help-0

[gray isle]

that's what they're trying to prove

derive it from angle sum on line

[pulsar moat]

my teacher told me angle 1 and 3 are equal

[fading eagle]

but back to his question, he's tryin to prove that 5 == 3

[pulsar moat]

yes

she told me i was wrong

but i believe im right

[fading eagle]

well, it is not neccessarily congruent unless stated if this is the only given

[gray isle]

well <5 and <3 are indeed congruent, but the attempt at a proof is nonsense

[pulsar moat]

what's a better way to prove it?

[gray isle]

if you're allowed to consider vertical angles, you can simply quote that

if you're not, then derive it yourself

from angle sum on line

[pulsar moat]

ok thank you

[gray isle]

are all those long lines supposed to be straight lines?

[pulsar moat]

yes

[lone heart]

@pulsar moat Has your question been resolved?

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[steep lion]

How come the velocity vector (derivative) of the positional vector only differentiates the first and last parts? How come 2j remains unchanged?

[muted niche]

I dont think that should be the case.... if its a derivative with respecy to t, then the i and j components should be 0 as they are constants

[frigid geyser]

looks like a typo

[steep lion]

Right, that's what I was thinking. So it should be just -sin(t)k?

[muted niche]

Indeed

[steep lion]

Thanks, Khazali, have a good one:)

[muted niche]

No problem

[lone heart]

@steep lion Has your question been resolved?

[lone heart]

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[formal solar]

Hey guys

[alpine sable]

can someone help me with this?

[formal solar]

It's occupied love, use another channel

[formal solar]

#help-25

Hey guys. In a series

When we want to know if a series converges sometimes we use limit (an) and if the limit converges to 0 then an converges

But other times we use limit comparison test...

I really don't know when to use what?

[frigid geyser]

$\lim_{n \to \infty} a_n = \alpha$ only means that $a_n$ converges

[ocean seal]

xdk1235

[frigid geyser]

it doesnt say anything about the series $\sum_{n=1}^\infty a_n$

[ocean seal]

xdk1235

[frigid geyser]

ahh you mean like the limit being zero

Checking if a_n -> 0 is to see if the series stands a chance to converge, if a_n does not converge to zero, the series just diverges. this is all you can get

you never know for sure if the series converges or not just because the limit of the sequence is zero

now for the limit comparison test, say you have some series that you know it converges/diverges, and it looks like you can take the limit of the ratio

thats when you use it

[lone heart]

@formal solar Has your question been resolved?

[formal solar]

Ooooooh okk thanks

[lone heart]

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[vapid crypt]

This is the formula for a parallelepiped. Where did the cos theta go and how come it's possible to have that dot product?

[abstract fractal]

v • w = |v||w|cosθ, where θ is the angle between v and w

[vapid crypt]

thanks!

.close

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[north hemlock]

the vector w is 6i-12j+8k

I'm not sure how this is wrong, perhaps I'm adding wrong?

[charred heron]

i got

4i - 9k + 21k

so (4, -9, 21)

[north hemlock]

I-

[north hemlock]

hold on, why are u subtracting?

[vapid crypt]

the formula of a position vector from two points PQ are <Q1-P1, Q2-P2, Q3-P3>

[north hemlock]

okay uhm

I'm R2, i drew this

R2 being the xy-plane here

OHHH

I see now

[charred heron]

does this make sene

[north hemlock]

Yeah, I jsut drew something similar

yep yep

[charred heron]

nice

[north hemlock]

what I tried to do was uhh

I guess

P + Q = w but Q starts at P but it should start at the origin

or something like that, idk

in any case

.close

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[alpine sable]

How do I differentiate y=arcsin2x

[pliant dune]

you shouldn't treat derivatives like fractions in general, but dy/dx = 1/(dx/dy)

[alpine sable]

I get that

It’s the first few steps

[pliant dune]

so if you take sin of both sides and get sin(y)=2x

[alpine sable]

Yes I got that

[pliant dune]

and differentiate wrt to y

[alpine sable]

Do I divide by 2 first?

[buoyant kayak]

what

just

huh

just take the derivative

[buoyant kayak]

why this

no need for that

[alpine sable]

That’s how I was taught it

[buoyant kayak]

just use chain rule

[alpine sable]

So 2x = t?

[buoyant kayak]

i mean if you want to set up a sub and everything sure

[pliant dune]

I just assumed that the problem was more focused on finding the derivative of arcsin so assumed they couldn't take a standard result for it

[alpine sable]

I’ve never differentiated something where the subject is something like 2x

So I just assumed I needed it to be x =

[buoyant kayak]

you've never used the chain rule?

[alpine sable]

I have

First time seeing a question like this though

[buoyant kayak]

this is just the chain rule

$\dv{x}(\arcsin{(x)})=\frac{1}{\sqrt{1-x^2}}$

[ocean seal]

a disappointing son

[alpine sable]

Arcsin is simple

[buoyant kayak]

this is the standard derivative for arcsin, you can just apply the chain rule from here

[alpine sable]

Arcsin2x looks confusing

[buoyant kayak]

only difference is replacing x with (2x) and applying chain rule

[alpine sable]

Oh I see

I’ll try that

[pale kestrel]

tbh people say this, but the chain rule says you can right?

$$\frac{d^2y}{dx^2}$$

[ocean seal]

Shuri2060

[pale kestrel]

^it makes no sense to flip something like this anyways

$$\frac{dx^2}{d^2y}$$

[ocean seal]

Shuri2060

[pale kestrel]

this expression is meaningless

I guess it doesn't necessarily hold for partials though.

So 'flipping' is only possible with the chain rule

[alpine sable]

Is this good?

[buoyant kayak]

looks good

[alpine sable]

I gotta use this answer to differentiate 1/(1-4x^2)^1/2

I don’t know how to even start

[pliant dune]

The chain rule would probably be the way

[alpine sable]

I could integrate without part a

But it tells me to use part a

[pliant dune]

wait are you integrating or differentiating it?

[alpine sable]

Integrating

[pliant dune]

well dy/dx is almost what you want to integrate

[alpine sable]

Except for the fact it’s multiplied by 2

[pliant dune]

what would dy/dx be if y=0.5*arcsin(2x)?

[alpine sable]

Probably just take out the 1/2?

[pliant dune]

for constant factors you can take them outside of the differentiation

so if you differentiate 0.5*y you get 0.5*dy/dx

[alpine sable]

Oh I see

So

2y = arcsinx?

[pliant dune]

well you know that differentiating y gives you 2 times what you want

so differentiating 0.5*y removes the factor of 2 you don't want

[alpine sable]

So I differentiate again but with 0.5y as the subject?

Wait

Is it

2(arcsin(2x))

@pliant dune

[pliant dune]

no, I mean since you've already differentiated arcsin(2x) it should be easy to find what the derivative of 0.5(arcsin(2x)) is

[alpine sable]

That’s 1/(1-4x^2)^1/2 right

[pliant dune]

yes

then since you know 0.5(arcsin(2x)) differentiates to 1/(1-4x^2)^1/2, you can conclude that 1/(1-4x^2)^1/2 integrates to 0.5(arcsin(2x))+C

[pliant dune]

I'm just saying it because lots of people say it tbh. Like you say it doesn't work with partials but I also imagine there's lots of creative ways people could find to treat derivatives like fractions that don't work

[pale kestrel]

👌

[alpine sable]

I’ve been told to use the derivative of ln(cosx) to find the integral of tanx

The derivative of ln(cosx) is -tanx

[pale kestrel]

alternatively, you can spot that tanx = -(-sinx)/cosx

[alpine sable]

But the integral of tanx is sec^2(x) right

[pale kestrel]

uhhhh doubts?

[alpine sable]

Oh that’s dy/dx

[pale kestrel]

it should be -ln|cos x| + C

the modulus sign is important.

[pliant dune]

This is the same sort of idea as the second part of the last question, you can multiply both by some factor to find what you differentiate to get tan(x)

[alpine sable]

So it’s -ln|cosx| + c

[pale kestrel]

did i miss the minus

my bad

[alpine sable]

You said this

[alpine sable]

^

[pale kestrel]

yes

[alpine sable]

So I just multiply -1 by the integral of -tanx

[pale kestrel]

I was lazy and looked at the first result

https://i.imgur.com/nlO0EcL.png

which seems to be wrong lol

,w integrate tan x

[ocean seal]

Results provided by WolframAlpha

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

Requested by Shuri2060#9525

[pale kestrel]

wolfram doesnt do mod sign smh

cus complex ig? not sure

[pliant dune]

pretty sure that's why

[alpine sable]

The integral of cos^2(x) isnt sin^2(x) right

[pale kestrel]

doubts.

no

integral cos is sin

but almost certainly not for squared

[pale kestrel]

Any idea how to do it?

[alpine sable]

Nope

[pale kestrel]

i would suggest

adding 0

integrate

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

[alpine sable]

That’s just an identity right

[pale kestrel]

?

[alpine sable]

A trig identity

[pale kestrel]

👌

If you're thinking pythagoras, that doesn't help though.

[alpine sable]

Oh it’s cos^2(2x)

[pale kestrel]

oops i wrote it slightly wrong

$$(\cos^2(x) - 0.5) + 0.5$$

[ocean seal]

Shuri2060

[alpine sable]

I don’t see how that helps

[pale kestrel]

double angle formula

[alpine sable]

Cos2x?

[pale kestrel]

👌

try to see how it helps

[alpine sable]

Does it work if theta is 2x?

[pale kestrel]

you can either

substitute u = 2x

Or think about what happens when you substitute 2x for x in integrals (like reverse differentiating)

Like if I gave you sin(2x), what does that differentiate to

[alpine sable]

1/2cos(2x)

[pale kestrel]

2cos2x

[alpine sable]

Oh yeah

I’m trying to use cos2u where u = 2x

[pale kestrel]

you can do that to, as long as you substitute properly...

Have you learnt substitution for integrals?

[alpine sable]

Nope

[pale kestrel]

ok, then dont'

[alpine sable]

What approach should I take?

[pale kestrel]

d/dx (sin(2x))

=2cos2x

reverse differentiate, basically

[pale kestrel]

what do you have right now?

[alpine sable]

Just the question

[pale kestrel]

$$\int \cos^2x - \frac{1}{2} + \frac{1}{2}\dd{x}$$

[ocean seal]

Shuri2060

[pale kestrel]

After the double angle formula?

[alpine sable]

You told me not to use it

[pale kestrel]

eh?

don't use subtitution

u = 2x

keep everything in x

[alpine sable]

Oh I see

[pale kestrel]

double angle formula, you need it

cos 2a = 2cos^2 a - 1

[alpine sable]

Cos^2(2x) - sin^2(2x)

[pale kestrel]

there's only 1 that will help

[pale kestrel]

$$=\int\frac{1}{2}(2\cos^2x - 1) + \frac{1}{2}\dd{x}$$

[ocean seal]

Shuri2060

[alpine sable]

Man I don’t even know where that 1/2 in front of the bracket came from

[pale kestrel]

.

These 2 expressions are identical

[alpine sable]

Oh yeah

[pale kestrel]

$$\cos(2x) = 2\cos^2x-1$$

[ocean seal]

Shuri2060

[pale kestrel]

^ we want this form

The point is, we know how to integrate cos x

But we don't know how to integrate cos^2 x

[alpine sable]

Oh you want cos(2x) instead?

[pale kestrel]

That's the point of using the double angle formula

We have no idea of how to integrate the original thing

We don't know any functions that differentiate to cos^2 x

[alpine sable]

What’s the point of having the 0?

[pale kestrel]

where

[alpine sable]

The two constants cancel out

[pale kestrel]

I write it like that

because that is the form of the double angle formula

$$\int\cos^2x \dd{x}$$

[ocean seal]

Shuri2060

[pale kestrel]

This is the original thing you have no idea how to integrate

Writing it in this form was my hint

Because you can use this

Integration is 'hard', and basically boils down to trying to reverse differentiating. If we don't know an integral, we try to change it to something we DO know.

[alpine sable]

I’m sorry but I’m not understanding any of this

This is the first time for me integrating cos^2(x)

[pale kestrel]

$$\int\cos^2x \dd{x}$$
$$=\int \cos^2x - \frac{1}{2} + \frac{1}{2}\dd{x}$$
$$=\int\frac{1}{2}(2\cos^2x - 1) + \frac{1}{2}\dd{x}$$

[ocean seal]

Shuri2060

[pale kestrel]

$$=\int\frac{1}{2}\cos(2x) + \frac{1}{2}\dd{x}$$

[ocean seal]

Shuri2060

[pale kestrel]

If this helps?

That is the process

And now we know how to integrate the final thing

[alpine sable]

That makes sense but I could never see that by myself

[pale kestrel]

No, it takes experience

A lot of ppl find integration hard at first

cus of this

You don't see the 'tricks' that are meant to be used first time

And need a lot of hints

It's completely different from trying to differentiate

[pale kestrel]

So this final thing, we can integrate, because

$$\dv{x}\sin(2x) = 2\cos(2x)$$

[ocean seal]

Shuri2060

[pale kestrel]

We need a factor of 1/2 outside

$$\dv{x}\left(\frac{1}{4}\sin(2x)\right) = \frac{1}{2}\cos(2x)$$

[ocean seal]

Shuri2060

[pale kestrel]

and this is the exact form we need

[alpine sable]

Man differentiation was simple

Why’s integration soo complex

[pale kestrel]

Exactly.

It turns out 'most' of the integrals out there are hard/impossible to do

The ones 'chosen' for school exercises

Are handpicked

to be doable

If you picked a bunch of random functions and combined them, you can still easily differentiate with the chain rule

Not so, for integration - most likely it would result in something impossible to integrate by hand.

$$\int e^{-x^2}\dd{x}$$

[ocean seal]

Shuri2060

[pale kestrel]

^ In terms of the functions you know

There is none that satisfy this (an example of something you definitely wouldn't be given to integrate)

[alpine sable]

Guess I’ll just need a lot of practice

[lone heart]

@alpine sable Has your question been resolved?

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[pale kestrel]

?

[fossil valley]

I have no clue how to do that

Need help

[pale kestrel]

ok, what do those symbols mean?

in words?

[fossil valley]

evaluate f(x) from -2 to 5

You have to subtract values

[pale kestrel]

integrate

not evaluate?

[fossil valley]

Yes

[pale kestrel]

In words

it should be

'The integral from -2 to 5 of f(x) with respect to x'

ok?

Do you know what the integral is, graphically?

What does it represent?

[fossil valley]

so am I supposed to subtract f(x) at -2 from f(x) at 5

[pale kestrel]

no.

[fossil valley]

antiderivative

[pale kestrel]

That's not graphically

That's algebraically

[fossil valley]

area under the curve

[pale kestrel]

👌

[pale kestrel]

More precisely

[glass lichen]

signed area

[pale kestrel]

The integral of f(x) is the signed area between curve of f(x) and the x-axis between the given limits

[fossil valley]

what does signed area mean?

[pale kestrel]

If it is under the x-axis

The area is negative

[fossil valley]

so we are finding both

positive area and negative?

[glass lichen]

you find all 3 components, then sum them.

[pale kestrel]

https://i.imgur.com/wai9zwU.png

For example in this picture

I would add the blue area

and subtract the purple area

If I'm integrating that random function from 1 to 8

[fossil valley]

and you do that by using the formula that comes with solving those right

[glass lichen]

you find areas, yes.

[fossil valley]

Got it

[fossil valley]

Damn I thought I had to integrate the function or something

[pale kestrel]

You aren't given a function

[glass lichen]

how would you integrate that...

you're not given a function

[pale kestrel]

So how can you find the antiderivative?

You are just given a graph.

[fossil valley]

Lmao

I have no clue

hence why I posted in here

[glass lichen]

I mean, you look at the page and see no function

[fossil valley]

thank you

[glass lichen]

Hence why you shouldn't be thinking "Oh I need to actually integrate"

[fossil valley]

.close

[lone heart]

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[still crest]

How do I prove that this equation has exactly one root in the given interval?

[alpine sable]

ladies and gentlemen, i need quick help please!

[little drum]

Graph

Do you know derivatives? and it's uses?

[still crest]

Yes I'm supposed to solve this with derivatives

[little drum]

Like to find minima/maxima and all..

Yeah

[still crest]

Kind of yea

[little drum]

So first of all, imagine or try to make a rough sketch of how the graph for e^(-x) would look like

Go for an unoccupied channel please?

5, 19 and 25 are available!?

[alpine sable]

oh my, i'm sorry..

sorry to bother.

[still crest]

One sec

Is it approaching 0?

[little drum]

uhh .. yeah that's about right

so your clue is that sin 0 = 0 < e^0, sin 1 > 1 / e

you just need to show that the graph don't intersect more than once in that interval

[still crest]

I don't think I'm supposed to do this with graphs

[little drum]

i-

Umm but the visualization comes that way

anyways, sin x is increasing in that interval while e^-x is continuously decreasing

So they can only intersect at most once, or not .. but sin 0 < e ^ 0 and sin 1 > e^-1 so they must've intersected once

hence only one solution

[still crest]

I understand this solution but that's not what the exercise wants. This set of exercises has exercises with derivatives

[little drum]

Uhh, you need the derivatives to show that the graph for sin and e^-x is increasing/decreasing in the interval!?

[still crest]

No I probably need some kind of theorem

[little drum]

Ig if you're curious about using a theorem... you could use the IVT THEOREM

Have you heard of Intermediate value theorem?

[still crest]

I found the solution in my high school book xD

I'm studying in uni

With bolzano's theorem you prove there is at least one root and with rolle's theorem you prove there can't be 2 solutions so there is only one

.close

[lone heart]

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[little drum]

smh so round about way to doing easy stuff

[lone heart]

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[median python]

I can’t figure out how to do this problem

[wary stream]

,rotate

[pale kestrel]

the top right root 2

is not root 2

[gray isle]

you fked up the last application of pythag you did

[median python]

Ye I know

I got the root of 5 would it be correct?

[gray isle]

for x?

[median python]

Ye

[gray isle]

yes

[median python]

Ok thx

.close

[lone heart]

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[median python]

.reopen

[lone heart]

✅

[median python]

I’m having trouble with this problem too

[pale kestrel]

,rotate

[pale kestrel]

There's a line you can draw

that will help you.

[median python]

Ye I found the answer

Tnx for the help

.close

[lone heart]

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[knotty saddle]

help =((((

[lone heart]

@knotty saddle Has your question been resolved?

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[marble bloom]

Can I please get help with number 17

[wary stream]

,rotate

[wary stream]

Like it says, use law of cosines to find the angles

[marble bloom]

I’m so confused man

[wary stream]

Don't ask in an occupied channel
Read #❓how-to-get-help

[wary stream]

Do you know what law of cosines is?

[marble bloom]

Yeah it’s where you have to set stuff equal to cosines like side side side and angle side angle right

[wary stream]

But the formula for it, do you know it?

[marble bloom]

Ngl no

There’s so many different formulas

Isn’t it like two of the sides squared minus the side across squared?

[wary stream]

There's literally a formula for law of cosines

[marble bloom]

What’s the formula

Do you know

[wary stream]

You can Google it

[marble bloom]

C= square root of a squared plus b squared minus two ab times cos angle opposite c

But the problem is there is not angle opposite c

There’s all sides

[wary stream]

You're finding the angle

[gray isle]

there an angle opposite every side in a triangle

[marble bloom]

How do I find the angle though

That’s what I’m wondering

[gray isle]

from the cosine rule/law

[marble bloom]

The Who now?

[gray isle]

there are 4 variables in the cosine law that represent,
3 sides (which are all known)
and angle (which you want to find)

[marble bloom]

Exactly

[wary stream]

Use the law of cosines

[marble bloom]

So now I need to find an angle

[wary stream]

To find that angle

[marble bloom]

Could you explain what that means

What is the law of cosines

[wary stream]

That formula you just looked up

That's law of cosines

[marble bloom]

But for that forumula

I need an angle

And I don’t have an angle so I can’t use that formula

[gray isle]

wdym

[marble bloom]

It says it needs angle opposite of c

[gray isle]

yes

[wary stream]

You have everything else but the angle, you're finding the value of the angle

[marble bloom]

This is the formula

What’s that symbol at the end telling me then

[wary stream]

That's just a variable

[marble bloom]

So what do I put next to cosine

[gray isle]

that's a pretty suboptimal representation of the cosine law for this

[marble bloom]

DOSENT an angle need to go next to the cosine

[gray isle]

in this case the gamma would be the opposite side c

[marble bloom]

Yes but there is none

Because it’s just sides

[gray isle]

which can simply be represented by the capital C

[wary stream]

That is the value you're looking for

The angle

[gray isle]

by conventional labelling

and/or explicitly introduce one yourself

[marble bloom]

Ok so I would put a C after the the cosine

But there’s already a little c in the problem

The little c is what we’re solving for

[gray isle]

no

you already know what little c is

[marble bloom]

Oh yes

It’s 17

So I would put 17 there then

[gray isle]

well yeh...since c=17,
you'd replace c with 17

[marble bloom]

So it’d look like this

[wary stream]

,rotate

[gray isle]

extend that radical bar pls

[marble bloom]

Ye I will

So now I just simplify

[gray isle]

and as mentioned before

that's a pretty suboptimal representation of the cosine law for this
which requires you to do more work when solving for (capital) C

[marble bloom]

So what I gotta do

[lone heart]

@marble bloom Has your question been resolved?

[alpine sable]

Uhh

Solve it

[marble bloom]

How do I solve it

U guys are supposed to help Aren’t u

Like what

[gray isle]

algebraic manipulation

consider first isolating cos(C)

(there is also a form of the cosine rule that starts with that already isolated)

[lone heart]

@marble bloom Has your question been resolved?

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[silent hornet]

i need to complete the rhombus, im not sure how, idk what formulas to use. P is the interception point of the diagonals

[silent hornet]

<@&286206848099549185>

.close

[lone heart]

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[woeful jasper]

I'm stuck. I'm trying to solve the thermal diffusion equation in spherical polar coordinates, where the diffusion equation is written in terms of r (point in space) and time t only.

[tacit arch]

What have you tried and mathematically what are you trying to prove? Pics are ok

[woeful jasper]

this is what im trying to get to , and ive got close but im out by a factor of 1/r and I can't figure out why

Because I havent worked with polars much I'm thinking I must've done something wrong during the separation of variables

[tacit arch]

Do you know the laplacian in spherical coordinates?

[woeful jasper]

Yes

for just r

[tacit arch]

Does the rest match the last eqn here
http://www.thphys.nuim.ie/Notes/MP469/Laplace.pdf

[woeful jasper]

oh sorry let me see

it doesnt match

[woeful jasper]

in my notes it says this

versus yours saying

yours gives 1/r

[tacit arch]

Cool

And since you don't need the angle coordinates, that should simplify pretty easily

[tacit arch]

Yea I don't know where this comes from, but you can multiply it by r or 1/r and it'll still be true

[woeful jasper]

How can I relate it into the rest of my working ?

[tacit arch]

Did you calculate both sides of the equation already and show that they're equal?

[tacit arch]

First eqn here

[woeful jasper]

yeah i separated the variables and made them equal to a constant

let me show you

[tacit arch]

I think you're doing too much

[woeful jasper]

i end with this

[tacit arch]

All you have to calculate is 2 things. The laplacian of T and the time derivative of T. Then you multiply the latter by 1/h^2 and set that equal to the Laplacian. This gives you an equation for h in terms of k and alpha

Somewhere you should be able to divide by T and only have constants remain

[woeful jasper]

Im not sure i understand

I've gotten this , where I get the variables separated and make them equal to a const

[tacit arch]

I don't think any of that is right

I suggest starting over

[woeful jasper]

Yeah I dont thjink so either as I've done it the way i would with cartesians

But I'm not sure where to begin with polars

[tacit arch]

Your derivatives look right

What did you get for the time derivative of T?

dT/dt

[woeful jasper]

ohh

i havent really got a time derivative of T

[tacit arch]

Stop doing separation of variables

All you're really doing is plugging in the second equation into the first

[woeful jasper]

I'm not sure what I ought to do

How I should start different

I don't understand why the laplacian is = 0 in the beginning

del2 T = 1/h2 dT/dt

suddenly making it = 0 doesnt make sense

[tacit arch]

This is very different from what you're initially asking

Read your question again and again

[woeful jasper]

my initial question is getting from the diffusion eqn in polars to a general solution

i got very close to the shown equation however I am out by a factor of 1/r

i do not understand where the 1/r comes from , so i figured it may come from some part of the polar coordinates

[tacit arch]

Did you calculate the right hand side of this?

[woeful jasper]

like this ?

[tacit arch]

Oh boy

Do this step by step

Write down T

Then multiply by r

[woeful jasper]

I dont have T

oh ok

right

[tacit arch]

What's rT?

[tacit arch]

Good

What's d(rT)/dr ?

[tacit arch]

Looks good. Now d^2(rT)/dr^2

[tacit arch]

After dividing by r, can you factor out T?

[tacit arch]

Close, the exponential is part of T

So just erase that altogether

[woeful jasper]

ah yes!

[tacit arch]

So you have the left hand side now. The right hand side is a little simpler

[tacit arch]

Wait wait

Start with T again

And calculate dT/dt

[tacit arch]

Now factor out a T again

[tacit arch]

Actually I think the 1/r is part of T here

[tacit arch]

Same here

[woeful jasper]

good point

[tacit arch]

Looks good. Multiply by 1/h^2

[tacit arch]

Yeaaaa

Set that equal to your laplacian

[tacit arch]

Divide by T and solve for h

[tacit arch]

Me thinks you're done

[woeful jasper]

my guy this is insane

[tacit arch]

[woeful jasper]

ive spent the past like 6 hours on this

maybe more

[tacit arch]

It's easy to get lost if your first path is in the wrong direction. We've all been there

[woeful jasper]

thank you

next i gotta create the general solution , but now i have the equations i needed

i hope this wont take as long 🥵

[tacit arch]

I think your calculation skills are pretty good. You just need to review this solution again and learn from it. Why did I tell you to calculate those things in that order? Why did separation of variables not help you?

[woeful jasper]

I think in my case I've been taught to begin with separation of variables and work that way, with no indication of what a general solution may be looking like.

So with this question it was generous for them to give T in that form

[tacit arch]

Yea if you do enough pde and ode problems, you get used to "Show this function satisfies the differential equation" and it becomes plug and chug. That's vastly different than showing "show there exists functions that satisfy this differential equation and that it's unique."

[woeful jasper]

Yeah that's impression I'm getting. I want to get more comfortable with spherical polars

[lone heart]

@woeful jasper Has your question been resolved?

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[fossil grail]

ok so, im not 100% sure but

I'm pretty sure it's 7

because

25000=a(r)^n-1

a is 100

25000=100(3)^n-1

250=3^n-1

3^5=243, so 3^6>250

6=n-1

n=7

not sure if this is correct though

[lone heart]

@fossil grail Has your question been resolved?

[tacit arch]

Where does 3 come from?

[fossil grail]

oh it says that it trebles

[tacit arch]

Your work looks right

[fossil grail]

oh ok

thanks

.close

[lone heart]

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[fleet bolt]

Just need to understand the method for these two

[fleet bolt]

https://cdn.discordapp.com/attachments/810625768685436968/932870029249245214/20220118_002928.jpg

https://cdn.discordapp.com/attachments/810625768685436968/932870168147820664/20220118_003249.jpg

3 and 4

[alpine sable]

It’s the centroid

[fleet bolt]

sorry should've said it more clearly, need to get to the values of x in both of those problems

@alpine sable

[alpine sable]

Ik

It’s directly related to property of centroid

Do u know how a centroid divides a median

In what ratio

[fleet bolt]

1:2?

[alpine sable]

2:1

[fleet bolt]

ah

so 3x+10 would be double that of 2x in question 3

[alpine sable]

Yes

[fleet bolt]

Thanks a lot for your help mate, it was right in front of me

[alpine sable]

No no

Wait

Double of 2x = 3x + 10

@fleet bolt

4x = 3x + 10

[fleet bolt]

oh no worries, I got that

just needed that slight reminder

[alpine sable]

Ok

[fleet bolt]

thanks again

.close

[lone heart]

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[gloomy portal]

<@&286206848099549185> I need help in math can somebody help please

[alpine sable]

What u need help with

[gloomy portal]

Factoring polynomials

[alpine sable]

Also don’t ping helpers immediately

[gloomy portal]

Oh okay

[alpine sable]

It clearly says after 15 minutes

[muted niche]

make sure to ping helpers with a minimum of 15 minute intervals

[alpine sable]

This is 4th time you have done this

Please be patient

[gloomy portal]

Okay

Um how do I factor these

I learnt the foil method the other day very grateful to learn to in first outer inner last

Hello epilson

[alpine sable]

I’m helping someone else rn

Wait

[gloomy portal]

Okay

[alpine sable]

Do u know middle term splitting or quadratic formula

@gloomy portal

[gloomy portal]

Yes

Hello

No

[alpine sable]

Bruh

[gloomy portal]

I am not familiar I tried learning that

[alpine sable]

Where are u getting these questions from

[gloomy portal]

I know a bit not much

Kumon

[alpine sable]

Did your teacher not teach you about quadratic equations before giving all this homework

[gloomy portal]

She does not teach well at all

[alpine sable]

Bruh

[gloomy portal]

Epilson your my teacher

Please help

[alpine sable]

Oof

Ok

This will be a long one to teach

[gloomy portal]

It’s fine I’m here all day

[alpine sable]

But I’m not

[gloomy portal]

Oh

[alpine sable]

I need to study myself

[gloomy portal]

Yes true

[alpine sable]

I can recommend u a text which u can read

[gloomy portal]

Ok

[alpine sable]

I can’t teach you

[gloomy portal]

Okay

Can you factor the first one as a example and show working out

I got photographic memory in photos

[gloomy portal]

Thanks I will try learning it

If I need help can I ask later

[alpine sable]

Don’t memorise

[gloomy portal]

Ok

I got quick question is the answer

(Q-3) (Q-15)

[lone heart]

@gloomy portal Has your question been resolved?

[gloomy portal]

No it’s hasn’t

[little drum]

Yes lol

Epsilon did write ( q - 3 )( q - 15 ) no?

Like there's, q² - 18q + 45
so, the gist "middle term splitting" goes like: you split the middle term, that is 18 here, into two numbers a, b such that a + b = 18 and a x b = 45

Unfortunately though, you'll have to practice a bit before you start immediately recognizing the numbers, like a = 15, b = 3 here..

And then you write q² - 18q + 45 = q² - 15q - 3q + 45 = q( q - 15 ) - 3( q - 15 ) = ( q - 3 )( q - 15 )

that's it

I'll give you b. too.. but the rest, you've gotta learn from your intuition :o
b. p² - 15p - 54 = p² - ( 18 - 3 )p + ( 18 x -3 ) = p( p - 18 ) + 3( p - 18 ) = ( p + 3 )( p - 18 )

[gloomy portal]

Is this correct guys

[proven rune]

ayo wy u spamming everywhere

choose one channel

.close

[lone heart]

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[muted island]

anyone kind enough to help me with this?

inequalities

i have no idea how its supposed to be done

[vale wigeon]

,rccw

[muted island]

Thank you

[vale wigeon]

god, the wording on this

[muted island]

Apologies though..

[vale wigeon]

you have nothing to apologize for

it's not your fault

it's the fault of whoever made this exercise sheet

anyway, the inequalities they give you take care of the left and right edges of your region.

[muted island]

Hahaa alrighty

[muted island]

How so

[vale wigeon]

do you know what the graph of just the inequality y ≥ x-8 looks like?

[muted island]

Im pretty new to graphs of inequality

So im not quite sure

[vale wigeon]

maybe you should review that before trying to assemble a region out of four inequalities

do you at least know what the graph of the equation y = x-8 looks like

[muted island]

No..

Mind walking me through?

[vale wigeon]

wait what

have you never graphed straight lines before

[muted island]

I had, a year ago. But it confuses me often. Is it okay if you explain it to me step by step ?

[vale wigeon]

idk how to explain it to you "step by step"

i'd need to basically teach you from scratch how to graph straight lines

[muted island]

If you have time for it.. sure

[vale wigeon]

no, i don't have time for it.

[muted island]

Alright. Thanks

.close

[lone heart]

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[alpine sable]

help

I know how to do the vector part, but i dont know how to identify it

[lone heart]

@alpine sable Has your question been resolved?

[alpine sable]

<@&286206848099549185>

[muted island]

the positive 3 means moving your object to the right side 3 units whereas the positive 1 you'll have to move your object upwards 1 unit

whereas negative integers does the opposite

@alpine sable

[alpine sable]

Thank U

[lone heart]

@alpine sable Has your question been resolved?

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[vapid crypt]

for cross product, axb and bxa, how do you know which cross product will give/result in a positive vector?

[vapid crypt]

for example, if I want the positive perpendicular vector, which order do I use? axb and bxa

[alpine sable]

Do you know about the right hand thumb rule?

@vapid crypt

[vapid crypt]

kind of

here's a picture

[alpine sable]

Assuming positive means up for you, axb is what you are looking for.

Because

Actually, tough to explain with text.

But try to totally grasp the thumb rule.

Or wait for someone else to reply.

[vapid crypt]

OK, do you have a video link for thumb rule

[alpine sable]

I haven't really watched any personally.

Only 1 but it was in my native language.

[vapid crypt]

OK i see

[lone heart]

@vapid crypt Has your question been resolved?

[vapid crypt]

also, the other way if I have the angle, which unit vector is n? (in the picture)

[lone heart]

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[willow gull]

Help

[sweet parrot]

i need help with inventory management.
so i have uncertain/unknown distribution for the demand: [100-20,100+20]

over period of 12 months i need to find:

best Q, r that will bring TC to minimum in the worst case of demand

if anyone has knowledge about this/guide/video/or even the correct name for this model that would be helpful , ping me please

[jagged imp]

this channel is taken

[jagged imp]

which question are you struggling with and what have you done so far?

[willow gull]

7a

[covert agate]

do you know what stationary points are

[vale wigeon]

how did you get 3e^(3x) = 0

and how did you go from that to e^(3x) = 1/3

[willow gull]

I genuinely dont know

[vale wigeon]

so you just wrote down that equation without knowing where it came from?

[lone heart]

@willow gull Has your question been resolved?

[willow gull]

I’m confused

[craggy salmon]

you have to take the derivative of $\frac{e^{3x-5}}{x^2}$

[ocean seal]

Dev Shah

[craggy salmon]

and equate that to 0

for finding stationary points

[lone heart]

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[uneven wedge]

How would I solve for a when a is a positive integer and:
2/3 = 1/a + 1/3a + 1/(a-1) ? Cross multiplication? Should I combine anything beforehand?

[gray isle]

you could consider multiplying both sides by the lcm of the denominators of all the fractions directly (or any common multiple, but lcm would be ideal)

combining your fractions into a single fraction first is fine too

there are multiple routes that can be taken, just make sure that the steps are mathematically valuid

[uneven wedge]

So combining the right side?

[gray isle]

you can do that first if you want

[hollow cairn]

(Here's how i solve it)

[lone heart]

@uneven wedge Has your question been resolved?

[lone heart]

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[covert agate]

do you know the 1st derivative test?

ok

do you know critical numbers?

what about derivatives..?

bruh

i need to know how much you know

before explaining

if you give up no one can help you

what’s that formula?

[lean spire]

ah

derivatives

[covert agate]

i dont think you’re best fit for learning stuff rn

maybe get some sleep first

[lone heart]

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[eager shale]

How does one get 12 1/2 from this

I'm supposed to solve length of the line DE

[gray isle]

you didn't set up your equation properly

x/40 is the ratio of the bases,
however 15/33 is NOT the ratio of the left sides of the triangles (as 33 is not the length of the leg of the big triangle)

[eager shale]

Yeah got it now

thanks for help

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[still crest]

I tried transfering it to the polar form but then I don't know what to do after I find the polar form

What do I do with the exponent?

[craggy salmon]

post that form too

[still crest]

Okay

One sec

[toxic pagoda]

Please help me

[craggy salmon]

aight, now turn the trig form into e^i(theta)

[craggy salmon]

#❓how-to-get-help

[still crest]

Oh, I have no idea how to that

Wait

e^(i*7π/18)?

[craggy salmon]

yess

now just distribute the exponent into both 4sqrt(2) and the exponent

[still crest]

(4sqrt(2))^120*(e^120 * 7pi/18)?

[craggy salmon]

yup

[still crest]

That was way easier than I thought

[craggy salmon]

now simplify if you wanna

[still crest]

Yep 1 sec

[craggy salmon]

it is pretty easy yeah!

[still crest]

(4sqrt(2))^120*(e^(140π/3))

[craggy salmon]

you can subtract 2npi from the exp without changing it's value

[still crest]

I have heard about 2kp

Is that the same

2kpi*

Ohhh I'm dumb

[craggy salmon]

yeah

:)

[still crest]

I thought pi as different letters and I thought it was imaginary unit

Lmao

Yeah I got it

Thank you very much

[craggy salmon]

I didn't have the symbol and was lazy for latex lol

No worries!

[still crest]

Yeah I'm on phone which also makes me lazy to use it

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[jaunty crane]

can someone help me with a simple math

[jaunty crane]

change the gradient x to (x+1)

[gray isle]

the instructions/goal is unclear

[proven rune]

what do u want m8 its unclear

[jaunty crane]

like they give 3logy=mx+n

hold up

change the 3logy=mx+n

into logy=...

any ideas?

find m and n is the question

[covert agate]

you get log y = m/3 (x) + n/3

and 10^(mx/3 + n/3) = y

[jaunty crane]

oh ok

what should i do next

[covert agate]

but thats a logarithmic equation?

[covert agate]

it’s a linear segment

wtf

[jaunty crane]

oh yea my y needs to be log y

the variables x and y are related by the equation 3logy = mx+n. The diagram below shows a straight line formed by plotting logy against x+1. Calculate the value of m and n

let me translate

there

(x, y) = (a - 1, 10^b)??

[covert agate]

ok how do i explain this

[covert agate]

3 log y = m (x + 1) + n - m right
so subbing those points,
3 (4) = m (4) + n - m, and
3 (10) = m (7) + n - m

there you have a system of equations

[jaunty crane]

oh i can just plug x+1 to the equation from the first place?

[covert agate]

they’re related in this fashion

so yea

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[earnest lodge]

I feel completely lost

[pale kestrel]

What have u tried

or what are you lost about

the entire thing?

[solar merlin]

(mainly) unrelated, but R.I.P Trifon Madas. he was truly a legend

[earnest lodge]

I've looked at the answer key

completely lost

I've been lost for the previous 12 questions

but I understood the answer and managed to solved it

but qn 13 is 😐😭

[pale kestrel]

Ok, notice anything about the Left hand side thing

[earnest lodge]

to infinity?

[pale kestrel]

Perhaps try writing out what the first few terms are...

But I think you are expected to recognise this is geometric series. (and use the formula for it)

[earnest lodge]

oh my god

I tried writing out the terms for the right side but not the left side

thank you

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@earnest lodge Has your question been resolved?

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