correct answer

[uncut mulch]

1/2 = 2^(-1)

[lilac storm]

sorry i don't get what does that mean?

[uncut mulch]

$\ln(\frac12) = \ln(2^{-1})$

[obsidian monolith]

ℝamonov

[lilac storm]

yes i got it

so after getting ln(2^-1), what do i do next?

so i have ln(2^-1)/-27

[uncut mulch]

power rule (for logs)

[lilac storm]

so how would i start, if you could help me?

[uncut mulch]

do you know your log laws?

[lilac storm]

yes

[uncut mulch]

specifically the

power rule (for logs)

and can you apply that to:
ln(2^(-1))

[lilac storm]

i know them but i just can't remember them at the moment

[uncut mulch]

if you know them, you should be remember them

if not you should look them up

[lilac storm]

do i do like ln(2^-1/27)

[uncut mulch]

as i've mentioned power rule/law (for logs) 5? times already

no

now you also need to review your order of operations

[unborn blade]

Can someone walk me through how to do something like this?

[uncut mulch]

you could start with
2x + b = x^2 -4
and then consider the value of the discriminant

[median ivy]

cos2A = 3/5 right?

[sick steppe]

no

[timber ice]

Hi, i just have a question about limits as they approach +/- infinity. I was wondering when you want to examine how a function behaves when it goes towards +/- infinity is there some condition that the function has to fulfill for *you * to examine how it behaves when it goes towards +/- infinity?

If this question should go under question channel lmk :p

Like, as i understood it, the values will still be the same whether it goes towards +/-

For example if we look at this function, i believe they have done some polynomial division to figure out the asymptote. Is this just some type of preference to write it as +/-?

if you'd answer please tag me

[lilac storm]

so for this question this is what i have done so far, so i think i have solved half of the question

so i have solved for k but now i don't know what to do next, like what is my next step going to be.

if anyone is there that can please help me?

i also have the answer key for this so the answer key says the answer k=ln(2)/27, which i have gotten already and then the actual question finding the time which is 54 years which i don't know how to find, like i don't know the next step to find it?

anyone there to help please?

anyone there...?

[viscid thistle]

https://gyazo.com/c2d210e880aad5749cb5df7d5aba8146

Does anyone know how to do this? I'm so confused

[hollow zenith]

U could start by modelling the question into an equation first!

[pliant locust]

You have three cases:

One where if x approaches a value c implies f(x) approaches value C then the limit is C however if the function approaches infinity when x approaches c then you have a vertical asymptote (lodrät asymptot). You also have to consider two-sided and one-sided limit when studying the limit as x approaches a singularity. (Singularities in calc 1 are for those x where the function isnt defined, for example, division by zero, negative argument to squareroot or logarithm , or x-values that are outside the domain of certain trig functions for example x<-1 is undefined for the explicit function arcsin(x).

The third case is when x approaches infinity, this is when you study if the function has a horizontal asymptote(vågräta asymptoter). If x -> infinity implies f(x) -> C then C is a horizontal asymptote of f(x). However if x->infinity implies f(x)->infinity then you dont have a horizontal asymptote.

To sum up:

1. When x->c you are checking for vertical asymptotes at x=c (You always check left and rightside limit.)

2. When x->infinity (from left and from right) you are checking for horizontal asymptotes.

As to how you compute certain left-side and right-side limit i will link a post where i explained an informal way of doing it. The right way is using epsilon-delta proofs.

@timber ice

#calculus message

[timber ice]

shit

[pliant locust]

We can begin by just doing one of the limits

[timber ice]

so when i check for vertical asymptotes i check for x as it approaches infinity from left and right

[pliant locust]

No

That is horizontal

[timber ice]

oh

when x approaches c

[pliant locust]

Yeah

[timber ice]

and for horizontal it's when it approaches infinity from left and right

[pliant locust]

And c in this case is the x that makes the function undefined

[pliant locust]

Exactly

[timber ice]

sorry i am a bit snurrig rn

lol

[pliant locust]

Ye np,

[timber ice]

That's very well written, i always just check for infinity

[timber ice]

If we look at this one, what does "+infinity då x ->-3^-"? if thats not a complex question to ask

does it mean it goes towards -3 from the left side?

[pliant locust]

Yes

3^- can be interpreted as
3 - d where d is a very small number

And 3^+ is 3 +d where d is a very small number

[timber ice]

and the values don't really change tho right? when you check for infinity from left and right, you could just add ^- and ^+ as soon as you have figured out what value it goes towards

[timber ice]

oh ok makes sense

[pliant locust]

For example, does e^x have horizontal asymptotes?

Solve limit as x approaches positive infinity

And now solve the limit as x approaches negative infinity

The limit is infinity when x approaches infinity right

But when you are going towars negative infinity you will have the limit equal 0

Therefore y=0 is a horizontal asymptote to e^x

Which is just the x-axis

[timber ice]

försöker bara förstå vad som står :p en sekund xd

oh

juste

because e is defined for e>0

so right side is infinity whiles as minus infinity is 0

so it can differ

[pliant locust]

e is defined for all x

Hold on i will draw

Ok let me ask, how do you evaluate the limit 1/x as x approaches 0 from left?

[timber ice]

let me get paper

it goes towards +/- infinity if x=/0 or x=/1

it goes towards 1

because like

[pliant locust]

solve specifically for x -> 0^-

What is 1/ (0 -0,01) equal to?

[timber ice]

it becomes -1000

it goes towards -infinity, i'm not sure how to explain it but it get's bigger and bigger

[pliant locust]

Exactly

[timber ice]

but negative

[pliant locust]

Or well 0-0,01 is -0,99 right

And the smaller i make d the larger a negative

And if i go from right
x->0^+

It is 1/(0+0,01) which is 1 divided by a positive very small number

Which gives you a positive infinity

Now

[timber ice]

yeah

[pliant locust]

To go back to e^x

How do you solve e^x when x approaches negative infinity

[timber ice]

well it goes towards +infinity right?

the numbers decrease but never reaches negative

no

it goes towards 0

[pliant locust]

e^-x = 1/e^x right

So it goes towards 0

Since x is approaching infinity we know that 0 is a horizontal asymptote

[timber ice]

yeah

damn

[pliant locust]

Lets skip horizontals for now

Lets just do verticals

[timber ice]

yeah please

[pliant locust]

Consider the function: 1/(x-5)

[timber ice]

let me actually check which is which one sec

[pliant locust]

What is it domain? Is it defined for all x?

[timber ice]

it's not defined for when x=5

At 5 we have a vertical asymptote

oh yeah i remember now, vertical asymptote is when the nämnare is 0 and horizontal is as it approaches infinity.

[pliant locust]

Yeah, so for x=5 the function isnt defined, x=5 is a singularity we want to study

usually it implies there is a vertical asymptote but that is not always true

You have to confirm it by solvind the limit

So lets solve what is the limit when we approach 5 from left respectivelt from right?

[timber ice]

it's 5^- and 5^+, no?

[pliant locust]

Ye

So what does the function approach when approaching 5 from left?

[timber ice]

it approaches minus infinity right?

if we plug in 1/(4.99-5) we get -100 and it continues to decrease

[pliant locust]

Exactly

Good

Looks like we are getting somewhere!

[timber ice]

yeah xD

i am starting to get the hang of it

[pliant locust]

Since we have infinity from left and right

We have confirmed that there is a vertical asymptote at x=5

Now lets see of our function has a horizontal asymptote

Lets start with positive infinity

Solve 1/(x+5) when x approaches positive infinity

[timber ice]

it's 0

vågrät asymptote at 0

[pliant locust]

Both from left and right ye

[timber ice]

since we can't breake out any 0's

[pliant locust]

One divided by a very big number gets us zero

And one divided by a very big negative number gives us 0 aswell

[timber ice]

and these asymptotes are basically these values that the function never touches

[pliant locust]

Yeah

So vertical asymptotes are at valdes where x approaches a value that is in the domain

While x when approaching infinity we are checking for horizontal asymptotes

[timber ice]

ah i see now

So that's basically what my teacher was doing in that question

[timber ice]

in this one

she is looking for when x=c which was our scenario one

[pliant locust]

Yeah and ske confirmed that we have a vertical asymptote at 3 and at x=5

[timber ice]

alright, tack så sjukt mycket igen

Du är fan tvär bra 😳

[pliant locust]

[floral remnant]

hmm

[lilac storm]

so for this question this is what i have done so far, i think i have solved half of the question

so i have solved for k but now i don't know what to do next, like what is my next step going to be

if anyone is there that can please help me?

i also have the answer key for this so the answer key says the answer k=ln(2)/27, which i have gotten already and then the actual question finding the time which is 54 years which i don't know how to find, like i don't know the next step to find it?

if anyone is there that can please help me out?

[pliant locust]

You didnt take into account that the substance is four times the safe level

[lilac storm]

so i don't get how would i do it?

like have i done everything correctly so far because i have only solved for k so far?

[pliant locust]

Let $N_{0}$ be the safe level of the substance. We know it is four times the safe level, so $Q_{0} = 4N_{0}$ We know that $Q(t)= Q_{0} \cdot\left(\frac{1}{2}\right)^\frac{t}{27}$
then solve the follwing equation.
$$Q(t) = N_{0}$$
$$ N_{0}= 4N_{0}\cdot\left(\frac{1}{2}\right)^\frac{t}{27}$$

[obsidian monolith]

rts

[lilac storm]

so i don't have the initial value which is No so how will i solve for t(time)?

[pliant locust]

$$ N_{0}= 4N_{0}\cdot\left(\frac{1}{2}\right)^\frac{t}{27}$$
$$ \frac{1}{4}=\left(\frac{1}{2}\right)^\frac{t}{27}$$
$$ \log{(\frac{1}{4})}=\log{\left(\left(\frac{1}{2}\right)^\frac{t}{27}\right)}$$

[obsidian monolith]

rts

[lilac storm]

sorry, where did No go?

[pliant locust]

I divided both sides with 4N

[lilac storm]

oh i got it

[pliant locust]

then i log both sides

[lilac storm]

yes your right

so after taking the log of both sides, are you left with log(1/4)=(1/2)^t/27?

[pliant locust]

Im left with the latest row in above pic

Should be straightforward now on how to isolate t

the t is your unknown

Recall that:
$$\log{x^a}= a\cdot \log{x}$$

[obsidian monolith]

rts

[lilac storm]

so is it log(1/4)=(t/27) x (1/2)?

[pliant locust]

what happened to the log on RHS?

[lilac storm]

doesn't it go away when i bring down the exponent?

[pliant locust]

nope

[lilac storm]

oh no, my bad