help-23

[polar iron]

.reopen

[safe radish]

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[lean otter]

Gotta find the derivative of the top function. Using chain rule I got to the solution at the bottom, but I'm not sure how to simplify it further.

[lean otter]

Wait, just realized I can put it into this. Now I'm stuck

[safe radish]

@lean otter Has your question been resolved?

[untold anchor]

going back to the second step you can put the derivative of the inside function as a fraction to make it simpler

@lean otter

[lean otter]

I'm not sure I follow

[untold anchor]

this is the derivative of 1/sqrt(v)

[lean otter]

I pretty much already have that

[untold anchor]

yeah then you just multiply the denominators straight across

you can multiply the v^-3/2 by what’s inside the square root

or v^3/2 i mean

if that makes sense

[lean otter]

I don't quite follow

[untold anchor]

in this picture you don’t need the negative in the v^-3/2

[lean otter]

Oh, whoops

[untold anchor]

and you can turn it into a radical

to then simplify further

[lean otter]

.close

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[slow merlin]

hi

[lean otter]

post ur question

@slow merlin

[safe radish]

@slow merlin Has your question been resolved?

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[naive shoal]

does any one know how to approach this

[tender meteor]

Where Is the problem?

[ionic nest]

for lim f(xn) u can use l'hopital rule

[tender meteor]

@naive shoal

[naive shoal]

lim n tends to infinity xn is zero..is that right

[ionic nest]

no it should be 1

or

no ur right

[naive shoal]

lim n tends to infinity sin(1/xn) does not exist right? as it oscillates between -1 and 1

[ionic nest]

cant you rewrite sin as

ok what about this

1/n (pi/2)

will be 0

as 1/inf =0

sin(0)=1

yeah

and then lim xn will be 0

[naive shoal]

sin (1/xn)

[ionic nest]

1st one is 0, 2nd one is 1, 3rd one is 0

wait no last one is different

3rd one will go to infinity

[naive shoal]

sin (npi/2) as n approaches infinity

[ionic nest]

that one goes goes to 1

its 1/n*pi/2

[naive shoal]

sin(1/xn)

thats sin (npi/2) isnt it

[ionic nest]

xn= 1/n*pi/2

oh ur right

yeah inf

[naive shoal]

so, whats the limit of sin (1/xn) here?

[ionic nest]

inf

[naive shoal]

no

[ionic nest]

oh i see ur question

u might need to use squeeze theorem

the answer would be

-1=<sin(npi/2)<=1

like that i suppose

[naive shoal]

yes, i was confused there

i thought sin (npi/2) doesnt have a limit as it oscillates

[ionic nest]

its been a while since i did this but im pretty sure you just use squeeze

for those cases

[naive shoal]

dont we use that for calculating lim f(x) ?

can any one else help..

let me repost the question here

[ocean crypt]

Hello

What Is the problem

[naive shoal]

My thoughts are..the answer to question 1 is zero because we have zero in the denominator

[ocean crypt]

So

[naive shoal]

the answer to question 2 is "does not exist" as sin oscillates between -1 and 1

is that correct

[ocean crypt]

$x_{n}=\frac{2}{n×pi}$

[naive shoal]

2/(npi)..right?

[ocean crypt]

Yes

So

What happens if you do the limit?

[flat frigate]

alee

[naive shoal]

zero is the answer..right?

[ocean crypt]

Yes

This Is the question Number 1

[naive shoal]

yes

[ocean crypt]

Why is it 0?

[naive shoal]

for question 1?

[ocean crypt]

Yes

[naive shoal]

its zero because as n approaches inf in the denominator, the limit tends to zero

no?

[ocean crypt]

Yes

$2/n*1/\pi$

[flat frigate]

BobTheBuilder

[ocean crypt]

If the term on the left approaches 0

0 for a constant is 0

Right?

[naive shoal]

yes..iam clear abt question 1

[ocean crypt]

Ok

The 2 ?

[naive shoal]

limit does not exist for 2?

[ocean crypt]

Why?

[naive shoal]

letmme repost the question

sin oscillates between -1 and 1

thats why

[ocean crypt]

So it doesn't exist?

[naive shoal]

not sure...what r ur thoughts

[ocean crypt]

Suppose you rewrite the sin argument as above

$sin(2/n*1/\pi)$

[flat frigate]

BobTheBuilder

[naive shoal]

sin (npi/2)

[ocean crypt]

If the term on the left approaches 0

What is the limit

[naive shoal]

can we apply limit law like this ?

yes..sin (npi/2)

[ocean crypt]

Let's try another way maybe it's more understandable

[flat frigate]

alee

[ocean crypt]

We said that question 1, the limit was 0

I know

[naive shoal]

sin (1/xn)

not sin (xn)

isnt it

[ocean crypt]

$\epsilon_{n}=0$

[flat frigate]

BobTheBuilder

[ocean crypt]

This is the limit of xn

You know

$\frac{sin(\epsilon_{n})}{\epsilon_{n}}=1$

[flat frigate]

BobTheBuilder

[ocean crypt]

You know this ??

[naive shoal]

does the above resemble question 2 that we are looking at?

[ocean crypt]

Wdym

[naive shoal]

sin(npi/2) as n approaches inf...isnt it?

[ocean crypt]

You do not have to calculate this limit

[tender meteor]

,w Limit[Sin[2/nπ],n->∞]

[flat frigate]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by alee31._

[ocean crypt]

You have to make this limit with n and π in the denominator

[naive shoal]

sin (npi/2) not sin (2/npi)

because thats sin (1/xn) not sin (xn)

[tender meteor]

,w Limit[Sin[nπ/2],n->∞]

[flat frigate]

Results provided by WolframAlpha

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Requested by alee31._

[ocean crypt]

But why do you have to do this?

[naive shoal]

yes...thats what iam saying

[tender meteor]

I wasn't following the discussion

[ocean crypt]

Oh yes....

[naive shoal]

lim n tends to inf sin (npi/2) doesnt exist

[ocean crypt]

Yes

[tender meteor]

.

[naive shoal]

i feel bad aby myself 🙂

am i missing something

[tender meteor]

Why?

[naive shoal]

whats the answer to question 2

[tender meteor]

Doesn't Wolfram say exactly what you're saying? If yes, then you're done

[ocean crypt]

Do you need to justify your answer?

[naive shoal]

yes

[ocean crypt]

Ok

So

[naive shoal]

i can just say..limit does not exist because sin value oscillated between -1 and 1 ?

[flat frigate]

alee

[naive shoal]

yes thx

lets proceed to question 3

here

[ocean crypt]

You could also write that it doesn't meet the definition of limit

Maybe even demonstrate it

[tender meteor]

It's not clear to me what question 3 is

[ocean crypt]

Instead of x you have to replace xn

And make it the limit

$\frac{n\pi}{2}*sin(\frac{n\pi}{2})$

[flat frigate]

BobTheBuilder

[naive shoal]

yes

[ocean crypt]

So ?

[tender meteor]

.

[ocean crypt]

How much is the limit?

[naive shoal]

again it comes indeterminate?

[ocean crypt]

Meaning what?

[naive shoal]

its of this form lim x sin x as x tends to inf isnt it?

[tender meteor]

How do I answer, if I don't know what the question is, if you want a clear answer the question must be clear

[naive shoal]

here

we are referring to the last part of the question

[tender meteor]

Okay

[ocean crypt]

Are you saying that the limit doesn't exist?

[naive shoal]

hmmm..not sure...confused

[tender meteor]

Yes

,w Limit[nπ/2Sin[nπ/2],n->∞]

[naive shoal]

doesnt exist..right?

[flat frigate]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by alee31._

[tender meteor]

Nop

[ocean crypt]

The sin does not admit limits

The left side goes to infinity

It cannot be solved in any way

[naive shoal]

cant be solved..means ...does not exist?

[ocean crypt]

Mmm

Yes

[naive shoal]

hmm...the answer to first question is zero and the other two..doesnt exist?

[ocean crypt]

Yes

[tender meteor]

What is the definition of limit?

[ocean crypt]

The last limit

[naive shoal]

limit means a point to which sequence converges

[ocean crypt]

Suppose the limit was something like: n^5/f(xn)

How much would the limit be?

[naive shoal]

we have to apply lhosptal?

not sure

[ocean crypt]

It could go well

No

Wait

[tender meteor]

,w Limit[n^5/Sin[x],n->∞]

[flat frigate]

Results provided by WolframAlpha

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

Requested by alee31._

[ocean crypt]

Does the limit comply with the conditions for using the hospital?

[tender meteor]

@ocean crypt What does this mean?

[naive shoal]

no doesnt

[ocean crypt]

So how is it solved?

[naive shoal]

there the product rule is applied

[tender meteor]

If you mean sin(n) then

,w Limit[n^5/Sin[n],n->∞]

[flat frigate]

Results provided by WolframAlpha

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Requested by alee31._

[naive shoal]

hmm..

lets wind up this question

[ocean crypt]

Why can't it be done?

[naive shoal]

0, doesnt exist and doesnt exist..right?

[ocean crypt]

?

[naive shoal]

i mean the original question..

[ocean crypt]

But that would be the domain

The x cannot be 0

D:(-∞,+∞) With the exception of {0}

D=domain

[tender meteor]

@naive shoal

Maybe crashed

[naive shoal]

what?

so, are my answers correct

0, doesnt exist and doesnt exist

[tender meteor]

The justifications or the resolution of the limits?

Or both

[naive shoal]

the answers

[tender meteor]

These are the answers

It wasn't very clear

[ocean crypt]

There are correct

[tender meteor]

Anyway Yes

[naive shoal]

i shall provide the justifications in the assignment..

.close

[safe radish]

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[safe radish]

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[tight void]

well well well BeanSauce

[crisp sluice]

i am confused abt the red point separate from the lines

[tight void]

what have you done so far

[tight void]

well open dot means that the function is not there at that point

while closed dot means it is

so F(-1) would not be 0 but rather would be ___

[crisp sluice]

oh aight so thats 1 then no?

what abt f(1) from left side if lines ends with open dot

[safe radish]

@crisp sluice Has your question been resolved?

[safe radish]

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[pallid flare]

im confused on how he goes from 4/x -1 to (4-x)/x

the second to third line

[fallen grotto]

turn -1 to -x/x

[clever ridge]

is that calc 2?

[fallen grotto]

end of precalc/beginning of calc 1

[clever ridge]

woah that is way harder then what im doing

and i just started calc 1

[tender meteor]

Just put a common denominator

[flat frigate]

alee

[pallid flare]

ok i think i get it

[safe radish]

@pallid flare Has your question been resolved?

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[graceful mantle]

The cost of 2 footballs and 3 tennis balls is £21.73. The cost of 5 footballs and 7 tennis balls is £53.20.

Work out the cost of

a) a football.

b) a tennis ball.

[light crypt]

Use systems of equations

You know how to do that?

[lean otter]

2 ways of doing it

1. The primary school way (the baby method)
2. Algebra

Oh wait nvm

I haven’t don’t this since P6

Done*

Ok I got it

53.20-21.73-21.73

You’ll get the cost of 1 of each

Then you probably work from there

Please forgive me, it has been years since I’ve done this

[light crypt]

Wdym

[lean otter]

Bc you minus 4 footballs and 6 tennis balls

[light crypt]

That isn’t 21.73

😭

@graceful mantle what’ve you tried so far?

[graceful mantle]

I don’t even know where to start

None of it makes sense to me

Oh wait no

I get it

[light crypt]

All good then?

[graceful mantle]

Nvm

I got the total of 1 football and 1 tennis ball

But how do I know how much each is?

[light crypt]

Walk me through what you did

[graceful mantle]

53.2 - 21.73 - 21.73 = 9.74

Which is one football and one tennis ball

Ohhhhh

Do I make it into a ratio?

Wait no

That doesn’t work

[light crypt]

Completely disregard that

What two equations can you make from the information given?

[graceful mantle]

7 footballs and 10 tennis balls = 74.93

And

3 footballs and 4 tennis balls = 31.47

[light crypt]

There's a much simpler set of equations already available

2 footballs + 3 tennis balls = 21.73 and 5 footballs and 7 tennis balls = 53.20

For the sake of making it easier to read, we can say "Let x be footballs" and "Let y be tennis balls"

So, 2x + 3y = 21.73

and, 5x + 7y = 53.20

[graceful mantle]

What do I do from there tho?

[light crypt]

You have two options from here, you can do substitution or elimination

Do you know how to do either?

[graceful mantle]

No

[light crypt]

Alright, I'll take you through both then, is that okay?

[graceful mantle]

Yes that would be nice

Thank you

[light crypt]

Alright, perfect

So, to do substitution, we need to isolate a variable

Which one do you want to isolate?

[graceful mantle]

Which ever one is least complex

[light crypt]

Neither one will be too complex

Let's go with x

[graceful mantle]

Alr

[light crypt]

For 2x + 3y = 21.73, what would x be when isolated?

[graceful mantle]

2x = 21.73 - 3y

[light crypt]

Yep, you'd just need to divide it by 2 to completely isolate it

So x = (21.73 -3y) / 2

Now that we know what x is in terms of y, we can substitute x into the other equation, 5x + 7y = 53.20

What do you think that would be?

[graceful mantle]

((21.73 - 3y) / 2) x 5 + 7y = 53.2

[light crypt]

Yep! Now that you have only one variable you're able to solve for y

That's substitution

[graceful mantle]

Ohhh I think I get it

[light crypt]

So what would y be then?

[graceful mantle]

Uhm

I don’t know how I would finish it

I thought I could just put it in a calculator but that doesn’t work

[light crypt]

No worries

I'll write down the steps for you

[graceful mantle]

I think I get it

[light crypt]

Alright

Elimination time?

[graceful mantle]

I guess so

Thank you for your help

[light crypt]

np

[graceful mantle]

Lemme try this before we start

Gimme one sec

[light crypt]

Alright, lmk when

[graceful mantle]

Alright let’s start elimination

[light crypt]

Alrighty

So, for elimination, we take our two equations and choose which term we want to get rid of

Want to do y this time?

[graceful mantle]

Alright

[light crypt]

Alright

So we have

2x + 3y = 21.73

5x + 7y = 53.20

Looking at x, we need to find a number that when we subtract the two equations, x will become 0

what can we multiply both equations by to get them at the same number?

Actually, let's do x again, it'll be simpler

[graceful mantle]

Ok

First equation by 5 second by 2

[light crypt]

Exactly

[graceful mantle]

Both x’s become 10

[light crypt]

So we would get

10x + 15y = 108.65

10x + 14y = 106.4

[graceful mantle]

Then we minus them?

[light crypt]

Now we just need to subtract the bottom from the top

[light crypt]

Yep!

[graceful mantle]

So 2.25

[light crypt]

Yep! And we can see that both methods gave us the same value for y

[graceful mantle]

Or y = 2.25

Elimination is much easier

[light crypt]

Now after finding one of the values, all we do is substitute it back into either original equation to get our x

[light crypt]

100% 😭

It's still good to know both but I hate using substitution for linear systems

[graceful mantle]

It is hard

[light crypt]

Agreed

[graceful mantle]

I’m going to stick to elimination unless the criculem requires otherwise

[light crypt]

Yeah, hopefully they don't ask you to use it specifically

Although it's good to build algebraic skills like that

[graceful mantle]

Agreed

[light crypt]

All good then?

[graceful mantle]

Yes

Thank you so much for your help

[light crypt]

Np

[graceful mantle]

It means a lot

[light crypt]

No worries

Don't forget to use .close to close the channel

[graceful mantle]

Yes

I will

.close

[safe radish]

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[lean otter]

Hello, I was wondering if I could get some help with this binomial expansion question

[tardy pivot]

worth memorising this one:

[lean otter]

So, I changed it to 1+x(1-x)^-1 and expanded out (1-x)^-1 and then times the 3 terms but 1+x but got the wrong answer

[lean otter]

Thank you

[tardy pivot]

when u expanded did ur terms match this one ^^?

[lean otter]

Yes

I got 1,x,x^2

Then I got:

1+x,x+x^2,x^2+x^3

[tardy pivot]

then collect like terms

to get 1 + 2x + 2x^2 + 2x^3 ...

[lean otter]

Ah didn’t know that was a thing 😂

Thank you

[tardy pivot]

imma sleep now

gl

[lean otter]

Thank you

[tardy pivot]

no prob

[lean otter]

Sleep well

.close

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[dire sedge]

how to do part d

it says $x'(t)=8t-t^2$ and $y'(t)=-t+\sqrt{t^{1.2}+20}$.

[flat frigate]

mXd

[lavish scaffold]

what does "moving towards the x-axis" mean mathematically in this case

[dire sedge]

y is decreasing

[lavish scaffold]

why y?

[dire sedge]

cuz like moving toward the x-axis is horizontal motion

[lavish scaffold]

you mean x then

[dire sedge]

wait but x motion is like left to right

[lavish scaffold]

my bad yeah

y is decreasing

[dire sedge]

why does it say particle remains in the first quadrant

[lavish scaffold]

i think they mean the top right part of the plane

[safe radish]

@dire sedge Has your question been resolved?

[safe radish]

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[earnest needle]

Please tell me if the following derivation is correct. I'm new to derivatives so I'm afraid of making mistakes. Thank you.

We know that $\lim_{x->0} \dfrac{sinx}{x} = 1$
So does $\lim_{dr->0} sindr=dr$

Also does $\lim_{dr->0} cos(r+dr)=cos(r)$

[hard crest]

use \lim

[flat frigate]

EphemeralEuphoria

[hard crest]

i'm not even sure if this really makes sense as a statement (i don't think it does) but no, $\lim_{dr\to0} \sin dr = 0$

[flat frigate]

hayley table

[safe radish]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[hard crest]

yeah no it just doesn't make sense tbh

you need two variables involved for d* to make sense, that's why we have like dy/dx

[earnest needle]

I mean like can I treat d* as a variable because it's just a part in the proof I'm trying to do?

[hard crest]

yes.... with some hesitation

dx is an infinitesimal, you can think of this as "the smallest positive number"

i don't think it makes sense to take the limit as dr goes to 0

[earnest needle]

I don't really understand how dx can't be intepret as $\lim_(x->0). Can you explain more since smallest number seems to be the same as approaching 0?

[hard crest]

that's not really what i said

i said i don't think it makes sense to take the limit as dx goes to 0

[earnest needle]

Oh sorry, so do you mean the part sindx=dx is wrong?

[hard crest]

yeah that just doesn't make sense to me

what are you trying to do with these?

[safe radish]

@earnest needle Has your question been resolved?

[earnest needle]

I want to prove $\dfrac{dx}{cosx}=\dfrac{sindx}{cos(x+dx)}$. Using trigonometry formula is pretty long so I wonder if I can just take "shortcut".

[flat frigate]

EphemeralEuphoria

[safe radish]

@earnest needle Has your question been resolved?

[safe radish]

@earnest needle Has your question been resolved?

[safe radish]

@earnest needle Has your question been resolved?

[safe radish]

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[proven fern]

for part ii can i use angle between 2 lines formula

but how do i get the slope of f-x-

[worthy estuary]

Derivative of f(x) at B

But since it is a parabola

U can also find the average rate of change

Around that point

With x values that are the same distance from B's x value

[proven fern]

would it not be at the derivative at point B BC thats where the basket is

[worthy estuary]

Oh ye

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[grim plover]

,rccw

[grim plover]

Q11, I got a few equations, but idk what to do with them to get something useful

(just substituted the corresponding zeroes in their equations)

[river field]

Either I'm blind or I don't see a Q11

[spiral imp]

Yeah I was also confused, I think they mean q5 lol

[grim plover]

Yes 💀

[grim plover]

Yeah mb I meant q5

[river field]

well you have a system of 4 equations and 4 unknowns so it's technically solvable

but like

[grim plover]

U aren't supposed to solve it

[river field]

try to remember some way you can rewrite quadratic

[grim plover]

Ur supposed to manipulate the equations to get something

[river field]

in terms of their zeroes

[grim plover]

Hmm let me try that good idea

[river field]

Hint : ax^2 + bx + c = a * (x - x_1) * (x - x_2)

uh

fixed

[grim plover]

Ye i remember

Alright got it

Tyvm

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[pulsar zealot]

Could anyone help me with the integration of ln(x-4)

[pseudo scroll]

Integrate by parts

[pulsar zealot]

Yeh I’ve tried but it’s kinda hard

[pseudo scroll]

It's alright we'll try it again

What did you choose to integrate and differentiate?

[pulsar zealot]

That’s where I got stuck

I tried a bunch of options but none of them worked out

[summer crow]

what did you try

[pulsar zealot]

I set u as x-4 and v as lnu

[summer crow]

wait so you let u = x-4, meaning ln(x-4) = ln(u), and then you let v = ln(u). Then you tried to differentiate u and integrate v? That's not integration by parts..

Hint: $\ln(x-4) = 1 \cdot \ln(x-4)$, can you now guess what to differentiate and integrate?

[flat frigate]

lgkoo

[pulsar zealot]

Wait so what was I doing then

[summer crow]

oh ok

[pulsar zealot]

I get now that u = ln(x-4) and v is 1

[summer crow]

what did you get

right

[pulsar zealot]

What did I get for what?

[summer crow]

what did you get after you have u = ln(x-4) and v' = 1

[pulsar zealot]

1/x-4 and x

I think I’m fine from here

[summer crow]

[pulsar zealot]

Thanks for the help

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[slow flax]

Elsa and Victoria each picked 600 rose bushes. Elsa worked for two hours longer than Victoria because she picked 15 fewer rose bushes per hour. How many hours did Elsa work, and how many hours did Victoria work?

i am not sure how to get to an answer.

[slow flax]

i am guessing the table looks somewhat like this, but i am not sure what numbers to input

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[slow flax]

figured out more and turned out like this

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[tardy oar]

after reducing how did we get that matrix?

[granite idol]

have you carried out the operations?

[tardy oar]

what operations?

i watched this video and he wrote this matrix immediately

[granite idol]

the answer to "how" is "by row reduction"

[tardy oar]

ok, i have to watch it and learn very well, thank you!

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[trim carbon]

does this symbol mean the number has to be a positive real number?

or just any real number

[granite idol]

yes

[trim carbon]

which one?

[lone arch]

first

[trim carbon]

oh ok tysm!

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[lean otter]

Any help would be much appreciated

<@&286206848099549185>

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[ivory flame]

Yep

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[rustic goblet]

yes, I believe so

oh wait

you have to check that the inner product of p with t is 0, and that the inner product of p with k is 0

it can’t just be either of them

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[rustic goblet]

oh I was about to construct a counterexample lol

but seems like you got it

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[pearl falcon]

hello

[pearl falcon]

how do you get the estimated value of 'y' in linear regressions

thats the paper and i think i know how to do the others im just not sure on how to get the expected y value

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[plain sage]

<@&286206848099549185>

[final halo]

one idea: you can see how the shaded area can be broken up into a big square and then a triangle on each side

[plain sage]

yep i tried doing that

however my answer was 11a

i think i did something wrong when calculating the length of the triangles

[sterile tapir]

Can you show your work of the triangles length?

[plain sage]

so since each side is a, i used pythagoras for this isoceles and this square root a^2+ a^2 to get 2a as the bottom length of the triangle

[sterile tapir]

$$\sqrt{a^2+a^2} \Rightarrow \sqrt{2a^2}$$

[flat frigate]

vrsth

[sterile tapir]

so

[flat frigate]

vrsth

[sterile tapir]

that should be the length right

[plain sage]

ooo thank youu

and then do i use the normal area of a triangle rule

[sterile tapir]

Yeah

[plain sage]

okk thank you!

[sterile tapir]

Your welcome good luck!

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[plain sage]

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[lean otter]

I'm stuck on the left side, but I have the following:

Then when I plug that in

Does this all look right so far?

<@&286206848099549185> I am stuck, could someome please help?

[fathom jewel]

The b cancels too

[lean otter]

how so?

[fathom jewel]

Imagine $z = \frac{b}{x_p}$ then $\frac{1}{z\ln(10)} \cdot -z \cdot \frac{1}{x_p}$

[flat frigate]

𝔸dωn𝓲²s

[fathom jewel]

Is that more clearer?

[lean otter]

This is what I ended up getting

so something cancels there?

[fathom jewel]

hold on

[lean otter]

wait

would it have cancelled here

[fathom jewel]

yea

if you flip b/x_p

[lean otter]

ah ok

Let me do that real quick

[fathom jewel]

,,\dd x_p = -\frac{\ln(10)}{x_p} \dd u

[flat frigate]

𝔸dωn𝓲²s

[lean otter]

yep i see that now

i wonder if that ends up doing anything in the final answer though

the initial condition is x_p(0)=1

[fathom jewel]

Let's see

[fathom jewel]

Also wouldn ya end up with a constant C on the right after integration?

[lean otter]

Aha

Yes, right?

[fathom jewel]

yea

[fathom jewel]

Oh there

[lean otter]

i have that

[fathom jewel]

okoko

[lean otter]

xD

sorry i may have not put it in everything

[fathom jewel]

it's fine

https://tenor.com/view/dinkelver-gif-21026349

[lean otter]

hmmmmm

[fathom jewel]

oh no

wait

[lean otter]

did i do something wrong maybe

[fathom jewel]

no me

hold on

[fathom jewel]

Since z cancels we would get 1/ln(10)x_p

[lean otter]

and how would that affect what i did

is z like another u sub or

[fathom jewel]

,,\dd u = -\frac{1}{\ln(10)x_p} \dd x_p \iff \dd x_p = -\ln(10)x_p \dd u

[flat frigate]

𝔸dωn𝓲²s

[lean otter]

yes

[fathom jewel]

oh right

,,-\ln(10) \int \frac{1}{u} \dd u = at + C

[lean otter]

-ln(10)?

[fathom jewel]

yes

[flat frigate]

𝔸dωn𝓲²s

[fathom jewel]

Ok yea

[lean otter]

ok so -ln(10) * ln | u | = at + C

and plug u back in

[fathom jewel]

yea

[fathom jewel]

this

[lean otter]

yep

[fathom jewel]

x_p(0) = 1

so x_p is a function of

[lean otter]

and if I have the initial condition of

[fathom jewel]

x_p = 1

but what is 0

[lean otter]

not sure

[fathom jewel]

HUH

I would guess t

[lean otter]

well a and t are constant

[fathom jewel]

seems like a differential equation

[lean otter]

it is a differential equation

[fathom jewel]

but you integrated by t

[lean otter]

well let me try t as one

[fathom jewel]

no

x_p = 1

[lean otter]

and t = 0

[fathom jewel]

if x_p(t)

yes

[lean otter]

let me try that

and see if it gets me the answer

[fathom jewel]

then you would get something in terms of b for C

[lean otter]

yes

i need to solve for C

and plug back in

[flat frigate]

𝔸dωn𝓲²s

[fathom jewel]

yea haha

[lean otter]

now plug that back in

[fathom jewel]

,,- \ln (10) \cdot \ln \abs{\log_{10}\left ( \frac{b}{x_p} \right )} = at - \ln (10) \cdot \ln \abs{\log_{10}(b)}

[flat frigate]

𝔸dωn𝓲²s

[lean otter]

uh

that looks nasty

[fathom jewel]

yea

but x_p(t) seems to make the most sense

[lean otter]

this was the original

[fathom jewel]

separation of variables

[lean otter]

and x_p(t) = b^(1-e^-at)

that is what i need to get

[fathom jewel]

there it is x_p(t) lmao

[fathom jewel]

So we need to solve for x_p

[lean otter]

how do we do that

[fathom jewel]

divide by -ln10

then apply e

[lean otter]

so they cancel

the -ln(10)?

[fathom jewel]

ye

,,\abs{\log_{10}\left ( \frac{b}{x_p} \right )} = e^{-\frac{at}{\ln(10)}} \cdot \abs{\log_{10}(b)}

oh no

forogto

[lean otter]

what]

[flat frigate]

𝔸dωn𝓲²s

[lean otter]

wait

-ln(10) cancels i thought

[fathom jewel]

yes

with the other

it's a sum

at - ln(10) * log...

[lean otter]

ok

ok

now that we are here... should we rewrite something?

[fathom jewel]

I think we dont need the absolute values

[lean otter]

nah

[fathom jewel]

because of the condiition

[lean otter]

since it's going to be positive anyways

yeah

[fathom jewel]

then i would subtract the left

,,\log_{10}(b) - \log_{10}(x_p) = e^{-\frac{at}{\ln(10)}} \cdot \log_{10}(b)

[lean otter]

ah yes yes

[fathom jewel]

non nvm

[lean otter]

wait

-e?

[flat frigate]

𝔸dωn𝓲²s

[fathom jewel]

I applied log(a/b) = loga - logb

[lean otter]

yep

[fathom jewel]

,,\log_{10}(x_p) = \log_{10}(b) - e^{-\frac{at}{\ln(10)}} \cdot \log_{10}(b) = \log_{10}(b) \left (1- e^{-\frac{at}{\ln(10)}} \right )

[flat frigate]

𝔸dωn𝓲²s

[fathom jewel]

Apply 10^(...)

,,x_p = 10^{\log_{10}(b) \left (1- e^{-\frac{at}{\ln(10)}} \right )}

[flat frigate]

𝔸dωn𝓲²s

[lean otter]

so log(10) and 10 cancel?

[fathom jewel]

yes kinda

[lean otter]

lg10 is just 1

what

[fathom jewel]

we have 10^(ab)

[lean otter]

correct

[fathom jewel]

,,x_p = \left ( 10^{\log_{10}(b)} \right )^{\left (1- e^{-\frac{at}{\ln(10)}} \right )}

[flat frigate]

𝔸dωn𝓲²s

[fathom jewel]

so now it cancels

[lean otter]

ok

[fathom jewel]

,,x_p = b^{\left (1- e^{-\frac{at}{\ln(10)}} \right )}

[flat frigate]

𝔸dωn𝓲²s

[lean otter]

hip hip hooray

[fathom jewel]

really?

[lean otter]

except

ln(10)

[fathom jewel]

they are missing the ln(10)

[lean otter]

it shouldnt be there

according to them

[fathom jewel]

hahahaha

maybe log was ln all along

[lean otter]

hm?

[fathom jewel]

not log_10

but log_e

[lean otter]

uh

how do i fix that

[fathom jewel]

I hate when they dont specify the base

[lean otter]

lol

so uh

how to fix that then

do we have to go back

[fathom jewel]

if it was ln(...) = u then we wouldn't have ln(10)

[lean otter]

so log is actually ln

in this

[fathom jewel]

We would have -1/x_p

[fathom jewel]

yes

[lean otter]

BRUH

[fathom jewel]

stupid author

otherwise our steps were valid

[lean otter]

i am so pissed

[fathom jewel]

how come we come so close

[lean otter]

so just replace everything with ln now?

[fathom jewel]

and differ by some factor

no not quiet

Let's do the trick quickly

[lean otter]

ok

[fathom jewel]

,,u = \ln \frac{b}{x_p} \implies \dd u = \frac{1}{\frac{b}{x_p}} \cdot -\frac{b}{x_p^2} \dd x

[flat frigate]

𝔸dωn𝓲²s

[fathom jewel]

So we end up with $\dd u = -\frac{1}{x_p} \dd x \iff \dd x = -x_p \dd u$

[flat frigate]

𝔸dωn𝓲²s

[fathom jewel]

So imagine

We substitute

Imagine we do all steps instead with -ln(10) we do it with (-1) only

So yea

we should get to their solution doing so

[lean otter]

WHY WOULD THEY NOT HAVE STATED THAT FIRST

AHHHHH

[fathom jewel]

Like here we would divide by (-1) and then do e^(...)

then we would have e^(-at) ...

[fathom jewel]

So eventually $x_p = b^{\left (1- e^{-at} \right )}$

[flat frigate]

𝔸dωn𝓲²s

[fathom jewel]

but it was a quick fix

the math is mathing

[lean otter]

wait

so let me show you

[fathom jewel]

which is pretty rare in such occasions

[lean otter]

wait

[fathom jewel]

Honestly if you gave the solutions with ln(10) then your prof should have take away points from himself lmao

[lean otter]

so the left side was just 1/u

[fathom jewel]

with minus

all steps are similiar

we just differ by some factor

replace it ln(10) = 1

haha looks wrong

ln(10) to 1

[fathom jewel]

yea

[lean otter]

then exponentiate

[fathom jewel]

yea

also

did you realize

whehter we did it before or after

we got it right

with the constant C

[lean otter]

wdym sorry my brain is fried rn

[fathom jewel]

You want first to solve for x_p

which is fine

previously we first evaluated C and then solved x_p

[lean otter]

so dont exponentiate yet

[fathom jewel]

both would work

[fathom jewel]

thats not what i am saying

[lean otter]

are you talking about before

when we assumed the base was 10

instead of e

[fathom jewel]

yes

[lean otter]

ah ok

[fathom jewel]

we first evaluated the constant

then solved for x_p

now we doing it backwards

my point

both work

just as a note

[lean otter]

so now we solving for C

firwst

first

[fathom jewel]

I would actually solve for x_p

[lean otter]

oh?

[fathom jewel]

yea

[lean otter]

so then is becomes

[fathom jewel]

because solving for C first usually results to lengthier terms

like previously

haha

[fathom jewel]

But first I would multiply by (-1)

no

[lean otter]

wah

wuhhhhhhhh

[fathom jewel]

,,\ln \abs{\ln \left ( \frac{b}{x_p} \right ) } = -at - C

[flat frigate]

𝔸dωn𝓲²s

[lean otter]

why no negative

OH

you multiply by negative 1

[fathom jewel]

you really fried haha

[lean otter]

it would be + C regardless though

right

[fathom jewel]

now let's e^(...)

[lean otter]

at least that's what my teacher says

[fathom jewel]

,,\ln \left ( \frac{b}{x_p} \right ) = e^{-at - C}

[fathom jewel]

yea but i dont want to

[lean otter]

xD

[fathom jewel]

your teacher screwed us already once

[lean otter]

LMAO

[fathom jewel]

might aswell leave the abs

[flat frigate]

𝔸dωn𝓲²s

[lean otter]

hmmm now we plugin x_p(0) = 1

so plugin 0 for a and 1 in for x_p?

[fathom jewel]

Now you wanna solve mid way for C 😂

[lean otter]

wuhhhhhhhhhh

[fathom jewel]

,,\ln(b) - \ln(x_p) = e^{-at - C}

[lean otter]

never heard of that

[flat frigate]

𝔸dωn𝓲²s

[lean otter]

yes

[fathom jewel]

yea you're fried lmao

[lean otter]

bye

that was too easy

why i not say that

[fathom jewel]

haha

yea

I am seeing things

[lean otter]

now we plugin? 🥺

[fathom jewel]

let's solve for x_p first

[lean otter]

or sh ould be add term over first

[fathom jewel]

haha

[lean otter]

ye

[fathom jewel]

get it to x_p = ...

,,\ln(x_p) = \ln(b) - e^{-at - C}

[flat frigate]

𝔸dωn𝓲²s

[lean otter]

correct

[fathom jewel]

Now the same bs as before 🥱

[lean otter]

but but but

[fathom jewel]

,,x_p = e^{\ln(b) - e^{-at - C}}

[flat frigate]

𝔸dωn𝓲²s

[lean otter]

yes exponenitate

i cant spell

[fathom jewel]

e^(a+b) = e^a * e^b

[lean otter]

now plugin

[fathom jewel]

,,x_p = e^{\ln(b)}\cdot e^{ - e^{-at - C}}

[flat frigate]

𝔸dωn𝓲²s

[fathom jewel]

at least stick to your word

let us finish

[lean otter]

e^1

wait wtf

poor C so high up

[fathom jewel]

hahaha

fine let's plug in

[lean otter]

YAY

[fathom jewel]

,,1 = e^{\ln(b)}\cdot e^{ - e^{- C}} = e^{\ln(b) - e^{- C}}

[flat frigate]

𝔸dωn𝓲²s

[fathom jewel]

We can write as 1 = e^0 and compare exponents

,,e^0 = e^{\ln(b) - e^{- C}} \implies 0 = \ln(b) - e^{- C}

[flat frigate]

𝔸dωn𝓲²s

[fathom jewel]

,,\implies e^{-C} = \ln(b) \iff C = -\ln (\ln (b))

[flat frigate]

𝔸dωn𝓲²s

[fathom jewel]

So

[lean otter]

NOW WE PLUGIN C

[fathom jewel]

,,x_p = e^{\ln(b)}\cdot e^{ - e^{-at + \ln (\ln (b))}}

[flat frigate]

𝔸dωn𝓲²s

[lean otter]

hmmmm

i see two ln(b))

[fathom jewel]

,,x_p = b \cdot e^{ - e^{-at} e^{\ln (\ln (b))} }

[flat frigate]

𝔸dωn𝓲²s

[lean otter]

twa

[fathom jewel]

yea

,,x_p = b \cdot e^{ - e^{-at} \ln(b) }

[flat frigate]

𝔸dωn𝓲²s

[lean otter]

wait

where that e come from

[fathom jewel]

power rules

[lean otter]

i thought e^ln(b) cancels to just b

[fathom jewel]

,, e^{a+b} = e^a \cdot e^b

[flat frigate]

𝔸dωn𝓲²s

[fathom jewel]

what about at?

[lean otter]

oh wait no i see it now

[fathom jewel]

,,x_p = b \cdot \left (e^{\ln(b) } \right )^{-e^{-at}}

[flat frigate]

𝔸dωn𝓲²s

[lean otter]

power rule again, right?

[fathom jewel]

yes

[lean otter]

then they cancel

the ln(b) and e

right

[flat frigate]

𝔸dωn𝓲²s

[fathom jewel]

,,x_p = b \cdot b^{-e^{-at}}

[lean otter]

so switch

[flat frigate]

𝔸dωn𝓲²s

[fathom jewel]

Last step is up to you :) Dinkelberg

https://tenor.com/view/-gif-4171790

[lean otter]

BYEEEEEEEEEEEEEEEEEEEEEEEEEEEEEE

like bases

[fathom jewel]

I was your biggest comeback

[lean otter]

😝

[fathom jewel]

yea continue

[lean otter]

so

x_p = b^1-e^(-at)

PERIOD