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[coarse elm]

mhm

[oak chasm]

[P'(t) = -500 e^{-\log(2)}]

[ocean seal]

Chai T. Rex

[coarse elm]

ya but now i get -370

😢

oh if i use log

if i use In then I get -250

[hushed locust]

i think when they say "rate of decay" they mean something other than dP/dt, probably dP/dt normalized by something like the initial value

[hushed locust]

for this formula "log" means natural log

[coarse elm]

Yupp

[oak chasm]

Hmm, I get -250.

Oh, I see.

We divide it by 100.

Because we want the rate of decay, not a percentage.

That gives us -2.5.

So:
c(t) = e^(-5t)
c'(t) = -5 e^(-5t)
c'(log(2)/5) = -5 e^(-5 log(2)/5)
c'(log(2)/5) = -5 e^(-log(2))
c'(log(2)/5) = -5/2
c'(log(2)/5) = -2.5.

[coarse elm]

ohhh

i see

that makes sense then

alright thank you so so so much

@oak chasm

[oak chasm]

You're welcome.

[coarse elm]

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[tiny rune]

A student is taking a MCQ test. The test has 100 questions, for each correct answer he is awarded +4 for each wrong answer he is given -1 and for each question skipped he is given -4. Find total number of ways a student can solve the test to score exactly 300 marks.

[lone heart]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[tiny rune]

2

[tawdry urchin]

What have you done so far?

[tiny rune]

x+y+z = 100 and 4x-y-4z = 300

after that i am unsure how do i do this

when my teacher gave it there was 0 marks for z so it became easy

i solved for values of y and z

for all values of x

[tawdry urchin]

what did you get when you did that

[tiny rune]

that was a different question

i thought what if there was -ve marks for skipping

how would i solve it

[wispy plaza]

?

[tiny rune]

the orginal question had 0 marks for skipping

so the second eqn had 2 variables

so i just found possible values of x

and for each i found y and z

and then it was easy to slove

but now my second eqn has 3 variables

[wispy plaza]

Sorry for the delay, I’m in the airport

Normally this system has infinite solutions

[idle thorn]

Wow so cool question

[wispy plaza]

But because your x y and z are all positive integers there are finite solutions

[tiny rune]

ya

[idle thorn]

Have we found a solution to solve that question yet btw

[tiny rune]

i sent this to my frnd and said its some diophantine eqns or something and sent me this image

[wispy plaza]

You should try and eliminate one of the negative only variables

[tiny rune]

ok

ok so if i take y = 100-x-z and put it in second eqn we get this

ok w8

so z must be a integral multiple of 5

for x to have a

integer value

so z = 5k

x = (400+15k)/5

y = 100-5k-80-3k = 20-8k

now what do i do

[silent zenith]

nga

[tiny rune]

nga???

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[oak chasm]

Still here?

c + w + s = 100
4c - w - 4s = 300

We have two equations, so we can solve two variables in terms of the third:

c + w = 100 - s
4c - w = 300 + 4s

-4c - 4w = -400 + 4s
4c - w = 300 + 4s
-5w = -100 + 8s
w = 20 - 8/5 s

c + w = 100 - s
c + 20 - 8/5 s = 100 - s
c = 80 + 3/5 s

Which is about where you were.

We know that s must be a multiple of 5, as you found out.

Now we also know that c and w must be nonnegative integers. With 0 <= c = 80 + 3/5 s, we can't really get anywhere because the inequality is true for all nonnegative s values, but with 0 <= w = 20 - 8/5 s, when s gets large enough, the the inequality is no longer true.

0 <= w = 20 - 8/5 s
0 <= 20 - 8/5 s
8/5 s <= 20
8s <= 100
s <= 100/8 = 12.5

So, since s must also be nonnegative as well as a multiple of 5, how many multiples of 5 are there in [0, 12.5]?

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[unkempt quiver]

I need help and learn with Calculas using for electronics and circuit designing

[thick basalt]

Is there a specific problem you need help with or are you trying to learn calculus

[unkempt quiver]

no, I need to learn from basic for electronics

[thick basalt]

if you need to learn the basics you could search up calculus videos on youtube

3blue1brown on youtube has a good overview series

it explains what calculus is, then you can go to more class-like videos to actually learn the specifics

anyways im pretty sure this channel is for specific problem helping

but yeah i would go to yt

[lavish cave]

https://ocw.mit.edu/courses/18-01sc-single-variable-calculus-fall-2010/

https://tutorial.math.lamar.edu/Classes/CalcI/CalcI.aspx

two good resources

[winged pawn]

better off with books on calculus as well 🤔

yeah that

[lavish cave]

also try OpenStax textbooks

[winged pawn]

for electronics and circuits i feel like calc 1 might not be sufficient 😔

[lavish cave]

they need to start somewhere

also MIT has calc 1 + calc 2 combined in that page

[winged pawn]

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[digital magnet]

any ideas as to how I submit this?

I've tried so many things and nothing can be understood

nvm lol

.close

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.reopen

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✅

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[native lagoon]

I solved for alpha and beta and got alpha=-2,+2 and beta=-sqrt(2),+sqrt(2)

[native lagoon]

I could answer the first three options, but I dont understand what the 4 option wants to say

[oak chasm]

It's asking for the count of integers that work as k.

Each k has to produce a root above and a root below beta_2.

And it's saying that the count of ks that work is 5.

[native lagoon]

Ohk okay

so like if k=1, x=-sqrt(10) and sqrt(10), so we have to coount K, right?

[oak chasm]

It beta_2 is strictly between -sqrt(10) and sqrt(10), then that k counts.

[native lagoon]

yh beta_2 is sqrt(2)

[oak chasm]

Oh, OK.

[native lagoon]

so, k=0,1,2,-1,-2

yh thats 5 integers

Thanks!

[oak chasm]

You're welcome.

[native lagoon]

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[hoary nimbus]

How do we do part b)?

The markscheme simply ||computed det(A)|| but i don’t know why that works?

[vale wigeon]

det(A) gives you the factor by which all areas are stretched

since your transformation just stretches along one direction and that's that, the area scale factor equals its stretch factor

(imagine a rectangle with one side parallel to y=mx being subjected to this)

[hoary nimbus]

What does ‘stretches along one direction’ mean?

I can’t visualise it

Is it a shear?

[vale wigeon]

no

do you know / can you visualize stretches along the x or y axes

[hoary nimbus]

Nvm ok i think i can visualise it

[hoary nimbus]

Is this what a square(blue) looks like after(yellow) being stretched in the direction y=x?

I’m not sure I visualise it properly

[vale wigeon]

nope

you do not just drag one corner out

[hoary nimbus]

Wait, will it be like enlargement wrt origin?

[vale wigeon]

you end up with a rhombus actually bc points on the y axis don't stay on the y axis, nor do points on the x axis stay on the x axis

[vale wigeon]

no

[hoary nimbus]

Um so, is it like this?

[vale wigeon]

kinda but the origin should stay at the origin.

also again. the axes are not stationary

[hoary nimbus]

Then how will it be a rhombus?

[vale wigeon]

you seem to be thinking that the points (0,1) and (1,0) are fixed by the stretch

they aren't

[hoary nimbus]

You mean the x and y axes themselves move?

[vale wigeon]

sure

[hoary nimbus]

I don’t get it

[hoary nimbus]

How would that look like?

[vale wigeon]

let me try to give you a sketch, one moment

[hoary nimbus]

If I wanna visualise it without the axes themselves moving, is this finally correct?

Or would the corners satisfying the equation y=-x squish out outward instead of inwards?

[vale wigeon]

they would not squish at all

the origin stays at the origin also

[hoary nimbus]

So like what kind of transformation do we say happens to the point (0,1)?

(0,1) —> (0.5, 1.5)
(1,0)—-> (1.5, 0.5) so what’s the rule for each coordinate ?

But (1,1), which was along the line y=x, had both it’s coordinates doubled, became (2,2)

[vale wigeon]

i could tell you the matrix for this particular one happens to be $\bmqty{1.5 & 0.5 \ 0.5 & 1.5}$ but honestly i wanted to just give you the visual here for what you were drawing

[ocean seal]

Ann

[hoary nimbus]

Oh wait, do we just take component of (0 1) along y=x and add THAT to (0 1)?

Component is 0.5root(2) and we just added that to the vector (0 1) right?

[hoary nimbus]

Yeah that makes sense

[hoary nimbus]

Thank you so much I think I got it

[hoary nimbus]

I don’t understand how we would understand that the factor by which area changes would be equal to factor by which length of the vector changes

Without calculating the individual areas

[vale wigeon]

ok then let me try to make a more illustrative picture...

the letter inside the rectangle is meant to be S to stand for its area. i bungled it.

[hoary nimbus]

Awesome, thank you so much again, I finally got it

.close

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[zenith urchin]

Hello! I'm currently proving something about topology in lean4, and I'm stuck on a small detail.
I've noticed that I can't do the proof manually either.

Let X be a set with two topologies: T1 and T2.
The topologies are homeomorphic, and T1 <= T2.
I need to show that T1 = T2.

I would appreciate some help!

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@zenith urchin Has your question been resolved?

[zenith urchin]

<@&286206848099549185> im still stuck

[solemn seal]

@old osprey your good at topology

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[restive mist]

whats the formula for calculating the derivative of fraction

[crimson tendon]

the quotient rule of derivatives?

[proper tangle]

if you have a new question, use an available channel, instead of dumping it in an occupied one

[zenith urchin]

The statement is impossible to prove because it is false

.close

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[scarlet talon]

This is Calculus 2, determining absolute convergence, conditional convergence, or divergence of a series

Does anyone see where I made my mistake? I'm confused why it is absolutely convergent and not conditionally convergent when my Ratio Test was inconclusive

[scarlet talon]

AST is the Alternating Series Test

[restive mist]

sybau u pmo icl

[hushed locust]

if the ratio test is inconclusive then it doesn't tell you anything. it could converge conditionally, it could converge absolutely, it could diverge

[scarlet talon]

If my ratio test is inconclusive, but still passes the alternating series test, then it could still converge absolutely?

[hushed locust]

at this point you know it converges, but you don't know whether it converges absolutely or not

[scarlet talon]

Is there another way to find absolute convergence if the ratio test is inconclusive? I only remember the root test, but I don't think that works here

oh wait, limit comparison test

that can tell absolute convergence right? because it's the limit of |an / bn|

If it fails the limit comparison test then can I say that it's conditionally convergent?

[hushed locust]

limit comparison test is only for positive series

here i think it is useful to remember the definition of absolute convergence: $\sum a_n$ is absolutely convergent if $\sum \abs{a_n}$ converges

[ocean seal]

cloud

[scarlet talon]

so I can just do a regular convergence test on the absolute value of my series, which is arctan(n)/n^2, and if that converges then our series converges absolutely

ok from there I can just use the Comparison Test with 1/n^2 since thats a convergent P series that behaves similar to that

and by the p series/integral test, P > 1, so that converges

meaning our |an| converges, so theres the proof for absolute convergence

ok i see now

[hushed locust]

i think limit comparison would be good for showing this formally

[scarlet talon]

lim n→∞ (arctan(n)/n^2) / (1/n^2)

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[magic snow]

how do i do this

im kind of lost in vectors

[hushed locust]

i would start by writing ma + nb as a single vector

[magic snow]

like this?

[vale wigeon]

this is pretty far off the mark sorry

[ivory igloo]

this is (m+n)(a+b)

[vale wigeon]

(also the brackets are missing around the vector components)

(you can't drop those)

[magic snow]

yeah i have no actual clue

unless you just mean

[vale wigeon]

(also the brackets are missing around the vector components)
(you can't drop those)

[ivory igloo]

somewhat

[small lance]

2a + b = 5
3a + 2b = 1
Just see it as two equations

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[vale wigeon]

m and n not a and b

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[rich jolt]

hi! i hope this is counted as a math question, but i've been trying to decode a message encrypted using the vigenere cipher. i've gotten to the part where i count the number of coincidences in order to find the key length. i'm struggling w finding the pattern though

[rich jolt]

sending the coincidence amt below

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[amber isle]

20: an earthquake was felt up to 86mi from its epicenter. Your house is located 62mi east and 48mi south from the epicenter. How far away from the outer boundary of the quake do you live

[amber isle]

ik the quake radius is x^2 + y^2 = 86^2

and the house is (62,-48)

but how do i calculate the distance from the point to the circle

[vale wigeon]

calculate the distance from you to the epicenter first

[amber isle]

ohhh

and minus that from 86 the radius?

[vale wigeon]

yes

[amber isle]

ok thx

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[waxen glen]

can someone tell me what 8e-05 is as a 1/blank?

[waxen glen]

<@&286206848099549185>

[twilit veldt]

1/8e05

[winter light]

!15min

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Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[waxen glen]

ok

like for example

0.01 is 1/100

what would 8e-05 be?

[shadow coral]

u sure about that?

oh yeah

[waxen glen]

im stupid

[winter light]

1/(1/(8e-5))

[waxen glen]

thats why im here

[shadow coral]

do u know what e means in this context?

[waxen glen]

no

[shadow coral]

e means 10^ something

here at least

[waxen glen]

oh ok

[shadow coral]

acc, lemme check smth rq

[waxen glen]

is there a % for that?

8e-05

[winter light]

What do you mean?

[waxen glen]

i failed alergy 1

[winter light]

8e-5$= 8\cdot 10^{-5}$

[waxen glen]

i must be very lucky

[ocean seal]

Alberto Z.

[shadow coral]

you can start by just finding out what 8 * 10^5 is

[winter light]

@waxen glen

[waxen glen]

0.00008

wow

[shadow coral]

and then use index laws to figure out what happens when u make the power (in this case 5) negative

oh, u already got the decimal answer

[waxen glen]

im in prealerga

[shadow coral]

ok, so if u have something like 2^2 = 4, if u make the power negative you get 2^-2 = 1/4

[waxen glen]

oh

ok

[shadow coral]

so if u find out what 8 * 10^5 is, you can just do the same thing

[waxen glen]

thanks a lot

im so fucking stupid

[shadow coral]

dw 😭

[waxen glen]

.close

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[spice python]

hey

[spice python]

why they use P(X >= 12)

I don't get how they got 12 from the question

[vale wigeon]

they calculated the tail probabilities for X until they found when it crosses through 0.05

[spice python]

MM

i have brain damage

so they did 1-P(X <= n) < 0.05

until they found a valid n

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[kind atlas]

If someone is moving 800 ft in 6 seconds how do you calculate the speed

[tacit arch]

speed = distance traveled / time spent traveling

[kind atlas]

Thanks

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[long spade]

Hello

pls help with question 5

[wary fable]

translate @long spade

[north rover]

x ^ 53 just means

?!???!

Latex...

$x^{53}$?

[ocean seal]

;(

[true rune]

no

it means gcd of x and 53

[north rover]

Oh

Well you can handle this one lol

Oh wait.

This isn’t too bad, is it?

[true rune]

arithmetic is the most horrible thing in maths

[true rune]

i have no idea lmao

[north rover]

Modular arithmetic works nice here, for a proof by contradiction maybe?

[true rune]

last time i made modular arithmetic was 3 years ago 😭

you’re alone on this one 🙏🏼

[true rune]

$2491 = 53 \times 47$ donc $x^{1198}-2x+1 = 0 \ \text{mod} 53$ et $x^{1198}-2x+1 = 0 \ \text{mod} 47$, puis suppose que $pgcd(x,53) \neq 1$ ça veut dire que $x$ est divisible par $53$ et abouti à une contradiction, fait de même pour $47$

[ocean seal]

tm

[true rune]

finally, yes lol

[long spade]

its okay I aleady solved it

[true rune]

a ok

[long spade]

I used Bezout

[true rune]

parle français mdr

[long spade]

theoreme de Bezout

[true rune]

comment t’as utilisé bezout la 💀💀

[grave matrix]

en reduisant l'indice ou qch ?

j'ai pas d'idée

[true rune]

pareil

[grave matrix]

Attends, tu parles du Théorème de Bachet-Bézout, ouais ?

(ffs pourquoi les noms ne sont pareils entre les langues )

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[lone heart]

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[wheat ridge]

Let X be any topological space, and Y be a contractible space, with contracting homotopy h. Prove that X and $X \times Y$ are homotopy equivalent by using h to construct a homotopy inverse to the inclusion X by $x \rightarrow (x, \ast)$. where $\ast$ is the point to which h contracts Y.

[ocean seal]

Zerofall

[wheat ridge]

I have an idea of what to do already, I made two maps

$i : X \rightarrow X \times Y$ and $p : X \times Y \rightarrow X$

[ocean seal]

Zerofall

[dense hornet]

So, what exactly are you asking for help on

[wheat ridge]

I am struggling with the final homotopy

[dense hornet]

So, I’m assuming i(x) = (x, c) where c = *

[wheat ridge]

I know that $p \circ i $ is homeomorphic to Id_x

[dense hornet]

And p(x, y) = x, right?

[wheat ridge]

And I know that i o p is homotopic to the Id_{X x Y}

[wheat ridge]

Yes

[wheat ridge]

And yes

So I have H((x,y), t) = (x, h(y,t))

But thats really just to show that X x Y is homotopic to the Id X x Y

[dense hornet]

So you know there’s a homotopy to the identity on both compositions?

[wheat ridge]

Yep

Im just struggling on the final homotopy equivalence between the spaces

[dense hornet]

So that’s a homotopy equivalence already?

[wheat ridge]

oh

[dense hornet]

That pair of homotopies is it?

[wheat ridge]

Yeah lol

[dense hornet]

Namely, p(i(x)) = x

[wheat ridge]

I got caught up thinking that because they were Id on different spaces they weren't homotopy equivalent 😭

[dense hornet]

and a homotopy for the other is indeed H((x,y), t) = (x, h(y, t)) I think

[dense hornet]

Obviously it’s the identity on different spaces

[wheat ridge]

Yeah that one was easy

[dense hornet]

You aren’t proving Id_X is homotopic to ID_{X x Y}

[wheat ridge]

Thank you so much

[dense hornet]

That doesn’t “make sense”

[wheat ridge]

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[barren tree]

Question: Use ratio test to determine whether the series is
convergent or divergent.

My work (in the second image): I know how to set up the ratio test, however, the notation for the denominator confuses me.

[hushed locust]

it's the product of all the odd numbers from 1 up to (2n-1)

which if you've ever heard of it is (2n-1)!!

[barren tree]

so is that just a factorial property i would just know/memorize?

[hushed locust]

!! is a double factorial, not factorial applied twice

you're not expected to have heard of it which is why they didn't use that notation

basically the double factorial is like the factorial but only multiplying the odd numbers, it skips all of the even numbers

[barren tree]

ah so if i wanted only even, would it just be (2n)!!

[hushed locust]

yes

[barren tree]

ok, thank you very much!

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[foggy vector]

Check that the following implicitly defined functions are solutions of the corresponding differential equations

i've done this but i've reached $2x^2y+y^2=c$, not $x^2y+y^2=c$

[ocean seal]

ransik (gmdn)

[foggy vector]

this is what chatgpt has to say, but i'm not understanding honestly

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@foggy vector Has your question been resolved?

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@foggy vector Has your question been resolved?

[north rover]

!nogpt

[lone heart]

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

[foggy vector]

no it was actually pretty helpful

it's been helping me a lot with my work

phyisics, chemistry and math

the only problem was it was trying to use partial derivatives which i have not studied yet

but its fine because it's helping me work it out

[north rover]

So is it know that it is a solution?

[foggy vector]

$x^2y+y^2=c$ is a solution, yes

[ocean seal]

ransik (gmdn)

[foggy vector]

by differentiating and integrating the solution i was able to reach the implicitly defined function

.close

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[raw jetty]

let $ABC$ be a triangle where $AB<AC$, let $H$ be the orthocenter and let the bisector of angle $BAC$ intersect the circumcircle with diameter $AA'$ at point $M\neq A$, let there be point $X$ on $AM$ such that $HX\perp AM$, let $P$ and $Q$ be the intersections of $HX$ and $MA'$ to $BC$ respectively, prove that $PMQ$ is an isoceles triangle

[ocean seal]

skissue.in.a.teacup

[cyan cargo]

Thank God, you've alr done the hardest part

[raw jetty]

i have some ideas that could possibly work:
-proving PC bisects XPM
-proving BP=CQ
-proving that if you extend AH so that it intersects the circle at D, DAM=A'AM (and also DA'//BC)

i couldnt find how to continue them all though

[buoyant minnow]

usually, when I see orthocenters, I think of the circle of nine points

[raw jetty]

i dont see how to use it?

[buoyant minnow]

sorry, when I get stuck I try to think of all formulas and things I can do, I thought maybe you could remember something out of it

[lone heart]

@raw jetty Has your question been resolved?

[lone heart]

@raw jetty Has your question been resolved?

[west girder]

if u reduced it to the last point it just follows from the fact that orthocenter and circumcenter are isogonal

circumcenter lies on AA', orthocenter lies on AH

[raw jetty]

isogonal?

[west girder]

yeah

like

<BAH=<CAO

[raw jetty]

how fo you know that

[west girder]

just angle chase

on a new triangle

or similar triangles

BAD similar A'AC

where D is AH intersect BC

[raw jetty]

oh

ok ty

.close

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[livid dust]

According to the WHO MONICA Project the mean blood pressure for people in China is 128 mmHg with a standard deviation of 23 mmHg. Assume that blood pressure is normally distributed.

Round the probabilities to four decimal places.

It is possible with rounding for a probability to be 0.0000.

a) State the random variable.

Select an answer

b) Find the probability that a randomly selected person in China has a blood pressure of 104.3 mmHg or more.

c) Find the probability that a randomly selected person in China has a blood pressure of 163.8 mmHg or less.

d) Find the probability that a randomly selected person in China has a blood pressure between 104.3 and 163.8 mmHg.

e) Find the probability that randomly selected person in China has a blood pressure that is at most 70.5 mmHg.

f) Is a blood pressure of 70.5 mmHg unusually low for a randomly selected person in China?

Why or why not?

Select an answer

g) What blood pressure do 73% of all people in China have less than?

Round your answer to two decimal places in the first box.

Put the correct units in the second box.

[tacit arch]

Try to ask a shorter question

[vale wigeon]

which part(s) of this are you struggling with

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@livid dust Has your question been resolved?

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[hushed spear]

can someone tell me how to solve integral with absolute value please!

[oak swallow]

sure mate

let's take a sample question shall we?

[hushed spear]

yes

[oak swallow]

alrighty

lemme see if I can find one online

$$\int_{-3}^4 |x^2 - 4| dx$$

[ocean seal]

Klein Bottle

[oak swallow]

does this work mate?

[hushed spear]

yes i see

[oak swallow]

,w plot |x^2 - 4|

[ocean seal]

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[oak swallow]

let's look at the graph shalll we?

okay as you can see, the graph has 2 x-axis intercepts

whatever is below the x-axis is shift above the x-axis

therefore we must integrate between the following intervals
--> -3 to -2
--> -2 to 2 (but we must add a negative sign so that this bit becomes positive since it was initially below the x-axis)
--> 2 to 4

I believe you must know how to integrate a polynomial function already?

therefore I am just going to use wolfram

,w integrate x^2 - 4 between -3 and -2

[ocean seal]

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[oak swallow]

,w integrate -(x^2 - 4) between -2 and 2

[ocean seal]

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[oak swallow]

,w integrate x^2 - 4 between 2 and 4

[ocean seal]

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[oak swallow]

notice how for this section, I used the negative sign since that area has kinda be "shifted" to being above the x-axis

now adding them up

,w 32/3 + 32/3 + 7/3

[ocean seal]

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[oak swallow]

let's verify

,w integrate |x^2 - 4| beween -3 and 4

[ocean seal]

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[oak swallow]

and voila! that's that!

they match

also i've been saying "shifted" above the x-axis.....I believe the correct term to be used here is "reflected" about the x-axis

[hushed spear]

i dont understand why -2-2 is negative and 2-4 is positive

<@&286206848099549185>

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[floral sky]

On my 30th Python script trying to fix this problem. Been two weeks. Any help will be greatly appreciated.

[floral sky]

<@&286206848099549185>

I have done everything, math model, a posteriori, everything and somehow I keep getting diverging answers in both excel and in python. I can provide further information and files upon request

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[limber prairie]

I’ve had a go at both of these questions but haven’t gotten the given answers. Where have I gone wrong?

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@limber prairie Has your question been resolved?

[undone goblet]

did you simplify the branches or try to uncomplicate the algebra?

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@limber prairie Has your question been resolved?

[limber prairie]

I'm not sure how to do that

[limber prairie]

.close

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[carmine topaz]

how do I solve part d?

[lone heart]

@carmine topaz Has your question been resolved?

[wind cloak]

What properties of asymptotic behaviour do you know?

[carmine topaz]

like there will be horizontal aysmptote at y=some number if lim x--> +-inf f(x) = that number

[wind cloak]

yeah

you want the x coordinate to approach infinity

or -infinity whatever

can you write an integral to express the x coordinate

[carmine topaz]

x coordinate?

[wind cloak]

well yeah we would like x to approach +/- infinity

and use that value on f

[carmine topaz]

do i use the y(t) or y'(t)?

$$\lim_{t \to \infty} y(t)$$

[ocean seal]

soo

[carmine topaz]

like this?

[wind cloak]

nah

there's gonna be some f such that y(t) = f(x(t))

[carmine topaz]

$$\lim_{t \to \infty} x(t)$$

[ocean seal]

soo

[wind cloak]

we need x(t) --> +/- infinity

infinity doesn't mean infinity here in the regular sense btw, just the ends of your function

[carmine topaz]

ok

whats next?

[wind cloak]

this is not necessary

again

[golden badger]

@carmine topaz

see if it is correct

i trief

[lone heart]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[golden badger]

tried

[carmine topaz]

$$\lim_{t \to \infty} x(t) = \infty$$

[ocean seal]

soo

[carmine topaz]

like this?

but why does x(t) have to approach inf?

[long sail]

it doesnt, im pretty sure we're evaluating the limiting behaviour of y(x) as it flattens out, not x(t)

[carmine topaz]

why is y expression like that?

[carmine topaz]

ok

where does the lower limit 2 comes from?

[long sail]

@golden badger

[golden badger]

hello

[long sail]

where did your lower limit 2 come from

if you got it from previous parts of the question i'm not sure if it's related

[golden badger]

bec at time t=2, the object is at the point (6,-3)

then we know the value of y(t) at t=2

and y(2) = -3

[carmine topaz]

so we are using the position given + the improper integral?

[golden badger]

Yep

[carmine topaz]

but i kinda dont get the idea of expressing horizontal aymptote y=c using improper integral

[golden badger]

Okay, so basically, the horizontal asymptote is where the graph chills out as time goes on forever, right?

[carmine topaz]

yea

[golden badger]

We know where the graph starts at t=2 (y=-3).

The integral is like adding up all the tiny little changes in the y-value from that starting point all the way to infinity.

So, the asymptote is just: “Where we started” + “All the changes that happen after that”

[carmine topaz]

ok so we are just expressing the lim t--> inf y(t) using improper integral cuz its heading to inf right?

[golden badger]

Which translates to: Asymptote = -3 + (Integral from 2 to infinity of dy/dt)

[golden badger]

yep

[carmine topaz]

ok i get it

thank you 😀

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[vivid breach]

can someone help?

[lone heart]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[vivid breach]

1

[wheat isle]

do u know how to approach improper integrals in general

[vivid breach]

no, its a new topic being taught in college and im behind the class

[zinc haven]

https://tutorial.math.lamar.edu/classes/calcii/improperintegrals.aspx

[long sail]

do you know how to integrate this normally

with completely the square + arctan form

[shell agate]

So

[long sail]

actually for the first integral you don't need to do that

[shell agate]

I think for the first one

[long sail]

notice where the asymptotes are

[shell agate]

We can get partial fractions

2nd one completing square?

[long sail]

yeah

[shell agate]

Yeah so 1st one partial fractions and 2nd one completing square

[long sail]

oops i forgot to mention partial fractions

[shell agate]

@vivid breach

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@vivid breach Has your question been resolved?

[lone heart]

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[slow island]

For question 2, would this be a correct triangle?

[slow island]

Please don’t leave me hanging like last time

[winter light]

Nope

The line (with the vertical).between the observer and the plane at point Q is not 50° in your drawing, rather it's 90°

[slow island]

So it would be this?

[winter light]

Huh? This is even worse

[slow island]

My apologies

[winter light]

How can that angle be 90°?

[slow island]

I misread it

[winter light]

Alright, don't worry

[slow island]

Would the 50 degree angle be at the top?

[winter light]

It depends on how you orient your drawing

[winter light]

Still wrong

You really have to imagine to stay at point O and watch the plane above you to the right and creating an angle of 50°

[winter light]

Because here the angle they tell you to be 50° is still 90° in this drawing 😅

[slow island]

The second part helped me realize what I was doing wrong

The plane is still gaining elevation

Is this better?

[winter light]

Yep now it's correct

[slow island]

How’s this?

Wait no that’s not right

[winter light]

Yeah because there's not a right triangle

[winter light]

Yep awesome

I haven't checked the calculations but the numbers seem pretty reasonable 👍

[slow island]

Got it

Thank You

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[next stratus]

guys my teacher told that (a) is the correct answer here

[next stratus]

but isn't that the general differential eqn for ellipses with centre at origin

how is the restriction passing through point(0,3) taken into account

[void nymph]

uhhh

show work i guess

[next stratus]

But like instead of b^2 he put 9

So if its wrong, do you know how to solve it @void nymph ?

[next stratus]

<@&286206848099549185>

[next stratus]

.

can smn pls help 😭

[tacit arch]

(0, 3) is a point on the ellipse so it satisfies the equation in blue. x = 0 so that means 3^2 / b^2 = 1

[next stratus]

yeah i get that but the differential eqn doesn't represent ellipses passing through (0,3)

it just represents and ellipse with centre as origin

[tacit arch]

why do you say this

[next stratus]

because I can do the same with b^2 instead of 9, and i'll still get (a)

[tacit arch]

right. solutions to differential equations aren't unique. they form a family of solutions.

the question gives just one solution in the family

[next stratus]

but the question specifies the family of ellipses which passes through (0,3), is it just not possible to do that??

like can we make such an eqn in the first place?

[tacit arch]

what

family refers to the ellipses

the "and passing through..." refers to the specific ellipse

[next stratus]

thats actually a family since you can vary a

btw i got the answer

[tacit arch]

oh yes

[next stratus]

i just had to eleminate a

(d) is what im getting

[next stratus]

thanks for the help anyways!

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[formal hornet]

Is there a way to do D without knowing how to do double integrals?

Ie the jacobian

[tacit arch]

definition of beta function

[formal hornet]

please elaborate

[tacit arch]

https://en.wikipedia.org/wiki/Beta_function

[formal hornet]

you cant just send a wikipedia and expect me to know what ur talking about

theres literally so many definitions and if you're talking about the first one then the way to prove the RHS is to do a double integral

[tacit arch]

why

[tacit arch]

it's given in the wiki

did you bother looking for it

if you know the definition of beta function and you know you're supposed to get something with gamma functions like it asks in part D, this is a <5 minute read

[vale crag]

you just need to use part c) really

[tacit arch]

you might not understand every step right away, but you should work that out yourself

[formal hornet]

bro i said i want to know a method without a double integral

i already read the wiki article and it has this

[formal hornet]

i litreally did and also its gvien in the problem too, i just want to know if theres a method without a double integral and you throw me some page and the only method involving it is a double integral

[formal hornet]

could you give me a hint or first step please?

[tacit arch]

right my bad i missed that you didn't know how to do double integrals

looks like you need to plug this into the right side of the eqn in d)

[formal hornet]

i see, ill try myself first then thank you

[tacit arch]

and probably use identities a)-c)

[lone heart]

@formal hornet Has your question been resolved?

[vale crag]

going fine or you want more hint ?

[formal hornet]

im going to eat rn but im having trouble getting the denominator

i dont mind if u just tyoe the solution here and i can look at it later

[vale crag]

go eat and type your solution afterwards, we'll see what you have first

[lone heart]

@formal hornet Has your question been resolved?

[formal hornet]

,tex Suppose I(a,b) = \frac{\Gamma(a+1)\Gamma(b+1)}{\Gamma(a+b+2)}

for a = 0, and b = 0:

I(0, 0) = 1

\frac{\Gamma(1)\Gamma(1)}{\Gamma(2)} = 1

assuming this is true for all of a, b >=1

prove for I(a, b+1) and I(a+1, b)
its so much to type but it does cancel out

and give us the identity

[ocean seal]

micha
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

Suppose $I(a,b) = \frac{\Gamma(a+1)\Gamma(b+1)}{\Gamma(a+b+2)}$

for $a = 0$, and $b = 0$:

$$I(0, 0) = 1$$

$$\frac{\Gamma(1)\Gamma(1)}{\Gamma(2)} = 1$$

assuming this is true for all of $a, b \geqslant 1$ 

prove for $I(a, b+1)$ and $I(a+1, b)$
its so much to type but it does cancel out

[formal hornet]

how did you do that

[tight pier]

the answer to that is contained among the infinite decimals of the number π

[formal hornet]

https://tenor.com/view/sus-funny-meme-emoji-jar-gif-6413585594020144302

[tight pier]

also is this a different task or still related to c)

[formal hornet]

d

[tight pier]

seems platypus is back anyway

see ya

[formal hornet]

bye, thanks!

[formal hornet]

so is this approach right?

[vale crag]

"assuming this is true for all a,b >= 1" well you have nothing to prove then, gg
I'm just being picky on language

well I did a different induction so I can't really tell if your approach would work rn

unless you spill the beans

[formal hornet]

i used B and it worked

[formal hornet]

ok ill send the writing one sec

[vale crag]

yeah a pic is fine, no worries

[formal hornet]

is this even allowed

[vale crag]

yeah it sounds super off

I don't see how your induction even makes sense here

[formal hornet]

i feel like im just using an identity we proved to

like write it different

yeah i think its wrong and idk how to do induction with 2 variables

[vale crag]

like ok you start with I(0, 0) why not

[formal hornet]

i think i need to do IBP

[vale crag]

but then what's your induction hypothesis that allows you to know I(a+1, b) and I(a, b+1) ?

[formal hornet]

this? and b)?

im not sure

perhaps induction inst the right approach?

[vale crag]

induction works, just a sneakier one, here's my approach

just do induction on b, then for all b you show it for every a

[vale crag]

i.e. the base case consists of checking that I(a, 0) fits that formula for every a [and I(a, 0) is easy to compute]

[formal hornet]

so i ignore a?

how does this work

[vale crag]

then for the induction case, you assume you know the I(a,k) for all a, then you want to show that I(a, k+1) fits the formula for all a

with a picture

this is the base case

then the induction case consists of computing the next row from the previous one

[formal hornet]

so its like a double induction? you prove for a first, then b

[vale crag]

you prove first row, then second from first, then third from second, etc...

so it's a lot like a single induction, but for each step you're showing a whole row

[formal hornet]

i get it

[vale crag]

part c) works really well with this induction cause c) essentially gives you info like this

[formal hornet]

alr thanks then

[vale crag]

you can compute I(0,1) from I(1,0), I(1, 1) from I(2, 0), etc...

the fact that you go from a to a+1 doesn't matter if you've shown the whole previous row

[formal hornet]

so first, i would prove that I(a, 0) and then I(a+1, 0) ... etc is true, then continue with I(a, k) and I(a, k+1) .... etc

[vale crag]

yeah

[formal hornet]

i see i see

thank you!

.close

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[tepid goblet]

So

Basically

I’m asked to state the range after finding whether it is a function or not

And function (h) is a function and I’m having trouble locating the range.

Am I wrong if I say this is a restricted function and it cannot continues to go upwards?

[tight shuttle]

You’re not wrong

[tepid goblet]

So

The range should be

-1<_y<_8

Am I correct?

[tight shuttle]

Yes that’s write

[tepid goblet]

Thanks

A lawt

☺️

[tight shuttle]

$-1 \leq y \leq 8$

[ocean seal]

plantsyavi

[tight shuttle]

There we go haha

[tepid goblet]

Is it ok if I have other questions?

[tight shuttle]

Yup that’s fine

[tepid goblet]

K 2 sec

Ima grab it

If a question tells u f(x)=2x+5 , X e r

What does the x e r means

[simple sonnet]

f(x
3
−x+4)=x
6
−2x+x
2
+1

[tight shuttle]

Start a new channel

[tepid goblet]

Hm k

.end

[tight shuttle]

Not youu

The other guy

[tepid goblet]

Oh ok

[tight shuttle]

But do you mean

[simple sonnet]

I'm in exam right now pls help plsss

I'm new here

[tepid goblet]

Sorry, I’m new there ^^

[tight shuttle]

$ x \in R $

[glacial patrol]

<@&268886789983436800>

[tepid goblet]

Hmm. We both joint today lol.

[glacial patrol]

No exam help :p

[tepid goblet]

Bro’s tryna cheat

[simple sonnet]

Okay, sorry

[tepid goblet]

I hate cheaters so bad

☺️All of my class cheats except me and I’m still first

[tight shuttle]

$ x \in R $ means the x values must be real

[polar root]

Flex

[tepid goblet]

Oh. Thanks a lot

[tight shuttle]

Dunno why the latex isn’t working but anyways

R is the symbol for reals

For example you could define a function only on natural numbers (1,2,3,…)

In this case it’s for all real numbers

[tardy stag]

please review the rules. do not cheat on exams. this is your last warning.

.close

[lone heart]

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[digital ledge]

Is anyone a teacher, proffesor or a student, or know someone who is? If so you can text me^^

[lone heart]

Please don't occupy multiple help channels.

[forest marsh]

https://discord.com/channels/268882317391429632/903481102591750184

Close this one or the other one

[void nymph]

eh

.close

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[tepid goblet]

Hi, I’ve done all the parts in 7. But pretty much clueless for the graph part where it asked to sketch.

[void nymph]

,rccw

[void nymph]

clueless?

[tepid goblet]

Yh. Cuz I haven’t see one example of how to do it

I want to know

How to find the points

I know that u have to use answers from the previous part probably.

[void nymph]

can you graph, like, $(x-2)(x-3)$?

[ocean seal]

Percy

[tepid goblet]

Yh

X=2 , x=3

Y=0

[forest marsh]

(-a,b) is the minimum/maximum of the graph by definition

[tepid goblet]

I already completed the square on ques 6 . But having trouble for d where it asked to sketch

[void nymph]

what

[forest marsh]

Yeah and this form is giving u the minimum of parabola

[tepid goblet]

The minimum is -2

But what do I do with it

Oh I can find the turning pt

U still can’t do anything with tht

[forest marsh]

You know its symetric now wrt that axis

[tepid goblet]

Do I just sketch

(2,-2)

Then I draw a line of symetry

Or am I missing some points

And y-intercept 2

[forest marsh]

You have the minimum

[void nymph]

bhai kya

[forest marsh]

Get something easy as f(0)

And get the symetric wrt to x = 2 axis

And then you link the points

Also having the minimum answer the range question

[forest marsh]

Theres nothing wierd in what i explain

But also using the factor form is not efficient here cuz the roots aren't cool

[void nymph]

no i said that to him

[forest marsh]

Oh ok

[lone heart]

@tepid goblet Has your question been resolved?

[exotic canopy]

the two hindi words i actually know 💪

#❓how-to-get-help

[lone heart]

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[polar chasm]

For what value of a will the equation (x1/a)(x-cosx1)+sinx1 always be the equation for the tangent line of a unit circle.

[polar chasm]

.close

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[sage maple]

can someone tell me what I am doing wrong??? I mean obviously the second r is a constant so it would be -pi but idk .-.

[wispy plaza]

Your bounds of integration are wrong

[sage maple]

I mean i set it as 0 to pi/2 and then doubled it ? can't I do that?

[wispy plaza]

No

[sage maple]

.-.

professor told me if they r symmetrical we can double lmao

[wispy plaza]

It’s not about the doubling (I’m not so sure, I’m kinda forgetting calc 1), It’s about your bounds of integration. Where do the two functions intersect?

[sage maple]

its calculus 2 lmao

and oh

that makes sense

lmao

sec

[wispy plaza]

Definitely not at pi/2 or 0

[sage maple]

OH

is there anything else that you see that is wrong? I changed the bound but it is still wrong lmaoooo

[wispy plaza]

I can’t see anything immediately, let me do it real quick to see what’s up

[sage maple]

Ty

[wispy plaza]

Wait a minute

I was tripping

Why is your lower bound 0

[sage maple]

well since u said I cant double I shoulve changed it lmao

should be -pi/2 if i am not *2

[wispy plaza]

I didn’t say you can’t double, I’m murky about the polar integral formulas, so I’m not sure. What I’m sure about is that your boundries of integration were wrong

Think about what a lower bound of 0 means for this equation

[wispy plaza]

The lower bound should not be -pi/2

[sage maple]

oh my

OH

[wispy plaza]

Actually, this formula is correct since cos^2 is an even function

Wait ignore what I just said

[sage maple]

I am confusion beyond means HAHAHA

do both of the integrals have different bounds?

[wispy plaza]

No wait, I’m right cos^2 is an even function, I’m just tripping because it’s too late in the night

[wispy plaza]

No

[wispy plaza]

This formula should be correct

[sage maple]

yeah I think I am just struggling with the bounds lmao

I cant figure out the LB

sorry got a pounding headache

[wispy plaza]

What’s LB

[sage maple]

lower bound

[wispy plaza]

Lowkey same

[sage maple]

HAHAHA

well my professor set hers as lower bound = -arccos(1/8)? is that right?

[wispy plaza]

This formula is correct. As for why your not getting the right answer, I’m not sure about

[wispy plaza]

Yes, but in this case she wouldn’t double it

[sage maple]

let me remove the double rq n try that then

OH

we got it

Idk why it didnt work before .-.

so just to make sure for future references, the bounds r essentially where the two functions intersect right?

[wispy plaza]

Yes

[sage maple]

bet bet this helped a LOT

Thank you so much!\

Oh one last question

so how would we solve for the lower bound or the upper bound if it is just one function? we set the equation to zero and solve for theta?

[wispy plaza]

By the way, the two methods are equivalent, so you probably just made a calculation error on your first try

[wispy plaza]

You don’t solve for lower bounds and upper bounds if you just have one function. At least not in the same sense as we do here.

You’ll either be given a boundary, some geometric boundary, or something else

[sage maple]

W at least I know it was a calculation error HAHA

[sage maple]

well from what I saw in my class, we are given r=cos_ or something of the sorts and we have to get the bounds, n go from there HAHA

Well the graph also certainly helps so

[wispy plaza]

Have you gotten to volumes of cylindrical shapes by revolution yet?

[sage maple]

no not yet

we just got introduced to polar coordinates on wednesday and today (now yesterday) calcualting area using polar coordinates or wtv

i cant spell

LOL

[wispy plaza]

Okay. Anyways, the boundary is kinda just whatever fits the bill for what they want. Sometimes they give it to you. Sometimes it’s an intersection.

My recommendation, go back and review polar functions a bit. It really helped me when I was doing limasons and circles and stuff

[sage maple]

got it

Ty so much lol this helped a ton

[wispy plaza]

I got really tripped up with boundaries and stuff because some of those limasons loop back upon themselves, I forget the exact rules.

[sage maple]

yeah im tripping over it because it moves so fast

(quarter system)

honestly enjoying it

but again ty lol

ima close this now

.close

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[charred jewel]

If I have a language such that all strings start AND end with ba, does the string ba itself start and end with ba

[wispy plaza]

yes?🤔

Depends on what you mean by start and end

[lunar saddle]

I would assume so

If you define start and end as "first two characters are ba" and "last two characters are ba" then yes ig

[wispy plaza]

Yeah, being the first two and last two characters are not mutually exclusive when you word’s only 2 letters long

[charred jewel]

i'll take it then

then the regular expression should be

$ba((a+b)^*ba + \lambda)$

[ocean seal]

Shioshi

[lunar saddle]

what the $\sum$

[ocean seal]

Carbonite

[charred jewel]

not that brainrot meme bro

[lunar saddle]

Kk

Mb

[charred jewel]

nah its just regular expressions

[stoic dragon]

Does anyone know how to solve this?
🙏
It would be a great help for me

[charred jewel]

!occupied

[lone heart]

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).

[stoic dragon]

Oops sorry

[charred jewel]

<@&286206848099549185> can someone verify my regex here ty

[lone heart]

@charred jewel Has your question been resolved?

[lone heart]

@charred jewel Has your question been resolved?

[lone heart]

@charred jewel Has your question been resolved?

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[mighty pasture]

in the rhs here, i dont understand why youre allowed to substitute dy/dx of a completely different graph

the thing im concerned about is the "show that" and a snippit of the answer is on the right

[lone heart]

@mighty pasture Has your question been resolved?

[wanton fiber]

pretty sure they subbed in y^2 = c^2/(x^2+a^2) then differentiated it

[mighty pasture]

right but why is that allowed, theyre two different functions no?

[pliant charm]

they are not

[mighty pasture]

wait what

[pliant charm]

function means y is different