help-23

[covert yoke]

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[burnt night]

is this proof valid?

QUESTION
Let A = QR, where Q is an orthonormal matrix and R is an upper triangular matrix. A, Q, and R are
all n×n matrices. Show that if A is invertible, then R is also invertible.

[burnt night]

Proof:
A = QR

A^-1 exists, so A has linearly indepdent columns

if we write A = [a1 a2 ... a_n] and r = [r1 r2 ... r_n] where these are column vectors

then we get a_i = Qr_i

Q is orthonormal and square, meaning it must be invertible. so r_i = (Q^-1)a_i

the inverse of any orthonormal matrix, however, is its transpose, and is also orthnormal. so r_i = (Q^T)a_i

to show R is invertible, we need to show that its columns are linearly independent (that r1, r2, .. r_n) are linearly independent

AKA show that B1(Q^T)a_1 + B2(Q^T)a_2 + ... + B_n(Q^T)a_n = 0 iff B1...Bn (scalars) equal zero

we can factor this into Q^T(B1a_1 + B2a_2 + ... + B_n*a_n) = 0

since Q^T is orthonormal, and is square, it is full-rank and thus has no nullspace. so (Q^T)v = 0 iff v = 0

this means that v = B1a_1 + B2a_2 + ... + B_n*a_n = 0

which we know is only possible if B1 ... B_n = 0 because a_1 ... a_n are linearly independent

end of proof

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[raven veldt]

Can I have verification on this question?

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[fickle yew]

Water is flowing over a Cippoletti weir of 4 meters long under a head of 1 meter. Compute the discharge, if the coefficient of discharge for the weir is 0.6.

[fickle yew]

anyone here good at solving weirs I kinda have an idea how to solve it but I just don't know why the problem didn't use the formula for cipoletti weirs?

also I thought Cipoletti weirs are trapezoidal so I don't know why it just assumed it's rectangular and didn't add the triangular part or am I getting this wrong🤔

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[jolly dust]

can anyone help please

[jolly dust]

<@&286206848099549185>

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<@&268886789983436800>

[polar lynx]

Don't ping mods for homework help

[polar lynx]

Make sure to read this and #❓how-to-get-help

[jolly dust]

<@&286206848099549185>

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[obsidian turtle]

hey i need help

[obsidian turtle]

<@&286206848099549185>

sorry its in french

this is my first exam

i ask any expert to finish it in under an hour then to tell me as a normal 16 y old not bad not good guy how can i do anything in the same difficulty within under an hour

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[worldly knot]

Hello. Would the answer be b=ha-3p/h so none of the above

[oblique bear]

Yes (if you mean $\frac{ha-3P}{h}$)

[flat frigate]

Calculustache

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[worldly knot]

Thank you

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[lethal hawk]

for part 3 of this question

can someone confirm my answer of 10

108=3^3*2^2 so the amount of combinations you can have to mix up AABBB must be 5C2 or 5C3 which are both 10

but I feel like im wrong

[peak siren]

looks right

[lethal hawk]

cool thanks

.close

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[buoyant shadow]

.reopen

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✅

[buoyant shadow]

wait i'm not convinced

[lethal hawk]

oh is it

[peak siren]

only thing i’m not sure about is if c values are the same

[lethal hawk]

do I gotta do 5P2 or summit

[peak siren]

it idk if that matters

[lethal hawk]

bare in mind I cant use a calc for this paper

[buoyant shadow]

yeah ok, you don't get higher powers, it's just normal x

[lethal hawk]

wdym?

[cold relic]

Yeah I think 10 should be right

Gotta apply A 2 times and B three times

[lethal hawk]

ye Im sure of that

[buoyant shadow]

you have ax + c, and both functions keep it ax + c

[lethal hawk]

oh ye

[buoyant shadow]

so 108 only happens from the multiplcation

just didn't see it

[lethal hawk]

ohh ye

so 10 should be right

[buoyant shadow]

yes

[lethal hawk]

cool thanks

.close

[buoyant shadow]

.close

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[gentle maple]

can someone help with this?

[gentle maple]

R(t) = 1 - F(t)

F(t) = integration of f(t)dt

1/50(70) - 1/50(20) = 1 ?

0 t<20
1 20<t<70
0 t>70

?

[lethal hawk]

what is f ment to represent

[gentle maple]

like any function

big F is failure rate

R is realibility

@lethal hawk

[plucky elk]

F(t) = 1 t>70

[gentle maple]

shouldnt it be 0

oh failure rate

realibility would be 0

?

<@&286206848099549185>

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[scenic bough]

the probem asks us to find the equation of any veritcal tangent lines for the grap y^2-12y=4x

[scenic bough]

is my work okay?

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[grim basin]

This is my work so far for #3. I've gotten pretty far but im not sure how to finish the problem. Any advice?

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[lean otter]

i need help with this financial math problem i cant find the answer to year three

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[lean otter]

<@&286206848099549185> anyone?

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[stiff vine]

What am I suppose to do now

Idk what to do with the inf signs

[sleek plank]

you don't directly plug in infinity for a, but rather take the limit

[stiff vine]

I don’t get what that means

So how do I do that

[sleek plank]

what's the limit of 2a^ea - 2e^a as a -> inf

[stiff vine]

Yeah idk how to solve that

I thought u plug it in??

[sleek plank]

you can't since inf isn't a number

[stiff vine]

But some problems they do

Like e^x lim goes to inf = e^inf which is inf

[sleek plank]

you're not supposed to plug in there, you just "notice" that it equals infinity

[cedar owl]

huh

[tall spade]

It is not 2a^ea, it is 2a*e^a, typo in the middle part. It changes from 2x e^x to 2x^(ex) whitout reason!

[cedar owl]

thanks

,w lim of 2a*e^a - 2e^a as a -> inf

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[stiff vine]

Ok i changed it to 2ae^a-2e^a

Now what

[tall spade]

Now you have: lim ( (0-2) - (2ae^a - 2e^a) ) = lim (-2 - 2e^a*(a-1) ) = - infinity , because e^a and (a-1) both go to +infinity and the factor of (-2) changes the sign.

[stiff vine]

In my question the limit goes to -inf

[stiff vine]

So it is -2 +inf -inf?

[tall spade]

That's what I wrote.

[stiff vine]

So I plug in -inf for a right

[tall spade]

It is -2 - 2*(+inf)*(+inf) = -inf

[stiff vine]

On the integral calculator it says the answer is just -2

[tall spade]

It is easier to decide the sign of the result, if you factorize like I did it above.

[stiff vine]

So how would I remove the inf signs to just get -2

[tall spade]

What do you mean by "remove the inf signs"?

[stiff vine]

Idk it says the final answer is suppose to just be -2

[tall spade]

Is the limit is for a -> - infinity???

Then the result would be -2 indeed!

[stiff vine]

Yes

How

So using ur thing it becomes -2-2(-inf)(-inf)?

[tall spade]

You really need to write the stuff more clearly! The limit of e^a for a -> -infinity is 0!

[stiff vine]

How

I thought e^-inf is -inf

[tall spade]

,w plot e^x

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[stiff vine]

Bruh this is so confusing

So I’m not suppose to plug in inf and -infs??

[tall spade]

"e^inf" = inf and "e^(-inf)" = 0. I put the quotes around it, because these are not valid expressions.

[stiff vine]

Ok so for 2ae^a wouldn’t that become -inf times 0 then

[tall spade]

In these cases, the e-function dominates the polynomial function, so the limit of a * e^a for a->-infinity is 0.

[stiff vine]

How does it dominate it

[tall spade]

e^a grows faster than any polynomial function a^n and from that follws, that e^(-a) goes faster towards zero, than a polynomial like a^n goes to infinity.

[stiff vine]

Ok so e^-inf also =0

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[brittle root]

What went wrong here when I was trying to solve this problem?

[stiff sleet]

The one u crossed out in the bottom left?

[brittle root]

Yeah the problem IVP is the top most sentence

The thing I got at the bottom is wrong

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[brittle root]

.reopen

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✅

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[zinc schooner]

wait what if i

.close

nope

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[fresh shell]

need help with 10th grade physics

[fresh shell]

What would the displacement time graphs of these two look like?

[lean otter]

,w rotate

Tf 🥲

,rotate

[lean otter]

,rotate

[lean otter]

Phew finally

!status

[safe radish]

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

[fresh shell]

1

1. I don't know where to begin.

[lean otter]

Use this is a reference

And try to break down the information given into multiple events

Like for instance for the first question, you have two events

1. move at constant speed of 8 ms towards lights which are 60m away
2. when 20m away, accelerate at constant rate of 2 ms^-2 to get past the lights

You can keep the traffic lights as displacement = 0

[fresh shell]

what about d?

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[sacred crane]

I. Which of the four vector can be replaced by w and still span same span.
II. Rank
III. All possible vector basis in {v1, v2, v3, v4, w}

[sinful glen]

Ich würde mir anschauen welche Vektoren mit w linear abhängig sind

Den Rang ausrechnen macht man dann in dem man schaut wieviele linear unabhängige Vektoren es in der Basis gibt

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[sacred crane]

Ich habe es versucht. Aber ich bin unsicher.

In b und c alle 4 vektoren sind voneinander unabhängig

Bedeutet es, dass ich keine von dieser Vektoren austauschen darf oder was?

[fathom jewel]

wie kommts?

[sacred crane]

v1 v2 v3 v4 sind voneinander unabhängig

[fathom jewel]

,w rref {{2,1,-2,0},{1,1,1,2},{1,-1,0,1},{2,-1,-1,1}}

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[fathom jewel]

Also nach dem bot, stimmt das nicht mal

Aber selbst wenn das der Fall wäre, könntest du trotzdem w mit einen Vektor austauschen, wenn du auf Zeilenstufenform reduzierst, dann wird anhand der Treppenstufenform ersichtlich, welche Vektoren linear abhängig sind

[sacred crane]

Ich darf nicht Zeilenstufenform nutzen denke ich

Nur austauschlemma

Du hast recht. Ich habe es falsch kopiert.

Bedeutet es, dass ich v4 durch w austauschen kann?

[fathom jewel]

Du meinst den von Steinitz?

Hä aber du musst ja trotzdem wissen, welche linear unabhänging und welche abhängig sind

[sacred crane]

Ich weiß nicht

[fathom jewel]

Der Satz von Steinitz sagt nur aus, angenommen für einen Vektorraum V hast du eine Basis B mit n l.u. Vektoren und eine andere Basis C mit m l.u. Vektoren und C ist eine echte Teilmenge von V, dann gibt es in B sozusagen m Vektoren die man mit den Vektoren von C austauschen kann

[sacred crane]

V1, v2, v3 sind unabhängig?

[fathom jewel]

ja

oder v1 v2 und v4

[sacred crane]

Könntest du mir ein von dieser lösen. 😭

Ich bin verwirrt

[fathom jewel]

,w rref {{2,1,-2,0,4},{1,1,1,2,0},{1,-1,0,1,1},{2,-1,-1,1,3}}

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[sacred crane]

Aber ohne das. Nur mit austauschlemma

[fathom jewel]

das macht keinen Sinn

[sacred crane]

Wirklich?

[fathom jewel]

Irgendwie muss du ja herausfinden wie du einen Vektor v_i als Linearkombination der restlichen Vektoren v_j mit w darstellen kannst

damit du v_i mit w austauschen kannst

[sacred crane]

Ja aber nicht jetzt 😂

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[lean otter]

show that for a suitable positive constant K there exist subsequences b_m and c_r of cos n such that b_k > K and c_r < -K.

(cos n) = cos1, cos2, cos3, cos4 .......

i said let K = 1/2
b_m = {cosm: m \in N, cosM>1/2}
and defined c_r similarly.

does this work or am i doing fake maths

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[dull warren]

bot

find probability using standard normal distribution P (z<1.73)

[obtuse jackal]

#❓how-to-get-help

[obtuse jackal]

yeah this is a simple question

[lean otter]

i see and it says to finally show that a_n diverges. can i just argue that if b_m converges itll be to some value >=1/2 and c_r for <=-1/2 meaning it'll be different limits so a_n must diverge

[obtuse jackal]

yes

because subsequences of convergent sequences must have the same limit

[lean otter]

thanks 😄

[obtuse jackal]

here's a much stronger but much harder to prove result: for every x in [-1, 1], there is a subsequence of (cos n) that converges to it

[lean otter]

i will think abt it after my problem set🥲

.close

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[obtuse jackal]

if you actually wanna prove it, I'd recommend proving the following lemma first:
Z + 2pi Z is dense in R

depending on how much you've seen, that may be a significant prereq issue

[safe radish]

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[lean otter]

how do i do this help

[lean otter]

is there some kind of formula

[austere cypress]

try drawing it

[lean otter]

yh but i dont get graph paper in my test

[honest perch]

,rotate

[honest perch]

distance formula

[lean otter]

oh ok ty

.close

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[tired ember]

Hi, I have a 2 sets of numbers.

321 -> 461 (increase of 140 raw stat points)

124 damage becomes 152

[tired ember]

So basically

Increasing 321 to 461 increases the 124 to 152.

I am trying to figure out how much an increase of 1% to the 321 number would increase the 124

...does that make any sense?

I don't know the exact way to approach solving for that.

It's a game where 321 is the raw value for Attack.

At 321 Attack, you do a corresponding 124 damage in HP.

So I need to know how much increasing Attack by 1% is, so I can make a chart for comparing values.

I'd like to learn how I should approach solving for this.

321 * 0.01 = 3.21

[austere cypress]

i don't know what you're saying but sounds like a proportion problem

[tired ember]

So if I increase the 321 by 1%

How much would that increase the 124 value?

[versed wave]

this is assuming the increase is linear

[tired ember]

it is

[versed wave]

first find the ratio of their increase

[tired ember]

I just really don't know what type of math to use / learn to solve for this

[versed wave]

increasing 140 attack will increase 28 in hp

so increasing 1 attack will increase ... in hp?

[tired ember]

YES!

That's the thought process I'm working towards solving.

[versed wave]

you should get a pretty good solution to this

[versed wave]

once you find the ratio of increase, find the 101% of 321

here, it is 324.21

its a 3.21 increase

since you got the ratio of increase, you should be able to find how many hp will increase if attack increase by 3.21

[tired ember]

so Ratio of Increase is a term I should remember and learn more about?

[versed wave]

not really, i made up the term for it

but everyone should understand you when you say it

[tired ember]

Ah okay, well it was enough to plug into a google search online and find some similar hits that I can learn more about.

Thank you so much.

My GF started playing a new videogame and it's fairly new so no one has gotten too into figuring out the battle system, so I'm trying haha

[versed wave]

glad for you two, keep it up

[tired ember]

.close

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[rustic mist]

hai

[river field]

haiyaaaa

How can I help you today?

[safe radish]

@rustic mist Has your question been resolved?

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[cedar lance]

Guys, I don't understand why "Power" means two different things. Is it referring to an JUST EXPONENT or a BASE & EXPONENT? Google is giving me both thus, I don't understand what my instructions mean and why I disregarded "2" in "2(4)^2x" in the equation to solve if anyone could explain it?

[lethal hawk]

what dont u get?

[cedar lance]

What does power exactly mean?

[lethal hawk]

its the exponent of the number

[cedar lance]

Google's like:

[lethal hawk]

the number in the top right is called the power

[lethal hawk]

the right image is right

2^n n is the power of 2

[cedar lance]

I can agree with that, but I'm confued why it's talking about "We need to make sure there is only one power on each side so we'll isolate the base with the variable in the exponent." in the instruction. 🥲

Ah wait, was it referring to 2^1 like "1" itself from 2?

[lethal hawk]

I think its justs saying that they want to make it so you have a power of 4 on both sides of the equation so that u can equate the powers

so they are turning that 1/64 into a 4^-3 so that it is a power of 4

and then because you have 4^2x=4^-3 you can make 2x=-3

and then get x=-3/2

[cedar lance]

When you say "power of" 4, you're referring to the exponent of 4. I get that. In that case, why did they say "First, we'll write both sides of the equation as powers of 2." 🥲 I'm sorry I'm really confused with how they worded it.

[lethal hawk]

ngl I dont know because they didnt do that

They wrote both sides as powers of 4

I mean you write both sides as powers of 2 but itd give u the same answer

you would get up with 2^4x=2^-6

[cedar lance]

Okay okay. Glad I wasn't the only one confused. Thanks so much!

.close

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[rustic mist]

.close

hi

.close

how to close cuh .....

.repoen

.reopen

.close

.reopen

hi

h

.reopen

man

smh

[cedar lance]

.reopen

[safe radish]

✅

[cedar lance]

.close

[safe radish]

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[safe radish]

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[spice torrent]

i need help

[spice torrent]

how to do part a?

<@&286206848099549185>

[lean otter]

!15mins

[safe radish]

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.

[lean otter]

i will try

[spice torrent]

thank you

[lean otter]

I have the answer

So basically

use kinematics equations of motion

yes

this thing

$$ S = ut +\frac{1}{2}at^2 $$

[flat frigate]

Cristiano Ronaldo Dos Santos

[lean otter]

yes

take A as the origin

yeah

[spice torrent]

how to form the two equations?

[lean otter]

well for P, a = 2 (constant)
The initial speed = 4 ms^-1
The time = t
just put the things in the equation

[spice torrent]

i dont understand why the two equations become (4t+t^2) and (3(t-1)+0.53.6(t-1)^2) and not (4t+t^2) and (3(t+1)+0.53.6(t+1)^2)

[red delta]

Note the velocity at point A

[lean otter]

well

[red delta]

Which will be the initial velocity

[lean otter]

Q has 1 second less time to travel

So the time is t-1

[spice torrent]

hu?

[lean otter]

since it starts 1 seconds later

t-1

[spice torrent]

but shouldnt t be t+1

for q

[lean otter]

No, why?

[spice torrent]

cuz it passes 1 second later

[lean otter]

No but like we consider the diplacement from the origin

[spice torrent]

i still am blind

[lean otter]

So it starts from the origin at t = 1 seconds and the time left for it to travel = t-1

[spice torrent]

i dont understand

thats the only hardest part in math for me lol

"why is it t-1 and not t+1"

[red delta]

Think about it
What is the displacement at t=1

[spice torrent]

in fact i got a world prize in mathematics for my igcses

[spice torrent]

0?

[lean otter]

Correct

[spice torrent]

so?

[red delta]

Yes now use this equation of motion and replace t with f(t)

[spice torrent]

ok and then?

[lean otter]

$$ you get 4t + t^2 $$

[flat frigate]

Cristiano Ronaldo Dos Santos

[red delta]

you will see that s=0 when f(t)=0
But t=1
Now find the relationship between f(t) and t

[lean otter]

you get $$4t + t^2 $$

[flat frigate]

anjali

[red delta]

Then replace f(t) with what you get

[spice torrent]

i got that equation for p

but i still dont get why it aint t+1

[lean otter]

it ain't t+1 because we don't consider the diplacement from where Q starts, but from point A (when it crosses A)

[nova shoal]

yall i need help.

[spice torrent]

so at point A it is t

[lean otter]

#❓how-to-get-help

[spice torrent]

?

[lean otter]

,rotate

[nova shoal]

its kinda blurry srry

[safe radish]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[spice torrent]

yawl help him

im too dumb to figure out t+1 and t-1 💀

will doing more questions on it help me?

[red delta]

put t=t+1
Then sub t=1 and see if s=0

[spice torrent]

wha?

[red delta]

You have to open your own channel
This is someone else's.

[nova shoal]

yea ik

[spice torrent]

david, is there a visual representation of this which u can find?

[red delta]

Huh

[spice torrent]

like a diagram explaining this

[red delta]

Nope

[spice torrent]

aight, im cooked

cya

[red delta]

All you need is this
If it is=0 when t=1 then t here has to be t-1

1-1=0

[spice torrent]

tf

1s and 0s aah

peace, im out

dont bother with me lol

[safe radish]

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).

[red delta]

Seems very AI

[safe radish]

@spice torrent Has your question been resolved?

[safe radish]

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

need help

[wicked tide]

Not sure how to solve the integral I circled

[ivory obsidian]

well you can pull 2y/10 sqrt(x_3) out and only have the integral from 8 to 12 of 1 dW2

because x and y dont depend on w2

no idea how you go from that to 2x^4 though...

[wicked tide]

it's 2 times 4

that's the integral

that I'm asking about

[ivory obsidian]

ahh i see i think

[wicked tide]

integral 8 to 12 (2)(dw2) is equal to 2x4 I just don't get it

I also don't get where 2/3 is from, shouldn't it be -1/2 not -2/3?

[ivory obsidian]

yeah they pull only y/10 sqrt(x3) out and you have integral from 8 to 12 of 2 dw2

uh i dont know the name of that how do you call F(x) if you have f(x)

parent function?

well whatever you have 2 dw2

one possible F(x) of f(x)=2 would be 2x+c

and then [2x]12,8 -> 12*2 - 8*2 = 8

[wicked tide]

okay thank you that makes sense

can you help me with the 2/3 I think that's straight up wrong

[ivory obsidian]

ugh how did you get there

where does it come from

no idea what you are calculating

[wicked tide]

4x3 + 1.55y/sqrtx3 wrt x3

it's just d/dx wrt x3

[ivory obsidian]

of what did you take d/dx what was the term

ahh the end result of that

bottom right?

[wicked tide]

it's a

yeah

it's just another part of the problem

but again

1/sqrt(x3)
x3^(-1/2)
(-1/2) (x3^ (-3/2)

-1/2 x3^ 3/2

so -1/2 is the constant

idk why it's -2/3 here

[ivory obsidian]

yeah

[wicked tide]

wrong?

[ivory obsidian]

i also get -1/2

[wicked tide]

okay thanks

[ivory obsidian]

should the 2/3 be correct?

4-1.55y / 2x^(3/2) is what i have

[wicked tide]

that's what I have too

and what mathway has

so i think the solution is wrong

[ivory obsidian]

https://www.derivative-calculator.net/
says the same

[wicked tide]

okay just before you go

[ivory obsidian]

yeah

[wicked tide]

[(expression)^(2/3)] ^ 1/2

is 2/3 times 1/2

1/3

?

[ivory obsidian]

you mean 1/3 * 1/2 or
(x^(2/3))^(1/2)

dont really get your question

[wicked tide]

1 second

[ivory obsidian]

ahh

yes

a^x^y = a^xy

if thats what you mean

[ivory obsidian]

yes you can multiply 2/3 and 1/2 to 1/3

if that was your question?

[wicked tide]

yes

and

(31/160)^(1/3) * (31/160)^ 2/3

is just

1/3 + 2/3

1?

[ivory obsidian]

yes

[wicked tide]

okay thanks again

I think I'm decently cracked at math

i have a midterm tomorrow

thanks alot

[ivory obsidian]

no problem 🙂

good luck

[wicked tide]

actually

can you just double check my math

because it's different from the solution

i think i did it right

also thank you

[ivory obsidian]

what exactly

[wicked tide]

I just simplified it

31/160y ^(1/3) to eliminate the denominator

And make the two expressions common so I can add them simply

[ivory obsidian]

has that x3 at the start a power or is that just some mark?

[wicked tide]

It's a star

Ie sub in

[ivory obsidian]

ahh you put in some value for x* at the first step?

[wicked tide]

Yeah

I just wanna know if I simplified it right

I think I did anyway

It looks right

The only step I'm wondering about is

To get rid of the denominator if I multiple both expressions by (31/160)^(1/3)

Which then makes it 4(31/160)^1 because 2/3+1/3

[ivory obsidian]

how did the y get a 2/3 power at the 3rd step?

[wicked tide]

That's

Y/y^1/3

Or y^ 1 - 1/3

Or 2/3

[ivory obsidian]

ahh did see its missing in the bottom

[wicked tide]

Does it seem right

[ivory obsidian]

should be good

[wicked tide]

Okay thanks

It's similar to the solutions but alot of steps were missing

Anyway thanks a million goodbye

.close

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[shell berry]

can someone help me with b and c

[shell berry]

a i have done it

[split ether]

For b, it's easy to see that f itself is odd. Combining that with the definition of f^{-1} produces that the inverse is also odd

For c, that can be rewritten as f(f(x)) = 0. Since f(x) is a bijection, so is f(f(x)), meaning there exists a single solution to the equation f(f(x)) = 0, can you think of one?

[safe radish]

@shell berry Has your question been resolved?

[shell berry]

which mean x= 0 ?

for c

[split ether]

Yeah

[shell berry]

tysm

[safe radish]

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[flat frigate]

Noah

Noah

Noah

Noah

[safe radish]

@scarlet tapir Has your question been resolved?

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[forest herald]

Can someone just help guide me for the steps to this problem? I just want to understand the thought process behind solving this. Thank you!

[indigo laurel]

Do you know how to apply implicit differentiation?

[forest herald]

A little, I'm not the best at it.

[indigo laurel]

That's fine!

So let's try solving for dy/dx first

[forest herald]

I missed a lot of precalc stuff, so I'm a bit behind in class

[indigo laurel]

What is the derivative of the LHS? (left hand side)

[forest herald]

1 = 2(dy/dx) - sinx?

[indigo laurel]

Almost

The second term (2(dy/dx)) is off

What is the derivative of y^2 with respect to y?

like

d/dy of y^2

[forest herald]

is it 2y(dy/dx)?

[indigo laurel]

yes

correct!

Now, let's isolate dy/dx

[forest herald]

OHH OKAY

[indigo laurel]

So it's based off of the chain rule here

[forest herald]

aw i got confused if i should keep the y or not

okay so it becomes 1+sinx= 2y(dy/dx)

[indigo laurel]

actually, no

[indigo laurel]

This is wrong for another reason

what's the derivative of cos x?

is it sin x?

[forest herald]

-sin?

but its -cos

so would i thought it was cancel out

[indigo laurel]

So

yes it would cancel out

so d/dx [cos x] = -sin x

[forest herald]

right

[indigo laurel]

but taht means in your expression, we get 1 = 2y(dy/dx) + sin x

so moving it over, it's now negative again

[forest herald]

ahhh

okay okay oops

[indigo laurel]

all good

[forest herald]

okay so moving on

it would be 1 - sinx = 2y(dy/dx)

thank you by the way lol

[indigo laurel]

yes

So now isolate dy/dx

and we should get...

you got this

[forest herald]

(1- sinx) / 2y

?

[indigo laurel]

$\dfrac{dy}{dx}=\dfrac{1-\sin x}{2y}$

yes!

[flat frigate]

Oreo

[indigo laurel]

So now, we need to get the second derivative

this requires the quotient rule (or the product rule if you're really finnicky with the quotient rule)

[forest herald]

aaa okay

give me one moment, im doing this on paper and I have the digital answer key up lol

[indigo laurel]

you're good

[forest herald]

the quotient rule is.. (f'g) - (g'f) / g^2

right?

[indigo laurel]

yes...

if it is d/dx (f/g)

[forest herald]

can i apply it here?

f' would be cosx? and g' would be 2y(dy/dx)

?

[indigo laurel]

uhhhh

f(x) would be the top

g(x) would be the bottom

so

no for f'

no for g'

Let's do f'(x) first

what's the derivative of sin x?

[forest herald]

cosx?

[indigo laurel]

yes

so would f' be positive cos or -cos?

[forest herald]

positive

?

[indigo laurel]

but isn't it 1 - sin x?

would the negative go away?

[forest herald]

oh yes

no

it would be -cos

sorry i keep forgettting about the sign

[indigo laurel]

All good

This is pretty normal

So yes, negative cosine

what about the denominator?

do we keep the y when we derive?

[forest herald]

no

[indigo laurel]

yeah, so in the previous case we did

since it was y^2

but now it's just y

[forest herald]

ohh

i was going to ask abt that

so it would just be 2?

[indigo laurel]

yes

[forest herald]

okay

so plugging it in

it should be ((-cosx)(2y) - (2)(1-sinx)) / (2y)^2

[indigo laurel]

so

huh?

oh wait I'm dumb

yes, you're right

[forest herald]

wait

[indigo laurel]

no I forgot about quotient rule

[forest herald]

ah okay

[indigo laurel]

Yeah, I think you're good

but where'd the dy/dx go in the denominator?

[forest herald]

huh?

we use dy/dx in the denominator?

[indigo laurel]

since we derived y?

oh wait

it would be in the numerator here

denominator part of hte original dy/dx

so

when deriving 2y with respect to x

for the (2)(1-sin x)

where did the dy/dx go

so it's

$\dfrac{d^2y}{dx^2}=\dfrac{(-\cos x)(2y) - \frac{d}{dx}[(2y)]\cdot(1-\sin x)}{(2y)^2}$

[flat frigate]

Oreo

[indigo laurel]

is that right?

[forest herald]

wait i thought the derivative of 2y would just be 2?

and we dont use dy/dx for it

[indigo laurel]

that's with respect to y

but this one is respect to x

[forest herald]

oh

[indigo laurel]

so, we technically have

$\dfrac{d}{dx}[2y]=2\cdot\dfrac{d}{dx}[y]=2\dfrac{dy}{dx}$

[flat frigate]

Oreo

[indigo laurel]

so we keep the dy/dx in it

[forest herald]

oh ok

[indigo laurel]

so now we have, in all the blob

$\dfrac{d^2y}{dx^2}=\dfrac{(-\cos x)(2y) - (2\frac{dy}{dx})\cdot(1-\sin x)}{(2y)^2}$

[flat frigate]

Oreo

[indigo laurel]

this isn't fully completed, since we know what dy/dx is right?

[forest herald]

yes

do we plug it in?

[indigo laurel]

so we need to plug that in

yaaaay

[forest herald]

yayy

lol

okay idk how it so the bot can do it

but

[indigo laurel]

just put it in, and I'll try to transcrbe it

[forest herald]

((-cosx)(2y) - (2(1-sinx)/2y)) (1-sinx)) / 2y^2

does that make sense

[indigo laurel]

yes

that is almost what I got

is our denominator correct?

is it 2y^2

or (2y)^2?

[forest herald]

oh oops yes

[indigo laurel]

$\dfrac{d^2y}{dx^2}=\dfrac{(-\cos x)(2y) - \left(2\cdot\dfrac{1-\sin x}{2y}\right)\cdot(1-\sin x)}{(2y)^2}$

[forest herald]

(2y)^2

[flat frigate]

Oreo

[indigo laurel]

does this seem right?

[forest herald]

yes!

thank you so much

[indigo laurel]

alright nice

now time to simplify it

[forest herald]

oh dear

[indigo laurel]

yep

[forest herald]

im going to have trouble with this part

[indigo laurel]

the bane of every calculus student

the algebra

[forest herald]

😭

umm

[indigo laurel]

What should we do first?

[forest herald]

the (2(1-sinx/2y))?

[indigo laurel]

sure

[forest herald]

okay okay

[indigo laurel]

the 2's cancel right?

[forest herald]

OH YAEH

i just noticed

woah

[indigo laurel]

WAITR ACTULALY

NO SIMPLIFYING

WE'RE GOOD

[forest herald]

OH OKAY

LOL

[indigo laurel]

so

it tells us

at point (0,-1)

and this is technically a correct form of the second derivative

so we can just plug in here

[forest herald]

OH RIGHT

OKAY

toh dear

unit circle..

[indigo laurel]

oh dear indeed

[forest herald]

i am

quite intimated

i dont know the unit circle that well

[indigo laurel]

you'll memorize it sooner or later

with... enough practice 😭

but for now, it's only 1 point of notice

$\dfrac{d^2y}{dx^2}=\dfrac{(-\cos 0)(-2) - \left(2\cdot\dfrac{1-\sin 0}{-2}\right)\cdot(1-\sin 0)}{(-2)^2}$

[forest herald]

okay im plugging in now

[flat frigate]

Oreo

[forest herald]

o wait

the points they're asking for is (π/6),(1/2)

[indigo laurel]

oh god

OH NO

AAAAAAAAA

I JUST SOLVED IT 😭

🥲

Time to do it again!

[forest herald]

IM SO SORY

😭

[indigo laurel]

WE RUN IT BACK

[forest herald]

i wasn tsur if u knew or not

YAYY

ITS OKAY

WE CAN DO THIS

[indigo laurel]

WE GOT THIS

[forest herald]

let me try too

[indigo laurel]

alright I have a solution

[forest herald]

okay wait

oh dear

[indigo laurel]

oh dear?

[forest herald]

okay im not sure if i did this right

[indigo laurel]

well that's math in a nutshell

so let's see it

[forest herald]

um

okay wait

I wasnt exactly sure

what to do for one of the parts

but i got

((√3/2) - (2 (1-(1/2)/1)) (1/2) / (1/2))

ih dear

[indigo laurel]

OH GOD

[forest herald]

LOL

[indigo laurel]

WHAT IS THAT

[forest herald]

IM SORRy

BAHH

IM SoRRY

OKAY WAIT

[indigo laurel]

BAHAHAHHAAH

[forest herald]

IM SDORRY

LOLL

NO I WAS LOOKING AT IT

AND

IN TEXT FORM iT LOOKS bad

OKAY WAIT TGE (1/2)'s CANCEL OUT

[indigo laurel]

lemme decipher th is

[forest herald]

so it should just be

okay ait

let me try ti somplify ieven more

(-√3/2) - (1)

?

[indigo laurel]

almost

i think

so

when we did the $1-\sin \pi/6$

[flat frigate]

Oreo

[indigo laurel]

what did we get?

[forest herald]

1 - (1/2)

[indigo laurel]

which is?

also

I think you messed up somewhere else maybe...

since I got

$-\dfrac{\sqrt{3}+1}{2}$

which is just

[flat frigate]

Oreo

[indigo laurel]

$-\dfrac{\sqrt{3}}{2}-\dfrac{1}{2}$

[flat frigate]

Oreo

[indigo laurel]

so very close to your answer

I might be wrong though

[forest herald]

no ur prob right

let me process this

[forest herald]

also if i ddont respond my wifi went out 😭 sorry about that in advance

[indigo laurel]

😭 you good

[forest herald]

wait i hav the answer key

wait

WHAAT

MY PACKET HAS DIFFERENT POINTS THAN THE ANSWER KEY??

[indigo laurel]

OH NO

[forest herald]

ill ask her abt it tmrw 😭

[indigo laurel]

HAHAHAHAHAHA

[forest herald]

BRO WHAT

[indigo laurel]

GETTING TROLLED BY THE TEACHER

[forest herald]

OKAY BUT UR PROBABLY RIGHT

NO GENUINELY

MY PACKET SAYS

oh dar

[indigo laurel]

Here is my work though

https://artofproblemsolving.com/texer/wtycvfhi

[forest herald]

okay

thank you so much

[indigo laurel]

no problem

good lukc with your calc class!

.close

[forest herald]

thank you!! is it okay if i add u to maybe ask for help for further questions..?

you expained it very well 😭

.close

[safe radish]

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[safe radish]

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Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[dusk gate]

I can't figure out what I'm doing wrong

[plucky elk]

,calc sqrt(11/41)

[flat frigate]

Result:

0.51796977028281

[plucky elk]

,calc 9* sqrt(451)/41

[flat frigate]

Result:

4.6617279325453

[dusk gate]

Even if it were just a matter of not reducing my fractions properly, I've got the wrong value for y

oh shit.. there it is..

I left y in the numerator when I was isolating lambda

.close

[safe radish]

Channel closed

Closed by @dusk gate

Use .reopen if this was a mistake.

[safe radish]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[limpid estuary]

is that correct

[severe pond]

well

why the -

[limpid estuary]

why not

[severe pond]

i asked you first

[limpid estuary]

i asked second

oh

gotcvha

[severe pond]

$\frac{\frac{2\pi}{3}}{\frac{\pi}{4}}$

[flat frigate]

knief

[severe pond]

no negative

and you already said to the left

[limpid estuary]

yes my mistake

thanks

wait

[severe pond]

yep

[limpid estuary]

thats what it is aking for right

i wasnt sure

[severe pond]

draw a triangle

[limpid estuary]

ok

[severe pond]

suppose y = arccos(x/5) then cos(y) = x/5

so adjacent = x

hypotenuse = 5

find the opposite side

then use that for csc(y)

[severe pond]

^

which you already did

in the answer

[limpid estuary]

sqrt25-x^2

[severe pond]

yep

[limpid estuary]

so no fraction

[severe pond]

that’s the opposite side

csc = hyp/opp

so it’s correct

[limpid estuary]

oh okay

thanks

Find the exact value of the expression without the use of a calculator.

csc(csc−1(−1))

hold

[severe pond]

✋🏻

[limpid estuary]

-1

right

hm

[severe pond]

is this supposed to be $\csc(\csc^{-1}(-1))$

[flat frigate]

knief

[limpid estuary]

yes

[severe pond]

then yes

[limpid estuary]

hazzah

[jaunty bobcat]

you divided wrong

1/2x not x/2

same thing another one. 9/2y and not 9y/2

[jaunty bobcat]

hmm

sounds fine to me

[limpid estuary]

this too

[jaunty bobcat]

well it seems fine

as long you know unit circle

[limpid estuary]

yes i do

[jaunty bobcat]

i mean ye theres no phase shift or stretch going on

[limpid estuary]

not verticxally

[jaunty bobcat]

well it looks like 3/2

[safe radish]

@limpid estuary Has your question been resolved?

Channel closed

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Use .reopen if this was a mistake.

[safe radish]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[heavy glade]

when a problem says f(x) < g(x) for all x in [a,b] what does that mean exactly 😭

[heavy glade]

does it mean only if the x are the same in both? like i cant do f(a) < g(b) right?

[noble kiln]

yes

[heavy glade]

:( ugh

thank you :(