help-49

[dreamy lichen]

The problem is that it would be fucking stupid

literally

[next rover]

like here, but maybe this one is stupider, i can't tell

[dreamy lichen]

this one looks like it actually makes sense

[sinful condor]

If it helps, this is for a discrete simulation model

Trying to simulate revenue for different types of pax at an airport

And theres F&B visit probabalities

For pax_regions, pax_flight_type, and pax_travel reasons

[dreamy lichen]

do you know how many people are in bussines? How many are asian

in percents

[sinful condor]

Yes I do, Asians are at 17%

People in business is 8%

[dreamy lichen]

okay, this could be useful

is the percentage of asian in bussines the same as the percentage of non-asians in bussines?

[sinful condor]

Yes you can assume that

[dreamy lichen]

and do we know what percentage visits F&B

[sinful condor]

5% of business travelers

And 60% of asians

[dreamy lichen]

Okay, so to sum up, we have:
P(A) = 17%
P(B) = 8%
P(C | A) = 60%
P(C | B) = 5%
A and B are independent
We need P(C | A & B)
If anyone competent in probability can solve it with this, then go ahead

i will try to solve it, but im not very good at this so idk if i can do it

it might still be unsolvable

[sinful condor]

Ahh okay, thanks for trying

Looks I'll have to survey each type of passenger as well

.close

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[lost sphinx]

am i on the right track?

[lost sphinx]

i kind of forgot how to solve equations with 5 variables too

and with powers like these

[midnight plank]

@lost sphinx Has your question been resolved?

[lost sphinx]

<@&286206848099549185>

[spring wave]

i would use a different method, but i guess this should work

[lost sphinx]

what method?

i learnt this as lagrange multipliers

[spring wave]

first i would note down x^2 <= 5
then id subsitute y^2 = 5 - x^2/2 in the second equation to get
x^2 + 8z^2 = 6
then id do
xdx + 8zdz = 0
and
dx+2dz = 0

from here
x = 4z

so from here it is pretty easy

[lost sphinx]

so its rly all just about substitution

are my 5 equations even correct tho

[livid sorrel]

from 1st condition substitue y^2

put it in 2nd condition

get z in x

so u got a function in x

use derivate for maxima minima

[lost sphinx]

does this mean “substitute y^2 into condition 1”?

[livid sorrel]

ye, sorry

get y^2 in terms of x

then z in terms of x

then f(x), so f'(x)=0

[lost sphinx]

i might be blind but how does y^2 = 1/(2a+b)^2 fit into x = 1/(2a+2b) ?

[livid sorrel]

just use x^2+2y^2=10

and then from here, x^2+y&2+4z^2=8

[lost sphinx]

wait my 3rd equation should be 0 = 0a + 8bz

but so ur saying i dont need the first 3 equations?

just the 4th and 5th?

[livid sorrel]

leave everything

restart it

[livid sorrel]

this

2y^2=10-x^2

[livid sorrel]

put it here

get z^2

now get z = in terms of x

[midnight plank]

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[twilit field]

$\int_{0}^{2 \pi} \int_{1}^{\sqrt{e}} \ln(r^2) dr dt$

Would this be right

[grand pond]

What a wonderful world !

[dawn dagger]

yes

[twilit field]

Cool

thanks

so we have $y+2x=v;y-x=u$

[grand pond]

What a wonderful world !

[twilit field]

We then have v ranging from 4 to 7 and u ranging from -2 to 1

Now to find the Jacobean of this transformation , $\det{ 1& 2 \ 1& -1}$

[grand pond]

What a wonderful world !
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[twilit field]

The determinant of $\begin{bmatrix} 1 & 2 \ 1& -1 \end{bmatrix}$

[grand pond]

What a wonderful world !

[twilit field]

Which is -3

Fine so far?

,w factorise 2x^2-xy-y^2

[grand pond]

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[twilit field]

so we have $\int_{4}^{7} \int_{-2}^{1} -\frac{1}{3} vu dvdu$

is this fine?

oops

[grand pond]

What a wonderful world !

[twilit field]

is this fine

I think the sign of the jacobian is wrong

[dawn dagger]

the jacobian is positive

[twilit field]

so I solve for x and y and then differentiate 'em

right

,w solve y+2x=v;y-x =u for x and y

[grand pond]

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[twilit field]

so I have 1/3,-1/3, 2/3, 1/3 as the entries of the det in that order

right

so 1/3

[twilit field]

so the integral is $\int_{4}^{7} \int_{-2}^{1} \frac{vu}{3} dv du$

[grand pond]

What a wonderful world !

[twilit field]

-33/4?

yeah

thanks

[dawn dagger]

,calc 33/4

[grand pond]

Result:

8.25

[dawn dagger]

ok but it's not supposed to be negative

[dawn dagger]

there is a minus missing, not from the jacobian but from your integrand

(2x+y)(x-y) = -uv

[midnight plank]

@twilit field Has your question been resolved?

[midnight plank]

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[sacred gate]

quikquestion

[sacred gate]

why sen^2(x)=0 is the same as sen(x)=0

[fallow scarab]

yes

oh why

[kindred dagger]

Because when you take th esquare root of both sides of the first equation, it turns into th esecond equation

[sharp coral]

x^2 = 0 is the same as x = 0, same logic applies

[kindred dagger]

And you dont have to worry about + because __+__0 = 0

[dusty portal]

$\pm 0=0$

[grand pond]

;(

[sacred gate]

oh

okok

thx to all of you

[dawn dagger]

try to prove it

[dusty portal]

I could technically use modulous function and whatnot

Anyways

[kindred dagger]

-0 = 0
+0 = 0
???

[sacred gate]

if you have 5 apples and they took zero apples they took 0 apples

[dusty portal]

Yes

[sacred gate]

if you have 5 and they gave you 0 they gave you 0

[dusty portal]

Ggez

[sacred gate]

5-0=5

5+0=5

means

if n=0

5-n=5
5+n=5

+-n=n

nvm gotta lock in

.close

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[glossy pagoda]

I have a problem with this code in R

you can see by how its ploted, where the mean is about at the 20 x axis

but its wrong because i see this in the answer sheet

[lyric charm]

can you show the directions

[glossy pagoda]

An explanation of how to do it

[lyric charm]

oh sounds like theirs is just shifted 20 to the right from yours then lol

[tired osprey]

did you have x <- 20:60?

instead of 0:40?

[glossy pagoda]

yes

it didnt work

i will show what happens

[sharp jolt]

try entering the full command? like maybe the arguments size and prob are not really in that order

[glossy pagoda]

i know full command will work

[sharp jolt]

then it must be that the function args are not ordered that way

let me see the docs online

[glossy pagoda]

like the answer sheet that im using?

[sharp jolt]

there are several tutorials, documentations for programming languages online

the order of arguments seem right, plot function seems the culprit here

[glossy pagoda]

yh thats what i thought

but maybe its because its a negative binomial distribution and the question is asking like this

range

of 20 to 60 even though i use 0 to 40

[sharp jolt]

got it, plot function uses the following syntax:
plot(x1:xn, y1:yn)

if only plot(x1:xn) is passed, it will plot this:
plot(x1:xn, x1:xn)

[glossy pagoda]

so i just gotta make sure i add the x axis values as well

[sharp jolt]

so, in the plot shown in the answer sheet, x values are from 20-60 and y values are from dbinom function

in the one u used, the x values and y values are same, that's why they appeared shifted

[sharp jolt]

yes

[glossy pagoda]

oh

cool

[sharp jolt]

BTW, https://www.w3schools.com/r/r_graph_plot.asp

I took help from here

[glossy pagoda]

thank. u

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[empty ivy]

Hi, how do you do part a?

[fallow scarab]

integrate 1/t from 1 to x^{alpha/2}

and bound it below by the rectangle formed by [1, x^{alpha/2}] and the minimum of the function on that interval

[empty ivy]

Wait I don't get the second part. Wouldn't it be this and not 0?

[fallow scarab]

i don't know what you're referring 0 as

[empty ivy]

Because in the question it says show that the function is > 0

[fallow scarab]

that still doesn't answer my question

[empty ivy]

Wait I don't get what you mean by the 'minimum of the function'

[empty ivy]

so it is the blue dot?

[fallow scarab]

the integrand 1/t

[empty ivy]

Yeah I did that

I got the middle term

[fallow scarab]

show your work

[empty ivy]

Okay

[fallow scarab]

and be more clear and precise what you're asking

[empty ivy]

Okay sorry it's because I did try it and didn't get the terms they wanted

[fallow scarab]

yea i can't tell what you're doing

[empty ivy]

Yeah idk what im doing either

[fallow scarab]

you should start with the integral then derive the inequality bounds, not the other way around

[fallow scarab]

what is the result of the integration?

[empty ivy]

Well, I got alpha/2 logx

[fallow scarab]

and if x > 1 and alpha > 0, what's the sign of (alpha / 2) * log(x)?

[empty ivy]

Positive

Is that what you mean by sign-

[fallow scarab]

yes

sign of something is either positive or negative

[empty ivy]

Okay yep

[fallow scarab]

if something is positive is it > 0 or < 0 ?

[empty ivy]

0

[fallow scarab]

so you've proven this inequality so far

[fallow scarab]

this will give you the other inequality

[empty ivy]

Ohh

Okay thanks! I'll try continue from here

.close

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[frail bloom]

Hey, basically I need to give examples of two functions (f and g) to satisfy these conditions:

Can someone help me?

[frail bloom]

(e stands for and btw)

[fallow scarab]

try thinking of lines

oh

yea lines should still work i think

[frail bloom]

Ok, imma try it rn

[subtle blaze]

There’s a classic example even

[midnight plank]

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[frail bloom]

Ty guys, Idk how but I solved it, lol

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[tribal wharf]

have i graphed the piecewise function properly?

[kindred dagger]

You didnt use the domain correctly for 0, t <= 2

[tribal wharf]

how should i have done it?

[dusk pier]

Why does y=0 have a slope?

[kindred dagger]

Also why didnt you graph between 10 < t < 11

[dusk pier]

Straight line from -infinity to x=2

[tribal wharf]

oh ok

[tribal wharf]

i thought i did it the right way

[slender walrus]

assuming t represents time, 0 to 2, not -inf

[tribal wharf]

how should i have graphed that region

yeah t is time

C is cost

[kindred dagger]

t > 10 doesnt mean you jump from 10 to 11

[tribal wharf]

so from 0 to 2 it should be aline straight up?

[slender walrus]

no

[dusk pier]

Straight line from t=0 to t=2 (really the x-axis)

[slender walrus]

not up

[tribal wharf]

oh im stupid

ok i got that now

[tribal wharf]

then what is the proper way to graph that

[dusk pier]

You graph from t=10 onwards?

[kindred dagger]

Have you used open holes/closed dots?

[dusk pier]

There is not much to it

[kindred dagger]

Guys is C cont at t = 10

Im too tired to check

[dusk pier]

Nope

,calc 0.3(10-7)+2.5

[grand pond]

Result:

3.4

[dusk pier]

Yeah it isn't

[tribal wharf]

so when i made the rest of the points

i just subbed in the number

but for above 10

what number would i sub in?

[dusk pier]

t=10?

[tribal wharf]

but the 4th part is for t > 10

why would t = 10 work

[alpine quiver]

you have to put 10 but you exclude the point 10 because it isn't included

[dusk pier]

You find 0.3(t-10)+2.4 at t=10 then graph it from there

[dusk pier]

As an endpoint

I'm not necessarily saying it is included

[tribal wharf]

im going to draw it

again

[dusk pier]

Ok

[tribal wharf]

tell me if i do it right

continuing from 10 onward

i realise now that the green dot should have been open not closed

but besides that is it correct?

[dusk pier]

Yes.

[dusk pier]

👍

[tribal wharf]

tyyy

.close

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[alpine quiver]

(2,0) should be closed since it's in the first thing (0 t≤2)

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[unique cliff]

So I have a few problems here that I thought I understood pretty well but none of what I input would work

[unique cliff]

Idk if this is like a website issue or if I’m so lost that I can’t even see how I’m wrong but either I’m confused

[dusk pier]

Nope

Your first answer is incorrect

[unique cliff]

Ok how would I do it then cause what I did was just invert the graph, then find the x values on that inverted graph

[dusk pier]

You can just find the y-values that correspond

$x=f^{-1}(-4)\implies f(x)=-4$

[grand pond]

mathisfun

[unique cliff]

On the graph that’s there or on the inverted graph

[dusk pier]

On the graph.

[unique cliff]

So I find y=-4, and then the corresponding x value is the answer?

Cause I swear I tried that too and it said it was wrong

[dusk pier]

Yes

[unique cliff]

So then for the first one it would be 3.5 right? But when I put that in it says it’s incorrect

[dusk pier]

It's 3

Read it again

[unique cliff]

Would it not be this?

[dusk pier]

Are we looking at the same thing

[unique cliff]

I think so?

[dusk pier]

Yes we are

Yeah I have bad eyesight

[dusk pier]

It should be 3.5 then

[unique cliff]

Ok maybe I have to put them all in to tell me if it’s right or not let me try that

This website sucks like that

[dusk pier]

Probably

[unique cliff]

It’s says that one of these is wrong

Won’t tell which

[dusk pier]

The last one

[unique cliff]

HOLY THANK YOU SO MUCH 😭😭😭😭😭😭

Ok now this one

I just straight up don’t understand what that error means

[dusk pier]

Both parts are wrong

The vertex of f(x) is at x=-4

The inverse should satisfy the relation $x=(y+4)^2$

[grand pond]

mathisfun

[unique cliff]

Ohhh yeah cause it’s inside the parentheses right

So than how would I do the “containing 100” bit

Would that not just be positive infinity?

[dusk pier]

Hmm

[dusk pier]

Yes

[unique cliff]

Cause it says the -4 part is right so I got that but it says the infinity is wrong

[dusk pier]

Try this first to at least get credit

[unique cliff]

Didn’t work either

[dusk pier]

$y+4=\sqrt{x}$

[grand pond]

mathisfun

[unique cliff]

I got rid of that error by putting parentheses around the x+4 but something else is still wrong

Ok actually I’m gonna ask my teacher about it tomorrow and hope that she knows how this stupid website works

Thank you tho you did help a lot

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[twilit field]

I was thinking that this would require me to set up an ODE

[urban widget]

DE is so painful 😭

[twilit field]

trying to figure out the ODE at this point tbh

Because the ODE is $\frac{dy}{dx} =\frac{-y}{x} =t$

[urban widget]

@brisk widget hi Vipey wipey

[compact wraith]

[grand pond]

What a wonderful world !

[urban widget]

Viper could probably help you with this one

[twilit field]

which gives me $y = -tx+C$

[grand pond]

What a wonderful world !

[urban widget]

How’d you get that? (Cause I don’t think it’s integrating)

The +C

[twilit field]

integrating

solving for the curve passing via (a,0), we get

[brisk widget]

wouldn't it be -x/y?

[twilit field]

oops

yea

[brisk widget]

butuh

[twilit field]

but I'm not using that here, so it doesn't really change much

[brisk widget]

it doesn't seem like you can express x or y as functions of just t, i'm not seein how this parameterization works 💀

[twilit field]

so solving for the curve passing via (0,a) and t=0, we find C=a

so $y=-tx+a$ is the parametrixation

[grand pond]

What a wonderful world !

[brisk widget]

oh

[twilit field]

Does that work?

https://math.stackexchange.com/questions/3059756/parametric-equations-for-semicircle

I was half way there

I have to solve $x^2+(-tx+a)^2=a^2$ for x and then y

[grand pond]

What a wonderful world !

[brisk widget]

this aint uh

correct

cuz its dy/dx = -x/y = t

oh wait nvm im trippin

but no yea uhh

[twilit field]

hmm?

[brisk widget]

how did you get y=-tx + a?

[twilit field]

The differential equation makes sense, right

[brisk widget]

y'= -x/y makes sense

[twilit field]

y'=t makes sense too

[brisk widget]

yes

[twilit field]

and I know that the curve passes via (0,a)

[brisk widget]

yes

[twilit field]

so that gives y =tx+a

[brisk widget]

how

💀

[twilit field]

$\frac{dy}{dx} =t$

[grand pond]

What a wonderful world !

[brisk widget]

oh

[twilit field]

so y=tx+C

[brisk widget]

LOL alr yea i see now

[twilit field]

thanks

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[twilit field]

for (a) I was thinking $f(t)= (0,t), f:[a,b] \to R^2$

[grand pond]

What a wonderful world !

[twilit field]

for (b) the same parametrization , but from [-1,2] \to R^2

[rose trout]

Well in the first case your domain interval is really R, but yeah for b it’s from -1 to 2

[twilit field]

Yeah, but the issue is by book defines a curve as follow

[untold sentinel]

it doesnt cover a plane

x is always 0

[twilit field]

R isn't an interval, is it

[rose trout]

Usually it’s assumed that R is an interval

[twilit field]

yeah, I know

[twilit field]

Ah

okay

That doesn't make much sense though

[rose trout]

I mean R has all the same properties that proper intervals have

Usually the defining property is that for any two points in the interval, every point in between is in the interval

In other words intervals are connected subsets of R

[twilit field]

Now for $(c)f(t)=(t^2,t^2-1)$

[grand pond]

What a wonderful world !

[rose trout]

Hmmm

Well with this parametrization, you have y = x-1

[twilit field]

I was thinking x-2=t

oops

yeah, that's a big oopse

x-2=t \implies x=t+2; y= t^2-1

$(t+2,t^2-1)$

[grand pond]

What a wonderful world !

[twilit field]

from R \to R^2 again

now for (d). (x-1)=t \implies $y = \sqrt{4-t^2}+2$

[grand pond]

What a wonderful world !
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[twilit field]

$f(t)=(t+1,\sqrt{4-t^2}+2)$

[grand pond]

What a wonderful world !

[twilit field]

This only plots the top half of the circle, which is a probelm

I suppose a trignometric sub would be better here

$x=2sin(t)+1;y=2\cos(t)+2$

[grand pond]

What a wonderful world !

[twilit field]

where f(t) is a function from [0,2\pi] to R^2

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[molten bay]

Hello can anyone help me with the definition of interior of q is empty

[carmine sigil]

@molten bay topology?

If so it means, roughly and intuitively, that the set is "all boundary."

[molten bay]

Real numbers

[molten bay]

Boundary i see

So (0,1) is interior of (0,1) am I right?

[olive matrix]

yes

[molten bay]

[0,1] interior will be (0,1)

Wow thanks guys

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[grizzled glade]

does anyone know how to solve this or have an example

[old rampart]

you can set up a triangle and use law of sines or similar to find the angle

[molten bay]

Can I interrupt for a second?

[old rampart]

might be an easier way but idk lol

[molten bay]

Above question

[lethal path]

1 radius = 3 units
-1/3 radius = -1 unit
..... radiuses = -1.33 units

[old rampart]

well make your own channel

[carmine sigil]

@molten bay !occupied

Just remake a new channel

!occupied

[midnight plank]

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).

[grizzled glade]

i might b a lil slow but idk what to do w this

😭

[lethal path]

that's how to do part a

called the unitary method

[grizzled glade]

im slow

[lethal path]

https://www.youtube.com/watch?v=AXkCfkVrjK8

[midnight plank]

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[midnight plank]

@grizzled glade Has your question been resolved?

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[twilit field]

(a) = $\int_{a}^{b} \sqrt{2}dt$?

[grand pond]

What a wonderful world !

[twilit field]

so $\sqrt{2} ( b-a)$

[grand pond]

What a wonderful world !

[midnight plank]

@twilit field Has your question been resolved?

[twilit field]

The second one would be $\int_{0}^{k} \sqrt{ 2e^{2t}} dt$

[grand pond]

What a wonderful world !

[twilit field]

so $\int_{0}^{k} \sqrt{2} e^t dt = \sqrt{2} ( e^{k}-1)$

[grand pond]

What a wonderful world !

[twilit field]

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[dry zodiac]

Hi.. I'm struggling with this one integral question..

[dry zodiac]

I tried doing it by parts but I'm stuck

It says the answer is (a)

[tawdry laurel]

,rccw

[dry zodiac]

I cancelled the root x in the denominator

Sorry if it wasn't clear

[rose geyser]

dont cancel and take root(x) = t

u get 1/(2root(x)) dx = dt

integral t^2/(1+t^2) dt

[rose geyser]

yup

[dry zodiac]

How do I continue-

[rose geyser]

write num as t^2 + 1 - 1 and split into 2 terms

[dry zodiac]

Yess I'm getting tysm!!!!

[rose geyser]

[dry zodiac]

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[dry zodiac]

.reopen

[midnight plank]

✅

[dry zodiac]

How do you tilt it-

[rain wasp]

,rccw

[dry zodiac]

Tyyy

The ninth one.. I thought of getting it in the format of e^x(f(x) + f'(x)) so that's it's integral would be e^x(f(x)) but how would you do that

with the square

[rain wasp]

just distribute the square across the fraction, and only concern yourself with the numerator

can you show your work so far

[dry zodiac]

Did what you said.. now what do I do?

[rose geyser]

if you look carefully it is in the form you wanted

[dry zodiac]

Ahh you're right

Got it thanks again

Please help, I'm clueless how to even begin

[rose geyser]

the options are a big hint

[dry zodiac]

So I divide by x^6?

Or 9..

I'll try with 6 first

[rose geyser]

nuh (4x^2 + 1) to something like (4+1/x^2) how do you do that

[dry zodiac]

Take x^2 common, bring it out of the brackett as x^12

[rose geyser]

yup

[dry zodiac]

Ok.. what do i now

[rose geyser]

think of some substitution that could work

||again options are a big hint||

[dry zodiac]

1/x^2 = t

[rose geyser]

yeah that works too but t = 4+1/x^2 makes it easier

[dry zodiac]

Okay

I'm getting it! Thanks

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[clever karma]

I need some help getting comfortable with changing the order of integration for triple integrals.

[clever karma]

If we have a solid described by: $0 \le x \le 2, 0 \le y \le 2-x, 0 \le z \le x + y$

[grand pond]

johnseymour20

[clever karma]

How do we go about changing the order to x -> z -> y?

The x limits should be in terms of y,z. The y limits in terms of z, and the z limits should be constants.

[midnight plank]

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[clever karma]

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[twilit field]

$(gard f)(x,y) = (\frac{x}{\sqrt{x^2+y^2}}, \frac{y}{\sqrt{x^2+y^2}})$

[grand pond]

What a wonderful world !

[twilit field]

as for the second function

uh

$(\frac{ 2x}{(x+y)^2}, \frac{-2y}{(x+y)^2})$

Is this fine

[grand pond]

What a wonderful world !

[dawn dagger]

gard

[twilit field]

https://tenor.com/view/lord-of-the-rings-gandalf-indeed-gif-18505269

thanks

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[twilit field]

for (a), I'm getting (-y,-z,-x)

It's thus not irrotational

[lament knoll]

its good

for b) what you have ?

[twilit field]

I haven't tried it yet

Give me 2 minutes

the second one is 0

[lament knoll]

yes so this field is … ?

[twilit field]

irrotational

[lament knoll]

yes 👍

[twilit field]

The third one doesn't look pleasent to compute

Lem'me think if there's a trick to it

[lament knoll]

yeah lol i didn’t try

[twilit field]

I'mma come back to this in a bit

This doesn't look pleasent in the slightest

[lament knoll]

it’s not that hard, after the calculation the components cancel each other out

[twilit field]

hmm, okay

so it's irrotational, right

[lament knoll]

yes it is

[twilit field]

the next one is (0,0,2x) and is thus not irrotational

[lament knoll]

for d)?

[twilit field]

ye

[lament knoll]

There is one more factor, what’s the result of $\frac{\partial}{\partial x} \frac{x^2}{2}$ ?

[grand pond]

tm

[twilit field]

x

[lament knoll]

yessir

so what’s the curl for d) ?

the 3rd component

[twilit field]

x

[lament knoll]

yes !

[twilit field]

Thanks!

[lament knoll]

i’ll let you try for e)

[twilit field]

I suspect (e) is irrotational

[lament knoll]

yesssirrr

The partial derivatives will all do 0

[twilit field]

Yea

Thanks so much!

[lament knoll]

don’t run away like that, did you calculate the curl for the c)?😂

I think they take the Euclidean norm?

[twilit field]

yea

It should be 0

[lament knoll]

yes, what’s the result of $\frac{\partial}{\partial y} \frac{z}{\sqrt{x^2+y^2+z^2}}$ ?

[grand pond]

tm

[lament knoll]

or $z \frac{\partial}{\partial y} \frac{1}{\sqrt{x^2+y^2+z^2}}$ if you prefer

[grand pond]

tm

[twilit field]

$\frac{-yz}{ 2\sqrt{x^2+y^2+z^2}}$

[lament knoll]

you forgot a } at the end

[grand pond]

What a wonderful world !

[lament knoll]

i have $\frac{-zy}{r^3}$

[grand pond]

tm

[twilit field]

hmm

[lament knoll]

$\frac{\partial}{\partial y} \frac{1}{r}= \frac{\partial}{\partial y} (x^2+y^2+z^2)^{-1/2})=…$?

[grand pond]

tm

[lament knoll]

apply the derivation rules for $x^n$

[grand pond]

tm

[twilit field]

yea

$-\frac{y}{(x^2+y^2+z^2)^{3/2}}$

[grand pond]

What a wonderful world !

[lament knoll]

yess so $\frac{\partial}{\partial y} \frac{z}{\sqrt{x^2+y^2+z^2}}=-\frac{zy}{r^3}$

[grand pond]

tm

[lament knoll]

then, what’s the result of $\frac{\partial}{\partial z } \frac{y}{\sqrt{x^2+y^2+z^2}}$

[grand pond]

tm

[twilit field]

-zy/r^3

[lament knoll]

yes, so the first component of the curl is $-\frac{zy}{r^3} - (- \frac{zy}{r^3})=0$

[twilit field]

got it

[grand pond]

tm

[twilit field]

similarly for all other components

[lament knoll]

yeah it’s practically the same method, a little boring to do😂

[twilit field]

Yea

Thanks

[lament knoll]

np you welcome !

[twilit field]

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[cinder rain]

Wtf does this mean

[midnight plank]

Please don't occupy multiple help channels.

[cinder rain]

The awnser says 89

[sullen jungle]

the minimum y you can get by plugging in some value of x

so basically yeah differentiate

[cinder rain]

Huh but then how do you even solve it

Anwser literally says 89

Like huh

<@&286206848099549185>

[last slate]

change it into a A^2 + B^2 + C^2 + ... form

[sullen jungle]

ye was thinking about that too

[cinder rain]

Wdym

[cinder rain]

I'm js curious on how 89 is the anwser

[last slate]

try completing the square

and see what you get

[paper inlet]

you can take the partial derivative of both and then solve the system of equations

[cinder rain]

💀 wut

[paper inlet]

wait this doesnt work

nvm

[dusk pier]

@cinder rain Conic section

[cinder rain]

?

[midnight plank]

@cinder rain Has your question been resolved?

[last slate]

set t = x + 3y

should get a quadratic

[cinder rain]

I'm rn using chatgpt and I'm getting confused

[last slate]

?

[cinder rain]

Nvm

[pearl hull]

@cinder rain have you solved it yet?

[cinder rain]

I guess?

Uh

[pearl hull]

Do you prefer English or 中文?

[cinder rain]

Honestly

English

I suck at chinese

[pearl hull]

alr, have you understood the logic behind this

[cinder rain]

Nop

For me i js squared it like they said and got

[pearl hull]

Ok so

Have you squared it yet?

[cinder rain]

(2x + 6y)²

• 164x + 492 y + 1770

Idk how to square this part

My first idea was like

164 (x + 3y)

• 1770

But then idk

[pearl hull]

You can also simplify 164x + 492 y

[pearl hull]

I don’t think that’s a correct simplification to 164x + 492y

[cinder rain]

😭

[pearl hull]

Try again

[cinder rain]

Uh

._.

Idk honestly

[pearl hull]

Just do it again

You just had a mistake

[cinder rain]

Uh 2(82x + 246y)

[pearl hull]

More, it’s not fully simplified yet

[cinder rain]

Can't it be js 164 ( x + 3y)

[pearl hull]

Sure, but it doesn’t match with this one

You should take back a 2 back into the parentheses

[jagged pilot]

try swaping 82 and 2

[cinder rain]

Uh

So

82 ( 2x + 6y)

[pearl hull]

Yes, that’s all you need

[cinder rain]

Huh

._.

[pearl hull]

Now, you get

(2x+6y)^2 + 82(2x+6y) + 1770

Find anything familiar?

[jagged pilot]

yea i can feel smth coming up

[cinder rain]

._.

Uh

Uhhhh

No 😭

[cinder rain]

Oh wait yea

82(2x+6y)³ + 1770

[jagged pilot]

no

[cinder rain]

TvT

[jagged pilot]

ax^2 + bx + c = 0

[cinder rain]

Oh so we js make 2x+6y into x

[pearl hull]

Do you know the standard form of a quadratic function?

[cinder rain]

Wdym? The x = -b +- square root 4 ac /2a?

[pearl hull]

Set 2x+6y as t, so as not to mess up

[pearl hull]

No, I mean how do you find the minimum or the maximum of the function

[cinder rain]

-b / 2a ?

[jagged pilot]

me personally i suggest u to try with this

[cinder rain]

But then I get 2 x square root of 89 i

Cus it becomes 6724 - 7080

Making x = square root of 89 x i

[jagged pilot]

this formula is little off tho

[cinder rain]

( -b + or - square root of b² - 4ac ) / 2a

[jagged pilot]

yeah i got (i ) as well...so i think this is the long way ..maybe the other guy knows the short answer

[cinder rain]

Yea so unless I'm able to uh make it so x is able to be times by 89 and divided by i

Xd

[jagged pilot]

yeah we got whole another page of math if u do that

[cinder rain]

Xd I think I maybe can get the awnser using chatgpt

[jagged pilot]

yeah do it and send me the answer as well

[cinder rain]

I think I got it?

V:

Uh

[cinder rain]

Basically make the 2x+6y as z

So it's z² + 82z + 1770

And then this can become

82 / 2 = 41

So

(Z + 41)² - 41² + 1770

So it's 1681

So -1681 + 1770 = 89

So it's (Z+41)² + 89

So 89 the anwser

Somehow

💀 I thought this was unsolvable 1 hr ago

[jagged pilot]

ohhh it did that

[cinder rain]

Yipee time to sleep cus it's 1 am

.close

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[jagged pilot]

dang bro aight gn

[grizzled wadi]

@everyone

im ge

[midnight plank]

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[modern shard]

$\int^{1/2}_{-1} \frac{e^x(2-x^2)}{(1-x)(\sqrt{1-x^2})}$

[grand pond]

Vansh

[modern shard]

well I didn't do anythig much, tried what I can do

but looked dead end everywhere

[grim vector]

, rotate

[modern shard]

what I thought:

tried to see if it was e^x(f(x) + f'(x)) form (it wasnt)

then

i tried to use property to see if e^x by chance get eliminated but nope

if anyone thinks of any idea, please lmk by pinging me

I will try that!

||PS: it would hurt me if anyone sees internet for direct solution, please try yourself||

[modern shard]

<@&286206848099549185>

[umbral timber]

see if its an odd or even function 🤔

also try to make a substitution

[polar star]

does not matter

the bounds are not symmetric

[polar star]

the first approach looks promising

[modern shard]

I just solved it

break it down

2-x² as 1+1-x²

and it will be converted to e^x(f(x) + f'(x))

thanks for your time tho, i appreciate both of you :))

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[main glade]

Can someone explain why 6mm^2 is 6m x 10^-6 m and not 10^-3 m ?

[arctic vapor]

hELP HOW TO SOLVE THE RADIUS HE JUMP!

<@&286206848099549185>

[glad compass]

bro 💀

[paper inlet]

seems too difficult

[glad compass]

1mm=$10^{-3}m$

[grand pond]

therealtdp

[glad compass]

just squaring both sides,

[arctic vapor]

i still haven't get the answer

[glad compass]

1mm$^2=10^{-6}m$

[grand pond]

therealtdp

[main glade]

Ok thanks

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[lunar pivot]

the solution says the singularities "cancel" but what does that mean?

because $\mathcal{F} g(0) = \int_{-\infty}^\infty g(t),dt = \frac{1}{2}$

[grand pond]

carburetor

[lunar pivot]

but then if the imaginary part is $(\sin(2\pi s) - 2\pi s) / (4\pi^2s^2)$, clearly the F.T. evaluated at 0 isn't equal to $1 / 2$

[grand pond]

carburetor

[lunar pivot]

<@&286206848099549185>

[midnight plank]

@lunar pivot Has your question been resolved?

[midnight plank]

@lunar pivot Has your question been resolved?

[midnight plank]

@lunar pivot Has your question been resolved?

[midnight plank]

@lunar pivot Has your question been resolved?

[midnight plank]

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[grave kernel]

dont understand where im going wrong
in the solution theyve given the reciprocal of the set of equations im using, but idt thatll play any role with that and also the notation is opposite
like theyve used lambda for M and mu for N
but other than that cant figure out where im going wrong cuz theyre getting
2 and 4 as answer while in mine i have a 1/3 factor

the relation of
u = 2L
is also fine
theyve got 2u = L
theres something wrong with the last part i suppose?

[gaunt imp]

Could you post the original problem?

[grave kernel]

its on the paper

on top

thats the given info

a line passes through (-1,2,3) and intersects 2 lines
L1: (x-1)/3 = (y-2)/2 = (z+1)/2 at M (alpha, beta, gamma) and
L2: (x+2)/(-3) = (y-2)/(-2) = (z-1)/4 at N (a, b, c)
Find ((alpha + beta + gamma)²)/((a + b + c)²)

[gaunt imp]

Well, that's quite the strange problem, they usually would want you to find the line or find a plane/another line satisfying some conditions, for which I'd try to form the direction vector and define the line through a point

But in this case that clearly won't work since your intersections are parametrized

[grave kernel]

yeah i mean thats me doing that
assuming a general point on the lines

and then applying the given condition

which then gives me two equations to solve for them but im going wrong in the solving part i suppose

[gaunt imp]

What are you doing here though? Mostly confused with the notation at first

[grave kernel]

well i "found" A and N and M

and then i find AN and AM

like the vectors

the vectors are parallel cuz the lie in the same line

and hence they should be proportionate

so which is why ive divided them

took the ratios

[gaunt imp]

How do you define parallelism between vectors?

[grave kernel]

a = kb where a and b are vectors

and k is a constant

[gaunt imp]

Isn't that linear dependence?

[gaunt imp]

And also, dividing vectors?

[grave kernel]

dividing the coefficients i mean, sorry

[grave kernel]

kinda?
but im not using 2 vectors to write a 3rd one
its more like
if i have a vector
3i + 2j
then
6i + 4j is also parallel to it right

[gaunt imp]

If two vectors lie in the same line, there'll exist such a constant k, yeah

But a priori, you don't know which scalar it is, how did you find such coefficients to divide them?

[grave kernel]

well if a = kb
a/b = k
all of the coefficients' ratios will be equal to k and so i equate all the coeffs

[gaunt imp]

It's more a thing of me never hearing of anyone define parallel vectors but well, guess we'll go with that

[grave kernel]

i mean, lets just go with that they lie in the same line for now

and hence im establishing that

[gaunt imp]

Didn't you say a and b are vectors? They're each linearly dependent to a third one by a unique coefficient

Well, lemme check this out rq

[grave kernel]

i mean yeah but does that matter?

[gaunt imp]

Yeah, don't worry about that, I think I got you wrong, I think I now get what you're doing

[grave kernel]

cuz like if i define a as ai + bj + ck and b as xi + yj + zk
and theyre parallel(lying in the same line)
so its just
ai + bj + ck = K(xi + yj + zk)
now if i take ratios of the coeffs or i and j and k
theyre equatl to K
a/x = b/y = c/z = K

thats what im doing for this

except ive just assumed a general point in a weird way and finding the vectors for them and then ratio-ing

[gaunt imp]

Alright, so you've assumed the line where the intersections lie of the form (-1,2,3)+λ(a,b,c), where (a,b,c) is the direction vector, written the intersections in the form N=P+uv and M=P+zv

Am I right?

[grave kernel]

its either the solution is wrong or i am, i dont understand

wait lemme check

[grave kernel]

well yes, youre right but ive not found them this way though

ive found the general points from the given lines itself, not using the point given

[gaunt imp]

Oh, alright, so you've written them in terms of the general point of the line they lie on

[grave kernel]

mhm

and then if i apply the given conditions, theyll match up

[gaunt imp]

Then taken the vectors PN and PM and equated their coordinates?

[grave kernel]

not equated

proportionated, idk

but yeah

and then from then i solve the parametrics

[gaunt imp]

Well, equated the coordinates of each one to a scalar k times the other

[grave kernel]

but im getting a 1/3 factor

[grave kernel]

yes

[gaunt imp]

Alright, I've finally figured out what you've done, ok, the idea seems correct

[grave kernel]

right

now the solving part

[gaunt imp]

When it comes to that the mistake is almost always just doing a sum wrong or unintentionally swapping a sign

I'd check the calculations first, then, if they're all right, I'd check how the solution went about the problem, and if their solution makes sense

[grave kernel]

they got this(notations a bit different)

[grave kernel]

well i did check the calculations

and i couldnt figure out what else to check, so i asked

lemme check again

[gaunt imp]

Check the last factor, the -2+4λ

You've taken the vectors in the opposite direction the solution chose, so you should be getting exactly what they're getting multiplied by -1 in the system

That -2+4λ is the only thing you calculated that doesn't match what the solution got multiplied by -1

[grave kernel]

my god.

it should be +2

wait no

1 + 4λ - 3

[gaunt imp]

Well, I'm fairly sure the mistake is there

Now, why you've gotten something totally different, I don't really know since I'm not sure which choice of direction you've taken

Or whether you've confused the direction vectors just for that one specific factor

[grave kernel]

wait i believe the solution is wrong

[gaunt imp]

Well, that may be possible too

[gaunt imp]

Is this the whole solution?

[grave kernel]

look at M
the last point is -2λ - 1
but if we take the general point
the last point is -1 + 2λ

its that or i copied the question wrong

[grave kernel]

yeah i mean you just need λ and u
after that its just adding and subtracting and squaring

[gaunt imp]

Well, guess you can't check whether the solution makes or doesn't make sense then

And yeah, if the question is as you've copied it

[grave kernel]

yep
i copied the question wrong🤡

in L1
its (z+1)/-2
not (z+1)/2
😭

sorry to bother you sm😭

[gaunt imp]

Welp, it's always stuff like this, don't worry about it

[grave kernel]

thank you so much though

[gaunt imp]

Don't mention it, hope I could help you

[grave kernel]

mhm

.close

[midnight plank]

Channel closed

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[midnight plank]

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[dusk pier]

$\int_0^{2\pi}\sqrt{1+\sin(x)}dx$

[grand pond]

mathisfun

[dusk pier]

Someone tell me if my method is right

$\int_0^{2\pi}\sqrt{1+\cos(x-\frac{\pi}{2})}dx$

[grand pond]

mathisfun

[dusk pier]

$\int_{-\frac{\pi}{2}}^{\frac{3\pi}{2}}\sqrt{1+\cos(u)}du$

[grand pond]

mathisfun

[dusk pier]

$\sqrt{2}\int_{-\frac{\pi}{2}}^{\frac{3\pi}{2}}\cos(\frac{u}{2})du$

[grand pond]

mathisfun

[dusk pier]

$2\sqrt{2}\sin(\frac{u}{2})\bigg\vert_{-\frac{\pi}{2}}^{\frac{3\pi}{2}}$

[grand pond]

mathisfun

[dusk pier]

So just $2\sqrt{2}\qty(\frac{1}{\sqrt{2}}-(-\frac{1}{\sqrt{2}}))$

[grand pond]

mathisfun

[dusk pier]

Which is $2(1+1)=2(2)=4$

[grand pond]

mathisfun

[dusk pier]

Yes or no?

@tribal temple Something wrong?

[tribal temple]

Remember sqrt{y^2} is |y|, not just y

[pine thicket]

,w fnInt(sqrt(1+sin(x)),x,0,2pi)

[grand pond]

Results provided by WolframAlpha

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Requested by locallunatic8

[dusk pier]

Bruh

[lament knoll]

how u went from this

[lament knoll]

to this

[pine thicket]

half angle

[dusk pier]

I'm cooked

[lament knoll]

oh yeah mb

[tribal temple]

(you can of course split the integral up into the "positive" and "negative" bits to take account of that, then easy-ish steps )

[dusk pier]

Ok

New method I'm learning

I will do this on own

.close

[midnight plank]

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• Show your work, and if possible, explain where you are stuck.
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[plucky prairie]

how do i turn this into a system of eqations

[slender walrus]

start by introducing variables to represent
number of adults
number of senior citizens

its recommended that you use ones that resemble/relate to what they represent

[plucky prairie]

so a and s

$20a and $10s

[slender walrus]

what do $20a and $10s represent

[plucky prairie]

20 dollars per adult and 10 dollars per senior

[slender walrus]

that would be incorrect

[plucky prairie]

what would be correct

[slender walrus]

$20/adult is the given price,
20a is different from that

$20a is
the price for 1 adult ticket
multiplied by
the number of adults

which will be the total from adult tickets

(you'll need this value, you weren't clear about what it actually represented)

same idea for $10s

can you try setting up your equations now?

[plucky prairie]

yeah i dont knwo how to

[slender walrus]

which will be the total from adult tickets
do you understnad that?

[plucky prairie]

no

is there soppoussed to be text on that grey line

[slender walrus]

no, i'm just quoting part what i typed earlier

[plucky prairie]

oh

[slender walrus]

$20a is
the price for 1 adult ticket
multiplied by
the number of adults
which will be the total from adult tickets

[plucky prairie]

yeah i get that

and 10s is the price of the senior tickets times the number of seniors?

[slender walrus]

yes

[plucky prairie]

so is it 20a + 10s = 5730 or something like that

[slender walrus]

yes

and your other equation will be for the total number of tickets

[plucky prairie]

and so if i wanna find only the amount of adults and senior tickets withoiut the price is it a + s = 477

[slender walrus]

yeh

now that you have your equations,
take your pick of sub or elim

[plucky prairie]

so
20a + 10s = 5730
a + s = 477

lemme elim and see if i get the right answer

did i write the equation wrong

i just got 381

oh yeah

447

not 477

ok so i still did it right

thanks

.close

[midnight plank]

Channel closed

Closed by @plucky prairie

Use .reopen if this was a mistake.