Giving me a migraine

[quasi olive]

Please help, anything helps

[red dew]

which one are you on?

[quasi olive]

a

[red dew]

for A you just plug the definition of I(A) in where it says I in the definition of L(I)

[quasi olive]

I don't know how to do that lol

https://tenor.com/view/sweating-nervous-paranoid-gif-4974019

[red dew]

well you see
write down the defintion of L but instead of writing I, you write what I(A) is defined as above

[quasi olive]

Ok

what next

[red dew]

then that's the answer

for (a)

[quasi olive]

Could you answer the first 2 questions then show me the solution? please

[red dew]

for (b) what you do is set L(A) the definition you wrote down just now for (a)
set it equal to 30 dB
then solve for A, as that's what it asks for

[quasi olive]

so what are the final answers

[red dew]

the typical practice is to not give you the answer

you wont learn that way

here's L(A)
$10log_{10}\left(\frac{2(\pi f A)^2 \rho c}{10^{-12}}\right)$

[jolly nest]

zfnQRZJT (not anemic)

[quasi olive]

wtf

thats the answer to A?

[red dew]

as i said, i simply plugged

into

[quasi olive]

So l is 10log10 and A is that fraction

[red dew]

wdym
A isn't a fraction

[quasi olive]

i mean the 2pifa^2pc/10-^12

[red dew]

I = 2pifa^2pc

I was divided by 10^-12

(btw, why are they writing it as I/10^-12, write it as I10^12, lol)

[quasi olive]

RIght i think i'm getting it

So what's the final answer

[quasi olive]

is it this

[red dew]

so what you do is set this equal to 30 and solve for A
the confusing part is there are a bunch of other letters so you just have to think of them as constants like pi

and then what you do is plug in the numbers they give for density of air and speed of light

and this results in a final very weird equation with some quite large numbers

[quasi olive]

Right

Can you tell me the answer, then tell me how you got it

so i can like learn backwards

[red dew]

mmmmm

[red dew]

$10log_{10}\left(\frac{2(\pi f A)^2 \rho c}{10^{-12}}\right) = 30\
log_{10}\left(\frac{2(\pi f A)^2 \rho c}{10^{-12}}\right) = 3\
\left(\frac{2(\pi f A)^2 \rho c}{10^{-12}}\right) = 1000\
2(\pi f A)^2 \rho c = \frac{1000}{10^{12}} = 10^{-9}\
2\pi^2 f^2 A^2 \rho c = 10^{-9}\
A^2 = \frac{10^{-9}}{2 \pi^2 f^2 \rho c}\
A = \frac{10^{-4.5}}{\pi f \sqrt{2 \rho c}}$

i think that's right

[quasi olive]

woah

[red dew]

no it's not

clearly

[quasi olive]

is that the elimination process

[jolly nest]

zfnQRZJT (not anemic)

[quasi olive]

Is that right?

[red dew]

do you understand each step

[quasi olive]

and thats the process

It looks easier that way

[red dew]

yes i just did the operation to undo the outermost function on both sides until A was left

[quasi olive]

What would the amplitude of a wave sounding at 30dB?

[red dew]

they give you values for rho and c

so A will be a function of frequency

[quasi olive]

wait so did you just answer a or b?

[quasi olive]

this

[red dew]

i already answered a for you

a is just plugging in

b is just solving for A

[quasi olive]

So you solved b aswell?

[red dew]

yes i did

[quasi olive]

So do i graph that and find the domain and range

[red dew]

graph the function from (a)

since it asks for the domain and range of L

[quasi olive]

Which one is from question a and b?

[red dew]

[quasi olive]

https://tenor.com/view/bh187-spongebob-sweating-stressed-panic-gif-21499965

[red dew]

if you just want to write down the answers wait for someone who will do that for you

[quasi olive]

so the A=10-4.5/pifsqrt2pc is B?

[red dew]

yes

[quasi olive]

Yeah i got that

but which one is the question A?

[red dew]

v2

[red dew]

.

[quasi olive]

Alright so i'm gonna put that in desmos and get the range and domain

@red dew

[red dew]

h

[quasi olive]

It's not working

[red dew]

try making a definition for c, rho, f, and A

actually no

only c, rho, f

[quasi olive]

How do I make the log10 smaller

What should it be?

[red dew]

desmos considers "log" as log base 10

just remove the 10 after the log

[quasi olive]

Just 10log(

Still not working

[red dew]

what does the error say

[quasi olive]

Subscripts cannot be empty

[red dew]

wherever theres a grey box put your cursor in it and press delete

[quasi olive]

Now is says too many variables

[red dew]

make a definition for those ones

since f is not given, put some random number

[quasi olive]

Could you try?

@red dew

[red dew]

let's try finding the domain without desmos

[quasi olive]

Alright how so

[red dew]

what amplitude could you plug in to make this undefined

[quasi olive]

Infinity

[red dew]

not a real number

[quasi olive]

-10

[red dew]

plugging in -10 would leave it still defined

[quasi olive]

Fuck idk then

I usually just look at graphs

[red dew]

where is log defined

[quasi olive]

Get the highest and lowest points

On the left

Outside the bracket

[red dew]

"on the left" it's where the argument is positive

[quasi olive]

Yeah

[red dew]

so how would you make the argument nonpositive

ie

0 or negative

[quasi olive]

I thought by taking 10 from log

No?

Negative log

[red dew]

what?

how would you make 2(pifA)^2 rho c not positive

[quasi olive]

Well anything squared is positive so idk

[red dew]

not quite

[quasi olive]

Explain

[red dew]

there are things whose square is not positive

aka 0

grr i got confused

i am doing too many things at once

[quasi olive]

but anything to the power of 0 = 1

which is positive

@red dew

[red dew]

why is it to the power of 0

0^2 is not positive
that's what i meant

[quasi olive]

My friend did it a different way

for question B

Simplify by 2(pifA)^21.225310^8=10^-9
(pifA)^2=.13610^-18
pifA=1.1710^-9
A=(1.1710^-9)/(pif), f not = 0

is he wrong or right

@red dew

[red dew]

yes hes right

he just plugged in the numbers

[quasi olive]

But his answer is diff to yours?

[red dew]

i think hes right, at least

[quasi olive]

? @red dew

[red dew]

.

the problem gives you rho and c

as numbers

i personally find the non-plugged form much nicer

but anyway

[quasi olive]

And for question D he put

There is no restriction you can apply to the domain as the range is undefined.

[red dew]

the range is undefined? wha

and the domain does have a restriction

[quasi olive]

Can you tell me please

[red dew]

yes the inside of the log function can't be 0

which means (pi*f*A) can't be 0

[quasi olive]

So that's the domain restrictiom

whats the range defined as

[red dew]

the domain and range are different

domain is simply where f and A are nonzero

range is all real