help-39

[shy ginkgo]

reread

[blissful cloak]

ah there

[vital estuary]

Ok yea its good then

[blissful cloak]

then yes

[light helm]

then yes

was just answering the question as initially stated

[shy ginkgo]

ty

.close

[pearl pond]

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[blissful cloak]

[shy ginkgo]

.reopen

[pearl pond]

✅

[shy ginkgo]

btw why is 2 plugged into x and not 4

[blissful cloak]

that's a mistake but the final answer is right

[shy ginkgo]

oh

[buoyant stump]

Just put x = 4.

[shy ginkgo]

so itd be correct if it said x = 2?

[buoyant stump]

No. Answer is given according to x = 4.

[shy ginkgo]

oh bcuz f(x) = sqrt(x) and x = 4 so f(x) = 2. Then 2 is plugged into 1/2(sqrt(x))?

[brazen vector]

no follow me

f'(x)=1/(2sqrt(x)) then f'(4)=??

[shy ginkgo]

oh

1/4

wait

1/8?

yeah 1/8 i think

[brazen vector]

why 1/8

show me your calculation steps

[shy ginkgo]

f(4) = sqrt(x) so if 4 = sqrt(x) then x = 16 so 1/(2*sqrt(16)) = 1/(2 x 4) = 1/8

[brazen vector]

wrong

[blissful cloak]

f'(4) means you put 4 in for every x

[brazen vector]

f(4)=sqrt(4)

you just replace each x by 4

same for f'

f'(x)=1/(2sqrt(x)) so f'(4)=1/(2sqrt(4))

[shy ginkgo]

so its 1/2(sqrt(4)) = 1/2*2 = 1/4

[brazen vector]

exactly

[shy ginkgo]

o

[blissful cloak]

$f(\textcolor{red}{4})=\frac{1}{2\sqrt{\textcolor{red}{4}}}$

oh cmon stupid thing

[shy ginkgo]

lmao

ok i get it now thx

[blissful cloak]

sorry was trying to do something with the colors

lmao

[jolly parrot]

MrFancy

[blissful cloak]

ah ha!

[shy ginkgo]

lol

[buoyant stump]

!redtex

[shy ginkgo]

tex

.close

[pearl pond]

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[merry steeple]

would I just equate the equation for y to 0 for the first part

and then that gives me many values for t and I just choose the second one?

[pearl pond]

@merry steeple Has your question been resolved?

[merry steeple]

<@&286206848099549185>

[desert oyster]

Yeah

[merry steeple]

I’m confused about the second part

[desert oyster]

Whenever it is travelling horizontally, its vertical velocity would be zero

At that point

Its vertical velocity is the rate at which its distance in y-direction is changing

Which is equal to dy/dt

[merry steeple]

so I’m getting 0

could you check if these are correct

[desert oyster]

Did you find dy/dt ?

[desert oyster]

They're not

[merry steeple]

5cos(5t)

[desert oyster]

Right, now what are you supposed to do next

We're supposed to find the points where dy/dt=0

[desert oyster]

If you put t=0 in this, you dont get 0 so it is wrong

Correct! But there might be more number of points other than these

SInce the stone doesnt sink until 7 pi / 10

Yes!

Hmm good question

But I think I kinda would XD cos it's still boundary condition

But not super sure about that

I personally think it's better and safe to include it than to exclude it

[pearl pond]

@merry steeple Has your question been resolved?

[pearl pond]

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[sweet radish]

Can someone explain this to me the solution does not make sense

[static helm]

!status

[pearl pond]

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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[grim fractal]

I'd just guess and check

[toxic lichen]

status already indicated as 5

[grim fractal]

Like think of some pairs (x,y) that satisfy the first inequality and then plug them into the 2nd and 3rd

it's simpler

[static helm]

Oop misread

[sweet radish]

Ah well the only combination I can think of 2+3
Where y is 2

So is that it ?

[grim fractal]

well test and see if it's right

if that satisfies the other 2 then you're good

[odd holly]

Wait isn't that brute force and is that the way?

[sweet radish]

Yep it workss

[static helm]

Plug and chug is a way yeah

[odd holly]

Although it feels like OP got the answer, I'd love if someone could tell how do I get the answer when options are not provided.

[static helm]

Simply, after the difference of squares, x + y is most likely 5 or 6. Checking for x + y = 6, we see no solutions are less than 7, so it must be x + y = 5. We have 4 and 1 or 2 and 3. Out of these 2 option, only (3, 2) works giving y = 2

[sweet radish]

I have another question may I ask it here or a new channel?

[static helm]

Pretty sure here is fine

[sweet radish]

How do I chose between b & c

[grim fractal]

It's a way, and honestly brute force is really quick and easy with problems like these because the numbers are small

If I had bigger numbers I'd solve it algebraically

[static helm]

Plug in x = 3?

And see what the value is

[sweet radish]

Ahh okayy

Could someone pls also explain what this negative/positive infinity in the solution is

[static helm]

Think it just is another way of solving it

x^3 will be the biggest or small number in something like this usually

So when a negative number is cubed, it is still negative

So it’ll approach negative infinity

And vise versa

(Probably takes more steps then the x = 3 thing but still something you could have used)

[limpid kindle]

anyone could help my brother

[sweet radish]

Ah okay thank u very muchh

[static helm]

!occupied

[pearl pond]

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).

[static helm]

@limpid kindle

[static helm]

Gotchu

[sweet radish]

.close

[pearl pond]

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[limpid kindle]

?

[pearl pond]

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[alpine imp]

how to prove? z1, z2 are complex numbers (a1, b1), (a2, b2), how to "multiply"? why z1z2 = (a1a2 - b1b2, a1b2 + a2b1)?

[undone quail]

hey bro where u got that pfp?

[grim fractal]

google images I think

though it looked cool

[undone quail]

i have the same pfp in my whatsapp pfp lol

[grim fractal]

oh lmao

[alpine imp]

please help

[surreal ocean]

(a1+b1i)*(a2+b2i)=a1a2-b1b2+a2b1i+a1b2i

[pearl pond]

@alpine imp Has your question been resolved?

[alpine imp]

its prove?

[surreal ocean]

oh sorry not really a formal proof.

though it does illustr why it's true in a way

[alpine imp]

yes, am agree, its actually, we can actually illustrate in polar form, or mb some other

but im interested in proof

[pearl pond]

@alpine imp Has your question been resolved?

[pearl pond]

@alpine imp Has your question been resolved?

[pearl pond]

@alpine imp Has your question been resolved?

[alpine imp]

<@&286206848099549185>

(sorry for ping😭)

[pearl pond]

@alpine imp Has your question been resolved?

[elder haven]

Wdym "why"? It's how it's defined. Also you mixed up the x and ys with a and bs

Try to do multiplication just as u would with real numbers

And make sure to "differentiate" the imaginary part and the real one

[alpine imp]

i mixed up 😞

[alpine imp]

so i can’t prove it?

[elder haven]

U can

Alr so

Let z1 = x + yi and z2 = a + bi
z1z2 = (x + yi)(a + bi)

Now expand, like, using the same algebra thingies you're used to

[alpine imp]

(a1, b1)(a2, b2i)
a1a2 + a1b2 + b1a2 + b1b2
but this, that with both imaginary parts, turns to, with the reverse sign, and so now this product is real
we still remember to separate the imaginary parts from the now available real parts
we now have
((a1a2 - b1b2), (a1b2 + b1a2))

can i ask another question?

[elder haven]

Yeah. But I think it's clearer if you did the arithmetic with the imaginary units and representing it as a+ bi and not (a,b), and then just remember to regroup the numbers back after multiplying

[elder haven]

[alpine imp]

complex number and its conjugate, by what operation *, they are inverse elements for each other in most of cases? (C, *)

[elder haven]

Not inverses. Rather, when a complex number z is multiplied with its conjugate, it returns |z|^2 (the bars denote abs value)

Inverses mean that if you multiplied it with its inverse it returns u the identity- in this case, 1 (or rather, 1 + 0i)

[alpine imp]

when multiplie both between, the imaginary unit escapes

but product not identity element

[alpine imp]

ok it's not inverses at all

[elder haven]

Yeah

[pearl pond]

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[midnight haven]

How do i find this limt? $\lim_{x \to \infty} (1 - \frac{1}{x+1})^x$

[jolly parrot]

What should I do: Physics

[midnight haven]

it was orginally

[midnight haven]

are you familiar with the limit of e

$\lim_{x \to \infty} (\frac{x}{x+1})^x$

[jolly parrot]

What should I do: Physics

[midnight haven]

maybe

because this is basically the limit of e in disguise

[
\lim_{n \to \infty} \parens [\bigg]{1 + \f x n}^n = e^x
]

[midnight haven]

So what you have can be written as [
\lim_{x \to \infty} \parens [\bigg]{1 + \f{-1}{x+1}}^x
]

[midnight haven]

the +1 in the denominator is negligible as x approaches infinity

so all in all we can conclude...?

1/e ?

yes

desmos is showing a value lower than 1/e

idk

[steel dagger]

Input is not infinity

[midnight haven]

Okay

TY

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[hard flint]

How can I prove by induction 14 b) ??

[dense goblet]

!status

[pearl pond]

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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[hard flint]

2. I tried to put n+1 on both sides not sure if it’s the 1st step to do

[dense goblet]

what do you mean put

[hard flint]

I changed n to n+1 to see if it happens to all the numbers

[dense goblet]

right, so proving $2^{n+1}>(n+1)^3$

[jolly parrot]

chlamydia

[dense goblet]

and we have the assumption $2^n>n^3$

[jolly parrot]

chlamydia

[hard flint]

yes but I couldn’t take a conclusion from the final equation

[dense goblet]

we need to get that to look more like the assumption

and we can try to simplify from there

[hard flint]

what can I do now to simplify it?

[dense goblet]

$2^{n+1}\neq2+2^n$

[jolly parrot]

chlamydia

[dense goblet]

it's 2 multiplied n+1 times

[hard flint]

yes mb

now should I solve (n+1)^3?

it doesn’t look like it will help

[dense goblet]

using this might be better

to prove that $2^{n+1}>2n^3>(n+1)^3$

[jolly parrot]

chlamydia

[dense goblet]

and then it's just a polynomial

[hard flint]

I don’t understand the 2n^3

[dense goblet]

$2^n>n^3$ from assumption $\Rightarrow 2\times2^n>2n^3$

[jolly parrot]

chlamydia

[hard flint]

makes sense I’ll try to solve it now

[dense goblet]

$2n^3>(n+1)^3$ yeah?

[jolly parrot]

chlamydia

[hard flint]

well idk if we can say that

[dense goblet]

i mean proving it

[hard flint]

but why is 2n^3 > (n+1)^3 and not <

[dense goblet]

why would it be <

[hard flint]

I just don’t understand why you assume that 2n^3 is bigger

[dense goblet]

if we can prove it then it proves the induction

thats why i'm assuming that

[hard flint]

ok makes sense

[dense goblet]

-1 not +1

[hard flint]

where?

[dense goblet]

at the end

so from here because n3-3n2-3n-1 isn't convenient to factor, the next best method is using calculus to show that it's always positive

[hard flint]

idk how can I do it

[dense goblet]

if the curve starts from above 0 and only increases from there then it must be positive

[hard flint]

so should I check the graph?

[dense goblet]

no need to sketch, but check the curve

using calculus

[hard flint]

to see that I need to calculate the derivative see the 0s and then see where it is > 0 right?

@dense goblet

or in this case make to times the derivative to see the curve

[dense goblet]

the question just asks for n>=10, so check that it's >0 from there, and that the derivative is positive

[hard flint]

it is positive so we prove this

[dense goblet]

yeah

[hard flint]

but couldn’t the curve be positive but the function still negative and that would made the affirmation negative?

[dense goblet]

what do you mean

[hard flint]

the function was rising when n>10 but in n=11 f(x) still negative

[dense goblet]

that's not true? what are you using for f(x)

[hard flint]

n^3 - 3n^2 -3n - 1

in this case it is positive but the curve can be positive and that the function still negative no?

[dense goblet]

,wolf calc 11^3-311^2-311-1

f'(n)=3n2-6n-3

,wolf calc 3*121-66-3

[jolly parrot]

Results provided by WolframAlpha

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

Requested by chlamydia.3844

[dense goblet]

no it's still positive

[dense goblet]

did you get this for the derivative

[hard flint]

no

the derivative was 6n^2 - 6 to see the curves

what I’m asking is if it was another equation which in n=11 the result was negative but the curve for n>10 was always positive. We can’t just see the curve right?

because in this case f(x)>0 when x>10 is false

and the curve was positive to x>10

[dense goblet]

are you trying to say that if f(x) was different then this method wouldn't work?

[hard flint]

yes

[dense goblet]

then yes

it happens that this works for this question

[hard flint]

so what would be a the right method to do all the time?

check first the equation on calculator see if it positive and then check the curve?

[dense goblet]

i mean i don't know all the questions, but i know that for normal induction questions, there's usually a simple way out like this

[hard flint]

ok thanks for your help

have a nice day

[dense goblet]

np

[hard flint]

.close

[pearl pond]

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[worn turret]

hi can someone help me with these discrete math python questions?

[worn turret]

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[low portal]

Hello guys, I need to simplify this set. Сan I get help with this?

[sly wadi]

xd

[low portal]

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[humble tinsel]

Why is this not correct?

Apperantly the answer is 2pi/3

[light locust]

tan(pi/3)=sqrt3 but ur looking for -sqrt3. so you need to find where the slope is negative and has that steepness. that happens at 120degrees which is 2pi/3

[humble tinsel]

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[shy ledge]

How on earth can you solve the following system of equations?

(f-a)^2 + (g-b)^2 = r^2
(g-b)/(f-a) = -(c-f)(d-g)

[shy ledge]

Solve for f

I tried wolfram mathematica and it gave me an equation bigger than my last essay

[strange quiver]

yea I think it's suppose to be complicated

[shy ledge]

Hmm yeah do you think you can help me?

I really want to solve this problem, I'll visualize it for you in photoshop hold on

There are two equations, which are the contraints for the problem.

I'm trying to find the two points that coincide with some circle A, and point B is the intersection of the two tangent lines that the two points sit on.

I'm trying to generalize it so I can compute it in desmos

The two circles are the contraints, where they meet is the solution, which are always two. If the tangent intersection point is within the circle there are no solutions.

Here are the equations so far

[shy ledge]

The ONLY thing I have left to do is solve this system of equations, nothing else.

[golden orbit]

What have you attempted?

It does give you a large answer

[shy ledge]

Yes, I have attempted to solve for f in the equation {(f-a)^2 + (g-b)^2=r^2}, and then substituted f expression in terms of g for the f's in the equation (g-b)/(f-a) = -(c-f)(d-g), I then subsequently have trouble solving for g.

[golden orbit]

Aren't you trying to solve for f?

[shy ledge]

Doesn't matter, either f or g, they are x & y coordinates of the points I'm seeking for

Here is f in terms of g

Do you understand?

[golden orbit]

Isn't that what you want?

[shy ledge]

Well I need to then substitute that expression for the f's in the other equation, and then after that, I need to solve for g

Solving system of equation via substituion method

Hold on let me just give you the expression, perhaps that would be better

Solve for g

[golden orbit]

Why not just sub in $g-b=\sqrt{r^2-(f-a)^2}$

[golden orbit]

Into your second eq here

[jolly parrot]

Shubh

[shy ledge]

I think it will still be as complicated, let me try

Subsitute that for f in the equation you sent?

[golden orbit]

Yea i think

Or substitute f-a version of the first eq.

To get g

[shy ledge]

Well, the equation is definitely smaller by a large amount so that's good

I believe this is correct then?

It's just so large bruh

Ok so I tried to reduce it and i concluded with this

[golden orbit]

lemme actually try on paper

[shy ledge]

I wonder if wolfram completes the square

Surely it does

[golden orbit]

No way

I got a simpler solution

for g

[golden orbit]

Just do it

It's not that painful

Manually

[shy ledge]

Ok let me try

[shy ledge]

wait what equation?

[golden orbit]

Sub this into your second equation you mentioned earlier: (g-b)/(f-a) = -(c-f)(d-g)

It should take like max 10 lines for you to do this with normal algebra,

@shy ledge

[shy ledge]

I'm trying to solve it right now hold up

Did you get two answers?

Like did you use both plus and minus for the equation?

[golden orbit]

I didn't even get a quadratic

Wait check my work then

[golden orbit]

Wait you said you wanted simultaneous of those two

[shy ledge]

Wdym simultaneous?

[golden orbit]

Solve the two system of eq in terms of f or g

[shy ledge]

in terms of either one

let's just say g

[golden orbit]

Would this be fine then?

[shy ledge]

wait let me check if it is

[golden orbit]

I used your second eq from here

[golden orbit]

Sqrt might be +/-

I might have made a mistake there

[shy ledge]

Okay so you did everything right but now you have to use your solved expression, and substitute that for the g variables in the other equation, and then solve for f

So the final expression cannot have either f or g, because they are the unknowns.

[golden orbit]

Ahh

[shy ledge]

Yep

[golden orbit]

You want without f

For g

[shy ledge]

Yes, I want an equation that says "F is equal to some expression that does not contain F or G".

Because I do not know what f or g is

[shy ledge]

I guess this is just the solution then, it's hairy as hell but it works, I tried in desmos.

@golden orbit Thanks for your help! If you have something to add just DM me or add me 🙂

.close

[pearl pond]

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[finite gulch]

I need help on 1e

[dusty patio]

,rccw

[clear trench]

Are you familiar with the cosine rule @finite gulch ?

[finite gulch]

Yes but im not sure how to find the angle

For the cos

[clear trench]

Are you allowed to use a calculator?

[finite gulch]

Yup

[clear trench]

You have 1 angle and 2 sides

oh 1e

sorry

ignore that

Maybe you can use the fact that the angles should add up to 360

[finite gulch]

Oh yeah all of them combined

[clear trench]

mhm

[finite gulch]

Thank you a lot

.close

[pearl pond]

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[daring valley]

,rotate

[strange quiver]

for these types of question you square them

[daring valley]

is it 4sqrt(5)-4sqrt(2)

i mean sqrt(5)-sqrt(2)

[strange quiver]

yeah I think so

[daring valley]

,rotate

[strange quiver]

hmm

maybe use (a+b)(a-b)=a^2-b^2

by grouping them

[daring valley]

i got a very weird

answer

,rotate

[strange quiver]

ok I'm probably wrong

wait

you can apply the formula again

[daring valley]

what formula

[strange quiver]

(a+b)(a-b)=a^2-b^2

[daring valley]

oh okay

4-7n^2?

[strange quiver]

yeah

I got the same answer

[daring valley]

,rotate

[daring valley]

@strange quiver my answer for this is 1152

it's unclear but find x^4+y^4+(x+y)^4

[strange quiver]

can you show your work please

I'm too lazy to calculate myself 😛

[daring valley]

it's

messy

but ok

,rotate

[daring valley]

the bottom is just x^4+y^4 = 527

@strange quiver

[strange quiver]

I think it's correct

[daring valley]

how do i factorise

a cube roote

roote

root

like cube root(4) + cube root(2)

how do i factorise out the cuberoot

[strange quiver]

you can't

[daring valley]

oh you can't?

,rotate

[daring valley]

there must be a way to factorise a

[strange quiver]

idk how to do this sry

[daring valley]

<@&286206848099549185>

[harsh silo]

rewrite it with fractional exponents and factor

and also rewrite 4 as 2² @daring valley

[daring valley]

so a = (2^2)^(1/3)+2^(1/3)+1?

[harsh silo]

yes

[daring valley]

what do I do from there

[lucid idol]

U can cube both sides making a-1 in the left
so that u can vanish the 3 in the exponent's fraction

From there make an equation in the left side and 0 in the right

Then factor the equation

[daring valley]

it still doesn't vanish the 3 in the exponent's fraction

@lucid idol

[steel dagger]

@daring valley hi

Are you there

[daring valley]

yes

i am here

[steel dagger]

Let a= cuberoot(4) and b= cuberoot(2)

The only way to get rid of root is by using

a³+b³ = (a+b) ( a² -ab + b²)

Brute force

Oh wait there is also a constant as well

So we need to obtain a³ + b³ + c by multiplying (a+b+1) with something

Where c is an integer we dont know

[daring valley]

bruh this is confusing i don't understand

the previous person told me to do cube this equation

[steel dagger]

Consider $$ \frac{1}{\sqrt 2 - 1 }$$

How do you get rid of irrational values in the denominator?

[jolly parrot]

Cyrenux

[daring valley]

a-1 = 4^(1/3) + 2^(1/3)

[steel dagger]

You multiply by the conjugate right

[daring valley]

yeah

[steel dagger]

We will do the same here

[daring valley]

you're everywhere

[steel dagger]

Here it is 3/a + 3/a² + 1/a³

[daring valley]

yeah

[steel dagger]

Lets first consider 3/a

a is irrational right

[daring valley]

why would a be irrational

oh yeah it's irrational ok

[steel dagger]

$$\frac{3}{a} = \frac{3}{ \sqrt[3]{4} + \sqrt[3]{2} + 1}$$

[jolly parrot]

Cyrenux

[steel dagger]

We need to multiply by conjugate

[daring valley]

yeah

[steel dagger]

But we dont know what the conjugate is

We need to figure

[daring valley]

we do

wdym we do know the conjugate

[steel dagger]

I said we dont

[daring valley]

we do?

[steel dagger]

What is it then

Tell me conjugate of a

[daring valley]

on second thought idk

[steel dagger]

Hmm

Do you know how formula
x³+y³= (x+y) ( x² -xy + y²) is derived

[daring valley]

yes

[steel dagger]

If you do, we can find a formula for x³ + y³ + z³ similiarly

But its painful lol

[daring valley]

i have notes which say a^3+b^3+c^3-3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ac)

||i feel like this method is going to be so confusing||

[steel dagger]

Yes use this

c = 1 in our question

Get rid of irrational numbers from the denominators, then its free

[daring valley]

but how

[steel dagger]

Oh wait -3abc on LHS

Yeah wont work

Check to see if there is a formula for a³+b³+c³ which can be factored as multiplication of (a+b+c) and something else

I need to go for now

[daring valley]

ok

[pearl pond]

@daring valley Has your question been resolved?

[pearl pond]

@daring valley Has your question been resolved?

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[flint wyvern]

am i right with D?

[fathom saffron]

yes

[flint wyvern]

ty

[pearl pond]

@flint wyvern Has your question been resolved?

[pearl pond]

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[willow kelp]

help - for the question this how i proceeded and stuck at the last step, and it looks like a cube eq ...how to solve it?

[pearl pond]

@willow kelp Has your question been resolved?

[rough stream]

"solve it"?

The inner integral should multiply by 1/2, not 1/3

[willow kelp]

i got 1/3 when integrating w.r.t. y and takin x as const. and yh i made mistake while integrating "-e^2/2" w.r.t. x now the final eq is like 1/3 [-3e^2/2 +e^3 +1/2 ]

should i take out cube roots for e? and its -0.5 and 1

[willow kelp]

.close

[pearl pond]

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[stiff flicker]

How do i get the time with only acceleration, initial velocity and displacement?

[stiff flicker]

a= 2m/s^2
vi=0m/s
d=500m

t=?

[rough stream]

It's 1:11PM where I live

Did you have more context?

[stiff flicker]

Imma type the word question

A sports car starting from rest accellerates uniformly at 2m/s^2. After how long will the car be 500m from its starting point.

[pearl pond]

@stiff flicker Has your question been resolved?

[stiff flicker]

<@&286206848099549185>

[pearl pond]

@stiff flicker Has your question been resolved?

[pearl pond]

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[long flame]

Can someone please explain what I’m supposed to do here?

What does it mean by “show that …” ?

[nimble cave]

it means to write down a series of logically deductive steps, beginning with the data the problem gives you, such that you end up with the conclusion

[long flame]

I’m sorry I don’t understand, could you explain that in more simple terms?

[nimble cave]

sorry, yes i can try :). you are given that $y = \int (2x^3 - x) dx$ and that the point $(2, -2)$ is on the curve $C$. now you have to show that the conclusion is true

[jolly parrot]

ves

[nimble cave]

and you have to do this with valid arguments and true statements, i.e., you can't just write nonsense and say "therefore, 2y = x^4 - x^2 - 16"

[long flame]

Lol yeah

[nimble cave]

i actually don't like the way it's formatted though. i wish it said "you are given that dy/dx = 2x^3 - x"

but oh well.

[long flame]

Yeah, it really confused me

I’m trying to solve this past paper for my exam tomorrow

[nimble cave]

what do you think that we should do?

to solve the problem

[long flame]

Try to trace down how the conclusion was found?

Or basically prove the conclusion to be true?

[nimble cave]

yeah. one thing that i notice is that $y = \int (2x^3 - x)dx$ can be written in a more explicit way

[jolly parrot]

ves

[nimble cave]

or a simplified way, to use another word

[long flame]

I didn’t know that

[analog peak]

thats so wrong, why would u say its a curve

[nimble cave]

yeah its pretty badly written

since $\int (2x^3 - x) dx$ is technically a family of curves

[jolly parrot]

ves

[long flame]

Ahh

[analog peak]

well it doesnt even mean anything like that

[long flame]

What does it mean?

[nimble cave]

but, what is $\int (2x^3 - x)dx$ equal to? @long flame

[jolly parrot]

ves

[analog peak]

like that its just a primitive

[long flame]

Hold on

Uhh so I think

Its

1/2x^4 - 1/2 x^2 + c

Oh wait

Is that correct?

[nimble cave]

and then +C

[long flame]

Yeah I almost forgot about that, sorry

[nimble cave]

no problem. (well it is on a test. and for this question i guess, lol)

[long flame]

Haha yeah

[nimble cave]

so now we know that y = 1/2 x^4 - 1/2 x^2 + C, where C is some unspecified constant

[long flame]

Yeah

[nimble cave]

we want to determine C how should we do that

[long flame]

Oh umm

Do we

Substitute

X for a value?

Like either 2 or -2?

[nimble cave]

sort of, yeah. the other information we are given is the point (2,-2) is on the curve C

do you know what it means for (2, -2) to lie on the curve?

[long flame]

Sorry, what does that mean? 😅

[nimble cave]

i'm going to write it in 2 ways, the first slightly more general, and then the second for this specific case

[long flame]

Thank you so much 🙏

[nimble cave]

if you have a (probably continuous) function f, then the equation y = f(x) determines some set of points (x,y) in the plane (the set of points that "solve" the equation, or the set of points where the equation is true). we can call this curve C. a point (r,s) lies on the curve C if s = f(r).

so in this case, f(x) = 1/2 x^4 - 1/2 x^2 + C, where C is some constant. the equation y = 1/2 x^4 - 1/2 x^2 defines the curve C

so a point (r,s) lies on the curve if s = 1/2 r^4 - 1/2 r^2 + C

so like, even more concretely, (0,1) would lie on the curve if (0,1) "solves" the equation, which would mean, in this case, that 1 = 1/2 0^4 - 1/2 0^2 + C

but i'm not saying it does, just that this could be an example :)

[long flame]

Ouuuu

[nimble cave]

if you have y = x^2, then (1,1) lies on the curve, (2, 4) lies on the curve, etc

so the problem tells you that you know (2, -2) lies on the curve. you have to write down what this means and see what it tells you :)

[long flame]

Ohhhh

Thank you so much !!

[nimble cave]

i assume you solved it

so to "show" it, you would write like "since y = \int (2x^3 - x)dx" ,we know ... [do integration] ... so the curve C is defined by y = 1/2 x^4 - x^2 + C. since (2, -2) lies on the curve ... [plug in values] ... thus C = <whatever you get>" and then maybe multiply by 2

and then you'll have what they wanted you to show

[long flame]

That’s perfect

Thank you so much again

I appreciate everything

Take care of yourself 🙏

[nimble cave]

no problem

[long flame]

Idk how to close this channel sorry 😅

Nvm I think I got it

.close

[pearl pond]

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[echo nest]

Pls help I’ve been stuck for so long and feel it’s so obvious

[tropic saddle]

write down the condition for convergence

[echo nest]

For all epsilon >0 there exists N such that for all n greater than or equal to n mod Xn-x less epsilon

Is all I have to do state |xn+m-x|<|xn-x|<€

Pretty sure not true tho

As xn+m can be bigger or less than xn

[tropic saddle]

you have to show that condition for y_n

you have to show that there exists such a N for the y's

and you know that there exists another N_2 which works for the x's

[echo nest]

Yes I know

I still don’t get how to show it

Thoughts something like this cld work

[tropic saddle]

if lets say eps=0.5 and m=7. lets say that the N that works for the x's is 15

so all x_n after 15 are at most 0.5 away

so x_15, x_16, ...

which y's are those

[echo nest]

So can I state without further justification

|yn-x|<=|xn-x|

[tropic saddle]

not at all what I am saying

[echo nest]

II understand what you are saying

I just don’t get how to formalise it yk

[tropic saddle]

answer this

[echo nest]

Y N+1 y N+2 etc

[tropic saddle]

I used explicit numbers for a reason

[echo nest]

Y1,y2 etc

[tropic saddle]

no

which y is equal to x_15

[echo nest]

Y15+m

Y15

[tropic saddle]

no

y_n=x_15

x_n+7=x_15

so n=...?

[echo nest]

8

[tropic saddle]

yes

so we know that for all y8, y9, y10, ... you are at most 0.5 away

so for the y's you can choose the N as 8

[echo nest]

So N_2=N_1-M

I still have zero intuition for it formally

[tropic saddle]

yes

because then |y_n-x| = |x_n+m-x| which is smaller than eps because n+m is bigger than N_1

[echo nest]

How is it bigger than n_1

[tropic saddle]

because n is bigger than N_2

so n+m >= N_2+m =N_1-m+m=N_1

[echo nest]

Yeah whatever I don’t. Get it

It’s bcus of the subscripts that confuses me

I don’t get how you go from x_n+m to x_n +m

[tropic saddle]

I don't

sry I am on mobile I don't wanna tex this

I suppose I should

$|y_n-x|=|x_{n+m}-x|<\varepsilon$ if $n\geq N_2$, because $n+m\geq N_1$

[jolly parrot]

Denascite

[echo nest]

Ohhhh ur notation was confusing that makes more sense

[tropic saddle]

sry bout that

[echo nest]

Ty tho

I feel slightly better about it

Feel as if it’s a weird question tho

All other problems I’ve been find with the subscripts were just kinda fucking with my head

And it’s one of those things of being simpler than it seems

[tropic saddle]

yeah subscript battles like this are hard

(how my prof used to call stuff like this)

[pearl pond]

@echo nest Has your question been resolved?

[pearl pond]

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[sweet radish]

Can someone explain this to me no idea where to start and the solution isn't making sense either

[sweet bone]

it's using viete's formulae

they're a set of results that relate the roots of a polynomial together in terms of its coefficients

they're just using them to set up simultaneous equations in n and m

[sweet radish]

Never heard of this formula :/

Is there another method to solve this

[sweet bone]

you could do a lot of polynomial division

it'd be painful, but possible

[sweet radish]

Do the vietes formula is the dumm of all roots?

Or is it the product of all roots?

[sweet bone]

the sum, sum of pairwise products, sum of 3-way products, etc. all give different information about the polynomial

in this case they're just using the sum and full product form

[sweet radish]

Alright thank uu

.close

[pearl pond]

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[ocean osprey]

I need help finding angle AEB

[pearl pond]

@ocean osprey Has your question been resolved?

[pearl pond]

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[still fiber]

how do I find velocity in part a

i feel like im missing sometihng obvious

im assuming it has to do with rev/min

im just forgetting a formula

[fast zealot]

u have rpm and radius

convert from rpm to angular velocity

[still fiber]

will do

[fast zealot]

u can go straight from w and r to F_c you dont have to do another calculation to find v either

[still fiber]

wdym by that

[fast zealot]

F=mw^2r

[still fiber]

what is w?

[fast zealot]

angular velocity

[still fiber]

right

[fast zealot]

got anywhere with that?

[still fiber]

sec

I did 40( 2pi/60 ) to get 4.19

then did this

which is not right

[fast zealot]

so you did mw^2/r

[still fiber]

is that not what I should do

[fast zealot]

no

going from mv^2/r to mw^2r is done through v=rw

[still fiber]

what is w

im confused

what does it stand for

[fast zealot]

angular velocity, I said earlier

[still fiber]

right

so

what is v then..

[fast zealot]

linear velocity

v=rw

but you don't need to calculate it if you just substitute that into the relationship f_c=mv^2/r and obtain f_c=mw^2r

[still fiber]

AH

yo

you're so smart

[pearl pond]

@still fiber Has your question been resolved?

[pearl pond]

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[unreal whale]

So do you add or multiply for this

🥹

[pearl pond]

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[olive lantern]

for the range of piecewise

u just

list the numbers?

[regal herald]

the range would be the set of values, but here its literally only 2 or 3 that you can have

so its just {2,3}

[olive lantern]

so i dont say yER

like domanin?

[regal herald]

2 and 3 belong to the reals, you dont need to state that

[olive lantern]

okkk

tyty

.close

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[still fiber]

What does part c even mean

A and B are easy but I don’t even know what c wants from me

[rough stream]

Probably trying to get you to realize gravity is very strong

[still fiber]

Yes but I’m a bit confused on what this actually means like

Is there a calculation or something??

[rough stream]

Probably not. Just say "wow kid heavy"

[still fiber]

.close

[pearl pond]

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[still fiber]

The answer is 15N but I got 02.45N

Where does my mistake lie?

[pearl pond]

@still fiber Has your question been resolved?

[still fiber]

<@&286206848099549185>

[pearl pond]

@still fiber Has your question been resolved?

[still fiber]

.close

[pearl pond]

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[vague mesa]

Instead of u=lnz I used z-1

[vague mesa]

Why is my answer so different?

Wait I think I see what’s wrong

I set x and u equal…

Ok I think I got it

[pearl pond]

@vague mesa Has your question been resolved?

[pearl pond]

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[polar steppe]

Would I need to use the quadratic formula for this?

Or is there something I'm overlooking

[sweet bone]

the question is just asking for the form

where would you use the quadratic formula

[coarse dawn]

Why is the first letter of every word bolded?

[sweet bone]

it makes it easier to read for some people

[polar steppe]

well I need to factor the denom right?

[polar steppe]

It's a chrome extension

It's to help me stay focused when reading

otherwise I'll jump 😦

[sweet bone]

the denominator doesn't factor over the integers

[polar steppe]

For reference

hmm

[sweet bone]

the numerator has the same degree as the denominator

so you could do polynomial division

[polar steppe]

sooo

[sweet bone]

to obtain a constant, then have a linear over that quadratic left over

[polar steppe]

that's actually what I thought

wait

Oh I might need an explanation for that

because typically when I go down the route of using long division

I'm not using that method above anymore

[summer imp]

Long division will give you a constant term plus the remainder (linear) over the polynomial. However, x^2+x+4 doesn't factor in the reals, so unless you want to write it using complex roots you can maybe leave it in that form.

Like (constant) + (linear term)/(polynomial)

Maybe that's what they want

[polar steppe]

I see I see

I'm bringing up this example to help me

So, lets pretend this is the function instead

It would be 4 / x+ 1

right?>

You're saying that's what they would want?

[summer imp]

What they want is the most reduced form where the numerator is of lower degree than the denominator

[polar steppe]

OKAY

[summer imp]

In the image you just sent, you don't need to actually do partial fractions since your long division already gets rid of the factors.

[polar steppe]

Following so far

[summer imp]

So

In your original problem, you don't have that the degree of the numerator is lower, right?

So to make it lower than the denominator, you can do long division.

(Or, secret trick, you can add x + 4 and then subtract x + 4 to the numerator)

Well actually they just want the form, so you don't even need to actually carry out long division. Since they are the same degree, you will get a constant term, say A, and a linear remainder (Bx+C)/(x^2+x+4)

[polar steppe]

hmm

Just a constant term A?

wait so to be clear

in this case, since the degrees are the same, I'd get:
A + Bx + C / x^2 + x+ 4?

If so I will need to think about that

[summer imp]

But A is alone, not over the polynomial

A + (Bx + C)/(x^2+x+4)

It just comes from the fact that a quadratic over a quadratic will always give you a constant quotient and some (at most) linear remainder

[polar steppe]

makes sense

[summer imp]

But, in general, first thing when you do this is you make sure deg(numerator) < deg(denominator). If not, you make it using long division.

[polar steppe]

Right right!

[summer imp]

Then you can do partial fraction accordingly. In this case, there's no proper partial fraction because the denominator does not factor in R.

[polar steppe]

I just got confused because so far I haven't had to rewrite anything in that form when doing long division

I'd usually jsut integrate after that

[summer imp]

Yeah fair enough

I mean in a usual case you wouldn't really be thinking about the form of your partial fraction. Once you do long div and all it just falls into place

[polar steppe]

indeed

hmmm

Did I make a mistake?

[pearl pond]

@polar steppe Has your question been resolved?

[polar steppe]

.close

[pearl pond]

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[stuck widget]

INTERGRATION QUESTION

[stuck widget]

can I take out the 8 in the intergral -8+2^-2 dx ?

[last summit]

Please show the original question

-8+2^(-2) is just -31/4

I assume you wrote this wrong

take a picture of the question

[stuck widget]

this is those multi question questions

[last summit]

you can split integrals at subtraction

[stuck widget]

were looking at C

[last summit]

yes

is this for a test?

[stuck widget]

no

past test paper

no cheating here

[last summit]

Okay well have I answered your question?

You can split integrals at subtraction/addition yes

[stuck widget]

oh

so I can't bring it outside the integral sign

thats only for multiplication yes?

I'll take the silence as a yes

thanks for the help Austin

.close

[pearl pond]

Channel closed

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[sweet bone]

yes, you can't take a constant being added outside like that

the integral represents the area bounded under the function, and adding a constant certainly changes the area

[stuck widget]

I see thanks for the info Desync

appriciate it

[pearl pond]

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[cedar coyote]

can someone explain how he jumped from line 3 to 4

not sure where the v0 in the denominator came from

[last summit]

v0/v0=1

so really you have 1-(kx)/v0

if that helps

[cedar coyote]

i understand that

but i assumed line 3

was the full integral

[last summit]

evaluated from 0 to x

[cedar coyote]

if that makes sense

[last summit]

still needs to happen

it must still be evaluated at the endpoints

[cedar coyote]

oh ok

[feral sedge]

Yeah this notation is cringe 😬

[cedar coyote]

it is

[feral sedge]

x is being used as 2 different things in the problem

making it very confusing

[cedar coyote]

x is the time it takes to hit the resisting medium

[feral sedge]

$\int_0^x \frac{du}{v_0-ku}$

[jolly parrot]

thewizardofOU

[cedar coyote]

this is the full problem

[feral sedge]

Like this

x should not be the variable of integration when it is already the bound of the integral

[cedar coyote]

yeah

its just how they do it for some reason

ok like

for this

im still clueless

on where the v0 came from in the denominator

ohhh

wait

i see it

i forgot the ln can do funny things

[feral sedge]

ah ye

[cedar coyote]

.close

[pearl pond]

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[dawn sierra]

!claim

[dawn sierra]

When adding and subtracting polynomials, why do you multiply the numerators by each other?

Instead of just treating it like Normal addition or subtraction

[reef quiver]

Do you have a specific example in mind?

[dawn sierra]

Yeah

[reef quiver]

I'm not sure about where we would "multiply the numerators by each other"

[dawn sierra]

Question 31

[cedar coyote]

you need a common denominator

to add and subtract fractions

so you multiply them by the deonominator of the other

[dawn sierra]

This is what I did just for reference

[cedar coyote]

but what u do to the bottom

must be done for the top as well

otherwise u are making a different function

[reef quiver]

To add numerators in 2 fractions, we must make sure their denominators are the same

For example, 2/5 + 2/5 = 4/5 (2 + 2 = 4)

But we cannot say the same about 2/5 + 2/6

oh ur right

[dawn sierra]

No I get that, it's just when I did my problem and got the answer I got and try to check it, was wrong.

[reef quiver]

You've done nearly all steps correct in your work

[coarse dawn]

What, no

[dawn sierra]

Okay

What did I do wrong

[reef quiver]

You've found the common denominator (x+4)(x+5)

[coarse dawn]

And (2/5) + (2/6) doesn't not equal 4/11

[reef quiver]

The first term you should've multiplied by (x+5)/(x+5)

instead of (x+4)/(x+4)

[dawn sierra]

Why is that?

[coarse dawn]

Maybe you should

[reef quiver]

We want to make the denominator of the first term (x+4)(x+5)

So we multiply it by (x+5)/(x+5), which is equal to 1, so we can add the numerators after making the same denominator

[dawn sierra]

Ohhhhhhhhhh

[coarse dawn]

Because you "cross" multiply, the denominator of the one fraction multiplies with the numerator of the other

[dawn sierra]

I got it

[reef quiver]

What you did was multiply by (x+4)/(x+4)

Which makes (x+4)^2 in the denominator

[dawn sierra]

Okay, I get it

[reef quiver]

nice!

[coarse dawn]

Jump into #help-27 so you don't clutter this channel You were too late

[dawn sierra]

Because the way I did there would be no change

[reef quiver]

yes, the denominators of both terms wouldn't be the same